Algebra and Trigonometry [4 ed.] ISBN: 978-1-119-32086-9 (ePub), ISBN: 978-1-119-27345-5 (LLPC)

Table of contents :
Cover......Page 1
Title Page......Page 5
Copyright......Page 6
Dedication......Page 7
About the Author......Page 8
Preface......Page 9
Contents......Page 16
A Note from the Author: TO THE STUDENT......Page 20
LEARNING OBJECTIVES......Page 24
IN THIS CHAPTER......Page 25
0.1.1. The Set of Real Numbers......Page 26
0.1.3. Order of Operations......Page 28
0.1.4. Properties of Real Numbers......Page 31
SKILLS......Page 36
APPLICATIONS......Page 37
TECHNOLOGY......Page 38
0.2.1. Integer Exponents......Page 39
0.2.2. Scientific Notation......Page 43
SKILLS......Page 46
APPLICATIONS......Page 47
TECHNOLOGY......Page 48
0.3.1. Adding and Subtracting Polynomials......Page 49
0.3.2. Multiplying Polynomials......Page 50
0.3.3. Special Products......Page 51
SKILLS......Page 55
APPLICATIONS......Page 56
TECHNOLOGY......Page 57
0.4.1. Greatest Common Factor......Page 58
0.4.2. Factoring Formulas: Special Polynomial Forms......Page 59
0.4.3. Factoring a Trinomial as a Product of Two Binomials......Page 61
0.4.4. Factoring by Grouping......Page 64
0.4.5. A Strategy for Factoring Polynomials......Page 65
SKILLS......Page 66
TECHNOLOGY......Page 67
0.5.1. Rational Expressions and Domain Restrictions......Page 68
0.5.2. Simplifying Rational Expressions......Page 70
0.5.3. Multiplying and Dividing Rational Expressions......Page 71
0.5.4. Adding and Subtracting Rational Expressions......Page 73
0.5.5. Complex Rational Expressions......Page 76
[SECTION 0.5] SUMMARY......Page 78
SKILLS......Page 79
APPLICATIONS......Page 80
TECHNOLOGY......Page 81
0.6.1. Square Roots......Page 82
0.6.2. Other (nth) Roots......Page 84
0.6.3. Rational Exponents......Page 88
[SECTION 0.6] SUMMARY......Page 89
APPLICATIONS......Page 90
TECHNOLOGY......Page 91
0.7.1. The Imaginary Unit, i......Page 92
0.7.2. Adding and Subtracting Complex Numbers......Page 93
0.7.4. Dividing Complex Numbers......Page 94
0.7.5. Raising Complex Numbers to Integer Powers......Page 95
[SECTION 0.7] SUMMARY......Page 96
APPLICATION......Page 97
TECHNOLOGY......Page 98
[CHAPTER 0 REVIEW]......Page 99
[CHAPTER 0 REVIEW EXERCISES]......Page 101
[CHAPTER 0 PRACTICE TEST]......Page 103
LEARNING OBJECTIVES......Page 104
IN THIS CHAPTER......Page 105
1.1.1. Solving Linear Equations in One Variable......Page 106
1.1.2. Solving Rational Equations That Are Reducible to Linear Equations......Page 109
SKILLS......Page 112
APPLICATIONS......Page 113
TECHNOLOGY......Page 114
1.2.1. Solving Application Problems Using Mathematical Models......Page 115
1.2.2. Geometry Problems......Page 117
1.2.3. Interest Problems......Page 119
1.2.4. Mixture Problems......Page 121
1.2.5. Distance–Rate–Time Problems......Page 122
APPLICATIONS......Page 125
TECHNOLOGY......Page 128
1.3.1. Factoring......Page 129
1.3.2. Square Root Method......Page 131
1.3.3. Completing the Square......Page 132
1.3.4. Quadratic Formula......Page 134
1.3.5. Applications Involving Quadratic Equations......Page 137
[SECTION 1.3] SUMMARY......Page 138
SKILLS......Page 139
APPLICATIONS......Page 140
CONCEPTUAL......Page 141
TECHNOLOGY......Page 142
1.4.1. Radical Equations......Page 143
1.4.2. Equations Quadratic in Form: u-Substitution......Page 145
1.4.3. Factorable Equations......Page 147
SKILLS......Page 148
APPLICATIONS......Page 149
TECHNOLOGY......Page 150
1.5.1. Graphing Inequalities and Interval Notation......Page 151
1.5.2. Solving Linear Inequalities......Page 153
SKILLS......Page 157
APPLICATIONS......Page 158
TECHNOLOGY......Page 160
1.6.1. Polynomial Inequalities......Page 161
1.6.2. Rational Inequalities......Page 166
SKILLS......Page 169
CATCH THE MISTAKE......Page 170
1.7.1. Equations Involving Absolute Value......Page 171
1.7.2. Inequalities Involving Absolute Value......Page 174
SKILLS......Page 176
CATCH THE MISTAKE......Page 177
TECHNOLOGY......Page 178
[CHAPTER 1 REVIEW]......Page 179
[CHAPTER 1 REVIEW EXERCISES]......Page 180
[CHAPTER 1 PRACTICE TEST]......Page 184
[CHAPTER 1 CUMULATIVE TEST]......Page 185
LEARNING OBJECTIVES......Page 186
IN THIS CHAPTER......Page 187
2.1.1. Cartesian Plane......Page 188
2.1.2. Distance Between Two Points......Page 189
2.1.3. Midpoint of a Line Segment Joining Two Points......Page 190
[SECTION 2.1] SUMMARY......Page 191
SKILLS......Page 192
APPLICATIONS......Page 193
CATCH THE MISTAKE......Page 194
2.2.1. Point-Plotting......Page 195
2.2.2. Intercepts......Page 198
2.2.3. Symmetry......Page 199
2.2.4. Using Intercepts and Symmetry as Graphing Aids......Page 202
[SECTION 2.2] SUMMARY......Page 203
SKILLS......Page 204
CATCH THE MISTAKE......Page 206
TECHNOLOGY......Page 207
2.3.1. Graphing a Line......Page 208
2.3.2. Equations of Lines......Page 212
2.3.3. Parallel and Perpendicular Lines......Page 216
SKILLS......Page 219
APPLICATIONS......Page 221
TECHNOLOGY......Page 223
2.4.1. Standard Equation of a Circle......Page 224
2.4.2. Transforming Equations of Circles to the Standard Form by Completing the Square......Page 227
SKILLS......Page 228
APPLICATIONS......Page 229
CHALLENGE......Page 230
2.5.1. Scatterplots......Page 231
2.5.2. Identifying Patterns......Page 235
2.5.3. Linear Regression......Page 241
SKILLS......Page 247
CATCH THE MISTAKE......Page 249
APPLICATIONS......Page 250
CHALLENGE......Page 251
[CHAPTER 2 REVIEW]......Page 253
[CHAPTER 2 REVIEW EXERCISES]......Page 254
[CHAPTER 2 PRACTICE TEST]......Page 256
[CHAPTERS 1–2 CUMULATIVE TEST]......Page 257
LEARNING OBJECTIVES......Page 258
IN THIS CHAPTER......Page 259
3.1.1. Relations and Functions......Page 260
3.1.2. Functions Defined by Equations......Page 262
3.1.3. Function Notation......Page 264
3.1.4. Domain of a Function......Page 269
[SECTION 3.1] SUMMARY......Page 271
SKILLS......Page 272
APPLICATIONS......Page 274
TECHNOLOGY......Page 276
3.2.1. Recognizing and Classifying Functions......Page 277
3.2.2. Increasing and Decreasing Functions......Page 281
3.2.3. Average Rate of Change......Page 282
3.2.4. Piecewise-Defined Functions......Page 285
[SECTION 3.2] SUMMARY......Page 288
SKILLS......Page 289
APPLICATIONS......Page 292
CATCH THE MISTAKE......Page 294
TECHNOLOGY......Page 295
3.3.1. Horizontal and Vertical Shifts......Page 296
3.3.2. Reflection about the Axes......Page 300
3.3.3. Stretching and Compressing......Page 302
SKILLS......Page 305
APPLICATIONS......Page 307
TECHNOLOGY......Page 308
3.4.1. Adding, Subtracting, Multiplying, and Dividing Functions......Page 309
3.4.2. Composition of Functions......Page 311
[SECTION 3.4] SUMMARY......Page 315
SKILLS......Page 316
APPLICATIONS......Page 317
CATCH THE MISTAKE......Page 318
TECHNOLOGY......Page 319
3.5.1. Determine Whether a Function Is One-to-One......Page 320
3.5.2. Inverse Functions......Page 323
3.5.3. Graphical Interpretation of Inverse Functions......Page 325
3.5.4. Finding the Inverse Function......Page 326
SKILLS......Page 330
CATCH THE MISTAKE......Page 332
TECHNOLOGY......Page 333
3.6.1. Direct Variation......Page 334
3.6.2. Inverse Variation......Page 336
3.6.3. Joint Variation and Combined Variation......Page 337
SKILLS......Page 339
APPLICATIONS......Page 340
CONCEPTUAL......Page 341
TECHNOLOGY......Page 342
[CHAPTER 3 REVIEW]......Page 344
[CHAPTER 3 REVIEW EXERCISES]......Page 346
[CHAPTER 3 PRACTICE TEST]......Page 350
[CHAPTERS 1–3 CUMULATIVE TEST]......Page 351
LEARNING OBJECTIVES......Page 352
IN THIS CHAPTER......Page 353
4.1.1. Graphs of Quadratic Functions: Parabolas......Page 354
4.1.2. Finding the Equation of a Parabola......Page 362
[SECTION 4.1] SUMMARY......Page 365
SKILLS......Page 366
APPLICATIONS......Page 367
CATCH THE MISTAKE......Page 369
TECHNOLOGY......Page 370
4.2.1. Identifying Polynomial Functions......Page 371
4.2.2. Graphing Polynomial Functions Using Transformations of Power Functions......Page 373
4.2.3. Real Zeros of a Polynomial Function......Page 375
4.2.4. Graphing General Polynomial Functions......Page 377
SKILLS......Page 381
APPLICATIONS......Page 383
CHALLENGE......Page 384
4.3.1. Long Division of Polynomials......Page 385
4.3.2. Synthetic Division of Polynomials......Page 389
SKILLS......Page 391
CATCH THE MISTAKE......Page 392
TECHNOLOGY......Page 393
4.4.1. The Remainder Theorem and the Factor Theorem......Page 394
4.4.2. The Rational Zero Theorem and Descartes’ Rule of Signs......Page 397
4.4.3. Factoring Polynomials......Page 402
4.4.4. The Intermediate Value Theorem......Page 404
4.4.5. Graphing Polynomial Functions......Page 405
SKILLS......Page 407
APPLICATIONS......Page 408
TECHNOLOGY......Page 409
4.5.1. Complex Zeros......Page 410
4.5.2. Factoring Polynomials......Page 414
[SECTION 4.5] SUMMARY......Page 415
APPLICATIONS......Page 416
CATCH THE MISTAKE......Page 417
4.6.1. Domain of Rational Functions......Page 418
4.6.2. Vertical, Horizontal, and Slant Asymptotes......Page 420
4.6.3. Graphing Rational Functions......Page 426
SKILLS......Page 432
APPLICATIONS......Page 434
CATCH THE MISTAKE......Page 435
CONCEPTUAL......Page 436
TECHNOLOGY......Page 437
[CHAPTER 4 REVIEW]......Page 438
[CHAPTER 4 REVIEW EXERCISES]......Page 440
[CHAPTER 4 PRACTICE TEST]......Page 444
[CHAPTERS 1–4 CUMULATIVE TEST]......Page 445
LEARNING OBJECTIVES......Page 446
IN THIS CHAPTER......Page 447
5.1.1. Evaluating Exponential Functions......Page 448
5.1.2. Graphs of Exponential Functions......Page 449
5.1.3. The Natural Base e......Page 453
5.1.4. Applications of Exponential Functions......Page 454
SKILLS......Page 458
APPLICATIONS......Page 459
CATCH THE MISTAKE......Page 460
TECHNOLOGY......Page 461
5.2.1. Evaluating Logarithms......Page 462
5.2.2. Common and Natural Logarithms......Page 464
5.2.3. Graphs of Logarithmic Functions......Page 465
5.2.4. Applications of Logarithms......Page 468
[SECTION 5.2] SUMMARY......Page 472
SKILLS......Page 473
APPLICATIONS......Page 474
CHALLENGE......Page 476
TECHNOLOGY......Page 477
5.3.1. Properties of Logarithmic Functions......Page 478
5.3.2. Change-of-Base Formula......Page 482
SKILLS......Page 484
CATCH THE MISTAKE......Page 485
5.4. EXPONENTIAL AND LOGARITHMIC EQUATIONS......Page 486
5.4.1. Exponential Equations......Page 487
5.4.2. Solving Logarithmic Equations......Page 490
5.4.3. Applications......Page 491
SKILLS......Page 493
APPLICATIONS......Page 494
TECHNOLOGY......Page 495
5.5. EXPONENTIAL AND LOGARITHMIC MODELS......Page 496
5.5.1. Exponential Growth Models......Page 497
5.5.2. Exponential Decay Models......Page 498
5.5.3. Gaussian (Normal) Distribution Models......Page 499
5.5.4. Logistic Growth Models......Page 500
5.5.5. Logarithmic Models......Page 501
SKILLS......Page 502
APPLICATIONS......Page 503
CONCEPTUAL......Page 505
TECHNOLOGY......Page 506
[CHAPTER 5 REVIEW]......Page 507
[CHAPTER 5 REVIEW EXERCISES]......Page 509
[CHAPTER 5 PRACTICE TEST]......Page 512
[CHAPTERS 1–5 CUMULATIVE TEST]......Page 513
LEARNING OBJECTIVES......Page 514
IN THIS CHAPTER......Page 515
6.1.1. Angles and Degree Measure......Page 516
6.1.2. Triangles......Page 519
6.1.3. Special Right Triangles......Page 521
6.1.4. Similar Triangles......Page 524
SKILLS......Page 528
APPLICATIONS......Page 529
CONCEPTUAL......Page 531
TECHNOLOGY......Page 532
6.2.1. Trigonometric Functions: Right Triangle Ratios......Page 533
6.2.2. Reciprocal Identities......Page 535
6.2.3. Cofunctions......Page 537
6.2.4. Evaluating Trigonometric Functions Exactly for Special Angle Measures: 30°, 45°, and 60°......Page 539
6.2.5. Using Calculators to Evaluate (Approximate) Trigonometric Functions......Page 541
SKILLS......Page 542
APPLICATIONS......Page 543
CATCH THE MISTAKE......Page 545
TECHNOLOGY......Page 546
6.3.1. Accuracy and Significant Digits......Page 547
6.3.2. Solving a Right Triangle Given the Measure of an Acute Angle and a Side Length......Page 548
6.3.3. Solving a Right Triangle Given the Length of Two Sides......Page 550
SKILLS......Page 554
APPLICATIONS......Page 555
CATCH THE MISTAKE......Page 558
TECHNOLOGY......Page 559
6.4.1. Angles in Standard Position......Page 560
6.4.2. Coterminal Angles......Page 563
6.4.3. Trigonometric Functions: The Cartesian Plane......Page 565
SKILLS......Page 571
APPLICATIONS......Page 572
TECHNOLOGY......Page 573
6.5.1. Algebraic Signs of Trigonometric Functions......Page 574
6.5.2. Ranges of the Trigonometric Functions......Page 577
6.5.3. Reference Angles and Reference Right Triangles......Page 578
6.5.4. Evaluating Trigonometric Functions for Nonacute Angles......Page 581
SKILLS......Page 585
APPLICATIONS......Page 587
TECHNOLOGY......Page 588
6.6.1. The Radian Measure of an Angle......Page 589
6.6.2. Converting Between Degrees and Radians......Page 591
6.6.3. Arc Length......Page 594
6.6.4. Area of a Circular Sector......Page 595
6.6.5. Linear Speed......Page 597
6.6.6. Angular Speed......Page 598
6.6.7. Relationship Between Linear and Angular Speeds......Page 599
SKILLS......Page 601
APPLICATIONS......Page 602
CONCEPTUAL......Page 605
6.7. DEFINITION 3 OF TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH......Page 606
6.7.1. Trigonometric Functions and the Unit Circle (Circular Functions)......Page 607
6.7.2. Properties of Circular Functions......Page 609
[SECTION 6.7] SUMMARY......Page 611
SKILLS......Page 612
APPLICATIONS......Page 613
CONCEPTUAL......Page 614
TECHNOLOGY......Page 615
6.8.1. The Graphs of Sinusoidal Functions......Page 616
6.8.2. Graphing a Shifted Sinusoidal Function: y = A sin (Bx + C) + D and y = A cos (Bx + C) + D......Page 628
6.8.3. Harmonic Motion......Page 631
6.8.4 Graphing Sums of Functions: Addition of Ordinates......Page 636
SKILLS......Page 639
APPLICATIONS......Page 641
CONCEPTUAL......Page 643
TECHNOLOGY......Page 644
6.9.1. Graphing the Tangent, Cotangent, Secant, and Cosecant Functions......Page 645
6.9.2. Translations of Circular Functions......Page 655
[SECTION 6.9] SUMMARY......Page 658
SKILLS......Page 659
CATCH THE MISTAKE......Page 661
TECHNOLOGY......Page 662
[CHAPTER 6 REVIEW]......Page 663
[CHAPTER 6 REVIEW EXERCISES]......Page 669
[CHAPTER 6 PRACTICE TEST]......Page 672
[CHAPTERS 1–6 CUMULATIVE TEST]......Page 673
LEARNING OBJECTIVES......Page 674
IN THIS CHAPTER......Page 675
7.1.1. Reciprocal Identities......Page 676
7.1.2. Quotient Identities......Page 678
7.1.3. Pythagorean Identities......Page 679
SKILLS......Page 683
CATCH THE MISTAKE......Page 685
TECHNOLOGY......Page 686
7.2.1. Trigonometric Identities......Page 687
SKILLS......Page 693
APPLICATIONS......Page 694
CONCEPTUAL......Page 695
TECHNOLOGY......Page 696
7.3. SUM AND DIFFERENCE IDENTITIES......Page 697
7.3.1. Sum and Difference Identities for the Cosine Function......Page 698
7.3.2. Sum and Difference Identities for the Sine Function......Page 702
7.3.3. Sum and Difference Identities for the Tangent Function......Page 705
SKILLS......Page 707
APPLICATIONS......Page 709
CHALLENGE......Page 710
7.4.1. Applying Double-Angle Identities......Page 711
SKILLS......Page 716
APPLICATIONS......Page 717
CONCEPTUAL......Page 718
7.5.1. Applying Half-Angle Identities......Page 719
SKILLS......Page 726
CATCH THE MISTAKE......Page 728
TECHNOLOGY......Page 729
7.6.1. Product-to-Sum Identities......Page 730
7.6.2. Sum-to-Product Identities......Page 732
[SECTION 7.6] SUMMARY......Page 734
SKILLS......Page 735
APPLICATIONS......Page 736
TECHNOLOGY......Page 737
7.7.1. Inverse Sine Function......Page 738
7.7.2. Inverse Cosine Function......Page 742
7.7.3. Inverse Tangent Function......Page 745
7.7.4. Remaining Inverse Trigonometric Functions......Page 747
7.7.5. Finding Exact Values for Expressions Involving Inverse Trigonometric Functions......Page 750
SKILLS......Page 754
APPLICATIONS......Page 756
CATCH THE MISTAKE......Page 757
TECHNOLOGY......Page 758
7.8.1. Solving Trigonometric Equations by Inspection......Page 759
7.8.2. Solving Trigonometric Equations Using Algebraic Techniques......Page 763
7.8.3. Solving Trigonometric Equations That Require the Use of Inverse Functions......Page 764
7.8.4. Using Trigonometric Identities to Solve Trigonometric Equations......Page 766
SKILLS......Page 771
APPLICATIONS......Page 772
CATCH THE MISTAKE......Page 774
TECHNOLOGY......Page 775
[CHAPTER 7 REVIEW]......Page 776
[CHAPTER 7 REVIEW EXERCISES]......Page 779
[CHAPTER 7 PRACTICE TEST]......Page 784
[CHAPTERS 1–7 CUMULATIVE TEST]......Page 785
LEARNING OBJECTIVES......Page 786
IN THIS CHAPTER......Page 787
8.1.1. Solving Oblique Triangles......Page 788
[SECTION 8.1] SUMMARY......Page 798
APPLICATIONS......Page 799
CONCEPTUAL......Page 801
8.2.1. Solving Oblique Triangles......Page 802
SKILLS......Page 808
APPLICATIONS......Page 809
CATCH THE MISTAKE......Page 811
TECHNOLOGY......Page 812
8.3.1. The Area of a Triangle (SAS Case)......Page 813
8.3.2. The Area of a Triangle (SSS Case)......Page 815
[SECTION 8.3] SUMMARY......Page 816
APPLICATIONS......Page 817
TECHNOLOGY......Page 819
8.4.1. Vectors: Magnitude and Direction......Page 820
8.4.2. Vector Operations......Page 823
8.4.4. Unit Vectors......Page 825
8.4.5. Resultant Vectors......Page 826
[SECTION 8.4] SUMMARY......Page 830
APPLICATIONS......Page 831
TECHNOLOGY......Page 834
8.5.1. Multiplying Two Vectors: The Dot Product......Page 835
8.5.2. Angle Between Two Vectors......Page 836
8.5.3. Work......Page 838
SKILLS......Page 840
APPLICATIONS......Page 841
CHALLENGE......Page 842
8.6.1. Complex Numbers in Rectangular Form......Page 843
8.6.2. Complex Numbers in Polar Form......Page 844
SKILLS......Page 848
CONCEPTUAL......Page 849
TECHNOLOGY......Page 850
8.7.1. Products of Complex Numbers......Page 851
8.7.2. Quotients of Complex Numbers......Page 852
8.7.3. Powers of Complex Numbers......Page 854
8.7.4. Roots of Complex Numbers......Page 855
[SECTION 8.7] SUMMARY......Page 859
SKILLS......Page 860
CATCH THE MISTAKE......Page 861
TECHNOLOGY......Page 862
8.8.1. Polar Coordinates......Page 863
8.8.2. Converting Between Polar and Rectangular Coordinates......Page 864
8.8.3. Graphs of Polar Equations......Page 865
[SECTION 8.8] SUMMARY......Page 872
SKILLS......Page 874
APPLICATIONS......Page 875
CONCEPTUAL......Page 876
TECHNOLOGY......Page 877
[CHAPTER 8 REVIEW]......Page 878
[CHAPTER 8 REVIEW EXERCISES]......Page 881
[CHAPTER 8 PRACTICE TEST]......Page 884
[CHAPTERS 1–8 CUMULATIVE TEST]......Page 885
LEARNING OBJECTIVES......Page 886
IN THIS CHAPTER......Page 887
9.1.1. Solving Systems of Linear Equations......Page 888
9.1.2. Three Methods and Three Types of Solutions......Page 897
SKILLS......Page 900
APPLICATIONS......Page 901
CONCEPTUAL......Page 902
9.2.1. Solving Systems of Linear Equations in Three Variables......Page 903
9.2.2. Types of Solutions......Page 906
[SECTION 9.2] SUMMARY......Page 910
APPLICATIONS......Page 911
CHALLENGE......Page 913
9.3.1. Performing Partial-Fraction Decomposition......Page 914
SKILLS......Page 923
CHALLENGE......Page 924
9.4.1. Linear Inequalities in Two Variables......Page 925
9.4.2. Systems of Linear Inequalities in Two Variables......Page 927
SKILLS......Page 933
APPLICATIONS......Page 934
TECHNOLOGY......Page 935
9.5.1. Solving an Optimization Problem......Page 936
APPLICATIONS......Page 940
CATCH THE MISTAKE......Page 941
TECHNOLOGY......Page 942
[CHAPTER 9 REVIEW]......Page 943
[CHAPTER 9 REVIEW EXERCISES]......Page 944
[CHAPTER 9 PRACTICE TEST]......Page 946
[CHAPTERS 1–9 CUMULATIVE TEST]......Page 947
LEARNING OBJECTIVES......Page 948
IN THIS CHAPTER......Page 949
10.1.1. Matrices......Page 950
10.1.2. Augmented Matrices......Page 952
10.1.3. Row Operations on a Matrix......Page 953
10.1.4. Row–Echelon Form of a Matrix......Page 954
10.1.5. Gaussian Elimination with Back-Substitution......Page 955
10.1.6. Gauss–Jordan Elimination......Page 957
10.1.7. Inconsistent and Dependent Systems......Page 961
[SECTION 10.1] SUMMARY......Page 966
SKILLS......Page 967
APPLICATIONS......Page 969
CONCEPTUAL......Page 971
10.2.1. Equality of Matrices......Page 972
10.2.2. Matrix Addition and Subtraction......Page 974
10.2.3. Scalar and Matrix Multiplication......Page 976
SKILLS......Page 982
APPLICATIONS......Page 983
TECHNOLOGY......Page 985
10.3.1. Matrix Equations......Page 986
10.3.2. Finding the Inverse of a Matrix......Page 987
10.3.3. Solving Systems of Linear Equations Using Matrix Algebra and Inverses of Square Matrices......Page 992
[SECTION 10.3] SUMMARY......Page 994
SKILLS......Page 995
APPLICATIONS......Page 996
CHALLENGE......Page 997
10.4.1. Determinant of a 2 × 2 Matrix......Page 998
10.4.2. Determinant of an n × n Matrix......Page 999
10.4.3. Cramer’s Rule: Systems of Linear Equations in Two Variables......Page 1002
10.4.4. Cramer’s Rule: Systems of Linear Equations in Three Variables......Page 1004
[SECTION 10.4] SUMMARY......Page 1007
SKILLS......Page 1008
APPLICATIONS......Page 1009
CATCH THE MISTAKE......Page 1010
TECHNOLOGY......Page 1011
[CHAPTER 10 REVIEW]......Page 1012
[CHAPTER 10 REVIEW EXERCISES]......Page 1015
[CHAPTER 10 PRACTICE TEST]......Page 1018
[CHAPTERS 1–10 CUMULATIVE TEST]......Page 1019
LEARNING OBJECTIVES......Page 1020
IN THIS CHAPTER......Page 1021
11.1.1. Three Types of Conics......Page 1022
SKILLS......Page 1024
11.2.1. Parabola with a Vertex at the Origin......Page 1025
11.2.2. Parabola with a Vertex at the Point (h, k)......Page 1029
11.2.3. Applications......Page 1032
[SECTION 11.2] SUMMARY......Page 1033
SKILLS......Page 1034
APPLICATIONS......Page 1035
TECHNOLOGY......Page 1036
11.3.1. Ellipse Centered at the Origin......Page 1037
11.3.2. Ellipse Centered at the Point (h, k)......Page 1041
11.3.3. Applications......Page 1043
[SECTION 11.3] SUMMARY......Page 1045
SKILLS......Page 1046
APPLICATIONS......Page 1047
TECHNOLOGY......Page 1049
11.4.1. Hyperbola Centered at the Origin......Page 1050
11.4.2. Hyperbola Centered at the Point (h, k)......Page 1054
11.4.3. Applications......Page 1056
[SECTION 11.4] SUMMARY......Page 1057
SKILLS......Page 1058
APPLICATIONS......Page 1059
TECHNOLOGY......Page 1060
11.5.1. Solving a System of Nonlinear Equations......Page 1061
SKILLS......Page 1069
CATCH THE MISTAKE......Page 1070
11.6.1. Nonlinear Inequalities in Two Variables......Page 1071
11.6.2. Systems of Nonlinear Inequalities......Page 1073
SKILLS......Page 1076
APPLICATIONS......Page 1077
TECHNOLOGY......Page 1078
11.7.1. Rotation of Axes Formulas......Page 1079
11.7.2. The Angle of Rotation Necessary to Transform a General Second-Degree Equation into an Equation of a Conic......Page 1082
[SECTION 11.7] SUMMARY......Page 1086
CONCEPTUAL......Page 1087
TECHNOLOGY......Page 1088
11.8.1. Equations of Conics in Polar Coordinates......Page 1089
SKILLS......Page 1096
APPLICATIONS......Page 1097
TECHNOLOGY......Page 1098
11.9.1. Parametric Equations of a Curve......Page 1100
11.9.2. Applications of Parametric Equations......Page 1102
SKILLS......Page 1105
APPLICATIONS......Page 1106
TECHNOLOGY......Page 1107
[CHAPTER 11 REVIEW]......Page 1108
[CHAPTER 11 REVIEW EXERCISES]......Page 1111
[CHAPTER 11 PRACTICE TEST]......Page 1115
[CHAPTERS 1–11 CUMULATIVE TEST]......Page 1117
LEARNING OBJECTIVES......Page 1118
IN THIS CHAPTER......Page 1119
12.1.1. Sequences......Page 1120
12.1.2. Factorial Notation......Page 1122
12.1.3. Recursion Formulas......Page 1123
12.1.4. Sums and Series......Page 1124
SKILLS......Page 1127
APPLICATIONS......Page 1128
CATCH THE MISTAKE......Page 1129
TECHNOLOGY......Page 1130
12.2.1. Arithmetic Sequences......Page 1131
12.2.2. The General (nth) Term of an Arithmetic Sequence......Page 1132
12.2.3. The Sum of an Arithmetic Sequence......Page 1133
SKILLS......Page 1136
APPLICATIONS......Page 1137
CONCEPTUAL......Page 1138
12.3.1. Geometric Sequences......Page 1139
12.3.2. The General (nth) Term of a Geometric Sequence......Page 1140
12.3.3. Geometric Series......Page 1141
[SECTION 12.3] SUMMARY......Page 1146
APPLICATIONS......Page 1147
CATCH THE MISTAKE......Page 1148
TECHNOLOGY......Page 1149
12.4.1. Mathematical Induction......Page 1150
[SECTION 12.4] SUMMARY......Page 1152
APPLICATIONS......Page 1153
TECHNOLOGY......Page 1154
12.5.1. Binomial Coefficients......Page 1155
12.5.2. Binomial Expansion......Page 1157
12.5.3. Pascal’s Triangle......Page 1158
12.5.4. Finding a Particular Term of a Binomial Expansion......Page 1160
APPLICATIONS......Page 1161
TECHNOLOGY......Page 1162
12.6.1. The Fundamental Counting Principle......Page 1163
12.6.2. Permutations......Page 1165
12.6.3. Combinations......Page 1167
12.6.4. Permutations with Repetition......Page 1168
APPLICATIONS......Page 1170
TECHNOLOGY......Page 1172
12.7.1. Sample Space......Page 1173
12.7.2. Probability of an Event......Page 1174
12.7.3. Probability of an Event Not Occurring......Page 1175
12.7.4. Mutually Exclusive Events......Page 1176
12.7.5. Independent Events......Page 1178
SKILLS......Page 1179
APPLICATIONS......Page 1180
TECHNOLOGY......Page 1181
[CHAPTER 12 REVIEW]......Page 1182
[CHAPTER 12 REVIEW EXERCISES]......Page 1184
[CHAPTER 12 PRACTICE TEST]......Page 1188
[CHAPTERS 1–12 CUMULATIVE TEST]......Page 1189
Section 0.3......Page 1191
Section 0.5......Page 1192
Review Exercises......Page 1193
Section 1.3......Page 1194
Section 1.4......Page 1195
Section 1.5......Page 1196
Section 1.7......Page 1197
Review Exercises......Page 1198
Section 2.1......Page 1199
Section 2.2......Page 1200
Section 2.3......Page 1202
Section 2.5......Page 1203
Review Exercises......Page 1207
Section 3.1......Page 1208
Section 3.2......Page 1209
Section 3.3......Page 1210
Section 3.4......Page 1212
Review Exercises......Page 1215
Practice Test......Page 1216
Section 4.1......Page 1217
Section 4.2......Page 1218
Section 4.4......Page 1220
Section 4.5......Page 1222
Section 4.6......Page 1223
Review Exercises......Page 1224
Practice Test......Page 1226
Section 5.1......Page 1227
Section 5.2......Page 1228
Section 5.4......Page 1230
Review Exercises......Page 1231
Section 6.1......Page 1232
Section 6.3......Page 1233
Section 6.4......Page 1234
Section 6.6......Page 1235
Section 6.8......Page 1236
Section 6.9......Page 1239
Review Exercises......Page 1241
Cumulative Test......Page 1243
Section 7.3......Page 1244
Section 7.5......Page 1246
Section 7.6......Page 1247
Section 7.8......Page 1248
Review Exercises......Page 1250
Section 8.2......Page 1251
Section 8.4......Page 1252
Section 8.6......Page 1253
Section 8.7......Page 1254
Section 8.8......Page 1255
Review Exercises......Page 1257
Practice Test......Page 1258
Section 9.1......Page 1259
Section 9.4......Page 1260
Review Exercises......Page 1262
Section 10.1......Page 1263
Section 10.2......Page 1264
Section 10.3......Page 1265
Review Exercises......Page 1266
Section 11.2......Page 1267
Section 11.3......Page 1268
Section 11.4......Page 1269
Section 11.6......Page 1270
Section 11.7......Page 1271
Section 11.8......Page 1273
Section 11.9......Page 1274
Review Exercises......Page 1276
Practice Test......Page 1278
Section 12.2......Page 1279
Section 12.5......Page 1280
Review Exercises......Page 1281
Cumulative Test......Page 1282
Applications Index......Page 1283
Subject Index......Page 1287
EULA......Page 1305

Citation preview

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Algebra and Trigonometry [F O U R T H E D I T I O N ]

Cynthia Y. Young PROFESSOR OF MATHEMATICS University of Central Florida

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VICE PRESIDENT & DIRECTOR Laurie Rosatone

FREELANCE PROJECT EDITOR Anne Scanlan-Rohrer

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COVER PHOTO Jupiter Images/Getty Images

This book was set in 10/12 Times by Cenveo® Publisher Services, and printed and bound by Quad/Graphics, Inc. The cover was printed by Quad/Graphics, Inc. This book is printed on acid free paper. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website: www.wiley.com/go/citizenship. Copyright © 2017, 2013, 2009 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201) 748-6011, fax (201) 748-6008, website www.wiley.com/go/permissions. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return mailing label are available at www.wiley.com/go/returnlabel. If you have chosen to adopt this textbook for use in your course, please accept this book as your complimentary desk copy. Outside of the United States, please contact your local sales representative. The inside back cover will contain printing identification and country of origin if omitted from this page. In addition, if the ISBN on the back cover differs from the ISBN on this page, the one on the back cover is correct. ISBN: 978-1-119-32086-9 (ePub) ISBN: 978-1-119-27345-5 (LLPC)

Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

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for

Christopher and Caroline

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University of Central Florida

University of Central Florida

About the Author

Cynthia Y. Young is the Pegasus Professor of Mathematics and the Vice Provost for Faculty Excellence and UCF Global at the University of Central Florida (UCF) and the author of College Algebra, Trigonometry, Algebra and Trigonometry, and Precalculus. She holds a BA in Secondary Mathematics Education from the University of North Carolina (Chapel Hill), an MS in Mathematical Sciences from UCF, and both an MS in Electrical Engineering and a PhD in Applied Mathematics from the University of Washington. She has taught high school in North Carolina and Florida, developmental mathematics at Shoreline Community College in Washington, and undergraduate and graduate students at UCF. Dr. Young joined the faculty at UCF in 1997 as an assistant professor of mathematics, and her primary research area was the mathematical modeling of the atmospheric effects on propagating laser beams. Her atmospheric propagation research was recognized by the Office of Naval Research Young Investigator Award, and in 2017 she was selected as a fellow of the International Society for Optical Engineering. Her secondary area of research centers on improvement of student learning in mathematics. She has authored or co-authored over 60 books and articles and has served as the principal investigator or co-principal investigator on projects with more than $2.5 million in federal funding. Dr. Young was on the team at UCF that developed the UCF EXCEL program, which was originally funded by the National Science Foundation to support the increase in the number of students graduating with a degree in science, technology, engineering, and mathematics (STEM). The EXCEL learning community approach centered around core mathematics courses has resulted in a significant increase in STEM graduation rates and has been institutionalized at UCF. Dr. Young has been the recipient of many of UCF’s awards (Excellence in Undergraduate Teaching, Excellence in Research, Teaching Incentive Program, Research Incentive Program, Scholarship of Teaching and Learning award, and UCF’s highest honor, UCF Pegasus Professor). She has shared her techniques and experiences with colleagues around the country through talks at colleges, universities, and conferences.

VI

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Preface As a mathematics professor, I would hear my students say, “I understand you in class, but when I get home I am lost.” When I would probe further, students would continue with “I can’t read the book.” As a mathematician, I always found mathematics textbooks quite easy to read—and then it dawned on me: Don’t look at this book through a mathematician’s eyes; look at it through the eyes of students who might not view mathematics the same way that I do. What I found was that the books were not at all like my class. Students understood me in class, but when they got home they couldn’t understand the book. It was then that the folks at Wiley lured me into writing. My goal was to write a book that is seamless with how we teach and is an ally (not an adversary) to student learning. I wanted to give students a book they could read without sacrificing the rigor needed for conceptual understanding. The following quote comes from a reviewer when asked about the rigor of the book: I would say that this text comes across as a little less rigorous than other texts, but I think that stems from how easy it is to read and how clear the author is. When one actually looks closely at the material, the level of rigor is high.

DISTINGUISHING FEATURES

Four key features distinguish this book from others, and they came directly from my classroom.

PARALLEL WORDS AND MATH Have you ever looked at your students’ notes? I found that my students were only scribbling down the mathematics that I would write—never the words that I would say in class. I started passing out handouts that had two columns: one column for math and one column for words. Each example would have one or the other; either the words were there and students had to fill in the math or the math was there and students had to fill in the words. If you look at the examples in this book, you will see that the words (your voice) are on the left and the mathematics is on the right. In most math books, when the author illustrates an example, the mathematics is usually down the center of the page, and if the students don’t know what mathematical operation was performed, they will look to the right for some brief statement of help. That’s not how we teach; we don’t write out an example on the board and then say, “Class, guess what I just did!” Instead we lead our students, telling them what step is coming and then performing that mathematical step together—and reading naturally from left to right. Student reviewers have said that the examples in this book are easy to read; that’s because your voice is right there with them, working through problems together.

VII

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VIII 

PREFACE

SKILLS AND CONCEPTS (LEARNING OBJECTIVES AND EXERCISES) In my experience as a mathematics teacher/instructor/professor, I find skills to be on the micro level and concepts on the macro level of understanding mathematics. I believe that too often skills are emphasized at the expense of conceptual understanding. I have purposely separated learning objectives at the beginning of every section into two categories: skills objectives—what students should be able to do—and conceptual objectives—what students should understand. At the beginning of every class, I discuss the learning objectives for the day—both skills and concepts. These are reinforced with both skills exercises and conceptual exercises. Each subsection has a corresponding skills objective and conceptual objective.

CATCH THE MISTAKE Have you ever made a mistake (or had a student bring you his or her homework with a mistake) and you’ve gone over it and over it and couldn’t find the mistake? It’s often easier to simply take out a new sheet of paper and solve it from scratch than it is to actually find the mistake. Finding the mistake demonstrates a higher level of understanding. I include a few Catch the Mistake exercises in each section that demonstrate a common mistake. Using these in class (with individuals or groups) leads to student discussion and offers an opportunity for formative assessment in real time.

LECTURE VIDEOS BY THE AUTHOR I authored the videos to ensure consistency in the students’ learning experience. Throughout the book, wherever a student sees the video icon, that indicates a video. These videos provide mini lectures. The chapter openers and chapter summaries act as class discussions. The “Your Turn” problems throughout the book challenge the students to attempt a problem similar to a nearby example. The “worked-out example” videos are intended to come to the rescue for students if they get lost as they read the text and work problems outside the classroom.

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PREFACE 

IX

NEW TO THE FOURTH EDITION In the fourth edition, the main upgrades are updated applications throughout the text; new Skills and Conceptual objectives mapped to each subsection; new Concept Check questions in each subsection; and the substantially improved version of WileyPLUS, including ORION adaptive practice and interactive animations.

SKILLS AND CONCEPTUAL OBJECTIVES

CONCEPT CHECK QUESTIONS

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APPLICATIONS TO BUSINESS, ECONOMICS, HEATH SCIENCES, AND MEDICINE

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X 

PREFACE

FEATURE

BENEFIT TO STUDENT

Chapter-Opening Vignette

Piques the student’s interest with a real-world application of material presented in the chapter. Later in the chapter, the concept from the vignette is reinforced.

Chapter Overview, Flowchart, and Learning Objectives

Allows students to see the big picture of how topics relate, and overarching learning objectives are presented.

Skills and Conceptual Objectives

Skills objectives represent what students should be able to do. Conceptual objectives emphasize a higher-level, global perspective of concepts.

Clear, Concise, and Inviting Writing Style, Tone, and Layout

Enables students to understand what they are reading, which reduces math anxiety and promotes student success.

Parallel Words and Math

Increases students’ ability to read and understand examples with a seamless representation of their instructor’s class (instructor’s voice and what they would write on the board).

Common Mistakes

Addresses a different learning style: teaching by counterexample. Demonstrates common mistakes so that students understand why a step is incorrect and reinforces the correct mathematics.

Color for Pedagogical Reasons

Particularly helpful for visual learners when they see a function written in red and then its corresponding graph in red or a function written in blue and then its corresponding graph in blue.

Study Tips

Reinforces specific notes that you would want to emphasize in class.

Author Videos

Gives students a mini class of several examples worked by the author.

Your Turn

Engages students during class, builds student confidence, and assists instructor in real-time assessment.

Concept Checks

Reinforces concept learning objectives, much as Your Turn features reinforce skill learning objectives.

Catch the Mistake Exercises

Encourages students to assume the role of teacher—demonstrating a higher mastery level.

Conceptual Exercises

Teaches students to think more globally about a topic.

Inquiry-Based Learning Project (online only)

Lets students discover a mathematical identity, formula, and the like that is derived in the book.

Modeling Our World (online only)

Engages students in a modeling project of a timely subject: global climate change.

Chapter Review

Presents key ideas and formulas section by section in a chart. Improves study skills.

Chapter Review Exercises

Improves study skills.

Chapter Practice Test

Offers self-assessment and improves study skills.

Cumulative Test

Improves retention.

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PREFACE 

XI

INSTRUCTOR SUPPLEMENTS INSTRUCTOR’S SOLUTIONS MANUAL (ISBN: 978-1-119-27343-1)

• Contains worked-out solutions to all exercises in the text. INSTRUCTOR’S MANUAL

Authored by Cynthia Young, the manual provides practical advice on teaching with the text, including:

• sample lesson plans and homework assignments • suggestions for the effective utilization of additional resources and supplements • sample syllabi • Cynthia Young’s Top 10 Teaching Tips & Tricks • online component featuring the author presenting these Tips & Tricks ANNOTATED INSTRUCTOR’S EDITION (ISBN: 978-1-119-27346-2)

• Displays answers to the vast majority of exercise questions in the back of the book. • Provides additional classroom examples within the standard difficulty range of the in-text exercises, as well as challenge problems to assess your students’ mastery of the material. POWERPOINT SLIDES

• For each section of the book, a corresponding set of lecture notes and worked- out examples are presented as PowerPoint slides, available on the Book Companion Site (www.wiley.com/college/young) and WileyPLUS. TEST BANK

• Contains approximately 900 questions and answers from every section of the text. COMPUTERIZED TEST BANK

Electonically enhanced version of the Test Bank that • contains approximately 900 algorithmically generated questions. • allows instructors to freely edit, randomize, and create questions. • allows instructors to create and print different versions of a quiz or exam. • recognizes symbolic notation. BOOK COMPANION WEBSITE (WWW.WILEY.COM/COLLEGE/YOUNG)

• Contains all instructor supplements listed plus a selection of personal response system questions. WILEYPLUS

• WileyPLUS online homework features a full-service, digital learning environment, including additional resources for students, such as lecture videos by the author, self-practice exercises, tutorials, integrated links between the online text and supplements, and new interactive animations and ORION adaptive practice. • WileyPLUS has been substantially revised and improved since the third edition of Algebra and Trigonometry. It now includes ORION, an adaptive practice engine built directly into WileyPLUS that can connect directly into the WileyPLUS gradebook, or into your campus Learning Management System gradebook if you select that option. Wiley has been incorporating ORION into WileyPLUS courses for over five years, including the Young Precalculus program. ORION brings the power of adaptive learning, which will continue to help students and instructors “bridge the gap.”

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XII 

PREFACE

STUDENT SUPPLEMENTS STUDENT SOLUTIONS MANUAL (ISBN: 978-1-119-27342-4)

• Includes worked-out solutions for all odd problems in the text. BOOK COMPANION WEBSITE (WWW.WILEY.COM/COLLEGE/YOUNG)

• Provides additional resources for students to enhance the learning experience.

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PREFACE 

XIII

ACKNOWLEDGMENTS

I want to express my sincerest gratitude to the entire Wiley team. I’ve said this before, and I will say it again: Wiley is the right partner for me. There is a reason that my dog is named Wiley—she’s smart, competitive, a team player, and most of all, a joy to be around. There are several people within Wiley to whom I feel the need to express my appreciation: first and foremost to Laurie Rosatone, who convinced Wiley Higher Ed to invest in a young assistant professor’s vision for a series and who has been unwavering in her commitment to student learning. To my editor Joanna Dingle, whose judgment I trust in both editorial and preschool decisions; thank you for surpassing my greatest expectations for an editor. To the rest of the math editorial team (Jennifer Lartz, Anne Scanlan-Rohrer, and Ryann Dannelly), you are all first class! This revision was planned and executed exceptionally well thanks to you. To the math marketing manager John LaVacca, thank you for helping reps tell my story: you are outstanding at your job. To product designer David Dietz, many thanks for your role in developing the online course and digital assets. To Mary Sanger, thank you for your attention to detail. To Maureen Eide, thank you for the new design! And finally, I’d like to thank all of the Wiley reps: thank you for your commitment to my series and your tremendous efforts to get professors to adopt this book for their students. I would also like to thank all of the contributors who helped us make this an even better book. I’d first like to thank Mark McKibben. He is known as the author of the solutions manuals that accompany this series, but he is much more than that. Mark, thank you for making this series a priority, for being so responsive, and most of all for being my “go-to” person to think through ideas. I’d also like to thank Jodi B.A. McKibben, who is a statistician and teamed with Mark to develop the regression material. I’d like to thank Steve Davis, who was the inspiration for the Inquiry-Based Learning Projects (IBLPs) and a huge thanks to Lyn Riverstone, who developed all of the IBLPs. Special thanks to Laura Watkins for finding applications that are real and timely and to Ricki Alexander for updating all of the Technology Tips. I’d also like to thank Becky Schantz for her environmental problems (I now use AusPens because of Becky). Many thanks to Marie Vanisko, Jennifer Blue, and Diane Cook for accuracy checking of the text, exercises, and solutions. I’d also like to thank the following reviewers, whose input helped make this book even better.

Text Reviewers Sandy Carlson, Pennsylvania College of Technology Ze-Li Dou, Texas Christian University Deborah Lynn Doucette, Erie Community College – North Campus Scott Gentile, Hunter College, CUNY Mike Kirby, Tidewater Community College

Bridgett Lee, Georgia Southern University Lebbie Lee Ligon, Piedmont Technical College Kelly Pearson, Murray State University Colleen Quinn, Erie Community College Rachel Rader, Ohio Northern University Sean Smith, California State Polytechnic University, Pomona

Animations Reviewers Monika H. Champion, Ivy Tech Community College Kim Christensen, Metropolitan Community College Yamir DeJesus-Decena, Dutchess Community College Marissa Ford, Ivy Tech Community College Barbara Hess, California University of Pennsylvania Christina K. Houston, Community College of Allegheny County Phyllis Lefton, Manhattanville College

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Carrie McCammon, Ivy Tech Community College Holly J. Middleton, University of North Florida Becky Moening, Ivy Tech Community College, Warsaw Denise Race, Eastfield College Edward Schwartz, Manhattanville College Mike Shirazi, Germanna Community College Misty Vorder, Hillsborough Community College, Brandon

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Contents

Hero Images/Getty Images, Inc.

fotografixx/Getty Images

[0]  Prerequisites and Review  0.1 0.2 0.3 0.4 0.5 0.6 0.7

2

REAL NUMBERS  4 INTEGER EXPONENTS AND SCIENTIFIC NOTATION  17 POLYNOMIALS: BASIC OPERATIONS  27 FACTORING POLYNOMIALS  36 RATIONAL EXPRESSIONS  46 RATIONAL EXPONENTS AND RADICALS  60 COMPLEX NUMBERS  70

Review 77 | Review Exercises 79 | Practice Test 81

[1]  Equations and Inequalities  1.1 1.2 1.3 1.4 1.5 1.6 1.7

82

LINEAR EQUATIONS  84 APPLICATIONS INVOLVING LINEAR EQUATIONS  93 QUADRATIC EQUATIONS  107 OTHER TYPES OF EQUATIONS  121 LINEAR INEQUALITIES  129 POLYNOMIAL AND RATIONAL INEQUALITIES  139 ABSOLUTE VALUE EQUATIONS AND INEQUALITIES  149

Review  157 | Review Exercises  158 | Practice Test  162 | Cumulative Test  163

Fuse/Getty Images

[2]  Graphs 164 2.1 2.2 2.3 2.4 2.5*

BASIC TOOLS: CARTESIAN PLANE, DISTANCE, AND MIDPOINT  166 GRAPHING EQUATIONS, POINT-PLOTTING, INTERCEPTS, AND SYMMETRY  173 LINES 186 CIRCLES 202 LINEAR REGRESSION: BEST FIT  209

SuperStock/Alamy Stock Photo

Review  231 | Review Exercises  232 | Practice Test  234 | Cumulative Test  235

[3]  Functions and Their Graphs  3.1 3.2 3.3 3.4 3.5 3.6

236

FUNCTIONS 238 GRAPHS OF FUNCTIONS; PIECEWISE-DEFINED FUNCTIONS; INCREASING AND DECREASING FUNCTIONS; AVERAGE RATE OF CHANGE  255 GRAPHING TECHNIQUES: TRANSFORMATIONS  274 OPERATIONS ON FUNCTIONS AND COMPOSITION OF FUNCTIONS  287 ONE-TO-ONE FUNCTIONS AND INVERSE FUNCTIONS  298 MODELING FUNCTIONS USING VARIATION  312

Review  322 | Review Exercises  324 | Practice Test  328 | Cumulative Test  329

XIV

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CONTENTS 

4.1 4.2 4.3 4.4 4.5 4.6

330

Focus On Sport/Getty Images, Inc.

[4]  Polynomial and Rational Functions 

XV

QUADRATIC FUNCTIONS  332 POLYNOMIAL FUNCTIONS OF HIGHER DEGREE  349 DIVIDING POLYNOMIALS: LONG DIVISION AND SYNTHETIC DIVISION  363 THE REAL ZEROS OF A POLYNOMIAL FUNCTION  372 COMPLEX ZEROS: THE FUNDAMENTAL THEOREM OF ALGEBRA  388 RATIONAL FUNCTIONS  396

Review  416 | Review Exercises  418 | Practice Test  422 | Cumulative Test  423

5.1 5.2 5.3 5.4 5.5

424

Francois Gohier/Science Source

[5]  Exponential and Logarithmic Functions  EXPONENTIAL FUNCTIONS AND THEIR GRAPHS  426 LOGARITHMIC FUNCTIONS AND THEIR GRAPHS  440 PROPERTIES OF LOGARITHMS  456 EXPONENTIAL AND LOGARITHMIC EQUATIONS  464 EXPONENTIAL AND LOGARITHMIC MODELS  474

Review  485 | Review Exercises  487 | Practice Test  490 | Cumulative Test  491

6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9

492

Richard Cummins / Getty Images, Inc.

[6]  Trigonometric Functions 

ANGLES, DEGREES, AND TRIANGLES  494 DEFINITION 1 OF TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE RATIOS  511 APPLICATIONS OF RIGHT TRIANGLE TRIGONOMETRY: SOLVING RIGHT TRIANGLES  525 DEFINITION 2 OF TRIGONOMETRIC FUNCTIONS: CARTESIAN PLANE  538 TRIGONOMETRIC FUNCTIONS OF NONACUTE ANGLES  552 RADIAN MEASURE AND APPLICATIONS  567 DEFINITION 3 OF TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH  584 GRAPHS OF SINE AND COSINE FUNCTIONS  594 GRAPHS OF OTHER TRIGONOMETRIC FUNCTIONS  623

Review  641 | Review Exercises  647 | Practice Test  650 | Cumulative Test  651

7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8

652

BASIC TRIGONOMETRIC IDENTITIES  654 VERIFYING TRIGONOMETRIC IDENTITIES  665 SUM AND DIFFERENCE IDENTITIES  675 DOUBLE-ANGLE IDENTITIES  689 HALF-ANGLE IDENTITIES  697 PRODUCT-TO-SUM AND SUM-TO-PRODUCT IDENTITIES  708 INVERSE TRIGONOMETRIC FUNCTIONS  716 TRIGONOMETRIC EQUATIONS  737

Review  754 | Review Exercises  757 | Practice Test  762 | Cumulative Test  763

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laurentiu iordache / Alamy Stock Photo

[7]  Analytic Trigonometry 

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XVI 

CONTENTS

[8]  Additional Topics in Trigonometry  8.1 8.2 8.3 8.4 8.5 8.6 8.7

Bermuda Atlantic Ocean

Miami, Florida

San Juan, Puerto Rico

8.8

764

OBLIQUE TRIANGLES AND THE LAW OF SINES  766 THE LAW OF COSINES  780 THE AREA OF A TRIANGLE  791 VECTORS 798 THE DOT PRODUCT  813 POLAR (TRIGONOMETRIC) FORM OF COMPLEX NUMBERS  821 PRODUCTS, QUOTIENTS, POWERS, AND ROOTS OF COMPLEX NUMBERS; DE MOIVRE’S THEOREM  829 POLAR EQUATIONS AND GRAPHS  841

Review  856 | Review Exercises  859 | Practice Test  862 | Cumulative Test  863

[9]  Systems of Linear Equations and Inequalities  PAUL J. RICHARDS/ Getty Images, Inc.

9.1 9.2 9.3 9.4 9.5

864

SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES  866 SYSTEMS OF LINEAR EQUATIONS IN THREE VARIABLES  881 PARTIAL FRACTIONS  892 SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES  903 THE LINEAR PROGRAMMING MODEL  914

Review  921 | Review Exercises  922 | Practice Test  924 | Cumulative Test  925

[10]  Matrices 926

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10.1 10.2 10.3 10.4

MATRICES AND SYSTEMS OF LINEAR EQUATIONS  928 MATRIX ALGEBRA  950 MATRIX EQUATIONS; THE INVERSE OF A SQUARE MATRIX  964 THE DETERMINANT OF A SQUARE MATRIX AND CRAMER’S RULE  976

Review  990 | Review Exercises  993 | Practice Test  996 | Cumulative Test  997

receives

encrypts

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CONTENTS 

XVII

Ribeiro antonio/Shutterstock

[11]  Analytic Geometry and Systems of Nonlinear Equations and Inequalities  998 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9

CONIC BASICS  1000 THE PARABOLA  1003 THE ELLIPSE  1015 THE HYPERBOLA  1028 SYSTEMS OF NONLINEAR EQUATIONS  1039 SYSTEMS OF NONLINEAR INEQUALITIES  1049 ROTATION OF AXES  1057 POLAR EQUATIONS OF CONICS  1067 PARAMETRIC EQUATIONS AND GRAPHS  1078

Review  1086 | Review Exercises  1089 | Practice Test  1093 | Cumulative Test  1095

SEQUENCES AND SERIES  1098 ARITHMETIC SEQUENCES AND SERIES  1109 GEOMETRIC SEQUENCES AND SERIES  1117 MATHEMATICAL INDUCTION  1128 THE BINOMIAL THEOREM  1133 COUNTING, PERMUTATIONS, AND COMBINATIONS  1141 PROBABILITY 1151

Review  1160 | Review Exercises  1162 | Practice Test  1166 | Cumulative Test  1167

Answers to Odd-Numbered Exercises  1169 Applications Index  1261 Subject Index  1265

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A BASIC STRATEGY FOR BLACKJACK Dealer’s Up Card 2 S 17+ S 16 S 15 I S 14 S 13 H 12 D 11 D 10 II H 9 H 5-8 A, 8 - 10 S S A, 7 H A, 6 H III A, 5 H A, 4 H A, 3 H A, 2 A, A; 8, 8 SP S 10, 10 SP 9, 9 IV SP 7, 7 6, 6 H 5, 5 D 4, 4 H 3, 3 H 2, 2 H HIT

3 S S S S S H D D D H S D D H H H H SP S SP SP SP D H H H

4 S S S S S S D D D H S D D D D H H SP S SP SP SP D H SP SP

5 S S S S S S D D D H S D D D D D D SP S SP SP SP D H SP SP

6 S S S S S S D D D H S D D D D D D SP S SP SP SP D H SP SP

7 S H H H H H D D H H S S H H H H H SP S S SP H D H SP SP

8 S H H H H H D D H H S S H H H H H SP S SP H H D H H H

9 S H H H H H D D H H S H H H H H H SP S SP H H D H H H

10 S H H H H H D H H H S H H H H H H SP S S H H H H H H

A S H H H H H H H H H S H H H H H H SP S S H H H H H H

When surrender is allowed, surrender 9, 7 or 10, 6 vs 9, 10, A; 9, 6 or 10, 5 vs 10 When doubling down after splitting is allowed, split: 2’s, 3’s, 7’s vs 2-7; 4’s vs 5 or 6; 6’s vs 2-6

12.1 12.2 12.3 12.4 12.5 12.6 12.7

1096

Your Hand

[12]  Sequences, Series, and Probability 

STAND DOUBLE DOWN SPLIT

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A Note from the Author TO THE STUDENT

I wrote this text with careful attention to ways in which to make your learning experience more successful. If you take full advantage of the unique features and elements of this textbook, I believe your experience will be fulfilling and enjoyable. Let’s walk through some of the special book features that will help you in your study of College Algebra. Prerequisites and Review [Chapter 0] A comprehensive review of prerequisite knowledge (intermediate algebra topics) in Chapter 0 provides a brushup on knowledge and skills necessary for success in the course.

Clear, Concise, and Inviting Writing Special attention has been paid to presenting an engaging, clear, precise narrative in a layout that is easy to use and designed to reduce any math anxiety you may have.

Chapter Introduction, Flowchart, Section Headings, and Objectives An opening vignette, flowchart, list of chapter sections, and chapter learning objectives give you an overview of the chapter.

Skills and Conceptual Objectives For every section, objectives are further divided by skills and concepts so you can see the difference between solving problems and truly understanding concepts.

XVIII

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A NOTE FROM THE AUTHOR TO THE STUDENT  

XIX

Examples Examples pose a specific problem using concepts already ­presented and then work through the solution. These serve to enhance your understanding of the subject matter.

Your Turn Immediately following many examples, you are given a similar ­problem to reinforce and check your understanding. This helps build confidence as you progress in the chapter. These are ideal for in-class activity or for preparing for homework later. Answers are provided in the margin for a quick check of your work.

Common Mistake/Correct Versus Incorrect In addition to standard examples, some problems are worked out both correctly and incorrectly to highlight common errors. Counterexamples like these are often an effective learning approach.

Parallel Words and Math This text reverses the common textbook presentation of examples by placing the explanation in words on the left and the mathematics in parallel on the right. This makes it easier to read through examples as the material flows more naturally from left to right and as commonly presented in class.

Study Tips and Caution Notes These marginal reminders call out important hints or warnings to be aware of related to the topic or problem.

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XX 

A NOTE FROM THE AUTHOR TO THE STUDENT

Video Icons Video icons appear on all chapter introductions, chapter and section reviews, as well as selected examples throughout the chapter to indicate that the author has created a video segment for that element. These video clips help you work through the selected examples with the author as your “private tutor.”

Six Types of Exercises Every text section ends with Skills, Applications, Catch the Mistake, Conceptual, Challenge, and Technology exercises. The exercises gradually increase in difficulty and vary in skill and conceptual emphasis. Catch the Mistake exercises increase the depth of understanding and reinforce what you have learned. Conceptual and Challenge exercises specifically focus on assessing conceptual understanding. Technology exercises enhance your understanding and ability using ­scientific and graphing calculators.

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A NOTE FROM THE AUTHOR TO THE STUDENT  

1

Concept Check Questions Similar to how Your Turn features reinforce skill learning objectives, Concept Checks reinforce concept learning objectives.

Chapter Review, Review Exercises, Practice Test, Cumulative Test At the end of every chapter, a summary review chart organizes the key learning concepts in an easy-to-use one- or two-page layout. This feature includes key ideas and formulas, as well as indicating relevant pages and review exercises so that you can quickly summarize a chapter and study smarter. Review Exercises, arranged by section heading, are provided for extra study and practice. A Practice Test, without section headings, offers even more self-practice before moving on. A new Cumulative Test feature offers study questions based on all previous chapters’ content, thus helping you build upon previously learned concepts.

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[0[ CHAPTER

Prerequisites and Review

probably would say no because the foundation is “shaky.” Would you be able to walk successfully along a beam (4 inches wide)? Most people would probably say yes—even though for some of us it is still challenging. Think of this chapter as the foundation for your walk. The more solid your foundation is now, the more successful your walk through College Algebra will be. The purpose of this chapter is to review concepts and skills that you already have learned in a previous course. Mathematics is a cumulative subject in that it requires a solid foundation to proceed to the next level. Use this chapter to reaffirm your current knowledge base before jumping into the course.

fotografixx/Getty Images

kasto80/Getty Images, Inc.

Would you be able to walk successfully along a tightrope? Most people

LEARNING OBJECTIVES ■■ Understand

that rational and irrational numbers together constitute the real numbers. ■■ Apply properties of exponents.

Young_AT_6160_ch00_pp02-45.indd 2

■■ Perform

operations on polynomials. ■■ Factor polynomials. ■■ Simplify expressions that contain rational exponents.

■■ Simplify

radicals. complex numbers in standard form.

■■ Write

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[IN THIS CHAPTER] Real numbers, integer exponents, and scientific notation will be discussed, followed by rational exponents and radicals. Simplification of radicals and rationalization of denominators will be reviewed. Basic operations such as addition, subtraction, and multiplication of polynomials will be discussed followed by a review of how to factor polynomials. Rational expressions will be discussed, and a brief overview of solving simple algebraic equations will be given. After reviewing all of these aspects of real numbers, this chapter will conclude with a review of complex numbers.

PR E R E QUI S I T E S AN D R E V I E W 0.1

0.2

0.3

0.4

0.5

0.6

0.7

REAL NUMBERS

INTEGER EXPONENTS AND SCIENTIFIC NOTATION

POLYNOMIALS: BASIC OPERATIONS

FACTORING POLYNOMIALS

RATIONAL EXPRESSIONS

RATIONAL EXPONENTS AND RADICALS

COMPLEX NUMBERS

• The Set of Real Numbers • Approxima­ tions: Round­ ing and Truncation • Order of Operations • Properties of Real Numbers

• Integer Exponents • Scientific Notation

• Adding and Subtracting Polynomials • Multiplying Polynomials • Special Products

• Greatest Common Factor • Factoring Formulas: Special Polynomial Forms • Factoring a Trinomial as a Product of Two Binomials • Factoring by Grouping • A Strategy for Factoring Polynomials

• Rational Expressions and Domain Restrictions • Simplifying Rational Expressions • Multiplying and Dividing Rational Expressions • Adding and Subtracting Rational Expressions • Complex Rational Expressions

• Square Roots • Other (nth) Roots • Rational Exponents

• The Imaginary Unit, i • Adding and Subtracting Complex Numbers • Multiplying Complex Numbers • Dividing Complex Numbers • Raising Complex Numbers to Integer Powers

3

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4 

CHAPTER 0  Prerequisites and Review

0.1 REAL NUMBERS SKILLS OBJECTIVES ■■ Classify real numbers as rational or irrational. ■■ Round or truncate real numbers. ■■ Simplify expressions and evaluate algebraic expressions using the correct order of operations. ■■ Apply properties of real numbers and basic rules of algebra in simplifying and evaluating expressions.

CONCEPTUAL OBJECTIVES ■■ Understand that rational and irrational numbers are mutually exclusive and complementary subsets of real numbers. ■■ Understand the difference between rounding and truncating decimal values and that the resulting approximations may or may not be equal. ■■ Learn the order of operations for real numbers. ■■ Know and understand the basic properties of real numbers and the basic rules of algebra.

0.1.1  The Set of Real Numbers 0.1.1 S K I L L

Classify real numbers as rational or irrational. 0.1.1 C O N C E P T U A L

Understand that rational and irrational numbers are mutually exclusive and complementary subsets of real numbers.

A set is a group or collection of objects that are called members or elements of the set. If every member of set B is also a member of set A, then we say B is a subset of A and denote it as B ( A. For example, the starting lineup on a baseball team is a subset of the entire team. The set of natural numbers, 51, 2, 3, 4, . . .6, is a subset of the set of whole numbers, 50, 1, 2, 3, 4, . . .6, which is a subset of the set of integers, 5. . . , 24, 23, 22, 21, 0, 1, 2, 3, . . .6, which is a subset of the set of rational numbers, which is a subset of the set of real ­numbers. The three dots, called an ellipsis, indicate that the pattern continues indefinitely. If a set has no elements, it is called the empty set, or null set, and is denoted by the symbol [. The set of real numbers consists of two main subsets: rational and ­irrational numbers. DEFINITION

Rational Number

A rational number is a number that can be expressed as a quotient (ratio) of two a integers, ,where the integer a is called the numerator and the integer b is called b the denominator and where b 2 0 . Rational numbers include all integers or all fractions that are ratios of integers. Note that any integer can be written as a ratio whose denominator is equal to 1. In decimal form, the rational numbers are those that terminate or are nonterminating with a repeated decimal pattern, which is represented with an overbar. Those decimals that do not repeat and do not terminate are irrational numbers. The numbers 1 19 5,  217,   ,  !2,  p,  1.37,  0,  2 ,  3.666,  3.2179. . . 3 17 are examples of real numbers, where 5, 217, 13, 1.37, 0, 2  11 79 , and 3.666 are rational numbers, and !2, p, and 3.2179 . . . are irrational numbers. It is important to note that the ellipsis following the last decimal digit denotes continuing in an irregular fashion, whereas the absence of such dots to the right of the last decimal digit implies that the decimal expansion terminates.

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RATIONAL NUMBER (FRACTION)

CALCULATOR DISPLAY

DECIMAL REPRESENTATION

DESCRIPTION

7 2

3.5

3.5

Terminates

15 12

1.25

1.25

Terminates

2 3

0.666666666

0.6

Repeats

1 11

0.09090909

0 .0 9

Repeats

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0.1  Real Numbers 

5

Notice that the overbar covers the entire repeating pattern. The following figure and table illustrate the subset relationship and examples of different types of real numbers. Real Numbers

Irrational Numbers

Rational Numbers

Integers

Negative Counting Numbers

STUD Y T I P

Fractions Ratio of two nonzero integers that does not reduce to an integer

Every real number is either a rational number or an irrational number.

Whole Numbers

Zero

Natural Numbers

SYMBOL

NAME

DESCRIPTION

N

Natural numbers

Counting numbers

1, 2, 3, 4, 5, . . .

W

Whole numbers

Natural numbers and zero

0, 1, 2, 3, 4, 5, . . .

Z

Integers

Whole numbers and negative natural numbers

. . . , 25, 24, 23, 22, 21, 0, 1, 2, 3, 4, 5, . . .

a Ratios of integers: 1 b 2 0 2 b •  Decimal representation terminates, or

Q

Rational numbers

I

Irrational numbers

Numbers whose decimal representation does not terminate or repeat

R

Real numbers

Rational and irrational numbers

EXAMPLES

1 217, 2 19 7 , 0, 3 , 1.37, 3 .6 6 6 , 5

•  Decimal representation repeats !2, 1.2179. . . , p

p, 5, 2 32 , 17.25, !7

Since the set of real numbers can be formed by combining the set of rational numbers and the set of irrational numbers, then every real number is either rational or irrational. The set of rational numbers and the set of irrational numbers are mutually exclusive (no shared elements) and complementary sets. The real number line is a graph used to represent the set of all real numbers.

Classify the following real numbers as rational or irrational:

Solution:

1 4,

!3, p, 7.51,

1 3,

258,

6.66666

Rational:  23, 0, 14 , 7.51, 13 ,  258 ,  26.66666 Irrational: !3, p

▼

Y O U R T U R N   Classify the following real numbers as rational or irrational: 273, 5.9999, 12, 0, 25.27, !5, 2.010010001 . . .

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–3 –2 –1

√2 0

1

2

3

[ CONCEPT CHECK ]

EXAMPLE 1  Classifying Real Numbers 23, 0,

– 19 17

TRUE OR FALSE  All integers are rational numbers.

▼ ANSWER True

▼ ANSWER

Rational: 273 , 5.9999, 12, 0, 25.27 Irrational: !5 , 2.010010001 . . .

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6 

CHAPTER 0  Prerequisites and Review

0.1.2  Approximations: Rounding and Truncation 0.1.2 S K I L L

Round or truncate real numbers. 0.1.2 C O N C E P T U A L

Understand the difference between rounding and truncating decimal values and know that the resulting approximations may or may not be equal.

Every real number can be represented by a decimal. When a real number is in decimal form, it can be approximated by either rounding off or truncating to a given decimal place. Truncation is “cutting off” or eliminating everything to the right of a certain decimal place. Rounding means looking to the right of the specified decimal place and making a judgment. If the digit to the right is greater than or equal to 5, then the specified digit is rounded up, or increased by one unit. If the digit to the right is less than 5, then the specified digit stays the same. In both of these cases all decimal places to the right of the specified place are removed. EXAMPLE 2  Approximating Decimals to Two Places

[ CONCEPT CHECK ]

Approximate 17.368204 to two decimal places by a.  truncation   b. rounding

TRUE OR FALSE  Truncating and rounding always have the same result.

Solution:

▼

b. To round, look to the right of the 6.

ANSWER False

Because “8” is greater than 5, round up (add 1 to the 6).

▼ ANSWER a.  Truncation: 23.02

17.36

a. To truncate, eliminate all digits to the right of the 6.

17.37

▼ Y O U R T U R N   Approximate 23.02492 to two decimal places by a.  truncation   b. rounding

b.  Rounding: 23.02

EXAMPLE 3  Approximating Decimals to Four Places

ST U DY TIP When rounding, look to the right of the specified decimal place and use that digit (do not round that digit first). 5.23491 rounded to two ­decimal places is 5.23 (do not round the 4 to a 5 first).

▼ ANSWER a. Truncation: 22.3818 b. Rounding: 22.3819

Approximate 7.293516 to four decimal places by a.  truncation   b. rounding Solution:

The “5” is in the fourth decimal place. a. To truncate, eliminate all digits to the right of 5. b. To round, look to the right of the 5. Because “1” is less than 5, the 5 remains the same.

7.2935 7.2935

▼ Y O U R T U R N   Approximate 22.381865 to four decimal places by a.  truncation   b. rounding

It is important to note that rounding and truncation sometimes yield the same approximation (Example 3), but not always (Example 2).

0.1.3  Order of Operations Addition, subtraction, multiplication, and division are called arithmetic operations. The results of these operations are called the sum, difference, product, and quotient, respectively. These four operations are summarized in the following table. OPERATION

RESULT

Addition

a1b

Sum

Subtraction

a2b

Difference

Multiplication

a # b or ab or  1a2 1b2

Product

Division

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NOTATION

a  or a /b  1 b 2 0 2 b

Quotient (Ratio)

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0.1  Real Numbers 

Since algebra involves variables such as x, the t­ raditional multiplication sign 3 is not used. Three alternatives are shown in the preceding table. Similarly, the arithmetic sign for division 4 is often represented by vertical or slanted fractions. The symbol 5 is called the equal sign and is pronounced “equals” or “is.” It implies that the expression on one side of the equal sign is equivalent to (has the same value as) the expression on the other side of the equal sign.

7

0.1.3 S K I L L

Simplify expressions and evaluate algebraic expressions using the correct order of operations. 0.1.3 C O N C E P T U A L

WORDS MATH Learn the order of operations for The sum of seven and eleven equals eighteen: 7 1 11 5 18 real numbers.

Three times five is fifteen: Four times six equals twenty-four: Eight divided by two is four: Three subtracted from five is two:

3 # 5 5 15 4 162 5 24 8 54 2 5 2 3 5 2

When evaluating expressions involving real numbers, it is important to remember the correct order of operations. For example, how do we simplify the expression 3 1 2 # 5? Do we multiply first and then add, or do we add first and then multiply? In mathematics, conventional order implies multiplication first and then addition: 3 1 2 # 5 5 3 1 10 5 13. Parentheses imply grouping of terms, and the necessary operations should always be performed inside them first. If there are nested parentheses, always start with the innermost parentheses and work your way out. Within parentheses follow the conventional order of operations. Exponents are an important part of the order of operations and will be discussed in Section 0.2. ORDER OF OPERATIONS

1. Start with the innermost parentheses (grouping symbols) and work outward. 2. Perform all indicated multiplications and divisions, working from left to right. 3. Perform all additions and subtractions, working from left to right. EXAMPLE 4  S  implifying Expressions Using the Correct Order of Operations

Simplify the expressions. a. 4 1 3⋅ 2 2 7⋅ 5 1 6   b. 

Solution (a):

726 2⋅3 1 8 f

f

Perform multiplication first.            4 1 3 ⋅ 2 2 7⋅ 5 1 6 35 6 Then perform the indicated additions and subtractions.  5 4 1 6 2 35 1 6 5 219 Solution (b):

The numerator and the denominator are similar to expressions in parentheses. Simplify these separately first, following the correct order of operations.

TRUE OR FALSE  In Example 4(a), we could have also started with adding 4 1 3.

▼ ANSWER False

f

726 Perform multiplication in the denominator first.     2⋅ 3 1 8 6 Then perform subtraction in the numerator 726 1 5 5 and addition in the denominator. 618 14

[ CONCEPT CHECK ]

▼ Y O U R T U R N   Simplify the expressions. a.  27 1 4 ⋅ 5 2 2⋅ 6 1 9   b. 

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926 2⋅ 5 1 6

▼ ANSWER 3

a. 10   b.  16

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8 

CHAPTER 0  Prerequisites and Review

Parentheses ( ) and brackets [ ] are the typical notations for grouping and are often used interchangeably. When nesting (groups within groups), use parentheses on the innermost and then brackets on the outermost. EXAMPLE 5  S  implifying Expressions That Involve Grouping Signs Using the Correct Order of Operations

Simplify the expression 3 35 ⋅ 14 2 22 2 2 ⋅ 74. Solution:

Simplify the inner parentheses. 3 35 ⋅  14 2 22 2 2 ⋅ 74 5 3 35 ⋅ 2 2 2⋅ 74 Inside the brackets, perform the multiplication 5 3 310 2 144 5 ⋅ 2 5 10 and 2⋅ 7 5 14. Inside the brackets, perform the subtraction. 5 3 3244 Multiply. 5 212 ▼ ANSWER

224

▼ Y O U R T U R N   Simplify the expression 2 323 ⋅  113 2 52 1 4 ⋅ 34.

Algebraic Expressions Everything discussed until now has involved real numbers (explicitly). In algebra, however, numbers are often represented by letters (such as x and y), which are called variables. A constant is a fixed (known) number such as 5. A coefficient is the constant that is ­multiplied by a variable. Quantities within the algebraic expression that are separated by addition or subtraction are referred to as terms. DEFINITION

Algebraic Expression

An algebraic expression is the combination of variables and constants using basic operations such as addition, subtraction, multiplication, and division. Each term is separated by addition or subtraction. Algebraic Expression  Variable Term  Constant Term  Coefficient 5x 3 5     5x 1 3 When we know the value of the variables, we can evaluate an algebraic expression using the substitution principle:   Algebraic expression:   Value of the variable:  Substitute x 5 2:

5x 1 3 x52 5 122 1 3 5 10 1 3 5 13

EXAMPLE 6  Evaluating Algebraic Expressions

Evaluate the algebraic expression 7x 1 2 for x 5 3. Solution:

Start with the algebraic expression. 7x 1 2 Substitute x 5 3. 7 132 1 2 Perform the multiplication. 5 21 1 2 Perform the addition. 5 23 ▼ ANSWER

16

Young_AT_6160_ch00_pp02-45.indd 8

▼ Y O U R T U R N   Evaluate the algebraic expression 6y 1 4 for y 5 2.

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0.1  Real Numbers 

In Example 6, the value for the variable was specified in order for us to evaluate the algebraic expression. What if the value of the variable is not specified; can we simplify an expression like 3 12x 2 5y2? In this case, we cannot subtract 5y from 2x. Instead, we rely on the basic properties of real numbers, or the basic rules of algebra.

0.1.4  Properties of Real Numbers You probably already know many properties of real numbers. For example, if you add up four numbers, it does not matter in which order you add them. If you multiply five ­numbers, it does not matter in what order you multiply them. If you add 0 to a real number or multiply a real number by 1, the result yields the original real number. Basic properties of real numbers are summarized in the following table. Because these properties are true for variables and algebraic expressions, these properties are often called the basic rules of algebra.

9

0.1.4 S K I L L

Apply properties of real numbers and basic rules of algebra in simplifying and evaluating expressions. 0.1.4 C O N C E P T U A L

Know and understand the basic properties of real numbers and the basic rules of algebra.

PROPERTIES OF REAL NUMBERS (BASIC RULES OF ALGEBRA) NAME

DESCRIPTION

MATH (LET a, b, AND c EACH BE ANY REAL NUMBER)

EXAMPLE

Commutative property of addition

Two real numbers can be added in any order.

a1b5b1a

3x 1 5 5 5 1 3x

Commutative property of multiplication

Two real numbers can be multiplied in any order.

ab 5 ba

y ⋅ 3 5 3y

Associative property of addition

When three real numbers are added, it does not matter which two numbers are added first.

1a 1 b2 1 c 5 a 1  1b 1 c2

1x 1 52 1 7 5 x 1  15 1 72

Associative property of multiplication

When three real numbers are multiplied, it does not matter which two numbers are multiplied first.

1ab2 c 5 a 1bc2

123x2 y 5 23 1xy2

Distributive property

Multiplication is distributed over all the terms of the sums or differences within the parentheses.

Additive identity property

Adding zero to any real number yields the same real number.

a 1b 1 c2 5 ab 1 ac a 1b 2 c2 5 ab 2 ac

5 1x 1 22 5 5x 1 10 5 1x 2 22 5 5x 2 10

Multiplicative identity property

Multiplying any real number by 1 yields the same real number.

a⋅1 5 a 1⋅a 5 a

Additive inverse property

The sum of a real number and its additive inverse (opposite) is zero.

Multiplicative inverse property

The product of a nonzero real number and its multiplicative inverse (reciprocal) is 1.

a 1  12a2 5 0

18x2 112 5 8x

a105a 01a5a

1 a⋅ 5 1 a

7y 1 0 5 7y

4x 1 (24x2 5 0

a20

1x 1 22 ⋅ a

x 2 22

1 b 51 x12

The properties in the previous table govern addition and multiplication. Subtraction can be defined in terms of addition of the additive inverse, and division can be defined in terms of multiplication by the multiplicative inverse (reciprocal). SUBTRACTION AND DIVISION

Let a and b be real numbers. MATH

Subtraction

a 2 b 5 a 1  12b2

Division

a 4 b 5 a⋅ b20

Young_AT_6160_ch00_pp02-45.indd 9

1 b

TYPE OF INVERSE

2b is the additive inverse or opposite of b 1 is the multiplicative inverse or b reciprocal of b

WORDS

Subtracting a real number is equal to adding its opposite. Dividing by a real number is equal to multiplying by its reciprocal.

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10 

CHAPTER 0  Prerequisites and Review

EXAMPLE 7  Using the Distributive Property

Use the distributive property to eliminate the parentheses. a. 3  1x 1 52   b. 2 1y 2 62 Solution (a):

➤

➤

Use the distributive property.  31x 1 52 5 3 1x2 1 3 152 Perform the multiplication. 5 3x 1 15 Solution (b):

▼ ANSWER a. 2x 1 6   b. 5y 2 15

➤

➤

Use the distributive property. 2 1 y 2 62 5 2 1 y2 2 2 162 Perform the multiplication. 5 2y 2 12

▼ Y O U R T U R N   Use the distributive property to eliminate the parentheses. a. 2 1x 1 32   b. 5 1y 2 32

You also probably know the rules that apply when multiplying a negative real number. For example, “a negative times a negative is a positive.” PROPERTIES OF NEGATIVES MATH (LET a AND b BE POSITIVE REAL NUMBERS)

DESCRIPTION

A negative quantity times a positive quantity is a negative quantity. A negative quantity divided by a positive quantity is a negative quantity. or A positive quantity divided by a negative quantity is a negative quantity. A negative quantity times a negative quantity is a positive quantity. A negative quantity divided by a negative quantity is a positive quantity. The opposite of a negative quantity is a positive quantity (subtracting a negative quantity is equivalent to adding a positive quantity). A negative sign preceding an expression is distributed throughout the expression.

 12a2 1b2 5 2ab 2a a 52 b b a a 52 2b b

EXAMPLE

 1282 132 5 224

216 5 24 or 4 15 5 25 23  122x2 1252 5 10x

or

 12a2 12b2 5 ab 2a a 5 2b b

212 54 23 2 1292 5 9

2 12a2 5 a

2 1a 1 b2 5 2a 2 b 2 1a 2 b2 5 2a 1 b

23 1x 1 52 5 23x 2 15 23 1x 2 52 5 23x 1 15

EXAMPLE 8  Using Properties of Negatives

Eliminate the parentheses and perform the operations. a.  25 1 7 2  1222  b.  2 1232  1242  1262 Solution:

f

a. Distribute the negative.            25 1 7 2 122 2

12

Combine the three quantities.

5 25 1 7 1 2 5 4

b. Group the terms.                   32 12324 3 1242 12624 Perform the multiplication inside the 3  4. 5 334 3244

Multiply.

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5 72

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0.1  Real Numbers 

11

We use properties of negatives to define the absolute value of any real number. The absolute value of a real number a, denoted  a , is its magnitude. On a number line this is the distance from the origin, 0, to the point. For example, algebraically, the absolute value of 5 is 5, that is,  5  5 5; and the absolute value of 25 is 5, or  25  5 5. Graphically, the distance on the real number line from 0 to either 25 or 5 is 5. 5 –5

5 0

5

Notice that the absolute value does not change a positive real number, but it changes a negative real number to a positive number. A negative number becomes a positive number if it is multiplied by 21. a

IF a IS A…

EXAMPLE

Positive real number

 a 5 a

 5 5 5

Negative real number

 a  5 2a

Zero

 a 5 a

 25  5 2 1252 5 5  0 5 0

EXAMPLE 9  Finding the Absolute Value of a Real Number

TRUE OR FALSE   a 2 b  5  b 2 a 

Evaluate the expressions. a.  23 1 7 

b.  2 2 8 

▼ ANSWER True

Solution: a.  23 1 7  5  4 

b.  2 2 8  5  26 

5 4



[ CONCEPT CHECK ]

5 6

Properties of the absolute value will be discussed in Section 1.7. EXAMPLE 10  Using  Properties of Negatives and the Distributive Property

Eliminate the parentheses 2 12x 2 3y2.

common mistake

A common mistake is applying a negative only to the first term. ✓C O R R E C T

✖INCORRECT

2 12x 2 3y2

5 2 12x2 2  123y2 5 22x 1 3y

Error: 22x 2 3y The negative  122 was not distributed through the second term.

▼ Y O U R T U R N   Eliminate the parentheses. a.  22 1x 1 5y2  b.  2 13 2 2b2

▼ ANSWER a.  22x 2 10y b.  23 1 2b

What is the product of any real number and zero? The answer is zero. This property also leads to the zero product property, which is the basis for factoring (one of the methods used to solve quadratic equations, which will be discussed in Section 1.3).

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12 

CHAPTER 0  Prerequisites and Review

PROPERTIES OF ZERO DESCRIPTION

MATH (LET a BE A REAL NUMBER)

A real number multiplied by zero is zero.

a⋅0 5 0

0⋅x 5 0

Zero divided by a nonzero real number is zero.

0 50 a20 a a is undefined 0

0 50 32x

A real number divided by zero is undefined.

EXAMPLE

x23

x12 is undefined 0

ZERO PRODUCT PROPERTY DESCRIPTION

MATH

If the product of two real numbers is zero, then one of those numbers has to be zero.

If ab 5 0, then a 5 0 or b 5 0

EXAMPLE

If x 1x 1 22 5 0, then x 5 0 or x 1 2 5 0 therefore x 5 0  or  x 5 22

Note: If a and b are both equal to zero, then the product is still zero.

Fractions always seem to intimidate students. In fact, many instructors teach students to eliminate fractions in algebraic equations. It is important to realize that you can never divide by zero. Therefore, in the following table of fractional properties it is assumed that no denominators are zero. FRACTIONAL PROPERTIES DESCRIPTION

MATH

ZERO CONDITION

EXAMPLE

Equivalent fractions

a c 5 if and only if ad 5 bc b d

b 2 0 and d 2 0

y 6y 5 since 12y 5 12y 2 12

Multiplying two fractions

a c ac ⋅ 5 b d bd

b 2 0 and d 2 0

3 x 3x ⋅ 5 5 7 35

Adding fractions that have the same denominator

a c a1c 1 5 b b b a c a2c 2 5 b b b

b20

x 2 x12 1 5 3 3 3

b20

Adding fractions with different denominators using a common denominator

a c ad cb ad 1 bc 1 5 1 5 b d bd bd bd

b 2 0 and d 2 0

Subtracting fractions with different denominators using a common denominator

a c ad cb ad 2 bc 2 5 2 5 b d bd bd bd

b 2 0 and d 2 0

7 5 725 2 2 5 5 3 3 3 3 1 1 21 3 2 1 1 5 21 2 2 1 5 13 1 5 5 1 2 21 3 2 2 3 6

Dividing by a fraction is equivalent to multiplying by its reciprocal

a c a d 4 5 ⋅ b d b c

b 2 0, c 2 0, and d 2 0

Subtracting fractions that have the same denominator

1 1 21 4 2 2 1 1 21 3 2 1 1 1 2 5 5 1 3 21 4 2 3 4 12 x 2 x 7 7x 4 5 ⋅ 5 3 7 3 2 6

The least common multiple of two or more integers is the smallest integer that is evenly divisible by each of the integers. For example, the least common multiple (LCM) of 3 and 4 is 12. The LCM of 8 and 6 is 24. The reason the LCM of 8 and 6 is not 48 is that 8 and 6 have a common factor of 2. When adding and subtracting fractions, a common denominator can be found by multiplying the denominators. When there are common factors in the denominators, the LCM is the least common denominator (LCD) of the original denominators.

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0.1  Real Numbers 

13

EXAMPLE 11  Performing Operations with Fractions

Perform the indicated operations involving fractions and simplify. a. 

2 1 2 x 3 2   b.  4 4  c.  1 3 4 3 2 5

Solution (a):

Determine the LCD. Rewrite fractions applying the LCD.

3 ⋅ 4 5 12 2 1 2⋅4 1⋅3 2 5 2 3 4 3⋅4 4⋅3

2142 2 1132   5 3142 Eliminate the parentheses.

5

823 12

Combine terms in the numerator.

5

5 12

Solution (b):

Rewrite 4 with an understood 1 in the denominator.       

5

2 4 4 3 1

Dividing by a fraction is equivalent to multiplying by its reciprocal.

5

2 1 ⋅ 3 4

Multiply numerators and denominators, respectively.

5

2 12

Reduce the fraction to simplest form.

5

1 6

Solution (c):

Determine the LCD.                2 # 5 5 10 Rewrite fractions in terms of the LCD.          

5x 1 3 1 2 2 x 3 1 5 122 152 2 5

Simplify the numerator.                   5

5x 1 6 10

▼ Y O U R T U R N   Perform the indicated operations involving fractions. a. 

3 1 1 3 2 x 1   b.  4   c.  2 5 2 5 10 3 5

Young_AT_6160_ch00_pp02-45.indd 13

▼ ANSWER a. 

11 2 10 2 3x   b.    c.  10 3 15

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14 

CHAPTER 0  Prerequisites and Review

[ S E C T I O N 0 .1]     S U M M A R Y In this section, real numbers were defined as the set of all rational Subtraction and division can be defined in terms of addition and and irrational numbers. Decimals are approximated by either multiplication. truncating or rounding. • Subtraction: a 2 b 5 a 1  12b2 (add the opposite) • Truncating: Eliminate all values after a particular digit. 1 • Rounding: Look to the right of a particular digit. If the • Division: a 4 b 5 a⋅ , where b 2 0 (multiply by the b ­number is 5 or greater, increase the digit by 1; otherwise, reciprocal) leave it as is and eliminate all digits to the right. Properties of negatives were reviewed. If a and b are positive real The order in which we perform operations is numbers, then: 1. parentheses (grouping); work from inside outward. •  12a2  1b2 5 2ab 2. multiplication/division; work from left to right. •  12a2  12b2 5 ab 3. addition/subtraction; work from left to right. • 2 12a2 5 a The properties of real numbers are employed as the basic rules of • 2 1a 1 b2 5 2a 2 b and 2 1a 2 b2 5 2a 1 b algebra when dealing with algebraic expressions. 2a a • 52 • Commutative property of addition: a 1 b 5 b 1 a b b • Commutative property of multiplication: ab 5 ba 2a a • 5 • Associative property of addition: 2b b  1a 1 b2 1 c 5 a 1  1b 1 c2 Absolute value of real numbers:  a  5 a if a is nonnegative, and • Associative property of multiplication:  1ab2  c 5 a  1bc2  a  5 2a if a is negative. • Distributive property: Properties of zero were reviewed. a  1b 1 c2 5 ab 1 ac or a  1b 2 c2 5 ab 2 ac

• Additive identity: a 1 0 5 a • Multiplicative identity: a⋅1 5 a • Additive inverse (opposite): a 1  12a2 5 0 1 a

• Multiplicative inverse (reciprocal): a⋅ 5 1

0 a

• a ⋅ 0 5 0  and   5 0    a 2 0 •

a20

a is undefined 0

• Zero product property: If ab 5 0, then a 5 0 or b 5 0 Properties of fractions were also reviewed.

•

a c ad 6 bc 6 5   b 2 0 and d 2 0 b d bd

•

a c a d 4 5 ⋅   b 2 0, c 2 0, and d 2 0 b d b c

[ S E C T I O N 0 .1]   E X E R C I S E S • SKILLS In Exercises 1–8, classify the following real numbers as rational or irrational. 11

22

1. 3 2. 2.07172737. . . 4. p 3 3. 5. 2.7766776677 6. 5.222222 7. !5 8. !17

In Exercises 9–16, approximate the real number to three decimal places by (a) rounding and (b) truncation.

9. 7.3471 10. 9.2549 11. 2.9949 12. 6.9951 13. 0.234492 14. 1.327491 15. 5.238473 16. 2.118465

In Exercises 17–40, perform the indicated operations in the correct order. 17. 5 1 2 ⋅ 3 2 7

18. 2 1 5 ⋅ 4 1 3 ⋅ 6

21. 2 2 3 34 12 ⋅ 3 1 524

22. 4 ⋅ 6 15 2 92

25. 23 2 1262

Young_AT_6160_ch00_pp02-45.indd 14

26. 25 1 2 2  1232

19. 2 ⋅  15 1 7 ⋅ 4 2 202 23. 8 2  1222 1 7 27. x 2  12y2 2 z

20. 23 ⋅  12 1 72 1 8 ⋅  17 2 2 ⋅ 12 24. 210 2  1292

28. 2a 1 b 2 12c2

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0.1  Real Numbers 

23 1 5 2 121 2

15

12 123 2 124 2

29. 2 13x 1 y2

30. 2 14a 2 2b2

31.

33. 24 26 3 15 2 82  1424

34.

35. 2 16x 2 4y2 2  13x 1 5y2

36.

37. 2 13 2 4x2 2  14x 1 72

38. 2 2 3 3 14x 2 52 2 3x 2 74

39.

24 1 5 2 2 5 25

40. 26 12x 1 3y2 2 33x 2  12 2 5y24

214 5 2 122 2

In Exercises 41–56, write as a single fraction and simplify.

32. 2

24x 6 2 122 2

41.

1 5 1 3 4

42.

1 1 2 2 5

43.

5 1 2 6 3

44.

7 1 2 3 6

45.

3 5 1 2 12

46.

1 5 1 3 9

47.

1 2 2 9 27

48.

124 2 3 5 2 2 7 3 6

49.

x 2x 1 5 15

50.

y y 2 3 6

51.

x 2x 2 3 7

52.

y y 2 10 15

53.

123y 2 4y 2 15 4

54.

6x 7x 2 12 20

55.

3 7 1 40 24

56.

23 27 2 a b 10 12

In Exercises 57–68, perform the indicated operation and simplify, if possible. 57.

2 14 ⋅ 7 3

58.

2 9 ⋅ 3 10

59.

2 10 4 7 3

60.

4 7 4 5 10

61.

4b a 4    a 2 0 9 27

62.

3a b 4   b 2 0 7 21

63.

3x 6x 4   x 2 0 10 15

64. 4

1 1 47 5 20

65.

3x 9 4   y 2 0 4 16y

66.

14m 4 ⋅ 2 7

67.

3y 6x 4   y 2 0 7 28

68. 2

1 5 ⋅7 3 6

In Exercises 69–72, evaluate the algebraic expression for the specified values. 69.

2c  for c 5 24, d 5 3 2d

71.

m1 ⋅ m2  for m1 5 3, m2 5 4, r 5 10 r2

70. 2l 1 2w for l 5 5, w 5 10

72.

x2m   for  x 5 100, m 5 70, s 5 15 s

• A P P L I C AT I O N S On November 15, 2015, the United States debt was estimated at $18,152,809,942,589, and at that time the estimated population was 322,162,446 citizens. 73. U.S. National Debt. Round the debt to the nearest million. 74. U.S. Population. Round the number of citizens to the nearest

thousand. 75. U.S. Debt. If the debt is distributed evenly to all citizens, what is the national debt per citizen? Round your answer to the nearest dollar.

Young_AT_6160_ch00_pp02-45.indd 15

76. U.S. Debt. If the debt is distributed evenly to all citizens,

what is the national debt per citizen? Round your answer to the nearest cent.

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16 

CHAPTER 0  Prerequisites and Review

• C AT C H T H E M I S TA K E In Exercises 77–80, explain the mistake that is made. 79. Simplify the expression 3 1x 1 52 2 2 14 1 y2.

77. Round 13.2749 to two decimal places.

Solution:

Solution:

The 9, to the right of the 4, causes the 4 to round to 5.          13.275



Simplify.

The 5, to the right of the 7, causes the 7 to be rounded to 8.        13.28



This is incorrect. What mistake was made? 2 1 78. Simplify the expression 3 1 9 .



3x 1 7 1 y

This is incorrect. What mistake was made?

Solution:

Solution: 211 3 5 319 12

Reduce.                    5

3x 1 15 2 8 1 y

80. Simplify the expression 23 1x 1 22 2  11 2 y2.



Add the numerators and denominators.

Eliminate parentheses.

Eliminate parentheses.

Simplify.

23x 2 6 2 1 2 y 23x 2 7 2 y

This is incorrect. What mistake was made?

1 4

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 81–84, determine whether each of the following statements is true or false. 81. Student athletes are a subset of the students in the honors

program. 82. The students who are members of fraternities or sororities are a subset of the entire student population. 83. Every integer is a rational number. 84. A real number can be both rational and irrational.

85. What restrictions are there on x for the following to be true?

3 5 3 4 5 x x 5 86. What restrictions are there on x for the following to be true?

x x 4 53 2 6

• CHALLENGE In Exercises 87 and 88, simplify the expressions. 87. 22 33 1x 2 2y2 1 74 1 33 12 2 5x2 1 104 2 7 322 1x 2 32 1 54

88. 22525 1y 2 x2 2 2 33 12x 2 52 1 7 122 2 44 1 36 1 7

• TECHNOLOGY 89. Use your calculator to evaluate !1260. Does the answer

91. Use your calculator to evaluate !4489. Does the answer

90. Use your calculator to evaluate

92. Use your calculator to evaluate

appear to be a rational or an irrational number? Why?

144 . Does the answer Å 25 appear to be a rational or an irrational number? Why?

Young_AT_6160_ch00_pp02-45.indd 16

appear to be a rational or an irrational number? Why?

882 . Does the answer Å 49 appear to be a rational or an irrational number? Why?

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17

0.2  Integer Exponents and Scientific Notation 

0.2 INTEGER EXPONENTS AND SCIENTIFIC NOTATION SKILLS OBJECTIVES ■■ Evaluate expressions using integer exponents and properties of exponents. ■■ Convert numbers to and from scientific notation.

CONCEPTUAL OBJECTIVES ■■ Understand that positive integer exponents represent repeated multiplication and that negative exponents are reciprocals. ■■ Understand that scientific notation is an effective way to represent very large or very small real numbers.

0.2.1  Integer Exponents Exponents represent repeated multiplication. For example, 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 5 25. The 2 that is repeatedly multiplied is called the base, and the small number 5 above and to the right of the 2 is called the exponent. DEFINITION

Natural-Number Exponent

f

Let a be a real number and n be a natural number (positive integer); then an is defined as an 5 a ? a ? a ca  1 a appears as a factor n times 2 n factors

where n is the exponent, or power, and a is the base.

0.2.1 S K I L L

Evaluate expressions using integer exponents and properties of exponents. 0.2.1 C O N C E P T U A L

Understand that positive integer exponents represent repeated multiplication and that negative exponents are reciprocals.

EXAMPLE 1  E  valuating Expressions Involving Natural-Number Exponents

Evaluate the expressions. 1 a. 43   b. 81   c. 54   d.  A 2 B 5

STUDY T I P an: “a raised to the nth power” a2: “a squared” a3: “a cubed”

Solution:

a. 43 5 4⋅ 4 ⋅ 4 5 64

b. 81 5 8 1 5

4

c. 5 5 5 ⋅ 5 ⋅ 5 ⋅ 5 5 625

1 1 1 1 1

1

d. A 2 B 5 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 5 32

▼ Y O U R T U R N   Evaluate the expressions. 1 4

a. 63    b.  A 3 B

▼ ANSWER 1

a. 216   b.  81

We now include exponents in our order of operations: 1.  Parentheses 2.  Exponents

3.  Multiplication/Division­­ 4.  Addition/Subtraction

EXAMPLE 2  Evaluating Expressions Involving Natural-Number Exponents

Evaluate the expressions. a.   12324   b.  234   c.   12223 ⋅ 52 Solution:

a. 123 2 4 5 123 2 123 2 123 2 123 2 5 81

f

b. 234 5 2 1 3 ⋅ 3 ⋅ 3 ⋅ 3 2 5 281

81

c. 122 2 ⋅ 5 5 122 2 122 2 122 2 ⋅ 5 ⋅ 5 5 2200 3

2

f

f 28

25

▼ 3

3

Y O U R T U R N   Evaluate the expression 24 ⋅ 2 .

Young_AT_6160_ch00_pp02-45.indd 17

▼ ANSWER

2512

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18 

CHAPTER 0  Prerequisites and Review

ST U DY TIP A negative exponent implies a ­reciprocal.

So far, we have discussed only exponents that are natural numbers (positive integers). When the exponent is a negative integer, we use the following property. NEGATIVE-INTEGER EXPONENT PROPERTY

Let a be any nonzero real number and n be a natural number (positive integer); then a2n 5

1 an

a20

In other words, a base raised to a negative-integer exponent is equivalent to the reciprocal of the base raised to the opposite (positive) integer exponent.

[ CONCEPT CHECK ]

EXAMPLE 3  E  valuating Expressions Involving Negative-Integer Exponents

1 24 Simplify the expression: a b . a

Evaluate the expressions.

ANSWER a4

a. 224  b. 

▼

Solution: a. 224 5

b.

1 5 323

c. 423 ⋅

1 33 1 5 1 4 a 3 b 5 1 ⋅ 5 33 5 27 1 1 3 a 3b 3

1 5 223 ⋅ 126 2 2 5 128 2 1 36 2 5 2288 126 2 22 28 36

f

f

1 a.  225  b.  4

▼

1 1 5 16 24

1 1 4 16 1 5 24 5 3 ⋅ 2 5 64 4 2 4

d. 223 ⋅

▼ ANSWER

1 1 1   c.  423 ⋅ 24   d.  223 ⋅ 126 2 22 323 2

Y O U R T U R N   Evaluate the expressions. a.  2

1 1    b.  22 ⋅ 622 522 3

Now we can evaluate expressions involving positive and negative exponents. How do we evaluate an expression with a zero exponent? We define any nonzero real number raised to the zero power as 1.

ZERO-EXPONENT PROPERTY

Let a be any nonzero real number; then a0 5 1   a Þ 0

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0.2  Integer Exponents and Scientific Notation 

19

EXAMPLE 4  Evaluating Expressions Involving Zero Exponents

Evaluate the expressions. a. 50  b. 

Solution:

1   c.   12320  d. 240 20 b.

1 1 5 1 0 5 1 2

c.  12320 5 1

d. 240 5 21 f

a. 50 5 1

1

We now can evaluate expressions involving integer (positive, negative, or zero) exponents. What about when expressions involving integer exponents are multiplied, divided, or raised to a power? WORDS MATH

f

f

When expressions with the same base are multiplied, the exponents are added.              23 ⋅ 24 5 2⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 5 2314 5 27 23 24 When expressions with the same 5 2 2⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 2⋅2 base are divided, the exponents 5 5 22 3 5 2 ⋅ 2 ⋅ 2 1 2 are subtracted. or 2523 5 22 1 23 2 2 5  1822 5 64 When an expression involving an exponent is raised to a power, the exponents are multiplied.        or 1 23 2 2 5 23⋅2 5 26 5 64

The following table summarizes the properties of integer exponents.

PROPERTIES OF INTEGER EXPONENTS MATH (LET a AND b BE NONZERO REAL NUMBERS AND m AND n BE INTEGERS)

EXAMPLE

NAME

DESCRIPTION

Product property

When multiplying exponentials with the same base, add exponents.

am ⋅ an 5 am1n

x2 ⋅ x5 5 x215 5 x7

Quotient property

When dividing exponentials with the same base, subtract the exponents (numerator 2 denominator).

am 5 am2n an

x5 5 x 523 5 x 2  x 2 0 x3

Power property

When raising an exponential to a power, multiply exponents.

1 am 2 5 amn

Product to a power property

A product raised to a power is equal to the product of each factor raised to the power.

1ab2n 5 anbn

12x23 5 23 ⋅ x3 5 8x3

Quotient to a power property

A quotient raised to a power is equal to the quotient of the factors raised to the power.

a n an a b 5 n b b

x 4 x4 a b 5 4  y 2 0 y y

Young_AT_6160_ch00_pp02-45.indd 19

n

4

1 x 2 2 5 x 2⋅4 5 x 8

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20 

CHAPTER 0  Prerequisites and Review

Common Errors Made Using Properties of Exponents INCORRECT 4

3

12

x ⋅x 5 x

CORRECT 4

3

7

x ⋅x 5 x

ERROR

Exponents should be added (not multiplied).

x 5 x3 x6

x 5 x 12; x 2 0 x6

Exponents should be subtracted (not divided).

1 x2 2 3 5 x8

1 x2 2 3 5 x6

Exponents should be multiplied (not raised to a power).

18

3

18

3

12x2 5 2x 3

4

7

3

5

8

2 ⋅2 5 4 2 ⋅3 5 6

3

3

Both factors (the 2 and the x) should be cubed.

7

The original common base should be retained.

12x2 5 8x 3

4

3

5

2 ⋅2 5 2 2 ⋅3

The properties of integer exponents require the same base.

We will now use properties of integer exponents to simplify exponential expressions. An exponential expression is simplified when: All parentheses (groupings) have been eliminated. A base appears only once. ■■ No powers are raised to other powers. ■■ All exponents are positive. ■■ ■■

EXAMPLE 5  Simplifying Exponential Expressions

Simplify the expressions (assume all variables are nonzero). 3 a. 122x 2 y 3 2 1 5x 3 y 2   b.  1 2x 2 yz3 2   c. 

25x 3 y 6 25x 5 y 4

Solution (a):

Parentheses imply multiplication. Group the same bases together. Apply the product property.

122x 2 y 3 2 1 5x 3 y 2 5 122 2 1 5 2 x 2 x 3 y 3 y f

f

5 122 2 1 5 2 x 2 x 3 y 3 y x213 y311

5 210x5y4

Multiply the constants. Solution (b):

Apply the product to a power property. Apply the power property.

ST U DY TIP

Simplify.

It is customary not to leave negative exponents. Instead we use the negative exponent property to write exponential expressions with only positive exponents.

Solution (c):

5 8x2 ⋅ 3y1 ⋅ 3z3 ⋅ 3

5 8x6y3z9 25x 3 y 6 25 x3 y6 5 a b a ba b 25 x 5 y 4 25x 5 y 4

Group the same bases together.



5  1252x325y624

Apply the negative exponent property.

5

Apply the quotient property.

5 25x22y2 25y 2 x2

▼

▼ ANSWER

24x 3 a.  212x  y    b.  227x  y  z   c. y4 4 5

1 2x 2 yz3 2 3 5 1 2 2 3 1 x 2 2 3 1 y 2 3 1 z3 2 3

3 9 6

Young_AT_6160_ch00_pp02-45.indd 20

Y O U R T U R N   Simplify the expressions (assume all variables are nonzero). 3 a.  123x 3 y 2 2 1 4xy 3 2   b.  123xy 3 z2 2   c. 

216x 4 y 3 4xy 7

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21

0.2  Integer Exponents and Scientific Notation 

EXAMPLE 6  Simplifying Exponential Expressions

Write each expression so that all exponents are positive (assume all variables are nonzero). 1 x 2 y 23 2 2 122xy 2 2 3 23 a. 1 3x 2 z24 2   b.  21 4 23   c.  1x y 2 2 1 6xz3 2 2 Solution (a):

Apply the product to a power property.     1 3x 2 z24 2 23 5 1 3 2 23 1 x 2 2 23 1 z24 2 23 Apply the power property. 5 323x26z12 Apply the negative-integer exponent property.

5

z12 33 x 6

Evaluate 33.

5

z12 27x 6

Solution (b):

1 x 2 y 23 2 2 x 4 y 26 Apply the product to a power property.        21 4 23 5 3 212 1x y 2 x y

Apply the quotient property. Simplify. Solution (c):

Apply the product to a power property on both the numerator and denominator.

5 x423 y26212122 5 xy6

122xy 2 2 3 122 2 3 1 x 2 3 1 y 2 2 3 5 2 2 1 6xz3 2 2 1 6 2 2 1 x 2 2 1 z3 2 2 28x 3 y 6 236x 2 z6

Apply the power property.

5

Group constant terms and x terms.

5 a

Apply the quotient property.

5 a

Simplify.

5

28 x3 y6 b a 2b a 6 b 236 x z y6 8 b 1 x 322 2 a 6 b 36 z

2xy 6 9z6

▼ Y O U R T U R N   Simplify the exponential expression and express it in terms of

1 tv 2 2 23 p­ ositive exponents       4 3 21 . 1 2t v 2

0.2.2  Scientific Notation

You are already familiar with base 10 raised to positive-integer powers. However, it can be inconvenient to write all the zeros out, so we give certain powers of 10 particular names: thousand, million, billion, trillion, and so on. For example, we say there are 322 million U.S. citizens as opposed to writing out 322,000,000 citizens. Or we say that the national debt is $18 trillion as opposed to writing out $18,000,000,000,000. The following table contains scientific notation for positive exponents and examples of some common prefixes and abbreviations. One of the fundamental applications of scientific notation is measurement.

Young_AT_6160_ch00_pp02-45.indd 21

▼ ANSWER

2t v3

0.2.2 S K I L L

Convert numbers to and from scientific notation. 0.2.2 C O N C E P T U A L

Understand that scientific notation is an effective way to represent very large or very small real numbers.

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22 

CHAPTER 0  Prerequisites and Review

EXPONENTIAL FORM

NUMBER OF ZEROS FOLLOWING THE 1

REAL NUMBER

1

10

1

2

100

2

103

1000 (one thousand)

3

104

10,000

4

10 10

5

100,000

5

106

1,000,000 (one million)

6

107

10,000,000

7

10

8

100,000,000

8

10

9

1,000,000,000 (one billion)

9

10,000,000,000

10

10

1010 11

100,000,000,000

11

1012

1,000,000,000,000 (one trillion)

12

10

PREFIX

ABBREVIATION

EXAMPLE

kilo-

k

The relay-for-life team ran a total of 80 km (kilometers).

mega-

M

Modern high-­powered diesel–electric railroad locomotives typically have a peak power output of 3 to 5 MW (megawatts).

giga-

G

A flash drive typically has 1 to 4 GB (gigabytes) of storage.

tera-

T

Laser systems offer higher frequencies on the order of THz (terahertz).

Notice that 108 is a 1 followed by 8 zeros; alternatively, you can start with 1.0 and move the decimal point eight places to the right (insert zeros). The same type of table can be made for negative-integer powers with base 10. To find the real number associated with exponential form, start with 1.0 and move the decimal a certain number of places to the left (fill in missing decimal places with zeros).

EXPONENTIAL FORM

REAL NUMBER

NUMBER OF PLACES DECIMAL (1.0) MOVES TO THE LEFT

10

0.1

1

1022

0.01

2

0.001 (one thousandth)

3

1024

0.0001

4

1025

0.00001

5

0.000001 (one millionth)

6

21

10

10

23

26

0.0000001

7

10

28

0.00000001

8

10

29

0.000000001 (one billionth)

9

1027

0.0000000001

10

10

211

0.00000000001

11

10

212

0.000000000001 (one trillionth)

12

10210

PREFIX

ABBREVIATION

EXAMPLE

milli-

m

Excedrin Extra Strength tablets each have 250 mg (milligrams) of acetaminophen.

micro-

m

A typical laser has a wavelength of 1.55 mm (micrometers*).

nano-

n

PSA levels less than 4 ng/ml (nanogram per milliliter of blood) represent low risk for prostate ­cancer.

pico-

p

A single yeast cell weighs 44 pg (picograms).

*In optics a micrometer is called a micron.

Young_AT_6160_ch00_pp02-45.indd 22

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23

0.2  Integer Exponents and Scientific Notation 

SCIENTIFIC NOTATIONIN

STUD Y T I P

A positive real number can be written in scientific notation with the form c 3 10n, where 1 # c , 10 and n is an integer.

Scientific notation is a ­number between 1 and 10 that is ­multiplied by 10 to a power.

Note that c is a real number between 1 and 10. Therefore, 22.5 3 103 is not in scientific notation, but we can convert it to scientific notation: 2.25 3 104. For example, there are approximately 50 trillion cells in the human body. We write 50 trillion as 50 followed by 12 zeros 50,000,000,000,000. An efficient way of writing such a large number is using scientific notation. Notice that 50,000,000,000,000 is 5 followed by 13 zeros, or in scientific notation, 5 3 1013. Very small numbers can also be written using scientific notation. For example, in laser communications a pulse width is 2 femtoseconds, or 0.000000000000002 second. Notice that if we start with 2.0 and move the decimal point 15 places to the left (adding zeros in between), the result is 0.000000000000002, or in scientific notation, 2 3 10215.

Real numbers greater than 1 correspond to positive ­exponents in scientific notation, whereas real numbers greater than 0 but less than 1 correspond to negative exponents in ­scientific notation.

[ CONCEPT CHECK ]

EXAMPLE 7  E  xpressing a Positive Real Number in Scientific Notation

The U.S. national debt can be represented in scientific notation using an exponent that is positive or negative?

Express the numbers in scientific notation. a.  3,856,000,000,000,000   b. 0.00000275 Solution:

▼

a. Rewrite the number with the implied decimal point.   3,856,000,000,000,000.

Move the decimal point to the left 15 places.         b. Move the decimal point to the right six places. 

STUDY T I P

ANSWER Positive

5 3.856 3 1015

0.00000275 5 2.75 3 1026

▼ Y O U R T U R N   Express the numbers in scientific notation. a.  4,520,000,000   b. 0.00000043

▼ ANSWER a. 4.52 3 109 b. 4.3 3 1027

EXAMPLE 8  Converting from Scientific Notation to Decimals

Write each number as a decimal. a. 2.869 3 105   b. 1.03 3 1023 Solution: a. Move the decimal point 5 places to the

right (add zeros in between).          286,900. or 286,900

b. Move the decimal point 3 places to the

left (add zeros in between).           0.00103

▼ Y O U R T U R N   Write each number as a decimal. a.  8.1 3 104   b. 3.7 3 1028

▼ ANSWER a. 81,000 b. 0.000000037

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24 

CHAPTER 0  Prerequisites and Review

[ S E C T I O N 0 . 2 ]    S U M M A R Y In this section we discussed properties of exponents. INTEGER EXPONENTS

The following table summarizes integer exponents. Let a be any real number and n be a natural number. DESCRIPTION

MATH

Natural-number exponent

Multiply n factors of a.

an 5 a⋅a⋅a Na

f

NAME

n factors 1   a 2 0 an

Negative-integer exponent property

A negative exponent implies a reciprocal.

a2n 5

Zero-exponent property

Any nonzero real number raised to the zero power is equal to 1.

a0 5 1   a 2 0

PROPERTIES OF INTEGER EXPONENTS

The following table summarizes properties of integer exponents. Let a and b be nonzero real numbers and m and n be integers. NAME

DESCRIPTION

MATH

Product property

When multiplying exponentials with the same base, add exponents.

am ⋅ an 5 am1n

Quotient property

When dividing exponentials with the same base, subtract the exponents (numerator 2 denominator).

am 5 am2n an

Power property

When raising an exponential to a power, multiply exponents.

Product to a power property

A product raised to a power is equal to the product of each factor raised to the power.

1 am 2 5 amn

Quotient to a power property

A quotient raised to a power is equal to the quotient of the factors raised to the power.

n

1ab2n 5 anbn a n an a b 5 n b b

S C I E N T I F I C N O TAT I O N

Scientific notation is a convenient way of using exponents to represent either very small or very large numbers. Real numbers greater than 1 correspond to positive exponents in scientific notation, whereas real numbers greater than 0 but less than 1 correspond to negative exponents in scientific notation. Scientific notation offers the convenience of multiplying and dividing real numbers by applying properties of exponents. REAL NUMBER (DECIMAL FORM)

PROCESS

SCIENTIFIC NOTATION

2,357,000,000

Move the implied decimal point to the left 9 places

2.357 3 109

0.00000465

Move the decimal point to the right 6 places

4.65 3 1026

[SEC TION 0. 2]  E X E R C I S E S • SKILLS In Exercises 1–20, evaluate each expression. 1. 44

2. 53 2

2

6. 27

7. 22 ⋅4

11. 1021

12. a21

16. 22⋅42

17. 8⋅223 ⋅5

Young_AT_6160_ch00_pp02-45.indd 24

3. 12325 2

4. 12422 0

5. 252

9. 9

10. 28x0

13. 822

14. 324

15. 26⋅52

18. 5⋅224 ⋅32

19. 26⋅322 ⋅81

20. 6⋅42 ⋅424

8. 23 ⋅ 5

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0.2  Integer Exponents and Scientific Notation 

25

In Exercises 21–50, simplify and write the resulting expression with only positive exponents. 21. x2 ⋅x3

22. y3 ⋅y5

23. x2x23

24. y3 ⋅y27

25. 1x223

26. 1y322

27. 14a23

28. 14x223

29. 122t23 33.

x5 y3

34.

x7 y b 2

30. 123b24

24

y5 x2

y 22 x 25 c 3

22

31. 15xy222 13x3y2 35.

32. 14x2y2 12xy322

1 2xy 2 2

36.

1 22xy 2 3

123x 3 y 2 3 24 1 x 2 y 3 2

37. a b

38. a b

39. 19a22b3222

40. 129x23y2224

41.

42.

43.

44.

45.

49.

a22 b3 a4 b5 3 3 1 x2 y 2 4 12 1 x 22 y 2

c

3 a2 12xy 4 2 3 2d x 4 12a3 y 2 2

46.

50.

x 23 y 2 y 24 x 5 124x 22 2 2 y 3 z 1 2x 3 2 22 1 y 21z 2 4

23 1 3 224 5 c b 2 2x2 y5 3 d

y 12b x 2

47.

1 x 3 y 21 2 2 1 xy 2 2 22

1 x 24 y 5 2 22 322 1 x 3 2 2 y 24 4 5

1 x 3 y 22 2 2 1 x 4 y 3 2 23

48. 22x2 122x325

51. Write 28 ⋅163 ⋅ 1642 as a ?

power of 2 : 2

52. Write 39 ⋅815 ⋅ 192 as a power

of 3 : 3?

In Exercises 53–60, express the given number in scientific notation. 53. 27,600,000

54. 144,000,000,000

55. 93,000,000

56. 1,234,500,000

57. 0.0000000567

58. 0.00000828

59. 0.000000123

60. 0.000000005

63. 2.3 3 104

64. 7.8 3 1023

In Exercises 61–66, write the number as a decimal. 61. 4.7 3 107

62. 3.9 3 105

65. 4.1 3 1025

66. 9.2 3 1028

• A P P L I C AT I O N S In Exercises 67 and 68, refer to the following:

b. The Moon traces an elliptical path around the Earth, with

It is estimated that there are currently 5.0 3 109 cell phones being used worldwide. Assume that the average cell phone measures 5 inches in length and there are 5280 feet in a mile. 67. Cell Phones Spanning the Earth. a. If all of the cell phones currently in use were lined up next to each other tip to tip, how many feet would the line of cell phones span? Write the answer in scientific notation. b. The circumference of the Earth (measured at the equator) is approximately 25,000 miles. If the cell phones in part (a) were to be wrapped around the Earth at the equator, would they circle the Earth completely? If so, approximately how many times? 68. Cell Phones Reaching the Moon. a. If all of the cell phones currently in use were to be lined up next to each other tip to tip, how many miles would the line of cell phones span? Write the answer in scientific notation.

the average distance between them being approximately 239,000 miles. Would the line of cell phones in part (a) reach the Moon? Astronomy. The distance from Earth to Mars on a particular day can be 200 million miles. Express this distance in ­scientific notation. Astronomy. The distance from Mars to the Sun on a particular day can be 142 million miles. Express this distance in scientific notation. Lasers. The wavelength of a typical laser used for ­communication systems is 1.55 microns 1or 1.55 3 1026 meters2. Express the wavelength in decimal representation in terms of meters. Lasers. A ruby-red laser has a wavelength of 694 nanometers 1or 6.93 3 1027 meters2. Express the wavelength in decimal representation in terms of meters.

Young_AT_6160_ch00_pp02-45.indd 25

69.

70.

71.

72.

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26 

CHAPTER 0  Prerequisites and Review

• C AT C H T H E M I S TA K E In Exercises 73–76, explain the mistake that is made. 73. Simplify  122y32  13x2y22.



Group like factors together.



Use the product property.

1222  132  x 2 y 3y 2

26x2y6



This is incorrect. What mistake was made? 2 75. Simplify  122xy32   15x2y22.

Apply the product to a power property.      5  12222x2 1y32 2 1522 1x222y2 Apply the power rule.          5 4x2y925x4y2



Group like factors.       5  142  1252x2x4y9y2

Apply the product property.            5 100x6y11



This is incorrect. What mistake was made?

3 74. Simplify 12xy2 2  .

12xy2 2 3 5 2x3y6



Eliminate the parentheses.



This is incorrect. What mistake was made?

76. Simplify

24x 16 y 9 8x 2 y 3

.

Group like factors.       5 a

2 4 x 16 y 9 b a 2 b a 3b 8 x y



1 Use the quotient property.      5 2 x 8 y 3 2



This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 77–80, determine whether each of the following statements is true or false.

In Exercises 83–86, evaluate the expression for the given value.

77. 22n 5  1222n, if n is an integer.

84. 2a3 2 7a2 for a 5 4

78. Any nonzero real number raised to the zero power is one. 79.

x

83. 2a2 1 2ab for a 5 22, b 5 3 85. 216t2 1 100t for t 5 3

n11

x

n

5 x for x 5 any real number.

86.

a3 2 27 for a 5 22 a24

80. x21 1 x22 5 x23 n k

81. Simplify A 1 am 2 B . 82. Simplify A 1 a2m 2

2n 2k

B .

• CHALLENGE

87. The Earth’s population is approximately 7.3 3 109 people, 8

88. The population of the United States is approximately 3.3 3 108

people, and there are approximately 3.79 3 106 square miles of land in the United States. If one square mile is approximately 640 acres, how many acres per person are there in the United States? Round to the nearest tenth of an acre.

and there are approximately 1.5 3 10 square kilometers of land on the surface of the Earth. If one square kilometer is approximately 247 acres, how many acres per person are there on Earth? Round to the nearest tenth of an acre. 89. Evaluate:

1 4 3 10223 2 1 3 3 1012 2 . Express your answer in 1 6 3 10210 2

both scientific and decimal notation.

90. Evaluate:

12 3 10217 2 1 5 3 1013 2 . Express your answer in 1 1 3 1026 2

both scientific and decimal notation.

• TECHNOLOGY Scientific calculators have an EXP button that is used for scientific notation. For example, 2.5 3 103 can be input into the calculator by pressing 2.5 EXP 3. 91. Repeat Exercise 87 and confirm your answer with a calculator. 92. Repeat Exercise 88 and confirm your answer with a calculator.

In Exercises 93 and 94, use a graphing utility or scientific ­calculator to evaluate the expression. Express your answer in scientific notation. 93.

94.

Young_AT_6160_ch00_pp02-45.indd 26

1 7.35 3 10226 2 1 2.19 3 1019 2 1 3.15 3 10221 2 1 1.6849 3 1032 2 1 8.12 3 1016 2 1 3.32 3 1029 2

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27

0.3  Polynomials: Basic Operations 

0.3 POLYNOMIALS: BASIC OPERATIONS SKILLS OBJECTIVES ■■ Add and subtract polynomials. ■■ Multiply polynomials. ■■ Recognize special products and use them to perform operations on binomials.

CONCEPTUAL OBJECTIVES ■■ Recognize like terms. ■■ Understand how the distributive property is applied. ■■ Learn formulas for special products.

0.3.1  Adding and Subtracting Polynomials Polynomials in Standard Form 0.3.1 S K I L L

The expressions 2

3

3x 2 7x 2 1  4y 2 y  5z are all examples of polynomials in one variable. A monomial in one variable, axk, is the product of a constant and a variable raised to a nonnegative-integer power. The constant a is called the coefficient of the monomial, and k is called the degree of the monomial. A polynomial is the sum of monomials. The monomials that are part of a polynomial are called terms.

Add and subtract polynomials. 0.3.1 C O N C E P T U A L

Recognize like terms.

Polynomial

DEFINITION

A polynomial in x is an algebraic expression of the form anx n 1 an21x n21 1 an22x n22 1 c1 a2x 2 1 a1x 1 a0 where a0, a1, a2, . . . , an are real numbers, with an 2 0, and n is a nonnegative integer. The polynomial is of degree n, an is the leading coefficient, and a0 is the constant term. Polynomials with one, two, and three terms are called monomials, binomials, and trinomials, respectively. Polynomials are typically written in standard form in order of decreasing degrees, and the degree of the polynomial is determined by the highest degree (exponent) of any single term. POLYNOMIAL 3

7

STANDARD FORM

4x 2 5x 1 2x 2 6 3

7

3

25x 1 4x 1 2x 2 6 3

SPECIAL NAME

DEGREE

DESCRIPTION

Polynomial

7

A seventh-degree polynomial in x

5 1 2y 2 4y

2y 2 4y 1 5

Trinomial

3

A third-degree polynomial in y

7z2 1 2

7z2 1 2

Binomial

2

A second-degree polynomial in z

Monomial

5

A fifth-degree monomial in x

5

217x

5

217x

EXAMPLE 1  Writing Polynomials in Standard Form

Write the polynomials in standard form and state their degree, leading coefficient, and constant term. a. 4x 2 9x5 1 2 b. 3 2 x2 2 3 8 c. 3x 2 8 1 14x 2 20x 1 x d. 27x3 1 25x Solution:

Standard Form Degree  Leading Coefficient  Constant Term 5 a. 29x 1 4x 1 2 5 29 2 b. 2x2 1 3 2 21 3 8 3 2 c. 220x 1 14x 1 3x 1 x 2 8 8 220 28 3 d. 27x 1 25x 3 27 0 ▼ ANSWER

▼ Y O U R T U R N   Write the polynomial in standard form and state its degree, leading

coefficient, and constant term. 17x 2 2 4x 3 1 5 2 x

Young_AT_6160_ch00_pp02-45.indd 27

24x3 1 17x2 2 x 1 5 Degree: 3 Leading coefficient: 24 Constant term: 5

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28 

CHAPTER 0  Prerequisites and Review

Adding and Subtracting Polynomials Polynomials are added and subtracted by combining like terms. Like terms are terms having the same variables and exponents. Like terms can be combined by adding their coefficients. WORDS MATH

Identify like terms.

3x2 1 2x 1 4x2 1 5

Add coefficients of like terms.

7x2 1 2x 1 5

Note: The 2x and 5 could not be combined because they are not like terms. EXAMPLE 2  Adding Polynomials

Find the sum and simplify 1 5x 2 2 2x 1 3 2 1 1 3x 3 2 4x 2 1 7 2 . Solution:

Eliminate parentheses.

5x2 2 2x 1 3 1 3x3 2 4x2 1 7

Identify like terms.

5x 2 2 2x 1 3 1 3x 3 2 4x 2 1 7

Combine like terms.

x2 2 2x 1 10 1 3x3

Write in standard form. 3x3 1 x2 2 2x 1 10 ▼ ANSWER

▼ Y O U R T U R N   Find the sum and simplify:

22x5 1 6x3 1 2x2 1 5x 1 11

[ CONCEPT CHECK ] Determine the two like terms in: ax3 2 dx 2 b and kx4 1 cx3

▼

1 3x 2 1 5x 2 2x 5 2 1 1 6x 3 2 x 2 1 11 2 EXAMPLE 3  Subtracting Polynomials

Find the difference and simplify 1 3x 3 2 2x 1 1 2 2 1 x 2 1 5x 2 9 2 .

common mistake

ANSWER ax3 and cx3

Distributing the negative to only the first term in the second polynomial. ✓C O R R E C T

✖INCORRECT

Eliminate the parentheses. 3

3x 2 2x 1 1 2 x 2 5x 1 9 Identify like terms.

ST U DY TIP When subtracting polynomials, it is important to distribute the negative through all of the terms in the second polynomial.

▼ ANSWER

x3 2 7x2 2 4x 1 3

0.3.2 S K I L L

Multiply polynomials.

Error: 3x3 2 2x 1 1 2 x2 1 5x 2 9

2

Don’t forget to distribute the ­negative through the entire second polynomial.

3x 3 2 2x 1 1 2 x 2 2 5x 1 9 Combine like terms.

3x3 2 x2 2 7x 1 10

▼ Y O U R T U R N   Find the difference and simplify:

127x 2 2 x 1 5 2 2 1 2 2 x 3 1 3x 2

0.3.2  Multiplying Polynomials

0.3.2 C O N C E P T U A L

The product of two monomials is found by using the properties of exponents (Section 0.2). For example,

Understand how the distributive property is applied.

To multiply a monomial and a polynomial, we use the distributive property (Section 0.1).

Young_AT_6160_ch00_pp02-45.indd 28

125x3 2 1 9x2 2 5 125 2 1 9 2 x312 5 245x5

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29

0.3  Polynomials: Basic Operations 

EXAMPLE 4  Multiplying a Monomial and a Polynomial

Find the product and simplify 5x2 1 3x 5 2 x 3 1 7x 2 4 2 . Solution:

Use the distributive property.      5x2 1 3x 5 2 x 3 1 7x 2 4 2 ➤

➤

➤

➤

5 5x2 1 3x 5 2 2 5x2 1 x 3 2 1 5x2 17x2 2 5x2 142



Multiply each set of monomials.           5 15x7 2 5x5 1 35x3 2 20x2

▼ Y O U R T U R N   Find the product and simplify 3x2 14x2 2 2x 1 12.

▼ ANSWER

12x4 2 6x3 1 3x2

How do we multiply two polynomials if neither one is a monomial? For example, how do we find the product of a binomial and a trinomial such as  12x 2 52 1x2 2 2x 1 32? Notice that the binomial is a combination of two monomials. Therefore, we treat each monomial, 2x and 25, separately and then combine our results. In other words, use the distributive property repeatedly.

WORDS MATH

Apply the distributive property.   1 2x 2 5 2 1 x 2 2 2x 1 3 2 5 2x 1 x 2 2 2x 1 3 2 2 5 1 x 2 2 2x 1 3 2

Apply the distributive property.                5 1 2x 2 1 x 2 2 1 1 2x 2 1 22x 2 1 1 2x 2 1 3 2 2 5 1 x 2 2 2 5 122x2 2 5 132

Multiply the monomials.

5 2x3 2 4x2 1 6x 2 5x2 1 10x 2 15

Combine like terms.

5 2x3 2 9x2 1 16x 2 15

[ CONCEPT CHECK ] EXAMPLE 5  Multiplying Two Polynomials

Multiply and simplify 1 2x 2 3x 1 1 2 1 x 2 5x 1 7 2 . 2

2

Solution:

Multiply and Simplify (ax 1 b)(cx 2 d )

▼ ANSWER acx2 1 (bc 2 ad   )x 2 bd

 ultiply each term M of the first trinomial by the entire second trinomial.    5 2x 2 1 x 2 2 5x 1 7 2 2 3x 1 x 2 2 5x 1 7 2 1 1 1 x 2 2 5x 1 7 2

Identify like terms.        5 2x 4 2 10x 3 1 14x 2 2 3x 3 1 15x 2 2 21x 1 x 2 2 5x 1 7 Combine like terms.           5 2x4 2 13x3 1 30x2 2 26x 1 7

▼

Y O U R T U R N   Multiply and simplify 12x 3 1 2x 2 4 2 1 3x 2 2 x 1 5 2 .

▼ ANSWER

23x5 1 x4 1 x3 2 14x2 1 14x 2 20

0.3.3  Special Products The method outlined for multiplying polynomials works for all products of polynomials. For the special case when both polynomials are binomials, the FOIL method can also be used.

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30 

CHAPTER 0  Prerequisites and Review

0.3.3 S K I L L

WORDS MATH

Recognize special products and use them to perform operations on binomials.

Apply the distributive property.  15x 2 12 12x 1 32 5 5x 12x 1 32 2 1 12x 1 32 Apply the distributive property.             5 5x 12x2 1 5x 132 2 1 12x2 2 1 132 Multiply each set of monomials.                  5 10x2 1 15x 2 2x 2 3 Combine like terms.                    5 10x2 1 13x 2 3

0.3.3 C O N C E P T U A L

Learn formulas for special products.

 he FOIL method finds the T products of the First terms, Outer terms, Inner terms, and Last terms.

S TU DY TIP When the binomials are of the form (ax 1 b)(cx 1 d ), the outer and inner terms will be like terms and can be combined.

Product of First Terms 15x2 12x2

Product of Last Terms 1212 132

Product of Outer Terms  15x2 132

Product of Inner Terms 1212 12x2

15x 2 12 12x 1 32 5 10x2 1 15x 2 2x 2 3

EXAMPLE 6  Multiplying Binomials Using the FOIL Method

Multiply  13x 1 12 12x 2 52 using the FOIL method. Solution:

 13x2 12x2 5 6x2

Multiply the first terms.

 13x2 1252 5 215x

Multiply the outer terms.

 112 12x2 5 2x

Multiply the inner terms.

 112 1252 5 25

Multiply the last terms.

 dd the first, outer, inner, and last terms, A and identify the like terms.        1 3x 1 1 2 1 2x 2 5 2 5 6x 2 2 15x 1 2x 2 5 5 6x2 2 13x 2 5

Combine like terms. ▼ ANSWER

▼ Y O U R T U R N   Multiply  12x 2 32 15x 2 22.

2

10x 2 19x 1 6

Some products of binomials occur frequently in algebra and are given s­pecial names. Example 7 illustrates the difference of two squares and perfect squares. EXAMPLE 7  Multiplying Binomials Resulting in Special Products

Find the following:   a.   1x 2 52 1x 1 52  b.   1x 1 522  c.   1x 2 522 Solution:

Inner

First

Difference of two squares

f

f

f

Outer

x 2 2 52

5

x 2 2 25

f

f

a. 1 x 2 5 2 1 x 1 5 2 5 x 2 1 5x 2 5x 2 52 5 Last First

Inner

f

f

First

Outer

f

f

b. 1 x 1 5 2 2 5 1 x 1 5 2 1 x 1 5 2 5 x 2 1 5x 1 5x 1 52 5 x 2 1 2 1 5x 2 1 52 5 x 2 1 10x 1 25 Inner

Last

f

f

Outer

f

f

c. 1 x 2 5 2 2 5 1 x 2 5 2 1 x 2 5 2 5 x 2 2 5x 2 5x 1 52 5 x 2 2 2 1 5x 2 1 52 5 x 2 2 10x 1 25 Last

Let a and b be any real number, variable, or algebraic expression in the following special products.

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0.3  Polynomials: Basic Operations 

31

STUD Y T I P

DIFFERENCE OF TWO SQUARES

(a 1 b)(a 2 b) 5 a2 2 ab 1 ab 2 b2 5 a2 2 b2

1 a 1 b 2 1 a 2 b 2 5 a2 2 b2 PERFECT SQUARES

Square of a binomial sum:  1a 1 b22 5  1a 1 b2 1a 1 b2 5 a2 1 2ab 1 b2 Square of a binomial difference:  1a 2 b22 5  1a 2 b2 1a 2 b2 5 a2 2 2ab 1 b2

EXAMPLE 8  Finding the Square of a Binomial Sum

Find  1x 1 322.

common mistake Forgetting the middle term, which is twice the product of the two terms in the binomial. ✓C O R R E C T

✖INCORRECT 2



 1x 1 32 5  1x 1 32 1x 1 32 2

5 x 1 3x 1 3x 1 9 5 x2 1 6x 1 9

Error: 1 x 1 3 2 2 2 x 2 1 9

Don’t forget the middle term, which is twice the product of the two terms in the binomial.

EXAMPLE 9  Using Special Product Formulas

Find the following: a.   12x 2 122  b.   13 1 2y22  c.   14x 1 32 14x 2 32 Solution (a):

 rite the square of a binomial W difference formula.      1a 2 b22 5 a2 2 2ab 1 b2

 12x 2 122 5  12x22 2 2 12x2 112 1 12

Let a 5 2x and b 5 1.

5 4x2 2 4x 1 1

Simplify.

Solution (b):

 rite the square of a binomial W sum formula.     1a 1 b22 5 a2 1 2ab 1 b2

 13 1 2y22 5  1322 1 2 132 12y2 1  12y22

Let a 5 3 and b 5 2y.

5 9 1 12y 1 4y2

Simplify.

5 4y2 1 12y 1 9

Write in standard form. Solution (c):

 rite the difference of two W squares formula.       1a 1 b2 1a 2 b2 5 a2 2 b2

 14x 1 32 14x 2 32 5  14x22 2 32

Let a 5 4x and b 5 3.

5 16x2 2 9

Simplify.

▼ Y O U R T U R N   Find the following: 2

2

a.   13x 1 12   b.   11 2 3y2   c.   13x 1 22 13x 2 22

Young_AT_6160_ch00_pp02-45.indd 31

▼ ANSWER a. 9x2 1 6x 1 1 b. 9y2 2 6y 1 1 c. 9x2 2 4

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32 

CHAPTER 0  Prerequisites and Review

EXAMPLE 10  Cubing a Binomial

▼ CAUTION

1x 1 2 23 2 x3 1 8 1x 2 2 23 2 x3 2 8

Find the following: a.   1x 1 223  b.   1x 2 223 Solution (a):

f

 rite the cube as a product of    1 x 1 2 2 3 5 1 x 1 2 2 1 x 1 2 2 1 x 1 2 2 W three binomials. 2  1x 1 22

Apply the perfect square formula.               5 1 x 1 2 2 1 x 1 4x 1 4 2 2

Apply the distributive property.           5 x 1 x 2 1 4x 1 4 2 1 2 1 x 2 1 4x 1 4 2

Apply the distributive property.                5 x3 1 4x2 1 4x 1 2x2 1 8x 1 8 Combine like terms.              5 x3 1 6x2 1 12x 1 8 Solution (b):

                  1 x 2 2 2 3 5 1 x 2 2 2 1 x 2 2 2 1 x 2 2 2

f

 rite the cube as a product of   W three binomials.

2

 1x 2 222

Apply the perfect square formula.                5  1x 2 22 1x 2 4x 1 42

Apply the distributive property.            5 x 1x2 2 4x 1 42 2 2 1x2 2 4x 1 42

Apply the distributive property.            5 x3 2 4x2 1 4x 2 2x2 1 8x 2 8 Combine like terms.                   5 x3 2 6x2 1 12x 2 8

PERFECT CUBES

Cube of a binomial sum: Cube of a binomial difference:

[ CONCEPT CHECK ] Multiply and simplify: (ax 1 by)(ax 2 by)

▼ ANSWER a2x2 2 b2y2

 1a 1 b23 5 a3 1 3a2b 1 3ab2 1 b3  1a 2 b23 5 a3 2 3a2b 1 3ab2 2 b3

EXAMPLE 11  Applying the Special Product Formulas

Find the following: a.  12x 1 123  b.   12x 2 523 Solution (a):

 rite the cube of a binomial W sum formula.    1a 1 b23 5 a3 1 3a2b 1 3ab2 1 b3

Let a 5 2x and b 5 1.     12x 1 123 5  12x23 1 3 12x22 112 1 3 12x2 1122 1 13 Simplify.   5 8x3 1 12x2 1 6x 1 1 Solution (b):

 rite the cube of a binomial W difference formula.    1a 2 b23 5 a3 2 3a2b 1 3ab2 2 b3

Let a 5 2x and b 5 5.     12x 2 523 5  12x23 2 3 12x22 152 1 3 12x2 1522 2 53 Simplify. ▼ ANSWER

27x3 2 108x2 1 144x 2 64

Young_AT_6160_ch00_pp02-45.indd 32

5 8x3 2 60x2 1 150x 2 125

▼ Y O U R T U R N   Find  13x 2 423.

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33

0.3  Polynomials: Basic Operations 

EXAMPLE 12  A  pplying the Special Product Formulas for Binomials in Two Variables

Find  12x 2 3y22. Solution:

 rite the square of a binomial W difference formula.             1a 2 b22 5  1a 2 b2 1a 2 b2 5 a2 2 2ab 1 b2

Let a 5 2x and b 5 3y.

 12x 2 3y22 5  12x22 2 2 12x2 13y2 1  13y22 5 4x2 2 12xy 1 9y2

Simplify.

▼

▼ ANSWER

Y O U R T U R N   Find  13x 2 2y22.

9x2 2 12xy 1 4y2

[SECTION 0.3]  S U M M A RY In this section, polynomials were defined. Polynomials with P E R F E C T S Q U A R E S one, two, and three terms are called monomials, binomials, and Square of a binomial sum. trinomials, respectively. Polynomials are added and subtracted by  1a 1 b22 5  1a 1 b2 1a 1 b2 5 a2 1 2ab 1 b2 combining like terms. Polynomials are multiplied by distributing the monomials in the first polynomial throughout the second Square of a binomial difference. polynomial. In the special case of the product of two binomials,  1a 2 b22 5  1a 2 b2 1a 2 b2 5 a2 2 2ab 1 b2 the FOIL method can also be used. The following are special products of binomials. PERFECT CUBES

DIFFERENCE OF TWO SQUARES

 1a 1 b2 1a 2 b2 5 a2 2 b2

Cube of a binomial sum.  1a 1 b23 5 a3 1 3a2b 1 3ab2 1 b3 Cube of a binomial difference.  1a 2 b23 5 a3 2 3a2b 1 3ab2 2 b3

[SEC TI O N 0. 3]   E X E R C I S E S • SKILLS In Exercises 1–8, write the polynomial in standard form and state the degree of the polynomial. 1. 5x2 2 2x3 1 16 2 7x4

2. 7x3 2 9x2 1 5x 2 4

3. 4x 1 3 2 6x3

4. 5x5 2 7x3 1 8x4 2 x2 1 10

5. 15

6. 214

7. y 2 2

8. x 2 5

In Exercises 9–24, add or subtract the polynomials, gather like terms, and write the simplified expression in standard form. 9.  12x2 2 x 1 72 1  123x2 1 6x 2 22 10.  13x2 1 5x 1 22 1  12x2 2 4x 2 92

11.  127x2 2 5x 2 82 2  124x 2 9x2 1 102 12.  18x3 2 7x2 2 102 2  17x3 1 8x2 2 9x2 13.  12x4 2 7x2 1 82 2  13x2 2 2x4 1 92 14.  14x2 2 9x 2 22 2  15 2 3x 2 5x22

15.  17z2 2 22 2  15z2 2 2z 1 12 16.  125y3 2 7y2 1 9y2 2  114y2 2 7y 1 22 17.  13y3 2 7y2 1 8y 2 42 2  114y3 2 8y 1 9y22 18.  12x2 1 3xy2 2  1x2 1 8xy 2 7y22

19.  16x 2 2y2 2 2 15x 2 7y2 20. 3a 2 32a2 2  15a 2 4a2 1 324 21.  12x2 2 22 2  1x 1 12 2  1x2 2 52 22.  13x3 1 12 2  13x2 2 12 2  15x 2 32

23.  14t 2 t2 2 t32 2  13t2 2 2t 1 2t32 1  13t3 2 12 24.  12z3 2 2z22 1  1z2 2 7z 1 12 2  14z3 1 3z2 2 3z 1 22

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34 

CHAPTER 0  Prerequisites and Review

In Exercises 25–64, multiply the polynomials and write the expressions in standard form. 25. 5xy2 17xy2

26. 6z 14z32

27. 2x3 11 2 x 1 x22

28. 24z2 12 1 z 2 z22

33. 2ab2 1a2 1 2ab 2 3b22

34. bc 3d 2 1b2c 1 cd 3 2 b2d 42

35.  12x 1 12 13x 2 42

36.  13z 2 12 14z 1 72

29. 22x2 15 1 x 2 5x22

37.  1x 1 22 1x 2 22

41.  12x 2 12 11 2 2x2

45.  17y 2 2y22 1  y 2 y2 1 12 49.  1t 2 222

53. 3 1x 1 y2 2 342

57. y 13y 1 42 12y 2 12

61.  1b 2 3a2 1a 1 2b2 1b 1 3a2

1

30. 22 z 1 2z 1 4z2 2 10 2

38.  1  y 2 52 1y 1 52

42.  14b 2 5y2 14b 1 5y2

46.  14 2 t22 16t 1 1 2 t22 50.  1t 2 322

54.  12x2 1 3y22

58. p2 1p 1 12 1p 2 22

62.  1x 2 2y2 1x2 1 2xy 1 4y22

31.  1x2 1 x 2 222x3

32.  1x2 2 x 1 223x3

39.  12x 1 32 12x 2 32

40.  15y 1 12 15y 2 12

43.  12x2 2 32 12x2 1 32

44.  14xy 2 92 14xy 1 92

47.  1x 1 12 1x2 2 2x 1 32

48.  1x 1 32 1x2 2 3x 1 92

51.  1z 1 222

52.  1z 1 322

55.  15x 2 222

56.  1x 1 12 1x2 1 x 1 12

59.  1x2 1 12 1x2 2 12

63.  1x 1 y 2 z2 12x 2 3y 1 5z2

60.  1t 2 522 1t 1 522

64.  15b2 2 2b 1 12 13b 2 b2 1 22

• A P P L I C AT I O N S In Exercises 65–68, profit is equal to revenue minus cost: P 5 R 2 C. 65. Profit. Donna decides to sell fabric cord covers on eBay for

71. Geometry. Suppose a running track is constructed of a

$20 apiece. The material for each cord cover costs $9, and it costs her $100 a month to advertise on eBay. Let x be the number of cord covers sold. Write a polynomial representing her monthly profit. Profit. Calculators are sold for $25 each. Advertising costs are $75 per month. Let x be the number of calculators sold. Write a polynomial representing the monthly profit earned by selling x calculators. Profit. If the revenue associated with selling x units of a product is R 5 2x2 1 100x, and the cost associated with producing x units of the product is C 5 2100x 1 7500, find the polynomial that represents the profit of making and ­selling x units. Profit. A business sells a certain quantity x of items. The ­revenue generated by selling x items is given by the equation R 5 212x 2 1 50x. The costs are given by C 5 8000 2 150x. Find a polynomial representing the net profit of this business when x items are sold. Volume of a Box. A rectangular sheet of cardboard is to be used in the construction of a box by cutting out squares of side length x from each corner and turning up the sides. Suppose the dimensions of the original rectangle are 15 inches by 8 inches. Determine a polynomial in x that would give the ­volume of the box. Volume of a Box. Suppose a box is to be constructed from a square piece of material of side length x by cutting out a 2-inch square from each corner and turning up the sides. Express the volume of the box as a polynomial in the variable x.

r­ectangular portion that measures 2x feet wide by 2x 1 5 feet long. Each end of the rectangular portion consists of a ­semicircle whose diameter is 2x. Write a polynomial that determines the a. perimeter of the track in terms of the variable x. b. area of the track in terms of x.

66.

67.

68.

69.

70.

Young_AT_6160_ch00_pp02-45.indd 34

2x

2x 2x + 5

72. Geometry. A right circular cylinder whose radius is r and

whose height is 2r is surmounted by a hemisphere of radius r. a. Find a polynomial in the variable r that represents the

volume of the “silo” shape. b. Find a polynomial in r that represents the total surface

area of the “silo.” r 2r r

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35

0.3  Polynomials: Basic Operations 

73. Engineering. The force of an electrical field is given by the

equation F 5 k

q1q2 r

2

74. Engineering. If a football (or other projectile) is thrown

upward, its height above the ground is given by the equation s 5 16t 2 1 v0t 1 s0, where v0 and s0 are the initial velocity and initial height of the football, respectively, and t is the time in ­seconds. Suppose the football is thrown from the top of a building that is 192 feet tall, with an initial speed of 96 feet per second. a. Write the polynomial that gives the height of the football in terms of the variable t (time). b. What is the height of the football after 2 sec­onds have elapsed? Will the football hit the ground after 2 seconds?

. Suppose q1 5 x, q2 5 3x, and r 5 10x.

Find a polynomial representing the force of the electrical field in terms of the variable x.

• C AT C H T H E M I S TA K E In Exercises 75 and 76, explain the mistake that is made. 75. Subtract and simplify 1 2x 2 2 5 2 2 1 3x 2 x 2 1 1 2 .

Solution:

2x2 2 5 2 3x 2 x2 1 1

Eliminate the parentheses.

2

Collect like terms.

x 2 3x 2 4

This is incorrect. What mistake was made?

76. Simplify  12 1 x22.

Solution:

Write the square of the binomial as the sum of the squares.    12 1 x22 5 22 1 x2 5 x2 1 4

Simplify.

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 77–80, determine whether each of the following statements is true or false.

In Exercises 81 and 82, let m and n be real numbers and m + n.

77. All binomials are polynomials.

81. What degree is the product of a polynomial of degree n and a

78. The product of two monomials is a binomial.

polynomial of degree m? 82. What degree is the sum of a polynomial of degree n and a polynomial of degree m?

79.  1x 1 y23 5 x 3 1 y3

2 80.  1x 2 y2  5 x 2 1 y2

• CHALLENGE In Exercises 83–86, perform the indicated operations and simplify. 2 83.  17x 2 4y2  2   17x 1 4y 2  2 2

84.  13x 2 5y222  13x 1 5y222

85.  1x 2 a2 1x2 1 ax 1 a22

86.  1x 1 a2 1x2 2 ax 1 a22

• TECHNOLOGY 87.  Use a graphing utility to plot the graphs of the three

1 2x 1 3 2 1 x 2 4 2 , 2x 2 1 5x 2 12, and expressions 2 2x 2 5x 2 12. Which two graphs agree with each other?

Young_AT_6160_ch00_pp02-45.indd 35

88. Use a graphing utility to plot the graphs of the three

expressions 1 x 1 5 2 2, x 2 1 25, and x 2 1 10x 1 25. Which two graphs agree?

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36 

CHAPTER 0  Prerequisites and Review

0.4 FACTORING POLYNOMIALS SKILLS OBJECTIVES ■■ Factor out the greatest common factor. ■■ Factor polynomials using special forms. ■■ Factor a trinomial as a product of binomials. ■■ Factor polynomials by grouping. ■■ Follow a strategy to factor polynomials.

CONCEPTUAL OBJECTIVES ■■ Understand that factoring has its basis in the distributive property. ■■ Learn formulas for factoring special forms. ■■ Know the general strategy for factoring a trinomial as a product of binomials. ■■ Be able to identify polynomials that may be factored by grouping. ■■ Develop a general strategy for factoring polynomials.

In Section 0.3, we discussed multiplying polynomials. In this section, we examine the reverse of that process, which is called factoring. Consider the following product: 1 x 1 3 2 1 x 1 1 2 5 x 2 1 4x 1 3

To factor the resulting polynomial, you reverse the process to undo the multiplication: x 2 1 4x 1 3 5 1 x 1 3 2 1 x 1 1 2

The polynomials  1x 1 32 and  1x 1 12 are called factors of the polynomial x2 1 4x 1 3. The process of writing a polynomial as a product is called factoring. In Chapter 1 we will solve quadratic equations by factoring. In this section, we will restrict our discussion to factoring polynomials with integer coefficients, which is called factoring over the integers. If a polynomial cannot be factored using integer coefficients, then it is prime or irreducible over the integers. When a polynomial is written as a product of prime polynomials, then the polynomial is said to be factored completely.

0.4.1  Greatest Common Factor 0.4.1 S K I L L

Factor out the greatest common factor. 0.4.1 C O N C E P T U A L

Understand that factoring has its basis in the distributive property.

The simplest type of factoring of polynomials occurs when there is a factor common to every term of the polynomial. This common factor is a monomial that can be “factored out” by applying the distributive property in reverse: ab 1 ac 5 a 1 b 1 c 2

For example, 2x2 2 6x can be written as 2x 1x2 2 2x 132. Notice that 2x is a common factor to both terms, so the distributive property tells us we can factor this polynomial to yield 2x 1x 2 32. Although 2 is a common factor and x is a common factor, the monomial 2x is called the greatest common factor. GREATEST COMMON FACTOR

The monomial ax k is called the greatest common factor (GCF) of a polynomial in x with integer coefficients if both of the following are true: ■■ a is the greatest integer factor common to all of the polynomial coefficients. ■■ k is the smallest exponent on x found in all the terms of the polynomial.

POLYNOMIAL

7x 1 21 2

7

3x 1 12x

3x

4x3 1 2x 1 6

2

6x4 2 9x3 1 12x2

3x2

25x4 1 25x3 2 20x2

Young_AT_6160_ch00_pp02-45.indd 36

GCF

25x2

WRITE EACH TERM AS A PRODUCT OF GCF AND REMAINING FACTOR

FACTORED FORM

7 1x2 1 7 132

7 1x 1 32

3x 1x2 1 3x 142

3x 1x 1 42

2 12x32 1 2x 1 2 132

2 12x3 1 x 1 32

25x2 1x22 2 5x2 125x2 2 5x2 142

25x2 1x2 2 5x 1 42

3x2 12x22 2 3x2 13x2 1 3x2 142

3x2 12x2 2 3x 1 42

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0.4  Factoring Polynomials 

EXAMPLE 1  F  actoring Polynomials by Extracting the Greatest Common Factor

Factor: a. 6x5 2 18x4

b. 6x5 2 10x4 2 8x3 1 12x2

37

[ CONCEPT CHECK ] Find the greatest common factor of the polynomial a3x6 1 a4x4

▼ ANSWER a3x4

Solution (a):

Identify the greatest common factor. Write each term as a product with the GCF as a factor. Factor out the GCF.

6x4 6x5 2 18x4 5 6x4 1x2 2 6x4 132 5 6x4 1x 2 32

Solution (b):

Identify the greatest common factor. 2x2 Write each term as a product with the GCF as a factor. 6x5 2 10x4 2 8x3 1 12x2 5 2x2 13x32 2 2x2 15x22 2 2x2 14x2 1 2x2 162 Factor out the GCF.

▼ Y O U R T U R N   Factor:

2 3 2 5 2x  13x 2 5x 2 4x 1 62

a. 12x3 2 4x   b. 3x5 2 9x4 1 12x3 2 6x2

0.4.2  Factoring Formulas: Special Polynomial Forms The first step in factoring polynomials is to look for a common factor. If there is no common factor, then we look for special polynomial forms that we learned were special products in Section 0.3 and reverse the process. Difference of two squares Perfect squares Sum of two cubes Difference of two cubes

a2 2 b2 5  1a 1 b2 1a 2 b2

a2 1 2ab 1 b2 5  1a 1 b22 a2 2 2ab 1 b2 5  1a 2 b22

▼ ANSWER a.  4x 13x2 2 12

b.  3x2 1x3 2 3x2 1 4x 2 22

0.4.2 S K I L L

Factor polynomials using special forms. 0.4.2 C O N C E P T U A L

Learn formulas for factoring special forms.

a3 1 b3 5  1a 1 b2 1a2 2 ab 1 b22 a3 2 b3 5  1a 2 b2 1a2 1 ab 1 b22

EXAMPLE 2  F  actoring the Difference of Two Squares

Factor: a.  x2 2 9   b. 4x2 2 25   c.  x4 2 16 Solution (a): Rewrite as the difference of two squares. x2 2 9 5 x2 2 32 Let a 5 x and b 5 3 in a2 2 b2 5  1a 1 b2  1a 2 b2.

Solution (b):

[ CONCEPT CHECK ] Factor: a2x4 2 b2y2

▼ ANSWER (ax2 1 by)(ax2 2 by)

5  1x 1 32  1x 2 32

Rewrite as the difference of two squares. 4x2 2 25 5  12x22 2 52

Let a 5 2x and b 5 5 in a2 2 b2 5  1a 1 b2  1a 2 b2. 5  12x 1 52  12x 2 52 Solution (c):

Rewrite as the difference of two squares.

2 x4 2 16 5 1 x 2 2 2 42

Let a 5 x2 and b 5 4 in a2 2 b2 5  1a 1 b2  1a 2 b2. 5 1 x 2 1 4 2 1 x 2 2 4 2 Note that x2 2 4 is also a difference of two 2 squares (part a of the following Your Turn). 5  1x 1 22 1x 2 22 1x 1 42

▼

Y O U R T U R N   Factor:

a.  x2 2 4   b. 9x2 2 16   c.  x4 2 81

Young_AT_6160_ch00_pp02-45.indd 37

▼ ANSWER a.  1x 1 22  1x 2 22

b.  13x 1 42  13x 2 42

c.  1x2 1 92  1x 2 32  1x 1 32

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38 

CHAPTER 0  Prerequisites and Review

A trinomial is a perfect square if it has the form a2 6 2ab 1 b2. Notice that: The first term and third term are perfect squares. The middle term is twice the product of the bases of these two perfect squares. ■■ The sign of the middle term determines the sign of the factored form: ■■ ■■

a2 6 2ab 1 b2 5 1 a 6 b 2 2 EXAMPLE 3  Factoring Trinomials That Are Perfect Squares

Factor: a. x2 1 6x 1 9   b.  x2 2 10x 1 25   c. 9x2 2 12x 1 4 Solution (a):

Rewrite the trinomial so that the first and third terms are perfect squares.        x2 1 6x 1 9 5 x2 1 6x 1 32 Notice that if we let a 5 x and b 5 3 in a2 1 2ab 1 b2 5  1a 1 b22, then the middle term 6x is 2ab.     x2 1 6x 1 9 5 x2 1 2 13x2 1 32 5  1x 1 322

Solution (b):

Rewrite the trinomial so that the first and third terms are perfect squares.      x2 2 10x 1 25 5 x2 2 10x 1 52

Notice that if we let a 5 x and b 5 5 in a2 2 2ab 1 b2 5  1a 2 b22, then the middle term 210x is 22ab.   x2 2 10x 1 25 5 x2 2 2 15x2 1 52 5  1x 2 522 Solution (c):

Rewrite the trinomial so that the first and third terms are perfect squares.       9x2 2 12x 1 4 5  13x22 2 12x 1 22

Notice that if we let a 5 3x and b 5 2 in a2 2 2ab 1 b2 5  1a 2 b22, then the middle term 212x is 22ab.    9x2 2 12x 1 4 5  13x22 2 2 13x2 122 1 22 5  13x 2 222

▼

▼ ANSWER a.   1x 1 42

Y O U R T U R N   Factor: 2

b.   1x 2 22

a.  x2 1 8x 1 16   b.  x2 2 4x 1 4   c. 25x2 2 20x 1 4

2

c.   15x 2 222

EXAMPLE 4  Factoring the Sum of Two Cubes

Factor x3 1 27. Solution:

Rewrite as the sum of two cubes.

x3 1 27 5 x3 1 33

Write the sum of two cubes formula. a3 1 b3 5  1a 1 b2 1a2 2 ab 1 b22

Let a 5 x and b 5 3. x3 1 27 5 x3 1 33 5  1x 1 32 1x2 2 3x 1 92

Young_AT_6160_ch00_pp02-45.indd 38

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0.4  Factoring Polynomials 

39

EXAMPLE 5  Factoring the Difference of Two Cubes

Factor x3 2 125. Solution:

Rewrite as the difference of two cubes.    x3 2 125 5 x3 2 53 Write the difference of two cubes formula.       a3 2 b3 5  1a 2 b2 1a2 1 ab 1 b22

Let a 5 x and b 5 5.      x3 2 125 5 x3 2 53 5  1x 2 52 1x2 1 5x 1 252

▼

Y O U R T U R N   Factor: a.  x3 1 8   b.  x3 2 64

▼ ANSWER a.  1x 1 22 1x2 2 2x 1 42

b.  1x 2 42 1x2 1 4x 1 162

0.4.3 Factoring a Trinomial as a Product of Two Binomials The first step in factoring is to look for a common factor. If there is no common factor, look to see whether the polynomial is a special form for which we know the factoring formula. If it is not of such a special form and if it is a trinomial, then we proceed with a general ­factoring strategy. We know that 1x 1 32 1x 1 22 5 x2 1 5x 1 6, so we say the factors of x2 1 5x 1 6 are 1x 1 32 and 1x 1 22. In factored form we have x2 1 5x 1 6 5  1x 1 32  1x 1 22. Recall the FOIL method from Section 0.3. The product of the last terms  13 and 22 is 6, and the sum of the products of the inner terms  13x2 and the outer terms  12x2 is 5x. Let’s pretend for a minute that we didn’t know this factored form but had to work with the general form:

0.4.3 S K I L L

Factor a trinomial as a product of binomials. 0.4.3 C O N C E P T U A L

Know the general strategy for factoring a trinomial as a product of binomials.

x 2 1 5x 1 6 5 1 x 1 a 2 1 x 1 b 2

The goal is to find a and b. We start by multiplying the two binomials on the right. x 2 1 5x 1 6 5 1 x 1 a 2 1 x 1 b 2 5 x 2 1 ax 1 bx 1 ab 5 x 2 1 1 a 1 b 2 x 1 ab

Compare the expression we started with on the left with the expression on the far right x2 1 5x 1 6 5 x2 1 1a 1 b2 x 1 ab. We see that ab 5 6 and 1a 1 b2 5 5. Start with the possible combinations of a and b whose product is 6, and then look among those for the combination whose sum is 5. ab 5 6 a1b

a, b:

1, 6

21, 26

2, 3

22, 23

7

27

5

25

All of the possible a, b combinations in the first row have a product equal to 6, but only one of those has a sum equal to 5. Therefore, the factored form is x 2 1 5x 1 6 5 1 x 1 a 2 1 x 1 b 2 5 1 x 1 2 2 1 x 1 3 2 EXAMPLE 6  Factoring a Trinomial

Factor x2 1 10x 1 9. Solution:

 rite the trinomial as a product of two W binomials in general form.

Young_AT_6160_ch00_pp02-45.indd 39

x2 1 10x 1 9 5  1x 1 u2 1x 1 u2

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40 

CHAPTER 0  Prerequisites and Review

Write all of the integers whose product is 9. Integers whose product is 9

1, 9

21, 29

3, 3

23, 23

Determine the sum of the integers. Integers whose product is 9

1, 9

21, 29

3, 3

23, 23

Sum

10

210

6

26

Select 1, 9 because the product is 9 (last term of the trinomial) and the sum is 10 (middle term coefficient of the trinomial). x2 1 10x 1 9 5  1x 1 92 1x 1 12 ▼ ANSWER a.  1x 1 42 1x 1 52

Check:  1x 1 92 1x 1 12 5 x2 1 1x 1 9x 1 9 5 x2 1 10x 1 9

3

▼

Y O U R T U R N   Factor x2 1 9x 1 20.

In Example 6, all terms in the trinomial are positive. When the constant term is negative, then (regardless of whether the middle term is positive or negative) the ­factors will be opposite in sign, as illustrated in Example 7.

EXAMPLE 7  Factoring a Trinomial

Factor x2 2 3x 2 28. Solution:

 rite the trinomial as a product of two W binomials in general form. x2 2 3x 2 28 5  1x 1 u2  1x 2 u2 Write all of the integers whose product is 228. Integers whose product is 228

1, 228

21, 28

2, 214

22, 14

4, 27

24, 7

1, 228

21, 28

2, 214

22, 14

4, 27

24, 7

227

27

212

12

23

3

Determine the sum of the integers. Integers whose product is 228 Sum

Select 4, 27 because the product is 228 (last term of the trinomial) and the sum is 23 (middle term coefficient of the trinomial). x2 2 3x 2 28 5  1x 1 42 1x 2 72 ▼ ANSWER

 1x 1 62 1x 2 32

Young_AT_6160_ch00_pp02-45.indd 40

Check:  1x 1 42 1x 2 72 5 x2 2 7x 1 4x 2 28 5 x2 2 3x 2 28  3

▼

Y O U R T U R N   Factor x2 1 3x 2 18.

When the leading coefficient of the trinomial is not equal to 1, then we consider all possible factors using the following procedure, which is based on the FOIL method in reverse.

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0.4  Factoring Polynomials 

41

FACTORING A TRINOMIAL WHOSE LEADING COEFFICIENT IS NOT 1 Factors of a

  ax2 1 bx 1 c 5  1hx 1 h2 1hx 1 h2 Factors of c

Step 1:  Find two First terms whose product is the first term of the trinomial. Step 2:  Find two Last terms whose product is the last term of the trinomial. Step 3: Consider all possible combinations found in Steps 1 and 2 until the sum of the Outer and Inner products is equal to the middle term of the trinomial.

EXAMPLE 8  F  actoring a Trinomial Whose Leading Coefficient Is Not 1

Factor 5x2 1 9x 2 2. Solution: STEP 1  Start

STEP 2  The

with the first term. Note that 5x ? x 5 5x2.

product of the last terms should yield 22.

STEP 3  Consider

 15x 6 h2 1x 6 h2

all possible factors based on Steps 1 and 2.  

Since the outer and inner products must sum to 9x, the factored form must be:

21, 2 or 1, 22  15x 2 12  1x 1 22  15x 1 12  1x 2 22  15x 1 22  1x 2 12  15x 2 22  1x 1 12

5x2 1 9x 2 2 5  15x 2 12  1x 1 22

Check:  15x 2 12  1x 1 22 5 5x2 1 10x 2 1x 2 2 5 5x2 1 9x 2 2  

3

▼

Y O U R T U R N   Factor 2t2 1 t 2 3.

▼ ANSWER

 12t 1 32  1t 2 12

EXAMPLE 9  F  actoring a Trinomial Whose Leading Coefficient Is Not 1

Factor 15x2 2 x 2 6. Solution: STEP 1  Start

STEP 2  The

with the first term.

product of the last terms should yield 26.

Young_AT_6160_ch00_pp02-45.indd 41

 15x 6 h2  13x 6 h2 or  115x 6 h2  1x 6 h2 21, 6 or 1, 26 or 2, 23 or 22, 3

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CHAPTER 0  Prerequisites and Review

S TU DY TIP In Example 9, Step 3, we can eliminate any factors that have a common factor since there is no common factor to the terms in the trinomial.

 15x 2 12  13x 1 62   115x 2 12  1x 1 62  15x 1 62  13x 2 12   115x 1 62  1x 2 12  15x 1 12  13x 2 62   115x 1 12  1x 2 62  15x 2 62  13x 1 12   115x 2 62  1x 1 12  15x 1 22  13x 2 32   115x 1 22  1x 2 32  15x 2 32  13x 1 22   115x 2 32  1x 1 22  15x 2 22  13x 1 32   115x 2 22  1x 1 32  15x 1 32  13x 2 22   115x 1 32  1x 2 22

STEP 3  Consider

all possible factors   based on Steps 1 and 2.

Since the outer and inner products must sum to 2x, the factored form must be:

▼ ANSWER

 13x 2 42 12x 1 32

[ CONCEPT CHECK ] TRUE OR FALSE  A trinomial that is degree 2 and has integer coefficients with the leading coefficient equal to 1 is always factorable.

▼ ANSWER False

15x2 2 x 2 6 5  15x 1 32  13x 2 22

Check:  15x 1 32 13x 2 22 5 15x2 2 10x 1 9x 2 6 5 15x2 2 x 2 6 

3

▼

Y O U R T U R N   Factor 6x2 1 x 2 12.

EXAMPLE 10  Identifying Prime (Irreducible) Polynomials

Factor x2 1 x 2 8. Solution:

 rite the trinomial as a product of W two binomials in general form.

x2 1 x 2 8 5  1x 1 h2  1x 2 h2

Write all of the integers whose product is 28.

Integers whose product is 28

1, 28

21, 8

4, 22

24, 2

Determine the sum of the integers.

Integers whose product is 28

1, 28

21, 8

4, 22

24, 2

27

7

2

22

Sum

The middle term of the trinomial is 2x, so we look for the sum of the integers that equals 21. Since no sum exists for the given combinations, we say that this polynomial is prime (irreducible) over the integers.

0.4.4  Factoring by Grouping 0.4.4 S K I L L

Factor polynomials by grouping. 0.4.4 C O N C E P T U A L

Be able to identify polynomials that may be factored by grouping.

Much of our attention in this section has been on factoring trinomials. For polynomials with more than three terms we first look for a common factor to all terms. If there is no common factor to all terms of the polynomial, we look for a group of terms that have a common factor. This strategy is called factoring by grouping. EXAMPLE 11  Factoring a Polynomial by Grouping

Factor x3 2 x2 1 2x 2 2. Solution:

Group the terms that have a common factor.     5  1x3 2 x22 1  12x 2 22

Factor out the common factor in each pair of parentheses.                     5 x2 1x 2 12 1 2 1x 2 12 Use the distributive property.                     5  1x2 1 22  1x 2 12

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43

0.4  Factoring Polynomials 

EXAMPLE 12  Factoring a Polynomial by Grouping 2

Factor 2x 1 2x 2 x 2 1.

TRUE OR FALSE  Trinomials can­ not be factored by grouping

Solution:

Group the terms that have a common factor. 5  12x2 1 2x2 1  12x 2 12 Factor out the common factor in each pair of parentheses.  5 2x 1x 1 12 2 1 1x 1 12 Use the distributive property.

▼ Y O U R T U R N   Factor x3 1 x2 2 3x 2 3.

[ CONCEPT CHECK ]

5  12x 2 12  1x 1 12

▼ ANSWER True

▼ ANSWER

 1x 1 12  1x2 2 32

0.4.5  A Strategy for Factoring Polynomials The first step in factoring a polynomial is to look for the greatest common factor. When specifically factoring trinomials, look for special known forms: a perfect square or a ­difference of two squares. A general approach to factoring a trinomial uses the FOIL method in reverse. Finally, we look for factoring by grouping. The following strategy for factoring polynomials is based on the techniques discussed in this ­section.

0.4.5 S K I L L

Follow a strategy to factor polynomials. 0.4.5 C O N C E P T U A L

Develop a general strategy for factoring polynomials.

STRATEGY FOR FACTORING POLYNOMIALS

1. Factor out the greatest common factor (monomial). 2. Identify any special polynomial forms and apply factoring formulas. 3. Factor a trinomial into a product of two binomials:  1ax 1 b2  1cx 1 d2. 4. Factor by grouping.

EXAMPLE 13  Factoring Polynomials

Factor: a. 3x2 2 6x 1 3  b.  24x3 1 2x2 1 6x  c. 15x2 1 7x 2 2  d.  x3 2 x 1 2x2 2 2

[ CONCEPT CHECK ]

Solution (a):

▼

Factor out the greatest common factor. 3x2 2 6x 1 3 5 3 1x2 2 2x 1 12

Factor ax2 2 ay2 ANSWER a(x 1 y)(x 2 y)

The trinomial is a perfect square.  5 3 1x 2 122

Solution (b):

Factor out the greatest common factor. 24x3 1 2x2 1 6x 5 22x 12x2 2 x 2 32 Use the FOIL method in reverse to factor the trinomial.  5   22x 12x 2 32  1x 1 12 Solution (c):

There is no common factor.

15x2 1 7x 2 2

Use the FOIL method in reverse to factor the trinomial.  5  13x 1 22  15x 2 12 Solution (d):

Factor by grouping. x3 2 x 1 2x2 2 2 5  1x3 2 x2 1  12x2 2 22   5 x 1x2 2 12 1 2 1x2 2 12   5  1x 1 22  1x2 2 12 Factor the difference of two squares.  5    1x 1 22  1x 2 12  1x 1 12

Young_AT_6160_ch00_pp02-45.indd 43

STUD Y T I P When factoring, always start by factoring out the GCF.

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44 

CHAPTER 0  Prerequisites and Review

[ S E C T I O N 0 . 4 ]    S U M M A R Y In this section, we discussed factoring polynomials, which is FA C T O R I N G A T R I N O M I A L A S A P R O D U C T the reverse process of multiplying polynomials. Four main O F T W O B I N O M I A L S techniques were discussed. x2 1 bx 1 c 5  1x 1 ?2  1x 1 ?2

G R E AT E S T C O M M O N FA C T O R : ax k

• a is the greatest common factor for all coefficients of the polynomial. • k is the smallest exponent found on all the terms in the polynomial. FA C T O R I N G F O R M U L A S : S P E C I A L P O LY N O M I A L F O R M S

Difference of two squares: Perfect squares: Sum of two cubes: Difference of two cubes:

a2 2 b2 5  1a 1 b2  1a 2 b2 a2 1 2ab 1 b2 5  1a 1 b22 a2 2 2ab 1 b2 5  1a 2 b22 a3 1 b3 5  1a 1 b2  1a2 2 ab 1 b22 a3 2 b3 5  1a 2 b2  1a2 1 ab 1 b22

1. Find all possible combinations of factors whose product is c. 2. Of the combinations in Step 1, look for the sum of factors that equals b. ax2 1 bx 1 c 5  1?x 1 ?2  1?x 1 ?2

1. Find all possible combinations of the first terms whose product is ax2. 2. Find all possible combinations of the last terms whose product is c. 3. Consider all possible factors based on Steps 1 and 2. FA C T O R I N G B Y G R O U P I N G

• Group terms that have a common factor. • Use the distributive property.

[SEC T I O N 0. 4]   E X E R C I S E S • SKILLS In Exercises 1–12, factor each expression. Start by finding the greatest common factor (GCF). 1. 5x 1 25

2. x2 1 2x

3. 4t2 2 2

4. 16z2 2 20z

5. 2x3 2 50x

6. 4x2y 2 8xy2 1 16x2y2

7. 3x3 2 9x2 1 12x

8. 14x4 2 7x2 1 21x

9. x3 2 3x2 2 40x

10. 29y2 1 45y

11. 4x2y3 1 6xy

12. 3z3 2 6z2 1 18z

In Exercises 13–20, factor the difference of two squares. 13. x2 2 9 2

17. 2x 2 98

14. x2 2 25

15. 4x2 2 9 2

18. 144 2 81y

2

16. 1 2 x4 2

19. 225x 2 169y

20. 121y2 2 49x2

In Exercises 21–32, factor the perfect squares. 21. x2 1 8x 1 16

22. y2 2 10y 1 25

23. x4 2 4x2 1 4

24. 1 2 6y 1 9y2

25. 4x2 1 12xy 1 9y2

26. x2 2 6xy 1 9y2

27. 9 2 6x 1 x2

28. 25x2 2 20xy 1 4y2

29. x4 1 2x2 1 1

30. x6 2 6x3 1 9

31. p2 1 2pq 1 q2

32. p2 2 2pq 1 q2

35. y3 2 64

36. x3 2 1

In Exercises 33–42, factor the sum or difference of two cubes. 33. t3 1 27 3

34. z3 1 64 3

37. 8 2 x

38. 27 2 y

41. 27 1 x3

42. 216x3 2 y3

3

39. y 1 125

40. 64x 2 x4

In Exercises 43–52, factor each trinomial into a product of two binomials. 43. x2 2 6x 1 5

44. t2 2 5t 2 6

45. y2 2 2y 2 3

46. y2 2 3y 2 10

47. 2y2 2 5y 2 3

48. 2z2 2 4z 2 6

49. 3t2 1 7t 1 2

50. 4x2 2 2x 2 12

51. 26t2 1 t 1 2

52. 26x2 2 17x 1 10

In Exercises 53–60, factor by grouping. 53. x3 2 3x2 1 2x 2 6

54. x5 1 5x3 2 3x2 2 15

55. a4 1 2a3 2 8a 2 16

56. x4 2 3x3 2 x 1 3

57. 3xy 2 5rx 2 10rs 1 6sy

58. 6x2 2 10x 1 3x 2 5

59. 20x2 1 8xy 2 5xy 2 2y2

60. 9x5 2 a2x3 2 9x2 1 a2

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0.4  Factoring Polynomials 

45

In Exercises 61–92, factor each of the polynomials completely, if possible. If the polynomial cannot be factored, state that it is prime. 61. x2 2 4y2 2

62. a2 1 5a 1 6

63. 3a2 1 a 2 14

2

2

64. ax 1 b 1 bx 1 a 1

65. x 1 16

66. x 1 49

67. 4z 1 25

69. 6x2 1 10x 1 4

70. x2 1 7x 1 5

71. 6x2 1 13xy 2 5y2

68. 16 2 b4 72. 15x 1 15xy

73. 36s2 2 9t2

74. 3x3 2 108x

75. a2b2 2 25c2

76. 2x3 1 54

2

77. 4x 2 3x 2 10

78. 10x 2 25 2 x

79. 3x 2 5x 2 2x

80. 2y3 1 3y2 2 2y

81. x3 2 9x

82. w3 2 25w

83. xy 2 x 2 y 1 1

84. a 1 b 1 ab 1 b2

85. x4 1 5x2 1 6

86. x6 2 7x3 2 8

87. x2 2 2x 2 24

88. 25x2 1 30x 1 9

4

89. x 1 125x

2

4

3

2

4

90. x 2 1

91. x 2 81

92. 10x2 2 31x 1 15

• A P P L I C AT I O N S 93. Geometry. A rectangle has a length of 2x 1 4 and a width

96. Business. The break-even point for a company is given

of x. Express the perimeter of the rectangle as a factored ­polynomial in x. 94. Geometry. The volume of a box is given by the expression x3 1 7x2 1 12x. Express the volume as a factored polynomial in the variable x. 95. Business. The profit of a business is given by the expression P 5 2x2 2 15x 1 4x 2 30. Express the profit as a factored ­polynomial in the variable x.



by solving the equation 3x2 1 9x 2 4x 2 12 5 0. Factor the ­polynomial on the left side of the equation. 97. Engineering. The height of a projectile is given by the ­equation s 5 216t2 2 78t 1 10. Factor the expression on the right side of the equal sign. 98. Engineering. The electrical field at a point P between two charges is given by k 5 this expression.

10x 2 x 2 . Factor the numerator of 100

• C AT C H T H E M I S TA K E In Exercises 99 and 100, explain the mistake that is made. 99. Factor x3 2 x2 2 9x 1 9.

Solution:

Solution: 3

2



Group terms with common factors.  1x 2 x 2 1  129x 1 92



Distributive property.



Factor out common factors.

Factor x2 2 9.

100. Factor 4x2 1 12x 2 40.

x2 1x 2 12 2 9 1x 2 12

 1x 2 12  1x2 2 92

Factor the trinomial into a product of binomials.

 12x 2 42  12x 1 102

5 2 1x 2 22  1x 1 52



Factor out a 2.



This is incorrect. What mistake was made?

 1x 2 12  1x 2 322

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 101–104, determine whether each of the following statements is true or false. 101. All trinomials can be factored into a product of two binomials. 102.  All polynomials can be factored into prime factors with

respect to the integers.

103. x2 2 y2 5  1x 2 y2  1x 1 y2 104. x2 1 y2 5  1x 1 y22

• CHALLENGE 105. Factor a2n 2 b2n completely, assuming a, b, and n are positive

integers.

106. Find all the values of c such that the trinomial x2 1 cx 2 14

can be factored.

• TECHNOLOGY 107. Use a graphing utility to plot the graphs of the three expressions

8x3 1 1,  12x 1 12  14x2 2 2x 1 12, and  12x 2 12  14x2 1 2x 1 12. Which two graphs agree with each other?

Young_AT_6160_ch00_pp02-45.indd 45

108. Use a graphing utility to plot the graphs of the three expressions

27x3 2 1,  13x 2 123, and  13x 2 12  19x2 1 3x 1 12. Which two graphs agree with each other?

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46 

CHAPTER 0  Prerequisites and Review

0.5 RATIONAL EXPRESSIONS SKILLS OBJECTIVES ■■ Find the domain of an algebraic expression. ■■ Reduce a rational expression to lowest terms. ■■ Multiply and divide rational expressions. ■■ Add and subtract rational expressions. ■■ Simplify complex rational expressions.

CONCEPTUAL OBJECTIVES ■■ U  nderstand why rational expressions have domain restrictions. ■■ Understand that we can cancel a common nonzero factor shared by the numerator and denominator because that ratio of a nonzero factor divided by itself is equal to 1. ■■ Understand why we use the same rules to multiply or divide rational expressions that we use to multiply or divide rational numbers. ■■ Understand the least common denominator method for rational expressions. ■■ Know the two methods for simplifying complex rational expressions.

0.5.1  Rational Expressions and Domain Restrictions 0.5.1 S K I L L

Find the domain of an algebraic expression. 0.5.1 C O N C E P T U A L

Understand why rational expressions have domain restrictions.

Recall that a rational number is the ratio of two integers with the denominator not equal to zero. Similarly, the ratio, or quotient, of two polynomials is a rational expression. Rational Numbers:

9 3 5       11 7 9

Rational Expressions:

5x 2 3x 1 2 9    2    x25 3x 22 x 11

As with rational numbers, the denominators of rational expressions are never equal to zero. In the first and third rational expressions, there are values of the variable that would correspond to a denominator equal to zero; these values are not permitted: 3x 1 2 x25

x25

  

9 3x 2 2

x2

2 3

5x 2 In the second rational expression, 2 , there are no real numbers that will correspond x 11 to a zero denominator. The set of real numbers for which an algebraic expression is defined is called the domain. Since a rational expression is not defined if its denominator is zero, we must eliminate from the domain those values of the variable that would result in a zero denominator. To find the domain of an algebraic expression, we ask the question, “What can x (the variable) be?” For rational expressions, the answer in general is “any values except those that make the denominator equal to zero.” EXAMPLE 1  Finding the Domain of an Algebraic Expression

Find the domain of the expressions. a. 2x2 2 5x 1 3   b. 

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2 x 11 x 3 x 11   c.  2   d.  x x 24 x 11

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0.5  Rational Expressions 

[ CONCEPT CHECK ]

Solution: ALGEBRAIC EXPRESSION

TYPE

DOMAIN

NOTE

a. 2x2 2 5x 1 3

Polynomial

All real numbers

 he domain of all T polynomials is the set of all real numbers.

1 does not have a 11 1 domain restrictions, but 2 a 21 does have domain restrictions.

Rational expression

All real numbers except x 5 4

When x 5 4, the rational expression is undefined.

▼

Rational

All real numbers

There are no real numbers that will result in the denominator being equal to zero.

All real numbers except x 5 0

When x 5 0, the rational expression is undefined.

b. 

2x 1 1 x24

x c.  2 x 11

d. 

3x 1 1 x

expression

Rational expression

▼ Y O U R T U R N   Find the domain of the expressions. a. 

47

3x 2 1 5x 2 1 2x 1 5   b.    c. 3x2 1 2x 2 7   d.  2 x x11 x 14

Explain why

2

ANSWER There are no real values that ­correspond to the denominator a 2 1 1 being equal to zero, but there are real values (a 5 21 or a 5 1) that correspond to the denominator a 2 2 1 being equal to zero.

▼ ANSWER

a. x ∙ 21 b. x ∙ 0 c. all real numbers d. all real numbers

In this text, it will be assumed that the domain is the set of all real numbers except the real numbers shown to be excluded.

EXAMPLE 2  E  xcluding Values from the Domain of Rational Expressions

Determine what real numbers must be excluded from the domain of the following rational expressions. a.

7x 1 5 3x 1 2   b.  2 x 2 5x x2 2 4

Solution (a):

Factor the denominator.

7x 1 5 7x 1 5 5 2 1 x 1 22 1x 2 22 x 24

x 5 22 and    must Determine the values of x that will     be x52 make the denominator equal to zero. excluded from the domain. Solution (b):

Factor the denominator.

3x 1 2 3x 1 2 5 2 x1x 2 52 x 2 5x

x 5 0  and     x 5 5 must be Determine the values of x that will    make the denominator equal to zero. excluded from the domain. In this section, we will simplify rational expressions and perform operations on rational expressions such as multiplication, division, addition, and subtraction. The resulting expressions may not have explicit domain restrictions, but it is important to note that there are implicit domain restrictions because the domain restrictions on the original rational expression still apply.

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48 

CHAPTER 0  Prerequisites and Review

0.5.2  Simplifying Rational Expressions 0.5.2 S K I L L

Recall that a fraction is reduced when it is written with no common factors. 4 4 4 16 4#4 4 16 4#4 4 5 # 5 a b # a b 5 1 1 2 # a b 5 or 5 # 5 4 3 12 4 3 4 3 3 3 12 3

Reduce a rational expression to lowest terms. 0.5.2 C O N C E P T U A L

Understand that we can ­cancel a common nonzero factor shared by the numerator and ­denominator because that ratio of a nonzero factor divided by itself is equal to 1.

Similarly, rational expressions are reduced to lowest terms, or simplified, if the numerator and denominator have no common factors other than 61. As with real numbers, the ability to write fractions in reduced form is dependent on your ability to factor. REDUCING A RATIONAL EXPRESSION TO LOWEST TERMS (SIMPLIFYING)

1. Factor the numerator and denominator completely. 2. State any domain restrictions. 3. Cancel (divide out) the common factors in the numerator and denominator. EXAMPLE 3  Reducing a Rational Expression to Lowest Terms

Simplify

x2 2 x 2 2 and state any domain restrictions. 2x 1 2

Solution:

1x 2 22 1x 1 12 x2 2 x 2 2 Factor the numerator and denominator.   5 2x 1 2 21x 1 12 x 2 21

State any domain restrictions.

▼ ANSWER

x12    x 2 1 2

ST U DY TIP Factors can be divided out ­(canceled). Factor the numerator and denominator first, and then divide out common factors.

ST U DY TIP Determine domain restrictions of a rational expression before dividing out (canceling) common factors.

Young_AT_6160_ch00_pp46-81.indd 48

Cancel (divide out) the common factor, x 1 1.

5

The rational expression is now in lowest terms (simplified).

5

▼ Y O U R T U R N   Simplify

1x 2 22 1x 1 12 21x 1 12 x22 2

x 2 21

x2 1 x 2 2 and state any domain restrictions. 2x 2 2

The following table summarizes two common mistakes made with rational expressions. CORRECT

INCORRECT

x15 is already simplified. y15

Error:

x x 1x

COMMENT

Factors can be divided out (canceled). Terms or parts of terms cannot be divided out. Remember to factor the numerator and denominator first, and then divide out common factors.

x15 x 5 y15 y

Error:

2

x   x 2 0, x 2 21 x 1x 1 12 1 5 x 2 0, x 2 21 x11 5

x 1 5 x11 x2 1 x

x 2 21

Determine the domain restrictions before dividing out common factors.

Note: Missing x 2 0.

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0.5  Rational Expressions 

[ CONCEPT CHECK ]

EXAMPLE 4  Simplifying Rational Expressions

If a simplified rational expression 1 where x is not results in 1x 2 12 equal to 1 or 3, what was the common factor that was divided out?

x2 2 x 2 6 Reduce 2 to lowest terms and state any domain restrictions. x 1x22 Solution:

 actor the numerator and F 1x 2 32 1x 1 22 x2 2 x 2 6 5 denominator.   2 1x 2 12 1x 1 22 x 1x22 5

Simplify.

5 

Y O U R T U R N   Reduce

restrictions.

▼ ANSWER (x 2 3)

x 2 22, x 2 1

State domain restrictions. Divide out the common factor, x 1 2.

▼

49

1x 2 32 1x 1 22 1x 2 12 1x 1 22 x23 x21

x 2 22, x 2 1

x2 1 x 2 6 to lowest terms and state any domain x 2 1 2x 2 3

▼ ANSWER

x22   x 2 23, x 2 1 x21

EXAMPLE 5  Simplifying Rational Expressions

Reduce

x2 2 4 to lowest terms and state any domain restrictions. 22x

Solution:

Factor the numerator and denominator. State domain restrictions.

1x 2 22 1x 1 22 x2 2 4 5 12 2 x2 22x

Factor out a negative in the denominator. Cancel (divide out) the common factor, x 2 2.

x22 1x 2 22 1x 1 22 5 21x 2 22 5

1x 2 22 1x 1 22 21x 2 22

5 21x 1 22

Simplify.

▼

x22

x 2 2 25 Y O U R T U R N   Reduce to lowest terms and state any domain restrictions. 52x

▼ ANSWER

21x 1 5 2  x ∙ 5

0.5.3  Multiplying and Dividing Rational Expressions

0.5.3 S K I L L

The same rules that apply to multiplying and dividing rational numbers also apply to rational ­expressions.

Multiply and divide rational expressions.

PROPERTY

RESTRICTION

DESCRIPTION

a c ac ⋅ 5 b d bd

b ∙ 0, d ∙ 0

 ultiply numerators and denominators, M respectively.

a c a d 4 5 ⋅ b d b c

b ∙ 0, d ∙ 0, c ∙ 0

 ividing is equivalent to multiplying D by a reciprocal.

Young_AT_6160_ch00_pp46-81.indd 49

0.5.3 C O N C E P T U A L

Understand why we use the same rules to multiply or divide rational expressions that we use to multiply or divide rational numbers.

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50 

CHAPTER 0  Prerequisites and Review

MULTIPLYING RATIONAL EXPRESSIONS

1.  Factor all numerators and denominators completely. 2.  State any domain restrictions. 3.  Divide the numerators and denominators by any common factors. 4.  Multiply the remaining numerators and denominators, respectively. DIVIDING RATIONAL EXPRESSIONS

1.  Factor all numerators and denominators completely. 2.  State any domain restrictions. 3.  Rewrite division as multiplication by a reciprocal. 4.  State any additional domain restrictions. 5.  Divide the numerators and denominators by any common factors. 6.  Multiply the remaining numerators and denominators, respectively.

EXAMPLE 6  Multiplying Rational Expressions

Multiply and simplify

3x 1 1 x 3 1 3x 2 1 2x ⋅ . 9x 1 3 4x 2 1 4x

Solution:

1 3x 1 1 2 x 1 x 1 1 2 1 x 1 2 2 ⋅ 3 1 3x 1 1 2 4x 1 x 1 1 2

 actor the numerators and F denominators.

5

State any domain restrictions.

x 2 0, x 2 21, x 2 2

 ivide the numerators and D denominators by common factors.

5

Simplify.

5 

1 3

1 3x 1 1 2 x 1 x 1 1 2 1 x 1 2 2 ⋅ 3 1 3x 1 1 2 4x 1 x 1 1 2 x12 12

x 2 0, x 2 21, x 2 2

1 3

▼ ▼ ANSWER

x13  12

Y O U R T U R N   Multiply and simplify 

2x 1 1 x 3 1 2x 2 2 3x ⋅ . 8x 1 4 3x 2 2 3x

x 2 0, x 2 1, x 2 2 21

EXAMPLE 7  Dividing Rational Expressions

Divide and simplify

x2 2 4 3x 3 2 12x 4 . x 5x 3

Solution:

Factor numerators and denominators. State any domain restrictions. Write the quotient as a product. State any additional domain restrictions.

Young_AT_6160_ch00_pp46-81.indd 50

1x 2 22 1x 1 22 3x 1 x 2 2 2 1 x 1 2 2 4 x 5x 3 x20 1x 2 22 1x 1 22 5x 3 5 ⋅ x 3x 1 x 2 2 2 1 x 1 2 2 x 2 22, x 2 2

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51

0.5  Rational Expressions 

1x 2 22 1x 1 22 5x 3 ⋅ x 3x 1 x 2 2 2 1 x 1 2 2

Divide out the common factors.

5

Simplify.

5x 5  3

x 2 22, x 2 0, x 2 2

[ CONCEPT CHECK ] Divide and simplify 1 1 4 2 x2a x 2a 2

▼

▼ 2

ANSWER

3

x 29 2x 2 18x 4 Y O U R T U R N   Divide and simplify . x 7x 4

1 x11

x26a

▼ ANSWER

7x 2  2

0.5.4  Adding and Subtracting Rational Expressions The same rules that apply to adding and subtracting rational numbers also apply to rational expressions.

x 2 23, x 2 0, x 2 3

0.5.4 S K I L L

Add and subtract rational expressions.

PROPERTY

RESTRICTION

DESCRIPTION

a c a6c 6 5 b b b

b20

 dding or subtracting rational A expressions when the denominators are the same

a c ad 6 bc 6 5 b d bd

b 2 0 and d 2 0

 dding or subtracting rational A expressions when the denominators are different

0.5.4 C O N C E P T U A L

Understand the least common denominator method for rational expressions.

EXAMPLE 8  A  dding and Subtracting Rational Expressions: Equal Denominators

[ CONCEPT CHECK ]

Perform the indicated operation and simplify.

Add

a.

x17 3x 1 1 6x 1 7 2x 1 9 1   b.  2 1x 1 222 1x 1 222 2x 2 1 2x 2 1

▼ ANSWER

Solution (a):

Write as a single expression.

5

Combine like terms in the numerator.

5

Factor out the common factor in the numerator.

5

Cancel (divide out) the common factor, x 1 2.

5 

x 1 7 1 3x 1 1 1x 1 222 4x 1 8 1x 1 222

41x 1 22 1x 1 222 4 x12

x 2 22

Solution (b):

State any domain restrictions.

Young_AT_6160_ch00_pp46-81.indd 51

a1b12 1a 1 121b 1 12

x 2 22

State any domain restrictions.

 rite as a single expression. Use parentheses around W the second numerator to ensure that the negative will be distributed throughout all terms.

1 1 1 a11 b11

STUD Y T I P 6x 1 7 2 1 2x 1 9 2 5 2x 2 1 x2

1 2

When subtracting a rational ­expression, distribute the negative of the quantity to be subtracted over all terms in the numerator.

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52 

CHAPTER 0  Prerequisites and Review

Eliminate parentheses. Distribute the negative.

5

6x 1 7 2 2x 2 9 2x 2 1

Combine like terms in the numerator.

5

4x 2 2 2x 2 1

Factor out the common factor in the numerator.

5

2 1 2x 2 1 2 2x 2 1

Divide out (cancel) the common factor, 2x 2 1.

5 2

x2

1 2

EXAMPLE 9  A  dding and Subtracting Rational Expressions: No Common Factors in Denominators

Perform the indicated operation and simplify. a.

32x x 1 2 1   b.  2 2 2x 1 1 x21 x11 x

Solution (a):

The common denominator is the product of the denominators.



13 2 x2 1x 2 12 x 1 2x 1 1 2 32x x 1 5 1 1 2x 1 1 2 1 x 2 1 2 1 x 2 1 2 1 2x 1 1 2 2x 1 1 x21 5

1 3 2 x 2 1 x 2 1 2 1 x 1 2x 1 1 2 1 2x 1 1 2 1 x 2 1 2

Eliminate parentheses in 3x 2 3 2 x 2 1 x 1 2x 2 1 x the numerator.  5 1 2x 1 1 2 1 x 2 1 2 Combine like terms in the numerator.  5

ST U DY TIP When adding or subtracting rational expressions whose denominators have no ­common factors, the least common denominator is the product of the two denominators.

 he common denominator T is the product of the denominators. Eliminate parentheses in the numerator.  rite the numerator in W factored form to ensure no further simplification is possible.

▼ ANSWER

5x 2 1 3x 2 1 1 x 1 3 21 2x 1 1 2 x 2 23, x 2 2 21

a. 

21x 2 122 b.  x 1x2 1 1 2

Solution (b):

x20

Young_AT_6160_ch00_pp46-81.indd 52

x 2 1 5x 2 3 1 2x 1 1 2 1 x 2 1 2

1 x22 ,x21 2

1 1 2 1 x 1 1 2 2 2 1 x2 2 1 2 2 5 x11 x2 x2 1 x 1 1 2 5

5

2 1 2x 2 2 x 2 1 2 x 1 1 2 2x 2 22x 2 1 x 1 1 5 5 x2 1 x 1 1 2 x2 1 x 1 1 2 x2 1 x 1 1 2 2 1 2x 1 1 2 1 x 2 1 2 x2 1 x 1 1 2

x 2 21, x 2 0

▼ Y O U R T U R N   Perform the indicated operation and simplify. a. 

2x 2 1 x 2 1 1   b.  2 2 x x13 2x 1 1 x 11

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53

0.5  Rational Expressions 

When combining two or more fractions through addition or subtraction, recall that the least common multiple, or least common denominator (LCD), is the smallest real number that all of the denominators divide into evenly (that is, the smallest of which all are factors). For example, 2 1 4 1 2 3 6 9 To find the LCD of these three fractions, factor the denominators into prime factors: 353 2 1 4 12 1 3 2 8 7 6 5 3⋅ 2 1 2 5 5 9 5 3 ⋅3 3 6 9 18 18 3 ⋅ 2⋅ 3 5 18 Rational expressions follow this same procedure, only now variables are also considered: 1 1 1 1 1 35 1 x⋅x⋅x 2x 2x x 32x x 2 x 32x 2 1 2 1 2 2 5 1 1x 1 12 1x 1 12 1 2x 1 1 2 1x 1 12 x11 2x 1 1 x 1 12



LCD 5 2x3 LCD 5 1x 1 122  12x 1 12

The following box summarizes the LCD method for adding and subtracting rational expressions whose denominators have common factors. THE LCD METHOD FOR ADDING AND SUBTRACTING RATIONAL EXPRESSIONS

1. Factor each of the denominators completely. 2. The LCD is the product of each of these distinct factors raised to the highest power to which that factor appears in any of the denominators. 3. Write each rational expression using the LCD for each denominator. 4. Add or subtract the resulting numerators. 5. Factor the resulting numerator to check for common factors. EXAMPLE 10  S  ubtracting Rational Expressions: Common Factors in Denominators (LCD)

Perform the indicated operation and write in simplified form. 5x 7x 2 2 2 2 2x 2 6 x 2x26 Solution:

7x 2 2 5x 2 1x 2 32 1x 1 22 21x 2 32

Factor the denominators.

5

Identify the LCD. Write each expression using the LCD as the denominator.

LCD 5 2 1x 2 3 2 1x 1 2 2 5x 1 x 1 2 2 2 1 7x 2 2 2 5 2 1 1 2 1 2 2 x 2 32 1x 1 22 2x23 x12

 ombine into one expression. C Distribute the negative through the entire second numerator.

5

Simplify.

5

▼

5x 2 1 10x 2 14x 1 4 21x 2 32 1x 1 22 5x 2 2 4x 1 4 21x 2 32 1x 1 22

x 2 22, x 2 3

Y O U R T U R N   Perform the indicated operation and write in simplified form.

2x 5x 1 1 2 2 3x 2 6 x 1 2x 2 8

Young_AT_6160_ch00_pp46-81.indd 53

▼ ANSWER

2x 2 2 7x 2 3 3 1 x 2 2 21 x 1 4 2

x 2 24, x 2 2

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54 

CHAPTER 0  Prerequisites and Review

0.5.5  Complex Rational Expressions 0.5.5 S K I L L

Simplify complex rational expressions.

A rational expression that contains another rational expression in either its numerator or denominator is called a complex rational expression. The following are examples of complex rational expressions. 1 25 x 21x

0.5.5 C O N C E P T U A L

Know the two methods for simplifying complex rational expressions.

3 27 x 6 21 2x 2 5

22x 3 41 x21

TWO METHODS FOR SIMPLIFYING COMPLEX RATIONAL EXPRESSIONS

Procedure 1: Write a sum or difference of rational expressions that appear in either the numerator or denominator as a single rational expression. Once the complex rational expression contains a single rational expression in the ­numerator and one in the denominator, then rewrite the division as multiplication by the reciprocal. OR Procedure 2: Find the LCD of all rational expressions contained in both the numerator and denominator. Multiply the numerator and denominator by this LCD and simplify. EXAMPLE 11  Simplifying a Complex Rational Expression

Write the rational expression in simplified form. 2 11 x 1 11 x11 Solution: 2 State the domain restrictions.  1 1 x 2 0 , x 2 21 , and x 2 22 x

11

Procedure 1:

 dd the expressions in both the A numerator and denominator.

Simplify.

Express the quotient as a product.

Young_AT_6160_ch00_pp46-81.indd 54

1 x11 2 x 21x 1 x x x 5 5 1x 1 12 1 1 x11 1 1 x11 x11 x11 21x x 5 x12 x11 21x x11 5 ⋅ x x12

Divide out the common factors.

5

Write in simplified form.

5

21x x11 ⋅ x x12 x11 x

x 2 22, x 2 21, x 2 0

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0.5  Rational Expressions 

Procedure 2:

 ind the LCD of the numerator F and denominator.

2 11 x 1 11 x11

Identify the LCDs.  ultiply both numerator and M denominator by their combined LCD.

 ultiply the numerators and M denominators, respectively, applying the distributive property.

Divide out common factors.

55

Numerator LCD: Denominator LCD: Combined LCD: 2 11 x x1x 1 12 ⋅ 5 1 x1x 1 12 11 x11

x x11 x 1x 1 1 2

2 ⋅ x 1 x 1 1 2 1 1x 1 x 1 1 2 x 5 1 1 ⋅ x1x 1 12 1 ⋅ x1x 1 12 x11 2 ⋅ x 1 x 1 1 2 1 1x 1 x 1 1 2 x 5 1 x1x 1 12 1 ⋅ x1x 1 12 x11 21x 1 12 1 x1x 1 12 x1x 1 12 1 x

Simplify.

5

Apply the distributive property.

5

Combine like terms.

5

Factor the numerator and denominator.

5

Divide out the common factor.

5

Write in simplified form.

5  

2x 1 2 1 x 2 1 x x2 1 x 1 x x 2 1 3x 1 2 x 2 1 2x 1x 1 22 1x 1 12 x1x 1 22 1x 1 22 1x 1 12 x1x 1 22 x11 x

x 2 22, x 2 21, x 2 0

EXAMPLE 12  Simplifying a Complex Rational Expression

Write the rational expression in simplified form. 1 13 2 x 29 x 2 26, 23, 3 x 12 2x 1 6 Solution:   Using Procedure 1 Factor the respective denominators.

Young_AT_6160_ch00_pp46-81.indd 55

1 13 1x 2 32 1x 1 32 x 12 21x 1 32

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CHAPTER 0  Prerequisites and Review

Identify the LCDs. Multiply both the numerator and the denominator by the combined LCD.

 ultiply the numerators and M denominators, respectively, applying the distributive property.

[ CONCEPT CHECK ] Simplify:

▼ ANSWER

1 1 1a 1 1 b1 b1a 1 12 a1b 1 12

a 2 0, b 2 21, 0

Numerator LCD: 1x 2 32 1x 1 32 Denominator LCD: 2 1x 1 32 Combined LCD: 21x 2 32 1x 1 32

1 13 1x 2 32 1x 1 32 21x 1 32 1x 2 32 ⋅ 5 x 21x 1 32 1x 2 32 12 21x 1 32

Simplify.

21x 1 32 1x 2 32 1 3⋅ 2 1 x 1 3 2 1 x 2 3 2 1x 2 32 1x 1 32 5 x ⋅ 21x 1 32 1x 2 32 21x 1 32 1x 2 32 2 21x 1 32

Eliminate the parentheses.

5

2 1 6x 2 2 54 2x 2 2 18 2 x 2 1 3x

Combine like terms.

5

6x 2 2 52 x 2 1 3x 2 18

5

 actor the numerator and F denominator to make sure there are no common factors.

5

2 1 61x 1 32 1x 2 32 21x 1 32 1x 2 32 2 x1x 2 32

2 1 3x 2 2 26 2 1x 1 62 1x 2 32

x 2 26, x 2 23, x 2 3

[ S E C T I O N 0 . 5 ]    S U M M A R Y In this section, rational expressions were defined as quotients of polynomials. The domain of any polynomial is the set of all real numbers. Since rational expressions are ratios of ­polynomials, the domain of rational expressions is the set of all real numbers except those values that make the denominator equal to zero. In this section, rational expressions were simplified (written with no common ­factors), multiplied, divided, added, and subtracted. OPERATION

EXAMPLE

Multiplying rational expressions

2 3x 6x ⋅ 5 2 x11 x21 x 21

Dividing rational expressions

2 3x 4 x 2 61 x11 x21 2 x21    5 ⋅ x20 x 1 1 3x 21x 2 12    5 x 2 0, x 2 61 3x 1 x 1 1 2

Adding/subtracting rational expressions with no common f­ actors

x 2 61

2 3x 1 x 2 61 x11 x21 LCD 5 1 x 1 1 21 x 2 1 2

Young_AT_6160_ch00_pp46-81.indd 56

NOTE

5

5

When dividing rational ­expressions, remember to check for additional domain restrictions once the division is rewritten as multiplication by a reciprocal.

The LCD is the product of the two denominators.

2 1 x 2 1 2 1 3x 1 x 1 1 2 1 x 1 1 21 x 2 1 2

2x 2 2 1 3x 2 1 3x 1 x 1 1 21 x 2 1 2 3x 2 1 5x 2 2 5 1 x 1 1 21 x 2 1 2 5

State domain restrictions.

1 3x 2 1 21 x 1 2 2 1 x 1 1 21 x 2 1 2

x 2 61

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57

0.5  Rational Expressions 

OPERATION

EXAMPLE

Adding/subtracting rational expressions with common factors

3 2 2 x1x 1 12 x1x 1 22 LCD 5 x 1 x 1 1 21 x 1 2 2

5



5



5

NOTE

The LCD is the product of each of these distinct factors raised to the highest power that appears in any of the ­denominators.

x 2 22, 21, 0

31x 1 22 2 21x 1 12 x 1 x 1 1 21 x 1 2 2

3x 1 6 2 2x 2 2 x 1 x 1 1 21 x 1 2 2 x14 x 1 x 1 1 21 x 1 2 2

x 2 22, 21, 0

Complex rational expressions are simplified in one of two ways: 1. Combine the sum or difference of rational expressions in a numerator or denominator as a single rational expression. The result is a rational expression in the numerator and a rational expression in the denominator. Then write the division as multiplication by the reciprocal. 2. Multiply the numerator and denominator by the overall LCD (LCD for all rational expressions that appear). The result is a ­single rational expression. Then simplify, if possible.

[SEC TION 0.5]  E X E R C I S E S • SKILLS In Exercises 1–10, state any real numbers that must be excluded from the domain of each rational expression. 1.  7. 

3 x 2p2 2

p 21

2. 

5 x

3. 

8. 

3t t 29

9. 

2

3 x21 3p 2 1 p2 1 1

6 y21 2t 2 2 10.  2 t 14  4. 

5. 

5x 2 1 x11

6. 

2x 32x

In Exercises 11–30, reduce the rational expression to lowest terms and state any real numbers that must be excluded from the domain. 11. 

15. 

1x 1 32 1x 2 92 21x 1 32 1x 1 92

2 1 3y 1 1 2 1 2y 2 1 2 3 1 2y 2 1 2 1 3y 2

1 3x 1 7 2 1 x 2 4 2 4x 2 16 x17 23.  x17 19. 

27. 

x 2 1 5x 1 6 x 2 2 3x 2 10

12. 

16. 

4y 1 y 2 8 2 1 y 1 7 2 8y 1 y 1 7 2 1 y 1 8 2

7 1 2y 1 1 2 1 3y 2 1 2 5 1 3y 2 1 2 1 2y 2

1 t 2 7 2 1 2t 1 5 2 3t 2 21 2y 1 9 24.  2y 1 9 20. 

28. 

x 2 1 19x 1 60 x 2 1 8x 1 16

13. 

1x 2 32 1x 1 12 21x 1 12

14. 

1 2x 1 1 2 1 x 2 3 2 31x 2 32

1 5y 2 1 2 1 y 1 1 2 25y 2 5

18. 

21. 

x2 2 4 x22

22. 

t3 2 t t21

25. 

x2 1 9 2x 1 9

26. 

x2 1 4 2x 1 4

29. 

6x 2 2 x 2 1 2x 2 1 9x 2 5

30. 

15x 2 2 x 2 2 5x 2 1 13x 2 6

17. 

1 2t 2 1 2 1 t 1 2 2 4t 1 8

In Exercises 31–48, multiply the rational expressions and simplify. State any real numbers that must be excluded from the domain. 31. 34. 37. 40.

x 2 2 3x 1 5 4x 1 5 3x 1 4 5x 1 6 2x ⋅ 32. ⋅ 33. ⋅ x x11 x22 x 2 2 4x 1 5 5x 2 6 41x 2 22 1x 1 52 16x 2x 2 2 x 2 1 x 5x 2 5 x 2 1 x ⋅ 35. 36. ⋅ 2 ⋅ 2 1x 2 52 1x 1 52 8x 3x 10x x 21 x 21 3x 2 2 12 x 2 1 5x 4x 2 2 32x x 2 1 3x t 1 2 t 2 2 6t 1 9 38. ⋅ 2 39. ⋅ 2 ⋅ x x 3t 2 9 t 2 1 4t 1 4 x 1 3x 2 10 x 2 5x 2 24

y 1 3 y 2 2 10y 1 25 t2 1 4 3t 7a2 1 21a a 1 3 ⋅ 2 41. ⋅ 42. ⋅ 3y 1 9 y 1 3y 2 40 t23 t12 7 14 1 a2 2 9 2

Young_AT_6160_ch00_pp46-81.indd 57

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58 

CHAPTER 0  Prerequisites and Review

43.

y2 2 4 3y t 2 1 t 2 6 8t 3x 2 2 15x 2x 2 2 7x 2 15 ⋅ 44. ⋅ 45. ⋅ y23 y12 t2 2 4 2t 2 2x 3 2 50x 3x 2 1 15x

46.

5t 2 1 4t 2 1 3t 6x 2 2 11x 2 35 4x 2 2 49 3x 2 2 2x x 2 2 7x 2 18 ⋅ 47. ⋅ 2 48. ⋅ 2 2 3 2 4t 16t 2 9 8x 2 22x 2 21 9x 2 25 12x 2 8x 2x 2 2 162

In Exercises 49–66, divide the rational expressions and simplify. State any real numbers that must be excluded from the domain. 49. 

3 12 4 2 x x

50. 

5 10 4 3 2 x x

51. 

53. 

1 5 4 2 x21 x 21

54. 

5 10 4 2 3x 2 4 9x 2 16

55. 

2

22p 2

p 21

4

52. 

2p 2 4 p11

56. 

71y 2y 1 10

59. 

3t 3 2 6t 2 2 9t 6 1 6t 4 5t 2 10 4t 2 8

60. 

y 2 2 3y y 3 2 3y 2 4 2y 8y

63. 

x 2 1 4x 2 21 x 2 2 2x 2 63 4 2 2 x 1 3x 2 10 x 1 x 2 20

64. 

66. 

x 2 2 6x 2 27 2x 2 2 15x 2 27 4 2x 2 1 13x 2 7 2x 2 1 9x 2 5

57. 

36 2 n n16 4 n13 n2 2 9

58. 

49 2 y

61. 

w2 2 w w3 2 w 4 w 5w3

62. 

65. 

20x 2 2 3x 2 2 12x 2 1 23x 1 5 4 25x 2 2 4 3x 2 1 5x

2

6 12 4 1x 2 22 1x 1 22 x22

2

y 2 25

4

20 1 x 1 6 2 51x 1 62 4 8 10 1 x 2 6 2 42x 12 2 3x 4 x24 x 2 2 16

x 3 1 8x 2 1 12x 4x 1 8 4 2 5x 2 2 10x x 24 2y 2 2 5y 2 3 2

2y 2 9y 2 5

4

3y 2 9 y 2 2 5y

In Exercises 67–82, add or subtract the rational expression and simplify. State any real numbers that must be excluded from the domain. 67. 

3 2 2 x 5x

68. 

5 3 2 x 7x

69. 

5p 3 1 p22 p11

70. 

4 5x 2 91x x22

71. 

2x 1 1 3 2 2x 2 5x 2 1 1 2 5x

72. 

7 5 2 2x 2 1 1 2 2x

73. 

3y 2 1 2 2y 1 y11 y21

74. 

3 4 1 12x x21

75. 

3x 31x 1 x12 x 24

76. 

x21 x11 2 2 21x 42x

77. 

x21 x26 1 2 x22 x 24

78. 

2 7 1 y23 y12

79. 

5a 7 2 b2a a 2 b2

80. 

1 4 2 2 2 1 2 y y 24 y 2 2y

81.  7 1

1 x23

82. 

y2 2 y 3 4 2 1 2 5y 1 6 y22 5y 2 4y 2 12

2

2

In Exercises 83–90, simplify the complex rational expressions. State any real numbers that must be excluded from the domain. 1 21 x 83.  2 12 x

3 25 y 84.  2 42 y

1 x 85.  1 92 2 x

1 2 1 2 x x 86.  9 5 2 2 x x

1 11 x21 87.  1 12 x11

7 y17 88.  1 1 2 y y17

1 11 x21 89.  1 11 x11

3 3 2 x11 x21 90.  5 x2 2 1

31

• A P P L I C AT I O N S 91. Finance. The amount of payment made on a loan is given by the

pi , where p is the ­principal (amount 1 2 1/ 1 1 1 i 2 n r borrowed), and i 5 , where r is the interest rate expressed as n a decimal and n is the number of payments per year. Suppose n 5 5. Simplify the formula as much as possible. formula A 5



pi

92. Finance. Use the formula A 5

to calculate the 1 1 1 1 i 2 nt amount your monthly payment will be on a loan of $150,000 at an interest rate of 6.5% for 30 years 1nt 5 360 2. 12



Young_AT_6160_ch00_pp46-81.indd 58

93. Circuits. If two resistors are connected in parallel, the

1 , 1/R1 1 1/R2 where R1 and R2 are the individual resistances. Simplify the ­formula. 94. Optics. The focal length of a lens can be calculated by

­combined resistance is given by the formula R 5



applying the formula ƒ 5



1 , where p is the distance 1/p 1 1/q that the object is from the lens and q is the distance that the image is from the lens. Simplify the formula.

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0.5  Rational Expressions 

59

• C AT C H T H E M I S TA K E In Exercises 95 and 96, explain the mistake that is made. 95. Simplify

x 2 1 2x 1 1 . x11

96. Simplify

x11 . x 1 2x 1 1 2

Solution: Cancel the common 1s.

Solution: 1x 1 12 1x 1 12 x 2 1 2x 1 1 5 1x 1 12 x11

Factor the numerator.

Cancel the common factor, x 1 1. Write in simplified form.

5

1x 1 12 1x 1 12 1x 1 12

Factor the denominator. Cancel the common x. Write in simplified form.

5x11

This is incorrect. What mistake was made?

x11 x 2 1 2x 1 1 x 5 x1x 1 22 x 5 1 x x 1 22 1 5    x 2 22, x 2 0 x12

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 97–100, determine whether each of the statements is true or false. 97.

x 2 2 81 5x19 x29

 99. When adding or subtracting rational expressions, the LCD is

x29 1 98. 2 5     x 2 29, 9 x19 x 2 81

always the product of all the denominators. 100.

x2c 5 21 for all values of x. c2x

• CHALLENGE 101. Perform the operation and simplify (remember to state domain

102. Write the numerator as the product of two binomials. Divide

restrictions). x1a x1c 4 x1b x1d





out any common factors of the numerator and denominator. a2n 2 b2n an 2 bn

• TECHNOLOGY 103. Utilizing a graphing technology, plot the expression

x17 y5 . Zoom in near x 5 27. Does this agree with x17 what you found in Exercise 23?

104. Utilizing a graphing technology, plot the expression



x2 2 4 y5 . Zoom in near x 5 2. Does this agree with x22 what you found in Exercise 21?

Young_AT_6160_ch00_pp46-81.indd 59

In Exercises 105 and 106, for each given expression: (a) simplify the expression, (b) use a graphing utility to plot the expression and the answer in (a) in the same viewing window, and (c) determine the domain restriction(s) where the graphs will agree with each other. 1 2 12 x22 x13 105.   106.  1 1 12 11 x12 x14 11

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60 

CHAPTER 0  Prerequisites and Review

0.6 RATIONAL EXPONENTS AND RADICALS SKILLS OBJECTIVES ■■ Evaluate and simplify square roots. ■■ Simplify and combine radical expressions involving roots other than squares. ■■ Simplify and factor expressions involving rational exponents.

CONCEPTUAL OBJECTIVES ■■ Understand that a radical implies one number (the ­principal root), not two (6 the principal root). ■■ Know the properties of radicals. ■■ Understand that radicals are equivalent to rational exponents.

In Section 0.2, we discussed integer exponents and their properties. For example, 42 5 16 and x2 ? x3 5 x5. In this section, we expand our discussion of exponents to include any rational numbers. For example, 161/2 5 ? and 1 x 1/2 2 3/4 5 ?. We will first start with a more familiar notation (roots) and discuss operations on radicals, and then rational exponents will be discussed.

0.6.1  Square Roots 0.6.1 S K I L L

Evaluate and simplify square roots. 0.6.1 C O N C E P T U A L

Understand that a radical implies one number (the principal root), not (6 the principal root).

ST U DY TIP Incorrect:  !25 5 65 Correct:  !25 5 5 The principal square root, ! , is defined as a nonnegative real number.

Principal Square Root

DEFINITION

Let a be any nonnegative real number; then the nonnegative real number b is called the principal square root of a, denoted b 5 !a, if b2 5 a. The symbol ! is called a radical sign, and a is called the radicand. It is important to note that the principal square root b is nonnegative. “The principal square root of 16 is 4” implies 42 5 16. Although it is also true that 124 2 2 5 16, the ­principal square root is defined to be nonnegative. It is also important to note that negative real numbers do not have real square roots. For example, !29 is not a real number because there are no real numbers that when squared yield 29. Since principal square roots are defined to be nonnegative, this means they must be zero or positive. The square root of zero is equal to zero: !0 5 0. All other nonnegative ­principal square roots are positive. EXAMPLE 1  Evaluating Square Roots

Evaluate the square roots, if possible. 4   c.  !236 Å9

a. !169  b. 

Solution:

a. What positive real number squared results in 169?

Check: 132 5 169 4

b. What positive real number squared results in 9 ? 2

Check: A 23 B 5

4 9

c. What positive real number squared results in 236?

!169 5 13 4 2 5 Å9 3 No real number

SQUARE ROOTS OF PERFECT SQUARES

Let a be any real number; then: !a2 5 ∙ a ∙

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61

0.6  Rational Exponents and Radicals 

EXAMPLE 2  Finding Square Roots of Perfect Squares

Evaluate the following: a. "62  b. " 127 2 2  c. "x 2

Solution:

a. "62 5 !36 5  6  b. " 127 2 2 5 !49 5  7  c. "x 2 5 0 x 0

Simplifying Square Roots So far, only square roots of perfect squares have been discussed. Now we consider how to simplify square roots such as !12. We rely on the following properties. PROPERTIES OF SQUARE ROOTS

Let a and b be nonnegative real numbers, then: Property

Description



!a⋅ b 5 !a⋅ !b

The square root of a product is the product of the square roots.

!20 5 !4⋅ !5 5 2 !5

a !a 5 Åb "b

b20

The square root of a quotient is the quotient of the square roots.

Example

40 !40 !4⋅ !10 2!10 5 5 5 Å 49 7 7 !49

EXAMPLE 3  Simplifying Square Roots

[ CONCEPT CHECK ]

Simplify:

TRUE OR FALSE  "a2 5 ∙ a ∙ and 3 3 " a 5a

a. "48x 2  b. "28x 3  c.  !12x ⋅ !6x  d. 

"45x 3

Solution:

"5x

▼

ANSWER True

U

U

a. "48x 2 5 !48⋅ "x 2 5 !16⋅ 3⋅ "x 2 5 !16⋅ !3⋅ "x 2 5 4 0 x 0 !3

4 0x0

U

U

b. "28x 3 5 !28⋅ "x 3 5 !4⋅ 7⋅ "x 2 ⋅ x 5 !4⋅ !7⋅ "x 2 ⋅ !x

2 0x0

5 2 0 x 0 !7 !x 5 2 0 x 0 !7x 5  2x !7x  since x $ 0

U

U

c. ­!12x ⋅ !6x 5 "72x 2 5 "36⋅ 2⋅ x 2 5 !36⋅ !2⋅ "x 2 5 6 0 x 0 !2

6 0x0 STUDY T I P 5 6x !2  since x $ 0 If you have a single odd power

U

U

45x 3 "45x 3 d. 5 5 "9x 2 5 !9⋅ "x 2 5 3 0 x 0 5 3x since x . 0 Å 5x !5x 0x0 3 Note: x 2 0,  x . 0

▼

Y O U R T U R N   Simplify: a. "60x 3  b. 

Young_AT_6160_ch00_pp46-81.indd 61

"125x 5 "25x 3

under the square root, like "x 3, the ­variable is forced to be nonnegative because a ­negative cubed results in a negative inside the radical; therefore, the ­absolute value is not necessary.

▼ ANSWER a. 2x !15x  b. x !5

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CHAPTER 0  Prerequisites and Review

0.6.2  Other (nth) Roots 0.6.2 S K I L L

We now expand our discussion from square roots to other nth roots.

Simplify and combine radical expressions involving roots other than squares. 0.6.2 C O N C E P T U A L

Know the properties of radicals.

DEFINITION

Principal nth Root

Let a be a real number and n be a positive integer. Then the real number b is called n the principal nth root of a, denoted b 5 ! a, if bn 5 a. If n is even, then a and b are nonnegative real numbers. The positive integer n is called the index. The square root corresponds to n 5 2, and the cube root corresponds to n 5 3. n

a

b

EXAMPLE

Even

Positive

Positive

Even

Negative

Not a real number

Odd

Positive

Positive

Odd

Negative

Negative

4 ! 16 5 2 because 24 5 16

4 ! 216 is not a real number

3 ! 27 5 3 because 33 5 27

3 ! 2125 5 25 because 125 2 3 5 2125

A radical sign, " , combined with a radicand is called a radical. PROPERTIES OF RADICALS

Let a and b be real numbers, then PROPERTY

DESCRIPTION

EXAMPLE

"ab 5 "a ⋅ "b

The nth root of a product is the product of the nth roots.

3 3 3 3 ! 16 5 ! 8⋅ ! 2 5 2! 2

n

The nth root of a quotient is the quotient of the nth roots.

4 81 " 81 3 5 4 5 Å 16 2 "16

The nth root of a power is the power of the nth root.

3 2 3 " 8 5 A! 8B 5 1 2 2 2 5 4

n

n

n

n

n

if "a and "b both exist a "a 5 n     b 2 0 Åb "b n

n

n

if "a and "b both exist n

4

n

"am 5 A ! a B

m

2

n

When n is odd, the nth root of a raised to the nth power is a.

n

When n is even, the nth root of a raised to the nth power is the absolute value of a.

"an 5 a  n is odd " an 5 ∙ a ∙  n is even

3 3 " x 5x

4 4 " x 5 ∙ x∙

EXAMPLE 4  Simplifying Radicals

Simplify: 3

4

a. " 224x 5  b.  "32x 5

Solution: 3

3

3

3

3

3

3

u

u

a. " 224x 5 5 " 128 2 1 3 2 x 3x 2 5 ! 28 ⋅ ! 3 ⋅ " x 3 ⋅ " x 2 5 22x" 3x 2 u

u

22 x 4 4 4 4 4 4 4 4 4 5 4 b. "32x 5 "16⋅2⋅x ⋅x 5 !16⋅ !2⋅"x ⋅ !x 5 2 0 x 0 !2x 5 2x !2x  since x $ 0

0x0 2

Young_AT_6160_ch00_pp46-81.indd 62

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0.6  Rational Exponents and Radicals 

63

Combining Like Radicals We have already discussed properties for multiplying and dividing radicals. Now we focus on combining (adding or subtracting) radicals. Radicals with the same index and radicand are called like radicals. Only like radicals can be added or subtracted. EXAMPLE 5  Combining Like Radicals

Combine the radicals if possible. a. 4!3 2 6!3 1 7!3    b. 2!5 2 3 !7 1 6 !3 4 3 c. 3!5 1 !20 2 2!45  d.  ! 10 2 2! 10 1 3 !10 Solution (a):

4 !3 2 6 !3 1 7 !3 5 1 4 2 6 1 7 2 !3 5 5 !3

Use the distributive property. Eliminate the parentheses. Solution (b):

 one of these radicals are alike. N 2 !5 2 3!7 1 6!3 The expression is in simplified form.   Solution (c):

All three radicals are now like radicals. Use the distributive property. Simplify. Solution (d):

 one of these radicals are alike because N they have different indices. The expression is in simplified form.

u

Write the radicands as 3 !5 1 !20 2 2 !45 5 3 !5 1 !4 ⋅ 5 2 2 !9 ⋅ 5 products with a factor of 5. The square root of a product is the product of square roots. 5 3 !5 1 !4⋅ !5 2 2 !9⋅ !5 Simplify the square roots of perfect squares. 5 3 !5 1 2 !5 2 2 1 3 2 !5 6 5 3 !5 1 2 !5 2 6 !5 5 1 3 1 2 2 6 2 !5 5  2 !5

▼

4 3 ! 10 2 2 ! 10 1 3 !10

Y O U R T U R N   Combine the radicals. 3

3

3

a.  4 ! 7 2 6 ! 7 1 9! 7  b. 5!24 2 2 !54

▼ ANSWER 3

a. 7 ! 7  b. 4 !6

Rationalizing Denominators

When radicals appear in a quotient, it is customary to write the quotient with no ­radicals in the denominator. This process is called rationalizing the denominator and involves multiplying by an expression that will eliminate the radical in the denominator. 1 For example, the expression contains a single radical in the denominator. In !3 a case like this, multiply the numerator and denominator by an appropriate radical expression, so that the resulting denominator will be radical free:

U

1 1 !3 2 !3 !3 ⋅ 5 5 3 !3 1 !3 2 !3 ⋅ !3 1

If the denominator contains a sum of the form a 1 !b, multiply both the numerator and the denominator by the conjugate of the denominator, a 2 !b, which uses the difference of two squares to eliminate the radical term. Similarly, if the ­denominator

Young_AT_6160_ch00_pp46-81.indd 63

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64 

CHAPTER 0  Prerequisites and Review

contains a difference of the form a 2 !b, multiply both the numerator and the denominator by the conjugate of the denominator, a 1 !b. For example, to 1 rationalize , take the conjugate of the denominator, which is 3 1 !5: 3 2 !5 f

A3 1 !5B 1 3 1 !5 3 1 !5 3 1 !5 ⋅ 5 2 5 2 5 9 2 5 4 A3 2 !5B A3 1 !5B 3 1 3!5 2 3!5 2 A!5B



like terms

In general, we apply the difference of two squares: 2

2

A!a 1 !bB A!a 2 !bB 5 A!aB 2 A!bB 5 a 2 b

Notice that the product does not contain a radical. Therefore, to simplify the expression 1 A!a 1 !bB

multiply the numerator and denominator by A!a 2 !bB: A!a 2 !bB 1 ⋅ A!a 1 !bB A!a 2 !bB

The denominator now contains no radicals:

[ CONCEPT CHECK ] Simplify:

▼ ANSWER

1 1 1 "a

1 2 "a 12a

A!a 2 !bB 1a 2 b2 EXAMPLE 6  Rationalizing Denominators

Rationalize the denominators and simplify. a.

2 5 !5   b.    c.  3 !10 3 2 !2 !2 2 !7

Solution (a):

 ultiply the numerator and M denominator by !10.

5

Simplify.

5

 ivide out the common 2 in the D numerator and denominator.

5

Solution (b):

 ultiply the numerator and denominator M by the conjugate, 3 1 !2.

Young_AT_6160_ch00_pp46-81.indd 64

5



5

The denominator now contains no radicals.

5

Simplify.

5

2 !10 ⋅ 3 !10 !10

2 !10 2!10 2!10 5 5 3A!10B 2 30 3 1 10 2 !10 15

A3 1 !2B 5 ⋅ A3 2 !2B A3 1 !2B 5A3 1 !2B

A3 2 !2B A3 1 !2B

15 1 5!2 922

15 1 5 !2 7

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0.6  Rational Exponents and Radicals 

65

Solution (c):

 ultiply the numerator and denominator M by the conjugate, !2 1 !7.

5

A !2 1 !7B !5 ⋅ A !2 2 !7B A !2 1 !7B !5A !2 1 !7B

 ultiply the numerators and M denominators, respectively.

5

The denominator now contains no radicals.

5

Simplify.

5 2

▼ Y O U R T U R N   Write the expression

A !2 2 !7B A !2 1 !7B !10 1 !35 227

!10 1 !35 5

7 in simplified form. 1 2 !3

▼ ANSWER

2

7A1 1 !3B 2

SIMPLIFIED FORM OF A RADICAL EXPRESSION

A radical expression is in simplified form if No factor in the radicand is raised to a power greater than or equal to the index. The power of the radicand does not share a common factor with the index. ■■ The denominator does not contain a radical. ■■ The radical does not contain a fraction. ■■ ■■

EXAMPLE 7  Expressing a Radical Expression in Simplified Form

Express the radical expression in simplified form: Solution:

16x 5     x $ 0, y . 0 Å 81y 7 3

3 16x 5 " 16x 5 5 7 3 Ç 81y " 81y 7 3

Rewrite the expression so that the radical does not contain a fraction. Let 16 5 24 and 81 5 34.

5

Factors in both radicands are raised to powers greater than the index (3). Rewrite the expression so that each power in the radicand is less than the index. The denominator contains a radical. In order to eliminate the radical in the denominator, we 3 multiply the numerator and denominator by "9 y2.



3 4 " 2 ⋅ x5

3 4 " 3 ⋅ y7 3

5

2x" 2x 2

3 3y 2" 3y

3

3 2x" 2x 2 "9 y2 5 ⋅ 3 3 3y 2" 3y "9 y2

5

3 2x"18x 2y 2 3 3y 2" 27y 3



5

3 2x" 18x 2y 2 9y 3

The radical expression now satisfies the conditions for simplified form.

5

3 2x" 18x 2y 2 9y 3

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66 

CHAPTER 0  Prerequisites and Review

0.6.3  Rational Exponents 0.6.3 S K I L L

We now use radicals to define rational exponents.

Simplify and factor expressions involving rational exponents.

RATIONAL EXPONENTS: 1 – n

Let a be any real number and n be a positive integer, then

0.6.3 C O N C E P T U A L

n

a1/n 5 ! a 1 where is the rational exponent of a. n n ■■ When n is even and a is negative, then a1/n and !a are not real numbers. ■■ Furthermore, if m is a positive integer with m and n having no common factors, then m n am/n 5 1 a1/n 2 5 1 am 2 1/n 5 " am Note: Any of the four notations can be used.

Understand that radicals are equivalent to rational exponents.

EXAMPLE 8  Simplifying Expressions with Rational Exponents

Simplify: a. 163/ 2  b.  128 2 2/ 3 Solution:

3 3 a. 163/2 5 1 161/2 2 5 A !16B 5 43 5 64

▼ ANSWER

b. 128 2 2/3 5 3 128 2 1/3 4 2 5 122 2 2 5 4

▼

Y O U R T U R N   Simplify 272/3.

9

The properties of exponents that hold for integers also hold for rational numbers: 1 1 a21/n 5 1/n  and a2m/n 5 m/n   a 2 0 a a EXAMPLE 9  S  implifying Expressions with Negative Rational Exponents

Simplify

1 9x 2 21/2 4x 23/2

  x . 0.

Solution:

Negative exponents correspond to positive exponents in the reciprocal.

1 9x 2 21/2 4x

23/2

5 5

Apply the quotient property on x.

5

u

Eliminate the parentheses.



5

Simplify.

5

▼ ▼ ANSWER

18x2

Young_AT_6160_ch00_pp46-81.indd 66

Y O U R T U R N   Simplify

x 3/2 4 ⋅ 1 9x 2 1/2 x 3/2 4 ⋅ 91/2x 1/2 x 3/221/2 4 ⋅ 91/2 3 1 x 4⋅3 x 12

9x 3/2   x . 0. 1 4x 2 21/2

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67

0.6  Rational Exponents and Radicals 

EXAMPLE 10  S  implifying Algebraic Expressions with Rational Exponents

Simplify

1 28x 2 y 2 1/3   x . 0, y . 0. 1 9xy 4 2 1/2

Solution:

128x 2 y 2 1/3 128 2 1/3 1 x 2 2 1/3 y 1/3 128 2 1/3 x 2/3 y 1/3 22 2 5 5 a b x 2/321/2 y 1/322 5 2 x 1/6 y 25/3 1/2 1/2 4 1/2 5 1/2 1/2 2 1 9xy 4 2 1/2 1 2 3 3 9 x y 9 x y

Write in terms of positive exponents.

5 2

2x 1/6 3y 5/3

▼ Y O U R T U R N   Simplify

exponents.

1 16x 3 y 2 1/2 and write your answer with only positive 1 27x 2 y 3 2 1/3

▼ ANSWER

4x 5/6 3y 1/2

[ CONCEPT CHECK ]

EXAMPLE 11  F  actoring Expressions with Rational Exponents

Write

Factor completely x 8/3 2 5x 5/3 2 6x 2/3. Solution:

▼

using rational exponents

ANSWER a22/3

u

u

Factor out the greatest common x 8/3 2 5x 5/3 2 6x 2/3 5 x 2/3 1 x 2 2 5x 2 6 2 2/3 factor x . x2/3 x6/3 x2/3 x3/3

1 3 2 " a

5 x 2/3 1 x 2 6 2 1 x 1 1 2

Factor the trinomial.

▼ Y O U R T U R N   Factor completely x 7/3 2 x 4/3 2 2x 1/3.

▼ ANSWER

x 1/3 1 x 2 2 21 x 1 1 2

[ S E C T I O N 0 . 6 ]    S U M M A R Y n In this section, we defined radicals as “b 5 ! a means a 5 bn” Radicals can be combined only if they are like radicals (same for a and b positive real numbers when n is a positive even inte- radicand and index). Quotients with radicals in the denominator ger, and a and b any real numbers when n is a positive odd integer. are usually rewritten with no radicals in the denominator. Rational exponents were defined in terms of radicals: Properties of Radicals n a1/n 5 ! a. The properties for integer exponents we learned in Section 0.2 also hold true for rational exponents:

PROPERTY n

n

EXAMPLE

n

! ab 5 ! a⋅ ! b n

n

n

! am 5 A! aB n

! an 5 a n

! an 5 ∙ a ∙

Young_AT_6160_ch00_pp46-81.indd 67

b20

m

n

3

3

3

!16 5 !8⋅ !2 5 2!2

4 3 81 ! 81 5 5 4 2 Å 16 !16 4

m m am/n 5 1 a1/n 2 5 A! aB

and

n

am/n 5 1 am 2 1/n 5 ! am

1 , for m and n am/n p­ ositive integers with no common factors, a 2 0. Negative rational exponents: a2m/n 5

2

n is odd

3 3 2 5 A! 8B 5 1 2 2 2 5 4 " 8

3 5 3 3 3 2 3 2 " x 5" x ⋅"x 5 x" x x

E

a !a 5 n Åb !b n

3

n is even

4

4

4

4

"x 6 5 "x 4 ⋅"x 2 5 ∙ x ∙ "x 2

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CHAPTER 0  Prerequisites and Review

[ S E C T I O N 0 . 6]   E X E R C I S E S • SKILLS In Exercises 1–24, evaluate each expression or state that it is not a real number. 3

3

  1.  !100

 2.  !121

 3. 2 !144

 4.  !2169

 5.  ! 2216

 6.  ! 2125

13.  !216

14.  !21

5

15.  1227 2 1/3

16.  1264 2 1/3

17.  82/3

18.  1 264 2 2/3

3

 7.  ! 343

19.  1232 2 1/5

3

 8. 2 ! 227 20.  12243 2 1/3

8

 9.  !1

21.  121 2 1/3

7 21 10.  !

22.  15/2

3

5

11.  ! 0

12.  !0

24.  1 27 2 2/3

23.  93/2

In Exercises 25–40, simplify (if possible) the radical expressions. 25.  !2 2 5!2

26.  3!5 2 7!5

33.  !3!7

34.  !5!2

27.  3!5 2 2!5 1 7!5 3

3

29.  !12⋅ !2

30.  2!5⋅3!40

37.  "4x 2y

38.  "16x 3y

39.  "281x 6y 8

2 Ç5 2 46.  1 1 !3

43. 

35.  8"25x 2 3

In Exercises 41–56, rationalize the denominators. 1 Ç3 3 45.  1 2 !5 41. 

49. 

53. 

3 !2 2 !3

4 1 !5 3 1 2!5

42. 

50.  54. 

31.  ! 12⋅ ! 4

2 3!11

1 1 !2 1 2 !2 4 51.  3!2 1 2!3

6 3!2 1 4

55. 

4 4 32.  ! 8⋅ ! 4

36.  16"36y 4 5 232x 10y 8 40.  "

44. 

47. 

5 !2 1 !5

28.  6!7 1 7!7 2 10!7

!7 1 3 !2 2 !5

5 3!2

3 2 !5 3 1 !5 7 52.  2!3 1 3!2 48. 

56. 

!y

!x 2 !y

In Exercises 57–64, simplify by applying the properties of rational exponents. Express your answers in terms of positive exponents. 23 1 x 1/3 y 1/2 2 1 x 22/3y 23/4 2 22 2/3 1/4 12 1/2 2/3 6 1 2 1 2 58.  y y 59.  21/2 1/4 2 60.  57.  x y 1x y 2 1 x 1/3y 1/4 2 4 61. 

x 1/2 y 1/5 22/3 29/5 x y

12

62. 

1 y 23/4x 22/3 2 1 y 1/4x 7/3 2 24

3

63. 

In Exercises 65–68, factor each expression completely. 65. x7/3 2 x4/3 2 2x1/3

66. 8x1/4 1 4x5/4

1 2x 2/3 2 1 4x 21/3 2 2

67. 7x3/7 2 14x6/7 1 21x10/7

64. 

1 2x 22/3 2 3 1 4x 24/3 2 2

68. 7x21/3 1 70x

• A P P L I C AT I O N S 69. Gravity. If a penny is dropped off a building, the time it

d . If a penny is Ç 16 dropped off a 1280-foot-tall building, how long will it take until it hits the ground? Round to the nearest second. 70. Gravity. If a ball is dropped off a building, the time it takes

takes (seconds) to fall d feet is given by



(seconds) to fall d meters is approximately given by



d . Ç5 If a ball is dropped off a 600-meter-tall building, how long will it take until it hits the ground? Round to the nearest second.

Young_AT_6160_ch00_pp46-81.indd 68

71. Kepler’s Law. The square of the period p (in years) of a

planet’s orbit around the Sun is equal to the cube of the planet’s maximum distance from the Sun, d (in astronomical units, or AU). This relationship can be expressed mathematically as p2 5 d 3. If this formula is solved for d, the resulting equation is d 5 p2/3. If Saturn has an orbital period of 29.46 Earth years, calculate Saturn’s maximum distance from the Sun to the nearest hundredth of an AU. 72. Period of a Pendulum. The period (in seconds) of a pendulum

L 1/2 b . If a 9.8 certain pendulum has a length of 19.6 meters, determine the period P of this pendulum to the nearest tenth of a second. of length L (in meters) is given by P 5 2⋅p ⋅ a

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69

0.6  Rational Exponents and Radicals 

• C AT C H T H E M I S TA K E In Exercises 73 and 74, explain the mistake that is made. 73. Simplify 1 4x

1/2 1/4 2

Solution:

y 2

2 2 4 1 x 1/2 2 1 y 1/4 2 Use properties of exponents. Simplify. 4xy1/2 This is incorrect. What mistake was made?

74. Simplify

2 . 5 2 !11

Solution: Multiply numerator and denominator by 5 2 !11. Multiply numerators

and denominators. Simplify.

A5 2 !11B 2 ⋅ 5 2 !11 A5 2 !11B 2A5 2 !11B 25 2 11

2A5 2 !11B 14

5

5 2 !11 7

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 75–78, determine whether each statement is true or false.

In Exercises 79 and 80, a, m, n, and k are any positive real numbers.

75. !121 5 611

n k 21/k 79. Simplify 1 1 am 2 2 . 80. Simplify 1 a2k 2 .

76. "x 2 5 x, where x is any real number.

In Exercises 81 and 82, evaluate each algebraic expression for the specified values.

77. "a2 1 b2 5 "a 1 "b 78. !24 5 22

81.

"b2 2 4ac for a 5 1, b 5 7, c 5 12 2a

82. "b2 2 4ac for a 5 1, b 5 7, c 5 12

• CHALLENGE 83. Rationalize the denominator and simplify:

1 2. A !a 1 !bB

84. Rationalize the denominator and simplify:

!a 1 b 2 !a . !a 1 b 1 !a

• TECHNOLOGY 85. Use a calculator to approximate !11 to three decimal places. 3

86. Use a calculator to approximate !7 to three decimal places. 87. Given

4 5!2 1 4!3

a.  Rationalize the denominator. b. Use a graphing utility to evaluate the expression and the answer. c. Do they agree?

Young_AT_6160_ch00_pp46-81.indd 69

88. Given

2 4!5 2 3!6

a.  Rationalize the denominator. b. Use a graphing utility to evaluate the expression and the answer. c. Do they agree?

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CHAPTER 0  Prerequisites and Review

0.7

COMPLEX NUMBERS

SKILLS OBJECTIVES ■■ Write radicals with negative radicands as imaginary numbers. ■■ Add and subtract complex numbers. ■■ Multiply complex numbers. ■■ Divide complex numbers. ■■ Raise complex numbers to powers.

CONCEPTUAL OBJECTIVES ■■ Understand that real numbers and imaginary numbers are subsets of complex numbers. ■■ Recognize the real and imaginary parts of a complex number. ■■ Recognize that the square of i is 21. ■■ Understand how to eliminate imaginary numbers in denominators. ■■ Understand why i raised to a positive integer power can be reduced to 1, 21, i, or 2i .

0.7.1  The Imaginary Unit, i 0.7.1 S K I L L

Write radicals with negative ­radicands as imaginary numbers. 0.7.1 C O N C E P T U A L

Understand that real numbers and imaginary numbers are subsets of complex numbers.

In Section 1.3, we will be studying equations whose solutions sometimes involve the square roots of negative numbers. In Section 0.6, when asked to evaluate the square root of a negative number, like !216, we said “it is not a real number” because there is no real number such that x 2 5 216. To include such roots in the number system, mathematicians created a new expanded set of numbers, called the complex numbers. The foundation of this new set of numbers is the imaginary unit i. DEFINITION

The Imaginary Unit i

The imaginary unit is denoted by the letter i and is defined as i 5 !21

2

where i 5 21.

Recall that for positive real numbers a and b we defined the principal square root as b 5 !a  which means  b2 5 a

S TU DY T IP

Similarly, we define the principal square root of a negative number as !2a 5 i!a, 2 since Ai!aB 5 i 2 a 5 2a. If 2a is a negative real number, then the principal square root of 2a is

!2a 5 !21⋅ !a 5 i !a

!2a 5 i!a

where i is the imaginary unit and i2 5 21.

We write i!a instead of !a i to avoid any confusion as to what is included in the radical.

[ CONCEPT CHECK ] 2

Write "2a as an imaginary number.

▼

ANSWER i 0 a 0 ▼ ANSWER

12i

Young_AT_6160_ch00_pp46-81.indd 70

EXAMPLE 1  Using Imaginary Numbers to Simplify Radicals

Simplify using imaginary numbers. a. !29  b.  !28 Solution:

a. !29 5 i!9 5 3i    b.  !28 5 i!8 5 i ⋅ 2!2 5 2i!2

▼

Y O U R T U R N   Simplify !2144.

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0.7  Complex Numbers 

DEFINITION

71

Complex Number

A complex number in standard form is defined as a 1 bi where a and b are real numbers and i is the imaginary unit. We denote a as the real part of the complex number and b as the imaginary part of the complex number. A complex number written as a 1 bi is said to be in standard form. If a 5 0 and b 2 0, then the resulting complex number bi is called a pure imaginary number. If b 5 0, then a 1 bi is a real number. The set of all real numbers and the set of all ­imaginary numbers are both subsets of the set of complex numbers. Complex Numbers a 1bi

Real Numbers a 1b 5 0 2

Imaginary Numbers bi 1a 5 0 2

The following are examples of complex numbers.

DEFINITION

17  2 2 3i  25 1 i  3 2 i !11  29i Equality of Complex Numbers

The complex numbers a 1 bi and c 1 di are equal if and only if a 5 c and b 5 d. In other words, two complex numbers are equal if and only if both real parts are equal and both imaginary parts are equal.

[ CONCEPT CHECK ] Simplify: (a 2 bi ) 2 (a 1 bi )

▼ ANSWER 22bi

0.7.2  Adding and Subtracting Complex Numbers Complex numbers in the standard form a 1 bi are treated in much the same way as binomials of the form a 1 bx. We can add, subtract, and multiply complex numbers the same way we performed these operations on binomials. When adding or subtracting complex numbers, combine real parts with real parts and combine imaginary parts with imaginary parts. EXAMPLE 2  Adding and Subtracting Complex Numbers

0.7.2 S K I L L

Add and subtract complex numbers. 0.7.2 C O N C E P T U A L

Recognize the real and imaginary parts of a complex number.

Perform the indicated operation and simplify. a. 13 2 2i 2 1 121 1 i 2  b. 12 2 i 2 2 13 2 4i 2 Solution (a):

Eliminate the parentheses. 13 2 2i 2 1 121 1 i 2 5 3 2 2i 2 1 1 i Group real and imaginary numbers, respectively. 5 13 2 1 2 1 122i 1 i 2 Simplify. 5  2 2 i Solution (b):

Eliminate the parentheses (distribute the negative).

12 2 i 2 2 13 2 4i 2 5 2 2 i 2 3 1 4i

Group real and imaginary numbers, respectively. Simplify.

5 12 2 3 2 1 12i 1 4i 2 5   21 1 3i

▼ Y O U R T U R N   Perform the indicated operation and simplify: 14 1 i 2 2 13 2 5i 2.

Young_AT_6160_ch00_pp46-81.indd 71

▼ ANSWER

1 1 6i

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72 

CHAPTER 0  Prerequisites and Review

0.7.3  Multiplying Complex Numbers 0.7.3 S K I L L

Multiply complex numbers.

When multiplying complex numbers, you apply all the same methods as you did when multiplying binomials. It is important to remember that i2 5 21.

0.7.3 C O N C E P T U A L

WORDS MATH

Recognize that the square of i is 21.

Multiply the complex numbers. Multiply using the distributive property. Eliminate the parentheses. Let i2 5 21. Simplify. Combine real parts and imaginary parts, respectively.

ST U DY TIP When multiplying complex ­numbers, remember that i 2 5 21.

[ CONCEPT CHECK ] Multiply and simplify (a 1 bi )(a 2 bi )

▼ ANSWER a2 1 b2

15 2 i2 13 2 4i2 5 5 132 1 5 124i2 2 i  132 2 1i2 124i2 5 15 2 20i 2 3i 1 4i2 5 15 2 20i 2 3i 1 4  1212 5 15 2 20i 2 3i 2 4 5 11 2 23i

EXAMPLE 3  Multiplying Complex Numbers

Multiply the complex numbers and express the result in standard form, a 6 bi. a. 13 2 i2 12 1 i2  b.  i  123 1 i2 Solution (a):

Use the distributive property. Eliminate the parentheses. Substitute i2 5 21. Group like terms. Simplify. Solution (b):

Use the distributive property. Substitute i2 5 21.

13 2 i 2 12 1 i   2 5 3122 1 31i  2 2 i 122 2 i  1i  2 5 6 1 3i 2 2i 2 i2 5 6 1 3i 2 2i 2 1212 5 16 1 12 1 13i 2 2i  2 5   7 1 i i  123 1 i 2 5 23i 1 i2

Write in standard form. ▼ ANSWER

2 1 11i

5 23i 2 1

5 21 2 3i

▼ Y O U R T U R N   Multiply the complex numbers and express the result in standard form, a 6 bi: 14 2 3i2 121 1 2i2.

0.7.4  Dividing Complex Numbers 0.7.4 S K I L L

Divide complex numbers. 0.7.4 C O N C E P T U A L

Understand how to ­eliminate imaginary numbers in denominators.

Recall the special product that produces a difference of two squares, 1a 1 b2 1a 2 b2 5 a2 2 b2. This special product has only first and last terms because the outer and inner terms cancel each other out. Similarly, if we multiply complex numbers in the same manner, the result is a real number because the imaginary terms cancel each other out. COMPLEX CONJUGATE

The product of a complex number, z 5 a 1 bi, and its complex conjugate, z 5 a 2 bi, is a real number. zz 5 1 a 1 bi 2 1 a 2 bi 2 5 a2 2 b2 i 2 5 a2 2 b2 121 2 5 a2 1 b2

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73

0.7  Complex Numbers 

To write a quotient of complex numbers in standard form, a 1 bi, multiply the numerator and the denominator by the complex conjugate of the denominator. It is important to note that if i is present in the denominator, then the complex number is not in standard form.

[ CONCEPT CHECK ]

EXAMPLE 4  Dividing Complex Numbers

Write the quotient in standard form: Solution:

Write the quotient in standard 1 form: a 1 bi

22i . 1 1 3i

Multiply numerator and denominator by the complex conjugate of the denominator, 1 2 3i.

a

22i 1 2 3i ba b 1 1 3i 1 2 3i

5

Use the FOIL method (or distributive property).

5

2 2 6i 2 i 1 3i 2 1 2 3i 1 3i 2 9i 2

Combine imaginary parts.

5

2 2 7i 1 3i 2 1 2 9i 2

Substitute i2 5 21.

5

Simplify the numerator and denominator.

5

a1b a b 5 1 . c c c

ANSWER

1 2 2 i 2 1 1 2 3i 2 1 1 1 3i 2 1 1 2 3i 2

Multiply the numerators and denominators, respectively.

Write in standard form. Recall that

▼

a bi 2 2 1a2 1 b2 2 1a 1 b2 2

2 2 7i 2 3 1 2 9 121 2 21 2 7i 10

5 2

1 7 2 i 10 10

▼ Y O U R T U R N   Write the quotient in standard form:

3 1 2i . 42i

▼ ANSWER 10 17

1 11 17 i

0.7.5 Raising Complex Numbers to Integer Powers Note that i raised to the fourth power is 1. In simplifying imaginary numbers, we factor out i raised to the largest multiple of 4.

i 5 !21

2



i 5 21



i 3 5 i 2 ⋅ i 5 121 2 i 5 2i



Young_AT_6160_ch00_pp46-81.indd 73

i 4 5 i 2 ⋅ i 2 5 121 2 121 2 5 1 i5 5 i 4 ⋅ i 5 112 1i2 5 i

0.7.5 S K I L L

Raise complex numbers to powers. 0.7.5 C O N C E P T U A L

Understand why i raised to a positive integer power can be reduced to 1, 21, i, or 2i.

i 6 5 i 4 ⋅ i 2 5 1 1 2 121 2 5 21 i 7 5 i 4 ⋅ i 3 5 1 1 2 12i 2 5 2i i8 5 1i 422 5 1

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74 

CHAPTER 0  Prerequisites and Review

EXAMPLE 5  Raising the Imaginary Unit to Integer Powers

Simplify: a.  i7  b.  i13  c.  i100 Solution: a. i7 5 i4 ⋅ i3 5 112 12i 2 5 2i

b. i13 5 i12 ⋅ i 5 1i 4 2  3 ⋅ i 5 13 ⋅ i 5 i ▼ ANSWER

25 c. i100 5 1i42  5 125 5 1

▼

Y O U R T U R N   Simplify i27.

2i

EXAMPLE 6  Raising a Complex Number to an Integer Power

Write 12 2 i 23 in standard form. Solution:

[ CONCEPT CHECK ] Simplify i 4a where a 5 1, 2, 3, ...

▼ ANSWER 1

▼ ANSWER

2 1 11i

Recall the formula for cubing a binomial. 1a 2 b23 5 a3 2 3a2b 1 3ab2 2 b3 Let a 5 2 and b 5 i. 12 2 i 23 5 23 2 3122 2 1i 2 1 3122 1i 22 2 i3 Let i2 5 21 and i3 5 2i. 5 23 2 3122 2 1i 2 1 132 122 1212 2 12i 2 Eliminate parentheses. 5 8 2 6 2 12i 1 i Combine the real parts and imaginary parts, respectively. 5 2 2 11i

▼ 3 Y O U R T U R N   Write 12 1 i 2  in standard form.

[SEC TIO N 0.7 ]   S U M M A RY The Imaginary Unit i • i 5 !21 • i 2 5 21 Complex Numbers • Standard Form: a 1 bi, where a is the real part and b is the imaginary part. • The set of real numbers and the set of pure imaginary ­numbers are subsets of the set of complex numbers. Adding and Subtracting Complex Numbers

• 1a 1 bi 2 1 1c 1 di 2 5 1a 1 c2 1 1b 1 d  2 i • 1a 1 bi 2 2 1c 1 di 2 5 1a 2 c2 1 1b 2 d  2 i • To add or subtract complex numbers, add or subtract the real parts and imaginary parts, respectively.

Young_AT_6160_ch00_pp46-81.indd 74

Multiplying Complex Numbers • 1a 1 bi 2 1c 1 di 2 5 1ac 2 bd 2 1 1ad 1 bc2 i • Apply the same methods as for multiplying binomials. It is important to remember that i2 5 21. Dividing Complex Numbers • Complex conjugate of a 1 bi is a 2 bi. • In order to write a quotient of complex numbers in standard form, multiply the numerator and the denominator by the complex conjugate of the denominator: a 1 bi 1 c 2 di 2 ⋅ c 1 di 1 c 2 di 2

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75

0.7  Complex Numbers 

[SEC TIO N 0.7]   E X E R C I S E S • SKILLS In Exercises 1–12, write each expression as a complex number in standard form. Some expressions simplify to either a real number or a pure imaginary number. 1. !216

2. !2100

3. !220

9. 3 2 !2100

10. 4 2 !2121

11. 210 2 !2144

3

3

5. ! 264

4. !224

7. !264

6. ! 227

8. !227 3

In Exercises 13–40, perform the indicated operation, simplify, and express in standard form.

12. 7 2 ! 2125

13. 13 2 7i 2 1 121 2 2i 2

14. 11 1 i 2 1 19 2 3i 2

15. 13 2 4i 2 1 17 2 10i 2

16. 15 1 7i 2 1 1210 2 2i 2

25. 23116 2 9i 2

26. 5126i 1 32

27. 26 117 2 5i 2

28. 21218 1 3i 2

17. 14 2 5i 2 2 12 2 3i 2 21. 314 2 2i 2

29. 11 2 i 2 13 1 2i 2

18. 122 1 i 2 2 11 2 i 2 22. 417 2 6i 2

30. 123 1 2i 2 11 2 3i 2

33. 17 2 5i 2 16 1 9i 2

34. 123 2 2i 2 17 2 4i 2

19. 123 1 i 2 2 122 2 i 2 23. 1218 2 5i 2

31. 15 2 7i 2 123 1 4i 2

35. 112 2 18i 2 122 1 i 2

24. 23116 1 4i 2

32. 116 2 5i 2 122 2 i 2

36. 124 1 3i 2 124 2 3i 2

In Exercises 41–48, for each complex number z, write the complex conjugate z and find zz.

40. 123i 2 2 2 122 2 3i 2

41. z 5 4 1 7i

42. z 5 2 1 5i

43. z 5 2 2 3i

44. z 5 5 2 3i

45. z 5 6 1 4i

46. z 5 22 1 7i

47. z 5 22 2 6i

48. z 5 23 2 9i

37.

A 12

1

2iB A 49

38.

2 3iB

A234

9 2 16 iB A 3

1

1

4 9 iB

39. 12i 1 17 2 12 1 3i 2

20. 14 1 7i 2 2 15 1 3i 2

In Exercises 49–64, write each quotient in standard form. 49.

2 i

50.

3 i

51.

1 32i

52.

2 72i

53.

1 3 1 2i

54.

1 4 2 3i

55.

2 7 1 2i

56.

8 1 1 6i

57.

12i 11i

58.

32i 31i

59.

2 1 3i 3 2 5i

60.

21i 32i

61.

4 2 5i 7 1 2i

62.

7 1 4i 9 2 3i

63.

8 1 3i 9 2 2i

64.

10 2 i 12 1 5i

In Exercises 65–76, simplify. 65. i15

66. i99

67. i40

68. i18

71. 12 1 3i 2 2

72. 14 2 9i 2 2

73. 13 1 i 2 3

74. 12 1 i 2 3

69. 15 2 2i 2 2 75. 11 2 i 2 3

70. 13 2 5i 2 2 76. 14 2 3i 2 3

• A P P L I C AT I O N Electrical impedance is the ratio of voltage to current in ac circuits. Let Z represent the total impedance of an electrical circuit. If there are two resistors in a circuit, let Z1 5 3 2 6i ohms and Z2 5 5 1 4 i ohms. 77. Electrical Circuits in Series. When the resistors in the

78. Electrical Circuits in Parallel. When the resistors in the

circuit are placed in series, the total impedance is the sum of the two impedances Z 5 Z1 1 Z2. Find the total impedance of the electrical circuit in series.

­circuit are placed in parallel, the total impedance is given 1 1 1 by 5 1 . Find the total impedance of the electrical Z Z1 Z2 circuit in parallel.



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76 

CHAPTER 0  Prerequisites and Review

• C AT C H T H E M I S TA K E In Exercises 79 and 80, explain the mistake that is made. 79. Write the quotient in standard form:

2 . 42i

Solution: Multiply the numerator and the denominator by 4 2 i. Multiply the numerator using the distributive property and the denominator using the FOIL method. Simplify. Write in standard form.

80. Write the product in standard form: 12 2 3i 2 15 1 4i 2.

Solution:

2 42i ⋅ 42i 42i

8 2 2i 16 2 1 8 2 2i 15 8 2 2 i 15 15

Use the FOIL method to multiply the complex numbers.

10 2 7i 2 12i2

Simplify.

22 2 7i

This is incorrect. What mistake was made?

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 81–84, determine whether each statement is true or false. 81. The product is a real number: 1a 1 bi 2 1a 2 bi 2.

82. Imaginary numbers are a subset of the complex numbers.

83. Real numbers are a subset of the complex numbers. 84. There is no complex number that equals its conjugate.

• CHALLENGE 85. Factor completely over the complex numbers: x4 1 2x2 1 1.

86. Factor completely over the complex numbers: x4 1 18x2 1 81.

• TECHNOLOGY In Exercises 87–90, use a graphing utility to simplify the expression. Write your answer in standard form. 87. 1 1 1 2i 2 5         88.  1 3 2 i 2 6

Young_AT_6160_ch00_pp46-81.indd 76

89. 

1 1          90.  12 2 i23 1 4 1 3i 2 2

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Chapter Review 

77

[CHAPTER 0 REVIEW] SECTION

CONCEPT

0.1

Real numbers

KEY IDEAS/FORMULAS

a Rational: , where a and b are integers or a decimal that terminates or repeats. b Irrational: Nonrepeating/nonterminating decimal.

Approximations: Rounding and truncation

Rounding: Examine the digit to the right of the last desired digit. Digit , 5: Keep last desired digit as is. Digit $ 5: Round the last desired digit up 1. Truncating: Eliminate all digits to the right of the desired digit.

Order of operations

1.  Parentheses 2.  Multiplication and Division 3.  Addition and Subtraction

Properties of real numbers

■■ ■■ ■■ ■■

Integer exponents and scientific notation Integer exponents

Scientific notation

0.3

0.4

an 5 a ⋅ a ⋅ a . . . a n factors

am 5 am2n a0 5 1 an 1 n 1 a2n 5 n 5 a b   a 2 0 a a

am ⋅ an 5 am1n     1 am 2 n 5 amn

c 3 10n where c is a positive real number 1 # c , 10 and n is an integer.

Polynomials: Basic operations Adding and subtracting polynomials

Combine like terms.

Multiplying polynomials

Distributive property

Special products

Factoring polynomials

1x 1 a2 1x 1 b2 5 x2 1 1a 1 b2  x 1 ab Perfect Squares   1a 1 b2  2 5 1a 1 b2 1a 1 b2 5 a2 1 2ab 1 b2   1a 2 b2 2 5 1a 2 b2 1a 2 b2 5 a2 2 2ab 1 b2 Difference of Two Squares   1a 1 b2 1a 2 b2 5 a2 2 b2 Perfect Cubes   1a 1 b2 3 5 a3 1 3a2b 1 3ab2 1 b3   1a 2 b2 3 5 a3 2 3a2b 1 3ab2 2 b3

Greatest common factor

Factor out using distributive property: ax k

 actoring formulas: Special polynomial F forms

Difference of Two Squares   a2 2 b2 5 1a 1 b2 1a 2 b2 Perfect Squares  a2 1 2ab 1 b2 5 1a 1 b2 2   a2 2 2ab 1 b2 5 1a 2 b2 2 Sum of Two Cubes  a3 1 b3 5 1a 1 b2 1a2 2 ab 1 b22 Difference of Two Cubes   a3 2 b3 5 1a 2 b2 1a2 1 ab 1 b22

Factoring a trinomial as a product of two binomials

Young_AT_6160_ch00_pp46-81.indd 77

If xy 5 0, then x 5 0 or y 5 0 a c ad 6 bc 6 5 b 2 0 and d 2 0 b d bd a c a d 4 5 ⋅ b 2 0, c 2 0, and d 2 0 b d b c µ

0.2

a  1b 1 c2 5 ab 1 ac

CHAPTER 0 REVIEW

The set of real numbers

■■ ■■

x2 1 bx 1 c 5 1x 1 ?2 1x 1 ?2

ax2 1 bx 1 c 5 1?x 1 ?2 1?x 1 ?2

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78 

CHAPTER 0  Prerequisites and Review

SECTION

CONCEPT

KEY IDEAS/FORMULAS

Factoring by grouping

Group terms with common factors.

A strategy for factoring polynomials

1.  Factor out any common factors. 2.  Recognize any special products. 3.  Use the FOIL method in reverse for trinomials. 4.  Look for factoring by grouping.

0.5

Rational expressions

CHAPTER 0 REVIEW

Rational expressions and domain restrictions

Note domain restrictions when denominator is equal to zero.

Simplifying rational expressions Multiplying and dividing rational expressions

■■

Use properties of rational numbers.

■■

 tate additional domain restrictions once S division is rewritten as multiplication of a reciprocal.

Adding and subtracting rational expressions

Least common denominator (LCD)

Complex rational expressions

Two strategies: 1.  Write sum/difference in numerator/denominator as a rational expression. 2.  Multiply by the LCD of the numerator and denominator.

0.6

Rational exponents and radicals Square roots Other (nth) roots

!25 5 5 n

b 5 !a means a 5 bn for a and b positive real numbers and n a positive even integer, or for a and b any real numbers and n a positive odd integer. n

n n n a ! ab 5 ! a ⋅ ! b     n a 5 !     b 2 0 n Äb !b n n m ! am 5 A! aB n

! an 5 a    n is odd n

Rational exponents

! an 5  a     n is even n

a1/n 5 ! a n m m am/n 5 1 a1/n 2 5 A!aB a 2m/n 5

0.7

1 for m and n positive integers with no common factors, a 2 0. am/n

Complex numbers The imaginary unit, i

i 5 !21

Adding and subtracting complex numbers

Complex Numbers: a 1 bi where a and b are real numbers. Combine real parts with real parts and imaginary parts with imaginary parts.

Multiplying complex numbers

Use the FOIL method and i2 5 21 to simplify.

Dividing complex numbers

If a 1 bi is in the denominator, then multiply the numerator and the denominator by a 2 bi. The result is a real number in the denominator.

Raising complex numbers to integer powers

i 5 !21  i 2 5 21  i 3 5 2i  i 4 5 1

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Review Exercises 

79

[CHAPTER 0 REVIEW EXERCISES] 0.1  Real Numbers

Factor the trinomial into a product of two binomials.

Approximate to two decimal places by (a) rounding and (b) truncating.

31. 2x2 1 9x 2 5

 1. 5.21597

33. 16x2 2 25

 2. 7.3623

32. 6x2 2 19x 2 7 34. 9x2 2 30x 1 25

Simplify.  3. 7 2 2⋅5 1 4⋅3 2 5

Factor the sum or difference of two cubes.

 4. 22 15 1 32 1 713 2 2 ⋅52

35. x3 1 125

  5. 2

16 1 22 2 1 24 2

36. 1 2 8x3

37. 2x3 1 4x2 2 30x

Perform the indicated operation and simplify. y y y x x 2 1 2   7.   8. 4 3 3 5 6 a2 2a 10. 3 4 2 b b

12 21  9. ⋅ 7 4

38. 6x3 2 5x2 1 x

Factor into a product of two binomials by grouping. 39. x3 1 x2 2 2x 2 2 40. 2x3 2 x2 1 6x 2 3

0.2  Integer Exponents and Scientific Notation

0.5  Rational Expressions

Simplify using properties of exponents.

State the domain restrictions on each of the rational ­expressions.

2 3

12. 124z 2

3 11. 122z2 

1 3x 3 y 2 2 2 13. 4 2 1 x 2y 2

14.

1 2x 2 y 3 2 2 1 4xy 2

3

15. Express 0.00000215 in scientific notation.

41.

0.3  Polynomials: Basic Operations

Perform the indicated operation and write the results in ­standard form. 17. 1 14z2 1 2 2 1 13z 2 42

18. 1 27y 2 2 6y 1 2 2 2 1 y 2 1 3y 2 7 2

19. 1 36x 2 2 4x 2 5 2 2 1 6x 2 9x 2 1 10 2

20. 3 2x 2 1 4x 2 2 7x 2 4 2 3 3x 2 1 2x 2 1 5x 2 4 2 4 21. 5xy2 13x 2 4y2

22. 22st2 12t 1 s 2 2st2 23. 1x 2 72 1x 1 92

24. 12x 1 12 13x 2 22 2 25. 12x 2 32 

26. 15x 2 72 15x 1 72 2 27. 1x2 1 12 

28. 11 2 x 2 

Factor out the common factor. 29. 14x2y2 2 10xy3

1 x 11 2

43.

x2 2 4 x22

44.

x25 x25

45.

t2 1 t 2 6 t2 2 t 2 2

46.

z3 2 z z2 1 z

Perform the indicated operation and simplify. 47.

x 2 1 3x 2 10 x 2 1 x 2 2 ⋅ x 2 1 2x 2 3 x 2 1 x 2 6

48.

x2 2 x 2 2 x11 4 2 3 2 x 1 3x x 1 2x

49.

1 1 2 x11 x13

50.

1 1 1 2 1 x x11 x12

Simplify.

2 2

0.4  Factoring Polynomials

42.

Simplify.

16. Express 7.2 3 10 as a real number. 9

4x 2 2 3 x2 2 9

REVIEW EXERCISES

Factor into a product of three polynomials.

 6. 231x 2 y2 1 4 13x 2 2y2

21 51.

1 x23

1 14 5x 2 15



1 2 1 2 x x 52. 1 32 2 x

30. 30x4 2 20x3 1 10x2

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80 

CHAPTER 0  Prerequisites and Review

0.6  Rational Exponents and Radicals

Simplify. 53. !20 3

5 4

55. "2125x y

57. 3!20 1 5!80 59. A2 1 !5B A1 2 !5B

1 61. 2 2 !3 1 3x 2/3 2 2 63. 1 4x 1/3 2 2

REVIEW EXERCISES

65.

51/2 51/3

0.7  Complex Numbers

54. !80

4 56. " 32x 4 y 5

69. i 19

1 1.4805 3 1021 2 1 5.64 3 1026 2 1 1.68 3 1029 2

58. 4!27x 2 8!12x

Section 0.3

60. A3 1 !xB A4 2 !xB

89. Use a graphing utility to plot the graphs of the three expressions

1 62. 3 2 !x 1 4x 3/4 2 2 64. 1 2x 21/3 2 2

12 66. 1 x 22/3y 1/4 2

Simplify. 67. !2169

88.

68. !232 70. i9

Perform the indicated operation, simplify, and express in standard form. 71. 13 2 2i 2 1 15 2 4i 2

1 2x 1 3 2 3, 8x 3 1 27, and 8x 3 1 36x 2 1 54x 1 27. Which two graphs agree? 90. Use a graphing utility to plot the graphs of the three expressions 1 x 2 3 2 2, 8x 2 1 9, and x 2 2 6x 1 9. Which two graphs agree? Section 0.4 91. Use a graphing utility to plot the graphs of the three expressions

x 2 2 3x 1 18, 1 x 1 6 2 1 x 2 3 2 , and 1 x 2 6 2 1 x 1 3 2 . Which two graphs agree? 92. Use a graphing utility to plot the graphs of the three expressions x 2 2 8x 1 16, 1 x 1 4 2 2, and 1 x 2 4 2 2. Which two graphs agree? Section 0.5 For each given expression: (a) simplify the expression, (b) use a graphing utility to plot the expression and the answer in (a) in the same viewing window, and (c) determine the domain restriction(s) where the graphs will agree with each other.

72. 124 1 7i 2 1 122 2 3i 2 73. 112 2 i 2 2 122 2 5i 2 74. 19 1 8i 2 2 14 2 2i 2 75. 12 1 2i 2 13 2 3i 2

93.

76. 11 1 6i 2 11 1 5i 2 77. 14 1 7i 22

1 2 1 4/x 2 1 2 1 4/x 2 2

94.

Section 0.6

78. 17 2 i 22

Express the quotient in standard form. 1 79. 22i

1 80. 31i

7 1 2i 81. 4 1 5i

6 2 5i 82. 3 2 2i

10 83. 3i

7 84. 2i

Technology

Section 0.1 85. Use your calculator to evaluate "272.25. Does the answer

appear to be a rational or an irrational number? Why?

1053 . Does the answer Å 81 appear to be a rational or an irrational number? Why?

95. Given

1 2 1 3/x 2 1 1 1 9/x 2 2

6 !5 2 !2

a. Rationalize the denominator. b. Use a graphing utility to evaluate the expression and the

answer. c.  Do they agree? 96. Given

11 2!6 1 !13

a. Rationalize the denominator. b. Use a graphing utility to evaluate the expression and the

answer. c.  Do they agree? Section 0.7

86. Use your calculator to evaluate

In Exercises 97 and 98, use a graphing utility to simplify the expression. Write your answer in standard form.

Section 0.2

97. 13 1 5i 2 5 98.

Use a graphing utility to evaluate the expression. Express your answer in scientific notation. 1 8.2 3 1011 2 1 1.167 3 10235 2 87. 1 4.92 3 10218 2

Young_AT_6160_ch00_pp46-81.indd 80

1 1 1 1 3i 2 4

99. Apply a graphing utility to simplify the expression and write

your answer in standard form.

1 1 6 1 2i 2 4



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Practice Test 

81

[CHAPTER 0 PRACTICE TEST] Simplify.

Perform the indicated operations and simplify.

 1. !16

25.

2 3 1 x x21

26.

5x 4 2 2 x 2 2 7x 1 10 x 2 25

27.

x 2 1 x2 1 x 1 1 ⋅ x2 2 1 x3 2 1

28.

4x 2 2 9 x 2 2 16 ⋅ x 2 11x 2 60 2x 1 3

29.

x23 x2 2 9 4 2x 2 5 5 2 2x

10. A5!6 2 2!2B A !6 1 3!2B

30.

12t t 2 2 2t 1 1 4 3t 1 1 7t 1 21t 2

Perform the indicated operation and simplify.

Write the resulting expression in standard form.

11. 1 3y 2 2 5y 1 7 2 2 1 y 2 1 7y 2 13 2

31. 11 2 3i 2 17 2 5i 2

Factor.

33. Rationalize the denominator:

3

 2. " 54x 6

 3. 23 1 2 1 52 2 1 2 1 3 2 7 2 2 1 32 2 1 2 5  4. " 232

 5. "212x 2  6. i17  7.

1 x 2 y 23 z21 2 22 1 x 21 y 2 z3 2 1/2

 8. 3!x 2 4!x 1 5!x  9. 3!18 2 4!32

12. 12x 2 3 2 15x 1 7 2 13. x2 2 16

15. 4x2 1 12xy 1 9y2 16. x4 2 2x2 1 1 17. 2x2 2 x 2 1

2 2 11i 41i

7 2 2!3 . 4 2 5!3

34. Represent 0.0000155 in scientific notation.

1 2 2 x x11 35. Simplify and state any domain restrictions. x21

18. 6y2 2 y 2 1

36. For the given expression:

19. 2t 2 t 2 3t

5 x  25 12 2 x a. Simplify the expression. b. Use a graphing utility to plot the expression and the answer in (a) in the same viewing window. c. Determine the domain restriction(s) where the graphs will agree with each other. 37. Apply a graphing utility to evaluate the expression. Round your answer to three decimal places.

3

2

20. 2x 2 5x 2 3x 3

2

21. x2 2 3yx 1 4yx 2 12y2 22. x4 1 5x2 2 3x2 2 15 23. 81 1 3x3 24. 27x 2 x4

11

PRACTICE TEST

14. 3x2 1 15x 1 18

32.

2

!5 !13 2 !7



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[1[ CHAPTER

Equations and Inequalities

Hero Images/Getty Images, Inc.

Golf courses usually charge both greens fees (cost of playing the course) and cart fees (cost of renting a golf cart). Two friends who enjoy playing golf decide to investigate becoming members at a golf course. The course they enjoy playing the most charges $40 for greens fees and $15 for cart rental (per person), so it currently costs each of them $55 every time they play. The membership offered at that course costs $160 per month with no greens fees, but there is still the per person cart rental fee. How many times a month would they have to play golf in order for the membership option to be the better deal?* This is just one example of how the real world can be modeled with equations and ­inequalities.

*See Section 1.5, Exercises 109 and 110.

LEARNING OBJECTIVES ■■ Solve

linear equations. ■■ Solve application problems involving linear equations. ■■ Solve quadratic equations.

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■■ Solve

rational, polynomial, and radical equations. ■■ Solve linear inequalities.

■■ Solve

polynomial and rational inequalities. ■■ Solve absolute value equations and inequalities.

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[IN THIS CHAPTER] You will solve linear and quadratic equations. You will then solve more complicated equations (polynomial, rational, radical, and absolute value) by first transforming them into linear or quadratic equations. Then you will solve linear, quadratic, polynomial, rational, and absolute value inequalities. Throughout this chapter you will solve applications of equations and inequalities.

EQUAT I O N S AN D I N E QUAL I T I E S 1.1

1.2

1.3

1.4

1.5

1.6

1.7

LINEAR EQUATIONS

APPLICATIONS INVOLVING LINEAR EQUATIONS

QUADRATIC EQUATIONS

OTHER TYPES OF EQUATIONS

LINEAR INEQUALITIES

POLYNOMIAL AND RATIONAL INEQUALITIES

ABSOLUTE VALUE EQUATIONS AND INEQUALITIES

• Factoring • Square Root Method • Completing the Square • Quadratic Formula • Applications Involving Quadratic Equations

• Radical Equations • Equations Quadratic in Form: u-Substitution • Factorable Equations

• Graphing Inequalities and Interval Notation • Solving Linear Inequalities

• Polynomial Inequalities • Rational Inequalities

• Equations Involving Absolute Value • Inequalities Involving Absolute Value

• Solving Linear • Solving Application Equations in Problems One Variable Using • Solving Mathematical Rational Models Equations • Geometry That Are Problems Reducible • Interest to Linear Problems Equations • Mixture Problems • Distance– Rate–Time Problems

83

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84 

CHAPTER 1  Equations and Inequalities

1.1 LINEAR EQUATIONS SKILLS OBJECTIVES ■■ Solve linear equations in one variable. ■■ Solve rational equations that are reducible to linear equations.

CONCEPTUAL OBJECTIVES ■■ Understand the definition of a linear equation in one variable. ■■ Eliminate values that result in a denominator being equal to zero.

1.1.1  Solving Linear Equations in One Variable 1.1.1 S K I L L

Solve linear equations in one variable.

An algebraic expression (see Chapter 0) consists of one or more terms that are combined through basic operations such as addition, subtraction, multiplication, or division; for example: 3x 1 2  5 2 2y  x 1 y

1.1.1 C O N C E P T U A L

Understand the definition of a linear equation in one variable.

An equation is a statement that says two expressions are equal. For example, the following are all equations in one variable, x: x 1 7 5 11  x 2 5 9  7 2 3x 5 2 2 3x  4x 1 7 5 x 1 2 1 3x 1 5 To solve an equation means to find all the values of x that make the equation true. These values are called solutions, or roots, of the equation. The first of these statements shown above, x 1 7 5 11, is true when x 5 4 and false for any other values of x. We say that x 5 4 is the solution to the equation. Sometimes an equation can have more than one solution, as in x 2 5 9. In this case, there are actually two values of x that make this equation true, x 5 23 and x 5 3. We say the solution set of this equation is 523, 36. In the third equation, 7 2 3x 5 2 2 3x, no values of x make the statement true. Therefore, we say this equation has no solution. And the fourth equation, 4x 1 7 5 x 1 2 1 3x 1 5, is true for any values of x. An equation that is true for any value of the variable x is called an identity. In this case, we say the solution set is the set of all real numbers. Two equations that have the same solution set are called equivalent equations. For example, 3x 1 7 5 13  3x 5 6  x 5 2 are all equivalent equations because each of them has the solution set 526. Note that x  2 5 4 is not equivalent to these three equations because it has the solution set 522, 26. When solving equations, it helps to find a simpler equivalent equation in which the variable is isolated (alone). The following table summarizes the procedures for generating equivalent equations.

Generating Equivalent Equations ORIGINAL EQUATION

DESCRIPTION

31x 2 6 2 5 6x 2 x

n Eliminate

7x 1 8 5 29

Add (or subtract) the same quantity to (from) both sides of an equation.

parentheses.

EQUIVALENT EQUATION

3x 2 18 5 5x

n Combine

like terms on one or both sides of an equation. 7x 5 21

7x 1 8 2 8 5 29 2 8

Young_AT_6160_ch01_pp082-128.indd 84

5x 5 15

Multiply (or divide) both sides of an equation by the same nonzero 5x 15 quantity: 5 . 5 5

x53

27 5 x

Interchange the two sides of the equation.

x 5 27

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85

1.1  Linear Equations 

You probably already know how to solve simple linear equations. Solving a linear equation in one variable is done by finding an equivalent equation. In generating an equivalent equation, remember that whatever operation is performed on one side of an equation must also be performed on the other side of the equation. EXAMPLE 1  Solving a Linear Equation

Solve the equation 3x 1 4 5 16. Solution:

Subtract 4 from both sides of the equation.

3x 1 4 5 16 24 24 5 12 3x

Divide both sides by 3.

3x 12 5 3 3

The solution is x 5 4.

x54 The solution set is 546.

▼

▼ ANSWER

Y O U R T U R N   Solve the equation 2x 1 3 5 9.

Example 1 illustrates solving linear equations in one variable. What is a linear equation in one variable?

The solution is x 5 3. The solution set is 536.

Linear Equation

DEFINITION

A linear equation in one variable, x, can be written in the form ax 1 b 5 0 where a and b are real numbers and a 2 0.

What makes this equation linear is that x is raised to the first power. We can also classify a linear equation as a first-degree equation. EQUATION

x2750 2

x 2 6x 2 9 5 0 3

2

x 1 3x 2 8 5 0

DEGREE

GENERAL NAME

First

Linear

Second

Quadratic

Third

Cubic

EXAMPLE 2  Solving a Linear Equation

Solve the equation 5x 2 17x 2 42 2 2 5 5 2 13x 1 22. Solution:

Young_AT_6160_ch01_pp082-128.indd 85

▲

▲

▲

▲

Eliminate the parentheses.  5x 2 17x 2 42 2 2 5 5 2 13x 1 22 Don’t forget to distribute the negative sign through both terms inside the parentheses. 5x 2 7x 1 4 2 2 5 5 2 3x 2 2 Combine x terms on the left, constants 22x 1 2 5 3 2 3x 13x 1 3x on the right. Add 3x to both sides. x1253

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86 

CHAPTER 1  Equations and Inequalities

Subtract 2 from both sides.  

▼ ANSWER

The solution is x 5 2. The solution set is 526.

ST U DY TIP Prime Factors       2 5 2       6 5 2 ? 3       5 5         ? 5

LCD 5 2 ? 3 ? 5 5 30

2222 x51

Check to verify that x 5 1 is a      5 ? 1 2 17 ? 1 2 42 2 2 5 5 2 13 ? 1 1 22 solution to the original equation.    5 2 17 2 42 2 2 5 5 2 13 1 22      5 2 132 2 2 5 5 2 152 0 5 0 Since the solution x 5 1 makes the equation true, the solution set is 516.

▼

Y O U R T U R N   Solve the equation 41x 2 12 2 2 5 x 2 31x 2 22.

To solve a linear equation involving fractions, find the least common denominator (LCD) of all terms and multiply both sides of the equation by the LCD. We will first review how to find the LCD. To add the fractions 12 1 16 1 25, we must first find a common denominator. Some people are taught to find the lowest number that 2, 6, and 5 all divide evenly into. Others prefer a more systematic approach in terms of prime factors.

EXAMPLE 3  Solving a Linear Equation Involving Fractions

Solve the equation 12 p 2 5 5 34 p. Solution:

1 3 Write the equation.   p 2 5 5 p 2 4 1 3 142 p 2 1425 5 142 p Multiply each term in the equation 2 4 by the LCD, 4. The result is a linear equation with no fractions.

2p 2 20 5 3p

Subtract 2p from both sides.

22p     22p 220 5 p

     p 5 220

▼ ANSWER

The solution is m 5 218. The solution set is 52186.

[ CONCEPT CHECK ] Which one of these is a linear equation in one variable? (A) 3x 1 2 5 11 (B) x  2 5 9

Since p 5 220 satisfies the original equation, the solution set is 52206.

▼

1

Solving a Linear Equation in One Variable STEP

DESCRIPTION

1

Simplify the algebraic expressions on both sides of the equation.

23 1 x 2 2 2 1 5 5 7 1 x 2 4 2 2 1 23x 1 6 1 5 5 7x 2 28 2 1 23x 1 11 5 7x 2 29

2

Gather all variable terms on one side of the equation and all constant terms on the other side.

23x 1 11 5 7x 2 29 13x   13x 11 5 10x 2 29 129 129 40 5 10x

3

Isolate the variable.

(C) y 5 x 1 1

▼ ANSWER (A) 3x 1 2 5 11 (B) is nonlinear in one variable (C) is linear in two variables

Young_AT_6160_ch01_pp082-128.indd 86

1

Y O U R T U R N   Solve the equation 4 m 5 12 m 2 3.

EXAMPLE

 10x 5 40 x54

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1.1  Linear Equations 

87

1.1.2 Solving Rational Equations That Are Reducible to Linear Equations A rational equation is an equation that contains one or more rational expressions (Chapter 0). Some rational equations can be transformed into linear equations that you can then solve, but as you will see momentarily, you must be certain that the solution to the linear equation also satisfies the original rational equation.

1.1.2 S K I L L

Solve rational equations that are reducible to linear equations. 1.1.2 C O N C E P T U A L

Eliminate values that result in a denominator being equal to zero.

EXAMPLE 4  S  olving a Rational Equation That Can Be Reduced to a Linear Equation

Solve the equation

2 1 4 4 1 5 1 . x 3x 2 3

Solution:

2 1 4 4 1 5 1 x 3x 2 3

State the excluded values (those that make any denominator equal 0). 6x a

Eliminate fractions by multiplying each term by the LCD, 6x.

x20

2 1 4 4 b 1 6x a b 5 6x a b 1 6x a b x 3x 2 3

Simplify both sides. 4 1 3x 5 24 1 8x Subtract 4. 24      24 3x 5 20 1 8x Subtract 8x. 28x           28x

STUD Y T I P Since dividing by 0 is not defined, we exclude values of the variable that correspond to a denominator equaling 0.

25x 5 20 Divide by 25.

x 5 24

Since x 5 24 satisfies the original equation, the solution set is 5246.

▼

Y O U R T U R N   Solve the equation

3 7 125 . y 2y

Extraneous solutions are solutions that satisfy a transformed equation but do not satisfy the original equation. It is important to first state any values of the variable that must be eliminated based on the original rational equation. Once the rational equation is transformed to a linear equation and solved, remove any excluded values of the variable.

▼ ANSWER

The solution is y 5 14 . The solution set is U 14 V .

EXAMPLE 5  S  olving Rational Equations That Can Be Reduced to Linear Equations

Solve the equation

3x 3 125 . x21 x21

Solution:

State the excluded values (those that make any denominator equal 0).

3x 3 125   x21 x21 x21 Eliminate the fractions by 3x 3 ⋅ 1x 2 12 1 2 ⋅ 1x 2 12 5 ⋅ 1x 2 12 multiplying each term by x21 x21 the LCD, x 2 1. 3x 3 ⋅ 1x 2 12 1 2 ⋅ 1x 2 12 5 ⋅ 1x 2 12 Simplify. x21 x21     3x 1 21x 2 12 5 3

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CHAPTER 1  Equations and Inequalities

Distribute the 2. 3x 1 2x 2 2 5 3 Combine x terms on the left. 5x 2 2 5 3 Add 2 to both sides. 5x 5 5 Divide both sides by 5. x51 It may seem that x 5 1 is the solution. However, the original equation had the restriction x 2 1. Therefore, x 5 1 is an extraneous solution and must be eliminated as a possible solution. 3x 3 125 has no solution  . Thus, the equation x21 x21

ST U DY TIP  hen a variable is in the denomW inator of a fraction, the LCD will contain the variable. This sometimes results in an extraneous solution.

▼ ANSWER

▼ Y O U R T U R N   Solve the equation

no solution

2x 4 235 . x22 x22

We have reviewed finding the least common denominator (LCD) for real numbers. Now let us consider finding the LCD for rational equations that have different denominators. We multiply the denominators in order to get a common denominator. Rational expression:

1 2 1   LCD: x   1x 2 12 x x21

In order to find a least common denominator, it is useful to first factor the denominators to identify common multiples.

1 1 1 1 5 2 3x 2 3 2x 2 2 x 2x 1 1 1 1 5 Factor the denominators:   x1x 2 12 31x 2 12 21x 2 12 Rational equation:



LCD: 6x   1x 2 12

EXAMPLE 6  Solving Rational Equations

Solve the equation

1 1 1 2 5 2 . 3x 1 18 2x 1 12 x 1 6x

Solution:

Factor the denominators.

1 1 1 5 2 x 1x 1 62 21x 1 62 31x 1 62

State the excluded values. Multiply the equation by the LCD, 6x   1x 1 62.

x 2 0, 26

6x 1 x 1 6 2 ⋅

1 1 1 5 6x 1 x 1 6 2 ⋅ 2 6x 1 x 1 6 2 ⋅ 1 1 2 1 2 x x 1 62 2x16 3x16

6x 1 x 1 6 2 ⋅

1 1 1 5 6x 1 x 1 6 2 ⋅ 2 6x 1 x 1 6 2 ⋅ x 1x 1 62 21x 1 62 31x 1 62

Divide out the common factors.

Simplify. 2x 2 3x 5 6 Solve the linear equation. x 5 26 Since one of the excluded values is x 2 26, we say that x 5 26 is an extraneous solution. Therefore, this rational equation has  no solution  . ▼ ANSWER

no solution

Young_AT_6160_ch01_pp082-128.indd 88

▼ Y O U R T U R N   Solve the equation

2 1 1 52 1 . x x11 x 1x 1 12

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1.1  Linear Equations 

89

EXAMPLE 7  Solving Rational Equations

Solve the equation Solution:

23 2 5 . 12 2 x2 1x 2 32

What values make either denominator equal to zero? The values x 5 2 and x 5 3 must be excluded 23 2 5 from possible solutions to the equation. 12 2 x2 1x 2 32 Multiply the equation by the LCD, 1x 2 32 12 2 x2.

x 2 2, x 2 3

23 2 1x 2 32 12 2 x2 5 1x 2 32 12 2 x2 12 2 x2 1x 2 32

2 1 2 2 x 2 5 23 1 x 2 3 2 Divide out the common factors. Eliminate the parentheses. 4 2 2x 5 23x 1 9 Collect x terms on the left, x55 constants on the right. Since x 5 5 satisfies the original equation, the solution set is 556.

▼ ANSWER

▼

24 3 Y O U R T U R N   Solve the equation 5 . x18 x26

The solution is x 5 0. The solution set is 506.

EXAMPLE 8  Automotive Service

A car dealership charges for parts and an hourly rate for labor. If parts cost $273, labor is $53 per hour, and the total bill is $458.50, how many hours did the dealership spend working on your car? Solution:

Let x equal the number of hours the dealership worked on your car. total cost

e

e

parts

e

labor

53x 1 273 5 458.50 53x 5 185.50 x 5 3.5

Write the cost equation. Subtract 273 from both sides of the equation. Divide both sides of the equation by 53.

The dealership charged for 3.5 hours of labor. EXAMPLE 9  Grades

Dante currently has the following three test scores: 82, 79, and 90. If the score on the final exam is worth two test scores and his goal is to earn an 85 for his class average, what score on the final exam does Dante need to achieve his course goal? Solution:

Let x equal final exam grade.

final is worth two test scores

Simplify the numerator. Multiply the equation by 5 (or cross multiply). Solve the linear equation.

Which values of x must be eliminated as potential solutions of the rational equation 1 1 5 ? x2a x1b

▼ ANSWER x 5 a and x 5 2b

5

82 1 79 1 90 1      2x 5 85 5 5

Write the equation that determines the course grade.

5

e

scores 1, 2, and 3

[ CONCEPT CHECK ]

total of five test scores

average

251 1 2x 5 85 5 251 1 2x 5 425 x 5 87

Dante needs to score at least an 87 on the final exam.

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90 

CHAPTER 1  Equations and Inequalities

[ S E C T I O N 1 .1]   S U M M A R Y Linear equations, ax 1 b 5 0, are solved by:

Rational equations are solved by:

1. Simplifying the algebraic expressions on both sides of the equation. 2. Gathering all variable terms on one side of the equation and all constant terms on the other side. 3. Isolating the variable.

1. Determining any excluded values (denominator equals 0). 2. Multiplying the equation by the LCD. 3. Solving the resulting equation. 4. Eliminating any extraneous solutions.

[ S E C T I O N 1 .1]   E X E R C I S E S • SKILLS In Exercises 1–36, solve for the indicated variable. 1. 5x 5 35

2. 4t 5 32

3. 23 1 n 5 12

4. 4 5 25 1 y

5. 24 5 23x

6. 250 5 25t

1 7. 5

9. 3x 2 5 5 7

10. 4p 1 5 5 9

11. 9m 2 7 5 11

8. 6 5 3 p 12. 2x 1 4 5 5

13. 5t 1 11 5 18

14. 7x 1 4 5 21 1 24x

15. 3x 2 5 5 25 1 6x

16. 5x 1 10 5 25 1 2x

17. 20n 2 30 5 20 2 5n

18. 14c 1 15 5 43 1 7c

19. 4 1x 2 32 5 2 1x 1 62

20. 512y 2 12 5 214y 2 32

n53

1

22. 2 13n 1 42 5 21n 1 22

21. 2314t 2 52 5 516 2 2t2 23. 2 1x 2 12 1 3 5 x 2 31x 1 12

24. 4 1 y 1 62 2 8 5 2y 2 4 1 y 1 22

25. 5p 1 61 p 1 72 5 31 p 1 22

26. 31z 1 52 2 5 5 4z 1 71z 2 22

27. 7x 2 1 2x 1 3 2 5 x 2 2

28. 3x 2 1 4x 1 2 2 5 x 2 5

29. 2 2 1 4x 1 1 2 5 3 2 1 2x 2 1 2

31. 2a 2 91a 1 62 5 61a 1 32 2 4a 33. 32 2 34 1 6x 2 51x 1 424 5 4 13x 1 42 2 3613x 2 42 1 7 2 4x4

30. 5 2 1 2x 2 3 2 5 7 2 1 3x 1 5 2

32. 25 2 32 1 5y 2 31 y 1 224 5 2312y 2 52 2 351 y 2 12 2 3y 1 34

34. 12 2 33 1 4m 2 613m 2 224 5 2712m 2 82 2331m 2 22 1 3m 2 54 35. 20 2 4 3c 2 3 2 612c 1 324 5 513c 2 22 2 3217c 2 82 2 4c 1 74

36. 46 2 37 2 8y 1 916y 2 224 5 2714y 2 72 2 2 3612y 2 32 2 4 1 6y4

Exercises 37–48 involve fractions. Clear the fractions by first multiplying by the least common denominator, and then solve the resulting linear equation. a a 5 19 11 22

37.

1 1 m5 m11 5 60

38.

1 1 z5 z13 12 24

39.

x 2x 5 14 7 63

40.

41.

1 1 p532 p 3 24

42.

3x x 5 2x5 2 5 10 2

43.

5y 2y 5 2 2y 5 1 3 84 7

44. 2m 2

46.

c 5 c 2 2c 5 2 4 4 2

45. p 1 47.

p 5 5 4 2

x23 x24 x26 2 512 3 2 6

48. 1 2

5m 3m 4 5 1 8 72 3

x25 x12 6x 2 1 5 2 3 5 15

In Exercises 49–70, specify any values that must be excluded from the solution set and then solve the equation. 49.

4 5 255 y 2y

2 4 245 a 3a 2p 2 57. 531 p21 p21

53.

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50.

4 2 1 10 5 x 3x

4 5 225 x 2x 4t 8 58. 532 t12 t12

54.

51. 7 2

1 10 5 6x 3x

x 2 155 x22 x22 3x 2 59. 245 x12 x12

55.

52.

7 5 521 6t 3t

56.

n n 125 n25 n25

60.

5y 12 235 2y 2 1 2y 2 1

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1.1  Linear Equations 

61.

65.

69.

1 1 21 1 5 n n11 n1n 1 12

62.

n25 1 n23 5 2 6n 2 6 9 4n 2 4

66.

t21 3 5 12t 2

70.

1 1 1 1 5 x x21 x 1x 2 12

5 3 6 1 5 m m 2 2 m1m 2 22

63.

67.

3 2 9 2 5 a a13 a 1a 1 32

2 1 5 5x 1 1 2x 2 1

64.

68.

1 1 2 1 5 c c22 c1c 2 22 3 2 5 4n 2 1 2n 2 5

22x 3 5 x22 4

• A P P L I C AT I O N S 71.  Temperature.

To calculate temperature in degrees Fahrenheit, we use the formula F 5 95 C 1 32, where F is degrees Fahrenheit and C is degrees Celsius. Find the formula to convert from Fahrenheit to Celsius. 72.  Geometry. The perimeter P of a rectangle is related to the length L and width W of the rectangle through the equation P 5 2L 1 2W. Determine the width in terms of the perimeter and length. 73.  Costs: Cellular Phone. Your cell phone plan charges $15 a month plus 12 cents per minute. If your monthly bill is $25.08, how many minutes did you use? 74.  Costs: Rental Car. Becky rented a car on her Ft. Lauderdale vacation. The car was $25 a day plus 10 cents per mile. She kept the car for 5 days and her bill was $185. How many miles did she drive the car? 75.  Costs: Internet. When traveling in London, Charlotte decided to check her e-mail at an Internet café. There was a flat charge of $2 plus a charge of 10 cents a minute. How many minutes was she logged on if her bill was $3.70? 76.  Sales: Income. For a summer job, Dwayne decides to sell magazine subscriptions. He will be paid $20 a day plus $1 for each subscription he sells. If he works for 25 days and makes $645, how many subscriptions did he sell? 77. Business. The operating costs for a local business per year are a fixed amount of $15,000 and $2,500 per day. a. Find C that represents operating costs for the company, which depend on the number of days open, x. b.  If the business accrues $5,515,000 in annual operating costs, how many days did the business operate during the year? 78. Business. Negotiated contracts for a technical support provider produce monthly revenue of $5000 and $0.75 per minute per phone call. a. Find R that represents the revenue for the technical support provider, which depends on the number of minutes of phone calls x. b. In one month the provider received $98,750 in revenue. How many minutes of technical support were provided? In Exercises 79 and 80 refer to the following: Medications are often packaged in liquid form (known as a suspension) so that a precise dose of a drug is delivered within a volume of inert liquid—for example, 250 milligrams amoxicillin in 5 milliliters of a liquid suspension. If a patient is prescribed a

Young_AT_6160_ch01_pp082-128.indd 91

dose of a drug, medical personnel must compute the volume of the liquid with a known concentration to administer. The formula a5

d c

defines the relationship between the dose of the drug prescribed d, the concentration of the liquid suspension c, and the amount of the liquid administered a. 79. Medicine. A physician has ordered a 600-milligram dose of

amoxicillin. The pharmacy has a suspension of amoxicillin with a concentration of 125 milligrams per 5 milliliters. How much liquid suspension must be administered to the patient? 80. Medicine. A physician has ordered a 600-milligram dose

of carbamazepine. The pharmacy has a suspension of carbamazepine with a concentration of 100 milligrams per 5 milliliters. How much liquid suspension must be administered to the patient? 81.  Speed of Light. The frequency ƒ of an optical signal in hertz

(Hz) is related to the wavelength l in meters (m) of a laser c through the equation ƒ 5 , where c is the speed of light in l a vacuum and is typically taken to be c 5 3.0 3 108 meters per second (m/s). What values must be eliminated from the wavelengths?

82.  Optics. For an object placed near a lens, an image forms on

the other side of the lens at a distinct position determined by the distance from the lens to the object. The position of the image is found using the thin lens equation: 1 1 1 5 1 ƒ do di

where do is the distance from the object to the lens, di is the distance from the lens to the image, and ƒ is the focal length of the lens. Solve for the object distance do in terms of the focal length and image distance. Object 2f

do f

f f

2f

di Image

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CHAPTER 1  Equations and Inequalities

• C AT C H T H E M I S TA K E In Exercises 83–86, explain the mistake that is made. 83. Solve the equation 4x 1 3 5 6x 2 7.

Solution: Subtract 4x and add 7 to the equation. Divide by 3. This is incorrect. What mistake was made?

85. Solve the equation

4 2 235 . p 5p

Solution: 3 5 6x x52

84. Solve the equation 3 1x 1 12 1 2 5 x 2 3 1x 2 12.

Solution:

3x 1 3 1 2 5 x 2 3x 2 3 3x 1 5 5 22x 2 3 5x 5 28 8   x52 5 This is incorrect. What mistake was made?

1 p 2 322 5 415p2 Cross multiply.  2p 2 6 5 20p  26 5 18p 6 p52 18 1 p52 3 This is incorrect. What mistake was made? 1 1 1 86. Solve the equation 1 5 . x x21 x 1x 2 12 Solution: Multiply by the LCD, x   1x 2 12.

x 1x 2 12 x 1x 2 12 x 1x 2 12 1 5 x x21 x 1x 2 12

Simplify.   1x 2 12 1 x 5 1    x 2 1 1 x 5 1    2x 5 2   x 5 1

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 87–90, determine whether each of the statements is true or false. 1 87. The solution to the equation x 5 is the set of all real 1/x numbers.

90. x 5 1 is a solution to the equation

91. Solve for x, given that a, b, and c are real numbers and a 2 0:

ax 1 b 5 c

88. The solution to the equation



1 1 5 2 1x 2 12 1x 1 22 x 1x22

92. Solve for x, given that a, b, and c are real numbers and c 2 0:

a b 2 5c x x

is the set of all real numbers.

89. x 5 21 is a solution to the equation

x2 2 1 5 x 1 1. x21

x2 2 1 5 x 1 1. x21

• CHALLENGE b1c b2c 5 . Are there any x2a x1a restrictions given that a 2 0, x 2 0? 1 1 2 94. Solve the equation for y: 5 . 1 y2a y1a y21 Does y have any restrictions? 1 2 1/x 95. Solve for x: 5 1. 1 1 1/x 93. Solve the equation for x:

t 1 1/t 5 1. 1/t 2 1 97. Solve the equation for x in terms of y: a y5 1 1 b/x 1 c 96. Solve for t:

98.  Find the number a for which y 5 2 is a solution of the

equation y 2 a 5 y 1 5 2 3ay.

• TECHNOLOGY In Exercises 99–106, graph the function represented by each side of the equation in the same viewing rectangle and solve for x.   99. 31x 1 22 2 5x 5 3x 2 4 100. 251x 2 12 2 7 5 10 2 9x 101. 2x 1 6 5 4x 2 2x 1 8 2 2 102. 10 2 20x 5 10x 2 30x 1 20 2 10

Young_AT_6160_ch01_pp082-128.indd 92

x 1x 2 12 51 x2 2x 1 x 1 3 2 104. 52 x2 105. 0.035x 1 0.029 18706 2 x2 5 285.03 1 0.45 1 106. 2 5 x 0.75x 9 103.

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1.2  Applications Involving Linear Equations 

1.2 APPLICATIONS INVOLVING LINEAR EQUATIONS SKILLS OBJECTIVES ■■ Solve application problems involving common formulas. ■■ Solve geometry problems using linear equations. ■■ Solve simple interest problems. ■■ Solve mixture problems. ■■ Solve distance–rate–time problems.

CONCEPTUAL OBJECTIVES ■■ Understand the mathematical modeling process. ■■ Estimate a practical solution (guess) prior to solving a problem and then check solution at the end. ■■ Use intuition to confirm answers in multiple investment problems. ■■ Use intuition to confirm answers to mixture problems. ■■ Estimate distance–rate–time problem solutions prior to solving and then confirm with a check.

1.2.1 Solving Application Problems Using Mathematical Models In this section, we will use algebra to solve problems that occur in our day-to-day lives. You typically will read the problem in words, develop a mathematical model (equation) for the problem, solve the equation, and write the answer in words.

1.2.1 S K I L L

Solve application problems involving common formulas. 1.2.1 C O N C E P T U A L

Real-World Problem Translate Words to Math

Understand the mathematical modeling process.

Mathematical Model Solve Using Standard Methods Solve Mathematical Problem Translate from Math to Words Solution to Real-World Problem

You will have to come up with a unique formula to solve each kind of word problem, but there is a universal procedure for approaching all word problems. PROCEDURE FOR SOLVING WORD PROBLEMS

Step 1: Identify the question. Read the problem one time and note what you are asked to find. Step 2:  Make notes. Read until you can note something (an amount, a picture, anything). Continue reading and making notes until you have read the problem a second* time. Step 3:  Assign a variable to whatever is being asked for (if there are two choices, then let it be the smaller of the two). Step 4:  Set up an equation. Step 5: Solve the equation. Step 6:  Check the solution. Run the solution past the “common sense department” using estimation. *Step 2 often requires multiple readings of the problem.

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CHAPTER 1  Equations and Inequalities

EXAMPLE 1  How Long Was the Trip?

During a camping trip in North Bay, Ontario, a couple went one-third of the way by boat, 10 miles by foot, and one-sixth of the way by horse. How long was the trip? Solution: STEP 1  Identify



the question. How many miles was the trip?

STEP 2  Make

notes.

  Read . . . one-third of the way by boat . . . 10 miles by foot . . . one-sixth of the way by horse

Write BOAT: 13 of the trip FOOT: 10 miles HORSE: 16 of the trip

STEP 3  Assign



a variable. Distance of total trip in miles 5 x

STEP 4  Set up an equation. The total distance of the trip is the sum of all the distances by boat, foot, and horse.

Distance by boat 1 Distance by foot 1 Distance by horse 5 Total distance of trip

STEP 5  Solve

[ CONCEPT CHECK ] TRUE OR FALSE  In Example 1 we could have stopped at x 5 20.

▼

Multiply by the LCD, 6. Collect x terms on the right. Divide by 3. The trip was 20 miles.

The distance from their car to the gate is 1.5 miles.

Young_AT_6160_ch01_pp082-128.indd 94

5

5

1 1 x 1 10 1 x 5 x 3 6

1 1 x 1 10 1 x 5 x 3 6 2x 1 60 1 x 5 6x 60 5 3x 20 5 x x 5 20

STEP 6  Check



ANSWER False. A problem asked in words “How long was the trip?” needs an answer in words such as “The trip was 20 miles.”

▼ ANSWER

the equation.

boat foot horse total

5

Distance by boat 5 13x Distance by foot 5 10 miles Distance by horse 5 16 x

5



the solution. Estimate: The boating distance, 13 of 20 miles, is approximately 7 miles; the riding distance on horse, 16 of 20 miles, is approximately 3 miles. Adding these two distances to the 10 miles by foot gives a trip distance of 20 miles.

▼ Y O U R T U R N   A family arrives at the Walt Disney World parking lot. To get

from their car in the parking lot to the gate at the Magic Kingdom they walk 14 mile, take a tram for 13 of their total distance, and take a monorail for 12 of their total distance. How far is it from their car to the gate of Magic Kingdom?

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1.2  Applications Involving Linear Equations 

EXAMPLE 2  Find the Numbers

Find three consecutive even integers so that the sum of the three numbers is 2 more than twice the third. Solution: STEP 1  Identify



the question. What are the three consecutive even integers?

STEP 2  Make notes. Examples of three consecutive even integers are 14, 16, 18 or 28, 26, 24 or 2, 4, 6. STEP 3  Assign

a variable. Let n represent the first even integer. The next consecutive even integer is n 1 2 and the next consecutive even integer after that is n 1 4. n 5 1st integer n 1 2 5 2nd consecutive even integer n 1 4 5 3rd consecutive even integer STEP 4  Set

up an equation.

e

c

c

5

Read Write . . . sum of the three numbers n 1 1n 1 22 1 1n 1 42 . . . is 5 . . . two more than 12 . . . twice the third 21n 1 42 n 1 1 n 1 2 2 1 1 n 1 4 2    5   2  1   2 1 n 1 4 2 sum of the three numbers

is 2 more twice the than third

STEP 5  Solve

the equation. Eliminate the parentheses. Simplify both sides. Collect n terms on the left and constants on the right.

n 1 1n 1 22 1 1n 1 42 5 2 1 21n 1 42 n 1 n 1 2 1 n 1 4 5 2 1 2n 1 8 3n 1 6 5 2n 1 10 n54

The three consecutive even integers are 4, 6, and 8. STEP 6  Check the solution. Substitute the solution into the problem to see whether it makes sense. The sum of the three integers 14 1 6 1 82 is 18. Twice the third is 16. Since 2 more than twice the third is 18, the solution checks.

▼ Y O U R T U R N   Find three consecutive odd integers so that the sum of the three

integers is 5 less than 4 times the first.

1.2.2  Geometry Problems Some problems require geometric formulas in order to be solved. The following geometric formulas may be useful.

Young_AT_6160_ch01_pp082-128.indd 95

▼ ANSWER

The three consecutive odd integers are 11, 13, and 15.

1.2.2 S K I L L

Solve geometry problems using linear equations. 1.2.2 C O N C E P T U A L

Estimate a practical solution (guess) prior to solving a problem and then check solution at the end.

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CHAPTER 1  Equations and Inequalities GEOMETRIC FORMULAS

Rectangle w

Perimeter

Area

P 5 2l 1 2w

A 5 l ⋅w

Circumference

Area

C 5 2pr

A 5 pr 2

Perimeter

Area

l

Circle r

Triangle a

h height

c

P5a1b1c

A5

1 bh 2

           

b

base

EXAMPLE 3  Geometry

A rectangle 24 meters long has the same area as a square with 12-meter sides. What are the dimensions of the rectangle? Solution: STEP 1  Identify



the question. What are the dimensions (length and width) of the rectangle?

STEP 2  Make

notes. Read

A rectangle 24 meters long



Write/Draw w l = 24

area of rectangle 5 l ? w 5 24w . . . a square with 12-meter sides

12 m 12 m



[ CONCEPT CHECK ] If we were to guess an answer to Example 3, what would be our guess? (A) 24 m by 5 m (B) 24 m by 20 m (C) 24 m by 1 m

▼ ANSWER The answer is (A). Option (B) cannot be correct because that would have area much larger than a 12-m square, and (C) cannot be correct because that would have area much smaller than a 12-m square.

Young_AT_6160_ch01_pp082-128.indd 96

area of square 5 12 ? 12 5 144

STEP 3  Assign a variable. Let w 5 width of the rectangle. STEP 4  Set



up an equation. The area of the rectangle is equal to the area of the square. Substitute in known quantities.

STEP 5  Solve



rectangle area 5 square area 24w 5 144

the equation.

Divide by 24.

w5

144 56 24

The rectangle is 24 meters long and 6 meters wide.

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97

1.2  Applications Involving Linear Equations  STEP 6  Check



the solution. A 24 meter by 6 meter rectangle has an area of 144 square meters.

▼ Y O U R T U R N   A rectangle 3 inches wide has the same area as a square with

9-inch sides. What are the dimensions of the rectangle?

▼ ANSWER

The rectangle is 27 inches long and 3 inches wide.

1.2.3  Interest Problems In our personal or business financial planning, a particular concern we have is interest. Interest is money paid for the use of money; it is the cost of borrowing money. The total amount borrowed is called the principal. The principal can be the price of our new car; we pay the bank interest for loaning us money. The principal can also be the amount we keep in a CD or money market account; the bank uses this money and pays us interest. Typically interest rate, expressed as a percentage, is the amount charged for the use of the principal for a given time, usually in years. Simple interest is interest that is paid only on principal during a period of time. Later we will discuss compound interest, which is interest paid on both principal and the interest accrued over a period of time.

DEFINITION

1.2.3 S K I L L

Solve simple interest problems. 1.2.3 C O N C E P T U A L

Use intuition to confirm answers in multiple investment problems.

Simple Interest

If a principal of P dollars is borrowed for a period of t years at an annual interest rate r (expressed in decimal form), the interest I charged is I 5 Prt This is the formula for simple interest.

EXAMPLE 4  Simple Interest

Through a summer job Morgan is able to save $2500. If she puts that money into a 6-month certificate of deposit (CD) that pays a simple interest rate of 3% a year, how much money will she have in her CD at the end of the 6 months? Solution: STEP 1  Identify



the question. How much money does Morgan have after 6 months?

STEP 2  Make



notes. The principal is $2500. The annual interest rate is 3%, which in decimal form is 0.03. The time the money spends accruing interest is 6 months, or 12 of a year.

STEP 3  Assign



a variable. Label the known quantities.

STEP 4  Set



up an equation. Write the simple interest formula.

P 5 2500, r 5 0.03, and t 5 0.5

STEP 5  Solve the equation.

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I 5 Prt I 5 Prt I 5 125002 10.032 10.52 5 37.5

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CHAPTER 1  Equations and Inequalities

The interest paid on the CD is $37.50. Adding this to the principal gives a total of $2500 1 $37.50 5 $2537.50 STEP 6  Check

the solution. This answer agrees with our intuition. Had we made a mistake, say, of moving one decimal place to the right, then the interest would have been $375, which is much larger than we would expect on a principal of only $2500.

EXAMPLE 5  Multiple Investments

Theresa earns a full athletic scholarship for college, and her parents have given her the $20,000 they had saved to pay for her college tuition. She decides to invest that money with an overall goal of earning 11% interest. She wants to put some of the money in a low-risk investment that has been earning 8% a year and the rest of the money in a medium-risk investment that typically earns 12% a year. How much money should she put in each investment to reach her goal? Solution: STEP 1  Identify



the question. How much money is invested in each (the 8% and the 12%) account?

STEP 2  Make

notes.

Read Theresa has $20,000 to invest. If part is invested at 8% and the rest at 12%, how much should be invested at each rate to yield 11% on the total amount invested?

Write/Draw

STEP 3  Assign

a variable. If we let x represent the amount Theresa puts into the 8% investment, how much of the $20,000 is left for her to put in the 12% investment?

Amount in the 8% investment: x



Amount in the 12% investment: 20,000 2 x

STEP 4  Set

up an equation. Simple interest formula: I 5 Prt

INVESTMENT

PRINCIPAL

RATE

TIME (YR)

8% Account

INTEREST

x

0.08

1

0.08x

12% Account

20,000 2 x

0.12

1

Total

20,000

0.11

1

0.12  120,000 2 x2 0.11120,0002

Adding the interest earned in the 8% investment to the interest earned in the 12% investment should earn an average of 11% on the total investment. 0.08x 1 0.12  120,000 2 x2 5 0.11120,0002

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1.2  Applications Involving Linear Equations 

the equation. Eliminate the parentheses. Collect x terms on the left, constants on the right. Divide by 20.04. Calculate the amount at 12%.

99

STEP 5  Solve

0.08x 1 2400 2 0.12x 5 2200 20.04x 5 2200 x 5 5000 20,000 2 5000 5 15,000

Theresa should invest $5000 at 8% and $15,000 at 12% to reach her goal. the solution. If money is invested at 8% and 12% with a goal of averaging 11%, our intuition tells us that more should be invested at 12% than 8%, which is what we found. The exact check is as follows:

▼ ANSWER

$10,000 is invested at 18% and $14,000 is invested at 12%.

STEP 6  Check

0.08150002 1 0.12115,0002 5 0.11120,0002 400 1 1800 5 2200 2200 5 2200

▼

Y O U R T U R N   You win $24,000 and you decide to invest the money in two

different investments: one paying 18% and the other paying 12%. A year later you have $27,480 total. How much did you originally invest in each account?

[ CONCEPT CHECK ] If Teresa is going to invest in two accounts, one at 8% and one at 12% with a result of an average of 11% earnings, does your intuition think she should have (A) more at 8% than at 12% or (B) more at 12% than at 8%?

▼ ANSWER (B) because 11% is closer to 12% than 8%

1.2.4  Mixture Problems Mixtures are something we come across every day. Different candies that sell for different prices may make up a movie snack. New blends of coffees are developed by coffee connoisseurs. Chemists mix different concentrations of acids in their labs. Whenever two or more distinct ingredients are combined, the result is a mixture. Our choice at a gas station is typically 87, 89, and 93 octane. The octane number is the number that represents the percentage of iso-octane in fuel; 89 octane is significantly overpriced. Therefore, if your car requires 89 octane, it would be more cost effective to mix 87 and 93 octane.

1.2.4 S K I L L

Solve mixture problems. 1.2.4 C O N C E P T U A L

Use intuition to confirm answers to mixture problems.

EXAMPLE 6  Mixture Problem

The manual for your new car suggests using gasoline that is 89 octane. In order to save money, you decide to use some 87 octane and some 93 octane in combination with the 89 octane currently in your tank in order to have an approximate 89 octane mixture. Assuming you have 1 gallon of 89 octane remaining in your tank (your tank capacity is 16 gallons), how many gallons of 87 and 93 octane should be used to fill up your tank to achieve a mixture of 89 octane? Solution: STEP 1  Identify

the question. How many gallons of 87 octane and how many gallons of 93 octane should be used? STEP 2  Make

notes. Read Assuming you have one gallon of 89 octane remaining in your tank (your tank capacity is 16 gallons), how many 89 octane gallons of 87 and 93 [1 gallon] octane should you add?

Young_AT_6160_ch01_pp082-128.indd 99

Write/Draw

+ 87 octane + 93 octane [? gallons] [? gallons]

=

89 octane [16 gallons]

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100 

CHAPTER 1  Equations and Inequalities STEP 3  Assign

a variable. x 5 gallons of 87 octane gasoline added at the pump 15 2 x 5 gallons of 93 octane gasoline added at the pump 1 5 gallons of 89 octane gasoline already in the tank STEP 4  Set

up an equation. 0.89112 1 0.87x 1 0.93115 2 x2 5 0.891162

STEP 5  Solve the equation. 0.89112 1 0.87x 1 0.93115 2 x2 5 0.891162 Eliminate the parentheses. 0.89 1 0.87x 1 13.95 2 0.93x 5 14.24 Collect x terms on the left side. 20.06x 1 14.84 5 14.24 Subtract 14.84 from both sides of the equation. 20.06x 520.6 Divide both sides by 20.06. x 5 10 Calculate the amount of 93 octane. 15 2 10 5 5 ▼ ANSWER

Add 10 gallons of 87 octane and 5 gallons of 93 octane.

40 milliliters of 5% HCl and 60 milliliters of 15% HCl

[ CONCEPT CHECK ] If a chemistry student has HCl concentrations of 5% and 15% and the desired solution is 11% HCl, do we expect (A) more 15% than 5% or (B) more 5% than 15%?

▼ ANSWER (A) because 11% is closer to 15% than 5%.

STEP 6  Check the solution. Estimate: Our intuition tells us that if the desired mixture is 89 octane, then we should add approximately one part 93 octane and two parts 87 octane. The solution we found, 10 gallons of 87 octane and 5 gallons of 93 octane, agrees with this.

▼ Y O U R T U R N   For a certain experiment, a student requires 100 milliliters of a

solution that is 11% HCl (hydrochloric acid). The storeroom has only solutions that are 5% HCl and 15% HCl. How many milliliters of each available solution should be mixed to get 100 milliliters of 11% HCl?

1.2.5  Distance–Rate–Time Problems 1.2.5 S K I L L

Solve distance–rate–time problems. 1.2.5 C O N C E P T U A L

Estimate distance–rate–time problem solutions prior to solving and then confirm with a check.

The next example deals with distance, rate, and time. On a road trip, you see a sign that says your destination is 90 miles away, and your speedometer reads 60 miles per hour. Dividing 90 miles by 60 miles per hour tells you that your arrival will be in 1.5 hours. Here is how you know. If the rate, or speed, is assumed to be constant, then the equation that relates distance (d ), rate (r), and time (t) is given by d 5 r • t. In the above driving example, d 5 90 miles     r 5 60 Substituting these into d 5 r⋅ t, we arrive at

Solving for t, we get

Young_AT_6160_ch01_pp082-128.indd 100

miles hour

90 miles 5 c60 t5

miles d ⋅t hour

90 miles 5 1.5 hours miles 60 hour

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1.2  Applications Involving Linear Equations 

101

EXAMPLE 7  Distance–Rate–Time

It takes 8 hours to fly from Orlando to London and 9.5 hours to return. If an airplane averages 550 miles per hour in still air, what is the average rate of the wind blowing in the direction from Orlando to London? Assume the speed of the wind ( jet stream) is constant and the same for both legs of the trip. Round your answer to the nearest miles per hour. Solution: STEP 1  Identify



the question. At what rate in mph is the wind blowing?

STEP 2  Make

notes. Read It takes 8 hours to fly from Orlando to London and 9.5 hours to return.

Write/Draw

If the airplane averages 550 miles per hour in still air . . . STEP 3  Assign

a variable.

w 5 wind speed

STEP 4  Set

up an equation. The formula relating distance, rate, and time is d 5 r ? t. The distance d of each flight is the same. On the Orlando to London flight the time is 8 hours due to an increased speed from a tailwind. On the London to Orlando flight the time is 9.5 hours, and the speed is decreased due to the headwind. Let w represent the wind speed.

d 5 1550 1 w2 8 d 5 1550 2 w2 9.5

Orlando to London: London to Orlando:

These distances are the same, so set them equal to each other:

STEP 5  Solve

1550 1 w28 5 1550 2 w29.5

the equation. Eliminate the parentheses. Collect w terms on the left, constants on the right. Divide by 17.5.

4400 1 8w 5 5225 2 9.5w 17.5w 5 825 w 5 47.1429 < 47

The wind is blowing approximately  47 miles per hour  in the direction from Orlando to London. STEP 6  Check

the solution. Estimate: Going from Orlando to London, the tailwind is approximately 50 miles per hour, which added to the plane’s 550 miles per hour speed yields a ground speed of 600 miles per hour. The Orlando to London route took 8 hours. The distance of that flight is 1600 mph2 18 hr2, which is 4800 miles. The return trip experienced a headwind of approximately 50 miles per hour, so subtracting the 50 from 550 gives an average speed of 500 miles per hour. That route took 9.5 hours, so the distance of the London to Orlando flight was 1500 mph2 19.5 hr2, which is 4750 miles. Note that the estimates of 4800 and 4750 miles are close.

▼

Y O U R T U R N   A Cessna 150 averages 150 miles per hour in still air. With a tailwind

2 31

hours. Because of the headwind, it is it is able to make a trip in only able to make the return trip in 3 21 hours. What is the average wind speed?

Young_AT_6160_ch01_pp082-128.indd 101

▼ ANSWER

The wind is blowing 30 mph.

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CHAPTER 1  Equations and Inequalities

EXAMPLE 8  Work

[ CONCEPT CHECK ]

Connie can clean her house in 2 hours. If Alvaro helps her, they can clean the house in 1 hour and 15 minutes together. How long would it take Alvaro to clean the house by himself?

If Connie cleans her house alone it takes 2 hours, and if Alvaro helps her it takes 1 hour and 15 minutes. Which of the following is true?

Solution:

(A) Alvaro cleans faster than Connie.

STEP 1  Identify



(B) Alvaro and Connie clean at the same rate.

the question. How long would it take Alvaro to clean the house by himself?

STEP 2  Make

notes. 1 Connie can clean her house in 2 hours, so Connie can clean 2 of the house per hour. Together Connie and Alvaro can clean the house in 1 hour and 1 4 15 minutes, or 54 of an hour. Therefore, together they can clean 5 of 5/4 5 the house per hour.

(C) Alvaro cleans slower than Connie.

▼ ANSWER (C) is correct. (B) cannot be true or it would take them exactly one hour if they were working together. (A) cannot be true or the combined time would be less than one hour.

STEP 3  Assign

a variable. Let x 5 number of hours it takes Alvaro to clean the house by himself. So 1 Alvaro can clean of the house per hour. x STEP 4  Set

up an equation.



AMOUNT OF TIME TO DO ONE JOB

AMOUNT OF JOB DONE PER UNIT OF TIME

2

1 2

Alvaro

x

1 x

Together

5 4

4 5

Connie



Amount of house Alvaro can clean per hour

1

1 x

c

1 2

Amount of house they can clean per hour if they work together

c

c



5

▲

Amount of house Connie can clean per hour



1

5

4 5

STEP 5  Solve

the equation. Multiply 12 1 1x 5 45 by the LCD, 10x.

5x 1 10 5 8x 10 1 Solve for x. x5 53 3 3 It takes Alvaro 3 hours and 20 minutes to clean the house by himself.

STEP 6  Check

the solution.  Connie cleans the house in 2 hours. If Alvaro could clean it in 2 hours, then together it would take them 1 hour. Since together it takes them 1 hour and 15 minutes, we expect that it takes Alvaro more than 2 hours by himself.

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[S EC T I O N 1. 2]   S U M M A RY In the real world, many kinds of application problems can be solved through modeling with linear equations. The following six-step procedure will help you develop the model. Some problems require development of a mathematical model, while others rely on common formulas.

1. Identify the quantity you are to determine. 2. Make notes on any clues that will help you set up an equation. 3. Assign a variable. 4. Set up the equation. 5. Solve the equation. 6. Check the solution against your intuition.

[S EC T I O N 1. 2]   E X E R C I S E S • A P P L I C AT I O N S 1. Discount Price. Donna redeems a 10% off coupon at her

10.  Diet. A particular 1550-calories-per-day diet suggests

local nursery. After buying azaleas, bougainvillea, and bags of potting soil, her checkout price before tax is $217.95. How much would she have paid without the coupon? 2.  Discount Price. The original price of a pair of binoculars is $74. The sale price is $51.80. How much was the markdown? 3. Cost: Fair Share. Jeff, Tom, and Chelsea order a large pizza. They decide to split the cost according to how much they will eat. Tom pays $5.16, Chelsea eats 18 of the pizza, and Jeff eats 1 2 of the pizza. How much did the pizza cost? 4. Event Planning. A couple decide to analyze their monthly spending habits. The monthly bills are 50% of their takehome pay, and they invest 20% of their take-home pay. They spend $560 on groceries, and 23% goes to miscellaneous. How much is their take-home pay per month? 5. Discounts. A builder of tract homes reduced the price of a model by 15%. If the new price is $125,000, what was its original price? How much can be saved by purchasing the model? 6. Markups. A college bookstore marks up the price it pays the publisher for a book by 25%. If the selling price of a book is $79, how much did the bookstore pay for the book? 7.  Puzzle. Angela is on her way from home in Jersey City into New York City for dinner. She walks 1 mile to the train station, takes the train 34 of the way, and takes a taxi 16 of the way to the restaurant. How far does Angela live from the restaurant? 8.  Puzzle. An employee at Kennedy Space Center (KSC) lives in Daytona Beach and works in the vehicle assembly building (VAB). She carpools to work with a colleague. She drives 7 miles from her house to the park-and-ride. Then she rides with her colleague from the park-and-ride in Daytona Beach to the KSC headquarters building, and then takes the KSC shuttle from the headquarters building to the VAB. The drive from the park-and-ride to the headquarters building is 56 of her 1 total trip, and the shuttle ride is 20 of her total trip. How many miles does she travel from her house to the VAB on days when her colleague drives? 1 9. Puzzle. A typical college student spends 3 of her waking time 1 in class, 15 of her waking time eating, 10 of her waking time 1 working out, 3 hours studying, and 22 hours doing other things. How many hours of sleep does the typical college student get?

eating breakfast, lunch, dinner, and two snacks. Dinner is twice the calories of breakfast. Lunch is 100 calories more than breakfast. The two snacks are 100 and 150 calories. How many calories are each meal? 11.  Budget. A company has a total of $20,000 allocated for mon­ t­­ hly costs. Fixed costs are $15,000 per month and variable costs are $18.50 per unit. How many units can be manufactured in a month? 12.  Budget. A woman decides to start a small business making monogrammed cocktail napkins. She can set aside $1870 for monthly costs. Fixed costs are $1329.50 per month and variable costs are $3.70 per set of napkins. How many sets of napkins can she afford to make per month? 13.  Numbers. Find a number such that 10 less than 23 the number is 14 the number. 14.  Numbers. Find a positive number such that 10 times the number is 16 more than twice the number. 15.  Numbers. Find two consecutive even integers such that 4 times the smaller number is 2 more than 3 times the larger number. 16.  Numbers. Find three consecutive integers such that the sum of the three is equal to 2 times the sum of the first two ­integers. 17.  Geometry. Find the perimeter of a triangle if one side is 11 inches, another side is 15 the perimeter, and the third side is 1 4 the perimeter. 18.  Geometry. Find the dimensions of a rectangle whose length is a foot longer than twice its width and whose perimeter is 20 feet. 19.  Geometry. An NFL playing field is a rectangle. The length of the field (excluding the end zones) is 40 more yards than twice the width. The perimeter of the playing field is 260 yards. What are the dimensions of the field in yards? 20.  Geometry. The length of a rectangle is 2 more than 3 times the width, and the perimeter is 28 inches. What are the dimensions of the rectangle? 21.  Geometry. Consider two circles, a smaller one and a larger one. If the larger one has a radius that is 3 feet larger than that of the smaller circle and the ratio of the circumferences is 2:1, what are the radii of the two circles?















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22.  Geometry. The perimeter of a semicircle is doubled when

the radius is increased by 1. Find the radius of the semicircle.

3.2 pounds and the price was $17.56. How many pounds of each did she buy?

23.  Home Improvement. A man wants to remove a tall pine tree

33.  Chemistry. For a certain experiment, a student requires 100

from his yard. Before he goes to Home Depot, he needs to know how tall an extension ladder he needs to purchase. He measures the shadow of the tree to be 225 feet long. At the same time he measures the shadow of a 4-foot stick to be 3 feet. Approximately how tall is the pine tree?

milliliters of a solution that is 8% HCl (hydrochloric acid). The storeroom has only solutions that are 5% HCl and 15% HCl. How many milliliters of each available solution should be mixed to get 100 milliliters of 8% HCl? 34.  Chemistry. How many gallons of pure alcohol must be mixed with 5 gallons of a solution that is 20% alcohol to make a solution that is 50% alcohol? 35.  Automobiles. A mechanic has tested the amount of antifreeze in your radiator. He says it is only 40% antifreeze and the remainder is water. How many gallons must be drained from your 5 gallon radiator and replaced with pure antifreeze to make the mixture in your radiator 80% antifreeze? 36.  Costs: Overhead. A professor is awarded two research grants, each having different overhead rates. The research project conducted on campus has a rate of 42.5% overhead, and the project conducted in the field, off campus, has a rate of 26% overhead. If she was awarded $1,170,000 total for the two projects with an average overhead rate of 39%, how much was the research project on campus and how much was the research project off campus? 37.  Theater. On the way to the movies a family picks up a custom-made bag of candies. The parents like caramels 1$1.50/lb2 and the children like gummy bears 1$2.00/lb2. They bought a 1.25-pound bag of combined candies that cost $2.50. How much of each candy did they buy? 38.  Coffee. Joy is an instructional assistant in one of the college labs. She is on a very tight budget. She loves Jamaican Blue Mountain coffee, but it costs $12 a pound. She decides to blend this with regular coffee beans that cost $4.20 a pound. If she spends $14.25 on 2 pounds of coffee, how many pounds of each did she purchase? 39.  Communications. The speed of light is approximately 3.0 3 108 meters per second 1670,616,629 mph2. The distance from Earth to Mars varies because of the orbits of the planets around the Sun. On average, Mars is 100 million miles from Earth. If we use laser communication systems, what will be the delay between Houston and NASA astronauts on Mars? 40.  Speed of Sound. The speed of sound is approximately 760 miles per hour in air. If a gun is fired 12 mile away, how long will it take the sound to reach you? 41. Business. Because of holiday travel during the month of November, the average price of gasoline rose 4.7% in the United States. If the average price of gasoline at the end of November was $3.21 per gallon, what was the price of gasoline at the beginning of November? 42. Business. During the Christmas shopping season, the average price of a flat screen television fell by 40%. A shopper purchased a 42-inch flat screen television for $299 in late November. How much would the shopper have paid, to the nearest dollar, for the same television if it was purchased in September? 43. Medicine. A patient requires an IV of 0.9% saline solution, also known as normal saline solution. How much distilled water, to the nearest milliliter, must be added to

24.  Home Improvement. The same man in Exercise 23 realizes

he also wants to remove a dead oak tree. Later in the day he measures the shadow of the oak tree to be 880 feet long, and the 4-foot stick now has a shadow of 10 feet. Approximately how tall is the oak tree? 25.  Biology: Alligators. It is common to see alligators in ponds,

lakes, and rivers in Florida. The ratio of head size (back of the head to the end of the snout) to the full body length of an alligator is typically constant. If a 321-foot alligator has a head length of 6 inches, how long would you expect an alligator to be whose head length is 9 inches? 26.  Biology: Snakes. In the African rainforest there is a

snake called a Gaboon viper. The fang size of this snake is proportional to the length of the snake. A 3-foot snake typically has 2-inch fangs. If a herpetologist finds Gaboon viper fangs that are 2.6 inches long, how long a snake would she expect to find? 27.  Investing. Ashley has $120,000 to invest and decides to put

some in a CD that earns 4% interest per year and the rest in a low-risk stock that earns 7%. How much did she invest in each to earn $7800 interest in the first year? 28.  Investing. You inherit $13,000 and you decide to invest the

money in two different investments: one paying 10% and the other paying 14%. A year later your investments are worth $14,580. How much did you originally invest in each account? 29.  Investing. Wendy was awarded a volleyball scholarship to

the University of Michigan, so on graduation her parents gave her the $14,000 they had saved for her college tuition. She opted to invest some money in a privately held company that pays 10% per year and evenly split the remaining money between a money market account yielding 2% and a high-risk stock that yielded 40%. At the end of the first year she had $16,610 total. How much did she invest in each of the three? 30.  Interest. A high school student was able to save $5000 by

working a part-time job every summer. He invested half the money in a money market account and half the money in a stock that paid three times as much interest as the money market account. After a year he earned $150 in interest. What were the interest rates of the money market account and the stock? 31.  Budget: Home Improvement. When landscaping their yard,

a couple budgeted $4200. The irrigation system costs $2400 and the sod costs $1500. The rest they will spend on trees and shrubs. Trees each cost $32 and shrubs each cost $4. They plant a total of 33 trees and shrubs. How many of each did they plant in their yard? 32.  Budget: Shopping. At the deli Jennifer bought spicy tur-

key and provolone cheese. The turkey costs $6.32 per pound and the cheese costs $4.27 per pound. In total, she bought

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100 milliliters of a 3% saline solution to produce normal saline? 44. Medicine. A patient requires an IV of D5W, a 5% solution of dextrose (sugar) in water. To the nearest milliliter, how much D20W, a 20% solution of dextrose in water, must be added to 100 milliliters of distilled water to produce a D5W solution?

56.  Music. A minor chord in music is composed of notes whose

45.  Boating. A motorboat can maintain a constant speed of

57.  Grades. Danielle’s test scores are 86, 80, 84, and 90. The

16 miles per hour relative to the water. The boat makes a trip upstream to a marina in 20 minutes. The return trip takes 15 minutes. What is the speed of the current? 46.  Aviation. A Cessna 175 can average 130 miles per hour. If

a trip takes 2 hours one way and the return takes 1 hour and 15 minutes, find the wind speed, assuming it is constant. 47.  Exercise. A jogger and a walker cover the same distance.

The jogger finishes in 40 minutes. The walker takes an hour. How fast is each exerciser moving if the jogger runs 2 miles per hour faster than the walker? 48.  Travel. A high school student in Seattle, Washington,

a­ ttended the University of Central Florida. On the way to UCF he took a southern route. After graduation he returned to Seattle via a northern trip. On both trips he had the same average speed. If the southern trek took 45 hours and the northern trek took 50 hours, and the northern trek was 300 miles longer, how long was each trip? 49.  Distance–Rate–Time. College roommates leave for their

first class in the same building. One walks at 2 miles per hour and the other rides his bike at a slow 6 miles per hour pace. How long will it take each to get to class if the walker takes 12 minutes longer to get to class and they travel on the same path? 50.  Distance–Rate–Time. A long-distance delivery service

sends out a truck with a package at 7 a.m. At 7:30 a.m., the manager realizes there was another package going to the same location. He sends out a car to catch the truck. If the truck travels at an average speed of 50 miles per hour and the car travels at 70 miles per hour, how long will it take the car to catch the truck? 51.  Work. Christopher can paint the interior of his house in

15 hours. If he hires Cynthia to help him, they can do the same job together in 9 hours. If he lets Cynthia work alone, how long will it take her to paint the interior of his house? 52.  Work. Jay and Morgan work in the summer for a landscaper.

It takes Jay 3 hours to complete the company’s largest yard alone. If Morgan helps him, it takes only 1 hour. How much time would it take Morgan alone? 53.  Work. Tracey and Robin deliver soft drinks to local

convenience stores. Tracey can complete the deliveries in 4 hours alone. Robin can do it in 6 hours alone. If they decide to work together on a Saturday, how long will it take? 54.  Work. Joshua can deliver his newspapers in 30 minutes.

It takes Amber 20 minutes to do the same route. How long would it take them to deliver the newspapers if they worked together? 55.  Music. A major chord in music is composed of notes whose

frequencies are in the ratio 4:5:6. If the first note of a chord

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has a frequency of 264 hertz (middle C on the piano), find the frequencies of the other two notes. Hint: Set up two proportions using 4:5 and 4:6. frequencies are in the ratio 10:12:15. If the first note of a minor chord is A, with a frequency of 220 hertz, what are the frequencies of the other two notes? final exam will count as 23 of the final grade. What score does Danielle need on the final in order to earn a B, which requires an average score of 80? What score does she need to earn an A, which requires an average of 90? 58.  Grades. Sam’s final exam will count as two tests. His test scores are 80, 83, 71, 61, and 95. What score does Sam need on the final in order to have an average score of 80? 59.  Sports. In Super Bowl XXXVII, the Tampa Bay Buccaneers scored a total of 48 points. All of their points came from field goals and touchdowns. Field goals are worth 3 points and each touchdown was worth 7 points (Martin Gramatica was successful in every extra point attempt). They scored a total of 8 times. How many field goals and touchdowns were scored? 60.  Sports. A tight end can run the 100-yard dash in 12 seconds. A defensive back can do it in 10 seconds. The tight end catches a pass at his own 20-yard line with the defensive back at the 15-yard line. If no other players are nearby, at what yard line will the defensive back catch up to the tight end? 61.  Recreation. How do two children of different weights balance on a seesaw? The heavier child sits closer to the center and the lighter child sits farther away. When the product of the weight of the child and the distance from the center is equal on both sides, the seesaw should be horizontal to the ground. Suppose Max weighs 42 pounds and Maria weighs 60 pounds. If Max sits 5 feet from the center, how far should Maria sit from the center in order to balance the seesaw horizontal to the ground? 62.  Recreation. Refer to Exercise 61. Suppose Martin, who weighs 33 pounds, sits on the side of the seesaw with Max. If their average distance to the center is 4 feet, how far should Maria sit from the center in order to balance the seesaw horizontal to the ground? 63.  Recreation. If a seesaw has an adjustable bench, then the board can slide along the fulcrum. Maria and Max in Exercise 61 decide to sit on the very edge of the board on each side. Where should the fulcrum be placed along the board in order to balance the seesaw horizontally to the ground? Give the answer in terms of the distance from each child’s end. 64.  Recreation. Add Martin (Exercise 62) to Max’s side of the seesaw and recalculate Exercise 63. In Exercises 65–68, refer to this lens law. (See Exercise 82 in Section 1.1.)  The position of the image is found using the thin lens equation: 1 1 1 5 1 , ƒ do di

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where do is the distance from the object to the lens, di is the ­distance from the lens to the image, and ƒ is the focal length of the lens. Object 2f

do f

f f

2f

di Image

66.  Optics. If the focal length of the lens is 8 centimeters and

the image distance is 2 centimeters from the lens, what is the distance from the object to the lens? 67.  Optics. The focal length of a lens is 2 centimeters. If the image distance from the lens is half the distance from the object to the lens, find the object distance. 68.  Optics. The focal length of a lens is 8 centimeters. If the image distance from the lens is half the distance from the object to the lens, find the object distance.

65.  Optics. If the focal length of a lens is 3 centimeters and the

image distance is 5 centimeters from the lens, what is the distance from the object to the lens?

• CONCEPTUAL In Exercises 69–76, solve each formula for the specified variable. 69. P 5 2l 1 2w for w 70. P 5 2l 1 2w for l 1 71. A 5 2 bh for h

72. C 5 2pr for r 73. A 5 lw for w 74. d 5 rt for t 75. V 5 lwh for h 76. V 5 pr2h for h

• CHALLENGE 77.  Tricia and Janine are roommates and leave Houston on Interstate 10 at the same time to visit their families for a long weekend. Tricia travels west and Janine travels east. If Tricia’s average speed is 12 miles per hour faster than Janine’s, find the speed of each if they are 320 miles apart in 2 hours and 30 minutes.

78.  Rick and Mike are roommates and leave Gainesville on

Interstate 75 at the same time to visit their girlfriends for a long weekend. Rick travels north and Mike travels south. If Mike’s average speed is 8 miles per hour faster than Rick’s, find the speed of each if they are 210 miles apart in 1 hour and 30 minutes.

• TECHNOLOGY 79. Suppose you bought a house for $132,500 and sold it 3 years

later for $168,190. Plot these points using a graphing utility. Assuming a linear relationship, how much could you have sold the house for had you waited 2 additional years? 80. Suppose you bought a house for $132,500 and sold it 3 years later for $168,190. Plot these points using a graphing utility. Assuming a linear relationship, how much could you have sold the house for had you sold it 1 year after buying it? 81.  A golf club membership has two options. Option A is a $300 monthly fee plus a $15 cart fee every time you play.

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Option B has a $150 monthly fee and a $42 fee every time you play. Write a mathematical model for monthly costs for each plan and graph both in the same viewing rectangle using a graphing utility. Explain when Option A is the better deal and when Option B is the better deal. 82.  A phone provider offers two calling plans. Plan A has a $30 monthly charge and a $0.10 per minute charge on every call. Plan B has a $50 monthly charge and a $0.03 per minute charge on every call. Explain when Plan A is the better deal and when Plan B is the better deal.

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1.3 Q  UADRATIC EQUATIONS SKILLS OBJECTIVES ■■ Solve quadratic equations by factoring. ■■ Use the square root method to solve quadratic equations. ■■ Solve quadratic equations by completing the square. ■■ Use the Quadratic Formula to solve quadratic equations. ■■ Solve application problems using quadratic equations.

CONCEPTUAL OBJECTIVES ■■ Understand the zero product property in factoring. ■■ Understand that the square root method can only be used when there is no linear term in the quadratic equation. ■■ Understand that completing the square transforms a standard quadratic equation into a perfect square. ■■ Derive the Quadratic Formula. ■■ Understand why eliminated nonphysical answers should be eliminated.

1.3.1 Factoring In a linear equation, the variable is raised only to the first power in any term where it occurs. In a quadratic equation, the variable is raised to the second power in at least one term. Examples of quadratic equations, also called second-degree equations, are: x 2 1 3 5 7   5x 2 1 4x 2 7 5 0   x 2 2 3 5 0

1.3.1 S K I L L

Solve quadratic equations by factoring. 1.3.1 C O N C E P T U A L

Understand the zero product property in factoring. DEFINITION

Quadratic Equation

A quadratic equation in x is an equation that can be written in the standard form ax2 1 bx 1 c 5 0 where a, b, and c are real numbers and a 2 0.

There are several methods for solving quadratic equations: factoring, the square root method, completing the square, and the Quadratic Formula.

FACTORING METHOD

The factoring method applies the zero product property: WORDS MATH

If a product is zero, then at least one of its factors has to be zero.

If B ? C 5 0, then B 5 0 or C 5 0 or both.

Consider 1x 2 32 1x 1 22 5 0. The zero product property says that x 2 3 5 0 or x 1 2 5 0, which leads to x 5 22 or x 5 3. The solution set is 522, 36. When a quadratic equation is written in the standard form ax2 1 bx 1 c 5 0, it may be possible to factor the left side of the equation as a product of two first-degree polynomials. We use the zero product property and set each linear factor equal to zero. We solve the resulting two linear equations to obtain the solutions of the quadratic equation.

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EXAMPLE 1  Solving a Quadratic Equation by Factoring

Solve the equation x 2 2 6x 2 16 5 0. Solution:

The quadratic equation is already in standard form.

x 2 2 6x 2 16 5 0

Factor the left side into a product of two linear factors.   1x 2 82 1x 1 22 5 0 If a product equals zero, one of its factors has to be equal to zero.

x 2 8 5 0  or  x 1 2 5 0

Solve both linear equations.

▼ ANSWER

x 5 8  or  x 5 22

The solution set is 522, 86 .

▼

Y O U R T U R N   Solve the quadratic equation x 2 1 x 2 20 5 0 by factoring.

The solution is x 5 25, 4. The solution set is 525, 46.

EXAMPLE 2  Solving a Quadratic Equation by Factoring ▼ CAUTION

Solve the equation x 2 2 6x 1 5 5 24.

Don’t forget to put the quadratic equation in standard form first.

common mistake A common mistake is to forget to put the equation in standard form first and then use the zero product property incorrectly. ✖INCORRECT

✓C O R R E C T Write the original equation. 2

x 2 6x 1 5 5 24 Write the equation in standard form by adding 4 to both sides. x 2 2 6x 1 9 5 0

Factor the left side. 1 x 2 5 21 x 2 1 2 5 24

The error occurs here. x 2 5 5 24

or

x 2 1 5 24

Factor the left side. 1x 2 32 1x 2 32 5 0

Use the zero product property and set each factor equal to zero. x 2 3 5 0 or x 2 3 5 0 Solve each linear equation. x53

cDon’t forget to put the quadratic equation in standard form first.

Note: The equation has one solution, or root, which is 3. The solution set is 536. Since the linear factors were the same, or repeated, we say that 3 is a double root, or repeated root.

▼ ANSWER

▼ Y O U R T U R N   Solve the quadratic equation 9p2 5 24p 2 16 by factoring.

The solution is p 5 34 , which is a double root. The ­solution set is U 43 V .

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EXAMPLE 3  Solving a Quadratic Equation by Factoring ▼ CAUTION

Solve the equation 2x 2 5 3x.

common mistake The common mistake here is dividing both sides by x, which is not allowed because x might be zero. ✓C O R R E C T

✖INCORRECT

Write the equation in standard form by subtracting 3x.

Write the original equation. 2x 2 5 3x

2x 2 2 3x 5 0

The error occurs here when both sides are divided by x.

Factor the left side. x 1 2x 2 3 2 5 0

Use the zero product property and set each factor equal to zero. x50

or

2x 5 3

2x 2 3 5 0

Do not divide by a variable (because the value of that variable may be zero). Bring all terms to one side first and then factor.

[ CONCEPT CHECK ] If something times something is equal to zero, then _____ has to be zero: (A) One of those somethings; (B) Both of those somethings

▼ ANSWER (A) One of those somethings

Solve each linear equation. x50

or

x5

3 2

The solution set is U0, 32 V  .

In Example 3, the root x 5 0 is lost when the original quadratic equation is divided by x. Remember to put the equation in standard form first and then factor.

1.3.2  Square Root Method The square root of 16, !16, is 4, not 64. In the review (Chapter 0) the principal square root was discussed. The solutions to x2 5 16, however, are x 5 24 and x 5 4. Let us now investigate quadratic equations that do not have a first-degree term. They have the form ax 2 1 c 5 0 a 2 0 The method we use to solve such equations uses the square root property.

1.3.2 S K I L L

Use the square root method to solve quadratic equations. 1.3.2 C O N C E P T U A L

Understand that the square root method can only be used when there is no linear term in the quadratic equation.

SQUARE ROOT PROPERTY

WORDS MATH

If an expression squared is equal If x2 5 P, then x 5 6 !P . to a constant, then that expression is equal to the positive or negative square root of the constant. Note: The variable squared must be isolated first (coefficient equal to 1).

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[ CONCEPT CHECK ] Which of the following can be solved using the square root method? 2

(A) x   = x (B) x  2 = 9

▼ ANSWER (B) x  2 = 9 can be solved by the square root method because it is in the form of x  2 = constant. (A) cannot be solved using the square root method.

EXAMPLE 4  Using the Square Root Property

Solve the equation 3x 2 2 27 5 0. Solution:

Add 27 to both sides. 3x 2 5 27 Divide both sides by 3.   x 2 5 9 Apply the square root property.    x 5 6 !9 5 63 The solution set is 523, 36 .

If we alter Example 4 by changing subtraction to addition, we see in Example 5 that we get imaginary roots (as opposed to real roots), which we discussed in Chapter 0. EXAMPLE 5  Using the Square Root Property

Solve the equation 3x 2 1 27 5 0. Solution:

Subtract 27 from both sides. 3x 2 5 227 Divide by 3. x 2 5 29 Apply the square root property.  x 5 6 !29 Simplify.   x 5 6i !9 5 63i ▼ ANSWER

The solution is y 5 67 !3. The solution set is U 27 !3, 7 !3 V . The solution is v 5 68i. The solution set is 5 28i, 8i 6 .

▼

The solution set is 523i, 3i6 .

Y O U R T U R N   Solve the equations y2 2 147 5 0 and v2 1 64 5 0.

EXAMPLE 6  Using the Square Root Property

Solve the equation 1x 2 22 2 5 16. Solution:

Approach 1: If an expression squared is 16, then the expression equals 6 !16. Separate into two equations.

1 x 2 2 2 5 6 !16 x 2 2 5 !16 or x 2 2 5 2 !16 x 2 2 5 4 x 2 2 5 24 x 5 6 x 5 22

The solution set is 522, 66 .

Approach 2: It is acceptable notation to keep the equations together.

1 x 2 2 2 5 6 !16 x22564 x5264 x 5 22, 6

1.3.3  Completing the Square Factoring and the square root method are two efficient, quick procedures for solving many quadratic equations. However, some equations, such as x2 2 10x 2 3 5 0, cannot be solved directly by these methods. A more general procedure to solve this kind of

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equation is called completing the square. The idea behind completing the square is to transform any s­ tandard quadratic equation ax 2 1 bx 1 c 5 0 into the form 1x 1 A2 2 5 B, where A and B are constants and the left side, 1x 1 A22, has the form of a perfect square. This last equation can then be solved by the square root method. How do we transform the first equation into the second equation? Note that the above-mentioned example, x 2 2 10x 2 3 5 0, cannot be factored into expressions in which all numbers are integers (or even rational numbers). We can, however, transform this quadratic equation into a form that contains a perfect square.

1.3.3 S K I L L

Solve quadratic equations by completing the square. 1.3.3 C O N C E P T U A L

Understand that completing the square transforms a standard WORDS MATH quadratic equation into a perfect Write the original equation. x 2 2 10x 2 3 5 0 square.

Add 3 to both sides. x 2 2 10x 5 3 Add 25 to both sides.* x 2 2 10x 1 25 5 3 1 25 The left side can be written as a perfect square. 1x 2 52 2 5 28 Apply the square root method.  x 2 5 5 6 !28 Add 5 to both sides.  x 5 5 6 2 !7

*Why did we add 25 to both sides? Recall that 1x 2 c2 2 5 x 2 2 2xc 1 c2. In this case c 5 5 in order for 22xc 5 210x. Therefore, the desired perfect square 1x 2 522 results in x2 2 10x 1 25. Applying this product, we see that 125 is needed. A systematic approach is to take the coefficient of the first-degree term x2 2 10x 2 3 5 0, which is 210. Take half of 12102, which is 1252, and then square it 12522 5 25. SOLVING A QUADRATIC EQUATION BY COMPLETING THE SQUARE

[ CONCEPT CHECK ]

The quadratic equation x 2 2 4x 1 3 5 0 can be transformed WORDS MATH to which perfect square?

Express the quadratic equation in the following form. Divide b by 2 and square the result, then add the square to both sides. Write the left side of the equation as a perfect square.

x 2 1 bx 5 c 2

b b x 2 1 bx 1 a b 5 c 1 a b 2 2

Solve using the square root method.

2

(A) 1x 1 22 2 5 1

(B) 1x 2 22 2 5 1

▼

ANSWER (B)

b 2 b 2 ax 1 b 5 c 1 a b 2 2

EXAMPLE 7  Completing the Square

Solve the quadratic equation x 2 1 8x 2 3 5 0 by completing the square. Solution:

Add 3 to both sides. 2

Add Q 12 ⋅ 8R 5 42 to both sides.

Write the left side as a perfect square and simplify the right side.

x 2 1 8x 5 3 x 2 1 8x 1 42 5 3 1 42 1x 1 42 2 5 19

Apply the square root method to solve.  x 1 4 5 6 !19

Subtract 4 from both sides.  x 5 24 6 !19 The solution set is  U24 2 !19, 24 1 !19V .

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In Example 7, the leading coefficient (the coefficient of the x2 term) is 1. When the leading coefficient is not 1, start by first dividing the equation by that leading coefficient.

EXAMPLE 8  C  ompleting the Square When the Leading Coefficient Is Not Equal to 1

Solve the equation 3x2 2 12x 1 13 5 0 by completing the square. Solution:

Divide by the leading coefficient, 3.

x 2 2 4x 1

x 2 2 4x 5 2

Collect variables to one side of the equation and constants to the other side. Add A224 B 2 5 4 to both sides.

Solve using the square root method.

When the leading coefficient is not 1, start by first dividing the equation by that leading ­coefficient.

i !2 The solution is x 5 1 6 . The 2 i !2 i !2 solution set is e 1 2 ,11 f. 2 2

1.3.4 S K I L L

Use the Quadratic Formula to solve quadratic equations.

13 14 3 1 1x 2 222 5 2 3 x2256

1 2 Å 3

Simplify.

x526i

Rationalize the denominator (Chapter 0).

x526

Simplify.

x522

The solution set is  e 2 2 ▼ ANSWER

13 3

x2 2 4x 1 4 5 2

Write the left side of the equation as a perfect square and simplify the right side.

ST U DY TIP

13 50 3

1 Å3

i !3 ⋅ !3 !3

i !3 i !3 , x521 3 3

i !3 i !3 ,21 f  . 3 3

▼ Y O U R T U R N   Solve the equation 2x2 2 4x 1 3 5 0 by completing the square.

1.3.4  Quadratic Formula Let us now consider the most general quadratic equation: ax 2 1 bx 1 c 5 0 a 2 0 We can solve this equation by completing the square.

1.3.4 C O N C E P T U A L

Derive the Quadratic Formula.

WORDS MATH

Divide the equation by the leading coefficient a. c Subtract a from both sides. b Square half of a and add the result a

Young_AT_6160_ch01_pp082-128.indd 112

b 2 b to both sides. 2a

b c x2 1 a x 1 a 5 0 b c x2 1 a x 5 2 a b b 2 b 2 c x2 1 a x 1 a b 5 a b 2 a 2a 2a

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1.3  Quadratic Equations 

Write the left side of the equation as a perfect square and the right side as a single fraction.

ax 1

b 2 b2 2 4ac b 5 2a 4a2

b b2 2 4ac x1 56 2a Å 4a2

Solve using the square root method. b from both sides 2a and simplify the radical. Subtract

x52

Write as a single fraction.

x5

b "b2 2 4ac 6 2a 2a

2b6"b2 2 4ac 2a

We have derived the Quadratic Formula.

113

[ CONCEPT CHECK ] TRUE OR FALSE  Completing the square is still necessary in certain cases when the Quadratic Formula cannot be used to solve a particular Quadratic Equation.

▼ ANSWER False. The Quadratic Formula was derived by solving the general Quadratic Equation by completing the square and can be used for solving any quadratic equation.

STUD Y T I P

QUADRATIC FORMULA 2

If ax 1 bx 1 c 5 0, a 2 0, then the solution is x5

2b 6 "b2 2 4ac 2a Read as “negative b plus or minus the square root of the quantity b squared minus 4ac all over 2a.” x5

2b 6 "b2 2 4ac 2a

Note: The quadratic equation must be in standard form 1ax2 1 bx 1 c 5 0) in order to identify the parameters: a—coefficient of x2    b—coefficient of x    c—constant

We read this formula as negative b plus or minus the square root of the quantity b squared minus 4ac all over 2a. It is important to note that negative b could be positive (if b is negative). For this reason, an alternate form is “opposite b. . .” The Quadratic Formula should be memorized and used when simpler methods (factoring and the square root method) cannot be used. The Quadratic Formula works for any quadratic equation. EXAMPLE 9  U  sing the Quadratic Formula and Finding Two Distinct Real Roots

STUDY T I P The Quadratic Formula works for any quadratic equation.

STUDY T I P

Use the Quadratic Formula to solve the quadratic equation x2 2 4x 2 1 5 0.

Using parentheses as placeholders helps avoid 6 errors.

Solution:

x5

For this problem a 5 1, b 5 24, and c 5 21. 2

Write the Quadratic Formula.

x5

Use parentheses to avoid losing a minus sign.

x5

Substitute values for a, b, and c into the parentheses.

x5

2b 6 "b 2 4ac 2a

2 1 24 2 6 " 1 24 2 2 2 4 1 1 2 1 21 2 2112

▼

Y O U R T U R N   Use the Quadratic Formula to solve the quadratic equation 2

x 1 6x 2 2 5 0.

Young_AT_6160_ch01_pp082-128.indd 113

2 1 u 2 6 " 1 u 2 2 2 4 1 u 21 u 2 21 u 2

21 u 2 6 "1 u 22 2 41 u 2 1 u 2 21 u 2

4 6 "20 4 6 2"5 4 2"5 4 6 "16 1 4 5 5 5 6 5 2 6 "5 2 2 2 2 2 The solution set U2 2 !5, 2 1 !5V contains two distinct real numbers. Simplify. x 5

x5

2b 6 "b2 2 4ac 2a

▼ ANSWER

The solution is x 5 23 6 !11. The solution set is U 23 2 "11, 23 1 "11 V .

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CHAPTER 1  Equations and Inequalities

EXAMPLE 10  U  sing the Quadratic Formula and Finding Two Complex Roots

Use the Quadratic Formula to solve the quadratic equation x2 1 8 5 4x. Solution:

 rite this equation in standard form x2 2 4x 1 8 5 0 in order to identify a 5 1, W b 5 24, and c 5 8. Write the Quadratic Formula.

x5

Use parentheses to avoid overlooking a minus sign.

x5

Substitute the values for a, b, and c into the parentheses.

x5

Simplify.     x 5

2b 6 "b2 2 4ac 2a 21 u 2 6 "1 u 22 2 41 u 2 1 u 2 21 u 2

2 1 24 2 6 " 1 24 2 2 2 4 1 1 2 1 8 2 2112

4 6 !16 2 32 4 6 !216 4 6 4i 4 4i 5 5 5 6 5 2 6 2i 2 2 2 2 2

 he solution set 52 2 2i, 2 1 2i6 contains two complex numbers. Note that they T are complex conjugates of each other. ▼ ANSWER

▼ Y O U R T U R N   Use the Quadratic Formula to solve the quadratic equation

x2 1 2 5 2x.

The solution set is 51 2 i, 1 1 i6.

EXAMPLE 11  U  sing the Quadratic Formula and Finding One Repeated Real Root

Use the Quadratic Formula to solve the quadratic equation 4x2 2 4x 1 1 5 0. Solution:

Identify a, b, and c.

a 5 4, b 5 24, c 5 1

Write the Quadratic Formula.

x5

Use parentheses to avoid losing a minus sign.

x5

Substitute values a 5 4, b 5 24, and c 5 1.

x5

Simplify.

x5

2b 6 "b2 2 4ac 2a 21 u 2 6 "1 u 22 2 41 u 2 1 u 2 21 u 2 2 1 24 2 6 " 1 24 2 2 2 4 1 4 2 1 1 2 2142

4 6 "16 2 16 460 1 5 5 8 8 2

1 The solution set is a repeated real root e f . 2

▼ ANSWER 1 U3 V

Young_AT_6160_ch01_pp082-128.indd 114

Note: This quadratic equation also could have been solved by factoring: 12x 2 122 5 0.

▼

Y O U R T U R N   Use the Quadratic Formula to solve the quadratic equation

9x2 2 6x 1 1 5 0.

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1.3  Quadratic Equations 

115

TYPES OF SOLUTIONS

The term inside the radical, b2 2 4ac, is called the discriminant. The discriminant gives important information about the corresponding solutions or roots of ax2 1 bx 1 c 5 0, where a, b, and c are real numbers. b 2 2 4ac

SOLUTIONS (ROOTS)

Positive

Two distinct real roots

0

One real root (a double or repeated root)

Negative

Two complex roots (complex conjugates)

In Example 9, the discriminant is positive and the solution has two distinct real roots. In Example 10, the discriminant is negative and the solution has two complex (conjugate) roots. In Example 11, the discriminant is zero and the solution has one repeated real root.

1.3.5  Applications Involving Quadratic Equations In Section 1.2, we developed a procedure for solving word problems involving linear equations. The procedure is the same for applications involving quadratic equations. The only difference is that the mathematical equations will be quadratic, as opposed to linear.

1.3.5 S K I L L

Solve application problems using quadratic equations. 1.3.5 C O N C E P T U A L

Understand why nonphysical answers should be eliminated.

EXAMPLE 12  Stock Value

rom March 1 to May 1 F the price of Abercrombie & Fitch’s (ANF) stock was approximately given by P 5 23t2 1 6t 1 20, where P is the price of stock in dollars, t is in months, and t 5 0 corresponds to March 1. When was the value of the stock worth $22?

35 30 25 20 15 Feb

Mar

Apr

May

June

July

Aug

Sep

Oct

Nov

Solution: STEP 1  Identify



the question. When is the price of the stock equal to $22?

STEP 2  Make

notes. Stock price:

P 5 23t 2 1 6t 1 20 P 5 22

STEP 3  Set

23t2 1 6t 1 20 5 22

up an equation.

STEP 4  Solve



the equation. Subtract 22 from both sides.



Solve for t using the Quadratic Formula.

Simplify. t5

Young_AT_6160_ch01_pp082-128.indd 115

23t2 1 6t 2 2 5 0 2 1 6 2 6 "62 2 4 1 23 2 1 22 2 t5 2 1 23 2

26 6 "62 2 4 1 23 2 1 22 2 26 6 "12 < 1.57, 0.42 5 26 2 1 23 2

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CHAPTER 1  Equations and Inequalities

Rounding these two numbers, we find that t < 1.5 and t < 0.5. Since t 5 1 corresponds to March 1, the value of t 5 0.5 corresponds to March 15, and the value t 5 1.5 corresponds to April 15. STEP 5  Check the solution. Look at the figure. The horizontal axis represents the month, and the vertical axis represents the stock price. Estimating when the stock price is approximately $22, we find March 15 and April 15.

EXAMPLE 13  Pythagorean Theorem

 itachi makes a 60-inch HDTV that has a 60-inch diagonal. If the width of the H screen is approximately 52 inches, what is the approximate height of the screen? Solution: STEP 1  Identify



[ CONCEPT CHECK ]

the question. What is the approximate height of the HDTV screen?

STEP 2  Make

notes.

(B) height (C) width (D) temperature

▼ ANSWER (D) temperature. All others are defined as positive.

S TU DY T IP Dimensions such as length and width are distances, which are defined as positive quantities. Although the mathematics may yield both positive and negative values, the negative values are excluded.

hes

?

Which of the following would we NOT eliminate negative values for physical answers: (A) distance

52 inches nc 0i

6 STEP 3  Set



up an equation. Recall the Pythagorean theorem. Substitute in the known values.

a2 1 b2 5 c2 h2 1 522 5 602

STEP 4  Solve the equation. Simplify the constants. h2 1 2704 5 3600 Subtract 2704 from both sides. h2 5 896 h 5 6"896 < 630 Solve using the square root method. Distance is positive, so the negative value is eliminated. The height is approximately  30 inches . STEP 5  Check

the solution.

302 1 522 0 602 900 1 2704 0 3600 3604 < 3600

[S EC T I O N 1. 3]     S U M M A RY The four methods for solving quadratic equations

Formula and completing the square work for all quadratic equations can yield three types of solutions: two distinct real ax 2 1 bx 1 c 5 0 a20 roots, one real root (repeated), or two complex roots (conjugates are factoring, the square root method, completing the square, and of each other). the Quadratic Formula. Factoring and the square root method are 2b 6 "b2 2 4ac Quadratic Formula:    x 5 the quickest and easiest but cannot always be used. The Quadratic 2a

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1.3  Quadratic Equations 

117

[S EC T I O N 1. 3]   E X E R C I S E S • SKILLS In Exercises 1–22, solve by factoring. 1. x 2 2 5x 1 6 5 0

2. v 2 1 7v 1 6 5 0

3. p 2 2 8p 1 15 5 0

4. u2 2 2u 2 24 5 0

5. x 2 5 12 2 x

6. 11x 5 2x2 1 12

7. 16x 2 1 8x 5 21

8. 3x2 1 10x 2 8 5 0

9. 9y 2 1 1 5 6y

10. 4x 5 4x2 1 1

11. 8y 2 5 16y

12. 3A2 5 212A

13. 9p 2 5 12p 2 4

14. 4u 2 5 20u 2 25

15. x 2 2 9 5 0

16. 16v 2 2 25 5 0

17. x   1x 1 42 5 12

18. 3t 2 2 48 5 0

19. 2p 2 2 50 5 0

20. 5y 2 2 45 5 0

21. 3x 2 5 12

22. 7v 2 5 28

In Exercises 23–34, solve using the square root method. 23. p 2 2 8 5 0

24. y 2 2 72 5 0

25. x 2 1 9 5 0

26. v 2 1 16 5 0

2 27. 1x 2 32  5 36

2 28. 1x 2 12  5 25

2 29. 12x 1 32  5 24

2 30. 14x 2 12  5 216

2 31. 15x 2 22  5 27

2 32. 13x 1 82  5 12

2 33. 11 2 x2  59

In Exercises 35–44, what number should be added to complete the square of each expression? 35. x 2 1 6x 1

40. x 2 2 3 x

36. x 2 2 8x 2

41. x 2 1 5 x

37. x 2 2 12x 4

42. x 2 1 5 x

2 34. 11 2 x2  5 29

38. x 2 1 20x

39. x 2 2 2 x

43. x 2 2 2.4x

44. x 2 1 1.6x

1

In Exercises 45–56, solve by completing the square. 45. x 2 1 2x 5 3

46. y 2 1 8y 2 2 5 0

47. t  2 2 6t 5 25

48. x2 1 10x 5 221

49. y 2 2 4y 1 3 5 0

50. x2 2 7x 1 12 5 0

51. 2p2 1 8p 5 23

52. 2x2 2 4x 1 3 5 0

53. 2x 2 2 7x 1 3 5 0

54. 3x 2 2 5x 2 10 5 0

55.

x2 1 2 2x 5 2 4

56.

t2 2t 5 1 1 50 3 3 6

In Exercises 57–68, solve using the Quadratic Formula. 57. t  2 1 3t 2 1 5 0

58. t  2 1 2t 5 1

59. s 2 1 s 1 1 5 0

60. 2s 2 1 5s 5 22

61. 3x 2 2 3x 2 4 5 0

62. 4x 2 2 2x 5 7

63. x 2 2 2x 1 17 5 0

64. 4m2 1 7m 1 8 5 0

65. 5x 2 1 7x 5 3

66. 3x 2 1 5x 5 211

67. 4 x 2 1 3 x 2 2 5 0

1

2

1

1

2

1

68. 4 x 2 2 3 x 2 3 5 0

In Exercises 69–74, determine whether the discriminant is positive, negative, or zero, and indicate the number and type of root to expect. Do not solve. 69. x 2 2 22x 1 121 5 0

70. x 2 2 28x 1 196 5 0

71. 2y 2 2 30y 1 68 5 0

72. 23y 2 1 27y 1 66 5 0

73. 9x 2 2 7x 1 8 5 0

74. 23x 2 1 5x 2 7 5 0

In Exercises 75–94, solve using any method. 75. v 2 2 8v 5 20

76. v 2 2 8v 5 220

77. t  2 1 5t 2 6 5 0

78. t  2 1 5t 1 6 5 0

2 79. 1x 1 32  5 16

2 80. 1x 1 32  5 216

2 81. 1   p 2 22  5 4p

2 82. 1u 1 52  5 16u

83. 8w 2 1 2w 1 21 5 0

84. 8w2 1 2w 2 21 5 0

2 2 4 1 1 2 2 t 1 t5 88. x 2 1 x 5 3 3 5 2 3 5 5 3 41x 2 22 23 3 92. 541 91. 1 5 y14 y22 x x23 x1x 2 32 87.

Young_AT_6160_ch01_pp082-128.indd 117

85. 3p 2 2 9p 1 1 5 0 89. x 1

12 57 x

93. x 2 2 0.1x 5 0.12

86. 3p2 2 9p 2 1 5 0 90. x 2

10 5 23 x

94. y 2 2 0.5y 5 20.06

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CHAPTER 1  Equations and Inequalities

• A P P L I C AT I O N S 95.  Stock

Value. From November 2014 to November 2015, Amazon.com stock was approximately worth P 5 6.25t2 2 35t 1 336, where P is the price of the stock in dollars, t is months, and t 5 0 corresponds to November 2014. During what months was the stock equal to $310? $725.00 675.00 625.00 575.00

  99. Business/Economics. What is the smallest price increase that

will produce a weekly profit of $460? 100. Business/Economics. What is the smallest price increase that

will produce a weekly profit of $630? In Exercises 101 and 102 refer to the following: An epidemiological study of the spread of the flu in a small city finds that the total number P of people who contracted the flu t days into an outbreak is modeled by the function

525.00 475.00 450.00 425.00 400.00 375.00 350.00 325.00 300.00 275.00

P 5 2t 2 1 13t 1 130

101. Health/Medicine. After approximately how many days will

160 people have contracted the flu? 102. Health/Medicine. After approximately how many days will

172 people have contracted the flu? 103. Environment: Reduce Your Margins, Save a Tree.

Let’s define the usable area of an 8.5-inch by 11-inch piece of paper as the rectangular space between the margins of that piece of paper. Assume the default margins in a word processor in a college’s computer lab are set up to be 1.25 inches wide (top and bottom) and 1 inch wide (left and right). Answer the following questions using this information.

Nov Dec 2015 Feb Mar Apr May Jun Jul Aug Sep Oct Nov Source: www.nasdaq.com/symbol/amzn/stock-chart

Value. From November 2014 to November 2015, Apple stock was approximately worth P 5 20.39t2 1 4.29t 1 120.1, where P is the price of the stock in dollars, t is months, and t 5 0 corresponds to November 2014. During what months was the stock equal to $124?

 96.  Stock



$140.00 135.00 130.00 125.00 120.00 115.00 110.00 105.00 100.00 95.00 90.00 Nov Dec 2015 Feb Mar Apr May Jun Jul Aug Sep Oct Nov Source: www.nasdaq.com/symbol/aapl/stock-chart

In Exercises 97 and 98 refer to the following: Research indicates that monthly profit for Widgets R Us is modeled by the function P 5 2100 1 1 0.2q 2 3 2 q

where P is profit measured in millions of dollars and q is the quantity of widgets produced measured in thousands. 97. Business. Find the break-even point for a month to the nearest unit. 98. Business. Find the production level that produces a monthly profit of $40 million.

In Exercises 99 and 100 refer to the following: In response to economic conditions, a local business explores the effect of a price increase on weekly profit. The function P 5 25 1 x 1 3 2 1 x 2 24 2

models the effect that a price increase of x dollars on a bottle of

wine will have on the profit P measured in dollars.

Young_AT_6160_ch01_pp082-128.indd 118

1#t#6

a. Determine the amount of usable space, in square inches, on one side of an 8.5-inch by 11-inch piece of paper with the default margins of 1.25-inch and 1-inch. b.  The Green Falcons, a campus environmental club, has convinced their college’s computer lab to reduce the default margins in their word-processing software by x inches. Create and simplify the quadratic expression that represents the new usable area, in square inches, of one side of an 8.5-inch by 11-inch piece of paper if the default margins at the computer lab are each reduced by x inches.



c. Subtract the usable space in part (a) from the expression in part (b). Explain what this difference represents.



d. If 10 pages are printed using the new margins and as a result the computer lab saved one whole sheet of paper, then by how much did the computer lab reduce the margins? Round to the nearest tenth of an inch.

104. Environment: Reduce Your Margins, Save a Tree. Repeat

Exercise 103 assuming the computer lab’s default margins are 1 inch all the way around (left, right, top, and bottom). If 15 pages are printed using the new margins and as a result the computer lab saved one whole sheet of paper, then by how much did the computer lab reduce the margins? Round to the nearest tenth of an inch. 105. Television. A standard 32-inch television has a 32-inch diagonal and a 25-inch width. What is the height of the 32-inch television? 106. Television. A 42-inch LCD television has a 42-inch

diagonal and a 20-inch height. What is the width of the 42-inch LCD television? 107. Numbers. Find two consecutive numbers such that their sum

is 35 and their product is 306. 108. Numbers. Find two consecutive odd integers such that their

sum is 24 and their product is 143.

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1.3  Quadratic Equations 

109. Geometry. The area of a rectangle is 135 square feet. The

119

corner and folding up the edges will yield an open box (assuming these edges are taped together). If the desired volume of the box is 9 cubic feet, what are the dimensions of the original square piece of cardboard? 118. Volume. A rectangular piece of cardboard whose length is twice its width is used to construct an open box. Cutting a 1-foot by 1-foot square off of each corner and folding up the edges will yield an open box. If the desired volume is 12 cubic feet, what are the dimensions of the original rectangular piece of c­ ardboard? 119. Gardening. A landscaper has planted a rectangular garden that measures 8 feet by 5 feet. He has ordered 1 cubic yard (27 cubic feet) of stones for a border along the outside of the ­garden. If the border needs to be 4 inches deep and he wants to use all of the stones, how wide should the border be? 120. Gardening. A gardener has planted a semicircular rose ­garden with a radius of 6 feet, and 2 cubic yards of mulch (1 cubic yard 5 27 cubic feet) are being delivered. Assuming she uses all of the mulch, how deep will the layer of mulch be? 121. Work. Lindsay and Kimmie, working together, can balance the financials for the Kappa Kappa Gamma sorority in 6 days. Lindsay by herself can complete the job in 5 days less than Kimmie. How long will it take Lindsay to complete the job by herself? 122. Work. When Jack cleans the house, it takes him 4 hours. When Ryan cleans the house, it takes him 6 hours. How long would it take both of them if they worked together?

width is 6 feet less than the length. Find the dimensions of the ­rectangle. 110. Geometry. A rectangle has an area of 31.5 square meters. If the length is 2 more than twice the width, find the dimensions of the rectangle. 111. Geometry. A triangle has a height that is 2 more than 3 times the base and an area of 60 square units. Find the base and height. 112. Geometry. A square’s side is increased by 3 yards, which corresponds to an increase in the area by 69 square yards. How many yards is the side of the initial square? 113. Falling Objects. If a person drops a water balloon off the rooftop of a 100-foot building, the height of the water balloon is given by the equation h 5 216t2 1 100, where t is in seconds. When will the water balloon hit the ground? 114. Falling Objects. If the person in Exercise 113 throws the water balloon downward with a speed of 5 feet per second, the height of the water balloon is given by the equation h 5 216t2 2 5t 1 100, where t is in seconds. When will the water balloon hit the ground? 115. Gardening. A square garden has an area of 900 square feet. If a sprinkler (with a circular pattern) is placed in the center of the garden, what is the minimum radius of spray the sprinkler would need in order to water all of the garden? 116. Sports. A baseball diamond is a square. The distance from base to base is 90 feet. What is the distance from home plate to second base? 117. Volume. A flat square piece of cardboard is used to construct an open box. Cutting a 1-foot by 1-foot square off of each

• C AT C H T H E M I S TA K E In Exercises 123–126, explain the mistake that is made. 123. t2 2 5t 2 6 5 0

1t 2 32 1t 2 22 5 0 t 5 2, 3

124. 12y 2 322 5 25

125. 16a2 1 9 5 0







  2y 2 3 5 5  2y 5 8   y 5 54   y54



2

16a 5 29 9 a2 5 216

9 a 5 6"16

a 5 6 43



126.

2x 2 2 4x 5 3 2 1 x 2 2 2x 2 5 3 2 1 x 2 2 2x 1 1 2 5 3 1 1 21x 2 122 5 4 1x 2 122 5 2

x 2 1 5 6"2

x 5 1 6 !2

• CONCEPTUAL In Exercises 127–130, determine whether the following ­statements are true or false. 2 127. The equation 13x 1 12  5 16 has the same solution set as the equation 3x 1 1 5 4. 128. The quadratic equation ax2 1 bx 1 c 5 0 can be solved by

the square root method only if b 5 0. 129. All quadratic equations can be solved exactly. 130. The Quadratic Formula can be used to solve any quadratic

equation.

Young_AT_6160_ch01_pp082-128.indd 119

131. Write a quadratic equation in general form that has x 5 a as

a repeated real root. 132. Write a quadratic equation in general form that has x 5 bi as

a root. 133. Write a quadratic equation in general form that has the

solution set 52, 56.

134. Write a quadratic equation in general form that has the

solution set 523, 06.

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CHAPTER 1  Equations and Inequalities

In Exercises 135–138, solve for the indicated variable in terms of other variables. 135. Solve s 5

1 2 2 gt

for t. 2 136. Solve A 5 P 11 1 r2  for r. 2 2 2 137. Solve a 1 b 5 c for c. 138. Solve P 5 EI 2 RI 2 for I.

139. Solve the equation by factoring: x4 2 4x2 5 0. 140. Solve the equation by factoring: 3x 2 6x2 5 0. 141. Solve the equation using factoring by grouping:

x3 1 x2 2 4x 2 4 5 0. 142. Solve the equation using factoring by grouping: x3 1 2x2 2 x 2 2 5 0.

• CHALLENGE 143. Show that the sum of the roots of a quadratic equation is

b equal to 2 . a 144. Show that the product of the roots of a quadratic equation is c equal to . a 145. Write a quadratic equation in general form whose solution set is U 3 1 !5, 3 2 !5 V . 146. Write a quadratic equation in general form whose solution set

is 52 2 i, 2 1 i6. 147. Aviation. An airplane takes 1 hour longer to go a distance of 600 miles flying against a headwind than on the return trip with a tailwind. If the speed of the wind is 50 miles per hour, find the speed of the plane in still air. 148. Boating. A speedboat takes 1 hour longer to go 24 miles up a river than to return. If the boat cruises at 10 miles per hour in still water, what is the rate of the current?

149. Find a quadratic equation whose two distinct real roots are

the negatives of the two distinct real roots of the equation ax2 1 bx 1 c 5 0. 150. Find a quadratic equation whose two distinct real roots are the reciprocals of the two distinct real roots of the equation ax2 1 bx 1 c 5 0. 151. A small jet and a 757 leave Atlanta at 1 p.m. The small jet is traveling due west. The 757 is traveling due south. The speed of the 757 is 100 miles per hour faster than the small jet. At 3 p.m. the planes are 1000 miles apart. Find the average speed of each plane. 152. Two boats leave Key West at noon. The smaller boat is traveling due west. The larger boat is traveling due south. The speed of the larger boat is 10 miles per hour faster than the speed of the smaller boat. At 3 p.m. the boats are 150 miles apart. Find the average speed of each boat.

• TECHNOLOGY 153. Solve the equation x2 2 x 5 2 by first writing it in standard

155. a. Solve the equation x2 2 2x 5 b, b 5 8 by first writing it in

form and then factoring. Now plot both sides of the equation in the same viewing screen 1 y1 5 x2 2 x and y2 5 22. At what x-values do these two graphs intersect? Do those points agree with the solution set you found? 154. Solve the equation x2 2 2x 5 22 by first writing it in standard form and then using the quadratic formula. Now plot both sides of the equation in the same viewing screen 1 y1 5 x2 2 2x and y2 5 222. Do these graphs intersect? Does this agree with the solution set you found?

standard form. Now plot both sides of the equation in the same viewing screen 1 y1 5 x2 2 2x and y2 5 b2. At what x values do these two graphs intersect? Do those points agree with the solution set you found? b. Repeat part (a) for b 5 23, 21, 0, and 5. 156. a. Solve the equation x2 1 2x 5 b, b 5 8 by first writing it in standard form. Now plot both sides of the equation in the same viewing screen 1 y1 5 x2 1 2x and y2 5 b2. At what x values do these two graphs intersect? Do those points agree with the solution set you found? b. Repeat part (a) for b 5 23, 21, 0, and 5.

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121

1.4 OTHER TYPES OF EQUATIONS SKILLS OBJECTIVES ■■ Solve radical equations. ■■ Solve equations that are quadratic in form using u-substitutions. ■■ Solve equations that are factorable.

CONCEPTUAL OBJECTIVES ■■ Check for extraneous solutions. ■■ Recognize the u-substitution required to transform the equation into a simpler quadratic equation. ■■ Recognize when a polynomial equation or an equation with rational exponents can be factored either by grouping or by first factoring out a greatest common factor.

1.4.1  Radical Equations Radical equations are equations in which the variable is inside a radical (that is, under a square root, cube root, or higher root). Examples of radical equations follow. !x 2 3 5 2    !2x 1 3 5 x    !x 1 2 1 !7x 1 2 5 6

Until now your experience has been with linear and quadratic equations. Often you can transform a radical equation into a simple linear or quadratic equation. Sometimes the transformation process yields extraneous solutions, or apparent solutions that may solve the transformed problem but are not solutions of the original radical equation. Therefore, it is very important to check your answers.

1.4.1 S K I L L

Solve radical equations. 1.4.1 C O N C E P T U A L

Check for extraneous solutions.

EXAMPLE 1  Solving an Equation Involving a Radical

Solve the equation !x 2 3 5 2. Solution:

Square both sides of the equation. Simplify. Solve the resulting linear equation. Check:  !7 2 3 5 !4 5 2

▼

1 !x 2 3 2 2 5 22 x2354 x57

The solution set is 576.

Y O U R T U R N   Solve the equation !3p 1 4 5 5.

▼ ANSWER

p 5 7 or 576

STUD Y T I P

When both sides of an equation are squared, extraneous solutions can arise. For example, take the equation x52 If we square both sides of this equation, then the resulting equation, x2 5 4, has two ­solutions: x 5 22 and x 5 2. Notice that the value x 5 22 is not in the solution set of the original equation x 5 2. Therefore, we say that x 5 22 is an extraneous solution. In solving a radical equation, we square both sides of the equation and then solve the resulting equation. The solutions to the resulting equation can sometimes be extraneous in that they do not satisfy the original radical equation.

Extraneous solutions are common when we deal with radical equations, so remember to check your answers.

EXAMPLE 2  Solving an Equation Involving a Radical

Solve the equation !2x 1 3 5 x. Solution:

Square both sides of the equation. A!2x 1 3B 2 5 x 2 Simplify. 2x 1 3 5 x2 Write the quadratic equation in standard form. x2 2 2x 2 3 5 0 Factor. 1x 2 32 1x 1 12 5 0 Use the zero product property. x 5 3  or  x 5 21

Young_AT_6160_ch01_pp082-128.indd 121

[ CONCEPT CHECK ] Check the extraneous solution x 5 21 and explain why it does not satisfy the original equation.

▼ ANSWER !21212 1 3 0 21 !1 0 21 x 5 21 is extraneous.

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▼ ANSWER

t 5 4 or 546

▼ ANSWER

x 5 21 and x 5 23 or 523, 216

Check these values to see whether they both make the original equation statement true. x 5 3: !2 1 3 2 1 3 5 3 1 !6 1 3 5 3 1 !9 5 3 1 3 5 3    3 x 5 21:  !2 1212 1 3 5 21 1 !22 1 3 5 21 1 !1 5 21 1 1 2 21    X The solution is x 5 3 . The solution set is 536.

▼

Y O U R T U R N   Solve the equation !12 1 t 5 t.

Y O U R T U R N   Solve the equation !2x 1 6 5 x 1 3.

What happened in Example 2? When we transformed the radical equation into a quadratic equation, we created an extraneous solution, x 5 21, a solution that appears to solve the original equation but does not. When solving radical equations, answers must be checked to avoid including extraneous solutions in the solution set. EXAMPLE 3  Solving an Equation That Involves a Radical

Solve the equation 4x 2 2 !x 1 3 5 210. Solution:

Subtract 4x from both sides.

22 !x 1 3 5 210 2 4x !x 1 3 5 2x 1 5

Divide both sides by 22. Square both sides.

v

x 1 3 5 12x 1 52 12x 1 52



12x 1 522

x 1 3 5 4x2 1 20x 1 25

Eliminate the parentheses.

Rewrite the quadratic equation in standard form. 4x2 1 19x 1 22 5 0 Factor. 14x 1 112 1x 1 22 5 0 Solve.

▼ ANSWER

x 5 21 or 5216

x52

11 and x 5 22 4

11 The apparent solutions are 211 4 and 22. Note that 2 4 does not satisfy the original equation; therefore it is extraneous. The solution is x 522 . The solution set is 5226.

▼

Y O U R T U R N   Solve the equation 2x 2 4 !x 1 2 5 26.

common mistake ✓C O R R E C T

✖INCORRECT

Square the expression.

Square the expression. 2

A3 1 !x 1 2B Write the square as a product of two factors. A3 1 !x 1 2B A3 1 !x 1 2B

Use the FOIL method.

9 1 6!x 1 2 1 1 x 1 2 2

2

A3 1 !x 1 2B The error occurs here when only individual terms are squared. 2 9 1 1x 1 22

In Examples 1 through 3 each equation only contained one radical each. The next example contains two radicals. Our technique will be to isolate one radical on one side of the equation with the other radical on the other side of the equation.

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1.4  Other Types of Equations 

123

EXAMPLE 4  Solving an Equation with More Than One Radical

Solve the equation !x 1 2 1 !7x 1 2 5 6. Solution:

Subtract !x 1 2 from both sides. Square both sides.

Simplify. Multiply the expressions on the right side of the equation. I solate the term with the radical on the left side.

!7x 1 2 5 6 2 !x 1 2 2

A!7x 1 2B 5 A6 2 !x 1 2B

2

7x 1 2 5 A6 2 !x 1 2B A6 2 !x 1 2B 7x 1 2 5 36 2 12!x 1 2 1 1 x 1 2 2

12 !x 1 2 5 36 1 x 1 2 2 7x 2 2

Combine like terms on the right side. 12 !x 1 2 5 36 2 6x Divide by 6.

Square both sides.

2 !x 1 2 5 6 2 x

41x 1 22 5 16 2 x22

Simplify. 4x 1 8 5 36 2 12x 1 x2 Rewrite the quadratic equation in standard form.

x2 2 16x 1 28 5 0

STUDY T I P

Factor. 1x 2 142 1x 2 22 5 0 Solve.

Remember to check both solutions.

x 5 14  and  x 5 2

The apparent solutions are 2 and 14. Note that x 5 14 does not satisfy the original equation; therefore, it is extraneous. The solution is  x 5 2 . The solution set is 526.

▼

Y O U R T U R N   Solve the equation !x 2 4 5 5 2 !x 1 1.

▼ ANSWER

x 5 8 or 586

PROCEDURE FOR SOLVING RADICAL EQUATIONS

Step 1: Isolate the term with a radical on one side. Step 2: Raise both (entire) sides of the equation to the power that will eliminate this radical, and simplify the equation. Step 3: If a radical remains, repeat Steps 1 and 2. Step 4: Solve the resulting linear or quadratic equation. Step 5: Check the solutions and eliminate any extraneous solutions. Note: If there is more than one radical in the equation, it does not matter which radical is isolated first.

1.4.2  Equations Quadratic in Form: u -Substitution Equations that are higher order or that have fractional powers often can be transformed into a quadratic equation by introducing a u-substitution. When this is the case, we say that equations are quadratic in form. In the following table, the two original equations are quadratic in form because they can be transformed into a quadratic equation given the correct substitution. ORIGINAL EQUATION

SUBSTITUTION

NEW EQUATION

x 4 2 3x2 2 4 5 0

u 5 x2

u2 2 3u 2 4 5 0

t 2/3 1 2t1/3 1 1 5 0

u 5 t1/3

u2 1 2u 1 1 5 0

u 5 y21/2

2u2 2 u 1 1 5 0

2 1 2 1150 y !y

Young_AT_6160_ch01_pp082-128.indd 123

1.4.2 S K I L L

Solve equations that are quadratic in form using u-substitutions. 1.4.2 C O N C E P T U A L

Recognize the u-substitution required to transform the equation into a simpler quadratic equation.

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For example, the equation x4 2 3x2 2 4 5 0 is a fourth-degree equation in x. How did we know that u 5 x2 would transform the original equation into a quadratic equation? If we rewrite the original equation as 1x2   2 2 2 3 1x2  2 2 4 5 0, the expression in parentheses is the u-substitution. Let us introduce the substitution u 5 x2. Note that squaring both sides implies 2 u  5 x4. We then replace x2 in the original equation with u, and x4 in the original equation with u2, which leads to a quadratic equation in u: u2 2 3u 2 4 5 0. WORDS MATH

Solve for x. x4 2 3x2 2 4 5 0 Introduce u-substitution.   u 5 x2 3Note that u2 5 x4.4 Write the quadratic equation in u. u2 2 3u 2 4 5 0 Factor. 1u 2 42 1u 1 12 5 0 Solve for u.   u 5 4  or  u 5 21 2 x2 5 4  or  x2 5 21 Transform back to x, u 5 x .   Solve for x.  x 5 62  or  x 5 6i The solution set is 562, 6i   6.

It is important to correctly determine the appropriate substitution in order to arrive at an equation quadratic in form. For example, t 2/3 1 2t1/3 1 1 5 0 is an original equation given in the above table. If we rewrite this equation as 1t1/3  2  2 1 2 1t1/3 2 1 1 5 0, then it becomes apparent that the correct substitution is u 5 t1/3, which transforms the equation in t into a quadratic equation in u: u2 1 2u 1 1 5 0. PROCEDURE FOR SOLVING EQUATIONS QUADRATIC IN FORM

Step 1: Identify the substitution. Step 2: Transform the equation into a quadratic equation. Step 3: Solve the quadratic equation. Step 4: Apply the substitution to rewrite the solution in terms of the original variable. Step 5: Solve the resulting equation. Step 6: Check the solutions in the original equation. EXAMPLE 5  S  olving an Equation Quadratic in Form with Negative Exponents

Find the solutions to the equation x22 2 x21 2 12 5 0. Solution:

Rewrite the original equation. 1x2122 2 1x212 2 12 5 0 Determine the u-substitution.   u 5 x21 3Note that u2 5 x22.4 The original equation in x corresponds to a quadratic equation in u.   u2 2 u 2 12 5 0 Factor. 1u 2 42 1u 1 32 5 0 Solve for u. u 5 4  or  u 5 23 The most common mistake is forgetting to transform back to x. x21 5 4  or  x21 5 23 Transform back to x. Let u 5 x21. 1 1 1 Write x 21 as . 5 4 or 5 23 x x x 1 1 Solve for x. ­x 5   or   x 5 2 4 3 ▼ ANSWER

The solution is x 5 2 21 or x 5 13 . The solution set is U 2 21 , 31 V .

Young_AT_6160_ch01_pp082-128.indd 124

▼

The solution set is U231, 14 V .

Y O U R T U R N   Find the solutions to the equation x22 2 x21 2 6 5 0.

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1.4  Other Types of Equations 

EXAMPLE 6  S  olving an Equation Quadratic in Form with Fractional Exponents 2/3

Find the solutions to the equation x

1/3

2 3x

125

STUD Y T I P Remember to transform back to the original variable.

2 10 5 0.

Solution:

1 x 1/3 2 2 2 3x1/3 2 10 5 0

Rewrite the original equation. Identify the substitution as u 5 x1/3.

u2 2 3u 2 10 5 0

1u 2 52 1u 1 22 5 0

Factor. Solve for u.

u 5 5  or   u 5 22

Let u 5 x1/3 again.

x1/3 5 5

Cube both sides of the equations.

1/3 3

x1/3 5 22

1 x 2 5 1523  1 x 1/3 2 3 5 12223

Simplify.

x 5 125

x 5 28

The solution set is  528, 1256 , which a check will confirm.

[ CONCEPT CHECK ] What do we let u be equal to in order to transform the equation 2t 2 5t 1/2 2 3 5 0 into a quadratic equation?

▼

▼

Y O U R T U R N   Find the solution to the equation 2t 2 5t1/2 2 3 5 0.

ANSWER u 5 t1/2

▼ ANSWER

t 5 9 or 596.

1.4.3  Factorable Equations Some equations (both polynomial and with rational exponents) that are factorable can be solved using the zero product property.

1.4.3 S K I L L

Solve equations that are factorable. 1.4.3 C O N C E P T U A L

EXAMPLE 7  S  olving an Equation with Rational Exponents by Factoring

Solve the equation x7/3 2 3x4/3 2 4x1/3 5 0. Solution:

x1/31x2 2 3x 2 42 5 0

Factor the left side of the equation.

x1/31x 2 42 1x 1 12 5 0

Factor the quadratic expression. Apply the zero product property.

Recognize when a polynomial equation or an equation with rational exponents can be factored either by grouping or by first factoring out a greatest common factor.

x1/3 5 0   or   x 2 4 5 0   or   x 1 1 5 0

Solve for x.

x 5 0    or    x 5 4    or    x 5 21 The solution set is 521, 0, 46.

EXAMPLE 8  S  olving a Polynomial Equation Using Factoring by Grouping

Solve the equation x3 1 2x2 2 x 2 2 5 0. Solution:

Factor by grouping (Chapter 0). Identify the common factors. Factor. Factor the quadratic expression.

Young_AT_6160_ch01_pp082-128.indd 125

1 x 3 2 x 2 1 1 2x 2 2 2 2 5 0

x 1 x2 2 1 2 1 2 1 x2 2 1 2 5 0 1x 1 22 1 x 2 2 1 2 5 0

1x 1 22 1x 2 12 1x 1 12 5 0

[ CONCEPT CHECK ] How do we initially group the cubic equation in x 3 1 x 2 2 4x 2 4 5 0?

▼ ANSWER 1x3 2 4x2 1 1x2 2 42 5 0

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Apply the zero product property.

x 1 2 5 0   or   x 2 1 5 0   or   x 1 1 5 0

Solve for x. ▼ ANSWER

x 5 22    or    x 5 1    or    x 5 21 The solution set is 522, 21, 16.

▼

Y O U R T U R N   Solve the equation x3 1 x2 2 4x 2 4 5 0.

x 5 21 or x 5 62 or 522, 21, 26

[S EC T I O N 1. 4]   S U M M A RY Radical equations, equations quadratic in form, and factorable equations can often be solved by transforming them into simpler linear or quadratic equations. ■ Radical Equations: Isolate the term containing a radical and raise it to the appropriate power that will eliminate the radical. If there is more than one radical, it does not matter which radical is isolated first. Raising radical equations to powers may cause extraneous solutions, so check each solution.

■

■

Equations Quadratic in Form: Identify the u-substitution that transforms the equation into a q­ uadratic equa­tion. Solve the quadratic equation and then remember to transform back to the original variable. Factorable Equations: Look for a factor common to all terms or factor by grouping.

[S EC T I O N 1. 4]   E X E R C I S E S • SKILLS In Exercises 1–40, solve the radical equation for the given variable. 1. "t 2 5 5 2

5. "u 1 1 5 24 1/3

9. 14y 1 12

2. "2t 2 7 5 3

13. y 5 5!y

10. 15x 2 12

17. "2x 1 6 5 x 1 3

21. 3x 2 6!x 2 1 5 3

18. "8 2 2x 5 2x 2 2

25. 3!x 1 4 2 2x 5 9

26. 2!x 1 1 2 3x 5 25

5 21

4. 11 5 121 2 p21/2

11. "12 1 x 5 x 15. s 5 3"s 2 2

12. x 5 "56 2 x

19. "1 2 3x 5 x 1 1

20. "2 2 x 5 x 2 2

3 7. "5x 1253

6. 2"3 2 2u 5 9 1/3

3. 14p 2 721/2 5 5

54

y 14. !y 5 4

22. 5x 2 10!x 1 2 5 210

3 8. "1 2 x 5 22

23. 3x 2 6!x 1 2 5 3 2

27. "x 2 4 5 x 2 1

29. "x 2 2 2x 2 5 5 x 1 1

30. "2x 2 2 8x 1 1 5 x 2 3

31. !3x 1 1 2 !6x 2 5 5 1

16. 22s 5 "3 2 s

24. 2x 2 4!x 1 1 5 4 28. "25 2 x 2 5 x 1 1

32. !2 2 x 1 !6 2 5x 5 6

33. !x 1 12 1 !8 2 x 5 6

34. !5 2 x 1 !3x 1 1 5 4

35. "2x 2 1 2 "x 2 1 5 1

36. "8 2 x 5 2 1 "2x 1 3

37. "3x 2 5 5 7 2 "x 1 2

38. "x 1 5 5 1 1 "x 2 2

39. #2 1 "x 5 "x

40. #2 2 "x 5 "x

In Exercises 41–70, solve the equations by introducing a substitution that transforms these equations to quadratic form. 41. x2/3 1 2x1/3 5 0 4

2

42. x1/2 2 2x1/4 5 0 45. 2x 1 7x 1 6 5 0

46. x8 2 17x4 1 16 5 0

2 47. 12x 1 12  1 5 12x 1 12 1 4 5 0

48. 1x 2 322 1 6 1x 2 32 1 8 5 0

49. 4  1t 2 122 2 9 1t 2 12 5 22

56. 2x1/2 1 x1/4 2 1 5 0

5/3 57. 1x 1 32  5 32

58. 1x 1 224/3 5 16

22

53. 3y

21

1y

2450

2/3 59. 1x 1 12  54

Young_AT_6160_ch01_pp082-128.indd 126

2

43. x4 2 3x2 1 2 5 0

44. x 2 8x 1 16 5 0 50. 211 2 y22 1 5 11 2 y2 2 12 5 0

4

51. x28 2 17x24 1 16 5 0 22

54. 5a

21

1 11a

1250

4/3 60. 1x 2 72  5 81

52. 2u22 1 5u21 2 12 5 0 55. z2/5 2 2z1/5 1 1 5 0

61. 6t22/3 2 t21/3 2 1 5 0

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1.4  Other Types of Equations 

62. t22/3 2 t21/3 2 6 5 0

65. a

63. 3 5

2 1 1 b 1 a b 2 12 5 0 2x 2 1 2x 2 1

66.

68. u4/3 1 5u2/3 5 24

1 2 1 1x 1 12 1x 1 122

64.

5 3 52 2 1 2x 1 1 2 1 2x 1 1 2 2

67. u4/3 2 5u2/3 5 24

4 2 16 69. t 5 "t

In Exercises 71–86, solve by factoring.

1 4 1450 1 1x 1 12 1x 1 122

4 2 21 70. u 5 "22u

71. x3 2 x2 2 12x 5 0

72. 2y3 2 11y2 1 12y 5 0

73. 4p3 2 9p 5 0

74. 25x3 5 4x

75. u5 2 16u 5 0

76. t 5 2 81t 5 0

77. x3 2 5x2 2 9x 1 45 5 0

78. 2p3 2 3p2 2 8p 1 12 5 0

79. y  1 y 2 523 2 141 y 2 522 5 0 80. v 1v 1 323 2 40 1v 1 322 5 0 81. x9/4 2 2x5/4 2 3x1/4 5 0 5/3

21/3

83. t  2 25t

9/5

84. 4x

50

21/5

2 9x

50

3/2

85. y

1/2

2 5y

21/2

1 6y

50

82. u7/3 1 u4/3 2 20u1/3 5 0 86. 4p5/3 2 5p2/3 2 6p21/3 5 0

• A P P L I C AT I O N S In Exercises 87 and 88 refer to the following: An analysis of sales indicates that demand for a product during a calendar year is modeled by

94.  Grades. The average combined math and verbal SAT score

of incoming freshmen at a university is given by the equation S 5 1000 1 10!2t, where t is in years and t 5 0 corresponds to 1990. What year will the incoming class have an average SAT score of 1230?

d 5 3!t 1 1 2 0.75t where d is demand in millions of units and t is the month of the year where t 5 0 represents January. 87. Economics. During which month(s) is demand 3 million units? 88. Economics. During which month(s) is demand 4 million units?

 95. Speed of Sound. A man buys a house with an old well but

does not know how deep the well is. To get an estimate he decides to drop a rock in the opening of the well and time how long it takes until he hears the splash. The total elapsed time T given by T 5 t1 1 t2, is the sum of the time it takes for the rock to reach the water, t1, and the time it takes for the sound of the splash to travel to the top of the well, t2. The time (seconds) that it takes for the rock to reach the water is

In Exercises 89 and 90 refer to the following: Body Surface Area (BSA) is used in physiology and medicine for many clinical purposes. BSA can be modeled by the function BSA 5

wh Å 3600



where w is weight in kilograms and h is height in centimeters. 89. Health. The BSA of a 72-kilogram female is 1.8. Find the height of the female to the nearest centimeter. 90. Health. The BSA of a 177-centimeter-tall male is 2.1. Find the weight of the male to the nearest kilogram. 91.  Insurance: Health. Cost for health insurance with a private policy is given by C 5 !10 1 a, where C is the cost per day and a is the insured’s age in years. Health insurance for a 6-year-old, a 5 6, is $4 a day (or $1460 per year). At what age would someone be paying $9 a day (or $3285 per year)? 92.  Insurance: Life. Cost for life insurance is given by C 5 !5a 1 1, where C is the cost per day and a is the insured’s age in years. Life insurance for a newborn, a 5 0, is $1 a day (or $365 per year). At what age would someone be paying $20 a day (or $7300 per year)? 93.  Stock Value. The stock price of a certain pharmaceutical company from August to November can be approximately modeled by the equation P 5 5"t 2 1 1 1 50, where P is the price of the stock in dollars and t is the month with t 5 0 corresponding to August. Assuming this trend continues, when would the stock be worth $85? Stock Price

70 60 50 Sept

Young_AT_6160_ch01_pp082-128.indd 127

Oct

40



"d , where d is the depth of the well in feet. 4 Since the speed of sound is 1100 ft/s, the time (seconds) it d takes for the sound to reach the top of the well is t2 5 . 1100 If the splash is heard after 3 seconds, how deep is the well? given by t1 5

 96. Speed of Sound. If the owner of the house in Exercise 95

forgot to account for the speed of sound, what would he have calculated the depth of the well to be?  97. Physics: Pendulum. The period (T ) of a pendulum is related

to the length (L) of the pendulum and acceleration due to L . If gravity is 9.8 m/s2 Åg and the period is 1 second, find the approximate length of the pendulum. Round to the nearest centimeter. Note: 100 cm 5 1 m.

gravity (g) by the formula T 5 2p

 98. Physics: Pendulum. The period (T ) of a pendulum is related

to the length (L) of the pendulum and acceleration due to



gravity (g) by the formula T 5 2p



L . If gravity is 32 ft/s2 Åg and the period is 1 second, find the approximate length of the pendulum. Round to the nearest inch. Note: 12 in. 5 1 ft.

In Exercises 99 and 100, refer to the following: Einstein’s special theory of relativity states that time is relative: Time speeds up or slows down, depending on how fast one object is moving with respect to another. For example, a space probe traveling at a velocity v near the speed of light c will have “clocked” a time

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t hours, but for a stationary observer on Earth that corresponds to a time t0. The formula governing this relativity is given by t 5 t0

Å

12

v2 c2

 

 99. Physics: Special Theory of Relativity. If the time elapsed on

a space probe mission is 18 years but the time elapsed on Earth

during that mission is 30 years, how fast is the space probe traveling? Give your answer relative to the speed of light. 100. Physics: Special Theory of Relativity. If the time elapsed on a space probe mission is 5 years but the time elapsed on Earth during that mission is 30 years, how fast is the space probe traveling? Give your answer relative to the speed of light.

• C AT C H T H E M I S TA K E In Exercises 101–104, explain the mistake that is made. 101. Solve the equation !3t 1 1 5 24. Solution: 3t 1 1 5 16 3t 5 15 t55 This is incorrect. What mistake was made? 102. Solve the equation x 5 !x 1 2.

Solution: x2 5 x 1 2 x2 2 x 2 2 5 0 1x 2 22 1x 1 12 5 0 x 5 21, x 5 2 This is incorrect. What mistake was made?

103. Solve the equation x2/3 2 x1/3 2 20 5 0.

Solution:   u 5 x 1/3 2 u 2 u 2 20 5 0 1u 2 52 1u 1 42 5 0 x 5 5, x 5 24 This is incorrect. What mistake was made? 104. Solve the equation x 4 2 2x2 5 3. Solution: x 4 2 2x 2 2 3 5 0  u 5 x2 u2 2 2u 2 3 5 0 1u 2 32 1u 1 12 5 0 u 5 21, u 5 3 u 5 x2 x2 5 21, x2 5 3 x 5 61, x 5 63 This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 105–108, determine whether each statement is true or false. 105. The equation 12x 2 126 1 412x 2 123 1 3 5 0 is quadratic in form. 106. The equation t25 1 2t5 1 1 5 0 is quadratic in form.

107. If two solutions are found and one does not check, then the

other does not check. 108. Squaring both sides of !x 1 2 1 !x 5 !x 1 5 leads to x 1 2 1 x 5 x 1 5.

• CHALLENGE 109. Solve "x 2 5 x.

110. Solve "x 2 5 2x.

112. Solve the equation 3x7/12 2 x5/6 2 2x1/3 5 0. 2

2

111. Solve the equation 3x 1 2x 5 "3x 1 2x without squar-

ing both sides.

113. Solve the equation !x 1 6 1 !11 1 x 5 5!3 1 x. 4

3 114. Solve the equation # 2x" x !x 5 2.

• TECHNOLOGY 115. Solve the equation !x 2 3 5 4 2 !x 1 2. Plot both sides

of the equation in the same viewing screen, y1 5 !x 2 3 and y2 5 4 2 !x 1 2, and zoom in on the x-coordinate of the point of intersection. Does the graph agree with your solution? 116. Solve the equation 2!x 1 1 5 1 1 !3 2 x. Plot both sides of the equation in the same viewing screen, y1 5 2!x 1 1 and y2 5 1 1 !3 2 x, and zoom in on the x-coordinate of the points of intersection. Does the graph agree with your solution? 117. Solve the equation 24 5 !x 1 3. Plot both sides of the equation in the same viewing screen, y1 5 24 and y2 5 !x 1 3. Does the graph agree or disagree with your solution? 118. Solve the equation x1/4 5 24x1/2 1 21. Plot both sides of the equation in the same viewing screen, y1 5 x1/4 and

Young_AT_6160_ch01_pp082-128.indd 128

y2 5 24x1/2 1 21. Does the point(s) of intersection agree with your solution? 119. Solve the equation ­ x 1/2 5 24x 1/4 1 21. Plot both sides of the equation in the same viewing screen, y1 5 x 1/2 and y2 5 24x 1/4 1 21. Does the point(s) of intersection agree with your solution? 120. Solve the equation x 21 5 3x 22 2 10. Plot both sides of the equation in the same viewing screen, y1 5 x 21 and y2 5 3x 22 2 10. Does the point(s) of intersection agree with your solution? 121. Solve the equation x 22 5 3x 21 2 10. Plot both sides of the equation in the same viewing screen, y1 5 x 22 and y2 5 3x 21 2 10. Does the point(s) of intersection agree with your solution?

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129

1.5  Linear Inequalities 

1.5 LINEAR INEQUALITIES SKILLS OBJECTIVES ■■ Use interval notation. ■■ Solve linear inequalities in one variable.

CONCEPTUAL OBJECTIVES ■■ Apply intersection and union concepts. ■■ Understand that a linear inequality in one variable has an interval solution.

1.5.1  Graphing Inequalities and Interval Notation An example of a linear equation is 3x 2 2 5 7, whereas 3x 2 2 # 7 is an example of a linear inequality. One difference between a linear equation and a linear inequality is that the equation has at most only one solution, or value of x, that makes the statement true, whereas the inequality can have a range or continuum of numbers that make the statement true. For example, the inequality x # 4 denotes all real numbers x that are less than or equal to 4. Four inequality symbols are used. SYMBOL

1.5.1 S K I L L

Use interval notation. 1.5.1 C O N C E P T U A L

Apply intersection and union concepts.

IN WORDS

Less than Greater than Less than or equal to Greater than or equal to

, . # $

We call , and . strict inequalities. For any two real numbers a and b, one of three things must be true: a,b

or

a5b

or

a.b

This property is called the trichotomy property of real numbers. If x is less than 5 1x , 52 and x is greater than or equal to 22 1x $ 222, then we can represent this as a double (or combined) inequality, 22 # x , 5, which means that x is greater than or equal to 22 and less than 5. We will express solutions to inequalities in four ways: an inequality, a solution set, an interval, and a graph. The following are ways of expressing all real numbers greater than or equal to a and less than b. Inequality Notation

Solution Set

Interval Notation

a#x,b

5x  a # x , b6

3a, b2

In this example, a is referred to as the left endpoint and b is referred to as the right endpoint. If an inequality is a strict inequality 1, or .2, then the graph and interval notation use parentheses. If it includes an endpoint 1$ or #2, then the graph and interval notation use brackets. Number lines are drawn with either closed/open circles or brackets/parentheses. In this text, the brackets/parentheses notation will be used. Intervals are classified as follows: Open 1 , 2    Closed 3 , 4    Half open 1 , 4 or 3 , 2 LET x BE A REAL NUMBER. x IS...

INEQUALITY

SET NOTATION

INTERVAL

greater than a and less than b

a,x,b

5x  a , x , b6

1a, b2

greater than or equal to a and less than b

a#x,b

5x  a # x , b6

3a, b2

Young_AT_6160_ch01_pp129-163.indd 129

GRAPH

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CHAPTER 1  Equations and Inequalities

LET x BE A REAL NUMBER. x IS...

SET NOTATION

INEQUALITY

INTERVAL

greater than a and less than or equal to b

a,x#b

5x  a , x # b6

1a, b4

greater than or equal to a and less than or equal to b

a#x#b

5x  a # x # b6

3a, b4

less than a

x,a

less than or equal to a

x#a

5x  x , a6

12q, a2

greater than b

x.b

greater than or equal to b

x$b

5x  x . b6

1b, q2

R

R

all real numbers

5x  x # a6 5x  x $ b6

12q, a4 3b, q2

12q, q2

1. Infinity 1q2 is not a number. It is a symbol that means continuing indefinitely to the right on the number line. Similarly, negative infinity 12q2 means continuing indefinitely to the left on the number line. Since both are unbounded, we use a parenthesis, never a bracket. 2. In interval notation, the lower number is always written to the left. Write the inequality in interval notation: 21 # x , 3. ✓C O R R E C T

321, 32

✖INCORRECT

GRAPH

1

3

13, 214

EXAMPLE 1  E  xpressing Inequalities Using Interval Notation and a Graph

Express the following as an inequality, an interval, and a graph. a. x is greater than 23. b. x is less than or equal to 5. c. x is greater than or equal to 21 and less than 4. d. x is greater than or equal to 0 and less than or equal to 4. Solution:

Inequality

Interval

a. x . 23

123, q2

b. x # 5 c.  21 # x , 4 d. 0 # x # 4

Graph

12q, 54 321, 42 30, 44

Since the solutions to inequalities are sets of real numbers, it is useful to discuss two operations on sets called intersection and union.

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1.5  Linear Inequalities 

DEFINITION

131

Union and Intersection

The union of sets A and B, denoted A ∪ B, is the set formed by combining all the elements in A with all the elements in B. A ∪ B 5 5 x 0 x is in A or B or both 6

The intersection of sets A and B, denoted A ∩ B, is the set formed by the elements that are in both A and B. A ∩ B 5 5 x 0 x is in A and B 6

The notation “x  x is in” is read “all x such that x is in.” The vertical line represents “such that.”

As an example of intersection and union, consider the following sets of people:    A 5 5Austin, Brittany, Jonathan6    B 5 5Anthony, Brittany, Elise6

Intersection:  A ∩ B 5 5Brittany6

 Union: A ∪ B 5 5Anthony, Austin, Brittany, Elise, Jonathan6

EXAMPLE 2  D  etermining Unions and Intersections: Intervals and Graphs

If A 5 323, 24 and B 5 11, 72, determine A ∪ B and A ∩ B. Write these sets in interval notation, and graph. Solution:

Set A B A∪B A∩B

▼

Interval notation

Graph

323, 24

11, 72

323, 72

11, 24

Y O U R T U R N   If C 5 323, 32 and D 5 10, 54, find C ∪ D and C ∩ D. Express the

[ CONCEPT CHECK ] If A is the set of all of the students who are enrolled in a math class and B is the set of all students who are enrolled in a history class, then which set is larger? (A) the intersection of A and B or (B) the union of A and B? Assume that the two classes do not contain exactly the same students.

▼ ANSWER (B) the set of students who are enrolled in either a math class or a history class. (A) is the smaller set because it is all of the students who are enrolled in BOTH math and history.

▼ ANSWER

intersection and union in interval notation, and graph.

1.5.2  Solving Linear Inequalities As mentioned at the beginning of this section, if we were to solve the equation 3x 2 2 5 7, we would add 2 to both sides, divide by 3, and find that x 5 3 is the solution, the only value that makes the equation true. If we were to solve the linear inequality 3x 2 2 # 7, we would follow the same procedure: add 2 to both sides, divide by 3, and find that x # 3, which is an interval or range of numbers that make the inequality true. In solving linear inequalities, we follow the same procedures that we used in solving ­linear equations with one general exception: if you multiply or divide an inequality by a negative number, then you must change the direction of the inequality sign. For example, if 22 x , 210, then the solution set includes real numbers such as x 5 6 and x 5 7. Note that real numbers such as x 5 26 and x 5 27 are not included in the solution set. Therefore, when this inequality is divided by 22, the inequality sign must also be reversed: x . 5. If a , b, then ac , bc if c . 0 and ac . bc if c , 0.

Young_AT_6160_ch01_pp129-163.indd 131

1.5.2 S K I L L

Solve linear inequalities in one variable. 1.5.2 C O N C E P T U A L

Understand that a linear inequality in one variable has an interval solution.

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CHAPTER 1  Equations and Inequalities

ST U DY TIP If you multiply or divide an inequality by a negative number, remember to change the direction of the inequality sign.

The most common mistake that occurs when solving an inequality is forgetting to change the direction of, or reverse, the inequality symbol when the inequality is multiplied or divided by a negative number.

INEQUALITY PROPERTIES

Procedures That Do Not Change the Inequality Sign 1. Simplifying by eliminating parentheses    31x 2 62 , 6x 2 x and collecting like terms.            3x 2 18 , 5x 2. Adding or subtracting the same          7x 1 8 $ 29 quantity on both sides.                  7x $ 21 3. Multiplying or dividing by the           5x # 15 same positive real number.           x # 3 Procedures That Change (Reverse) the Inequality Sign 1. Interchanging the two sides of the inequality. 2. Multiplying or dividing by the same negative real number.

x # 4 is equivalent to 4 $ x. 25x # 15 is equivalent to x $ 23.

EXAMPLE 3  Solving a Linear Inequality

Solve and graph the inequality 5 2 3x , 23. Solution:

Write the original inequality. Subtract 5 from both sides. Divide both sides by 23 and reverse the inequality sign.

5 2 3x , 23 23x , 18 23x 18 . 23 23

Simplify.

▼ ANSWER

Solution set: 5x | x # 216   Interval notation: 12q, 214

x . 26

Solution set: 5x | x . 266    Interval notation: 126, q2

Graph:

▼

Y O U R T U R N   Solve the inequality 5 # 3 2 2x. Express the solution in set and

interval notation, and graph.

EXAMPLE 4  Solving Linear Inequalities with Fractions

Solve the inequality

5x 4 1 3x # . 3 2

common mistake A common mistake is using cross multiplication to solve inequalities. Cross multiplication should not be used because the expression by which you are multiplying might be negative for some values of x, and that would require the direction of the inequality sign to be reversed.

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1.5  Linear Inequalities 

▼ CAUTION

✓C O R R E C T

✖INCORRECT

Eliminate the fractions by multiplying by the LCD, 6.

Cross multiply. 314 1 3x2 # 215x2

Simplify.

133

Cross multiplication should not be used in solving inequalities.

The error is in cross multiplying.

4 1 3x 5x 6a b # 6a b 3 2 10x # 3 1 4 1 3x 2

Eliminate the parentheses.

10x # 12 1 9x Subtract 9x from both sides. x # 12

Although it is not possible to “check” inequalities since the solutions are often intervals, it is possible to confirm that some points that lie in your solution do satisfy the inequality. It is important to remember that cross multiplication cannot be used in solving inequalities.

EXAMPLE 5  Solving a Double Linear Inequality

Solve the inequality 22 , 3x 1 4 # 16.

f

Solution:

f

This double inequality can be written as two inequalities.   22 , 3x 1 4 # 16 Both inequalities must be satisfied.

22 * 3x 1 4 and 3x 1 4 " 16

Subtract 4 from both sides of each inequality.

26 * 3x

and

3x " 12

Divide each inequality by 3.

22 * x

and

x"4

Combining these two inequalities gives us 22 * x " 4 in inequality notation; in interval notation we have 122, H2 x 12H, 44 or 122, 44.

Notice that the steps we took in solving these inequalities individually were identical. This leads us to a shortcut method in which we solve them together: Write the combined inequality.

22 , 3x 1 4 # 16

Subtract 4 from each part. Divide each part by 3.

26 , 3x # 12 22 , x # 4

Interval notation: 122, 44

For the remainder of this section we will use the shortcut method for solving inequalities.

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CHAPTER 1  Equations and Inequalities

EXAMPLE 6  Solving a Double Linear Inequality

Solve the inequality 1 # notation, and graph.

22 2 3x , 4. Express the solution set in interval 7

Solution:

22 2 3x ,4 7

Write the original double inequality.

1#

Multiply each part by 7.

7 # 22 2 3x , 28

Add 2 to each part.

9 # 23x , 30

Divide each part by 23 and reverse the inequality signs.

23 $ x . 210

Write in standard form.

210 , x # 23



Interval notation: 1210, 234   Graph:

[ CONCEPT CHECK ] True or False: When expressing the solution to a linear equation or a linear inequality in one variable on a number line, a linear equation in one variable has a solution that is a single point, whereas a linear inequality in one variable has a solution that corresponds to a range or interval.

▼ ANSWER True

▼ ANSWER

EXAMPLE 7  Solving a Double Linear Inequality

Solve the inequality x 2 1 # 4x 2 4 # x 1 8. Express the solution in interval notation. Solution:

Subtract x from all three parts.

21 # 3x 2 4 # 8

Add 4 to all three parts.

3 # 3x # 12

Divide all three parts by 3.

1#x#4

Express the solution in interval notation.

▼

31, 44

Y O U R T U R N   Solve the inequality 2x 1 1 , 4x 1 2 , 2x 1 5. Express the

solution in interval notation.

A212 , 32 B

Applications Involving Linear Inequalities EXAMPLE 8  Temperature Ranges

New York City on average has a yearly temperature range of 23 degrees Fahrenheit to 86 degrees Fahrenheit. What is the range in degrees Celsius given that the conversion relation is F 5 32 1 59 C? Solution:

The temperature ranges from 238F to 868F.

 23 # F # 86

9 23 # 32 1 C # 86 5 9 Subtract 32 from all three parts. 29 # C # 54 5 5 Multiply all three parts by 9. 25 # C # 30 New York City has an average yearly temperature range of 258C to 308C . Replace F using the Celsius conversion.

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1.5  Linear Inequalities 

135

EXAMPLE 9  Comparative Shopping

Two car rental companies have advertised weekly specials on full-size cars. Hertz is advertising an $80 rental fee plus an additional $0.10 per mile. Thrifty is advertising $60 and $0.20 per mile. How many miles must you drive for the rental car from Hertz to be the better deal? Solution:

Let x 5 number of miles driven during the week. Write the cost for the Hertz rental.

80 1 0.1x

Write the cost for the Thrifty rental.

60 1 0.2x

Write the inequality if Hertz is less than Thrifty.

80 1 0.1x , 60 1 0.2x

Subtract 0.1x from both sides.

80 , 60 1 0.1x

Subtract 60 from both sides.

20 , 0.1x

Divide both sides by 0.1.

200 , x

You must drive more than 200 miles for Hertz to be the better deal.

[ S E C T I O N 1 . 5 ]     S U M M A R Y The solutions of linear inequalities are solution sets that can be Linear inequalities are solved using the same procedures as linear expressed four ways: equations with one exception: When you multiply or divide by a negative number, you must reverse the inequality sign. 1. Inequality notation a , x # b 2. Set notation 5x | a , x # b6 3. Interval notation 1a, b4 Note: Cross multiplication cannot be used with inequalities. 4. Graph (number line)

[S EC T I O N 1. 5]   E X E R C I S E S • SKILLS In Exercises 1–16, rewrite in interval notation and graph. 1. x $ 3

2. x , 22

3. x # 25

4. x . 27

5. 22 # x , 3

6. 24 # x # 21

7. 23 , x # 5

8. 0 , x , 6

9. 0 # x # 0

10. 27 # x # 27

11. x # 6 and x $ 4

12. x . 23 and x # 2

13. x # 26 and x $ 28

14. x , 8 and x , 2

15. x . 4 and x # 22

16. x $ 25 and x , 26

19. 127, 222

20. 323, 24

In Exercises 17–24, rewrite in set notation. 17. 30, 22

18. 10, 34

21. 12q, 64

22. 15, q2

In Exercises 25–32, write in inequality and interval notation.

24. 34, 44

23. 12q, q2

25. 26. 10

0

5

5

10

27. 28. 3

Young_AT_6160_ch01_pp129-163.indd 135

5

0

1 4

1 8

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CHAPTER 1  Equations and Inequalities

29.

30. 0

5

31.



5

10

8

0

5

32.

5

10

2

In Exercises 33–50, graph the indicated set and write as a single interval, if possible. 33. 1 25, 2 4 ∪ 1 21, 3 2

34. 1 2, 7 2 ∪ 3 25, 3 2

35. 3 26, 4 2 ∪ 3 22, 5 2

36. 3 23, 1 2 ∪ 3 26, 0 2

41. 3 25, 2 2 ∩ 3 21, 3 4

42. 3 24, 5 2 ∩ 3 22, 7 2

43. 1 2q, 4 2 ∪ 1 4, q 2

44. 1 2q, 23 4 ∪ 3 23, q 2

37. 1 2q, 1 4 ∩ 3 21, q 2

39. 1 2q, 4 2 ∩ 3 1, q 2

38. 1 2q, 25 2 ∩ 1 2q, 7 4

45. 1 2q, 23 4 ∪ 3 3, q 2

46. 1 22, 2 2 ∩ 3 23, 1 4

49. 1 26, 22 2 ∩ 3 1, 4 2

40. 1 23, q 2 ∩ 3 25, q 2

47. 1 2q, q 2 ∩ 1 23, 2 4

50. 1 2q, 22 2 ∩ 1 21, q 2

48. 1 2q, q 2 ∪ 1 24, 7 2

In Exercises 51–58, write in interval notation. 51.



2

53.

3

2

4

52.

5

54.

5

0

5

12

5

2

2

56.

55.

4

2

3

6

3

0



7

2

0

2

58.

57.



4

5

1

In Exercises 59–90, solve and express the solution in interval notation. 59. x 2 3 , 7

60. x 1 4 . 9

63. 25p $ 10

64. 24u , 12

65. 3 2 2x # 7

66. 4 2 3x . 217

67. 21.8x 1 2.5 . 3.4

68. 2.7x 2 1.3 , 6.8

71. 7 2 211 2 x2 . 5 1 31x 2 22

72. 4 2 312 1 x2 , 5

69. 3  1t 1 12 . 2t

70. 2  1 y 1 52 # 3 1 y 2 42

t25 # 24 3

75.

76.

2p 1 1 . 23 5

61. 3x 2 2 # 4

73. 77.

x12 x 22$ 3 2

5y 2 1 y 2 15 2 y2 , 2 1 2 1 y 2 3 2 3

62. 3x 1 7 $ 28

74.

y23 y 22# 5 4

78.

1s 2 32 s s 1 2 . 2 2 3 4 12

79. 22 , x 1 3 , 5

80. 1 , x 1 6 , 12

81. 28 # 4 1 2x , 8

83. 23 , 1 2 x # 9

84. 3 # 22 2 5x # 13

85. 0 , 2 2 3 y , 4

86. 3 , 2 A 2 3 , 7

89. 20.7 # 0.4x 1 1.1 # 1.3

90. 7.1 . 4.7 2 1.2x . 1.1

87.

11y 1 3 # # 2 3 4

88. 21 ,

22z 1 # 4 5

1

82. 0 , 2 1 x # 5 1

• A P P L I C AT I O N S 91. Weight. A healthy weight range for a woman is given by the

following formula: ■  110

pounds for the first 5 feet (tall)

■  2–6

pounds per inch for every inch above 5 feet

Write an inequality representing a healthy weight, w, for a 5 foot 9 inch woman. 92. Weight. NASA has more stringent weight allowances for its astronauts. Write an inequality representing allowable weight for a female 5 foot 9 inch mission specialist given 105 pounds

Young_AT_6160_ch01_pp129-163.indd 136

for the first 5 feet, and 1–5 pounds per inch for every additional inch. 93. Profit. A seamstress decides to open a dress shop. Her fixed costs are $4000 per month, and it costs her $20 to make each dress. If the price of each dress is $100, how many dresses does she have to sell per month to make a profit? 94. Profit. Labrador retrievers that compete in field trials typically cost $2000 at birth. Professional trainers charge $400 to $1000 per month to train the dogs. If the dog is a champion by age 2, it sells for $30,000. What is the range of profit for a champion at age 2?

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1.5  Linear Inequalities 

In Exercises 95 and 96 refer to the following:

137

103. Markups. Typical markup on new cars is 15–30%. If the

The annual revenue for a small company is modeled by R 5 5000 1 1.75x

where x is hundreds of units sold and R is revenue in thousands of dollars. 95. Business. Find the number of units (to the nearest 100) that must be sold to generate at least $10 million in revenue. 96. Business. Find the number of units (to the nearest 100) that must be sold to generate at least $7.5 million in revenue. In Exercises 97 and 98 refer to the following: The Target or Training Heart Rate (THR) is a range of heart rate (measured in beats per minute) that enables a person’s heart and lungs to benefit the most from an aerobic workout. THR can be modeled by the formula THR 5 1 HRmax 2 HRrest 2 3 I 1 HRrest

where HRmax is the maximum heart rate that is deemed safe for the individual, HRrest is the resting heart rate, and I is the intensity of the workout that is reported as a percentage. 97. Health. A female with a resting heart rate of 65 beats per minute has a maximum safe heart rate of 170 beats per minute. If her target heart rate is between 100 and 140 beats per minute, what percent intensities of workout can she consider? 98. Health. A male with a resting heart rate of 75 beats per minute has a maximum safe heart rate of 175 beats per minute. If his target heart rate is between 110 and 150 beats per minute, what percent intensities of workout can he consider? 99. Cost: Cell Phones. A cell phone company charges $50 for an 800-minute monthly plan, plus an additional $0.22 per minute for every minute over 800. If a customer’s bill ranged from a low of $67.16 to a high of $96.86 over a 6-month period, what were the most minutes used in a single month? What were the least? 100. Cost: Internet. An Internet provider charges $30 per month for 1000 minutes of DSL service plus $0.08 for each additional minute. In a one-year period the customer’s bill ranged from $36.40 to $47.20. What were the most and least ­minutes used? 101. Grades. In your general biology class, your first three test scores are 67, 77, and 84. What is the lowest score you can get on the fourth test to earn at least a B for the course? Assume that each test is of equal weight and the minimum score required to earn a B is an 80. 102. Grades. In your Economics I class there are four tests and a final exam, all of which count equally. Your four test grades are 96, 87, 79, and 89. What grade on your final exam is needed to earn between 80 and 90 for the course?

sticker price is $27,999, write an inequality that gives the range of the invoice price (what the dealer paid the manufacturer for the car). 104. Markups. Repeat Exercise 103 with a sticker price of $42,599. 105. Lasers. A circular laser beam with a radius rT is transmitted from one tower to another tower. If the received beam radius rR fluctuates 10% from the transmitted beam radius due to atmospheric turbulence, write an inequality representing the received beam radius. 106. Electronics: Communications. Communication systems are often evaluated based on their signal-to-noise ratio (SNR), which is the ratio of the average power of received signal, S, to average power of noise, N, in the system. If the SNR is required to be at least 2 at all times, write an inequality representing the received signal power if the noise can fluctuate 10%. 107. Real Estate. The Aguileras are listing their house with a real estate agent. They are trying to determine a listing price, L, for the house. Their realtor advises them that most buyers ­traditionally offer a buying price, B, that is 85–95% of the listing price. Write an inequality that relates the buying price to the listing price. 108. Humidity. The National Oceanic and Atmospheric Administration (NOAA) has stations on buoys in the oceans to measure atmosphere and ocean characteristics such as temperature, humidity, and wind. The humidity sensors have an error of 5%. Write an inequality relating the measured humidity hm, and the true humidity ht. 109. Recreation: Golf. Two friends enjoy playing golf. Their favorite course charges $40 for greens fees (to play the course) and a $15 cart rental (per person), so it currently costs each of them $55 every time they play. The ­membership offered at that course is $160 per month. The membership allows them to play as much as they want (no greens fees), but does still charge a cart rental fee of $10 every time they play. What is the least number of times they should play a month in order for the m ­ embership to be the better deal? 110. Recreation: Golf. The same friends in Exercise 109 have a second favorite course. That golf course charges $30 for greens fees (to play the course) and a $10 cart rental (per person), so it currently costs each of them $40 every time they play. The membership offered at that course is $125 per month. The membership allows them to play as much as they want (no greens fees), but does still charge a cart rental fee of $10. What is the least number of times they should play a month in order for the membership to be the better deal?

The following table is the 2015 Federal Tax Rate Schedule for people filing as single: TAX BRACKET #

I II III IV V VI VII

IF TAXABLE INCOME IS

$0 to $9,225 $9,226 to $37,450 $37,451 to $90,750 $90,751 to $189,300 $189,301 to $411,500 $411,501 to $413,200 $413,201 or more

THE TAX IS:

10% of amount over $0 $922.50 is 15% of the amount over $9,225 $5,156.25 plus 25% of the amount over $37,450 $18,481.25 plus 28% of the amount over $90,750 $46,075.25 plus 33% of the amount over $189,300 $119, 401.25 plus 35% of the amount over $411,500 $119,996.25 plus 39.6% of the amount over $413,200

111. Federal Income Tax. What is the range of federal income

taxes a person in tax bracket III will pay the IRS?

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112. Federal Income Tax. What is the range of federal income

taxes a person in tax bracket IV will pay the IRS?

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• C AT C H T H E M I S TA K E In Exercises 113–116, explain the mistake that is made. 113. Rewrite in interval notation.

115. Solve the inequality 2 2 3p # 24 and express the solution in

interval notation.

2

0

2

4

6

Solution:



This is incorrect. What mistake was made? 114. Graph the indicated set and write as a single interval if possible.





2 2 3p # 24



23p # 26



p#2



1 2q, 2 4

This is incorrect. What mistake was made?

116. Solve the inequality 3 2 2x # 7 and express the solution in

interval notation. 2

0

2

4

6

Solution:



This is incorrect. What mistake was made?





3 2 2x # 7



22x # 4



x $ 22



1 2q, 22 4

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 117 and 118, determine whether each statement is true or false. 117. If x , a, then a . x. 118.  If 2x $ a, then x $ 2 a.

119. mn . 0 120. mn , 0

In Exercises 119–122, select any of the statements a 2 d that could be true. a.  m . 0 and n . 0    b.  m , 0 and n , 0 c.  m . 0 and n , 0    d.  m , 0 and n . 0

In Exercises 123 and 124, select any of the statements a 2 c that could be true.

121.

m m . 0 122. ,0 n n

  a.  n 5 0    b.  n . 0    c. n , 0 123. m 1 n , m 2 n 124. m1n$m2n

• CHALLENGE 125. Solve the inequality x # 2x mentally (without doing any

127. Solve the inequality ax 1 b , ax 2 c, where 0 , b , c.

algebraic manipulation). 126. Solve the inequality x . 2x mentally (without doing any algebraic manipulation).

128. Solve the inequality 2ax 1 b , 2ax 1 c, where 0 , b , c.

• TECHNOLOGY 129. a.  Solve the inequality 2.7x 1 3.1 , 9.4x 22.5.

132. a.  Solve the inequality x 2 2 , 3x 1 4 # 2x 1 6.

                b.  Graph each side of the inequality in the same viewing

     b. Graph all three expressions of the inequality in the same

screen. Find the range of x-values when the graph of the left side lies below the graph of the right side.        c.  Do (a) and (b) agree? 130. a.  Solve the inequality 20.5x 1 2.7 . 4.1x 2 3.6.                 b.  Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies above the graph of the right side.                    c.  Do (a) and (b) agree? 131. a.  Solve the inequality x 2 3 , 2x 2 1 , x 1 4.                 b.  Graph all three expressions of the inequality in the same viewing screen. Find the range of x-values when the graph of the middle expression lies above the graph of the left side and below the graph of the right side.                    c.  Do (a) and (b) agree?

viewing screen. Find the range of x-values when the graph of the middle expression lies above the graph of the left side and on top of and below the graph of the right side.       c.  Do (a) and (b) agree? 133. a.  Solve the inequality x 1 3 , x 1 5.    b.  Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies below the graph of the right side.       c.  Do (a) and (b) agree? 1 2 134. a. Solve the inequality 2 x 2 3 . 23 x 1 1.    b.  Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies above the graph of the right side.       c.  Do (a) and (b) agree?

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139

1.6 POLYNOMIAL AND RATIONAL INEQUALITIES SKILLS OBJECTIVES ■■ Solve polynomial inequalities. ■■ Solve rational inequalities.

CONCEPTUAL OBJECTIVES ■■ Understand zeros and test intervals. ■■ Realize that a rational inequality has an implied domain restriction on the variable.

1.6.1  Polynomial Inequalities In this section we will focus primarily on quadratic inequalities, but the procedures outlined are also valid for higher degree polynomial inequalities. An example of a quadratic inequality is x2 1 x 2 2 , 0. This statement is true when the value of the polynomial on the left side is negative. For any value of x, a polynomial has a positive, negative, or zero value. A polynomial must pass through zero before its value changes from positive to negative or from negative to positive. Zeros of a polynomial are the values of x that make the polynomial equal to zero. These zeros divide the real number line into test intervals where the value of the polynomial is either positive or negative. For example, if we set the above polynomial equal to zero and solve:

1.6.1 S K I L L

Solve polynomial inequalities. 1.6.1 C O N C E P T U A L

Understand zeros and test intervals.

x2 1 x 2 2 5 0 1x 1 22 1x 2 12 5 0

x 5 22 or

x51

we find that x 5 22 and x 5 1 are the zeros. These zeros divide the real number line into three test intervals: 12q, 222, 122, 12, and 11, q2. Since the polynomial is equal to zero at x 5 22 and x 5 1, the value of the polynomial in each of these three intervals is either positive or negative. We select one real number that lies in each of the three intervals and test to see whether the value of the polynomial at each point is either positive or negative. In this example, we select the real numbers: x 5 23, x 5 0, and x 5 2. At this point, there are two ways we can determine whether the value of the polynomial is positive or negative on the interval. One approach is to substitute each of the test points into the polynomial x2 1 x 2 2. x 5 23 x50 x52

1 23 2 2 1 1 23 2 2 2 5 9 2 3 2 2 5 4

Positive

122 1 122 2 2 5 4 1 2 2 2 5 4

Positive

1 0 2 2 1 1 0 2 2 2 5 0 2 0 2 2 5 22 2

x2 + x – 2 > 0

2 y x +x–2>0

x

x = –2

x=1

x2 + x – 2 < 0

Negative

The second approach is to simply determine the sign of the result as opposed to actually calculating the exact number. This alternate approach is often used when the expressions or test points get more complicated to evaluate. The polynomial is written as the product 1x 1 22 1x 2 12; therefore, we simply look for the sign in each set of parentheses. 1x 1 22 1x 2 12





123 1 2 2 123 2 1 2 5 121 2 124 2 S 122 1 2 2 5 11 2

x 523:

1 0 1 2 2 1 0 2 1 2 5 1 2 2 121 2 S 112 122 5 122

  x 5 0:

1 2 1 2 2 1 2 2 1 2 5 1 4 2 1 1 2 S 112 112 5 112

  x 5 2:

3

0

2

In this second approach we find the same result: 12q, 222 and 11, q2 correspond to a positive value of the polynomial, and 122, 12 corresponds to a negative value of the polynomial.

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ST U DY TIP

In this example, the statement x2 1 x 2 2 , 0 is true when the value of the polynomial (in factored form), 1x 1 22 1x 2 12, is negative. In the interval 122, 12, the value of the ­polynomial is negative. Thus, the solution to the inequality x2 1 x 2 2 , 0 is 122, 12. To check the solution, select any number in the interval and substitute it into the original inequality to make sure it makes the statement true. The value x 5 21 lies in the interval 122, 12. Upon substituting into the original inequality, we find that x 5 21 satisfies the inequality 1212 2 1 1212 2 2 5 22 , 0. PROCEDURE FOR SOLVING POLYNOMIAL INEQUALITIES

Step 1:  Write the inequality in standard form. Step 2:  Identify zeros. Step 3:  Draw the number line with zeros labeled. Step 4:  Determine the sign of the polynomial in each interval. Step 5:  Identify which interval(s) make the inequality true. Step 6:  Write the solution in interval notation.

If the original polynomial is ,0, then the interval(s) that yield(s) negative products should be selected. If the original polynomial is .0, then the interval(s) that yield(s) positive products should be selected.

Note: Be careful in Step 5. If the original polynomial is , 0, then the interval(s) that corr­espond(s) to the value of the polynomial being negative should be selected. If the original polynomial is . 0, then the interval(s) that correspond(s) to the value of the polynomial being positive should be selected. EXAMPLE 1  Solving a Quadratic Inequality

Solve the inequality x2 2 x . 12. Solution:

x2 2 x 2 12 . 0 1x 1 32 1x 2 42 . 0

STEP 1  Write



the inequality in standard form. Factor the left side.

STEP 2  Identify

1x 1 32 1x 2 42 5 0

the zeros.

x 5 23  or x 5 4

STEP 3  Draw



the number line with the zeros labeled.

STEP 4  Determine the sign of 1x 1 32 1x 2 42 in each interval.

in which the value of the polynomial is positive make this inequality true.

4

0

5

STEP 5  Intervals

STEP 6  Write ▼ ANSWER

321, 64

Young_AT_6160_ch01_pp129-163.indd 140

the solution in interval notation.

▼

1 2`, 23 2   or   1 4, ` 2

1 2q, 23 2 ∪ 1 4, q 2

Y O U R T U R N   Solve the inequality x 2 2 5x # 6 and express the solution in interval

notation. The inequality in Example 1, x2 2 x . 12, is a strict inequality, so we use parentheses when we express the solution in interval notation 1 2q, 23 2 ∪ 1 4, q 2 . It is important to note that if we change the inequality sign from . to $, then the zeros x 5 23 and x 5 4 also make the inequality true. Therefore, the solution to x2 2 x $ 12 is 1 2q, 23 4 ∪ 3 4, q 2 .

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141

EXAMPLE 2  Solving a Quadratic Inequality

Solve the inequality x2 # 4. ▼ CAUTION

common mistake Do not take the square root of both sides. You must write the inequality in standard form and factor. ✓C O R R E C T

The square root method cannot be used for quadratic inequalities.

✖INCORRECT

STEP 1:  Write the inequality in standard form. x2 2 4 # 0

Error: Take the square root of both sides. x # 62

Factor. 1 x 2 2 21 x 1 2 2 # 0

STEP 2:  Identify the zeros.

1 x 2 2 21 x 1 2 2 5 0

x52

and

x 5 22

STEP 3:  Draw the number line with the zeros labeled. –2

2

STEP 4:  Determine the sign of 1 x 2 2 21 x 1 2 2 , in each interval. (–)(–) = (+) (–)(+) = (–) (+)(+) = (+) –3

–2

0

2

3

STEP 5:  Intervals in which the value of the polynomial is negative make the inequality true. 122, 22

The endpoints, x 5 22 and x 5 2, satisfy the inequality, so they are included in the solution. STEP 6:  Write the solution in interval notation. 322, 24

When solving quadratic inequalities, you must first write the inequality in standard form and then factor to identify zeros.

Not all inequalities have a solution. For example, x2 , 0 has no real solution. Any real number squared is always nonnegative, so there are no real values that when squared will yield a negative number. The zero is x 5 0, which divides the real number line into two intervals: 12q, 02 and 10, q2. Both of these intervals, however, correspond to the value of x2 being positive, so there are no intervals that satisfy the inequality. We say that this inequality has no real solution.

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ST U DY TIP

EXAMPLE 3  Solving a Quadratic Inequality

When solving quadratic inequalities, you must first write the inequality in standard form and then factor to identify zeros.

Solve the inequality x2 1 2x $ 23. Solution: STEP 1  Write

STEP 2  Identify



x2 1 2x 1 3 $ 0

the inequality in standard form.

x2 1 2x 1 3 5 0

the zeros.

Apply the quadratic formula.

Simplify.

x5

x5

22 6 "22 2 4 1 1 2 1 3 2 2112

22 6 !28 22 6 2i !2 5 5 21 6 i !2 2 2

Since there are no real zeros, the quadratic expression x2 1 2x 1 3 never equals zero; hence its value is either always positive or always negative. If we select any value for x, say, x 5 0, we find that 102  2 1 2  102 1 3 $ 0. Therefore, the quadratic expression is always ­positive, and so the solution is the set of all real numbers, 1 2q, q 2  . EXAMPLE 4  Solving a Quadratic Inequality

Solve the inequality x2 . 25x. ▼ CAUTION

Do not divide inequalities by a variable.

common mistake A common mistake is to divide by x. Never divide by a variable because the value of the variable might be zero. Always start by writing the inequality in standard form and then factor to determine the zeros. ✓C O R R E C T

✖INCORRECT

STEP 1:  Write the inequality in standard form. x 2 2 5x . 0 Factor. x1x 1 52 . 0

x 2 . 25x ERROR: Divide both sides by x.

STEP 2:  Identify the zeros.

x . 25

x 5 0, x 5 25 STEP 3:  Draw the number line with the zeros labeled. –5

Write the original inequality.

Dividing by x is the mistake. If x is negative, the inequality sign must be reversed. What if x is zero?

0

STEP 4:  Determine the sign of x 1 x 1 5 2 in each interval.

(–)(–) = (+) (–)(+) = (–) (+) (+) = (+) –6

–5

–1

0

1

STEP 5:  Intervals in which the value of the polynomial is positive satisfy the inequality. 12H, 252  and  10, 2H2

STEP 6:  Express the solution in interval notation. 1 2q, 25 2 ∪ 1 0, q 2

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143

EXAMPLE 5  Solving a Quadratic Inequality

Solve the inequality x2 1 2x , 1. Solution:

x2 1 2x 2 1 , 0 x2 1 2x 2 1 5 0

Write the inequality in standard form. Identify the zeros. Apply the quadratic formula.

x5

Simplify.

x5

22 6 "22 2 4 1 1 2 1 21 2 2112

22 6 "8 22 6 2"2 5 5 21 6 "2 2 2

Draw the number line with the intervals labeled. Note: 2 ­ 1 2 !2 < 22.41 21 1 !2 < 0.41

2

Test each interval.

1 2q, 21 2 !2 2

121 2 !2, 21 1 !2 2 121 1 !2, q 2

1 23 2 2 1 2 1 23 2 2 1 5 2 . 0

x 5 23:

1 0 2 2 1 2 1 0 2 2 1 5 21 , 0

x 5 0: x 5 1:

Intervals in which the value of the polynomial is negative make this inequality true.

1122 1 2112 2 1 5 2 . 0

▼

A21 2 !2, 21 1 !2B

Y O U R T U R N   Solve the inequality x2 2 2x $ 1.

▼ ANSWER

A2q, 1 2 !2 D ∪ C1 1 !2, qB

EXAMPLE 6  Solving a Polynomial Inequality

Solve the inequality x3 2 3x2 $ 10x. Solution:

Write the inequality in standard form.

x3 2 3x2 2 10x $ 0

Factor.

x 1x 2 52 1x 1 22 $ 0

Identify the zeros.  x 5 0, x 5 5, x 5 22 Draw the number line with the zeros (intervals) labeled.

[ CONCEPT CHECK ]

Test each interval.

Solve: (x 2 a) (x 1 b) < 0, where a > 0 and b > 0.

▼

Intervals in which the value of the polynomial is positive make this inequality true.

▼ 3

2

Y O U R T U R N   Solve the inequality x 2 x 2 6x , 0.

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ANSWER (2b, a)

3 22, 0 4 ∪ 3 5, q 2

▼ ANSWER

12q , 22 2 ∪ 1 0, 3 2

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1.6.2  Rational Inequalities 1.6.2 S K I L L S

Solve rational inequalities.

Rational expressions have numerators and denominators. Recalling the properties of ­negative real numbers (Chapter 0), we see that the following possible combinations ­correspond to either positive or negative rational expressions. 112 122 122 112 5 1 1 2     5 1 2 2     5 1 1 2     5 122 112 112 122 122

1.6.2 C O N C E P T U A L

Realize that a rational inequality has an implied domain restriction on the variable.

ST U DY TIP Values that make the denominator equal to zero are always excluded.

A rational expression can change signs if either the numerator or denominator changes signs. In order to go from positive to negative or vice versa, you must pass through x23 $ 0 we use a similar zero. Therefore, to solve rational inequalities such as 2 x 24 procedure to the one used for solving polynomial inequalities, with one exception. You must eliminate values for x that make the denominator equal to zero. In this example, we must eliminate x 5 22 and x 5 2 because these values make the denominator equal to zero. Rational inequalities have implied domains. In this example, x 2 62 is a domain restriction and these values 1x 5 22 and x 5 22 must be eliminated from a possible solution. We will proceed with a similar procedure involving zeros and test intervals that was outlined for polynomial inequalities. However, in rational inequalities once expressions are combined into a single fraction, any values that make either the numerator or the ­denominator equal to zero divide the number line into intervals. EXAMPLE 7  Solving a Rational Inequality

Solve the inequality

x23 $ 0. x2 2 4

Solution:

1x 2 32 $0 1x 2 22 1x 1 22

Factor the denominator.

[ CONCEPT CHECK ] Which of the following has an implied domain restriction on the variable: 1 1 (A) 2 or (B) 2 x 19 x 29

▼ ANSWER (B) x 2 63. (A) does not have a domain restriction because x 2 1 9 is never equal to zero when x is a real number.

x 2 2, x 2 22

State the domain restrictions on the variable.

Identify the zeros of numerator and denominator. x 5 22, x 5 2, x 5 3 Draw the number line and divide into intervals.

Test the intervals.

3

Intervals in which the value of the rational expression is positive satisfy this inequality.

0

2.5

4

122, 22  and  13, H2

Since this inequality is greater than or equal to, we include x 5 3 in our solution because it satisfies the inequality. However, x 5 22 and x 5 2 are not included in the solution because they make the denominator equal to zero.

▼ ANSWER

322, 12

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▼

The solution is  1 22, 2 2 ∪ 3 3, q 2 .

Y O U R T U R N   Solve the inequality

x12 # 0. x 21

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1.6  Polynomial and Rational Inequalities 

EXAMPLE 8  Solving a Rational Inequality

Solve the inequality

x25 , 0. x2 1 9

Solution:

Identify the zero(s) of the numerator.

x55

Note that the denominator is never equal to zero when x is any real number. Draw the number line and divide into intervals. 5

Test the intervals.

()  () ()

The denominator is always positive.

()  () ()

5

Intervals in which the value of the rational expression is negative satisfy the inequality.

12q, 5 2

Notice that x 5 5 is not included in the solution because of the strict inequality.

▼ Y O U R T U R N   Solve the inequality

x14 $ 0. x 2 1 25

EXAMPLE 9  Solving a Rational Inequality

Solve the inequality

▼ ANSWER

324, q2

x # 3. x12 ▼ CAUTION

common mistake Do not cross multiply. The LCD or expression by which you are multiplying might be negative for some values of x, and that would require the direction of the inequality sign to be reversed. ✓C O R R E C T Subtract 3 from both sides. x 23#0 x12 Write as a single rational expression.

Rational inequalities should not be solved using cross multiplication.

✖INCORRECT Error: Do not cross multiply. x # 31x 1 22

x 2 31x 1 22 #0 x12 Eliminate the parentheses. x 2 3x 2 6 #0 x12 Simplify the numerator. 22x 2 6 #0 x12 Factor the numerator. 22 1 x 1 3 2 #0 x12

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Identify the zeros of the numerator and the denominator. x 5 23

and

x 5 22

Draw the number line and test the intervals. 22 1 x 1 3 2 #0 x12

Intervals in which the value of the ­rational expression is negative satisfy the inequality, 12H, 234 and 122, H2. Note that x 5 22 is not included in the solution because it makes the denominator zero, and x 5 23 is included because it satisfies the inequality. The solution is: 1 2q, 23 4 ∪ 1 22, q 2

Applications EXAMPLE 10  Stock Prices

From June 2015 to November 2015, the price of Abercrombie and Fitch’s (ANF) stock was approximately given by P 5 0.5t2 2 2t 1 17, where P is the price of the stock in dollars, t is months, t 5 0 corresponds to June 2015, and t 5 5 corresponds to November 2015. During which months was the value of the stock worth no more than $21? 35 30 25 20 15 June

July

Aug

Solution:

Sep

Oct

Nov

Set the price less than or equal to 21:

0.5t2 2 3t + 25 # 21,  0 # t # 5

Write in standard form:

0.5t2 2 3t 1 4 # 0

Multiply by 2:

t2 − 6t 1 8 # 0

Factor.

1t − 42 1t 2 22 # 0

Identify zeros:

t 5 2 and t 5 4

(–)(–) = (+) (–)(+) = (–) (+)(+) = (+) 1

Test the intervals: Negative interval satisfies the inequality:

2

3

4

5

1t 2 42 1t 2 22 # 0 32, 44

The Abercrombie & Fitch price was no more than $21 during August 2015, September 2015, and October 2015.

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147

[ S E C T I O N 1 . 6]     S U M M A R Y The following procedure can be used for solving polynomial and rational inequalities. 1. Write in standard form—zero on one side. 2. Determine the zeros; if it is a rational function, note the domain restrictions. Polynomial Inequality – Factor if possible. – Otherwise, use the quadratic formula.

  Rational Inequality – Write as a single fraction. – Determine values that make the numerator or denominator equal to zero. Always exclude values that make the denominator 5 0. 3. Draw the number line labeling the intervals. 4. Test the intervals to determine whether they are positive or negative. 5. Select the intervals according to the sign of the inequality. 6. Write the solution in interval notation.

[ S E C T I O N 1 . 6]   E X E R C I S E S • SKILLS In Exercises 1–28, solve the polynomial inequality and express the solution set in interval notation. 1. x2 2 3x 2 10 $ 0

2. x2 1 2x 2 3 , 0

3. u2 2 5u 2 6 # 0

4. u2 2 6u 2 40 . 0

5. p2 1 4p , 23

6. p2 2 2p $ 15

7. 2t2 2 3 # t

8. 3t2 $ 25t 1 2

9. 5v 2 1 . 6v2

10. 12t2 , 37t 1 10

11. 2s2 2 5s $ 3

12. 8s 1 12 # 2s2

13. y 2 1 2y $ 4

14. y2 1 3y # 1

15. x2 2 4x , 6

16. x2 2 2x . 5

17. u2 $ 3u

18. u2 # 24u

19. 22x # 2x2

20. 23x # x2

21. x2 . 9

22. x2 $ 16

23. t2 , 81

24. t2 # 49

25. z2 . 216

26. z2 $ 22

27. y2 , 24

28. y2 #225

In Exercises 29–58, solve the rational inequality and graph the solution on the real number line. y 3 3 31. .0 29. 2 # 0 30. # 0 y13 x x

32.

y #0 22y

33.

t13 $0 t24

34.

2t 2 5 ,0 t26

35.

s11 $0 4 2 s2

36.

s15 #0 4 2 s2

37.

x23 $0 x 2 2 25

38.

12x #0 x2 2 9

39.

2u2 1 u ,1 3

40.

u2 2 3u $6 3

41.

3t 2 $ 5t t12

42.

22t 2 t 2 $t 42t

43.

2

2

3p 2 2p2 2

42p

,

31p 22p

44. 2

2

7p 2

p 2 100

#

p12 p 1 10

x2 1 2 ,0 x2 1 4

45.

x ,0 5 1 x2

46.

x #0 5 1 x2

47.

x 1 10 .0 x 2 1 16

48. 2

49.

v2 2 9 $0 v23

50.

v2 2 1 #0 v11

51.

2 1 1 $0 t23 t13

52.

53.

3 1 2 #0 x14 x22

54.

2 1 2 $0 x25 x21

55.

1 1 p2 2 48 1 1 56. 2 #2 1 . 2 p23 p13 p 1 4 p 2 4 p 2 16

57.

1 1 3 2 $ 2 p22 p12 p 24

58.

2 1 1 2 # 2 2p 2 3 p11 2p 2 p 2 3

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1 1 1 #0 t22 t12

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• A P P L I C AT I O N S 59.  Profit. A Web-based embroidery company makes mono-

grammed napkins. The profit associated with producing x orders of napkins is governed by the equation P 5 2x 2 1 130x 2 3000 Determine the range of orders the company should accept in order to make a profit. 60. Profit. Repeat Exercise 59 using P 5 x2 2 130x 1 3600. 61. Car Value. The term “upside down” on car payments refers to owing more than a car is worth. Assume you buy a new car and finance 100% over 5 years. The difference between the value of the car and what is owed on the car is governed t by the expression where t is age (in years) of the car. t23 Determine the time period when the car is worth more than t . 0b. you owe a t23 t , 0b ? When do you owe more than it’s worth a t23

66.  Stock

Value. From November 2014 to November 2015, Amazon.com stock was approximately worth P 5 6.25t2 2 25t 1 325, where P is the price of the stock in dollars, t is months, and t 5 0 corresponds to November 2014. During what months was the stock worth no more than $525?



22t 62.  Car Value. Repeat Exercise 61 using the expression 2 . 42t 63.  Bullet Speed. A .22-caliber gun fires a bullet at a speed of 1200 feet per second. If a .22-caliber gun is fired straight upward into the sky, the height of the bullet in feet is given by the equation h 5 216t2 1 1200t, where t is the time in seconds with t 5 0 corresponding to the instant the gun is fired. How long is the bullet in the air? 64.  Bullet Speed. A .38-caliber gun fires a bullet at a speed of

600 feet per second. If a .38-caliber gun is fired straight upward into the sky, the height of the bullet in feet is given by the equation h 5 216t2 1 600t. How many seconds is the bullet in the air? 65.  Geometry. A rectangular area is fenced in with 100 feet

of fence. If the minimum area enclosed is to be 600 square feet, what is the range of feet allowed for the length of the rectangle?

$725.00 675.00 625.00 575.00 525.00 475.00 450.00 425.00 400.00 375.00 350.00 325.00 300.00 275.00 Dec 2015 Feb Mar Apr May Jun Jul Aug Sep Oct Nov

www.nasdaq.com/symbol/amzn/stock-chart

In Exercises 67 and 68 refer to the following: In response to economic conditions, a local business explores the effect of a price increase on weekly profit. The function P 5 25 1 x 1 3 2 1 x 2 24 2

models the effect that a price increase of x dollars on a bottle of wine will have on the profit P measured in dollars. 67.  Economics. What price increase will lead to a weekly profit of less than $460? 68.  Economics. What price increases will lead to a weekly profit of more than $550? 69.  Real Estate. A woman is selling a piece of land that she advertises as 400 acres 167 acres2 for $1.36 million. If you pay that price, what is the range of dollars per acre you have paid? Round to the nearest dollar. 70.  Real Estate. A woman is selling a piece of land that she advertises as 1000 acres 1610 acres2 for $1 million. If you pay that price, what is the range of dollars per acre you have paid? Round to the nearest dollar.

• C AT C H T H E M I S TA K E In Exercises 71–74, explain the mistake that is made. 71. Solve the inequality 3x , x2. Solution: Divide by x. 3 , x Write the solution in interval notation. 13, q2

This is incorrect. What mistake was made?

72. Solve the inequality u2 , 25.

Solution: Take the square root of both sides. u , 25 Write the solution in interval notation. 12q, 252 This is incorrect. What mistake was made?

Young_AT_6160_ch01_pp129-163.indd 148

73. Solve the inequality

x2 2 4 . 0. x12

Solution:

1x 2 22 1x 1 22 ,0 Factor the numerator and 1x 1 22 denominator. Cancel the 1x 1 22 common factor. x22.0 Solve. x.2 This is incorrect. What mistake was made? 74. Solve the inequality

x14 1 ,2 . x 3

Solution: Cross multiply. 3 1x 1 42 , 211x2 Eliminate the parentheses. 3x 1 12 , 2x Combine like terms.                       4x , 212 Divide both sides by 4.              x , 23 This is incorrect. What mistake was made?

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149

• CONCEPTUAL In Exercises 75 and 76, determine whether each statement is true or false. Assume that a is a positive real number. 75. If x , a2, then the solution is 12q, a2. 76. If x $ a2, then the solution is 3a, q2.

77. Assume the quadratic inequality ax2 1 bx 1 c , 0 is true.

If b2 2 4ac , 0, then describe the solution. 78. Assume the quadratic inequality ax2 1 bx 1 c . 0 is true. If b2 2 4ac , 0, then describe the solution.

• CHALLENGE In Exercises 79–82, solve for x given that a and b are both positive real numbers. x 2 2 b2 80. ,0 79. 2x2 # a2 x1b

81.

x 2 1 a2 $0 x 2 1 b2

82.

a , 2b x2

• TECHNOLOGY In Exercises 83–90, plot the left side and the right side of each inequality in the same screen and use the zoom feature to determine the range of values for which the inequality is true. 83. 1.4x2 2 7.2x 1 5.3 . 28.6x 1 3.7 84. 17x2 1 50x 2 19 , 9x2 1 2 85. 11x2 , 8x 1 16 86. 0.1x 1 7.3 . 0.3x2 2 4.1 87. x , x2 2 3x , 6 2 2x

88. x2 1 3x 2 5 $ 2x2 1 2x 1 10 89.

2p .1 52p

90.

3p ,1 42p

1.7 ABSOLUTE VALUE EQUATIONS AND INEQUALITIES SKILLS OBJECTIVES ■■ Solve absolute value equations. ■■ Solve absolute value inequalities.

CONCEPTUAL OBJECTIVES ■■ Understand absolute value in terms of distance on the number line. ■■ Apply intersection and union concepts to expressing solutions to linear inequalities in one variable.

1.7.1  Equations Involving Absolute Value The absolute value of a real number can be interpreted algebraically and graphically. Algebraically, the absolute value of 5 is 5, or in mathematical notation, 0 5 0 5 5; and the absolute value of 25 is 5 or 0 25 0 5 5. Graphically, the absolute value of a real number is the distance on the real number line between the real number and the origin; thus, the ­distance from 0 to either 25 or 5 is 5. 5

5

Solve absolute value equations. 1.7.1 C O N C E P T U A L

Understand absolute value in terms of distance on the number line.

5

5

DEFINITION

1.7.1 S K I L L

Absolute Value

The absolute value of a real number a, denoted by the symbol 0 a 0 , is defined by 0a0 5 b

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a, 2a,

if a $ 0 if a , 0

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The absolute value of a real number is never negative. When a 5 25, this definition says 0 25 0 5 21252 5 5. PROPERTIES OF ABSOLUTE VALUE

For all real numbers a and b,

0a0 a 1.  0 a 0 $ 0   2.  0 2a 0 5 0 a 0   3.  0 ab 0 5 0 a 0 0 b 0   4.  ` ` 5 b 0b0

b20

Absolute value can be used to define the distance between two points on the real number line. DISTANCE BETWEEN TWO POINTS ON THE REAL NUMBER LINE

If a and b are real numbers, the distance between a and b is the absolute value of their difference given by 0 a 2 b 0 or 0 b 2 a 0 . EXAMPLE 1  F  inding the Distance Between Two Points on the Number Line

Find the distance between 24 and 3 on the real number line. Solution:

The distance between 24 and 3 is given by the absolute value of the difference.

0 24 2 3 0 5 0 27 0 5 7

Note that if we reverse the numbers the result is the same.

0 3 2 1 24 2 0 5 0 7 0 5 7

We check this by counting the units between 24 and 3 on the number line.

7

4

3

When absolute value is involved in algebraic equations, we interpret the definition of absolute value as follows. DEFINITION

Absolute Value Equation

If 0 x 0 5 a, then x 5 2a or x 5 a, where a $ 0.

In words, “if the absolute value of a number is a, then that number equals 2a or a.” For example, the equation 0 x 0 5 7 is true if x 5 27 or x 5 7. We say the equation 0 x 0 5 7 has the solution set 527, 76. Note: 0 x 0 5 23 does not have a solution because there is no value of x such that its absolute value is 23. EXAMPLE 2  Solving an Absolute Value Equation

Solve the equation 0 x 2 3 0 5 8 algebraically and graphically. Solution:

Using the absolute value equation definition, we see that if the absolute value of an expression is 8, then that expression is either 28 or 8. Rewrite as two equations: x 2 3 5 28   or   x 2 3 5 8 x 5 25         x 5 11 The solution set is 525, 116 .

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Graph: The absolute value equation 0 x 2 3 0 5 8 is interpreted as “what numbers are eight units away from 3 on the number line?” We find that eight units to the right of 3 is 11 and eight units to the left of 3 is 25. 8

8 3

5

11

▼ ANSWER

▼

x 5 2 or x 5 212. The solution set is 5212, 26.

Y O U R T U R N   Solve the equation 0 x 1 5 0 5 7.

EXAMPLE 3  Solving an Absolute Value Equation

STUD Y T I P

Solve the equation 0 1 2 3x 0 5 7.

Rewrite an absolute value ­equation as two equations.

Solution:

If the absolute value of an expression is 7,     1 2 3x 5 27 or then that expression is 27 or 7.         23x 5 28  8 x 5     3



▼

1 2 3x 5 7   23x 5 6   x 5 22

The solution set is U22, 83 V . ▼ ANSWER

Y O U R T U R N   Solve the equation 0 1 1 2x 0 5 5.

x 5 23 or x 5 2.  The solution set is 523, 26.

EXAMPLE 4  Solving an Absolute Value Equation

Solve the equation 2 2 3 0 x 2 1 0 5 24 0 x 2 1 0 1 7. Solution:

Isolate the absolute value expressions to one side. Add 4 0 x 2 1 0 to both sides.

2 1 0x 2 10 5 7

Subtract 2 from both sides.

If the absolute value of an expression

0x 2 10 5 5

x 2 1 5 25  or  x 2 1 5 5

is equal to 5, then the expression is  x 5 24    equal to either 25 or 5.

  x56

The solution set is 5 24, 6 6 .

▼

Y O U R T U R N   Solve the equation 3 2 2 0 x 2 4 0 5 23 0 x 2 4 0 1 11.

EXAMPLE 5  F  inding That an Absolute Value Equation Has No Solution

▼ ANSWER

x 5 24 or x 5 12.  The solution set is 524, 126.

Solve the equation 0 1 2 3x 0 5 27. Solution:

The absolute value of an expression is never negative. Therefore, no values of x make this equation true. No solution

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EXAMPLE 6  Solving a Quadratic Absolute Value Equation

Solve the equation 0 5 2 x2 0 5 1. Solution:

If the absolute value of an expression is 1, that expression is either 21 or 1, which leads to two equations.

[ CONCEPT CHECK ] |x 2 a| 5 b is interpreted on the number line as (A) b units from a or (B) a units from b?



1.7.2 S K I L L S

Solve absolute value inequalities. 1.7.2 C O N C E P T U A L

Apply intersection and union concepts to expressing solutions to linear inequalities in one variable.

   x 2 5 4

x 5 6"6   x 5 6"4 5 6 2 The solution set is U62, 6"6 V .

ANSWER (A ) b units from a

x 5 6"5 or x 5 63.     The solution set is U 6"5, 63 V .

2x 2 5 26   2x 2 5 24 x 2 5 6  



▼

▼ ANSWER

5 2 x 2 5 21 or 5 2 x 2 5 1

▼

Y O U R T U R N   Solve the equation 0 7 2 x2 0 5 2.

1.7.2  Inequalities Involving Absolute Value To solve the inequality 0 x 0 , 3, look for all real numbers that make this statement true. Some numbers that make it true are 22, 2 23, 21, 0, 15, 1, and 2. Some numbers that make it false are 27, 25, 23.5, 23, 3, and 4. If we interpret this inequality as distance, we ask what numbers are less than three units from the origin? We can represent the solution in the following ways.

Inequality notation:

23 , x , 3



Interval notation:

123, 32

Graph:

3

3

Similarly, to solve the inequality 0 x 0 $ 3, look for all real numbers that make the statement true. If we interpret this inequality as a distance, we ask what numbers are at least three units from the origin? We can represent the solution in the following three ways.

Inequality notation: x # 23 or x $ 3 1 2q, 23 4 ∪ 3 3, q 2 Interval notation:

Graph:

3

3

This discussion leads us to the following equivalence relations. PROPERTIES OF ABSOLUTE VALUE INEQUALITIES

1.  0 x 0 , a is equivalent to 2a , x , a 2.  0 x 0 # a is equivalent to 2a # x # a

3.  0 x 0 . a is equivalent to x , 2a or x . a 4.  0 x 0 $ a is equivalent to x # 2a or x $ a Note: a . 0.

It is important to realize that in the above four properties the variable x can be any algebraic expression.

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EXAMPLE 7  Solving an Inequality Involving an Absolute Value

Solve the inequality 0 3x 2 2 0 # 7. Solution:

We apply property (2) and squeeze the absolute value expression between 27 and 7. 27 # 3x 2 2 # 7 Add 2 to all three parts.

25 # 3x # 9

Divide all three parts by 3.

5 2 #x#3 3

The solution in interval notation is C 2 35, 3D .

Graph:   

 53

0

▼ ANSWER

Inequality notation: 26 , x , 5. Interval notation: 126, 52.

3

▼

Y O U R T U R N   Solve the inequality 0 2x 1 1 0 , 11.

STUD Y T I P Less than inequalities can be written as a single statement. Greater than inequalities must be written as two statements.

It is often helpful to note that for absolute value inequalities, less than inequalities can be written as a single statement (see Example 7). ■■ greater than inequalities must be written as two statements (see Example 8). ■■

EXAMPLE 8  Solving an Inequality Involving an Absolute Value

Solve the inequality 0 1 2 2x 0 . 5. Solution:

Apply property (3).

1 2 2x , 25  or  12 2x . 5

Subtract 1 from all expressions.

22x , 26

22x . 4

Divide by 22 and reverse the inequality sign. x . 3 x , 22 1 2q, 22 2 ∪ 1 3, q 2

Express the solution in interval notation. Graph: 2

3

▼

Y O U R T U R N   Solve the inequality 0 5 2 2x 0 $ 1.

Notice that if we change the problem in Example 8 to 0 1 2 2x 0 . 25, the answer is all real numbers because the absolute value of any expression is greater than or equal to zero. Similarly, 0 1 2 2x 0 , 25 would have no solution because the absolute value of an ­expression can never be negative.

Young_AT_6160_ch01_pp129-163.indd 153

[ CONCEPT CHECK ] If the solution to a linear inequality in one variable is all the values less than a OR greater than b (where a and b are both positive), then the appropriate notation is (A) intersection or (B) union.

▼ ANSWER (B) Union (2q, a) ∪ (b, q)

▼ ANSWER

Inequality notation: x # 2 or x $ 3. Interval notation: 1 2q, 2 4 ∪ 3 3, q 2 .

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EXAMPLE 9  Solving an Inequality Involving an Absolute Value

Solve the inequality 2 2 0 3x 0 , 1. Solution:

Subtract 2 from both sides.            2 0 3x 0 , 21 Multiply by 1212 and reverse the inequality sign.    0 3x 0 . 1

Apply property (3).

3x , 21  or  3x . 1 1 1 x , 2       x . 3 3

Divide both inequalities by 3.

1 1 a2q, 2 b ∪ a , qb 3 3

Express in interval notation.

Graph.

[ S E C T I O N 1 . 7 ]     S U M M A R Y Absolute value equations and absolute value inequalities are I N E Q U A L I T I E S solved by writing the equations or inequalities in terms of two 0 x 0 , A is equivalent to 2A , x , A equations or inequalities. Note: A . 0. E Q U AT I O N S

0 x 0 5 A is equivalent to x 5 2A or x 5 A

0 x 0 . A is equivalent to x , 2A or x . A

[S EC T I O N 1.7 ]   E X E R C I S E S • SKILLS In Exercises 1–38, solve the equation. 1. 0 x 0 5 3

2. 0 x 0 5 2

3. 0 x 0 5 24

4. 0 x 0 5 22

9. 0 4 2 y 0 5 1

10. 0 2 2 y 0 5 11

11. 0 3x 0 5 9

12. 0 5x 0 5 50

5. 0 t 1 3 0 5 2

13. 0 2x 1 7 0 5 9 17. 0 7 2 2x 0 5 9

21. 0 4.7 2 2.1x 0 5 3.3

6. 0 t 2 3 0 5 2

14. 0 2x 2 5 0 5 7

18. 0 6 2 3y 0 5 12

22. 0 5.2x 1 3.7 0 5 2.4

7. 0 p 2 7 0 5 3 15. 0 3t 2 9 0 5 3

19. 0 1 2 3y 0 5 1 2 23.  3 x

2

4 7

5

5 3

8. 0 p 1 7 0 5 3 16. 0 4t 1 2 0 5 2 20. 0 5 2 x 0 5 2 1

3

1

24.  2 x 1 4  5 16

25. 0 x 2 5 0 1 4 5 12

26. 0 x 1 3 0 2 9 5 2

27. 3 0 x 2 2 0 1 1 5 19

28. 2 0 1 2 x 0 2 4 5 2

33. 5 0 y 2 2 0 2 10 5 4 0 y 2 2 0 2 3

34. 3 2 0 y 1 9 0 5 11 2 3 0 y 1 9 0

35. 0 4 2 x 0 5 1

36. 0 7 2 x2 0 5 3

29. 5 5 7 2 0 2 2 x 0 37. 0 x2 1 1 0 5 5

Young_AT_6160_ch01_pp129-163.indd 154

30. 21 5 3 2 0 x 2 3 0 38. 0 x2 2 1 0 5 5

31. 2 0 p 1 3 0 2 15 5 5 2

32. 8 2 3 0 p 2 4 0 5 2

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155

In Exercises 39–70, solve the inequality and express the solution in interval notation. 39. 0 x 0 , 7

40. 0 y 0 , 9

41. 0 y 0 $ 5

42. 0 x 0 $ 2

47. 0 4 2 x 0 # 1

48. 0 1 2 y 0 , 3

49. 0 2x 0 . 23

50. 0 2x 0 , 23

43. 0 x 1 3 0 , 7

44. 0 x 1 2 0 # 4

51. 0 2t 1 3 0 , 5

45. 0 x 2 4 0 . 2

52. 0 3t 2 5 0 . 1

55. 0 4 2 3x 0 $ 0

53. 0 7 2 2y 0 $ 3

56. 0 4 2 3x 0 $ 1

59. 2 0 x 1 1 0 2 3 # 7

65. 0 1 2 2x 0 ,

68. 0 3.7 2 5.5x 0 . 4.3

73. Any real numbers at least

1 2

unit from

62. 7 2 3 0 x 1 2 0 $ 214

1 2

66. `

69. 0 x2 2 1 0 # 8

In Exercises 71–76, write an inequality that fits the description. 71. Any real numbers less than seven units from 2.

58. 5 0 x 21 0 1 2 # 7

61. 3 2 2 0 x 1 4 0 , 5

64. 4 2 0 x 1 1 0 . 1

67. 0 2.6x 1 5.4 0 , 1.8

54. 0 6 2 5y 0 # 1

57. 2 0 4x 0 2 9 $ 3

60. 3 0 x 2 1 0 2 5 . 4

63. 9 2 0 2x 0 , 3

46. 0 x 2 1 0 , 3

2 2 3x 2 ` $ 5 5

70. 0 x2 1 4 0 $ 29

72. Any real numbers more than three units from 22.

3 2.

5

75. Any real numbers no more than two units from a.

11

74. Any real number no more than 3 units from 3 . 76. Any real number at least a units from 23.

• A P P L I C AT I O N S 77.  Temperature. If the average temperature in Hawaii is 83°F

1615° 2 , write an absolute value inequality representing the temperature in Hawaii. 78.  Temperature. If the average temperature of a human is 97.8°F 161.22, write an absolute value inequality describing normal human body temperature. 79.  Sports. Two women tee off the green of a par-3 hole on a golf course. They are playing “closest to the pin.” If the first woman tees off and lands exactly 4 feet from the hole, write an inequality that describes where the second woman must land in order to win the hole. What equation would suggest a tie? Let d 5 the distance from where the second woman lands to the tee. 80. Electronics. A band-pass filter in electronics allows certain frequencies within a range (or band) to pass through to the receiver and eliminates all other frequencies. Write an absolute value inequality that allows any frequency ƒ within 15 Hertz of the carrier frequency ƒc to pass.

In Exercises 81 and 82 refer to the following: A company is reviewing revenue for the prior sales year. The model for projected revenue and the model for actual revenue are Rprojected 5 200 1 5x Ractual 5 210 1 4.8x where x represents the number of units sold and R represents the revenue in thousands of dollars. Since the two revenue models are not identical, an error in projected revenue occurred. This error is represented by E 5  Rprojected 2 Ractual  81. Business. For what number of units sold was the error in

projected revenue less than $5000? 82.  Business. For what number of units sold was the error in

projected revenue less than $3000?

• C AT C H T H E M I S TA K E In Exercises 83–86, explain the mistake that is made. 83. Solve the absolute value equation 0  x 2 3 0 5 7.

Solution:

84. Solve the inequality 0 x 2 3 0 , 7.

Solution:



Eliminate the absolute value symbols. x 2 3 5 7

Eliminate the absolute x 2 3 , 27  or  x 2 3 . 7 value symbols.



Add 3 to both sides.



Add 3 to both sides.



The solution is 1 2q, 24 2 ∪ 1 10, q 2 .

Check.

x 5 10 0 10 2 3 0 5 7

This is incorrect. What mistake was made?

Young_AT_6160_ch01_pp129-163.indd 155



x , 24         x . 10

This is incorrect. What mistake was made?

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85. Solve the inequality 0 5 2 2x 0 # 1.

86. Solve the equation 0 5 2 2x 0 5 21.

Solution:

Eliminate the absolute value symbols.

21 # 5 2 2x # 1

Solution:        5 2 2x 5 21  or  5 2 2x 5 1 22x 5 26     22x 5 24



Subtract 5.

26 # 22x # 24





Divide by 22.

3#x#2





Write the solution in interval notation.





1 2q, 2 4 ∪ 3 3, q 2

This is incorrect. What mistake was made?

x 5 3  

 x52

The solution is 52, 36.

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 87–90, determine whether each statement is true or false. 87. 2 0 m 0 # m # 0 m 0 88. 0 n2 0 5 n2 89. 0 m 1 n 0 5 0 m 0 1 0 n 0 is true only when m and n are both nonnegative. 90. For what values of x does the absolute value equation 0 x 2 7 0 5 x 2 7 hold?

In Exercises 91–96, assuming a and b are real positive numbers, solve the equation or inequality and express the solution in interval notation. 91. 0 x 2 a 0 , b 92. 0 a 2 x 0 . b 93. 0 x 0 $ 2a 94. 0 x 0 # 2b 95. 0 x 2 a 0 5 b 96. 0 x 2 a 0 5 2b

• CHALLENGE 97.  For what values of x does the absolute value equation

0 x 1 1 0 5 4 1 0 x 2 2 0 hold?

98. Solve the inequality 0 3x2 2 7x 1 2 0 . 8.

• TECHNOLOGY 99. Graph y1 5 0 x 2 7 0 and y2 5 x 2 7 in the same screen. Do

the x-values where these two graphs coincide agree with your result in Exercise 90? 100. Graph y1 5 0 x 1 1 0 and y2 5 0 x 2 2 0 1 4 in the same screen. Do the x-values where these two graphs coincide agree with your result in Exercise 97? 101. Graph y1 5 0 3x2 2 7x 1 2 0 and y2 5 8 in the same screen. Do the x-values where y1 lies above y2 agree with your result in Exercise 98?

Young_AT_6160_ch01_pp129-163.indd 156

102. Solve the inequality 0 2.7x2 2 7.9x 1 5 0 # 0 5.3x2 2 9.2 0 by

graphing both sides of the inequality and identify which x-values make this statement true.

x ` , 1 by graphing both sides of the x1 1 inequality, and identify which x-values make this statement true.

103. Solve the inequality `

x ` , 2 by graphing both sides of the x1 1 inequality, and identify which x-values make this statement true.

104. Solve the inequality `

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Chapter Review 

157

[CHAPTER 1 REVIEW] SECTION

CONCEPT

KEY IDEAS/FORMULAS

1.1

Linear equations

ax 1 b 5 0

Solving linear equations in one variable

Isolate variables on one side and constants on the other side.

Solving rational equations that are reducible to linear equations

Any values that make the denominator equal to 0 must be eliminated as possible solutions.

1.2

Solving application problems using mathematical models

Five-step procedure: Step 1: Identify the question. Step 2: Make notes. Step 3: Set up an equation. Step 4: Solve the equation. Step 5: Check the solution.

Geometry problems

Formulas for rectangles, triangles, and circles

Interest problems

Simple interest: I 5 Prt

Mixture problems

Whenever two distinct quantities are mixed, the result is a mixture.

Distance–rate–time problems

d 5 r ⋅t

Quadratic equations

ax 2 1 bx 1 c 5 0      a 2 0

Factoring Square root method

If 1x 2 h2 1x 2 k2 5 0, then x 5 h or x 5 k.

Completing the square

Find half of b; square that quantity; add the result to both sides.

Quadratic Formula

1.4

1.5

CHAPTER 1 REVIEW

1.3

Applications involving linear equations

If x2 5 P, then x 5 6"P.

x5

2b 6 "b2 2 4ac 2a

Other types of equations Radical equations

Check solutions to avoid extraneous solutions.

Equations quadratic in form: u-substitution

Use a u-substitution to write the equation in quadratic form.

Factorable equations

Extract common factor or factor by grouping.

Linear inequalities

Solutions are a range of real numbers.

Graphing inequalities and interval notation

n 

a , x , b is equivalent to 1a, b2. x # a is equivalent to 12q, a4. n  x . a is equivalent to 1a, q2. n 

Solving linear inequalities

1.6

1.7

If an inequality is multiplied or divided by a negative number, the inequality sign must be reversed.

Polynomial and rational inequalities Polynomial inequalities

Zeros are values that make the polynomial equal to 0.

Rational inequalities

 he number line is divided into intervals. The endpoints of these intervals are values T that make either the numerator or denominator equal to 0. Always exclude values that make the denominator 5 0.

Absolute value equations and inequalities

0 b 2 a 0 is the distance between points a and b on the number line.

Equations involving absolute value Inequalities involving absolute value

If 0 x 0 5 a, then x 5 2a or x 5 a.

n  n 

Young_AT_6160_ch01_pp129-163.indd 157

0 x 0 # a is equivalent to 2a # x # a. 0 x 0 . a is equivalent to x , 2a or x . a.

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[CHAPTER 1 REVIEW EXERCISES] 28. Geometry. Find the perimeter of a triangle if one side is

1.1  Linear Equations

Solve for the variable.  1. 7x 2 4 5 12

2. 13d 1 12 5 7d 1 6

 3. 20p 1 14 5 6 2 5p

4. 4  1x 2 72 2 4 5 4

  5. 31x 1 72 2 2 5 4 1x 2 22

  6. 7c 1 31c 2 52 5 2 1c 1 32 2 14

  7. 14 2 3 23 1 y 2 4 2 1 9 4 5 3 4 1 2 y 1 3 2 2 6 4 1 4

29.

30.

REVIEW EXERCISES

  8. 3 6 2 4x 1 2 1 x 2 7 2 4 2 52 5 3 1 2x 2 4 2 1 6 3 3 1 2x 2 3 2 1 6 4

g 7 1g5 3 9

  9.

12 6 2 3 5 1 4 b b

10.

11.

13x x 3 2x5 2 7 4 14

12. 5b 1

b b 29 5 2 6 3 6

31.

Specify any values that must be excluded from the solution set and then solve. 13. 15. 17.

1 3 2 4 5 2 5 x x 2 7 6 2 5 t14 t t1t 1 42 3 6 2 5 9 x 2x

14.

4 8 2 53 x11 x21

3 22 5 2x 2 7 3x 1 1 3 2 1 5/m 2 51 18. 2 1 1 5/m 2

32.

16.

33.

19. 7x 2 12 2 4x2 5 3 326 1 14 2 2x 1 724 1 12 20.

x x23 2 5 26 5 15

Solve for the specified variable.

34.

21. Solve for x in terms of y:

22. If y 5

3x 2 2 31 y 1 423 2 7 4 5 y 2 2x 1 6 1x 2 32 y12 x13 , find in terms of x. 1 1 2x 1 2 2y

10 inches, another side is 13 of the perimeter, and the third side is 16 of the perimeter. Investments. You win $25,000 and you decide to invest the money in two different investments: one paying 20% and the other paying 8%. A year later you have $27,600 total. How much did you originally invest in each account? Investments. A college student on summer vacation was able to make $5000 by working a full-time job every summer. He invested half the money in a mutual fund and half the money in a stock that yielded four times as much interest as the mutual fund. After a year he earned $250 in interest. What were the interest rates of the mutual fund and the stock? Chemistry. For an experiment, a student requires 150 milliliters of a solution that is 8% NaCl (sodium chloride). The storeroom has only solutions that are 10% NaCl and 5% NaCl. How many milliliters of each available solution should be mixed to get 150 milliliters of 8% NaCl? Chemistry. A mixture containing 8% salt is to be mixed with 4 ounces of a mixture that is 20% salt, in order to obtain a solution that is 12% salt. How much of the first solution must be used? Grades. Going into the College Algebra final exam, which will count as two tests, Danny has test scores of 95, 82, 90, and 77. If his final exam is higher than his lowest test score, then it will count for the final exam and replace the lowest test score. What score does Danny need on the final in order to have an average score of at least 90? Car Value. A car salesperson reduced the price of a model car by 20%. If the new price is $25,000, what was its original price? How much can be saved by purchasing the model?

1.3  Quadratic Equations

Solve by factoring.

1.2  Applications Involving Linear Equations

35. b2 5 4b 1 21

23. Transportation. Maria is on her way from her home near

36. x  1x 2 32 5 54

24.

25. 26.

27.

Orlando to the Sun Dome in Tampa for a rock concert. She drives 16 miles to the Orlando park-n-ride, takes a bus 34 of 1 of the way to a bus station in Tampa, and then takes a cab 12 the way to the Sun Dome. How far does Maria live from the Sun Dome? Diet. A particular 2000 calorie per day diet suggests eating breakfast, lunch, dinner, and four snacks. Each snack is 14 the calories of lunch. Lunch has 100 calories less than dinner. Dinner has 1.5 times as many calories as breakfast. How many calories are in each meal and snack? Numbers. Find a number such that 12 more than 14 the ­number is 13 the number. Numbers. Find four consecutive odd integers such that the sum of the four numbers is equal to three more than three times the fourth integer. Geometry. The length of a rectangle is one more than two times the width, and the perimeter is 20 inches. What are the dimensions of the rectangle?

Young_AT_6160_ch01_pp129-163.indd 158

37. x2 5 8x

38. 6y2 2 7y 2 5 5 0

Solve by the square root method. 39. q2 2 169 5 0 40. c2 1 36 5 0 2 41. 12x 2 42  5 264

2 42. 1d 1 72  2450

Solve by completing the square.

43. x2 2 4x 2 12 5 0 44. 2x2 2 5x 2 7 5 0

x2 x 541 2 2 46. 8m 5 m2 1 15 45.

30/11/16 10:20 AM

Review Exercises 

Solve by the Quadratic Formula.

4 78. 3 1x 2 42  2 111x 2 42 2 2 20 5 0

79. y22 2 5y21 1 4 5 0

47. 3t2 2 4t 5 7

80. p22 1 4p21 5 12

48. 4x2 1 5x 1 7 5 0 7

1

159

81. 3x1/3 1 2x2/3 5 5

49. 8ƒ 2 2 3 ƒ 5 6 50. x2 5 26x 1 6

82. 2x2/3 2 3x1/3 2 5 5 0 83. x 22/3 1 3x 21/3 1 2 5 0

Solve by any method.

84. y 21/2 2 2y 21/4 1 1 5 0

2

51. 5q 2 3q 2 3 5 0

85. x4 1 5x2 5 36

2 52. 1x 2 72  5 212

86. 3 2 4x21/2 1 x21 5 0

2

53. 2x 2 3x 2 5 5 0

Solve the equation by factoring.

55. 7x 5 219x 1 6

87. x3 1 4x2 2 32x 5 0

56. 7 5 2b2 1 1

88. 9t3 2 25t 5 0

2

Solve for the indicated variable. 57. S 5 p2h  for r

pr h   for r 3 h 5 vt 2 16t 2  for v A 5 2pr 2 1 2prh  for h Geometry. Find the base and height of a triangle with an area of 2 square feet if its base is 3 feet longer than its height. Falling Objects. A man is standing on top of a building 500 feet tall. If he drops a penny off the roof, the height of the penny is given by h 5 216t2 1 500, where t is in seconds. Determine how many seconds it takes until the penny hits the ground.

58. V 5

60. 61.

62.

89. p3 2 3p2 2 4p 1 12 5 0 90. 4x3 2 9x2 1 4x 2 9 5 0 91. p12p 2 522 2 312p 2 52 5 0

3

59.

REVIEW EXERCISES

54. 1  g 2 22 1  g 1 52 5 27

92. 2 1t2 2 923 2 20  1t2 2 922 5 0 93. y 2 81y21 5 0

94. 9x3/2 2 37x1/2 1 4x21/2 5 0 1.5  Linear Inequalities

Rewrite using interval notation. 95. x # 24 96. 21 , x # 7 97. 2 # x # 6 98. x . 21

1.4  Other Types of Equations

Rewrite using inequality notation.

Solve the radical equation for the given variable.

99. 126, q2 101. 323, 74

3

63. ! 2x 2 4 5 2

100. 12q, 04 102. 125, 24

64. !x 2 2 5 24

Express each interval using inequality and interval notation.

67. x 2 4 5 "x 2 1 5x 1 6

104.

65. 1 2x 2 7 2 1/5 5 3

103.

66. x 5 !7x 2 10

68. !2x 2 7 5 !x 1 3

69. !x 1 3 5 2 2 !3x 1 2 70. 4 1 !x 2 3 5 !x 2 5 71. x 2 2 5 "49 2 x

2

72. !2x 2 5 2 !x 1 2 5 3 73. 2x 5 !3 2 x

74. #15 1 2!x 2 4 1 !x 5 5

Solve the equation by introducing a substitution that transforms the equation to quadratic form. 75. 228 5 13x 2 222 2 1113x 2 22 76. x 4 2 6x2 1 9 5 0 77. a

2 x x b 5 15 2 2 a b 12x 12x

Young_AT_6160_ch01_pp129-163.indd 159

Graph the indicated set and write as a single interval, if ­possible. 105. 1 4, 6 4 ∪ 3 5, q 2

107. 1 3, 12 4 ∩ 3 8, q 2

106. 1 2q, 23 2 ∪ 3 27, 2 4

Solve and graph.

108. 1 2q, 22 2 ∩ 3 22, 9 2

109. 2x , 5 2 x

110. 6x 1 4 # 2

111. 4 1x 2 12 . 2x 2 7

112.

113. 6 , 2 1 x # 11 115.

2 11x 3 # # 3 6 4

x13 $6 3

114. 26 # 1 2 4 1x 1 22 # 16 116.

x x14 x 1 1 . 2 3 9 6 3

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160 

CHAPTER 1  Equations and Inequalities

148. Blood Alcohol Level. If a person registers a 0.08 blood

Applications

alcohol level, he will be issued a DUI ticket in the state of Florida. If the test is accurate within 0.007, write a linear inequality representing an actual blood alcohol level that will not be issued a ticket.

117. Grades. In your algebra class your first four exam grades

are 72, 65, 69, and 70. What is the lowest score you can get on the fifth exam to earn a C for the course? Assume that each exam is equal in weight and a C is any score greater than or equal to 70. 118. Profit. A tailor decided to open a men’s custom suit business. His fixed costs are $8500 per month, and it costs him $50 for the materials to make each suit. If the price he charges per suit is $300, how many suits does he have to tailor per month to make a profit?

Technology Exercises

Section 1.1

Graph the function represented by each side of the question in the same viewing rectangle, and solve for x. 149. 0.031x 1 0.01714000 2 x2 5 103.14

REVIEW EXERCISES

1.6  Polynomial and Rational Inequalities

Solve the polynomial inequality and express the solution set using interval notation. 119. x2 # 36

120. 6x2 2 7x , 20

121. 4x # x2

122. 2x2 $ 9x 1 14

123. 2x2 , 27x

124. x2 , 24

125. 4x2 2 12 . 13x

126. 3x # x2 1 2

Solve the rational inequality and express the solution set using interval notation. x x21 127. , 0 128. .0 x23 x24 129.

x 2 2 3x $ 18 3

130.

x 2 2 49 $0 x27

131.

3 1 2 # 0 x22 x24

132.

4 2 # x21 x13

2

133.

x 19 $ 0 x23

134. x ,

5x 1 6 x

1.7  Absolute Value Equations and Inequalities

Solve the equation. 135. 137.

0 x 2 3 0 5 24

0 3x 2 4 0 5 1.1

136. 0 2 1 x 0 5 5

138. 0 x 2 2 6 0 5 3

Solve the inequality and express the solution using interval notation. 139. 141. 143. 145.

0 x 0 , 4

140. 0 x 2 3 0 , 6

0 2x 0 . 6

144. `

0 x 1 4 0 . 7 0 2 1 5x 0 $ 0

Applications

142. 0 27 1 y 0 # 4

4 1 2x 1 ` $ 3 7

146. 0 1 2 2x 0 # 4

147. Temperature. If the average temperature in Phoenix is

150.

1 0.2 1 2 5 x 0.16x 4

Section 1.3 151. a. Solve the equation x2 1 4x 5 b, b 5 5 by first writing in

standard form and then factoring. Now plot both sides of the equation in the same viewing screen 1 y1 5 x2 1 4x and y2 5 b2. At what x-values do these two graphs intersect? Do those points agree with the solution set you found? b.  Repeat part (a) for b 5 25, 0, 7, and 12. 152. a. Solve the equation x2 2 4x 5 b, b 5 5 by first writing in standard form and then factoring. Now plot both sides of the equation in the same viewing screen 1 y1 5 x2 2 4x and y2 5 b2. At what x-values do these two graphs intersect? Do those points agree with the solution set you found? b.  Repeat part (a) for b 5 25, 0, 7, and 12. Section 1.4 153. Solve the equation 2x 1/4 5 2x 1/2 1 6. Round your answer

to two decimal places. Plot both sides of the equation in the same viewing screen, y1 5 2x 1/4 and y2 5 2x 1/2 1 6. Does the point(s) of intersection agree with your solution? 154. Solve the equation 2x 21/2 5 x 21/4 1 6. Plot both sides of the equation in the same viewing screen, y1 5 2x 21/2 and y2 5 x 21/4 1 6. Does the point(s) of intersection agree with your solution? Section 1.5 155. a. Solve the inequality 20.61x 1 7.62 . 0.24x 2 5.47.

Express the solution set using interval notation. b.  Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies above the graph of the right side. c. Do parts (a) and (b) agree? 3 1 156. a. Solve the inequality 2 2 x 1 7 , 4 x 2 5. Express the solution set using interval notation. b.  Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies below the graph of the right side. c.  Do parts (a) and (b) agree?

858F 1610° 2 , write an inequality representing the average temperature T in Phoenix.

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Review Exercises 

161

Section 1.6

Section 1.7

Plot the left side and the right side of each inequality in the same screen, and use the zoom feature to determine the range of values for which the inequality is true.

161. Solve the inequality 0 1.6x 2 2 4.5 0 , 3.2 by graphing both

157. 0.2x 2 2 2 . 0.05x 1 3.25 158. 12x 2 2 7x 2 10 , 2x 2 1 2x 2 1 159.

3p .1 7 2 2p

160.

7p ,1 15 2 2p

sides of the inequality, and identify which x-values make this statement true. Express the solution using interval ­notation and round to two decimal places. 162. Solve the inequality 0 0.8x 2 2 5.4x 0 . 4.5 by graphing both sides of the inequality, and identify which x-values make this statement true. Express the solution using interval ­notation and round to two decimal places.

REVIEW EXERCISES

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162 

CHAPTER 1  Equations and Inequalities

[CHAPTER 1 PRACTICE TEST] Solve the equation.

25. Puzzle. A piling supporting a bridge sits so that 4 of the

  1. 4p 2 7 5 6p 2 1



  2. 22 1z 2 12 1 3 5 23z 1 3 1z 2 12   3. 3t 5 t2 2 28

  4. 8x2 2 13x 5 6   5. 6x2 2 13x 5 8   6.

3 5 5 x21 x12

  7.

5 30 115 2 y23 y 29

  8. x4 2 5x2 2 36 5 0   9. !2x 1 1 1 x 5 7

10. 2x2/3 1 3x1/3 2 2 5 0 11. !3y 2 2 5 3 2 !3y 1 1

12. x  13x 2 523 2 2 13x 2 522 5 0 13. x7/3 2 8x4/3 1 12x1/3 5 0

Solve for the specified variable. 9

­piling is in the sand, 150 feet is in the water, and 35 of the piling is in the air. What is the total height of the piling? 26. Real Estate. As a real estate agent you earn 7% of the sale price. The owners of a house you have listed at $150,000 will entertain offers within 10% of the list price. Write an inequality that models the commission you could make on this sale. 27. Costs: Cell Phones. A cell phone company charges $49 for a 600-minute monthly plan, plus an additional $0.17 per minute for every minute over 600. If a customer’s bill ranged from a low of $53.59 to a high of $69.74 over a 6-month period, write an inequality expressing the number of monthly minutes used over the 6-month period. 28. Television. Television and film formats are classified as ratios of width to height. Traditional televisions have a 4:3 ratio (1.33:1), and movies are typically made in widescreen format with a 21:9 ratio (2.35:1). If you own a traditional 25-inch television (20 inch 3 15 inch screen) and you play a widescreen DVD on it, there will be black bars above and below the image. What are the dimensions of the movie and of the black bars? 25

Solve the inequality and express the solution in interval ­notation.

15

20

16. 7 2 5x . 218 17. 3x 1 19 $ 5 1x 2 32 18. 21 # 3x 1 5 , 26 19.

Andy Washnik

PRACTICE TEST

14. F 5 5 C 1 32  for C 15. P 5 2L 1 2W  for L

1

2 x18 1 , # 5 4 2

20. 3x $ 2x2 21. 3p2 $ p 1 4 22. 0 5 2 2x 0 . 1 23.

x23 #0 2x 1 1

24.

x14 $0 x2 2 9

Young_AT_6160_ch01_pp129-163.indd 162

1 0.45 1 2 5 . Graph the function x 0.75x 9  represented by each side in the same viewing rectangle and solve for x. 30. Solve the inequality 0.3 1 0 2.4x 2 2 1.5 0 # 6.3 by graphing both sides of the inequality, and identify which x-values make this statement true. Express the solution using interval notation. 29. Solve the equation

30/11/16 10:20 AM

Cumulative Test 

163

[CHAPTER 1 CUMULATIVE TEST] Simplify.

15. Tim can paint the interior of a condo in 9 hours. If Chelsea

 1. 5 ⋅ 1 7 2 3 ⋅ 4 1 2 2

Simplify and express in terms of positive exponents. 23   2. 1 4x 23b4 2

3.

16. 17.

1 x 2 y 22 2 3

18.

1 x 2 y 2 23

19.

Perform the operations and simplify.

20.

  4. 12x 4 1 2x32 1 1x3 2 5x 2 62 2 15x4 1 4x3 2 6x 1 82

is hired to help him, they can do a similar condo in 5 hours. Working alone, how long will it take Chelsea to paint a similar condo? Solve using the square root method: y 2 1 36 5 0. Solve by completing the square: x2 1 12x 1 40 5 0. Solve using the Quadratic Formula: x2 1 x 1 9 5 0. Solve and check: "4 2 x 5 x 2 4. Solve using substitution: 3x22 1 8x21 1 4 5 0.

  5. x2 1x 2 52 1x 2 32

Solve and express the solution in interval notation.

Factor completely.

21. 0 , 4 2 x # 7

  6. 3x3 2 3x2 2 60x

7. 2a3 1 2000

23.

Perform the operations and simplify.   8.

32x 5x 2 15 4 2 x11 x 21

9.

1

6x 5x 2 x22 x12 2

6x 8x 7x 2 542 5 3 15

14.

7

4 2 5x 3 ` $ 7 14

37

1

Solve for x.

2

24. `

26. Solve the equation x 6 1 8 x 3 5 27. Plot both sides of the 37

11. 7 x 5 8 x 1 9

12. Perform the operation and express in standard form: 

13.

x12 $ 0 9 2 x2

25. Solve for x:  5 x 1 3  5 15 .

Solve for x. 10. x3 2 x2 2 30x 5 0

22. 4x2 , 9x 2 11

45 . 6 2 3i

equation in the same viewing screen, y1 5 x 6 1 8 x 3 and y2 5 27. Does the point(s) of intersection agree with your solution? 3x 27. Solve the inequality ` ` , 1 by graphing both sides x22 of the inequality, and identify which x-values make this statement true.

x26 3 5 62x 2

CUMULATIVE TEST

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[2[ CHAPTER

Graphs

Graphs are used in many ways. There is only one temperature that yields the same number in degrees Celsius and degrees Fahrenheit. Do you know what it is?*­­The penguins are a clue.

Fuse/Getty Images

Antenatal clinic prevalence (%)

25

Emperor penguins walking in a line, Weddell Sea, Antarctica

ºF

20

South Africa

15 10 5 Thailand 1994 1998 2002 2006 2010 2014 Year

200

HIV infection rates (Data from http://data.worldbank.org/indicator/SH.DYN.AIDS.ZS/ countries/1W?page=4&display=default)

100

COKE 5–Minute

4:53 PM

ºC 50

100

55.00

The conversion between degrees Fahrenheit and degrees Celsius is a linear relationship. Notice that 08C corresponds to 328F.

54.75 10

the distance between two points and the midpoint of a line segment joining two points.

11

12

1

2

3

54.50

Stock prices fluctuating throughout the day

LEARNING OBJECTIVES ■■ Calculate

55.25

■■ Sketch

the graph of an equation using intercepts and symmetry as graphing aids.

■■ Find

the equation of a line. circles. ■■ Find the line of best fit for a given set of data.** ■■ Graph

*See Section 2.3, Exercise 107. **Optional Technology Required Section

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[IN THIS CHAPTER] You will review the Cartesian plane. You will calculate the distance between two points and find the midpoint of a line segment joining two points. You will then apply point-plotting techniques to sketch graphs of equations. Special attention is given to two types of equations: lines and circles.

G R AP H S 2.1

2.2

2.3

2.4

2.5*

BASIC TOOLS: CARTESIAN PLANE, DISTANCE, AND MIDPOINT

GRAPHING EQUATIONS: POINT-PLOTTING, INTERCEPTS, AND SYMMETRY

LINES

CIRCLES

LINEAR REGRESSION: BEST FIT

• Cartesian Plane • Distance Between Two Points • Midpoint of a Line Segment Joining Two Points

• Point-Plotting • Intercepts • Symmetry • Using Intercepts and Symmetry as Graphing Aids

• Graphing a Line • Equations of Lines • Parallel and Perpendicular Lines

• Scatterplots • Standard Equation • Identifying Patterns of a Circle • Linear Regression • Transforming Equations of Circles to the Standard Form by Completing the Square

*Optional Technology Required Section

165

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166 

CHAPTER 2  Graphs

2.1 BASIC TOOLS: CARTESIAN PLANE, DISTANCE, AND MIDPOINT SKILLS OBJECTIVES ■■ Plot points on the Cartesian plane. ■■ Calculate the distance between two points in the Cartesian plane. ■■ Find the midpoint of a line segment joining two points in the Cartesian plane.

CONCEPTUAL OBJECTIVES ■■ Expand the concept of a one-dimensional number line to a two-dimensional plane. ■■ Derive the distance formula using the Pythagorean theorem. ■■ Conceptualize the midpoint as the average of the x- and y-coordinates.

2.1.1  Cartesian Plane 2.1.1 S K I L L

Plot points on the Cartesian plane. 2.1.1 C O N C E P T U A L

Expand the concept of a one-dimensional number line to a two-dimensional plane. y-axis

II

I Origin

III

x-axis

IV

y

II x < 0, y > 0

I x > 0, y > 0

HIV infection rates, stock prices, and temperature conversions are all examples of relationships between two quantities that can be expressed in a two-dimensional graph. Because it is two dimensional, such a graph lies in a plane. Two perpendicular real number lines, known as the axes in the plane, intersect at a point we call the origin. Typically, the horizontal axis is called the x-axis, and the vertical axis is denoted as the y-axis. The axes divide the plane into four quadrants, numbered by Roman numerals and ordered counterclockwise. Points in the plane are represented by ordered pairs, denoted 1x, y2. The first number of the ordered pair indicates the position in the horizontal direction and is often called the x-coordinate or abscissa. The second number indicates the position in the vertical d­ irection and is often called the y-coordinate or ordinate. The origin is denoted 10, 02. Examples of other coordinates are given on the graph to the right. The point 12, 42 lies in quadrant I. To plot this point, start at the origin 10, 02 and move to the right two units y x-coordinate and up four units. All points in quadrant I have positive coordinates, (2, 4) and all points in quadrant III have negative coordinates. y-coordinate Quadrant II has negative x-coordinates and positive (–4, 1) x y-coordinates; quadrant IV has positive x-coordinates (0, 0) (5, 0) and negative y-coordinates. (0, –2) This representation is called the rectangular (–3, –2) coordinate system or Cartesian coordinate system, (3, –4) named after the French mathematician René Descartes (1596–1650).

x III x < 0, y < 0

IV x > 0, y < 0

EXAMPLE 1  Plotting Points in a Cartesian Plane a. Plot and label the points 121, 242, 12, 22, 122, 32, 12, 232, 10, 52, and 123, 02

in the Cartesian plane. b. List the points and corresponding quadrant or axis in a table. Solution: a.  (–2, 3)

[ CONCEPT CHECK ] If the point (a, b) lies in quadrant I, then the point (2a, 2b) lies in which quadrant?

▼ ANSWER Quadrant III.

Young_AT_6160_ch02_pp164-201.indd 166

   b. 

y (0, 5) (2, 2) x

(–3, 0) (2, –3) (–1, –4)

POINT

QUADRANT

12, 22

II

I

122, 32

III

12, 232

y-axis

121, 242 10, 52

123, 02

IV x-axis

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2.1  Basic Tools: Cartesian Plane, Distance, and Midpoint 

167

2.1.2  Distance Between Two Points Suppose you want to find the distance between any two points in the plane. In the previous graph, to find the distance between the points 12, 232 and 12, 22, count the units between the two points. The distance is 5. What if the two points do not lie along a horizontal or vertical line? Example 2 uses the Pythagorean theorem to help find the distance between any two points.

2.1.2 S K I L L

Calculate the distance between two points in the Cartesian plane. 2.1.2 C O N C E P T U A L

Derive the distance formula using the Pythagorean theorem.

EXAMPLE 2  Finding the Distance Between Two Points

Find the distance between the points 122, 212 and 11, 32. Solution:

STEP 1  Plot

and label the two points in the Cartesian plane and draw a line segment indicating the distance d between the two points.

y (1, 3) d

x

(–2, –1)

STEP 2  Form

a right triangle by connecting the points to a third point, 11, 212.

y (1, 3)

STEP 3  Calculate

the length of the horizontal segment. 3 5 |1 2 1222  | Calculate the length of the vertical segment. 4 5 |3 2 1212  |

(–2, –1)

d

4

3

(1, –1)

x

the Pythagorean theorem to d2 5 32 1 42 calculate the distance d.     d2 5 25        d 5 5

STEP 4  Use

WORDS

MATH

 or any two points, F 1x1, y12 and 1x2, y2 2:

y (x2, y2)

d

(x1, y1)

| y2 – y1|

x

|x2 – x1|

The distance along the horizontal segment is the absolute value of the difference between the x-values.        0 x2 2 x1 0

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168 

CHAPTER 2  Graphs

 he distance along the vertical T segment is the absolute value of the 0 y2 2 y1 0 difference between the y-values.           

 se the Pythagorean theorem to U calculate the distance d.          d 2 5 0 x2 2 x1 0 2 1 0 y2 2 y1 0 2

0 a 0 2 5 a2 for all real numbers a.

d 2 5 1x2 2 x122 1 1 y2 2 y122

d 56" 1 x2 2 x1 2 2 1 1 y2 2 y1 2 2 Use the square root property.            Distance can be only positive.            d 5 " 1 x2 2 x1 2 2 1 1 y2 2 y1 2 2

ST U DY TIP It does not matter which point is taken to be the first point or the second point.

DEFINITION

Distance Formula

The distance d between two points P1 5 1x1, y12 and P2 5 1x2, y22 is given by d 5 " 1 x2 2 x1 2 2 1 1 y2 2 y1 2 2

The distance between two points is the square root of the sum of the square of the distance between the x-coordinates and the square of the distance between the y-coordinates.

You will prove in the exercises that it does not matter which point you take to be the first point when applying the distance formula.

[ CONCEPT CHECK ] Which point is P1 and which point is P2?

▼ ANSWER It does not matter which point is labeled P1 and which point is labeled P2.

EXAMPLE 3  U  sing the Distance Formula to Find the Distance Between Two Points

Find the distance between 123, 72 and 15, 222. Solution:

Write the distance formula. Substitute 1x1, y12 5 123, 72 and 1x2, y22 5 15, 222. Simplify.

d 5 " 3 x2 2 x1 4 2 1 3 y2 2 y1 4 2

d 5 " 3 5 2 1 23 2 4 2 1 3 22 2 7 4 2 d 5 " 3 5 1 3 4 2 1 3 22 2 7 4 2

                d 5 "82 1 1 29 2 2 5 "64 1 81 5 "145

Solve for d.

▼

▼ ANSWER

d 5 "58

Y O U R T U R N   Find the distance between 14, 252 and 123, 222.

y

P1 = (x 1, y 1 ) d

d 5 "145

2.1.3  Midpoint of a Line Segment Joining Two Points (xm, ym) x d P2 = (x 2 , y 2 )

Young_AT_6160_ch02_pp164-201.indd 168

The midpoint, 1xm, ym 2 , of a line segment joining two points 1x1, y1  2 and 1x2, y2  2 is defined as the point that lies on the segment that has the same distance d from both points. In other words, the midpoint of a segment lies halfway between the given endpoints. The coordinates of the midpoint are found by averaging the x-coordinates and averaging the y-coordinates.

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169

2.1  Basic Tools: Cartesian Plane, Distance, and Midpoint 

DEFINITION

2.1.3 S K I L L

Midpoint Formula

Find the midpoint of a line segment joining two points in the Cartesian plane.

The midpoint, 1xm, ym 2 , of the line segment with endpoints 1x1, y12 and 1x2, y22 is given by x 1 x2 y1 1 y2 1 xm, ym 2 5 a 1 , b 2 2 The midpoint can be found by averaging the x-coordinates and averaging the y-coordinates.

2.1.3 C O N C E P T U A L

Conceptualize the midpoint as the average of the x- and y-coordinates.

EXAMPLE 4  Finding the Midpoint of a Line Segment

Find the midpoint of the line segment joining the points 12, 62 and 124, 222 . Solution:

Write the midpoint formula.

1 xm, ym 2 5 a

Substitute 1x1, y12 5 12, 62 and 1x2, y22 5 124, 222.

1 xm, ym 2 5 a

x1 1 x2 y1 1 y2 , b 2 2

2 1 1 24 2 6 1 1 22 2 , b 2 2

1xm, ym2 5 121, 22

Simplify.

One way to verify your answer is to plot the given points and the midpoint to make sure your answer looks reasonable.

[ CONCEPT CHECK ] Graph the two points (3, –4) and (5, 8) and draw the line segment connecting these points. Take a “guess” at the midpoint.

y (2, 6)

(–1, 2)

▼ ANSWER Note the midpoint is (4, 2).

x

(–4, –2)

▼ ANSWER

Midpoint 5 14, 22

▼ YOUR TURN  F  ind the midpoint of the line segment joining the points

13, 242 and 15, 82.

[ S E C T I O N 2 .1]     S U M M A R Y CARTESIAN PLANE

Plotting coordinates: 1x, y2 Quadrants: I, II, III, and IV n Origin: 10, 02 n n

D I S TA N C E B E T W E E N T W O P O I N T S

y-axis (–, +) II III (–, –)

(+, +) I Origin IV (+, –)

x-axis

d 5 " 1 x2 2 x1 2 2 1 1 y2 2 y1 2 2

MIDPOINT OF LINE SEGMENT JOINING TWO POINTS

Midpoint 5 1 xm, ym 2 5 a

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x1 1 x2 y1 1 y2 , b 2 2

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CHAPTER 2  Graphs

[ S E C T I O N 2 .1]   E X E R C I S E S • SKILLS In Exercises 1–6, give the coordinates for each point labeled. y 1. Point A 2. Point B B 3. Point C A 4. Point D C 5. Point E 6. Point F F

x

D E

In Exercises 7 and 8, plot each point in the Cartesian plane and indicate in which quadrant or on which axis the point lies. 7. A:  122, 32  B:  11, 42  C:  123, 232  D:  15, 212  E:  10, 222  F:  14, 02 8. A:  121, 22  B:  11, 32  C:  124, 212  D:  13, 222  E:  10, 52  F: 123, 02 9. Plot the points 123, 12, 123, 42, 123, 222, 123, 02, 123, 242. Describe the line containing points of the form 123, y2. 10. Plot the points 121, 22, 123, 22, 10, 22, 13, 22, 15, 22. Describe the line containing points of the form 1x, 22. In Exercises 11–32, calculate the distance between the given points, and find the midpoint of the segment joining them. 11. 11, 32 and 15, 32 12. 122, 42 and 122, 242 13. 121, 42 and 13, 02 14. 123, 212 and 11, 32 15. 1210, 82 and 127, 212 16. 122, 122 and 17, 152 17. 123, 212 and 127, 22 18. 124, 52 and 129, 272 19. 126, 242 and 122, 282 20. 10, 272 and 124, 252 7 10 1 1 21. A22 , 3 B and A 2 , 3 B 22. A 15, 73 B and A 95,223 B 2

1

7 1

1 1

1

7

23. A2 3 ,2 5 B and A 4 , 3 B 24. A 5 , 9 B and A 2 , 23 B 25. 121.5, 3.22 and 12.1, 4.72 26. 121.2, 22.52 and 13.7, 4.62 27. 1214.2, 15.12 and 116.3, 217.52 28. 11.1, 2.22 and 13.3, 4.42

1 3!5, 23!3 2 and 1 2 !5, 2 !3 2 29. 1 !3, 5!2 2 and 1 !3, !2 2 30. 1 2!5, 4 2 and 1 1, 2!3 2 31. 1 1, !3 2 and 1 2 !2, 22 2 32.

In Exercises 33 and 34, calculate (to two decimal places) the perimeter of the triangle with the following vertices: 33. Points A, B, and C y B 34. Points C, D, and E E x C A D

In Exercises 35–38, determine whether the triangle with the given vertices is a right triangle, an isosceles triangle, neither, or both. (Recall that a right triangle satisfies the Pythagorean theorem and an isosceles triangle has at least two sides of equal length.) 35. 10, 232, 13, 232, and 13, 52

36. 10, 22, 122, 222, and 12, 222

37. 11, 12, 13, 212, and 122, 242 38. 123, 32, 13, 32, and 123, 232

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2.1  Basic Tools: Cartesian Plane, Distance, and Midpoint 

• A P P L I C AT I O N S 39.  Cell Phones. A cellular phone company currently has three

43. NASCAR Revenue. Action Performance Inc., the leading

towers: one in Tampa, one in Orlando, and one in Gainesville to serve the central Florida region. If Orlando is 80 miles east of Tampa and Gainesville is 100 miles north of Tampa, what is the distance from Orlando to Gainesville? 40.  Cell Phones. The same cellular phone company in Exercise 39 has decided to add towers at each “halfway” between cities. How many miles from Tampa is each “halfway” tower? 41. Travel. A retired couple who live in Columbia, South Carolina, decide to take their motor home and visit two children who live in Atlanta and in Savannah, Georgia. Savannah is 160 miles south of Columbia, and Atlanta is 215 miles west of Columbia. How far apart do the children live from each other?

seller of NASCAR merchandise, recorded $260 million in revenue in 2002 and $400 million in revenue in 2004. Calculate the midpoint to estimate the revenue Action Performance Inc. recorded in 2003. Assume the horizontal axis represents the year and the vertical axis represents the revenue in millions.

Revenue in Millions

500

Greenville Marietta

400

(2002, 260)

200 100

Columbia

Atlanta

(2004, 400)

300

2001

2004

2007 Year

2010

Charleston

44.  Ticket Price. In 1993 the average Miami Dolphins ticket

Macon

price was $28, and in 2001 the average price was $56. Find the midpoint of the segment joining these two points to estimate the ticket price in 1997.

Columbus

Savannah

Average Cost of Miami Dolphins Ticket

Albany

42. Sports. In the 1984 Orange Bowl, Doug Flutie, the 5 foot 9 inch

quarterback for Boston College, shocked the world as he threw a “hail Mary” pass that was caught in the end zone with no time left on the clock, defeating the Miami Hurricanes 47–45. Although the record books have it listed as a 48-yard pass, what was the actual distance the ball was thrown? The following illustration depicts the path of the ball. End Zone

y

(2001, 56)

60 40 20 (1993, 28) 1995

2000 Year

(5, 50)

x

(–10, 2) Midfield

In Exercises 45 and 46, refer to the following: It is often useful to display data in visual form by plotting the data as a set of points. This provides a graphical display between the two variables. The following table contains data on the average monthly price of gasoline.

U.S. All Grades Conventional Retail Gasoline Prices, 2000–2015 (Dollars per Gallon) YEAR

JAN

FEB

MAR

APR

MAY

JUN

JUL

AUG

SEP

OCT

NOV

DEC

2000

1.319

1.409

1.538

1.476

1.496

1.645

1.568

1.480

1.562

1.546

1.533

1.458

2001

1.467

1.471

1.423

1.557

1.689

1.586

1.381

1.422

1.539

1.312

1.177

1.111

2002

1.134

1.129

1.259

1.402

1.394

1.380

1.402

1.398

1.403

1.466

1.424

1.389

2003

1.464

1.622

1.675

1.557

1.477

1.489

1.519

1.625

1.654

1.551

1.512

1.488

2004

1.595

1.654

1.728

1.794

1.981

1.950

1.902

1.880

1.880

1.993

1.973

1.843

2005

1.852

1.927

2.102

2.251

2.155

2.162

2.287

2.489

2.907

2.736

2.265

2.216

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CHAPTER 2  Graphs

YEAR

JAN

FEB

MAR

APR

MAY

JUN

JUL

AUG

SEP

OCT

NOV

DEC

2006

2.343

2.293

2.454

2.762

2.873

2.849

2.964

2.952

2.548

2.258

2.254

2.328

2007

2.237

2.276

2.546

2.831

3.157

3.067

2.989

2.821

2.858

2.838

3.110

3.032

2008

3.068

3.064

3.263

3.468

3.783

4.038

4.051

3.789

3.760

3.065

2.153

1.721

2009

1.821

1.942

1.987

2.071

2.289

2.645

2.530

2.613

2.530

2.549

2.665

2.620

2010

2.730

2.657

2.793

2.867

2.847

2.733

2.728

2.733

2.727

2.816

2.866

3.004

2011

3.109

3.219

3.561

3.796

3.900

3.678

3.665

3.664

3.624

3.454

3.385

3.277

2012

3.388

3.576

3.827

3.893

3.698

3.515

3.433

3.724

3.859

3.714

3.444

3.322

2013

3.324

3.668

3.713

3.566

3.621

3.634

3.582

3.583

3.542

3.358

3.263

3.288

2014

3.331

3.382

3.546

3.665

3.678

3.699

3.615

3.501

3.431

3.202

2.957

2.575

2015

2.136

2.235

2.433

2.454

2.661

2.783

2.750

2.608

2.369

2.325

2.188

2.052

Source: U.S. Energy Information Administration (Feb. 2016).

Dollars per Gallon (2000)

The following graph displays the data for the year 2000. 45. Economics. Create a graph displaying the price of gasoline

$2.00 1.80 1.60 1.40 1.20 1.00 0.80 0.60 0.40 0.20

for the year 2012. 46. Economics. Create a graph displaying the price of gasoline for the year 2014.

2

4

6 8 Month

10

12

• C AT C H T H E M I S TA K E In Exercises 47–50, explain the mistake that is made. 47. Calculate the distance between 12, 72 and 19, 102.

Solution:

Write the distance formula.  d 5 " 1 x2 2 x1 2 2 1 1 y2 2 y1 2 2

Substitute 12, 72 and 19, 102.        d 5 " 1 7 2 2 2 2 1 1 10 2 9 2 2

Simplify.  



                      d 5 " 1 5 2 2 1 1 1 2 2 5 "26

This is incorrect. What mistake was made? 48. Calculate the distance between 122, 12 and 13, 272.

Solution:

Write the distance formula.  d 5 " 1 x2 2 x1 2 2 1 1 y2 2 y1 2 2

Substitute 122, 12 and 13, 272.           d 5 " 1 3 2 2 2 2 1 1 27 2 1 2 2 Simplify.              d 5 " 1 1 2 2 1 1 28 2 2 5 "65

This is incorrect. What mistake was made?

49. Compute the midpoint of the segment with endpoints 123, 42

and 17, 92.

Solution:

Write the midpoint formula.

1 xm, ym 2 5 a

1 xm, ym 2 5 a Substitute 123, 42    and 17, 92.

x1 1 x2 y1 1 y2 , b 2 2

23 1 4 7 1 9 , b 2 2

1 16 1 Simplify.      1 xm, ym 2 5 a , b 5 a , 4b 2 2 2



This is incorrect. What mistake was made?

50. Compute the midpoint of the segment with endpoints 121, 222

and 123, 242.

Solution:

Write the midpoint    1 xm, ym 2 5 a formula.

x1 2 x2 y1 2 y2 , b 2 2

           1 xm, ym 2 5 a

21 2 1 23 2 22 2 1 24 2 , b 2 2

Substitute 121, 222 and 123, 242.

Simplify.        1xm, ym2 5 11, 12

Young_AT_6160_ch02_pp164-201.indd 172

This is incorrect. What mistake was made?

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2.2  Graphing Equations, Point-Plotting, Intercepts, and Symmetry 

173

• CONCEPTUAL In Exercises 51–54, determine whether each statement is true or false. 51. The distance from the origin to the point 1a, b2 is 54. The midpoint of any segment joining a point in quadrant I to a point in quadrant III also lies in either quadrant I or III. d 5 "a2 1 b2. 55. Calculate the length and the midpoint of the line segment 52. The midpoint of the line segment joining the origin and the joining the points 1a, b2 and 1b, a2. a a 56. Calculate the length and the midpoint of the line segment point 1a, a2 is a , b. 2 2 joining the points 1a, b2 and 12a, 2b2. 53. The midpoint of any segment joining two points in quadrant I also lies in quadrant I.

• CHALLENGE 57. Assume that two points 1x1, y12 and 1x2, y22 are connected by

a segment. Prove that the distance from the midpoint of the segment to either of the y two points is the same. (a, c) (a + b, c) 58. Prove that the diagonals of a parallelogram in the x figure intersect at their (0, 0) (b, 0) midpoints.

59. Assume that two points 1a, b2 and 1c, d 2 are the endpoints of a

line segment. Calculate the distance between the two points. Prove that it does not matter which point is labeled as the “first” point in the distance formula. 60. Show that the points 121, 212, 10, 02, and 12, 22 are collinear (lie on the same line) by showing that the sum of the distance from 121, 212 to 10, 02 and the distance from 10, 02 to 12, 22 is equal to the distance from 121, 212 to 12, 22.

• TECHNOLOGY In Exercises 61–64, calculate the distance between the two points. Use a graphing utility to graph the segment joining the two points and find the midpoint of the segment. 61. 122.3, 4.12 and 13.7, 6.22 62. 124.9, 23.22 and 15.2, 3.42 63. 11.1, 2.22 and 13.3, 4.42 64. 121.3, 7.22 and 12.3, 24.52

2.2 GRAPHING EQUATIONS, POINT-PLOTTING, INTERCEPTS, AND SYMMETRY SKILLS OBJECTIVES ■■ Sketch graphs of equations by plotting points. ■■ Find intercepts for graphs of equations. ■■ Determine if the graph of an equation is symmetric about the x-axis, y-axis, or origin. ■■ Use intercepts and symmetry as graphing aids.

CONCEPTUAL OBJECTIVES ■■ Understand that if a point 1a, b2 satisfies the equation, then that point lies on its graph. ■■ Understand that intercepts are points that lie on the graph and either the x-axis or y-axis. ■■ Relate symmetry graphically and algebraically. ■■ Understand that intercepts are good starting points, but not the only points, and that symmetry eliminates the need to find points in other quadrants.

In this section, you will learn how to graph equations by plotting points. However, when we discuss graphing principles in Chapter 3, you will see that other techniques can be more efficient.

2.2.1 Point-Plotting Most equations in two variables, such as y 5 x2, have an infinite number of ordered pairs as solutions. For example, 10, 02 is a solution to y 5 x2 because when x 5 0 and y 5 0, the equation is true. Two other solutions are 121, 12 and 11, 12.

Young_AT_6160_ch02_pp164-201.indd 173

2.2.1 S K I L L

Sketch graphs of equations by plotting points. 2.2.1 C O N C E P T U A L

Understand that if a point 1a, b2 satisfies the equation, then that point lies on its graph.

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CHAPTER 2  Graphs

The graph of an equation in two variables, x and y, consists of all the points in the xy-plane whose coordinates 1x, y2 satisfy the equation. A procedure for plotting the graphs of equations is outlined below and is illustrated with the example y 5 x2. WORDS MATH

Step 1: In a table, list several pairs of coordinates that make the equation true.

Step 2: Plot these points on a graph and connect the points with a smooth curve. Use arrows to indicate that the graph continues.

x

y 5 x 2

0

0

21

1

1

1

22

4

2

4

(x, y )

10, 02

121, 12 11, 12

122, 42 12, 42

y (–2, 4)

(2, 4)

(–1, 1)

(1, 1)

x

(0, 0)

In graphing an equation, first select arbitrary values for x and then use the equation to find the corresponding value of y, or vice versa.

EXAMPLE 1  Graphing an Equation of a Line by Plotting Points

Graph the equation y 5 2x 2 1. Solution: STEP 1  In

a table, list several pairs of coordinates that make the equation true.

x

y 5 2x 2 1

0

21

21

23

1

1

22

25

2

3

STEP 2  Plot

these points on a graph and connect the points, resulting in a line.

(x, y )

10, 212

121, 232 11, 12

122, 252 12, 32

y

(2, 3)

▼ ANSWER

(1, 1)

x

y

(0, –1) (–1, –3) (–1, 2)

(1, 0)

(–2, –5)

x

▼ Y O U R T U R N   The graph of the equation y 5 2x 1 1 is a line. Graph the line.

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2.2  Graphing Equations, Point-Plotting, Intercepts, and Symmetry 

175

EXAMPLE 2  Graphing an Equation by Plotting Points

Graph the equation y 5 x2 2 5. Solution: STEP 1  In

a table, list several pairs of coordinates that make the equation true.

these points on a graph and connect the points with a smooth curve, indicating with arrows that the curve continues.

x

y 5 x 2 2 5

(x, y )

0

25

21

24

10, 252

1

24

22

21

2

21

23

4

3

4

121, 242 11, 242

122, 212 12, 212 123, 42 13, 42

STEP 2  Plot

y (–3, 4)

(3, 4) x

(–2, –1)

(2, –1)

(–1, –4)

(1, –4) (0, –5)

This graph is called a parabola and will be discussed in further detail in Chapter 8.

▼

▼ ANSWER

Y O U R T U R N   Graph the equation y 5 x2 2 1.

y

(–2, 3)

EXAMPLE 3  Graph­­­­­­­­­­­­ing an Equation by Plotting Points

x (–1, 0) (0, –1)

Graph the equation y 5 x3. Solution: STEP 1  In

a table, list several pairs of coordinates that satisfy the equation.

these points on a graph and connect the points with a smooth curve, indicating with arrows that the curve continues in both the positive and negative directions.

x

(2, 3)

y 5 x 3

0

0

21

21

1

1

22

28

2

8

(1, 0)

(x, y )

10, 02

121, 212 11, 12

122, 282 12, 82

STEP 2  Plot

y (2, 8)

8 4 (–1, –1)

(1, 1) (0, 0) –4

(–2, –8)

–8

x

[ CONCEPT CHECK ] Show the points (1, 0) and (2, 3) satisfy the equation y 5 x  2 2 1.

▼ ANSWER (1, 0): 0 5 12 – 1 and (2, 3): 3 5 2 2 2 1

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CHAPTER 2  Graphs

2.2.2 Intercepts

2.2.2 S K I L L

Find intercepts for graphs of equations. 2.2.2 C O N C E P T U A L

Understand that intercepts are points that lie on the graph and either the x-axis or y-axis.

When point-plotting graphs of equations, which points should be selected? Points where a graph crosses (or touches) either the x-axis or y-axis are called intercepts, and identifying these points helps define the graph unmistakably. An x-intercept of a graph is a point where the graph intersects the x-axis. Specifically, an x-intercept is the x-coordinate of such a point. For example, if a graph intersects the x-axis at the point 13, 02, then we say that 3 is the x-intercept. Since the value for y along the x-axis is zero, all points corresponding to x-intercepts have the form 1a, 02. A y-intercept of a graph is a point where the graph intersects the y-axis. Specifically, a y-intercept is the y-coordinate of such a point. For example, if a graph intersects the y-axis at the point 10, 22, then we say that 2 is the y-intercept. Since the value for x along the y-axis is zero, all points corresponding to y-intercepts have the form 10, b2. It is important to note that graphs of equations do not have to have intercepts, and if they do have intercepts, they can have one or more of each type.

One x-intercept Two y-intercepts

No x-intercepts One y-intercept

y

y

x

x

No x-intercepts No y-intercepts

Three x-intercepts One y-intercept

y

y

x

x

Note: The origin 10, 02 corresponds to both an x-intercept and a y-intercept. The graph given to the left has two y-intercepts and one x-intercept. ■■ ■■

y

(0, 1) (–1, 0) (0, –1)

(3, 2)

x

(3, –2)

The x-intercept is 21, which corresponds to the point 121, 02. The y-intercepts are 21 and 1, which correspond to the points 10, 212 and 10, 12, respectively.

Algebraically, how do we find intercepts from an equation? The graph in the margin corresponds to the equation x 5 y2 2 1. The x-intercepts are located on the x-axis, which ­corresponds to y 5 0. If we let y 5 0 in the equation x 5 y2 2 1 and solve for x, the result is x 5 21. This corresponds to the x-intercept we identified above. Similarly, the y-intercepts are located on the y-axis, which corresponds to x 5 0. If we let x 5 0 in the equation x 5 y2 2 1 and solve for y, the result is y 5 61. These correspond to the y-intercepts we identified above.

ST U DY TIP

EXAMPLE 4  Finding Intercepts from an Equation

Identifying the intercepts helps define the graph unmistakably.

Given the equation y 5 x2 1 1, find the indicated intercepts of its graph, if any. a. x-intercept(s)  b.  y-intercept(s)

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2.2  Graphing Equations, Point-Plotting, Intercepts, and Symmetry 

[ CONCEPT CHECK ]

Solution (a):

0 5 x2 1 1 x2 5 21  no real solution

Let y 5 0. Solve for x.

If the graph passes through the origin, then the point (0, 0) is (A) an x-intercept, (B) a y-intercept, or (C) both an x-intercept and a y-intercept.

There are no x-intercepts. Solution (b):

y 5 02 1 1 y51

Let x 5 0. Solve for y.

▼ ANSWER (C) both

The y-intercept is located at the point  10, 12.

▼

177

▼ ANSWER

Y O U R T U R N   For the equation y 5 x2 2 4, a.  find the x-intercept(s), if any.   b.  find the y-intercept(s), if any.

a.  x-intercepts: 22 and 2 b.  y-intercept: 24

2.2.3 Symmetry The word symmetry conveys balance. Suppose you have two pictures to hang on a wall. If you  space them equally apart on the wall, then you prefer a symmetric décor. This is an example of symmetry about a line. The word (water) written below is identical if you rotate the word 180 degrees (or turn the page upside down). This is an example of symmetry about a point. Symmetric graphs have the characteristic that their mirror image can be obtained about a reference, typically a line or a point.

2.2.3 S K I L L

Determine if the graph of an equation is symmetric about the x-axis, y-axis, or origin. 2.2.3 C O N C E P T U A L

Relate symmetry graphically and algebraically.

In Example 2, the points 122, 212 and 12, 212 both lie on the graph of y 5 x2 2 5, as do the points 121, 242 and 11, 242. Notice that the graph on the right side of the y-axis is a mirror image of the part of the graph to the left of the y-axis. This graph illustrates ­symmetry with respect to the y-axis 1the line x 5 02. In the graph below of the equation x 5 y2 2 1, the points 10, 12 and 10, 212 both lie on the graph, as do the points 13, 22 and 13, 222. Notice that the part of the graph above the x-axis is a mirror image of the part of the graph below the x-axis. This graph illustrates symmetry with respect to the x-axis 1the line y 5 02. In Example 3, the points 121, 212 and 11, 12 both lie on the graph. Notice that rotating this graph 180 degrees (or turning your page upside down) results in an identical graph. This is an example of symmetry with respect to the origin 10, 02. Symmetry aids in graphing by giving information “for free.” For example, if a graph is symmetric about the y-axis, then once the graph to the right of the y-axis is y

y (–3, 4)

y = x2–5

y = x3

(2, –1)

(0, 1) (–1, 0) (0, –1)

(3, 2) (3, –2)

4 x (–1, –1)

(1, 1) (0, 0)

x

–4

(1, –4)

(–1, –4)

(2, 8)

8

(3, 4) x

(–2, –1)

y

x = y2–1

(–2, –8)

–8

(0, –5)

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CHAPTER 2  Graphs

found, the left side of the graph is the mirror image of that. If a graph is symmetric about the origin, then once the graph is known in quadrant I, the graph in quadrant III is found by rotating the known graph 180 degrees. It would be beneficial to know whether a graph of an equation is symmetric about a line or point before the graph of the equation is sketched. Although a graph can be symmetric about any line or point, we will discuss only symmetry about the x-axis, y-axis, and origin. These types of symmetry and the algebraic procedures for testing for symmetry are outlined below.

Types and Tests for Symmetry TYPE OF SYMMETRY

GRAPH y

Symmetric with respect to the x-axis

(a, b)

b

IF THE POINT (a, b ) IS ON THE GRAPH, THEN THE POINT . . .

ALGEBRAIC TEST FOR SYMMETRY

1a, 2b2 is on the graph.

Replacing y with 2y leaves the equation unchanged.

12a, b2 is on the graph.

Replacing x with 2x leaves the equation unchanged.

12a, 2b2 is on the graph.

Replacing x with 2x and y with 2y leaves the equation unchanged.

x a

ST U DY TIP

–b

Symmetry gives us information about the graph “for free.”

Symmetric with respect to the y-axis

(a, –b) y

(–a, b)

(a, b)

b

x –a

a

Symmetric with respect to the origin

y b

(a, b) x

(0, 0) –a (–a, –b)

a –b

EXAMPLE 5  Testing for Symmetry with Respect to the Axes

Test the equation y2 5 x3 for symmetry with respect to the axes. Solution:

Test for symmetry with respect to the x-axis. Replace y with 2y. 12y22 5 x3 Simplify. y2 5 x3 The resulting equation is the same as the original equation, y2 5 x3.     Therefore y2 5 x3 is symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis. Replace x with 2x. y2 5 12x23 Simplify. y2 5 2x3 The resulting equation, y2 5 2x3, is not the same as the original equation, y2 5 x3.     Therefore, y2 5 x2 is not symmetric with respect to the y-axis.

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2.2  Graphing Equations, Point-Plotting, Intercepts, and Symmetry 

179

When testing for symmetry about the x-axis, y-axis, and origin, there are five possibilities: ■■

No symmetry

Symmetry with respect to the x-axis Symmetry with respect to the y-axis ■■ Symmetry with respect to the origin ■■ Symmetry with respect to the x-axis, y-axis, and origin ■■ ■■

EXAMPLE 6  Testing for Symmetry

[ CONCEPT CHECK ]

Determine what type of symmetry (if any) the graphs of the equations exhibit. a. y 5 x2 1 1       b.  y 5 x3 1 1

If the point (a, b) and the, point (a, 2b) are on the graph then the graph is symmetric about (A) the y-axis, (B) the x-axis, or (C) the origin.

Solution (a):

Replace x with 2x. Simplify.

y 5 12x2 2 1 1 y 5 x2 1 1

The resulting equation is equivalent to the original equation, so the graph of the equation y 5 x2 1 1 is symmetric with respect to the y-axis. Replace y with 2y. 12y2 5 x2 1 1 Simplify. y 5 2x2 2 1

▼ ANSWER (B) the x-axis

The resulting equation y 5 2x2 2 1 is not equivalent to the original equation y 5 x2 1 1, so the graph of the equation y 5 x2 1 1 is not symmetric with respect to the x-axis. Replace x with 2x and y with 2y. Simplify.

12y2 5 12x2 2 1 1 2y 5 x2 1 1 y 5 2x2 2 1

The resulting equation y 5 2x2 2 1 is not equivalent to the original equation y 5 x2 1 1, so the graph of the equation y 5 x2 1 1 is not symmetric with respect to the origin. The graph of the equation y 5 x2 1 1 is symmetric with respect to the y-axis. Solution (b):

Replace x with 2x. Simplify.

y 5 12x2 3 1 1 y 5 2x3 1 1

The resulting equation y 5 2x3 1 1 is not equivalent to the original equation y 5 x3 1 1. Therefore, the graph of the equation y 5 x3 1 1 is not symmetric with respect to the y-axis. Replace y with 2y. 12y2 5 x3 1 1 Simplify. y 5 2x3 2 1 The resulting equation y 5 2x3 2 1 is not equivalent to the original equation y 5 x3 1 1. Therefore, the graph of the equation y 5 x3 1 1 is not symmetric with respect to the x-axis. Replace x with 2x and y with 2y. 12y2 5 12x2 3 1 1 Simplify. 2y 5 2x3 1 1 y 5 x3 2 1 3 The resulting equation y 5 x 2 1 is not equivalent to the original equation y 5 x3 1 1. Therefore, the graph of the equation y 5 x3 1 1 is not symmetric with respect to the origin.     The graph of the equation y 5 x3 1 1 exhibits no symmetry.

▼ 2

Y O U R T U R N   Determine the symmetry (if any) for x 5 y 2 1.

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▼ ANSWER

The graph of the equation is symmetric with respect to the x-axis.

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CHAPTER 2  Graphs

2.2.4  Using Intercepts and Symmetry as Graphing Aids 2.2.4 S K I L L

Use intercepts and symmetry as graphing aids. 2.2.4 C O N C E P T U A L

Understand that intercepts are good starting points, but not the only points, and that symmetry eliminates the need to find points in other quadrants.

How can we use intercepts and symmetry to assist us in graphing? Intercepts are a good starting point—though not the only one. For symmetry, look back at Example 2, y 5 x2 2 5. We selected seven x-coordinates and solved the equation to find the corresponding y-­coordinates. If we had known that this graph was symmetric with respect to the y-axis, then we would have had to find the solutions to only the positive x-coordinates, since we get the negative x-coordinates for free. For example, we found the point 11, 242 to be a solution to the equation. The rules of symmetry tell us that 121, 242 is also on the graph. EXAMPLE 7  Using Intercepts and Symmetry as Graphing Aids

For the equation x2 1 y2 5 25, use intercepts and symmetry to help you graph the equation using the point-plotting technique. Solution: STEP 1  Find



the intercepts. For the x-intercepts, let y 5 0. Solve for x.

The two x-intercepts correspond to the points 125, 02 and 15, 02.



x2 1 02 5 25 x 5 65

For the y-intercepts, let x 5 0. 02 1 y2 5 25 Solve for y. y 5 65 The two y-intercepts correspond to the points 10, 252 and 10, 52. y

STEP 2  Identify

the points on the graph corresponding to the intercepts.

5

x –5

5

–5

STEP 3  Test

for symmetry with respect to the y-axis, x-axis, and origin.

Test for symmetry with respect to the y-axis.     Replace x with 2x.     Simplify.

12x22 1 y2 5 25 x2 1 y2 5 25

        The resulting equation is equivalent to the original, so the graph of         x2 1 y2 5 25 is symmetric with respect to the y-axis. Test for symmetry with respect to the x-axis.     Replace y with 2y.     Simplify.

x2 1 12y22 5 25 x2 1 y2 5 25

        The resulting equation is equivalent to the original, so the graph of         x2 1 y2 5 25 is symmetric with respect to the x-axis.

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2.2  Graphing Equations, Point-Plotting, Intercepts, and Symmetry 

Test for symmetry with respect to the origin.     Replace x with 2x and y with 2y.     Simplify.

[ CONCEPT CHECK ]

12x22 1 12y22 5 25 x2 1 y2 5 25

        The resulting equation is equivalent to the original, so the graph of         x2 1 y2 5 25 is symmetric with respect to the origin.  ince the graph is symmetric with respect to the y-axis, x-axis, and origin, we S need to determine solutions to the equation on only the positive x- and y-axes and in quadrant I because of the following symmetries: Symmetry with respect to the y-axis gives the solutions in quadrant II. ■ Symmetry with respect to the origin gives the solutions in quadrant III. ■ Symmetry with respect to the x-axis yields solutions in quadrant IV. ■

Solutions to x2 1 y2 5 25.

(0, 5)

Additional points due to symmetry:

(–4, 3)

Quadrant II: 123, 42 , 124, 32

If the graph is symmetric about the x-axis, y-axis, and the origin, then once you find a point (a, b) in quadrant I that lies on the graph what points do you get for free from the symmetry? (A) (2a, b), (B) (a, 2b), (C) (2a, 2b), or (D) all of the above

▼ ANSWER (D) all of the above.

y (3, 4)

(–3, 4)

Quadrant I: 13, 42, 14, 32

181

(4, 3) (5, 0)

x

(–5, 0)

Quadrant III: 123, 242 , 124, 232

(–4, –3)

Quadrant IV: 13, 242 , 14, 232

(4, –3)

(–3, –4)

(3, –4) (0, –5)

 onnecting the points with a smooth curve yields a circle. We will discuss circles C in more detail in Section 2.4.

[ S E C T I O N 2 . 2 ]     S U M M A R Y Sketching graphs of equations can be accomplished using a Symmetry about the x-axis, y-axis, and origin is defined both point-plotting technique. Intercepts are defined as points where a algebraically and graphically. Intercepts and symmetry provide graph intersects an axis or the origin. much of the information useful for sketching graphs of equations. INTERCEPTS INTERCEPTS

POINT WHERE THE GRAPH INTERSECTS THE . . .

HOW TO FIND INTERCEPTS

x-intercept

x-axis

Let y 5 0 and solve for x.

y-intercept

y-axis

Let x 5 0 and solve for y.

TYPES AND TESTS FOR SYMMETRY

TYPE OF SYMMETRY

IF THE POINT (a, b ) IS ON THE GRAPH, THEN THE POINT . . .

GRAPH

Symmetric with respect to the x-axis

1a, 2b2 is on the graph.

y b

(a, b)

POINT ON GRAPH

1a, 02 10, b2

ALGEBRAIC TEST FOR SYMMETRY

Replacing y with 2y leaves the equation unchanged.

x a –b

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(a, –b)

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182 

CHAPTER 2  Graphs

Symmetric with respect to the y-axis

y (–a, b)

(a, b)

b

12a, b2 is on the graph.

Replacing x with 2x leaves the equation unchanged.

12a, 2b2 is on the graph.

Replacing x with 2x and y with 2y leaves the equation unchanged.

x –a

a

Symmetric with respect to the origin

y b

(a, b) x

(0, 0) –a (–a, –b)

a –b

[ SEXAMPLE7 EC TION 2 . 2]  E X E R C I S E S • SKILLS In Exercises 1–8, determine whether each point lies on the graph of the equation. 1. y 5 3x 2 5 2

3. y 5 5 x 2 4 5. y 5 x2 2 2x 1 1 7. y 5 "x 1 2

a. 11, 22

b. 122, 2112

2. y 5 22x 1 7

a. 17, 32

b. 126, 42

8. y 5 2 1 0 3 2 x 0

a. 15, 222 a. 121, 42

3

b. 125, 62

4. y 5 2 4 x 1 1 6. y 5 x3 2 1

b. 10, 212

a. 121, 92

b. 12, 242

a. 19, 242

b. 122, 72

a. 18, 52

a. 121, 02

b. 124, 42

b. 122, 292

In Exercises 9–14, complete the table and use the table to sketch a graph of the equation. 9.

11.

13.

x

y521x

(x, y )

10.

x

22

21

0

0

1

2

x

y 5 x 2 2 x

(x, y )

12.

x

21

23

0

22

1 2

21

1

0

2

1

x

1

y 5 "x 2 1

(x, y )

14.

x

22

2

21

5

2

10

7

Young_AT_6160_ch02_pp164-201.indd 182

y 5 3x 2 1

y 5 1 2 2x 2 x 2

y 5 2 "x 1 2

(x, y )

(x, y )

(x, y )

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2.2  Graphing Equations, Point-Plotting, Intercepts, and Symmetry 

183

In Exercises 15–22, graph the equation by plotting points. 15. y 5 23x 1 2

16. y 5 4 2 x

17. y 5 x2 2 x 2 2

19. x 5 y2 2 1

20. x 5 0 y 1 1 0 1 2

21. y 5 2 x 2 2

23. 2x 2 y 5 6

24. 4x 1 2y 5 10

25. y 5 x2 2 9

27. y 5 "x 2 4

28. y 5 "x 2 8

18. y 5 x2 2 2x 1 1 22. y 5 0.5 0 x 21 0

3

1

In Exercises 23–32, find the x-intercept(s) and y-intercepts(s) (if any) of the graphs of the given equations.

3

31. 4x2 1 y2 5 16

29. y 5

26. y 5 4x2 2 1

1 2 x 14

30. y 5

x 2 2 x 2 12 x

32. x2 2 y2 5 9

In Exercises 33–38, match the graph with the corresponding symmetry. a. No symmetry b. Symmetry with respect to the x-axis c. Symmetry with respect to the y-axis d. Symmetry with respect to the origin e. Symmetry with respect to the x-axis, y-axis, and origin 33.

34. y

35. y

y

(1, 1)

x

x

x

(–1, –1)

36.

37. y

38. y

y

(3, 3) x

(2, 2)

x (–2, –2)

x

(0, 0)

(–3, –3)

In Exercises 39–44, a point that lies on a graph is given along with that graph’s symmetry. State the other known points that must also lie on the graph. Point on a Graph 39. 121, 32

41. 17, 2102

43. 13, 222

The Graph Is Symmetric about the

x-axis



origin



x-axis, y-axis, and origin

Point on a Graph

The Graph Is Symmetric about the

   40. 122, 42



origin

   44. 121, 72



x-axis, y-axis, and origin

   42. 121, 212

y-axis

In Exercises 45–58, test algebraically to determine whether the equation’s graph is symmetric with respect to the x-axis, y-axis, or origin. 45. x 5 y2 1 4

46. x 5 2y2 1 3

47. y 5 x3 1 x

48. y 5 x5 1 1

49. x 5 0 y 0

50. x 5 0 y 0 2 2

51. x2 2 y2 5 100

52. x2 1 2y2 5 30

55. x2 1 y3 5 1

56. y 5 "1 1 x 2

53. y 5 x2/3 57. y 5

2 x

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54. x 5 y2/3 58. x y 5 1

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CHAPTER 2  Graphs

In Exercises 59–72, plot the graph of the given equation. 1 59. y 5 x 60. y 5 2 x 1 3 2 x3 63. y 5 64. x 5 y2 1 1 2 67. y 5 0 x 0

61. y 5 x2 2 1 65. y 5

68. 0 x 0 5 0 y 0

71. x2 2 y2 5 16

72. x 2 2

62. y 5 9 2 4x2

1 x

66. x y 5 21

69. x2 1 y2 5 16

70.

y2 x2 1 51 4 9

y2 51 25

• A P P L I C AT I O N S 73. Sprinkler. A sprinkler will water a grassy area in the shape

of x2 1 y2 5 9. Apply symmetry to draw the watered area, assuming the sprinkler is located at the origin. 74. Sprinkler. A sprinkler will water a grassy area in the shape y2 5 1. Apply symmetry to draw the watered area, 9 assuming the sprinkler is located at the origin. 75. Electronic Signals: Radio Waves. The received power of an electromagnetic signal is a fraction of the power transmitted. The relationship is given by 1 Preceived 5 Ptransmitted . 2 R where R is the distance that the signal has traveled in meters. Plot the percentage of transmitted power that is received for R 5 100 m, 1 km, and 10,000 km. 76. Electronic Signals: Laser Beams. The wavelength l and the frequency ƒ of a signal are related by the equation c f5 l of x 2 1

where c is the speed of light in a vacuum, c 5 3.0 3 108 meters per second. For the values, l 5 0.001, l 5 1, and l 5 100 mm, plot the points corresponding to frequency, ƒ. What do you notice about the relationship between frequency and wavelength? Note that the frequency will have units Hz 5 1/seconds. 77. Profit. The profit associated with making a particular product is given by the equation y 5 2x 2 1 6x 2 8 where y represents the profit in millions of dollars and x represents the number of thousands of units sold. 1x 5 1

corresponds to 1000 units and y 5 1 corresponds to $1M.2 Graph this equation and determine how many units must be sold to break even 1  profit 5 02. Determine the range of units sold that correspond to making a profit. 78. Profit. The profit associated with making a particular product is given by the equation y 5 2x 2 1 4x 2 3 where y represents the profit in millions of dollars and x represents the number of thousands of units sold. 1x 5 1 ­corresponds to 1000 units and y 5 1 corresponds to $1M.2 Graph this equation and determine how many units must be  sold  to break even 1  profit 5 02. Determine the range of units sold that correspond to making a profit. 79. Economics. The demand for an electronic device is modeled by p 5 2.95 2 "0.01x 2 0.01

where x is thousands of units demanded per day and p is the price (in dollars) per unit. a. Find the domain of the demand equation. Interpret your result. b.  Plot the demand equation. 80. Economics. The demand for a new electronic game is modeled by p 5 39.95 2 "0.01x 2 0.4

where x is thousands of units demanded per day and p is the price (in dollars) per unit. a. Find the domain of the demand equation. Interpret your result. b.  Plot the demand equation.

• C AT C H T H E M I S TA K E

In Exercises 81–84, explain the mistake that is made. 81. Graph the equation y 5 x2 1 1.

Solution:

y

x

y 5 x 2 1 1

(x, y )

0

1

1

2

10, 12

(0, 1)

11, 22



Young_AT_6160_ch02_pp164-201.indd 184

(1, 2)

x

This is incorrect. What mistake was made?

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2.2  Graphing Equations, Point-Plotting, Intercepts, and Symmetry 

82. Test y 5 2x2 for symmetry with respect to the y-axis.



Determine points that lie on the graph in quadrant I.

Solution: Replace x with 2x.

y 5 212x2

Simplify.

y5x

2

2

 The resulting equation is not equivalent to the original equation; y 5 2x2 is not symmetric with respect to the y-axis.

This is incorrect. What mistake was made? 83. Test x 5 0 y 0 for symmetry with respect to the y-axis.

y

x 2 5 y 2 1

(x, y )

1

0

2

1

10, 12

5

2

y

x 5 0y0

x –5

The resulting equation is equivalent to the original equation; x 5 0 y 0 is symmetric with respect to the y-axis.



This is incorrect. What mistake was made? 84. Use symmetry to help you graph x2 5 y 2 1.

Replace x with 2x. Simplify.

5

–5



Solution:

12, 52

5

x 5 0 2y 0

Simplify.

11, 22

Symmetry with respect to the x-axis implies that 10, 212, 11, 222, and 12, 252 are also points that lie on the graph.

Solution:

Replace y with 2y.

185

This is incorrect. What mistake was made?

12x22 5 y 2 1 x2 5 y 2 1

x2 5 y 2 1 is symmetric with respect to the x-axis.

• CONCEPTUAL In Exercises 85–88, determine whether each statement is true or false. 85. If the point 1a, b2 lies on a graph that is symmetric about the

x-axis, then the point 12a, b2 also must lie on the graph. 86. If the point 1a, b2 lies on a graph that is symmetric about the y-axis, then the point 12a, b2 also must lie on the graph.

87. If the point 1a, 2b2 lies on a graph that is symmetric about the

x-axis, y-axis, and origin, then the points 1a, b2, 12a, 2b2, and 12a, b2 must also lie on the graph. 88. Two points are all that is needed to plot the graph of an ­equation.

• CHALLENGE ax 2 1 b has any symcx 3 metry, where a, b, and c are real numbers.

89. Determine whether the graph of y 5

90. Find the intercepts of y 5 1x 2 a22 2 b2, where a and b are

real numbers.

• TECHNOLOGY In Exercises 91–96, graph the equation using a graphing utility and state whether there is any symmetry. 91. y 5 16.7x4 2 3.3x2 1 7.1

94. 3.2x2 2 5.1y2 5 1.3

92. y 5 0.4x5 1 8.2x3 2 1.3x

95. 1.2x2 1 4.7y2 5 19.4

93. 2.3x2 5 5.5 0 y 0

96. 2.1y2 5 0.8 0 x 1 1 0

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CHAPTER 2  Graphs

2.3 LINES SKILLS OBJECTIVES ■■ Calculate the slope of a line; graph a line; classify a line as rising, falling, vertical, or horizontal. ■■ Find the equation of a line given either slope and a point that lies on the line or two points that lie on the line. ■■ Find the equation of a line that is parallel or perpendicular to a given line.

CONCEPTUAL OBJECTIVES ■■ Understand slope as rate of change. ■■ Understand that the equation of a line given by y 5 mx 1 b corresponds to a line with slope m and y-intercept 10, b2. ■■ Understand that parallel lines have the same slope, whereas perpendicular lines have slopes that are negative reciprocals of one another.

2.3.1  Graphing a Line 2.3.1 S K I L L

Calculate the slope of a line; graph a line; classify a line as rising, falling, vertical, or horizontal. 2.3.1 C O N C E P T U A L

Understand slope as rate of change.

What is the shortest path between two points? The answer is a straight line. In this section, we will discuss characteristics of lines such as slope and intercepts. We will also discuss types of lines such as horizontal, vertical, falling, and rising, and recognize relationships between lines such as perpendicular and parallel. At the end of this section, you should be able to find the equation of a line when given two specific pieces of information about the line. First-degree equations, such as y 5 22x 1 4,  3x 1 y 5 6,  y 5 2,  and  x 5 23 have graphs that are straight lines. The first two equations given represent inclined or “slant” lines, whereas y 5 2 represents a horizontal line and x 5 23 represents a vertical line. One way of writing an equation of a straight line is called general form. EQUATION OF A STRAIGHT LINE: GENERAL* FORM

If A, B, and C are constants and x and y are variables, then the equation Ax 1 By 5 C is in general form, and its graph is a straight line. Note: A or B (but not both) can be zero. The equation 2x 2 y 5 22 is a first-degree equation, so its graph is a straight line. To graph this line, list two solutions in a table, plot those points, and use a straight edge to draw the line.

ST U DY TIP x-intercept: Set y 5 0. y-intercept: Set x 5 0.

x

y

22

22

1

4

y

(x, y )

122, 222

(1, 4)

11, 42

x (–2, –2)

y (1, 4)

2x – y = –2

(0, 2) x–intercept

y–intercept

(–1, 0) (–2, –2)

Intercepts x

The point where a line crosses, or intersects, the x-axis is called the x-intercept. The point where a line crosses, or intersects, the y-axis is called the y-intercept. By inspecting the graph of the previous line, we see that the x-intercept is 121, 02 and the y-intercept is 10, 22  . *Some books refer to this as standard form.

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187

2.3 Lines 

Intercepts were discussed in Section 2.2. There we found that the graphs of some equations could have no intercepts or one or multiple intercepts. Slant lines, however, have exactly one x-intercept and exactly one y-intercept. DETERMINING INTERCEPTS

x-intercept: y-intercept:

Coordinates 1a, 02

y

Axis Crossed Algebraic Method x-axis Set y 5 0 and solve for x.

10, b2

y-axis Set x 5 0 and solve for y.

(0, b)

x (a, 0)

Horizontal lines and vertical lines, however, each have only one intercept. TYPE OF LINE

Horizontal

EQUATION

x-INTERCEPT

y-INTERCEPT

y5b

None

b

GRAPH y (0, b) x

Vertical

x5a

a

None

y

(a, 0)

x

Note: The special cases of x 5 0 (y-axis) and y 5 0 (x-axis) have infinitely many y-intercepts and x-intercepts, respectively.

EXAMPLE 1  Determining x - and y -Intercepts

Determine the x- and y-intercepts (if they exist) for the lines given by the following equations. a. 2x 1 4y 5 10    b.  x 5 22 2x 1 4y 5 10 Solution (a):  To find the x-intercept, set y 5 0. 2x 1 4102 5 10 Solve for x. 2x 5 10 x55 The x-intercept corresponds to the point 15, 02  .

To find the y-intercept, set x 5 0. 2102 1 4y 5 10 Solve for y. 4y 5 10 y5                          

5 2

The y-intercept corresponds to the point A0, 52 B .

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CHAPTER 2  Graphs

Solution (b):  x 5 22 y

This vertical line consists of all points 122, y2. The graph shows that the x-intercept is 22. We also find that the line never crosses the y-axis, so the y-intercept does not exist.

x = –2 x (–2, 0)

▼

▼ ANSWER a.  x-intercept:

Y O U R T U R N   Determine the x- and y-intercepts (if they exist) for the lines given

by the following equations. a.  3x 2 y 5 2  b.  y 5 5

A 23 ,

0B; y-intercept: 10, 222 b.  x-intercept does not exist; y-intercept: 10, 52

Slope If the graph of 2x – y 5 22 represented an incline that you were about to walk on, would you classify that incline as steep? In the language of mathematics, we use the word slope as a measure of steepness. Slope is the ratio of the change in y over the change in x. An easy way to remember this is rise over run.

S TU DY TIP

SLOPE OF A LINE

To get the correct sign (6) for the slope, remember to start with the same point for both x and y.

A nonvertical line passing through two points 1x1, y12 and 1x2, y22 has slope m, given by the formula y m5 m5

y2 2 y1 , where x1 2 x2 or x2 2 x1 vertical change rise 5 run horizontal change

(x2, y2) rise

(x1, y1)

x

run

Note: Always start with the same point for both the x-coordinates and the y-coordinates. y (1, 4) 2x – y = –2

x

(–2, –2)

Let’s find the slope of our graph 2x 2 y 5 22. We’ll let 1x1, y12 5 122, 222 and 1x2, y22 5 11, 42 in the slope formula: m5

3 4 2 1 22 2 4 y2 2 y1 6 5 52 5 x2 2 x1 3 1 2 1 22 2 4 3

Notice that if we had chosen the two intercepts 1x1, y12 5 10, 22 and 1x2, y22 5 121, 02 instead, we still would have found the slope to be m 5 2.

common mistake ▼ CAUTION

Interchanging the coordinates can result in a sign error in the slope.

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The most common mistake in calculating slope is writing the coordinates in the wrong order, which results in the slope being opposite in sign.

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2.3 Lines 

189

Find the slope of the line containing the two points 11, 22 and 13, 42. ✓C O R R E C T

Label the points. 1 x1, y1 2 5 1 1, 2 2   1 x2, y2 2 5 1 3, 4 2 Write the slope formula.

m5

y2 2 y1 x2 2 x1

Substitute the coordinates.

m5

Simplify.  m 5

422 321

✖INCORRECT

The ERROR is interchanging the coordinates of the first and second points.

m5

422 123

The calculated slope is INCORRECT by a negative sign.

m5

2 5 21 22

2 5 1 2

When interpreting slope, always read the graph from left to right. Since we have ­determined the slope to be 2, or 21, we can interpret this as rising two units and running (to the right) one unit. If we start at the point 122, 222 and move two units up and one unit to the right, we end up at the x-intercept, 121, 02. Again, moving two units up and one unit to the right puts us at the y-intercept, 10, 22. Another rise of two and run of one takes us to the point 11, 42. See the figure on the right. Lines fall into one of four categories: rising, falling, horizontal, or vertical.

y (1, 4) Run = 1 (0, 2) Rise = 2

x

(–1, 0)

(–2, –2)

y

Line Slope Rising Positive 1m + 02 Falling Negative 1m * 02 Horizontal Zero 1m 5 02 , hence y 5 b Vertical Undefined, hence x 5 a

a

x

b

The slope of a horizontal line is 0 because the y-coordinates of any two points are the same. The change in y in the slope formula’s numerator is 0; hence m 5 0. The slope of a vertical line is undefined because the x-coordinates of any two points are the same. The change in x in the slope formula’s denominator is zero; hence m is undefined.

EXAMPLE 2  Graph, Classify the Line, and Determine the Slope

Sketch a line through each pair of points, classify the line as rising, falling, vertical, or ­horizontal, and determine its slope. y a.  121, 232 and 11, 12 b.  123, 32 and 13, 12 c.  121, 222 and 13, 222 d.  11, 242 and 11, 32 (1, 1)

Solution (a):  121, 232 and 11, 12

This line is rising, so its slope is positive. m5

x

(–1, –3)

1 2 1 23 2 4 2 5 5 52 2 1 1 2 1 21 2

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CHAPTER 2  Graphs

[ CONCEPT CHECK ] If two lines are rising from left to right, then the larger slope corresponds to the (A) steeper incline or (B) less steep incline.

Solution (b): 123, 32 and 13, 12

y

This line is falling, so its slope is negative. (–3, 3)

321 2 1 m5 52 52 23 2 3 6 3

x

(3, 1)

▼ ANSWER (A) steeper incline

y

Solution (c):  121, 222 and 13, 222

This is a horizontal line, so its slope is zero. m5

22 2 1 22 2 0 5 50 4 3 2 1 21 2

x (–1, –2)

(3, –2)

y

Solution (d):  11, 242 and 11, 32

This is a vertical line, so its slope is undefined. 3 2 1 24 2 7 m5 5 , which is undefined. 121 0

(1, 3) x

(1, –4)

▼ ANSWER a.  m 5 25, falling b.  m 5 2, rising c.  slope is undefined, vertical d.  m 5 0, horizontal

▼ Y O U R T U R N   For each pair of points classify the line that passes through them as

rising, falling, vertical, or horizontal, and determine its slope. Do not graph. a.  12, 02 and 11, 52 b.  122, 232 and 12, 52 c.  123, 212 and 123, 42 d.  121, 22 and 13, 22

2.3.2  Equations of Lines Slope–Intercept Form 2.3.2 S K I L L

Find the equation of a line given either slope and a point that lies on the line or two points that lie on the line. 2.3.2 C O N C E P T U A L

Understand that the equation of a line given by y 5 mx 1 b corresponds to a line with slope m and y-intercept 10, b2.

Young_AT_6160_ch02_pp164-201.indd 190

As mentioned earlier, the general form for an equation of a line is Ax 1 By 5 C. A more standard way to write an equation of a line is in slope–intercept form because it identifies the slope and the y-intercept. EQUATION OF A STRAIGHT LINE: SLOPE–INTERCEPT FORM

The slope–intercept form for the equation of a nonvertical line is y 5 mx 1 b Its graph has slope m and y-intercept b. For example, 2x 2 y 5 23 is in general form. To write this equation in slope–intercept form, we isolate the y variable: y 5 2x 1 3 The slope of this line is 2 and the y-intercept is 3.

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191

EXAMPLE 3  U  sing Slope–Intercept Form to Graph an Equation of a Line

Write 2x 2 3y 5 15 in slope–intercept form and graph it. Solution:



STEP 1  Write

in slope–intercept form. Subtract 2x from both sides.



Divide both sides by 23.

23y 5 22x 1 15 2 y5 x25 3

STEP 2  Graph.



2 Identify the slope and y-intercept. Slope: m 5   y-intercept: b 5 25 3

Plot the point corresponding to the y-intercept 10, 252.

y ▼ ANSWER

From the point 10, 252, rise two units and run (to the right) three units, which corresponds to the point 13, 232. Draw the line passing through the two points.

y 5 32 x 2 6 x

y

x

(3, –3) (0, –5) (2, –3)

▼ Y O U R T U R N   Write 3x 2 2y 5 12 in slope–intercept form and graph it.

(0, –6)

Instead of starting with equations of lines and characterizing them, let us now start with particular features of a line and derive its governing equation. Suppose that you are given the y-intercept and the slope of a line. Using the slope–intercept form of an equation of a line, y 5 mx 1 b, you could find its equation.

EXAMPLE 4  U  sing Slope–Intercept Form to Find the Equation of a Line

Find the equation of a line that has slope 23 and y-intercept 10, 12.

Solution:

Write the slope–intercept form of an equation of a line.   y 5 mx 1 b Label the slope.

m5

2 3

Label the y-intercept.                     b51

The equation of the line in slope–intercept form is y 5 23 x 1 1 .

▼

3

Y O U R T U R N   Find the equation of the line that has slope 22 and y-intercept 10, 22.

Young_AT_6160_ch02_pp164-201.indd 191

▼ ANSWER

y 5 2 23 x 1 2

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Point–Slope Form Now, suppose that the two pieces of information you are given about an equation are its slope and one point that lies on its graph. You still have enough information to write an equation of the line. Recall the formula for slope: m5

y2 2 y1 , x2 2 x1

where x2 2 x1

We are given the slope m, and we know a particular point that lies on the line 1x1, y12. We refer to all other points that lie on the line as 1x, y2. Substituting these values into the slope formula gives us y 2 y1 m5 x 2 x1 Cross multiplying yields y 2 y1 5 m 1 x 2 x1 2

This is called the point–slope form of an equation of a line.

EQUATION OF A STRAIGHT LINE: POINT–SLOPE FORM

The point–slope form for the equation of a line is y 2 y1 5 m 1 x 2 x1 2

Its graph passes through the point 1x1, y12, and its slope is m. Note: This formula does not hold for vertical lines since their slope is undefined.

EXAMPLE 5  U  sing Point–Slope Form to Find the Equation of a Line

Find the equation of the line that has slope 221 and passes through the point 121, 22. Solution:

Write the point–slope form of an equation of a line.        y 2 y1 5 m1x 2 x12

1 Substitute the values m 5 221 and 1x1, y12 5 121, 22.    y 2 2 5 2 1 x 2 1 21 2 2 2

1 1 Distribute.                                  y2252 x2 2 2 1 3 Isolate y.                         y52 x1 2 2 We can also express the equation in general form x 1 2y 5 3 . ▼ ANSWER

y 5 14 x 2 34 or x 2 4y 5 3

▼

1

YOUR TURN  D  erive the equation of the line that has slope 4 and passes through

the point A1,

221 B.

Suppose the slope of a line is not given at all. Instead, two points that lie on the line are given. If we know two points that lie on the line, then we can calculate the slope. Then, using the slope and either of the two points, the equation of the line can be derived.

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2.3 Lines 

193

EXAMPLE 6  Finding the Equation of a Line Given Two Points

Find the equation of the line that passes through the points 122, 212 and 13, 22. Solution:

Write the equation of a line. Calculate the slope.

Substitute 1x1, y12 5 122, 212 and 1x2, y22 5 13, 22.

y 5 mx 1 b

m5

m5

y2 2 y1 x2 2 x1 2 2 1 21 2 3 5 5 3 2 1 22 2

Substitute 35 for the slope.

3 y5 x1b 5

Let 1x, y2 5 13, 22. 1Either point satisfies the equation.2

25

Solve for b.

b5

Write the equation in slope–intercept form.

3 1 y5 x1 5 5

Write the equation in general form.

23x 1 5y 5 1

3 132 1 b 5

1 5

STUDY T I P When two points that lie on a line are given, first calculate the slope of the line, then use either point and the slope–intercept form (shown in Example 6) or the point–slope form: m 5 35 , 1 3, 2 2              y 2 y1 5 m 1 x 2 x1 2   y 2 2 5 35 1 x 2 3 2         5y 2 10 5 3 1 x 2 3 2      5y 2 10 5 3x 2 9              23x 1 5y 5 1

▼ Y O U R T U R N   Find the equation of the line that passes through the points 121, 32

and 12, 242.

▼ ANSWER

y 5 273 x 1 23  or 7x 1 3y 5 2

EXAMPLE 7  F  inding the Equation of a Horizontal or Vertical Line

y

Find the equation of each of the following lines given their slope and a point that the line passes through: a.  Slope: undefined; 123, 2 2     b.  Slope: m 5 0; 124, 5 2

(–3, 2) x

Solution (a):

A vertical line has undefined slope. The x-coordinate of the point the line passes through 123, 2 2 is 23. Graph of the line x 5 23 and the point 123, 2 2 indicated.

x 5 23

Solution (b):

A horizontal line has slope m 5 0. The y-coordinate of the point the line passes through 124, 5 2 is 5. Graph of the line y 5 5 with the point 124, 5 2 indicated.

x = 3

x5a

y

(–4, 5)

y5b y55

y=5

x

▼

YOUR TURN  F  ind the equation of each of the following lines given their slope

and a point that the line passes through: a.  Slope: m 5 0; 13, 72

b.  Slope: undefined; 1 5, 22 2

Young_AT_6160_ch02_pp164-201.indd 193

▼ ANSWER a.  y 5 7 b.  x 5 5

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CHAPTER 2  Graphs

2.3.3 Parallel and Perpendicular Lines 2.3.3 S K I L L

Find the equation of a line that is parallel or perpendicular to a given line. 2.3.3 C O N C E P T U A L

Understand that parallel lines have the same slope, whereas perpendicular lines have slopes that are negative reciprocals of one another.

[ CONCEPT CHECK ] In what order are the slope and y-intercept found? (A) y-intercept first, then the slope, or (B) the slope and then the y-intercept?

Two distinct nonintersecting lines in a plane are parallel. How can we tell whether the two lines in the graph on the left are parallel? Parallel lines must have the same steepness. In other words, parallel lines must have the same slope. The two lines shown on the right are parallel because they have the same slope, 2.

DEFINITION

y = 2x + 3

y = 2x – 1

x

Parallel Lines

Two distinct lines in a plane are parallel if and only if their slopes are equal. In other words, if two lines in a plane are parallel, then their slopes are equal, and if the slopes of two lines in a plane are equal, then the lines are parallel. WORDS MATH

Lines L1 and L2 are parallel. Two parallel lines have the same slope.

▼ ANSWER (B) the slope and then the y-intercept

y

L1 0 0 L2 m1 5 m2

EXAMPLE 8  Determining Whether Two Lines Are Parallel

Determine whether the lines 2x 1 3y 5 23 and y 5 13x 2 6 are parallel. Solution:

Write the first line in slope–intercept form. 2x 1 3y 5 23 Add x to both sides. 3y 5 x 2 3 Divide by 3. Compare the two lines.

1 y5 x21 3 1 1 y 5 x 2 1  and  y 5 x 2 6 3 3

Both lines have the same slope, 13. These are distinct lines because they have different y-intercepts. Thus, the two lines are parallel .

EXAMPLE 9  Finding an Equation of a Parallel Line

Find the equation of the line that passes through the point 11, 12 and is parallel to the line y 5 3x 1 1. Solution:

Write the slope–intercept equation of a line. Parallel lines have equal slope. Substitute the slope into the equation of the line. Since the line passes through 11, 12, this point must satisfy the equation. Solve for b. ▼ ANSWER

y 5 2x 1 5

Young_AT_6160_ch02_pp164-201.indd 194

y 5 mx 1 b m53 y 5 3x 1 b 1 5 3112 1 b b 5 22

The equation of the line is y 5 3x 2 2 .

▼ Y O U R T U R N   Find the equation of the line parallel to y 5 2x 2 1 that passes

through the point 121, 32.

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2.3 Lines 

Two perpendicular lines form a right angle at their point of intersection. Notice the slopes of the two perpendicular lines in the figure to the right. They are 212 and 2, negative reciprocals of each other. It turns out that almost all perpendicular lines share this property. Horizontal 1m 5 02 and vertical 1m undefined2 lines do not share this property. DEFINITION

195

y

1

y = – 2x

y = 2x x

Perpendicular Lines

Except for the special case of a vertical and a horizontal line, two lines in a plane are perpendicular if and only if their slopes are negative reciprocals.

In other words, if two lines in a plane are perpendicular, their slopes are negative reciprocals, provided their slopes are defined. Similarly, if the slopes of two lines in a plane are negative reciprocals, then the lines are perpendicular. WORDS

MATH

STUD Y T I P

Lines L1 and L2 are perpendicular. Two perpendicular lines have negative reciprocal slopes.

L 1' L 2

If a line has slope equal to 3, then a line perpendicular to it has slope 2 31 .

m1 5 2

1 m2

m1 2 0, m2 2 0

EXAMPLE 10  F  inding an Equation of a Line That Is Perpendicular to Another Line

Find the equation of the line that passes through the point 13, 02 and is perpendicular to the line y 5 3x 1 1. Solution:

Identify the slope of the given line y 5 3x 1 1.

m1 5 3

The slope of a line perpendicular to the given line is the negative reciprocal of the slope of the given line.

m2 5 2

1 1 52 m1 3

Write the equation of the line we are looking for in slope–intercept form.

y 5 m2x 1 b

Substitute m2 5 231 into y 5 m2x 1 b.

1 y52 x1b 3

Since the desired line passes through 13, 02, this point must satisfy the equation.

1 0 5 2 132 1 b 3

0 5 21 1 b Solve for b. b51

The equation of the line is y 5 231x 1 1 .

▼ Y O U R T U R N   Find the equation of the line that passes through the point 11, 252

and is perpendicular to the line y 5 221 x 1 4.

▼ ANSWER

y 5 2x 2 7

Applications Involving Linear Equations Slope is the ratio of the change in y over the change in x. In applications, slope can often be interpreted as the rate of change, as illustrated in the next example.

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CHAPTER 2  Graphs

EXAMPLE 11  Slope as a Rate of Change

The average age that a person first marries has been increasing over the last several decades. In 1970 the average age was 20, in 1990 it was 25 years old, and in 2030 it is expected that the average age will be 35 at the time of a person’s first marriage. Find the slope of the line passing through these points. Describe what that slope represents. I f we let x represent the year and y represent the age, then two points* that lie on the line are 11970, 202 and 12030, 352.

Age

Solution: 45 40 35 30 25 20 15 10 5 1930 1950 1970 1990 2010 2030 2050 Year

Write the slope formula.

m5

change in y change in x

 ubstitute the points into the S slope formula.

m5

35 2 20 15 1 5 5 2030 2 1970 60 4

 he slope is 14 and can be interpreted as the rate of change of the average age when a T person is first married. Every 4 years the average age at the first marriage is 1 year older.

EXAMPLE 12  Service Charges

Suppose that your two neighbors use the same electrician. One neighbor had a 2-hour job that cost her $100, and another neighbor had a 3-hour job that cost him $130. Assuming that a linear equation governs the service charge of this electrician, what will your cost be for a 5-hour job? Solution: STEP 1  Identify

the question.  etermine the linear equation for this electrician’s service charge and D calculate the charge for a 5-hour job.

STEP 2  Make

notes. A 2-hour job costs $100 and a 3-hour job costs $130.

STEP 3  Set up an equation. Let x equal the number of hours and y equal the service charge in dollars. Linear equation: y 5 mx 1 b Two points that must satisfy this equation are 12, 1002 and 13, 1302. STEP 4  Solve



the equation.

Calculate the rate of change (slope).

m5

130 2 100 30 5 5 30 322 1

Substitute the slope into the linear y 5 30x 1 b equation.                 *In Example 11 we chose to use the points 11970, 202 and 12030, 352. We could have also used the point 11990, 252 with either of the other points.

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2.3 Lines 

[ CONCEPT CHECK ]

Either point must satisfy              100 5 30122 1 b the equation. Use 12, 1002.                 100 5 60 1 b b 5 40 The service charge y is given by y 5 30x 1 40. Substitute x 5 5 into this equation for a 5-hour job.                          y 5 30152 1 40 5 190

197

The 5-hour job will cost $190 .

STEP 5  Check

the solution. The service charge y 5 30x 1 40 can be interpreted as a fixed cost of $40 for coming to your home and a $30 per hour fee for the job. A 5-hour job would cost $190. Additionally, a 5-hour job should cost less than the sum of a 2-hour job and a 3-hour job 1$100 1 $130 5 $2302, since the $40 fee is charged only once.

▼ YOUR TURN  Y  ou decide to hire a tutor that some of your friends recommended.

The tutor comes to your home, so she charges a flat fee per session and then an hourly rate. One friend prefers 2-hour sessions, and the charge is $60 per session. Another friend has 5-hour sessions that cost $105 per session. How much should you be charged for a 3-hour session?

Two parallel lines have the same slope and different y-intercepts, so the lines never intersect. Two perprendicular lines have negative reciprocal slopes (m and 21/m), and they intersect in (A) zero places, (B) one place, or (C) two places.

▼ ANSWER (B) one place

▼ ANSWER

$75

[ S E C T I O N 2 . 3 ]     S U M M A R Y In this section, we discussed graphs and equations of lines. Lines are often expressed in two forms:

n

n

We found equations of lines, given either two points or the slope and a point. We found the point–slope form, y 2 y1 5 m 1x 2 x12, useful when the slope and a point are given. We also discussed both parallel (nonintersecting) and perpendicular (forming a right angle) lines. Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes, provided their slopes are defined.

n

General Form: Ax 1 By 5 C Slope–Intercept Form: y 5 mx 1 b

All lines (except horizontal and vertical) have exactly one x-intercept and exactly one y-intercept. The slope of a line is a measure of steepness. n

Slope of a line passing through 1x1, y12 and 1x2, y22: m5

n

Horizontal lines: m 5 0 Vertical lines: m is undefined

y2 2 y1 rise 5 x2 2 x1 run

[SEC TI O N 2 . 3]   E X E R C I S E S • SKILLS In Exercises 1–10, find the slope of the line that passes through the given points. 1. 11, 32 and 12, 62

5. 127, 92 and 13, 2102 2

1

5

3

9. A 3 , 2 4 B and A 6 , 2 4 B

Young_AT_6160_ch02_pp164-201.indd 197

2. 12, 12 and 14, 92

6. 111, 232 and 122, 62 1 3

3 7

10. A 2 , 5 B and A24 , 5 B

3. 122, 52 and 12, 232

7. 10.2, 21.72 and 13.1, 5.22

4. 121, 242 and 14, 62

8. 122.4, 1.72 and 125.6, 22.32

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For each graph in Exercises 11–16, identify (by inspection) the x- and y-intercepts and slope if they exist, and classify the line as rising, falling, horizontal, or vertical. 11.

12.

13.

y

y

(2, 3)

y

(0, 3)

(–2, 3)

x

x

x (2, –1)

(–1, –3) (–4, –3)

15.

14.

16.

y

y

y (–4, 4)

(–3, 3)

(–1, 1)

(3, 1)

x

x

x

(2, –2) (–4, –3)

In Exercises 17–30, find the x- and y-intercepts if they exist and graph the corresponding line. 1

1

17. y 5 2x 2 3

18. y 5 23x 1 2

19. y 5 2 2 x 1 2

21. 2x 2 3y 5 4

22. 2x 1 y 5 21

23. 2 x 1 2 y 5 21

24. 3 x 2 4 y 5 12

25. x 5 21

26. y 5 23

27. y 5 1.5

28. x 5 27.5

7

1

20. y 5 3 x 2 1

1

1

1

1

5

29. x 5 2 2

30. y 5 3

In Exercises 31–42, write the equation in slope–intercept form. Identify the slope and the y-intercept. 31. 2x 2 5y 5 10

32. 3x 2 4y 5 12

33. x 1 3y 5 6

34. x 1 2y 5 8

35. 4x 2 y 5 3

36. x 2 y 5 5

37. 12 5 6x 1 3y

38. 4 5 2x 2 8y

39. 0.2x 2 0.3y 5 0.6

40. 0.4x 1 0.1y 5 0.3

41. 2 x 1 3 y 5 4

1

2

1

2

42. 4 x 1 5 y 5 2

In Exercises 43–50, write the equation of the line, given the slope and intercept. 43. Slope: m 5 2

y-intercept: 10, 32

47. Slope: m 5 0

y-intercept: 10, 22

44. Slope: m 5 22

y-intercept: 10, 12

48. Slope: m 5 0

y-intercept: 10, 21.52

1

1

45. Slope: m 5 2 3

46. Slope: m 5 2

49. Slope: undefined

50. Slope: undefined

y-intercept: 10, 02

x-intercept: A 32, 0B

y-intercept: 10, 232

x-intercept: 123.5, 02

In Exercises 51–60, write an equation of the line in slope–intercept form, if possible, given the slope and a point that lies on the line. 51. Slope: m 5 5

121, 232

55. Slope: m 5

11, 212

3 4



59. Slope: undefined

121, 42

Young_AT_6160_ch02_pp164-201.indd 198

52. Slope: m 5 2

11, 212

56. Slope: m 5

125, 32

217



60. Slope: undefined

14, 212

53. Slope: m 5 23

54. Slope: m 5 21

57. Slope: m 5 0

58. Slope: m 5 0

122, 22 122, 42

13, 242 13, 232

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199

In Exercises 61–80, write the equation of the line that passes through the given points. Express the equation in slope–intercept form or in the form x 5 a or y 5 b. 61. 122, 212 and 13, 22

65. 120, 2372 and 1210, 2422 1 3

3 9

62. 124, 232 and 15, 12

66. 128, 122 and 1220, 2122 2

1

7 1

63. 123, 212 and 122, 262

64. 125, 282 and 17, 222 72. 125, 222 and 125, 42

67. 121, 42 and 12, 252

69. A 2 , 4 B and A 2 , 4 B

70. A2 3 , 2 2 B and A 3 , 2 B

71. 13, 52 and 13, 272

77. 126, 82 and 126, 222

78. 129, 02 and 129, 22

79. A 5 , 2 4 B and A 5 , 2 B

73. 13, 72 and 19, 72

74. 122, 212 and 13, 212

75. 10, 62 and 125, 02 3

2

68. 122, 32 and 12, 232 76. 10, 232 and 10, 22 1 2

2 1

1 1

80. A 3 , 5 B and A 3 , 2 B

In Exercises 81–86, write the equation corresponding to each line. Express the equation in slope–intercept form. 81.

82. y

83. y

y

x

x

84.

x

85. y

86. y

y

x

x

x

In Exercises 87–100, find the equation of the line that passes through the given point and also satisfies the additional piece of information. Express your answer in slope–intercept form, if possible. 87. 123, 12; parallel to the line y 5 2x 2 1 88. 11, 32; parallel to the line y 5 2x 1 2 89. 10, 02; perpendicular to the line 2x 1 3y 5 12 90. 10, 62; perpendicular to the line x 2 y 5 7 91. 13, 52; parallel to the x-axis 92. 13, 52; parallel to the y-axis

93. 121, 22; perpendicular to the y-axis 94. 121, 22; perpendicular to the x-axis 1

1

2

3

95. 122, 272; parallel to the line 2 x 2 3 y 5 5 96. 11, 42; perpendicular to the line 23 x 1 2 y 5 22 2 2 97. A23 , 3 B; perpendicular to the line 8x 1 10y 5245 98. A 65 , 3B; perpendicular to the line 6x 1 14y 5 7 7

1

13

99. A 2 , 4B; parallel to the line 215x 1 35y 5 7 100. A24,2 9 B; parallel to the line 10x 1 45y 5 29

• A P P L I C AT I O N S

101. Budget: Home Improvement. The cost of having your

103. Budget: Monthly Driving Costs. The monthly costs associ-

bathroom remodeled is the combination of material costs and labor costs. The materials (tile, grout, toilet, fixtures, etc.) cost is $1200, and the labor cost is $25 per hour. Write an equation that models the total cost C of having your bathroom remodeled as a function of hours h. How much will the job cost if the worker estimates 32 hours? 102. Budget: Rental Car. The cost of a one-day car rental is the sum of the rental fee, $50, plus $0.39 per mile. Write an equation that models the total cost associated with the car rental.

ated with driving a new Honda Accord are the monthly loan payment plus $25 every time you fill up with gasoline. If you fill up 5 times in a month, your total monthly cost is $500. How much is your loan payment? 104. Budget: Monthly Driving Costs. The monthly costs associated with driving a Ford Explorer are the monthly loan payment plus the cost of filling up your tank with gasoline. If you fill up 3 times in a month, your total monthly cost is $520. If you fill up 5 times in a month, your total monthly cost is $600. How much is your monthly loan, and how much does it cost every time you fill up with gasoline?

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CHAPTER 2  Graphs

105. Business. The operating costs for a local business are a fixed

Height (inches)

amount of $1300 plus $3.50 per unit sold, while revenue is $7.25 per unit sold. How many units does the business have to sell in order to break even? 106. Business. The operating costs for a local business are a fixed amount of $12,000 plus $13.50 per unit sold, while revenue is $27.25 per unit sold. How many units does the business have to sell in order to break even? 107. Weather: Temperature. The National Oceanic and Atmospheric Administration (NOAA) has an online conversion chart that relates degrees Fahrenheit, 8F, to degrees Celsius, 8C. 778F is equivalent to 258C, and 688F is equivalent to 208C. Assuming the relationship is linear, write the equation relating degrees Celsius C to degrees Fahrenheit F. What temperature is the same in both degrees Celsius and degrees Fahrenheit? 108. Weather: Temperature. According to NOAA, a “standard day” is 158C at sea level, and every 500 feet elevation above sea level corresponds to a 18C temperature drop. Assuming the relationship between temperature and elevation is linear, write an equation that models this relationship. What is the expected temperature at 2500 feet on a “standard day”? 109. Life Sciences: Height. The average height of a man has increased over the last century. What is the rate of change in inches per year of the average height of men? 72 71 70 69 68 67 66 65 1900

1950 Year

2000

110. Life Sciences: Height. The average height of a woman has

Height (inches)

increased over the last century. What is the rate of change in inches per year of the average height of women? 67 66 65 64 63 62 61 60 1900

1950 Year

2000

111. L  ife Sciences: Weight. The average weight of a baby born in

1900 was 6 pounds 4 ounces. In 2000, the average weight of a newborn was 6 pounds 10 ounces. What is the rate of change of birth weight in ounces per year? What do we expect babies to weigh at birth in 2040? 112. Sports. The fastest a man could run a mile in 1906 was 4 minutes and 30 seconds. In 1957, Don Bowden became the first American to break the 4-minute mile. Calculate the rate of change in mile speed per year. 113. M  onthly Phone Costs. Mike’s home phone plan charges a flat monthly fee plus a charge of $0.05 per minute for long-distance calls. The total monthly charge is represented

Young_AT_6160_ch02_pp164-201.indd 200

by y 5 0.05x 1 35, x $ 0, where y is the total monthly charge and x is the number of long-distance minutes used. Interpret the meaning of the y-intercept. 114. C  ost: Automobile. The value of a Daewoo car is given by y 5 11,100 2 1850x, x $ 0, where y is the value of the car and x is the age of the car in years. Find the x-intercept and y-intercept and interpret the meaning of each. 115. Weather: Rainfall. The average rainfall in Norfolk, Virginia, for July was 5.2 inches in 2003. The average July rainfall for Norfolk was 3.8 inches in 2007. What is the rate of change of rainfall in inches per year? If this trend continues, what is the expected average rainfall in 2015? 116. Weather: Temperature. The average temperature for Boston in January 2005 was 438F. In 2007 the average January temperature was 44.58F. What is the rate of change of the temperature per year? If this trend continues, what is the expected average temperature in January 2016? 117. E  nvironment. In 2000, Americans used approximately 380 billion plastic bags. In 2005, approximately 392 billion were used. What is the rate of change of plastic bags used per year? How many plastic bags will be expected to be used in 2016? 118. Finance: Debt. According to financial reports, the average household credit card debt in 2010 was $7768 and in 2012 was $7117. What is the rate of change of the credit card debt per year? If this trend were to continue, how much would you expect the average household credit card debt to be in 2016? 119. B  usiness. A Web site that supplies Asian specialty foods to restaurants advertises a 64-ounce bottle of hoisin sauce for $16.00. Shipping cost for one bottle is $15.93. The shipping cost for two bottles is $19.18. The cost for five bottles, including shipping, is $111.83. Answer the following questions based on this scenario. Round to the nearest cent, when necessary. a.   Write the three ordered pairs where x represents the number of bottles purchased and y represents the total cost (including shipping) for one, two, or five bottles purchased. b.  Calculate the slope between the origin and the ordered pair that represents the purchase of one bottle of hoisin sauce. Explain what this amount means in terms of the sauce purchase. c.  Calculate the slope between the origin and the ordered pair that represents the purchase of two bottles of hoisin (including shipping). Explain what this amount means in terms of the sauce purchase. d.  Calculate the slope between the origin and the ordered pair that represents the purchase of five bottles of hoisin (including shipping). Explain what this amount means in terms of the sauce purchase. 120. Business. A Web site that supplies Asian specialty foods to restaurants advertises an 8-ounce bottle of plum sauce for $4.00, but shipping for one bottle is $14.27. The shipping cost for two bottles is $14.77. The cost for five bottles, including shipping, is $35.93. Answer the following questions based on this scenario. Round to the nearest cent, when necessary. a.   Write the three ordered pairs where x represents the number of bottles purchased and y represents the total cost, including shipping for one, two, or five bottles purchased. b.  Calculate the slope between the origin and the ordered pair that represents the purchase of one bottle of plum sauce. Explain what this amount means in terms of the sauce purchase.

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2.3 Lines 



c.  Calculate the slope between the origin and the ordered pair that represents the purchase of two bottles of plum sauce (including shipping). Explain what this amount means in terms of the sauce purchase.



201

d.  Calculate the slope between the origin and the ordered pair that represents the purchase of five bottles of plum sauce (including shipping). Explain what this amount means in terms of the sauce purchase.

• C AT C H T H E M I S TA K E In Exercises 121–124, explain the mistake that is made. 121. Find the x- and y-intercepts of the line with equation 2x 2 3y 5 6.

123. Find the slope of the line that passes through the points

123, 42 and 123, 72.

Solution: x-intercept: set x 5 0 and solve for y.

Solution:

23y 5 6

The x-intercept is 10, 222.

y 5 22

y-intercept: set y 5 0 and solve for x.   2x 5 6

This is incorrect. What mistake was made? 122, 32 and 14, 12.

y2 2 y1 x2 2 x1 23 2 1 23 2



a.  m 5 0  b.  m undefined  c.  m 5 2  d.  m 5 21

Solution:

Solution:

y2 2 y1 x2 2 x1 123 22 1 Substitute 122, 32 and 14, 12. m5 5 5 22 2 4 26 3 This is incorrect. What mistake was made? Write the slope formula.

m5

50 427 This is incorrect. What mistake was made? 124. Given the slope, classify the line as rising, falling, horizontal, or vertical.

122. F  ind the slope of the line that passes through the points



Write the slope formula.

Substitute 123, 42 and 123, 72. m 5

x 5 3 The y-intercept is 13, 02.



m5



a.  vertical line

b.  horizontal line



c.  rising

d.  falling



These are incorrect. What mistakes were made?

• CONCEPTUAL In Exercises 125–128, determine whether each statement is true or false. 125. A line can have at most one x-intercept.

128. If the slopes of two lines are 21 and 1, then the lines are

126. A line must have at least one y-intercept.

perpendicular. 129. If a line has slope equal to zero, describe a line that is perpendicular to it. 130. If a line has no slope, describe a line that is parallel to it.

127. If the slopes of two lines are

parallel.

215

and 5, then the lines are

• CHALLENGE 131. Find an equation of a line that passes through the point

12B, A 1 12 and is parallel to the line Ax 1 By 5 C. Assume that B is not equal to zero. 132. Find an equation of a line that passes through the point 1B, A 212 and is parallel to the line Ax 1 By 5 C. Assume that B is not equal to zero. 133.  Find an equation of a line that passes through the point 12A, B 212 and is perpendicular to the line Ax 1 By 5 C. Assume that A and B are both nonzero.

134. Find an equation of a line that passes through the point 1A, B 1 12

and is perpendicular to the line Ax 1 By 5 C. Assume that A and B are both nonzero. 135. Show that two lines with equal slopes and different y-intercepts have no point in common. Hint: Let y1 5 mx 1 b1 and y2 5 mx 1 b2 with b1 2 b2. What equation must be true for there to be a point of intersection? Show that this leads to a contradiction. 136. Let y1 5 m1x 1 b1 and y2 5 m2  x 1 b2 be two nonparallel lines 1 m1 2 m2 2 . What is the x-coordinate of the point where they intersect?

• TECHNOLOGY For Exercises 137–142, determine whether the lines are parallel, perpendicular, or neither, and then graph both lines in the same viewing screen using a graphing utility to confirm your answer. 137. y1 5 17x 1 22

1 y2 5 217 x

2 13

138. y1 5 0.35x 1 2.7

y2 5 0.35x 2 1.2

Young_AT_6160_ch02_pp164-201.indd 201

139. y1 5 0.25x 1 3.3

y2 5 24x 1 2 15 y2 5 2x 2 3

140. y1 5

1 2x

141. y1 5 0.16x 1 2.7

y2 5 6.25x 2 1.4

142. y1 5 23.75x 1 8.2

y2 5

4 15 x

1

5 6

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CHAPTER 2  Graphs

2.4 CIRCLES SKILLS OBJECTIVES ■■ Graph a circle. ■■ Transform general equations of circles to the standard form by completing the square.

CONCEPTUAL OBJECTIVES ■■ Understand algebraic and graphical representations of circles. ■■ Expand the concept of completing the square with quadratic equations to circles.

2.4.1  Standard Equation of a Circle 2.4.1 S K I L L

Graph a circle. 2.4.1 C O N C E P T U A L

Understand algebraic and ­graphical representations of circles.

Most people understand the shape of a circle. The goal in this section is to develop the equation of a circle.

DEFINITION

Circle

y

A circle is the set of all points in a plane that are a fixed distance from a point, the center. The center, C, is typically denoted by 1h, k2, and the fixed distance, or radius, is denoted by r.

(x, y) r (h, k)

x

What is the equation of a circle? We’ll use the distance formula from Section 2.1. Distance formula: d 5 " 1 x 2 2 x 1 2 2 1 1 y2 2 y1 2 2

The distance between the center 1h, k2 and any point 1x, y2 on the circle is the radius r. Substitute these values d 5 r, 1x1, y12 5 1h, k2 , and 1x2, y22 5 1x, y2 into the distance formula r 5 " 1 x 2 h 2 2 1 1 y 2 k 2 2 and square both sides: 1 x 2 h 2 2 1 1 y 2 k 2 2 5 r 2. All circles can be written in standard form, which makes it easy to identify the center and radius. EQUATION OF A CIRCLE

The standard form of the equation of a circle with radius r and center 1h, k2 is 1x 2 h22 1 1y 2 k22 5 r2

For the special case of a circle with center at the origin 10, 02, the equation simplifies to x 2 1 y 2 5 r 2. UNIT CIRCLE

A circle with radius 1 and center 10, 02 is called the unit circle: x2 1 y2 5 1

The unit circle plays an important role in the study of trigonometry. Note that if x2 1 y2 5 0, the radius is 0, so the “circle” is just a point.

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203

2.4 Circles 

EXAMPLE 1  Finding the Center and Radius of a Circle

Identify the center and radius of the given circle and graph. Solution:

1x 2 22 2 1 1 y 1 12 2 5 4

Rewrite this equation in standard form. Identify h, k, and r by comparing this equation with the standard form of a circle: 1x 2 h22 1 1y 2 k22 5 r 2.

3x 2 24 2 1 3 y 21212  4 2 5 22

 o draw the circle, label the center 12, 212. Label T four additional points two units (the radius) away from the center: 1 4, 21 2 , 1 0, 21 2 , 1 2, 1 2 , and 1 2, 23 2 . Note that the easiest four points to get are those obtained by going out from the center both horizontally and vertically. Connect those four points with a smooth curve.

y

h 5 2, k 5 21, and r 5 2



Center 12, 212 and r 5 2 (2, 1) x (0, –1)

(2, –1)

(4, –1)

(2, –3)

▼

▼ ANSWER

Y O U R T U R N   Identify the center and radius of the given circle and graph.

Center: 121, 222 Radius: 3

1x 1 12 2 1 1y 1 22 2 5 9

y

EXAMPLE 2  Graphing a Circle: Fractions and Radicals

Identify the center and radius of the given circle and sketch its graph. 2

Solution:

(–1, 1) x

2

1 1 ax 2 b 1 ay 1 b 5 20 2 3

(2, –2)

(–4, –2) (–1, –2) (–1, –5)

1 2 1 2 If r 2 5 20, then r 5 "20 5 2"5. ax 2 b 1 cy 2 a2 b d 5 A2 !5B 2 2 3 Write the equation in standard form.

1 1 Identify the center and radius. Center a , 2 b  and r 5 2"5 2 3  o graph the circle, we’ll use decimal T approximations of the fractions and radicals: 10.5, 20.32 for the center and 4.5 for the radius. Four points on the circle that are 4.5 units from the center are 124, 20.32, 15, 20.32, 10.5, 4.22, and 10.5, 24.82. Connect them with a smooth curve.

y

(0.5, 4.2)

(5.0, –0.3)

x

(0.5, –0.3) (–4.0, –0.3) (0.5, –4.8)

EXAMPLE 3  D  etermining the Equation of a Circle Given the Center and Radius

Find the equation of a circle with radius 5 and center 122, 32. Graph the circle. Solution:

Substitute 1h, k2 5 122, 32 and r 5 5 into the standard equation of a circle.

Simplify.

Young_AT_6160_ch02_pp202-235.indd 203

3x 2 1222 4 2 1 1 y 2 322 5 52

1x 1 222 1 1 y 2 322 5 25

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204 

CHAPTER 2  Graphs

 o graph the circle, plot the center 122, 32 T and four points 5 units away from the center: 127, 3 2 , 1 3, 3 2 , 122, 22 2 , and 122, 8 2 . Connect them with a smooth curve.

▼ ANSWER

x2 1 1 y 2 12 2 5 9

y

(–2, 8)

(–7, 3)

(–2, 3)

y

(3, 3)

(0, 4)

5 (–3, 1)

(0, 1)

3

(3, 1)

x

x

(–2, –2) (0, –2)

▼ Y O U R T U R N   Find the equation of a circle with radius 3 and center 10, 12 and graph.

[ CONCEPT CHECK ] A circle described algebraically by (x 2 h)2 + (y 2 k )2 5 r 2 can be graphed by first going to the center (h, k ) and then going to the points: (A) (2h 6 r, 2k 6 r ) or (B) (h 6 r, k 6 r )

Let’s change the look of the equation given in Example 1. In Example 1 the equation of the circle was given as: 1x 2 222 1 1y 1 122 5 4 Eliminate the parentheses. x2 2 4x 1 4 1 y2 1 2y 1 1 5 4 Group like terms and subtract 4 from both sides. x2 1 y2 2 4x 1 2y 1 1 5 0 We have written the general form of the equation of the circle in Example 1. The general form of the equation of a circle is x 2 1 y 2 1 ax 1 by 1 c 5 0 Suppose you are given a point that lies on a circle and the center of the circle. Can you find the equation of the circle?

▼ ANSWER (B) (h 6 r, k 6 r )

EXAMPLE 4  F  inding the Equation of a Circle Given Its Center and One Point

The point 110, 242 lies on a circle centered at 17, 282. Find the equation of the circle in general form. Solution:

y

2 4 6 8 10 12 14 x 2 4

(7, –3)

(10, –4)

(2, –8) 10 12 14

(7, –8)

(12, –8)

(7, –13)

▼ ANSWER 2

2

x 1 y 1 10x 2 6y 2 66 5 0

Young_AT_6160_ch02_pp202-235.indd 204

 his circle is centered at 17, 282, so its standard equation is 1x 2 72 2 1 1y 1 82 2 5 r 2. T All that remains is to find the radius. Approach 1: Since the point 110, 242 lies on the circle, it must satisfy the equation of the circle. 1 10 2 7 2 2 1 1 24 1 8 2 2 5 r 2 Substitute 1x, y2 5 110, 242. Simplify. 32 1 42 5 r 2 The distance from 110, 242 to 17, 282 is five units. r55 Approach 2: Find the distance between 110, 242 r 5 d 5 " 1 10 2 7 2 2 1 124 2 128 2 2 2 and 17, 282. 5 "32 1 42 5 "25 55 Substitute r 5 5 into the standard equation. 1x 2 722 1 1y 1 822 5 52 Eliminate the parentheses and simplify. x2 2 14x 1 49 1 y2 1 16y 1 64 5 25 x2 1 y2 2 14x 1 16y 1 88 5 0 Write in general form.

▼ YOUR TURN  T  he point 11, 112 lies on a circle centered at 125, 32. Find the

equation of the circle in general form.

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2.4 Circles 

205

2.4.2 Transforming Equations of Circles to the Standard Form by Completing the Square If the equation of a circle is given in general form, it must be rewritten in standard form in order to identify its center and radius. To transform equations of circles from general to standard form, complete the square (Section 1.3) on both the x- and y-variables.

2.4.2 S K I L L

Transform general equations of circles to the standard form by completing the square.

EXAMPLE 5  F  inding the Center and Radius of a Circle by Completing the Square

2.4.2 C O N C E P T U A L

Find the center and radius of the circle with the equation:

Expand the concept of ­completing the square with ­quadratic equations to circles.

2

2

x 2 8x 1 y 1 20y 1 107 5 0

Solution:

 ur goal is to transform this equation O into standard form

1x 2 h22 1 1 y 2 k22 5 r 2

Group x and y terms, respectively, on the left side of the equation; move constants to the right side.

1x2 2 8x2 1 1y2 1 20y2 5 2107



E

E

Complete the square on both the x and y expressions. 1x2 2 8x 1 u2 1 1 y2 1 20y 1 u2 5 2107 2 2 5 100 to both sides. Add A228 B 5 16 and A 20 2B 2 2  1x 2 8x 1 4 2 1 1 y2 1 20y 1 102 2 52107 1 16 1 100 16 100

Factor the perfect squares on the left side and simplify the right side.

1x 2 422 1 1 y 1 1022 5 9

1x 2 422 1 3 y 2 121024 2 5 32

Write in standard form.

The center is 14, 2102 and the radius is 3.

▼

Y O U R T U R N   Find the center and radius of the circle with the equation:

x 2 1 y 2 1 4x 2 6y 2 12 5 0

▼ ANSWER

Center: 122, 32  Radius: 5

common mistake A common mistake is forgetting to add both constants to the right side of the equation. Identify the center and radius of the circle with the equation: 2

2

x 1 y 1 16x 1 8y 1 44 5 0 ✓C O R R E C T

x 2 1 y 2 1 16x 1 8y 1 44 5 0

Ax 2 1 16xB 1 A y 2 1 8yB 5 244

Ax 2 1 16x 1 uB 1 A y 2 1 8y 1 uB 5 244 2 Ax 1 16x 1 64B 1 A y 2 1 8y 1 16B 5 244 1 64 1 16 1 x 1 8 2 2 1 1 y 1 4 2 2 5 36 Center: 128, 242  Radius: 6

Young_AT_6160_ch02_pp202-235.indd 205

✖INCORRECT

Ax 2 1 16x 1 64B 1 A y 2 1 8y 1 16B 5 244 ERROR: Don’t forget to add 16

and 64 to the right.

▼ CAUTION

Don’t forget to add both constants to each side of the equation when completing the square for x and y.

[ CONCEPT CHECK ] Show that (x 1 8)2 1 (  y 1 4)2 5 62 is equal to x 2 1 y 2 1 16x 1 8y 1 44 5 0

▼ ANSWER (x 1 8)2 1 (y 1 4)2 5 62 x 2 1 16x 1 64 1 y 2 1 8y 1 16 5 36 x 2 1 y 2 1 16x 1 8y 1 80 5 36 x 2 1 y 2 1 16x 1 8y 1 44 5 0

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206 

CHAPTER 2  Graphs

[ S E C T I O N 2 . 4 ]     S U M M A R Y The equation of a circle is given by n

n

Standard form: 1x 2 h22 1 1 y 2 k22 5 r 2. n Center: 1h, k2 n Radius: r

General form: x 2 1 y 2 1 ax 1 by 1 c 5 0. n Complete the square to transform the equation to standard form.

[SEC T I O N 2 . 4]   E X E R C I S E S • SKILLS In Exercises 1–20, write the equation of the circle in standard form.  1.  Center 11, 22

 2. Center 13, 42

r53

 5.  Center 15, 72

r53

13.  Center 10, 02

r 5 "2

17.  Center

r5

1 4

A 23,



r 5 "7

18. Center

r5

2 5

A213,

r58

11. Center 10, 22

12. Center 13, 02

r53

14. Center 121, 22

235 B

  8. Center 16, 272

r 5 13

10. Center 10, 02

r52

r 5 4

 7. Center 1211, 122

r56

 9.  Center 10, 02

 4. Center 121, 222

r 5 10

 6. Center 12, 82

r59



 3. Center 123, 242

r55

227 B

r52

15. Center 15, 232

16. Center 124, 212

r 5 2"3



19. Center 11.3, 2.72

r 5 3"5

20. Center 123.1, 4.22

r 5 3.2

r 5 5.5

In Exercises 21–32, find the center and radius of the circle with the given equations. 21. 1x 2 122 1 1 y 2 322 5 25 2

2

22. 1x 1 122 1 1 y 1 322 5 11

23. 1x 2 22 1 1 y 1 52 5 49 24. 1x 1 322 1 1 y 2 722 5 81 25. 1x 2 422 1 1 y 2 922 5 20 26. 1x 1 122 1 1 y 1 222 5 8 2 2

1 2

4

1 2

1 2

9

27. Ax 2 5 B 1 A y 2 7 B 5 9 28. Ax 2 2 B 1 A y 2 3 B 5 25 2 2 29. 1x 2 1.52 1 1 y 1 2.72 5 1.69 30. 1x 1 3.122 1 1 y 2 7.422 5 56.25

31. x2 1 y2 2 50 5 0 32. x2 1 y2 2 8 5 0

In Exercises 33–50, state the center and radius of each circle. 33. x2 1 y2 1 4x 1 6y 2 3 5 0 34. x2 1 y2 1 2x 1 10y 1 17 5 0 35. x2 1 y2 1 6x 1 8y 2 75 5 0 36. x2 1 y2 1 2x 1 4y 2 9 5 0 37. x2 1 y2 2 10x 2 14y 2 7 5 0 38. x2 1 y2 2 4x 2 16y 1 32 5 0 39. x2 1 y2 2 2y 2 15 5 0 40. x2 1 y2 1 2x 2 8 5 0 41. x2 1 y2 2 2x 2 6y 1 1 5 0 42. x2 1 y2 2 8x 2 6y 1 21 5 0 43. x2 1 y2 2 10x 1 6y 1 22 5 0 44. x2 1 y2 1 8x 1 2y 2 28 5 0 45. x2 1 y2 2 6x 2 4y 1 1 5 0 46. x2 1 y2 2 2x 2 10y 1 2 5 0

3y 1 x 3 5 0 48. x2 1 y2 2 2 1 50 4 2 2 8 49. x2 1 y2 2 2.6x 2 5.4y 2 1.26 5 0 50. x2 1 y2 2 6.2x 2 8.4y 2 3 5 0 47. x 2 1 y 2 2 x 1 y 1

In Exercises 51–56, find the equation of each circle. 51. Centered at 121, 222 and passing through the point 11, 02. 52. Centered at 14, 92 and passing through the point 12, 52.

53. Centered at 122, 32 and passing through the point 13, 72.

54. Centered at 11, 12 and passing through the point 128, 252.

55. Centered at 122, 252 and passing through the point 11, 292.

56. Centered at 123, 242 and passing through the point 121, 282.

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2.4 Circles 

207

• A P P L I C AT I O N S 57. Cell Phones. If a cellular phone tower has a reception radius

of 100 miles and you live 95 miles north and 33 miles east of the tower, can you use your cell phone while at home? 58. Cell Phones. Repeat Exercise 57, assuming you live 45 miles south and 87 miles west of the tower. 59. Construction/Home Improvement. A couple and their dog moved into a new house that does not have a fenced-in backyard. The backyard is square with dimensions 100 feet by 100 feet. If they put a stake in the center of the backyard with a long leash, write the equation of the circle that will map out the dog’s outer perimeter.

100 ft

62. Design. Repeat Exercise 61 for the outer circle with a d ­ iameter

of 6000 feet. 63. Cell Phones. A cellular phone tower has a reception radius of

200 miles. Assuming the tower is located at the origin, write the equation of the circle that represents the reception area. 64. Environment. In a state park, a fire has spread in the form of a circle. If the radius is 2 miles, write an equation for the circle. For Exercises 65 and 66, refer to the following:

A cell phone provider is expanding its coverage and needs to place four cell phone towers to provide complete coverage of a 100-square-mile area formed by a 10 mile by 10 mile square. This area can be represented by a region on the Cartesian coordinate system; see figure. y

100 ft

60. Construction/Home Improvement. Repeat Exercise 59

except that the couple put in a pool and a garden and want to restrict the dog to quadrant I. What coordinates represent the center of the ­circle? What is the radius? y

10

Cell Phone Coverage Area

50 x

r

10

(h, k) x

–50

50

–50

61. Design. A university designs its campus with a master plan

of two concentric circles. All of the academic buildings are within the inner circle (so that students can get between classes in less than 10 minutes), and the outer circle ­contains all the dormitories, the Greek park, cafeterias, the gymnasium, and intramural fields. Assuming the center of campus is the origin, write an equation for the inner circle if the diameter is 3000 feet. y

The placement of the four towers is very important in that the cell phone provider needs to provide coverage of the entire 100-square-mile area. The cell phone towers being installed can process signals from cell phones within a 3.5-mile radius. 65. Engineering. One plan under consideration is to place the four towers in locations that correspond to the points 12.5, 2.52, 12.5, 7.52, 17.5, 2.52, and 17.5, 7.52 on the graph. a. Write an equation that describes the perimeter of the cell phone coverage for each of the four towers. b. Draw the coverage provided by each of these towers. Will this placement of towers provide the needed coverage? 66. Engineering. One plan under consideration is to place the four towers in locations that correspond to the points 13, 32, 13, 72, 17, 32, and 17, 72 on the graph. a. Write an equation that describes the perimeter of the cell phone coverage for each of the four towers. b. Draw the coverage provided by each of these towers. Will this placement of towers provide the needed coverage?

x

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208 

CHAPTER 2  Graphs

• C AT C H T H E M I S TA K E In Exercises 67–70, explain the mistake that is made. 67.  Identify the center and radius of the circle with equation 1x 2 422 1 1 y 1 322 5 25. Solution:

The center is 14, 32 and the radius is 5.

This is incorrect. What mistake was made?

68. Identify the center and radius of the circle with equation

1x 2 222 1 1 y 1 322 5 2.



x2 1 y2 2 6x 1 4y 2 3 5 0.

Solution: Group like terms.    1x2 2 6x2 1 1 y2 1 4y2 5 3

Complete the   1x2 2 6x 1 92 1 1 y2 1 4y 1 42 5 12 2 square.     1 x 2 3 2 2 1 1 y 1 2 2 2 5 1 2!3 2



Solution:

70. Find the center and radius of the circle with the ­equation

The center is 13, 222 and the radius is 2!3. This is incorrect. What mistake was made?

The center is 12, 232 and the radius is 2.

This is incorrect. What mistake was made? 69. Graph the solution to the equation 1x 2 122 1 1 y 1 222 5 216. Solution:

The center is 11, 222 and the radius is 4. y

(1, 2) x (–3, –2) 4

(1, –2) (5, –2) (1, –6)

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 71–74, determine whether each statement is true or false. 71. The equation whose graph is depicted has infinitely many

solutions. (0, 5)

2 73. The equation 1x 2 22  1 1 y 1 522 5 220 has no solution.

2 74. The equation 1x 2 12  1 1 y 1 322 5 0 has only one solution.

y

75. Describe the graph (if it exists) of:

x2 1 y2 1 10x 2 6y 1 34 5 0

5 x

(–5, 0) (5, 0)

76. Describe the graph (if it exists) of:

x2 1 y2 2 4x 1 6y 1 49 5 0 77. Find the equation of a circle that has a diameter with ­endpoints

(0, –5)

72. The equation 1x 2 722 1 1 y 1 1522 5 264 has no solution.

• CHALLENGE

79.  For the equation x2 1 y2 1 ax 1 by 1 c 5 0, specify

c­ onditions on a, b, and c so that the graph is a single point. 80. For the equation x2 1 y2 1 ax 1 by 1 c 5 0, specify conditions on a, b, and c so that there is no corresponding graph.

Young_AT_6160_ch02_pp202-235.indd 208

15, 22 and 11, 262.

78. Find the equation of a circle that has a diameter with ­endpoints

13, 02 and 121, 242.

81. Determine the center and radius of the circle given by the

equation x2 1 y2 2 2ax 5 100 2 a2. 82. Determine the center and radius of the circle given by the equation x2 1 y2 1 2by 5 49 2 b2.

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209

• TECHNOLOGY In Exercises 83–86, use a graphing utility to graph each equation. Does this agree with the answer you gave in the Conceptual section? 83. 1x 2 222 1 1 y 1 522 5 220 (See Exercise 73 for comparison.) 84. 1x 2 122 1 1 y 1 322 5 0 (See Exercise 74 for comparison.) 85. x2 1 y2 1 10x 2 6y 1 34 5 0 (See Exercise 75 for comparison.) 86. x2 1 y2 2 4x 1 6y 1 49 5 0 (See Exercise 76 for comparison.)

In Exercises 87–88, (a) with the equation of the circle in ­standard form, state the center and radius, and graph; (b) use the quadratic formula to solve for y; and (c) use a graphing utility to graph each equation found in (b). Does the graph in (a) agree with the graphs in (c)? 87. x2 1 y2 2 11x 1 3y 2 7.19 5 0 88. x2 1 y2 1 1.2x 2 3.2y 1 2.11 5 0

2.5* LINEAR REGRESSION: BEST FIT SKILLS OBJECTIVES ■■ Draw a scatterplot. ■■ Use linear regression to determine the line of best fit associated with some data. ■■ Use the line of best fit to predict values of one variable from the values of another.

CONCEPTUAL OBJECTIVES ■■ Recognize positive or negative association. ■■ Recognize linear or nonlinear association. ■■ Understand what “best fit” means.

2.5.1 Scatterplots An important aspect of applied research across disciplines is to discover and understand relationships between variables, and often how to use such a relationship to predict values of one variable in terms of another. You have likely encountered such issues while watching TV, reading a magazine or newspaper, or simply talking with friends. Some questions include Is age predictive of texting speed? ■■ Is the level of pollution in a country related to the prevalence of asthma in that country? ■■ Do the ratings of car reliability necessarily increase with the price of the car? ■■

2.5.1 S K I L L

Draw a scatterplot. 2.5.1 C O N C E P T U A L

 ecognize positive or negative R association.

In this section, we focus on situations involving relationships between two variables x and y, so that the experimental data gathered consists of ordered pairs 1x1, y12, . . . , 1xn, yn2. A first step in understanding a data set of the form 5  1x1, y12, . . . , 1xn, yn2  6 is to create a pictorial representation of it. Identifying the first coordinates of these ordered pairs as values of an independent variable (or predictor variable) x and the ­second coordinates as the values of a dependent variable (or response variable) y, we ­simply plot them all on a single xy-plane. The resulting picture is called a scatterplot. EXAMPLE 1  Drawing a Scatterplot of Olympic Decathlon Data

 he 2012 Men’s Olympic Decathlon consisted of the following 10 events: 100 meter, T long jump, shot put, high jump, 400 meter, 110 meter hurdles, discus, pole vault, javelin throw, and 1500 meter. Actual scores are converted to a point system where points are assigned to each of these events based on performance. Events are equally weighted when converting to points. These points are then summed to obtain total scores and, in turn, medals are assigned based on these total scores. It would be interesting to know if certain events are more predictive of the total scores than are others. If someone does exceedingly well in the javelin throw, for example, is that person more likely to do well across all events and therefore obtain a large total points score? *Optional Technology Required Section.

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Data from the Men’s 2012 Olympic Decathlon are presented below. Let’s consider the paired data set 5  1x, y2  6 where x 5 score on the 400 m and y 5 total score. One scatterplot using the 400 m and total points information from this data set is shown in the following graph. Several natural questions arise: How are different pairs of these data related? Are there any discernible patterns present, and if so, how strong are they? Is there a single curve that can be used to describe the general trend present in the data? We shall answer these questions one by one in this section.

Men’s 2012 Olympic Decathalon— y 400 m vs. Total Points

Total Points

210 

9000 8750 8500 8250 8000 7750 7500 7250 7000 x 47

49 51 400 m

53

55

X100M

LONG JUMP

SHOT PUT

HIGH JUMP

X400M

X110M HURDLE

DISCUS

POLE VAULT

JAVELIN

RANK

TOTAL SCORE

Eaton

10.35

8.03

14.66

2.05

46.90

13.56

42.53

5.20

61.96

1

8869

Hardee

10.42

7.53

15.28

1.99

48.11

13.54

48.26

4.80

66.65

2

8671

OLYMPIANS

Suárez

11.27

7.52

14.50

2.11

49.04

14.45

4.70

76.94

3

8523

Alphen

11.05

7.64

15.48

2.05

49.18

14.89

  45.7 48.28

4.80

61.69

4

8447

Warner

10.48

7.54

13.73

2.05

48.20

14.38

45.90

4.70

62.77

5

8442

Freimuth

10.65

7.21

14.87

1.90

48.06

13.89

49.11

4.90

57.37

6

8320

Kasyanov

10.56

7.55

14.45

1.99

48.44

14.09

46.72

4.60

54.87

7

8283

Sviridov

10.78

7.45

14.42

1.99

48.91

15.42

47.43

4.60

68.42

8

8219

Coertzen

11.09

7.17

13.79

2.05

48.56

14.15

43.58

4.50

64.79

9

8173

Behrenbruch

11.06

7.15

15.67

1.96

50.04

14.33

44.71

4.70

64.80

10

8126

Sintnicolaas

10.85

7.37

14.18

1.93

48.85

14.43

32.26

5.30

58.82

11

8034

Newdick

11.10

7.36

15.09

1.96

50.22

15.02

46.15

4.70

59.82

12

7988

Barroilhet

11.18

6.80

14.49

2.05

51.07

14.12

41.27

5.40

57.25

13

7972

García

10.80

6.75

14.48

1.99

48.76

14.24

42.27

4.60

59.85

14

7956

Mayer

11.32

7.17

14.05

2.05

48.76

15.59

41.20

4.70

62.41

15

7952

Shkurenyov

11.01

7.25

12.89

2.02

49.81

14.39

43.51

5.10

53.81

16

7948

Mikhan

10.74

6.94

14.75

1.93

48.42

14.15

44.42

4.40

55.69

17

7928

Karpov

10.91

7.21

16.47

1.99

49.83

14.40

44.93

5.10

49.93

18

7926

Alberto de Araújo

10.70

7.16

13.52

1.93

48.25

14.79

44.76

4.60

51.59

19

7849

Ushiro

11.32

6.86

13.59

1.99

50.78

15.47

46.66

4.90

66.38

20

7842

Vos

10.98

7.27

13.77

1.96

49.62

14.61

42.26

4.50

61.60

21

7805

Erin. Š

10.99

6.98

13.45

1.93

50.62

15.22

45.10

4.50

57.35

22

7649

Addy

10.89

6.90

14.97

1.93

48.64

14.23

45.61

4.20

50.36

23

7586

Szabó

11.15

6.96

13.93

1.90

50.83

14.92

45.14

4.60

58.84

24

7581

Draudvila

10.95

7.12

15.17

1.96

50.13

14.87

46.43

4.20

50.16

25

7557

Artikov

11.37

6.41

14.11

1.93

51.91

14.74

43.53

4.40

56.62

26

7203

Creating a scatterplot by hand can be tedious, especially for large data sets. You can also very easily lose precision and detail. Using technology to create a scatterplot is very appropriate and quite easy. Below are the procedures for how you would create the ­scatterplot shown in Example 1 using the TI-831 (or TI-84) and Excel 2010.

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Creating a Scatterplot Using the TI-831 (or TI-84) INSTRUCTION

Entering the Data

SCREENSHOT

Step 1 Press STAT, followed by 1:Edit. . . . Clear any data already present in columns L1 and L2 so that the screen looks like the one to the right.

Step 2 Input the values of the x-variable (first entries in the ordered pairs) in column L1, pressing ENTER after each entry. Then, right arrow over to column L2 and input the values of the y-variable.The screen (starting from the beginning of the data set) should look like the one to the right when you are done. Plotting the Data

Step 3 Press Y5 and then select Plot1 in the top row of the screen. If either Plot2 or Plot3 is darkened, move the cursor onto it and press ENTER to undarken it. The screen should look like the one to the right when you are done.

Step 4 Press 2nd, followed by Y5 (for StatPlot). Select 1: and modify the entries to make the screen look like the one to the right.

Step 5 Press 2nd, followed by Y5 and make certain that both Plot2 and Plot3 are OFF. The screen should look like the one to the right.

Step 6 Make certain the ranges for the x and y values are appropriate for the given data set. Here, we use the window shown to the right.

Step 7 Press GRAPH and you should get the scatterplot shown to the right.

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Creating a Scatterplot Using Excel 2010 INSTRUCTION

Entering the Data

Step 1 Open a new Excel spreadsheet. Input the values of the x-variable (first entries in the ordered pairs) in column A, starting with cell 1A. Then, input the values of the y-variable in column B, starting with cell 1B. The screen should look like the one to the right when you are done.

Plotting the Data

Step 2 Highlight the data. The screen should look like the one to the right when you are done.

SCREENSHOT

Step 3 Go to the Insert tab and select the icon labeled Scatter. A window list of five possible choices pops up.

Step 4 Select the leftmost choice in the top row. Press ENTER. The scatterplot shown to the right should appear on the screen.

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INSTRUCTION

213

SCREENSHOT

Step 5 You can alter the format of the scatterplot with various bells and whistles by rightclicking anywhere near the data points and then selecting Format Plot Area at the bottom of the pop-up window.

Y O U R T U R N   Using the data in Example 1, identify x 5 score on the pole vault

and z 5 total score. Use technology to create a scatterplot for the data set consisting of the ordered pairs 1x, z2.

2.5.2  Identifying Patterns

While scatterplots show only clusters of ordered pairs, patterns of various types can emerge that can provide insight into how the variables x and y are related.

Men's 2012 Olympic Decathlon– Pole Vault and Total Points 9000 8750 8500 8250 8000 7750 7500 7250 7000 4.1 4.5 4.9 5.3 Pole Vault

Direction of Association This characteristic is analogous to the concept of slope of a line. If the cluster of points tends to rise from left to right, we say that x and y are positively associated, whereas if the cluster of points falls from left to right, we say that x and y are negatively associated. Certainly, the more closely packed together the points are to an identifiable curve, the easier it is to make such a determination. Some examples of scatterplots of varying degrees of positive and negative association are shown in the following table.

SCATTERPLOT

DIRECTION OF ASSOCIATION

Positive Association

y

y

▼ ANSWER

Total Points

▼

2.5.2 S K I L L

Use linear regression to determine the line of best fit associated with some data. 2.5.2 C O N C E P T U A L

Recognize linear or nonlinear association. VERBAL DESCRIPTION

A shape that increases from left to right is very discernible in each case.

x

x

The association is a bit loose, but you can still tell the data points tend to rise from left to right.

y

x

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CHAPTER 2  Graphs

SCATTERPLOT

DIRECTION OF ASSOCIATION y

VERBAL DESCRIPTION

No Association

Haphazard scattering of points suggests neither positive nor negative association.

Negative Association

The association is a bit loose, but you can still tell the data points tend to fall from left to right.

Negative Association

A shape that decreases from left to right is very discernible in each case.

x

y

x

y

y

x

x

[ CONCEPT CHECK ] If the points (x, y) of a scatterplot are tightly packed together and fall from left to right, we say that the relationship between x and y is a (A) strong negative linear relationship, (B) strong positive linear relationship, (C) strong negative nonlinear relationship, or (D) strong positive nonlinear relationship

Linearity Depending on the phenomena being studied and the actual sample being used, the data points comprising a scatterplot can conform very closely to an actual curve. If the curve is a line, we say that the relationship between x and y is linear; otherwise, we say the r­ elationship is nonlinear. Some illustrative examples follow.

SCATTERPLOT

LINEARITY

y

VERBAL DESCRIPTION

Linear

Perfect linear relationship; positive association

Linear

Perfect linear relationship; negative association

▼ ANSWER (A) strong negative linear relationship

x

y

x

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SCATTERPLOT

LINEARITY

y

215

VERBAL DESCRIPTION

Linear

Fairly tight linear relationship; positive association

Linear

Fairly tight linear relationship; negative association

Linear

Rather loose, but still somewhat discernible, linear relationship; positive association

Linear

Rather loose, but still somewhat discernible, linear relationship; negative association

Nonlinear

Perfect nonlinear relationship

Nonlinear

Fairly tight nonlinear relationship

No specifically identifiable relationship

No discernible linear relationship or degree of association

x

y

x

y

x

y

x

y

x

y

x

y

x

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EXAMPLE 2  Describing Patterns in a Data Set

Describe the patterns present in the paired data set 1x, y2 considered in Example 1, where x 5 points on the 400 m and y 5 total score. Is this intuitive? Solution:

We can surmise that the variable score on the 400 m has a relatively strong linear, negative association with the variable total score. This means that as the 400 m score decreases, the total score tends to increase. A negative relationship makes sense here in that 400 m scores reflect how much time it took to complete this race. So, lower scores (less time) reflect better performance and therefore more total points. ▼ ANSWER

The variable score the pole vault has a rather weak linear, positive association with the variable total score. This means that as the pole vault score increases, the total score tends to increase. In this case, a positive relationship makes sense in that pole vault scores reflect the height achieved. So, higher scores (greater height) reflect better performance and therefore more total points.

▼ Y O U R T U R N   Using the data in Example 1, identify x 5 score on the pole vault

and z 5 total score. Comment on the degree of association and linearity of the ­scatterplot consisting of the ordered pairs 1x, z2. Is this intuitive?

Strength of Linear Relationship The variability in the data can render it difficult to determine if there is a linear ­relationship between two variables. As such, it is useful to have a way of measuring how tightly a paired data set conforms to a linear shape. This measure is called the correlation coefficient, r, and is defined by the following formula: DEFINITION

For a paired data set 5  1x1, y12, . . . , 1xn, yn2  6, the correlation coefficient, r, is defined by r5

n a xy 2 Q a xRQ a yR

n x 2 2 Q a xR ~ n a y 2 2 Q a yR Å a Å 2

The symbol a z is a shorthand way of writing z1 1 2 2 a x 5 x1 1

c1

c1

2

zn. So, for instance,

x 2n.

This is tedious to calculate by hand but is easily computed using technology. Below are the procedures to compute the correlation coefficient for the data set introduced in Example 1 using the TI-83+ (or TI-84) and Excel 2010.

Computing a Correlation Coefficient Using the TI-831 (or TI-84) INSTRUCTION

Step 1

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SCREENSHOT

Enter the data following the procedure outlined earlier in this section. The screen should look like the one to the right.

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INSTRUCTION

Step 2

217

SCREENSHOT

Set up what will display! In order for the desired output to display once we execute the commands to follow, we must tell the calculator to do so. As such, do the following: i. Press 2nd, followed by 0 to get CATALOG. ii. Scroll down until you get to DiagnosticOn. Press ENTER. Then, this command will appear on the home screen. Press ENTER again. The resulting screen should look like the one to the right.

Step 3

Press STAT, followed by CALC, and then by 4:LinReg(ax1b). The resulting screen should look like the one to the right. Press ENTER.

Step 4

Press ENTER again. After a brief moment, your screen should look like the one to the right. The value we want is in the bottom row of the screen, about r 5 20.7045.

Note: Other information given will be pertinent once we define the best fit line in the next section.

Computing a Correlation Coefficient Using Excel 2010 INSTRUCTION

Step 1

Enter the data following the procedure outlined earlier in this section. The screen should look like the one to the right.

Step 2

Select the Formulas tab at the top of the screen and then choose the More Functions within the Function Library grouping (on the left). The pull-down menu should be as shown to the right.

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SCREENSHOT

Step 1

Step 2

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CHAPTER 2  Graphs

INSTRUCTION

Step 3

SCREENSHOT

From here, select Statistical, and then from this list, scroll down and choose CORREL. A pop-up window should appear, as shown to the right.

  Step 4

Enter A1:A26 in Array 1 and B1:B26 in Array 2, as shown to the right. Press OK. You will notice that the correlation coefficient appears directly beneath Array 2. In this case, r is about 20.7147.

 

The square of the correlation coefficient is interpreted as a signed percentage of the ­variability among the y-values that is actually explained by the linear relationship, where the sign corresponds to the direction of the association. For instance, an r-value of 11 means that 100% of the variability among the y-values is explained by a line with positive slope; in such a case, all of the points in the data set actually lie on a single line. An r-value of 21 means the same thing, but the line has a negative slope. As the r-values get closer to zero, the more dispersed the points become from a line describing the pattern, so that an r-value very close to zero suggests no linear relationship is discernible. The following sample of scatterplots with the associated ­correlation coefficients should give you a feel for the strength of linearity ­suggested by various values of r.

CORRELATION COEFFICIENT r

SCATTERPLOT

r 5 1.0

y

STRENGTH OF LINEARITY

Perfect positive linear relationship

x

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CORRELATION COEFFICIENT r

SCATTERPLOT y

219

STRENGTH OF LINEARITY

r 5 21.0

Perfect negative linear relationship

r 5 0.80

 easonably strong, though not perfect, R positive linear relationship

  r 5 20.45

 retty weak, barely discernible, P negative linear relationship

r 5 0.10

No discernible linear relationship

x

y

x

y

x

y

x

EXAMPLE 3  C  alculating the Correlation Coefficient Associated with a Data Set

Use technology to calculate the correlation coefficient r for the paired data set 1x, y2 considered in Example 1, where x 5 points on the 400 m and y 5 total score. Interpret the strength of the linear relationship. Solution:

We see that using either form of technology yields r 5 20.7147. This suggests that the data follow a relatively strong negative (i.e., negative slope) linear pattern.

▼ YOUR TURN  U  sing the data in Example 1, identify x 5 score on the pole vault

and z 5 total score, and calculate the correlation coefficient for the data set consisting of the ordered pairs 1x, z2 using technology. Interpret the strength of the linear relationship.

2.5.3  Linear Regression Determining the “Best Fit” Line Assuming that a data set follows a reasonably strong linear pattern, it is natural to ask which single straight line best describes this pattern. Having such a line would enable us not only to describe the relationship between the two variables x and y precisely, but also to predict values of y from values of x not present among the points of the data set.

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[ CONCEPT CHECK ] If the best-fit line for a set of data points (x, y) is horizontal, then we say (A) the relationship between x and y is positive linear, (B) the relationship between x and y is negative linear, or (C) there is no discernible relationship.

▼ ANSWER (C) There is no discernible relationship.

▼ ANSWER

The correlation coefficient is approximately r 5 0.45. This suggests that while the data follow a positive (i.e., positive slope) pattern, the degree to which an actual line describes the trend in the data is rather weak.

2.5.3 S K I L L

Use the line of best fit to predict values of one variable from the values of another. 2.5.3 C O N C E P T U A L

Understand what “best fit” means.

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CHAPTER 2  Graphs

Consider the paired data set 1x, y2 from Example 1, where x 5 points on the 400 m and y 5 total score. You learned in Section 2.3 that between any two points there is a unique line whose equation can be determined. Three such lines passing through ­various pairs of points in the data set are illustrated below. Men’s 2012 Olympic Decathalon— y 400 m vs. Total Points

Total Points

9000 8750 8500 8250 8000 7750 7500 7250 7000

line 1 line 2 line 3 x 47

49 51 400 m

53

55

The unavoidable shortcoming of all of these lines, however, is that not all of the data points lie on a single one of them. Each has a negative slope, which is characteristic of the data set, and each of the lines is close to some of the data points but not close to others. In fact, we could draw infinitely many such lines and make a similar assessment. But which one best fits the data? The answer to this question depends on how you define “best.” Reasonably, for the line that best fits the data, the error incurred in using it to describe all of the points in the data set should be as small as possible. The conventional approach is to define this error by summing the n ­distances di between the y-coordinates of the data points and the corresponding y-value on the line y 5 Mx 1 B (that is, the y-value of the point on the line corresponding to the same x-value). These distances are, in effect, the error in making the approximation. This is illustrated below: y (x6, y6) d6 (x3, y3) (x1, y1)

d5

(x2, Mx2+B) d 3 (x2, y2)

d1 y = Mx + B

(x5, y5)

x2

d7

(x3, Mx3+B) d4

(x1, Mx1+B) x1

d8 (x8, y8)

(x4, y4) x3 x4

x5 x6

(x7, y7) x7 x8

x

Using the distance formula, we find that di 5 " 1 xi 2 xi 2 2 1 1 yi 2 1 Mxi 1 B 2 2 2 5 0 yi 2 1 Mxi 1 B 2 0 .

Note that di 5 0 precisely when the data point 1xi, yi2 lies directly on the line y 5 Mx 1 B, and that the closer di is to 0, the closer the point 1xi, yi2 is to the line y 5 Mx 1 B. As such, the goal is to determine the values of the slope M and y-intercept B for which the sum d1 1 c 1 dn is as small as possible. Then, the resulting straight line y 5 Mx 1 B best fits the data set 5  1x1, y12, c, 1xn, yn2  6. This is fine, in theory, but it turns out to be inconvenient to work with a sum of absolute value expressions. It is actually much more convenient to work with the

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221

squared distances d 2i . The values of M and B that minimize d1 1 c 1 dn are precisely the same as those that minimize d 21 1 c 1 d 2n. Using calculus, it can be shown that the formulas for M and B are as follows:

M5

n a xy 2 A a xB A a yB n a x 2 2 A a xB 2

,  B 5

ay n

2M

ax n

The resulting line y 5 Mx 1 B is called the best fit least-squares regression line for the data set 5  1x1, y12, c, 1xn, yn2  6. Again, it is tedious to compute these by hand, but their values are produced easily using technology.

EXAMPLE 4  Finding the Line of Best Fit by Linear Regression

Find the line of best fit (best fit least-squares regression line) for the paired data set 1x, y2 from Example 1, where x 5 points on the 400 m and y 5 total score, using (a) the TI-831 (or TI-84) and (b) Excel 2010. Solution (a):

Determining the Best Fit Least-Squares Regression Line Using the TI-831 (or TI-84). INSTRUCTION

SCREENSHOT

STEP 1 Enter the data following the procedure

outlined earlier in this section. The screen should look like the one to the right.

STEP 2 Press STAT, followed by CALC, and

then by 4:LinReg(ax1b). The resulting screen should look like the one to the right. STEP 3 For this example, the data is stored in lists L1

and L2. And, since we will want to graph our best fit line on the scatterplot, it will need to be stored as a function of x, say as Y1. In order to do this, proceed as follows: Directly next to LinReg(ax1b) on the home screen, we need to type the following: L1, L2, Y1. Use the following key strokes: 2nd, 1, , 2nd, 2, , VARS, Y-VARS, 1:FUNCTION, Y1 The resulting screen should look like the one to the right. STEP 4 Press ENTER. The equation of the best

fit least-squares line with the slope (labeled a) and the y-intercept (labeled b) appears on the screen as shown to the right. So, the equation of the best fit least-squares regression line is approximately y 5 2206.9x 1 18317.03.

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CHAPTER 2  Graphs

STEP 5 In order to obtain a graph of the scatterplot with

the best fit line from Step 4 superimposed on it, press ZOOM, then 9:ZoomStat. The resulting screen should look like the one to the right.

Solution (b):

Determining the Best Fit Least-Squares Regression Line Using Excel 2010. INSTRUCTION

SCREENSHOT

STEP 1 Enter the data following the

procedure outlined earlier in this section. The screen should look like the one to the right.

STEP 2 Select the Formulas tab at the top

of the screen and then choose the More Functions within the Function Library grouping (on the left). The pull-down menu should be as shown to the right.

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2.5*  Linear Regression: Best Fit 

INSTRUCTION

223

SCREENSHOT

STEP 3 From here, select Statistical,

and then from this list, scroll down and choose LINEST. A pop-up window should appear, as shown to the right.

STEP 4 Enter B1:B26 in Known_y’s

and A1:A26 in Known_x’s, as shown to the right. You will notice that a set of two values occurs directly beneath the entry boxes—the output is about 52232.03, 19,473.076.

The first value is the slope M, and the second value is the y-intercept B of the best fit line. So, the equation of the best fit least-squares regression line is approximately y 5 2232.03x 1 19,437.07.

STEP 5 In order to obtain a graph of

the scatterplot with the best fit line from Step 4 superimposed on it, construct the scatterplot as before, right-click on the scatterplot near the data points, choose Add Trendline, and press ENTER.

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CHAPTER 2  Graphs

INSTRUCTION

SCREENSHOT

STEP 6 The best fit line will

appear on the scatterplot, along with a pop-up window allowing you to change the format of the line/ curve that is displayed.

▼ YOUR TURN  U  sing the data in Example 1, identify x 5 score on the pole vault and z 5 total score, and ­determine the

best fit least-squares regression line for the data set consisting of the ordered pairs 1x, z2.­Superimpose the graph of this line on the scatterplot.

▼ ANSWER

Total Points

Men’s 2012 Olympic Decathlon: Pole Vault and Total Points with Best Fit Line 9000 8600

It is important to realize that the correlation coefficient is NOT equal, or even related, to the actual slope of the best fit line. In fact, two distinct perfectly linear, ­positively associated scatterplots will both have r 5 1, even though the actual lines that fit the data might have slope M 5 15 and M 5 0.04.

8200 7800

Using the “Best Fit” Line for Prediction

7400 y = 545.58x + 5457.9

7000 4

4.4 4.8 5.2 5.6 Pole Vault

Here, the best fit line displays a positive relationship, and its equation is z 5 545.6x 1 5457.9.

It is important to realize that for any scatterplot, no matter how haphazardly ­dispersed the  data, a best fit least-squares regression line can be created. This is true even when the ­relationship between x and y is nonlinear. However, the utility of such a line in these instances is very limited. In fact, a best fit line should only be created when the linear ­relationship is reasonably strong, which means the correlation coefficient is “reasonably far away from 0.” This criterion can be made more precise using statistical methods, but for our present purposes, we shall make the blanket assumption that it makes sense to form the best fit line in all of the scenarios we present. Once we have the best fit line in hand, we can use it to predict y-values for values of x that do not correspond to any of the points in the data set. For instance, consider the following example.

EXAMPLE 5  Making Predictions Using the Line of Best Fit

Consider the data set from Example 1.    a. Use the best fit line to predict the total score given that an Olympian scored 50.05 points in the 400 m. Is it reasonable to use the best fit line to make such a prediction? b. Use the best fit line to predict the total score given that an Olympian scored 40.05 points in the 400 m. Is it reasonable to use the best fit line to make such a prediction? Solution: a. Using the line y 5 2232.03x 1 19,473.07, we see that y is approximately 7961

when x 5 50.05. This means that if an Olympian were to score 50.05 on the

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2.5*  Linear Regression: Best Fit 

400 m, then his predicted total score would be approximately 7860. Using the best fit line to predict the total score in this case is reasonable because the value 50.05 is well within the range of x-values already present in the data set. b. Using the line y 5 2232.03x 1 19473.07, we see that y is approximately 10,030 when x 5 40.05. This means that if an Olympian were to score 40.05 on the 400 m, then his predicted total score would be approximately 10,180. This prediction is questionable because the x-value at which you are using the best fit line to predict y is far away from the rest of the data points that were used to construct the line. As such, there is no reason to believe that the line is valid for such x-values.

225

▼ ANSWER

Using the line z 5 545.6x 1 5457.9, we see that z is approximately 7995 when x 5 4.65. This means that if an Olympian were to score 4.65 on the pole vault, then his predicted total score would be approximately 7995. Using the best fit line to predict the total score in this case is reasonable because the value 4.65 is well within the range of x-values already present in the data set.   Next, using the same line, we see that z is approximately 8,677 when x 5 5.9. This prediction is less reliable than the former one because 5.9 is outside of the range of x-values corresponding to the rest of the data points used to construct the line. But it isn’t too far outside this range, so there is a degree of validity to the prediction.

▼ Y O U R T U R N   Using the data in Example 1, identify x 5 score on the pole vault

and z 5 total score, and use the best fit line to predict the total score given that an Olympian scored 4.65 points on the pole vault and then, given that an Olympian scored 5.9 points on the pole vault. Comment on the validity of these predictions.

[ S E C T I O N 2 . 5 *]     S U M M A R Y Two variables x and y can be related in different ways. A paired data set 51x1, y12, . . . , 1xn, yn26 obtained experimentally can be illustrated using a scatterplot. Patterns concerning the direction of association and linearity can be used to describe the relationship between x and y, and the strength of the linear relationship can be

measured using the correlation coefficient r. If r is sufficiently far from 0, a best fit least-squares regression line can be formed to precisely describe the linear relationship and used for reasonable prediction purposes.

[ S E C T I O N 2 . 5 *]   E X E R C I S E S • SKILLS In Exercises 1–4, for each of the following scatterplots, identify the pattern as a. having a positive association, negative association, or no identifiable association. b. being linear or nonlinear. 1.

2.

y

3.

y

x

x

4.

y

x

x

In Exercises 5–8, match the following scatterplots with the following correlation coefficients. a. r 5 20.90 b. r 5 0.80 c. r 5 20.68

d. r 5 0.20

5.

8.

6.

y

x

Young_AT_6160_ch02_pp202-235.indd 225

7.

y

x

y

y

x

y

x

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CHAPTER 2  Graphs

For each of the following data sets, a. create a scatterplot. b. guess the value of the correlation coefficient r. c. use technology to determine the equation of the best fit line and to calculate r. d. give a verbal description of the relationship between x and y. 9.

10.

x

y

        14

  28

216

        8

  26

212

x

y

23 21

11.

x

12.

y

x

13.

y

x

y

14.

x

y

1

21

217

23

26

26

21

26

0

21/2

211

22

3

23

4

22

21/4

25

21

1

21

3

21/10

0

0

1

1

0

210

     0

          5

  24

28

  0

         1

        2

  22

24

  8

210

       3

 24

   1

   2

14

211

0

1

1

5

2

26

216

1/10

1

2

21

5

24

1/5

8

4

1

8

1

1

12

        5

   3

210

20

  6

In Exercises 15–18, for each of the data sets, a. use technology to create a scatterplot, to determine the best fit line, and to compute r. b.  indicate whether or not the best fit line can be used for predictive purposes for the following x-values. For those for which it can be used, give the predicted value of y: i.  x 5 0 ii. x 5 26 iii. x 5 12 iv. x 5 215 c. Using the best fit line, at what x-value would you expect y to be equal to 2? 15.

16.

17.

18.

x

y

0

220

15

5

3

218

10

0

6

3

214

3

1.5

7

6

214

8

5

4

7

9

213

3

7

2

8

9

28

0

10

8

8

15

28

23

9

9

25

26

9

15

1

211

10

15

1

215

10

18

x

y

28

5

23

0

22 2

x

y

25

For Exercises 19–22, a. use technology to create a scatterplot, to determine the best fit line, and to compute r for the entire data set. b. repeat (a), but with the data set obtained by removing the starred (***) data points. c. compare the r-values from (a) and (b), as well as the slopes of the best fit lines. Comment on any differences, whether they are substantive, and why this seems reasonable. 19.

20.

x

y

23

14

210

21

8

0

5

1

2

3 *** 5

21.

22.

x

y

215 215 215 213 210 210 210 25 22 22  2  2  4  4  4  7  7 10 10 10

  4 12 16   3   4   8 12   4   3 22   3   6 21   0   4 22   3 24 22   3

x

y

x

y

1

23

14

0

0

26

0

21

8

1

2

0

22

0

5

2

4

8

210

1

2

3

6

24

14

211

*** 3

24

4

8

210

*** 20

216

5

215

6

12

216

*** 7

25

x

y

*** 6

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2.5*  Linear Regression: Best Fit 

227

• CONCEPTUAL 23. Consider the data set from Exercise 17.

26. Consider the following data set.

a. Reverse the roles of x and y so that now y is the explanatory

variable and x is the response variable. Create a scatterplot for the ordered pairs of the form 1 y, x2 using this data set. b. Compute r. How does it compare to the r-value from ­Exercise 17? Why does this make sense? c. The best fit line for the scatterplot in (a) will be of the form x 5 my 1 b. Determine this line. d. Using the line from (c), find the predicted x-value for the following y-values, if appropriate. If it is not appropriate, tell why. i. y 5 23   ii. y 5 2   iii. y 5 216 24. Consider the data set from Exercise 16. Redo the parts in Exercise 23. 25. Consider the following data set. x

y

3

0

3

1

3

21

3

22

3

4

3

15

3

26

3

8

3

10

x

y

25

22

24

22

21

22

0

22

1

22

3

22

8

22

17

22

Guess the values of r and the best fit line. Then, check your answers using technology. What happens? Can you reason why this is the case?

Guess the values of r and the best fit line. Then, check your answers using technology. What happens? Can you reason why this is the case?

• C AT C H T H E M I S TA K E 27. The following screenshot was taken when using the TI-831

28. The following scatterplot was produced using the TI-831 for

paired data 1x, y2.

to determine the equation of the best fit line for paired data 1x, y2:



Using the regression line, we observe that there is a strong positive linear association between x and y, and that for every unit increase in x, the y-value increases by about 1.257 units.

Young_AT_6160_ch02_pp202-235.indd 227



 The equation of the best fit line was reported to be y 5 23.207x 1 0.971 with r2 5 0.9827. Thus, the correlation coefficient is given by r 5 0.9913, which indicates a strong linear ­association between x and y.

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CHAPTER 2  Graphs

• A P P L I C AT I O N S For Exercises 29 and 30, refer to the data set in Example 1. 29. a. Examine the relationship between each of the decathlon events and the total points by computing the correlation coefficient in each case. b. Using the information from part (a), which event has the strongest relationship to the total points? c. What is the equation of the best fit line that describes the relationship between the event from part (b) and the total points? d. Using the best fit line, if you had a score of 40 for this event, what would the predicted total points score be? For Exercises 31 and 32, refer to the following data set: Herd Immunity. According to the U.S. Department of Health and Human Services, herd immunity is defined as “a concept of protecting a community against certain diseases by having a high percentage of the community’s population immunized. Even if a few members of the community are unable to be immunized, the entire community will be indirectly protected because the disease has little opportunity for an outbreak. However, with a low percentage of population ­immunity, the disease would have great opportunity for an outbreak.”* Suppose a study is conducted that looks at the outbreak of ­Haemophilus influenzae type b in the winter of the previous year across 22 nursing homes. We might look at the percentage of residents in each of the nursing homes that were immunized and the percentage of residents who were infected with this type of influenza. The fictional data set is as follows. NURSING HOME

% RESIDENTS IMMUNIZED

% RESIDENTS WITH INFLUENZA

 1

70

11

 2

68

 9

 3

80

 8

 4

10

34

 5

12

30

 6

18

31

 7

27

22

 8

64

18

 9

73

 6

10

 9

31

11

35

19

12

56

16

13

57

22

14

83

10

15

74

13

16

64

15

17

16

28

18

23

25

19

29

24

20

33

20

21

82

28

22

67

 9

*http://archive.hhs.gov/nvpo/glossary.htm

Young_AT_6160_ch02_pp202-235.indd 228

30. a. Using the information from part (a), which event has the





second strongest relationship to the total points? b. What is the equation of the best fit line that describes the relationship between the event in part (b) and the total points? c. Is it reasonable to expect the best fit line from part (c) to produce accurate predictions of total points using this event? d. Using the best fit line, if you had a score of 40 for this event, what would the total points score be?

31. What is the relationship between the variables % residents

immunized and % residents with influenza? a. Create a scatterplot to illustrate the relationship between % residents immunized and % residents with influenza. b. What is the correlation coefficient between % residents immunized and % residents with influenza? c.  Describe the strength of the relationship between % residents immunized and % residents with influenza. d.  What is the equation of the best fit line that describes the relationship between % residents immunized and % residents with influenza? e.  Could you use the best fit line to produce accurate predictions of % residents with influenza using % residents immunized? 32. What is the impact of the outlier(s) on this data set? a. Identify the outlier in this data set. What is the nursing home number for this outlier? b. Remove the outlier and re-create the scatterplot to show the relationship between % residents immunized and % residents with influenza. c.  What is the revised correlation coefficient between % residents immunized and % residents with influenza? d. By removing the outlier, is the strength of the relationship between % residents immunized and % residents with influenza increased or decreased? e.  What is the revised equation of the best fit line that describes the relationship between % residents immunized and % residents with influenza? For Exercises 33–36, refer to the following data set: Amusement Park Rides. According to the International Association of Amusement Parks and Attractions (IAAPA), “There are more than 400 amusement parks and traditional attractions in the United States alone. . . . Amusement parks in the United States entertained 300 million ­visitors who safely enjoyed more than 1.7 billion rides.”** Despite the popularity of amusement parks, the wait times, especially for the most popular rides, are not so highly regarded. There are ­different approaches and tactics that people take to get the most rides in their visit to the park. Now, there are even apps for the iPhone and Android to track waiting times at various amusement parks. One might ask, “Are the wait times worth it? Are the rides with the longest wait times, the most enjoyable?” Consider the following fictional data. **http://www.iaapa.org/pressroom/Amusement_Park_Industry_Statistics.asp

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2.5*  Linear Regression: Best Fit 

AVG AVG RIDE WAIT ENJOYMENT PARK ID RIDE NAME TIME RATING PARK LOCATION

 1

Xoom

45

58

1

Florida

 2

Accentuator

35

40

1

Florida

 3

Wobbler

15

15

1

Florida

 4

Arctic_Attack

75

75

1

Florida

 5

Gusher

60

70

1

Florida

 6

Alley_Cats

 5

60

1

Florida

 7

Moon_Swing

10

15

1

Florida

 8

Speedster

70

50

1

Florida

 9

Hailstorm

80

90

1

Florida

10

DragonFire

70

88

1

Florida

 1

Xoom

50

10

2

California

 2

Accentuator

35

40

2

California

 3

Wobbler

20

75

2

California

 4

Arctic_Attack

70

60

2

California

 5

Gusher

70

80

2

California

 6

Alley_Cats

10

18

2

California

 7

Moon_Swing

15

80

2

California

 8

Speedster

80

35

2

California

 9

Hailstorm

95

40

2

California

10

DragonFire

55

60

2

California

229

The data shows 10 popular rides in two sister parks located in Florida and California. For each ride in each park, average wait times (in minutes) in the summer of 2010 and the average rating of ride enjoyment (on a scale of 1–100) are provided. 33. What is the relationship between the variables average wait times and average rating of enjoyment? a.  Create a scatterplot to show the relationship between average wait times and average rating of enjoyment. b. What is the correlation coefficient between average wait times and average rating of enjoyment? c. Describe the strength of the relationship between average wait times and average rating of enjoyment. d. What is the equation of the best fit line that describes the relationship between average wait times and average ­rating of enjoyment? e.  Could you use the best fit line to produce accurate predictions of average wait times using average rating of enjoyment? 34.  Examine the relationship between average wait times and average rating of enjoyment for Park 1 in Florida by repeating Exercise 33 for only Park 1. 35. Examine the relationship between average wait times and average rating of enjoyment for Park 2 in California by repeating Exercise 33 for only Park 2. 36. Compare the relationship between average wait times and average rating of enjoyment for Park 1 in Florida versus Park 2 in California.

• CHALLENGE For Exercises 37–40, refer to the following: Exploring other types of best fit curves

To describe the patterns that emerge in paired data sets, there are many more possibilities than best fit lines. Indeed, once you have drawn a scatterplot and are ready to identify the curve that best fits the data, there is a substantive collection of other curves that might more accurately describe the data. The following are listed among those in STATS/CALC on the TI-831, along with some comments: NAME OF REGRESSION CURVE

5: QuadReg 6: CubicReg

FORM OF THE CURVE 2

The data set must have at least 3 points to be able to select this option.

y 5 ax 1 bx 1 c 3

2

4

3

COMMENTS

y 5 ax 1 bx 1 cx 1 d 2

The data set must have at least 4 points to be able to select this option.

7: QuartReg

y 5 ax 1 bx 1 cx 1 dx 1 e

The data set must have at least 5 points to be able to select this option.

9: LnReg

y 5 a 1 b ln x

 he data set must have at least 2 points to be able to select this option, and x cannot T take on negative values.

0: ExpReg

y 5 a * bx

 he data set must have at least 2 points to be able to select this option, and y cannot T take on the value of 0.

A: PwrReg

y 5 a * x b

The data set must have at least 2 points to be able to select this option.

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CHAPTER 2  Graphs

For each of the following data sets, a. Create a scatterplot. b. Use LinReg(ax1b) to determine the best fit line and r. Does the line seem to accurately describe the pattern in the data? c. For each of the choices listed in the preceding chart, find the equation of the best fit curve and its associated r 2 value. Of all of the curves, which seems to provide the best fit? Note: The r2-value reported in each case is NOT the linear correlation coefficient reported when running LinReg(ax1B). ­ Rather, the value will typically change depending on the curve. The ­reason is that each time, the r 2-value is measuring how accurate the fit is between the data and that type of curve. A value of r 2 close to 1 still corresponds to a good fit with whichever curve you are fitting to the data. 37.

x

1 2

y

16.2  21

38.

x

y

0.5

1.20

1.0

0.760

3

23.7

1.5

0.412

4

24.8

2.1

0.196

5

23.9

2.9

0.131

6

20.7

3.3

0.071

7

15.8

8

 9.1

9

 0.3

Young_AT_6160_ch02_pp202-235.indd 230

39.

x

  1

y

 0.2

40.

x

y

1

32.3

   1.5

0.93

2

   8.12

 2

1.46

3

216.89

 3

2.25

5

10

4.51

6

   0.89

15

5.50

8

62.1

 245.2

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Chapter Review 

231

[CHAPTER 2 REVIEW] SECTION

CONCEPT

KEY IDEAS/FORMULAS

2.1

Basic tools: Cartesian plane, distance, and midpoint

Two points in the xy-plane: 1x1, y12 and 1x2, y22

Cartesian plane

x-axis, y-axis, origin, and quadrants

Distance between two points

d 5 " 1 x2 2 x1 2 2 1 1 y2 2 y1 2 2

2.2

2.3

Graphing equations: Pointplotting, intercepts, and symmetry

1 xm, ym 2 5 a

x1 1 x2 y1 1 y2 , b 2 2

Point-plotting

 ist a table with several coordinates that are solutions to the equation; L plot and connect.

Intercepts

x-intercept: let y 5 0   y-intercept: let x 5 0

Symmetry

The graph of an equation can be symmetric about the x-axis, y-axis, or origin.

Using intercepts and symmetry as ­graphing aids

If 1a, b2 is on the graph of the equation, then 12a, b2 is on the graph if symmetric about the y-axis, 1a, 2b2 is on the graph if symmetric about the x-axis, and 12a, 2b2 is on the graph if symmetric about the origin.

Lines

General form: Ax 1 By 5 C

Graphing a line

Vertical: x 5 a    Slant: Ax 1 By 5 C Horizontal: y 5 b   where A 2 0 and B 2 0

CHAPTER 2 REVIEW

Midpoint of a line segment joining two points

Intercepts x-intercept 1a, 02  y-intercept 10, b2 Slope m5 Equations of lines

y2 2 y1 x2 2 x1

, where x1 2 x2  

"rise" "run"

Slope–intercept form: y 5 mx 1 b  m is the slope and b is the y-intercept.

Parallel and perpendicular lines

Point–slope form: y 2 y1 5 m  1x 2 x12

L1 L2 if and only if m1 5 m2 (slopes are equal). L 1 ' L 2 if and only if m1 5 2

2.4

Circles Standard equation of a circle Transforming equations of circles to the standard form by completing the square

2.5*

Linear regression: Best fit Scatterplots

1x 2 h22 1 1y 2 k22 5 r 2   C: 1h, k2

General form: x2 1 y2 1 ax 1 by 1 c 5 0

Fitting data with a line Creating a scatterplot: Using Microsoft Excel 

n 

Identifying patterns

Linear regression

n 

Using a graphing calculator

Association

Linearity

n Positive  n Negative

n Linear  n Nonlinear



Correlation coefficient, r

n  n 

Young_AT_6160_ch02_pp202-235.indd 231

1 m1 2 0 (slopes are negative reciprocals). e  m2 m2 2 0

Determine the “best fit” line Making predictions

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CHAPTER 2  Graphs

[CHAPTER 2 REVIEW EXERCISES] 2.1  Basic Tools: Cartesian Plane, Distance,

Use symmetry as a graphing aid and point-plot the given equations.

Plot each point and indicate which quadrant the point lies in.

23. y 5 x2 2 3

1. 124, 22

24. y 5 0 x 0 2 4

and Midpoint

3 25. y 5 "x

2. 14, 72

26. x 5 y2 2 2

3. 121, 262

27. y 5 x"9 2 x 2

4. 12, 212

REVIEW EXERCISES

Calculate the distance between the two points.

5. 122, 02 and 14, 32 6. 11, 42 and 14, 42

7. 124, 262 and 12, 72 1 1

1

7

8. A 4 , 12 B and A 3 , 23 B

Calculate the midpoint of the segment joining the two points. 9. 12, 42 and 13, 82

10. 122, 62 and 15, 72

11. 12.3, 3.42 and 15.4, 7.22 12. 12a, 22 and 1a, 42

Applications

13. Sports. A quarterback drops back to pass. At the point

125, 2202 he throws the ball to his wide receiver located at 110, 302. Find the distance the ball has traveled. Assume the width of the football field is 3215, 15 4 and the length is 3250, 50 4  . Units of measure are yards. 14. Sports. Suppose that in the above exercise a defender was midway between the quarterback and the receiver. At what point was the defender located when the ball was thrown over his head? 2.2  Graphing Equations: Point-Plotting,

Intercepts, and Symmetry

Find the x-intercept(s) and y-intercept(s) if any. 15. x2 1 4y2 5 4 16. y 5 x2 2 x 1 2 17. y 5 "x 2 2 9 2

18. y 5

x 2 x 2 12 x 2 12

Use algebraic tests to determine symmetry with respect to the x-axis, y-axis, or origin. 2

3

19. x 1 y 5 4 2

20. y 5 x 2 2 21. xy 5 4 22. y2 5 5 1 x

Young_AT_6160_ch02_pp202-235.indd 232

28. x2 1 y2 5 36

Applications 29. Sports. A track around a high school football field is in the

shape of the graph 8x2 1 y2 5 8. Graph by using symmetry and plotting points. 30. Transportation. A “bypass” around a town follows the graph y 5 x3 1 2, where the origin is the center of town. Graph the equation. 2.3  Lines

Express the equation for each line in slope–intercept form. Identify the slope and y-intercept of each line. 31. 6x 1 2y 5 12 32. 3x 1 4y 5 9 1

1

1

2

1

1

33. 22 x 2 3 y 5 6 34. 23 x 2 4 y 5 8

Find the x- and y-intercepts and the slope of each line, if they exist, and graph. 35. y 5 4x 2 5 3

36. y 5 2 4 x 2 3 37. x 1 y 5 4 38. x 5 24 39. y 5 2 1

1

40. 2 2 x 2 2 y 5 3

Write the equation of the line, given the slope and the ­intercepts. 41. Slope: m 5 4 y-intercept: 10, 232 42. Slope: m 5 0

y-intercept: 10, 42

43. Slope: m is undefined x-intercept: 123, 02 22

44. Slope: m 5 3 3 y-intercept: A0, 4 B

30/11/16 10:26 AM

Review Exercises 

Write an equation of the line, given the slope and a point that lies on the line.

Find the center and the radius of the circle given by the ­equation.

45. m 5 22  123, 42

63. 1x 1 222 1 1y 1 322 5 81

Write the equation of the line that passes through the given points. Express the equation in slope–intercept form or in the form of x 5 a or y 5 b.

67. x2 1 y2 1 2y 2 4x 1 11 5 0

3 4 

64. 1x 2 422 1 1y 1 222 5 32

12, 162 47. m 5 0  124, 62 48. m is undefined  12, 252 46. m 5

3 2

49. 124, 222 and 12, 32

3 5

55. A2 4 , 2 B

56. 1a 1 2, b 2 12

parallel to the line 2x 2 3y 5 6 perpendicular to the line 5x 2 3y 5 0 perpendicular to the line 23 x 2 12 y 5 12 parallel to the line Ax 1 By 5 C

Applications

57. Grades. For a GRE prep class, a student must take a pretest

and then a posttest after the completion of the course. Two students’ results are shown below. Give a linear equation to represent the given data. PRETEST

POSTTEST

1020

1324

 950

1240

58. Budget: Car Repair. The cost of having the air conditioner

in your car repaired is the combination of material costs and labor costs. The materials (tubing, coolant, etc.) are $250, and the labor costs $38 per hour. Write an equation that models the total cost C of having your air conditioner repaired as a function of hours t. Graph this equation with t as the horizontal axis and C representing the vertical axis. How much will the job cost if the mechanic works 1.5 hours? 2.4  Circles

Write the equation of the circle in standard form. 59. center 122, 32 r 5 6

60. center 126, 282 r 5 3"6

3 5

61. center A 4 , 2 B r 5 25

68. 3x2 1 3y2 2 6x 2 7 5 0 69. 9x2 1 9y2 2 6x 1 12y 2 76 5 0

through 13, 62. 72. Find the equation of a circle that has the diameter with endpoints 122, 212 and 15, 52. Technology Exercises

Section 2.1

Determine whether the triangle with the given vertices is a right triangle, isosceles triangle, neither, or both. 73. 1210, 252, 120, 2452, 110, 102

74. 14.2, 8.42, 124.2, 2.12, 16.3, 210.52

Section 2.2

REVIEW EXERCISES

Find the equation of the line that passes through the given point and also satisfies the additional piece of information.

54. 15, 62

16

71. Find the equation of a circle centered at 12, 72 and passing

7 5

51. A24 , 2 B and A24 , 2 B 52. 13, 222 and 129, 22 53. 122, 212

1 2

65. Ax 1 4 B 1 Ay 2 2 B 5 36 66. x2 1 y2 1 4x 2 2y 5 0

70. x2 1 y2 1 3.2x 2 6.6y 2 2.4 5 0

50. 121, 42 and 122, 52 3 1

233

Graph the equation using a graphing utility and state whether there is any symmetry. 75. y 2 5 0 x 2 2 4 0

76. 0.8x 2 2 1.5y 2 5 4.8

Section 2.3

Determine whether the lines are parallel, perpendicular, or neither, then graph both lines in the same viewing screen using a graphing utility to confirm your answer. 77. y1 5 0.875x 1 1.5 9

8

y2 5 27x 2 14 78. y1 5 20.45x 2 2.1 5

9

y2 5 6 2 20 x

Section 2.4 79. Use the Quadratic Formula to solve for y, and use a graphing

utility to graph each equation. Do the graphs agree with the graph in Exercise 69? 9x 2 1 9y 2 2 6x 1 12y 2 76 5 0 80. Use the Quadratic Formula to solve for y, and use a graphing utility to graph each equation. Do the graphs agree with the graph in Exercise 70? x 2 1 y 2 1 3.2x 2 6.6y 2 2.4 5 0

62. center 11.2, 22.42 r 5 3.6

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234 

CHAPTER 2  Graphs

[CHAPTER 2 PRACTICE TEST] 1. Find the distance between the points 127, 232 and 12, 222. 2. Find the midpoint between 123, 52 and 15, 212.

12. Express the line in slope–intercept form: 4x 2 6y 5 12. 2

1

13. Express the line in slope–intercept form: 3 x 2 4 y 5 2.

3. Determine the length and the midpoint of a segment that

joins the points 122, 42 and 13, 62. 4. Research Triangle. The Research Triangle in North ­Carolina was established as a collaborative research center among Duke University (Durham), North Carolina State ­University (Raleigh), and the University of North Carolina (Chapel Hill).

Creedmoor Hillsborough

Gorman

Find the equation of the line that is characterized by the given information. Graph the line. 14. Slope 5 4; y-intercept 10, 32

15. Passes through the points 123, 22 and 14, 92

16. Parallel to the line y 5 4x 1 3 and passes through the

point 11, 72 17. Perpendicular to the line 2x 2 4y 5 5 and passes through the point 11, 12 18. x-intercept 13, 02; y-intercept 10, 62

For Exercises 19 and 20, write the equation of the line that corresponds to the graph.

Durham

19.

20. y

Chapel Hill

y

New Hope (1, 4)

Cary

(–2, 3)

Raleigh

x

x

PRACTICE TEST

(2, –1) (–2, –2)

 Durham is 10 miles north and 8 miles east of Chapel Hill,

and Raleigh is 28 miles east and 15 miles south of Chapel Hill. What is the perimeter of the research triangle? Round your answer to the nearest mile. 5. Determine the two values for y so that the point 13, y2 is five units away from the point 16, 52. 6. If the point 13, 242 is on a graph that is symmetric with respect to the y-axis, what point must also be on the graph? 7. Determine whether the graph of the equation x 2 y2 5 5 has any symmetry (x-axis, y-axis, and origin). 8. Find the x-intercept(s) and the y-intercept(s), if any: 4x2 2 9y2 5 36.



22. 23. 24.

Graph the following equations. 9. 2x2 1 y2 5 8 10. y 5

4 x 11 2

11. Find the x-intercept and the y-intercept of the line

x 2 3y 5 6.

Young_AT_6160_ch02_pp202-235.indd 234



21. Write the equation of a circle that has center 16, 272 and

25.

radius r 5 8. Determine the center and radius of the circle x2 1 y2 2 10x 1 6y 1 22 5 0. Find the equation of the circle that is centered at 14, 92 and passes through the point 12, 52. Solar System. Earth is approximately 93 million miles from the Sun. Approximating Earth’s orbit around the Sun as ­circular, write an equation governing Earth’s path around the Sun. Locate the Sun at the origin. Determine whether the triangle with the given vertices is a right triangle, isosceles triangle, neither, or both. 1 28.4, 16.8 2 , 1 0, 37.8 2 , 1 12.6, 8.4 2

26. Graph the given equation using a graphing utility and state

whether there is any symmetry. 0.25y 2 1 0.04x 2 5 1

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Cumulative Test 

235

[CHAPTERS 1–2 CUMULATIVE TEST] 1. Simplify

722 . 713

16. Solve using substitution:

2. Simplify and express in terms of positive exponents: 2

2

A 5x 3/4 B 4 25x 21/4

.

1 5 2 1 4 5 0. 1x 1 222 x12

Solve and express the solution in interval notation: 1

3. Perform the operation and simplify: 1x 2 42 1x 1 42 .

17. 6 , 4 x 1 6 , 9 18. x2 2 x $ 20

6. Solve for x: x3 2 5x2 2 4x 1 20 5 0.

20. Solve for x: 0 5 2 4x 0 5 23.

4. Factor completely 8x3 2 27y3.

5. Perform the operations and simplify:

1/x 2 1/5 . 1/x 1 1/5

7. Perform the operations and write in standard form: "236 1 5 2 2i 2 .

Solve for x.

8. 15 2 35 1 3x 2 4  12x 2 62  4

5 4  16x 2 72 2 3 5  13x 2 72 2 6x 1 10 4 5 3 9. 5 4x 1 1 4x 2 1 10. Ashley inherited $17,000. She invested some money in a

11. 12. 13. 14. 15.

CD that earns 5% and the rest in a stock that earns 8%. How much was invested in each account, if the interest for the first year is $1075? Solve by factoring: 5x2 5 45. Solve by completing the square: 3x2 1 6x 5 7. Use the discriminant to determine the number and type of roots: 5x2 1 2x 1 7 5 0. Solve for r: p2 1 q2 5 r2. Solve and check: "x 2 1 3x 2 10 5 x 2 2.

19. 0 2 2 x 0 , 4

21. Use algebraic tests to determine whether the graph of the

equation y 5 4x is symmetric with respect to the x-axis, y-axis, or origin. 22. Write an equation of a line in slope–intercept form with slope m 5 45 that passes through the point 15, 12. 23. Write an equation of a line that is perpendicular to the x-axis and passes through the point 15, 32 . 24. Write an equation of a line in slope–intercept form that passes through the two points A 17, 53 B and A267, 223 B. 25. Find the center and radius of the circle: 1x 1 522 1 1y 1 322 5 30. 26. Calculate the distance between the two points A2 !11, 5B and A2, !7B, and find the midpoint of the segment joining the two points. Round your answers to one decimal place. 27. Determine whether the lines y1 5 0.32x 1 1.5 and 5 y2 5 216 x 1 14 are parallel, perpendicular, or neither, then graph both lines in the same viewing screen using a ­graphing utility to confirm your answer.

CUMULATIVE TEST

Young_AT_6160_ch02_pp202-235.indd 235

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[3[

Functions and Their Graphs On a sales rack of clothes at a department store, you see a shirt you like. The original price of the shirt was $100, but it has been discounted 30%. As a preferred shopper, you get an automatic additional 20% off the sale price at the register. How much will you pay for the shirt? Naïve shoppers might be lured into thinking this shirt will cost $50 because they add the 20% and 30% to get 50% off, but they will end up paying more than that. Experienced shoppers know that they first take 30% off of $100, which results in a price of $70, and then they take an additional 20% off of the sale price, $70, which results in a final discounted price of $56. Experienced shoppers have already learned the concept of composition of functions. A composition of functions can be thought of as a function of a function. One function takes an input (original price, $100) and maps it to an output (sale price, $70), and then another function takes that output as its input (sale price, $70) and maps that to an output (checkout price, $56).

Getty Images

SuperStock/Alamy Stock Photo

CHAPTER

LEARNING OBJECTIVES ■■ Find

the domain and range of a function. ■■ Sketch the graphs of common functions.

Young_AT_6160_ch03_pp236-297.indd 236

■■ Sketch

graphs of general functions employing translations of common functions. ■■ Perform composition of functions.

■■ Find

the inverse of a function. ■■ Model applications with functions using variation.

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[IN THIS CHAPTER] You will find that functions are part of our everyday thinking: converting from degrees Celsius to degrees Fahrenheit, DNA testing in forensic science, determining stock values, and the sale price of a shirt. We will develop a more complete, thorough understanding of functions. First, we will establish what a relation is, and then we will determine whether a relation is a function. We will discuss common functions, domain and range of functions, and graphs of functions. We will determine whether a function is increasing or decreasing on an interval and calculate the average rate of change of a function. We will perform operations on functions and composition of functions. We will discuss one-to-one functions and inverse functions. Finally, we will model applications with functions using variation.

F U N C T I O N S AN D T H E I R G R AP H S 3.1

3.2

3.3

3.4

3.5

3.6

FUNCTIONS

GRAPHS OF FUNCTIONS; PIECEWISEDEFINED FUNCTIONS; INCREASING AND DECREASING FUNCTIONS; AVERAGE RATE OF CHANGE

GRAPHING TECHNIQUES:

OPERATIONS ON FUNCTIONS AND COMPOSITION OF FUNCTIONS

ONE-TO-ONE FUNCTIONS AND INVERSE FUNCTIONS

MODELING FUNCTIONS USING VARIATION

• Recognizing and Classifying Functions • Increasing and Decreasing Functions • Average Rate of Change • PiecewiseDefined Functions

• Horizontal and Vertical Shifts • Reflection about the Axes • Stretching and Compressing

• Adding, Subtracting, Multiplying, and Dividing Functions • Composition of Functions

• Direct Variation • Determine • Inverse Variation Whether a • Joint Variation Function Is and Combined One-to-One Variation • Inverse Functions • Graphical Interpretation of Inverse Functions • Finding the Inverse Function

• Relations and Functions • Functions Defined by Equations • Function Notation • Domain of a Function

TRANSFORMATIONS

237

Young_AT_6160_ch03_pp236-297.indd 237

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238 

CHAPTER 3  Functions and Their Graphs

3.1 FUNCTIONS SKILLS OBJECTIVES ■■ Determine whether a relation is a function. ■■ Determine whether an equation represents a function. ■■ Use function notation to evaluate functions for particular arguments. ■■ Determine the domain and range of a function.

CONCEPTUAL OBJECTIVES ■■ Understand that all functions are relations but not all relations are functions. ■■ Understand why the vertical line test determines if a relation is a function. ■■ Think of function notation as a placeholder or mapping. ■■ Understand the difference between implicit domain and explicit domain.

3.1.1  Relations and Functions 3.1.1 S K I L L

Determine whether a relation is a function. 3.1.1 C O N C E P T U A L

Understand that all functions are relations but not all relations are functions.

What do the following pairs have in common? Every person has a blood type. ■■ Temperature is some specific value at a particular time of day. ■■ Every working household phone in the United States has a 10-digit phone number. ■■ First-class postage rates correspond to the weight of a letter. ■■ Certain times of the day are start times of sporting events at a university. ■■

They all describe a particular correspondence between two groups. A relation is a correspondence between two sets. The first set is called the domain, and the corresponding second set is called the range. Members of these sets are called elements. DEFINITION

Relation

A relation is a correspondence between two sets where each element in the first set, called the domain, corresponds to at least one element in the second set, called the range.

A relation is a set of ordered pairs. The domain is the set of all the first components of the ordered pairs, and the range is the set of all the second components of the ordered pairs. PERSON

Michael

BLOOD TYPE

ORDERED PAIR

A

(Michael, A)

Tania

A

(Tania, A)

Dylan

AB

(Dylan, AB)

Trevor

O

(Trevor, O)

Megan

O

(Megan, O)

WORDS MATH

The domain is the set of all the first components.

5Michael, Tania, Dylan, Trevor, Megan6

The range is the set of all the 5A, AB, O6 second components. A relation in which each element in the domain corresponds to exactly one element in the range is a function.

Young_AT_6160_ch03_pp236-297.indd 238

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3.1 Functions 

DEFINITION

Function

Domain

A function is a correspondence between two sets where each element in the first set, called the domain, corresponds to exactly one element in the second set, called the range.

7:00 p.m.

MATH

The 1:00 start time corresponds to exactly one event, Football. The 2:00 start time corresponds to exactly one event, Volleyball. The 7:00 start time corresponds to two events, Soccer and Basketball.

(1:00 p.m., Football) (2:00 p.m., Volleyball) (7:00 p.m., Soccer) (7:00 p.m., Basketball)

Because an element in the domain, 7:00 p.m., corresponds to more than one element in the range, Soccer and Basketball, this is not a function. It is, however, a relation. EXAMPLE 1   Determining Whether a Relation Is a Function

Determine whether the following relations are functions. a. 5 123, 42 ,  12, 42 ,  13, 52 ,  16, 42 6 b. 5 123, 42 ,  12, 42 ,  13, 52 ,  12, 22 6 c. Domain 5 Set of all items for sale in a grocery store; Range 5 Price

BLOOD TYPE

Michael Megan Dylan Trevor Tania

A AB O B

Domain

Basketball

WORDS

Range

Function

PEOPLE

Note that the definition of a function is more restrictive than the definition of a relation. For a relation, each input corresponds to at least one output, whereas, for a function, each input corresponds to exactly one output. The blood-type example given is both a relation and a function. Also note that the range (set of values to which the elements of the domain correspond) is a subset of the set of all blood types. However, although all functions are relations, not all relations are functions. For example, at a university, four priTIME OF DAY COMPETITION mary sports typ­ically overlap in the late fall: 1:00 p.m. Football football, volleyball, soccer, and basketball. 2:00 p.m. Volleyball On a given Saturday, the accompanying table indicates the start times for the competitions. 7:00 p.m. Soccer

239

Range

Not a Function

START TIME

ATHLETIC EVENT

1:00 P.M. 2:00 P.M.

Football Volleyball Soccer Basketball

7:00 P.M.

[

STUD Y T I P

[

All functions are relations but not all relations are functions.

[ CONCEPT CHECK ] If the domain consists of all physical (home) addresses in a particular county and the range is the persons living in that county, does this describe a relation? And if so, is that relation a function?

▼ ANSWER This is a relation but not a function.

Solution:

a. No x-value is repeated. Therefore, each x-value corresponds to exactly one

y-value. This relation is a function. b. The value x 5 2 corresponds to both y 5 2 and y 5 4. This relation is not a function. c. Each item in the grocery store corresponds to exactly one price. This relation is a function.

▼ Y O U R T U R N   Determine whether the following relations are functions. a. 5 11, 22 ,  13, 22 ,  15, 62 ,  17, 62 6

b. 5 11, 22 ,  11, 32 ,  15, 62 ,  17, 82 6

c. 5 111:00 a.m., 838F2 ,  12:00 p.m., 898F2 ,  16:00 p.m., 858F2 6

Young_AT_6160_ch03_pp236-297.indd 239

▼ ANSWER a. function b. not a function c. function

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240 

CHAPTER 3  Functions and Their Graphs

All of the examples we have discussed thus far are discrete sets in that they represent a countable set of distinct pairs of 1x, y2. A function can also be defined algebraically by an equation.

3.1.2  Functions Defined by Equations 3.1.2 S K I L L

Determine whether an equation represents a function.

Let’s start with the equation y 5 x2 2 3x, where x can be any real number. This equation assigns to each x-value exactly one corresponding y-value.

3.1.2 C O N C E P T U A L

y 5 x 2 2 3x

x

Understand why the vertical line test determines if a relation is a function.

y 5  112 2 2 3 112

1

ST U DY TIP 2

We say that for x 5 y  , y is not a function of x. However, if we reverse the independent and dependent variables, then for x 5 y 2, x is a function of y.

10 22 9

2

y 5 A223 B 2 3A223 B

y 5  11.22 2 2 3 11.22

1.2

Not all equations are functions.

22

y 5  152 2 2 3 152

5 2 23

▼ CAUTION

y

22.16

Since the variable y depends on what value of x is selected, we denote y as the dependent variable. The variable x can be any number in the domain; therefore, we denote x as the independent variable. Although functions are defined by equations, it is important to recognize that not all equations are functions. The requirement for an equation to define a function is that each element in the domain corresponds to exactly one element in the range. Throughout the ensuing discussion, we assume x to be the independent variable and y to be the dependent variable. Equations that represent functions of x:

y 5 x2

Equations that do not represent functions of x:

x 5 y2

y 5 0 x0

y 5 x3

x2 1 y2 5 1 x 5 0 y 0

In the “equations that represent functions of x,” every x-value corresponds to exactly one y-value. Some ordered pairs that correspond to these functions are

y 5 x2:



y 5 0 x 0 :



3

y 5 x :

121, 12 10, 02 11, 12

121, 12 10, 02 11, 12

121, 212 10, 02 11, 12

The fact that x 5 21 and x 5 1 both correspond to y 5 1 in the first two examples does not violate the definition of a function. In the “equations that do not represent functions of x,” some x-values correspond to more than one y-value. Some ordered pairs that correspond to these equations are

RELATION

SOLVE RELATION FOR y

x 5 y2 x2 1 y2 5 1 x5 0 y0

Young_AT_6160_ch03_pp236-297.indd 240

POINTS THAT LIE ON THE GRAPH

y 5 6 !x

 11, 212  10, 02  11, 12

x 5 1 maps to both y 5 21 and y 5 1

y 5 6x

 11, 212  10, 02  11, 12

x 5 1 maps to both y 5 21 and y 5 1

y 5 6"1 2 x 2

 10, 212  10, 12  121, 02  11, 02

x 5 0 maps to both y 5 21 and y 5 1

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3.1 Functions 

241

Let’s look at the graphs of the three functions of x: y

y

y

x

x

x

y = |x|

y = x2

y = x3

Let’s take any value for x, say x 5 a. The graph of x 5 a corresponds to a vertical line. A function of x maps each x-value to exactly one y-value; therefore, there should be at most one point of intersection with any vertical line. We see in the three graphs of the functions above that if a vertical line is drawn at any value of x on any of the three graphs, the vertical line only intersects the graph in one place. Look at the graphs of the three equations that do not represent functions of x. y

y

y

[ CONCEPT CHECK ] Draw two vertical lines (x 5 0 and x 5 1) on the graph of x 5 y  2. What three points do these lines intersect on the graph of the equation? Explain why the equation x 5 y  2 cannot be a function.

▼ ANSWER (0, 0) (1, 21) and (1, 1): the vertical line x 5 1 intersects the graph at two places (fails the vertical line test).

x

x

x = y2

x2 + y2 = 1

x

x = |y|

A vertical line can be drawn on any of the three graphs such that the vertical line will intersect each of these graphs at two points. Thus, there are two y-values that correspond to some x-value in the domain, which is why these equations do not define y as a function of x. DEFINITION

Vertical Line Test

Given the graph of an equation, if any vertical line that can be drawn intersects the graph at no more than one point, the equation defines y as a function of x. This test is called the vertical line test. EXAMPLE 2   Using the Vertical Line Test

Use the vertical line test to determine whether the graphs of equations define functions of x. y y a. b.

x

Young_AT_6160_ch03_pp236-297.indd 241



x

09/12/16 2:48 PM

242 

CHAPTER 3  Functions and Their Graphs

Solution:

Apply the vertical line test. y a.



b.

y

x

x

[

S TU DY TIP If any x-value corresponds to more than one y-value, then y is not a function of x.

[

a. Because the vertical line intersects the graph of the equation at two points, this

equation does not represent a function. b. Because any vertical line will intersect the graph of this equation at no more than one point, this equation represents a function. ▼ ANSWER

The graph of the equation is a circle, which does not pass the vertical line test. Therefore, the equation does not define a function.

▼ Y O U R T U R N   Determine whether the equation  1x 2 32 2 1  1y 1 22 2 5 16 is a

function of x.

To recap, a function can be expressed one of four ways: verbally, numerically, algebraically, and graphically. This is sometimes called the Rule of 4. EXPRESSING A FUNCTION VERBALLY

Every real number has a corresponding absolute value.

NUMERICALLY

ALGEBRAICALLY

5 123, 32 ,  121, 12 ,  10, 02 ,  11, 12 ,  15, 52 6

y 5 0x0

GRAPHICALLY y

x

3.1.3  Function Notation 3.1.3 S K I L L

Use function notation to ­evaluate functions for particular arguments.

We know that the equation y 5 2x 1 5 defines y as a function of x because its graph is a nonvertical line and thus passes the vertical line test. We can select x-values (input) and determine unique corresponding y-values (output). The output is found by taking 2 times the input and then adding 5. If we give the function a name, say, “ƒ,” then we can use function notation: ƒ 1 x 2 5 2x 1 5

3.1.3 C O N C E P T U A L

Think of function notation as a placeholder or mapping.

The symbol ƒ1x2 is read “ƒ evaluated at x” or “ƒ of x” and represents the y-value that corresponds to a particular x-value. In other words, y 5 ƒ1x2. INPUT

x Independent variable

FUNCTION

OUTPUT

EQUATION

ƒ

ƒ1x2

ƒ1x2 5 2x 1 5

Mapping

Dependent variable

Mathematical rule

It is important to note that ƒ is the function name, whereas ƒ1x2 is the value of the ­function. In other words, the function ƒ maps some value x in the domain to some value ƒ1x2 in the range.

Young_AT_6160_ch03_pp236-297.indd 242

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243

3.1 Functions 

x

f (x) 5 2x 1 5

f (x)

Domain

0

ƒ102 5 2 102 1 5

ƒ102 5 5

x

ƒ122 5 2 122 1 5

ƒ122 5 9

ƒ112 5 2 112 1 5

1 2

ƒ112 5 7

Function

Range 2x  5

x0 x1 x2

f f f

f (0)  5 f (1)  7 f (2)  9

The independent variable is also referred to as the argument of a function. To e­ valuate functions, it is often useful to think of the independent variable, or argument, as a placeholder. For example, ƒ1x2 5 x2 2 3x can be thought of as ƒ1 2 5 1 2 2 2 31 2



In other words, “ƒ of the argument is equal to the argument squared minus 3 times the argument.” Any expression can be substituted for the argument:  ƒ 1 1 2 5 1 1 2 2 2 3 1 1 2

It is important to note:

 ƒ1x 1 12 5 1x 1 122 2 31x 1 12 ƒ 12x 2 5 12x 2 2 2 3 12x 2

[

STUD Y T I P

ƒ1x2 does not mean ƒ times x. ■■ The most common function names are ƒ and F since the word function begins with an “f.” Other common function names are g and G, but any letter can be used. ■■ The letter most commonly used for the independent variable is x. The letter t is also common because in real-world applications it represents time, but any letter can be used. ■■ Although we can think of y and ƒ1x2 as interchangeable, the function notation is useful when we want to consider two or more functions of the same independent variable. ■■

EXAMPLE 3 

It is important to note that ƒ(x) does not mean ƒ times x.

[

[ CONCEPT CHECK ] For the function in Example 3, find f A B.

▼

ANSWER f

Evaluating Functions by Substitution

A

B52

3

23

2

16

Given the function ƒ1x2 5 2x3 2 3x2 1 6, find ƒ1212. Solution:

Consider the independent variable x to be a placeholder. ƒ1 2 5 2 1 2 3 2 3 1 2 2 1 6 To find ƒ1212 , substitute x 5 21 into the function. ƒ1212 5 2 1212 3 2 3 1212 2 1 6 Evaluate the right side. ƒ1212 5 22 2 3 1 6 Simplify. ƒ1212 5 1

EXAMPLE 4   Finding Function Values from the Graph of a Function

The graph of ƒ is given on the right. a. Find ƒ102. b. Find ƒ112. c. Find ƒ122. d. Find 4ƒ132. e. Find x such that ƒ1x2 5 10. f. Find x such that ƒ1x2 5 2.

Young_AT_6160_ch03_pp236-297.indd 243

y 10

(5, 10)

(4, 5)

(0, 5) (1, 2)

(3, 2)

x

(2, 1) 5

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Solution: Solution (a):

The value x 5 0 corresponds to the value y 5 5. ƒ102 5 5 Solution (b): The value x 5 1 corresponds to the value y 5 2. ƒ112 5 2 Solution (c): The value x 5 2 corresponds to the value y 5 1. ƒ122 5 1 Solution (d): The value x 5 3 corresponds to the value y 5 2.  4ƒ132 5 4 ⋅ 2 5 8 Solution (e): The value y 5 10 corresponds to the value x 5 5 . Solution (f): The value y 5 2 corresponds to the values x 5 1 and x 5 3 . ▼ ANSWER a. ƒ1212 5 2 b. ƒ102 5 1 c. 3ƒ122 5 221

▼ Y O U R T U R N   For the following graph of a function, find: a. ƒ1212  b. ƒ102  c. 3ƒ122   d. the value of x that corresponds to ƒ1x2 5 0

d. x 5 1

y 10

(–2, 9)

(0, 1)

x

(–1, 2) (1, 0)

–5

5

(2, –7) –10



EXAMPLE 5   Evaluating Functions with Variable Arguments (Inputs)

For the given function ƒ1x2 5 x2 2 3x, evaluate ƒ1x 1 12 and simplify if possible.

common mistake A common misunderstanding is to interpret the notation ƒ1x 1 12 as a sum: ƒ1x 1 12 2 ƒ1x2 1 ƒ112. ✓C O R R E C T

✖INCORRECT

Write the original function. ƒ 1 x 2 5 x 2 2 3x

Replace the argument x with a placeholder. 2

ƒ1 2 5 1 2 2 31 2

The ERROR is in interpreting the notation as a sum. ƒ1x 1 12 2 ƒ1x2 1 ƒ112

2 x 2 2 3x 2 2

Substitute x 1 1 for the argument. ƒ1x 1 12 5 1x 1 122 2 31x 1 12

Eliminate the parentheses. ƒ 1 x 1 1 2 5 x 2 1 2x 1 1 2 3x 2 3

▼ CAUTION

Combine like terms. ƒ 1 x 1 1 2 5 x2 2 x 2 2

ƒ1x 1 12 2 ƒ1x2 1 ƒ112 ▼ ANSWER

g 1 x 2 1 2 5 x 2 2 4x 1 6

Young_AT_6160_ch03_pp236-297.indd 244

▼ Y O U R T U R N   For the given function g 1x2 5 x2 2 2x 1 3, evaluate g 1x 2 12.

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3.1 Functions 

245

EXAMPLE 6   Evaluating Functions: Sums

For the given function H 1x2 5 x2 1 2x, evaluate: a. H 1x 1 12  b. H 1 x2 1 H 112 Solution (a):

Write the function H in placeholder notation. Substitute x 1 1 for the argument of H. Eliminate the parentheses on the right side. Combine like terms on the right side. Solution (b):

H 1 2 5  1 22 1 2 1 2 H 1 x 1 12 5  1 x 1 122 1 2 1 x 1 12 H 1 x 1 12 5 x2 1 2x 1 1 + 2x 1 2 H 1 x 1 12 5 x2 1 4x 1 3

Write H 1 x2. H 1x2 5 x2 1 2x Evaluate H at x 5 1. H 112 5 112 2 1 2 112 5 3 Evaluate the sum H 1 x2 1 H 112. H 1x2 1 H 112 5 x2 1 2x 1 3 H 1x2 1 H 112 5 x 2 1 2x 1 3 Note: Comparing the results of part (a) and part (b), we see that H 1x 1 12 u H 1x2 1 H 112 .

EXAMPLE 7   Evaluating Functions: Negatives

For the given function G 1 t2 5 t 2 2 t, evaluate: a. G 12t2  b. 2G 1 t2 Solution (a):

Write the function G in placeholder notation.  G 1 2 5  1 22 2 1 2 Substitute 2t for the argument of G.   G 12t2 5  12t22 2  12t2 Eliminate the parentheses on the right side.   G 12t2 5  t 2 1  t Solution (b):

Write G 1 t2.   G 1 t2 5 t2 2 t Multiply by 21.   2G 1 t2 5 2 1 t2 2 t2 2G 1 t2 5 2t 2 1 t Eliminate the parentheses on the right side.   Note: Comparing the results of part (a) and part (b), we see that G 12t 2 u 2G 1 t 2 .

EXAMPLE 8   Evaluating Functions: Quotients

For the given function F 1 x2 5 3x 1 5, evaluate: 1 2

a. F a b   b.

Solution (a):

F 112 F 122

Write F in placeholder notation. Replace the argument with 12. Simplify the right side.

Young_AT_6160_ch03_pp236-297.indd 245

F 1 2 5 3 1 2 1 5

1 1 F a b 5 3a b 1 5 2 2 1 13 Fa b 5 2 2

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CHAPTER 3  Functions and Their Graphs

▼ CAUTION

ƒ1a2 a ƒa b 2 b ƒ1b2

▼ ANSWER a. G 1 t 2 22 5 3t 210

b. G 1 t2 2 G 1 22 5 3t 2 6 c.

G112 1 52 5 G132

Solution (b):

Evaluate F 112. F 112 5 3 112 1 5 5 8 Evaluate F 1 22. F 1 22 5 3 1 22 1 5 5 11 F 112 8 5 11 F 122

Divide F 112 by F 1 22.

F112 1 Note: Comparing the results of part (a) and part (b), we see that F a b u . 2 F122

▼

Y O U R T U R N   Given the function G 1 t2 5 3t 2 4, evaluate: a. G 1 t 2 22  b. G 1 t2 2 G 1 22  c.

G112 1   d. G a b 3 G132

Examples 6, 7, and 8 illustrate the following:

1 3

d. G a b 5 23

ƒ1a 1 b2 2 ƒ1a2 1 ƒ1b2

ƒ1a2 a ƒ 1 2t 2 2 2ƒ 1 t 2     ƒa b 2 b ƒ1b2

Now that we have shown that ƒ 1 x 1 h 2 2 ƒ 1 x 2 1 ƒ 1 h 2 , we turn our attention to one of the fundamental expressions in calculus: the difference quotient. ƒ1x 1 h2 2 ƒ1x2 h20 h Example 9 illustrates the difference quotient, which will be discussed in detail in Section 3.2. For now, we will concentrate on the algebra involved when finding the difference ­quotient. In Section 3.2, the application of the difference quotient will be the emphasis. EXAMPLE 9   Evaluating the Difference Quotient

For the function ƒ 1 x 2 5 x 2 2 x, find Solution:

ƒ1x 1 h2 2 ƒ1x2 , h 2 0. h

e

f

Use placeholder notation for the function ƒ1x2 5 x2 2 x. ƒ1 2 5  1 22 2  1 2 Calculate ƒ1x 1 h2. ƒ1x 1 h2 5  1 x 1 h22 2  1 x 1 h2 ƒ1x 1 h2 2 ƒ1x2 Write the difference quotient. h 2 2 Let ƒ1x 1 h2 5  1 x 1 h2 2  1 x 1 h2 and ƒ1x2 5 x 2 x. ƒ1x 1 h2 ƒ1x2

C 1 x 1 h 2 2 2 1 x 1 h 2 D 2 Cx2 2 xD ƒ1x 1 h2 2 ƒ1x2 5     h 2 0­­­­­ h h 3 x 2 1 2xh 1 h2 2 x 2 h 4 2 3 x 2 2 x 4 Eliminate the parentheses inside the 5 h first set of brackets. x 2 1 2xh 1 h2 2 x 2 h 2 x 2 1 x Eliminate the brackets in the numerator. 5 h 2 2xh 1 h 2 h Combine like terms. 5 h 1 h 2x 1 h 2 1 2 Factor the numerator. 5 h Divide out the common factor, h. 5 2x 1 h 2 1     h 2 0

▼ ANSWER

▼

Y O U R T U R N   Evaluate the difference quotient for ƒ1x2 5 x2 2 1.

2x 1 h

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3.1 Functions 

3.1.4  Domain of a Function

3.1.4 S K I L L

Sometimes the domain of a function is stated explicitly. For example, ƒ1x2 5 0 x 0

e

x,0

Determine the domain and range of a function.

domain

Here the explicit domain is the set of all negative real numbers 1 2q, 0 2 . Every ­negative real number in the domain is mapped to a positive real number in the range through the absolute value function. If the expression that defines the function is given but the domain is not stated explicitly, then the domain is implied. The implicit domain is the largest set of real numbers for which the function is defined and the output value ƒ1x2 is a real number. For example,

3.1.4 C O N C E P T U A L

Understand the difference between implicit domain and explicit domain. Domain (, 0)

ƒ 1 x 2 5 !x

Range (0, )

1 7 4

does not have the domain explicitly stated. There is, however, an implicit domain. Note that if the argument is negative, that is, if x , 0, then the result is an imaginary number. In order for the output of the function, ƒ1x2 , to be a real number, we must restrict the domain to ­nonnegative numbers, that is, if x $ 0. FUNCTION

f(x)  |x|

1 7 4

IMPLICIT DOMAIN

3 0, q 2

ƒ 1 x 2 5 !x

In general, we ask the question, “what can x be?” The implicit domain of a function excludes values that cause a function to be undefined or have outputs that are not real numbers.

EXPRESSION THAT DEFINES THE FUNCTION

[ CONCEPT CHECK ] EXCLUDED x-VALUES

Polynomial

None

Rational

x-values that make the denominator equal to 0

Radical

x-values that result in a square (even) root of a negative number

EXAMPLE

IMPLICIT DOMAIN

ƒ1x2 5 x3 2 4x2 g1x2 5

All real numbers x 263 or 1 2q, 23 2 ∪ 1 23, 3 2 ∪ 1 3, q 2

2 x2 2 9

Find the implicit domain for 1 f 1x2 5 !x 2 a

▼

ANSWER (a, q )

x $ 5 or 3 5, q 2

h 1 x 2 5 !x 2 5

EXAMPLE 10   Determining the Domain of a Function

State the domain of the given functions. a. F 1 x 2 5

3 4 3   b. H 1 x 2 5 ! 9 2 2x  c. G 1 x 2 5 ! x21 x 2 25 2



Solution (a):

Write the original equation. Determine any restrictions on the values of x. Solve the restriction equation. State the domain restrictions. Write the domain in interval notation.

Young_AT_6160_ch03_pp236-297.indd 247

F 1x2 5

3 x 2 25 2

x 2 2 25 2 0 x 2 2 25 or x 2 6 !25 5 65 x 2 65 12q, 25 2 ∪ 1 25, 5 2 ∪ 1 5, q 2

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CHAPTER 3  Functions and Their Graphs

Solution (b): 4 Write the original equation. H 1x2 5 ! 9 2 2x Determine any restrictions on the values of x. 9 2 2x $ 0 Solve the restriction inequality. 9 $ 2x

State the domain restrictions.

x#

Write the domain in interval notation. Solution (c):

▼ ANSWER a. x $ 3 or 3 3, q 2 b. x 2 62 or

1 2q, 22 2 ∪ 1 22, 2 2 ∪ 1 2, q 2

9 2

9 a2q, d 2 3 G1x2 5 ! x21

Write the original equation. Determine any restrictions on the values of x.

no restrictions

State the domain.

ℝ

Write the domain in interval notation.

12q, q 2

▼

Y O U R T U R N   State the domain of the given functions. a. ƒ 1 x 2 5 !x 2 3  b. g 1 x 2 5

1 x 24 2

Applications Functions that are used in applications often have restrictions on the domains due to physical constraints. For example, the volume of a cube is given by the function V1x2 5 x3, where x is the length of a side. The function ƒ1x2 5 x3 has no restrictions on x, and therefore the domain is the set of all real numbers. However, the volume of any cube has the restriction that the length of a side can never be negative or zero. EXAMPLE 11   Price of Gasoline

Following the capture of Saddam Hussein in Iraq in 2003, gas prices in the United States escalated and then finally returned to their precapture prices. Over a 6-month period, the average price of a gallon of 87 octane gasoline was given by the function C 1 x2 5 20.05x2 1 0.3x 1 1.7, where C is the cost function and x represents the number of months after the capture. a. Determine the domain of the cost function. b. What was the average price of gas per gallon 3 months after the capture? Solution (a):

Since the cost function C 1 x2 5 20.05x2 1 0.3x 1 1.7 modeled the price of gas only for 6 months after the capture, the domain is 0 # x # 6 or 30, 64 . Solution (b):

Write the cost function.  C 1x2 5 20.05x2 1 0.3x 1 1.7   0 # x # 6 Find the value of the function when x 5 3.  C 132 5 20.05 1322 1 0.3 132 1 1.7 Simplify.

C 1 3 2 5 2.15

The average price per gallon 3 months after the capture was $2.15 .

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3.1 Functions 

249

EXAMPLE 12   The Dimensions of a Pool

Express the volume of a 30 ft 3 10 ft rectangular swimming pool as a function of its depth. Solution:

The volume of any rectangular box is V 5 lwh, where V is the volume, l is the length, w is the width, and h is the height. In this example, the length is 30 ft, the width is 10 ft, and the height represents the depth d of the pool. Write the volume as a function of depth d. Simplify. Determine any restrictions on the domain.

V 1 d 2 5 1 30 2 1 10 2 d V 1 d 2 5 300d

d.0

[ S E C T I O N 3 .1]     S U M M A R Y Relations and Functions (Let x represent the independent variable and y the dependent variable.) TYPE

MAPPING/CORRESPONDENCE

Relation

Function

EQUATION

GRAPH

2

Every x-value in the domain maps to at least one y-value in the range.

x5y

Every x-value in the domain maps to exactly one y-value in the range.

y 5 x2

y x

y

x

Passes vertical line test

All functions are relations, but not all relations are functions. Functions can be represented by equations. In the following table, each column illustrates an alternative notation. INPUT

CORRESPONDENCE

OUTPUT

EQUATION

x

Function

y

y 5 2x 1 5

Independent Variable

Mapping

Dependent Variable

Mathematical Rule

Argument

ƒ

ƒ1x2

ƒ1x2 5 2x 1 5

The domain is the set of all inputs (x-values), and the range is the set of all corresponding outputs (y-values). Placeholder notation is useful when evaluating functions. ƒ 1 x 2 5 3x 2 1 2x

ƒ1 2 5 3 1 22 1 2 1 2

Explicit domain is stated, whereas implicit domain is found by excluding x-values that

• make the function undefined (denominator 5 0). • result in a nonreal output (even roots of negative real numbers).

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[ S E C T I O N 3 .1]   E X E R C I S E S • SKILLS In Exercises 1–24, determine whether each relation is a function. Assume that the coordinate pair (x, y) represents the independent variable x and the dependent variable y. 1.

Range

MONTH

AVERAGE TEMPERATURE

October January April

4.

2.

Domain

5.

Range

PERSON

DATE THIS WEEKEND Chris Alex Morgan

Jordan Pat

Range

PERSON Mary Jason Chester

78°F 68°F

Domain

Domain

3.

Domain

Range

10-DIGIT PHONE #

START TIME

NFL GAME

(202) 555–1212 (307) 123–4567 (878) 799–6504

1:00 P.M. 4:00 P.M. 7:00 P.M.

• Bucs/Panthers • Bears/Lions • Falcons/Saints • Rams/Seahawks • Packers/Vikings

Domain

Range

PERSON Carrie Michael Jennifer Sean

COURSE GRADE

6.

A B

7. 5 1 0, 232,  1 0, 32,  123, 02,  13, 026

Domain

Range

MATH SAT SCORE 500 650

PERSON Carrie Michael Jennifer Sean

8. 5 1 2, 222,  1 2, 22,  1 5, 252,  1 5, 526

9. 5 1 0, 02,  1 9, 232,  14, 222,  14, 22,  1 9, 326

10. 5 1 0, 02,  121, 212,  122, 282,  11, 12,  1 2, 826

15. x 5 y2

16. y 5 x3

12. 5 1 0, 12,  11, 12,  1 2, 12,  13, 126

11. 5 1 0, 12,  11, 02,  1 2, 12,  122, 12,  1 5, 42,  123, 426 13. x2 1 y2 5 9 17. y 5 0 x 2 1 0

14. x 5 0 y 0 18. y 5 3



19.

(0, 5)

y

20.

x

(–5, 0)

21.

y

y

x

x

(5, 0)

(0, –5)

22.

y

23.

x

Young_AT_6160_ch03_pp236-297.indd 250

24.

y

x

y

x

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3.1 Functions 

In Exercises 25–32, use the given graphs to evaluate the functions. 25. y 5 ƒ1x2

26. y 5 g1x2

27. y 5 p1x2

5

y

5

(2, 5)

(–3, 1)

x

x

(5, 0) –5

5

(–1, –1)

10

(–3, 5)

(0, 1)

y

7

(–5, 7)

(1, 3)

–5

28. y 5 r 1x2

y

y

(1, 5)

(–1, 3)

5 (3, –2) –7

–5

–5

  a. ƒ122  b. ƒ102  c. ƒ1222

    a. g 1232  b. g 102  c. g 152

y –5

(–4, –5)

(7, –3) (3, –5) –10

    a. r  1242  b. r  1212  c. r  132

31. y 5 S1x2

32. y 5 T1x2

y

4

5

10

(–6, –6)

    a. p 1212  b. p 102  c. p 112

y

x

–10

–3

30. y 5 q1x2

29. y 5 C1x2

x 3

(0, –5)

x

(–4, 0) (0, 2)

(–2, –3)

(–1, 4)

y

2

(6, 0) 6

–4

x

(6, 1)

–4

x

5

(–3, 4)

(4, 3)

6

(5, 1) –5

(4, –5)

x 5

(2, –5) (2, –4) –6

–10

   a. C122  b. C102  c. C1222

–8

   a. q1242  b. q102  c. q122

–5

    a. S1232  b. S102  c. S122

   a. T 1252  b. T 1222  c. T 142

34. Find x if g  1x2 5 22 in Exercise 26.

33. Find x if ƒ1x2 5 3 in Exercise 25.

36. Find x if C 1x2 5 27 in Exercise 29.

35. Find x if p1x2 5 5 in Exercise 27.

38. Find x if q 1x2 5 22 in Exercise 30.

37. Find x if C1x2 5 25 in Exercise 29.

40. Find x if T 1x2 5 4 in Exercise 32.

39. Find x if S1x2 5 1 in Exercise 31.

In Exercises 41–56, evaluate the given quantities applying the following four functions. ƒ1x2 5 2x 2 3 F 1t2 5 4 2 t 2 g1t2 5 5 1 t

G1x2 5 x2 1 2x 2 7

45. ƒ1222 1 g 112



41. ƒ1222

49.

ƒ 1 22 2 g112

53. ƒ1x 1 12 2 ƒ1x 2 12

42.  G 1232

43.  g 112

44.  F 1212

46.  G 1232 2 F 1212

47.  3ƒ1222 2 2g 112

48.  2F 1212 2 2G 1232

54.  F 1t 1 12 2 F 1t 2 12

55.  g 1x 1 a2 2 ƒ1x 1 a2

56.  G 1x 1 b2 1 F 1b2

50. 

G 1 23 2 F 1 21 2

51. 

ƒ 1 0 2 2 ƒ 1 22 2 g112

52. 

G 1 0 2 2 G 1 23 2 F 1 21 2

In Exercises 57–64, evaluate the difference quotients using the same ƒ, F, G, and g given for Exercises 41–56. 57.

ƒ1x 1 h2 2 ƒ1x2 h

58. 

F1t 1 h2 2 F1t2 h

59. 

g1t 1 h2 2 g1t2 h

60. 

G 1x 1 h2 2 G 1x2 h

61.

ƒ 1 22 1 h 2 2 ƒ 1 22 2 h

62. 

F 1 21 1 h 2 2 F 1 21 2 h

63. 

g11 1 h2 2 g112 h

64. 

G 1 23 1 h 2 2 G 1 23 2 h

In Exercises 65–96, find the domain of the given function. Express the domain in interval notation. 65. ƒ1x2 5 2x 2 5 69. P 1 x 2 5 73. F 1 x 2 5

x15 x25 1 x 11 2

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66.  ƒ1x2 5 22x 2 5 70.  Q 1 t 2 5 74.  G 1 t 2 5

2 2 t2 t13 2 t 14 2

67.  g(t) 5 t2 1 3t 71.  T 1 x 2 5

2 x 24 2

75.  q 1 x 2 5 !7 2 x

68. h(x) 5 3x4 2 1 72. R 1 x 2 5

1 x 21 2

76. k 1 t 2 5 !t 2 7

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77. ƒ 1 x 2 5 "2x 1 5 81. F 1 x 2 5 85. P 1 x 2 5 89. H 1 t 2 5

1 !x 2 3 1

82. G 1 x 2 5 86. Q 1 x 2 5

5 " x14 t 2

"t 2 t 2 6

2

78. g 1 x 2 5 "5 2 2x

93. r 1x2 5 x  13 2 2x2

21/2

90. ƒ 1 t 2 5

2 !5 2 x x

3 2 " x 29

t23

4 2

"t 1 9

79. G 1 t 2 5 "t 2 2 4

80. F 1 x 2 5 "x 2 2 25

3

5

83. ƒ 1 x 2 5 !1 2 2x

84. g 1 x 2 5 !7 2 5x

x11

87. R 1 x 2 5

88. p 1 x 2 5

4 " 3 2 2x

91. ƒ1x2 5  1x2 2 162 1/2

2

2

96. g 1 x 2 5

94. p 1x2 5  1x 2 12 2  1x2 2 92 23/5 95. ƒ 1 x 2 5 x 2 5 4

3

"25 2 x 2

92. g 1x2 5  12x 2 52 1/3

97. Let g 1x2 5 x2 2 2x 2 5 and find the values of x that correspond to g 1x2 5 3. 5

x2

2 2 1 3 x 2 x2 3 6 4

2

98. Let g 1 x 2 5 6 x 2 4 and find the value of x that corresponds to g 1 x 2 5 3 . 99. Let ƒ1x2 5 2x 1x 2 52 3 2 12 1x 2 52 2 and find the values of x that correspond to ƒ1x2 5 0.

100. Let ƒ1x2 5 3x 1x 1 32 2 2 6 1x 1 32 3 and find the values of x that correspond to ƒ1x2 5 0.

• A P P L I C AT I O N S 101. Budget: Event Planning. The cost associated with a catered

108. Volume. A cylindrical water basin will be built to harvest

wedding reception is $45 per person for a reception for more than 75 people. Write the cost of the reception in terms of the number of guests and state any domain restrictions. 102. Budget: Long-Distance Calling. The cost of a local home phone plan is $35 for basic service and $0.10 per minute for any domestic long-distance calls. Write the cost of monthly phone service in terms of the number of monthly long-distance minutes and state any domain restrictions. 103. Temperature. The average temperature in Tampa, Florida, in the springtime is given by the function T 1x2 5 20.7x2 1 16.8x 2 10.8, where T is the temperature in degrees Fahrenheit and x is the time of day in military time and is restricted to 6 # x # 18 (sunrise to sunset). What is the temperature at 6 a.m.? What is the temperature at noon? 104. Falling Objects: Firecrackers. A firecracker is launched straight up, and its height is a function of time, h 1t2 5 216t2 1 128t, where h is the height in feet and t is the time in ­seconds with t 5 0 corresponding to the instant it launches. What is the height 4 seconds after launch? What is the domain of this function? 105. Collectibles. The price of a signed Alex Rodriguez baseball card is a function of how many are for sale. When ­Rodriguez was traded from the Texas Rangers to the New York ­Yankees in 2004, the going rate for a signed baseball card on eBay was P 1 x 2 5 10 1 !400,000 2 100x, where x represents the number of signed cards for sale. What was the value of the card when there were 10 signed cards for sale? What was the value of the card when there were 100 signed cards for sale? 106. Collectibles. In Exercise 105, what was the lowest price on eBay, and how many cards were available then? What was the highest price on eBay, and how many cards were available then? 107. Volume. An open box is constructed from a square 10-inch piece of cardboard by cutting squares of length x inches out of each corner and folding the sides up. Express the volume of the box as a function of x, and state the domain.

rainwater. The basin is limited in that the largest radius it can have is 10 feet. Write a function representing the volume of water V as a function of height h. How many additional gallons of water will be collected if you increase the height by 2 feet? Hint: 1 cubic foot 5 7.48 gallons.

Young_AT_6160_ch03_pp236-297.indd 252

For Exercises 109–110, refer to the following: The weekly exchange rate of the U.S. dollar to the Japanese yen is shown in the graph as varying over an 8-week period. Assume the exchange rate E 1t2 is a function of time (week); let E 112 be the exchange rate during Week 1. Japanese Yen to One U.S. Dollar

E 90 89 88 87 86 85 84 83 82 t 1 2 3 4 5 6 7 8 9 10 Week

109. Economics. Approximate the exchange rates of the U.S.

d­ ollar to the nearest yen during Weeks 4, 7, and 8. 110. Economics. Find the increase or decrease in the number of Japanese yen to the U.S. dollar exchange rate, to the nearest yen, from (a) Week 2 to Week 3 and (b) Week 6 to Week 7. For Exercises 111–112, refer to the following: An epidemiological study of the spread of malaria in a rural area finds that the total number P of people who contracted malaria t days into an outbreak is modeled by the function 1 P 1 t 2 5 2 t 2 1 7t 1 180    1 # t # 14 4

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3.1 Functions 

Refer to the table below for Exercises 115 and 116. It illustrates the average federal funds rate for the month of January (2009 to 2017). YEAR

FED. RATE

2009

5.45

2010

5.98

2011

1.73

2012

1.24

2013

1.00

2014

2.25

2015

4.50

2016

5.25

2017

3.50

For Exercises 117 and 118, use the following figure: Employer-Provided Health Insurance Premiums for Single Plans 1999-2015

14 days into the outbreak? 112. Medicine/Health. How many people have contracted malaria 6 days into the outbreak? 113. Environment: Tossing the Envelopes. The average ­American adult receives 24 pieces of mail per week, ­usually of some combination of ads and envelopes with windows. Suppose each of these adults throws away a dozen ­envelopes per week. a.  The width of the window of an envelope is 3.375 inches less than its length x. Create the function A(x) that ­represents the area of the window in square inches. ­Simplify, if possible. b. Evaluate A(4.5) and explain what this value represents. c. Assume the dimensions of the envelope are 8 inches by 4 inches. Evaluate A(8.5). Is this possible for this particular envelope? Explain. 114. Environment: Tossing the Envelopes. Each month, Jack receives his bank statement in a 9.5 inch by 6 inch envelope. Each month, he throws away the envelope after removing the statement. a. The width of the window of the envelope is 2.875 inches less than its length x. Create the function A(x) that represents the area of the window in square inches. ­Simplify, if possible. b. Evaluate A(5.25) and explain what this value represents. c. Evaluate A(10). Is this possible for this particular e­ nvelope? Explain.

$12,000 10,000

Employee contribution Employer contribution

8,000 6,000 4,000 2,000 2000

2005 2010 Year

2015

Source: Kaiser Family Foundation Health Research and Education Trust.

117. Health Care Costs. Fill in the following table. Round

dollars to the nearest $1000. TOTAL HEALTH CARE COST FOR FAMILY PLANS

YEAR

1999 2003 2007 2011 2015

Write the five ordered pairs resulting from the table. 118. Health Care Costs. Using the table found in Exercise 115, let the years correspond to the domain and the total costs ­correspond to the range. Is this relation a function? Explain.

For Exercises 119 and 120, use the following information: Global Fossil Carbon Emissions Metric Tons of Carbon/Year (in millions)

111. Medicine/Health. How many people have contracted malaria

253

7,000 6,000 5,000 4,000 3,000

Total Petroleum Coal Natural Gas Cement Production

2,000 1,000 1800 1850 1900 1950 2000 Year

Source: http:/www.naftc.wvu.edu

115. Finance. Is the relation whose domain is the year and whose

range is the average federal funds rate for the month of ­January a function? Explain. 116. Finance. Write five ordered pairs whose domain is the set of even years from 2009–2017 and whose range is the set of corresponding average federal funds rate for the month of January.

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Let the functions ƒ, F, g, G, and H represent the number of tons of c­arbon emitted per year as a function of year corresponding to cement production, natural gas, coal, petroleum, and the total amount, ­respectively. Let t represent the year, with t 5 0 corresponding to 1900. 119. Environment: Global Climate Change. Estimate (to the nearest thousand) the value of a.  F 1502      b. g 1502      c. H 1502 120. Environment: Global Climate Change. Explain what the sum F 1 100 2 1 g 1 100 2 1 G 1 100 2 represents.

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• C AT C H T H E M I S TA K E In Exercises 121–126, explain the mistake that is made. 121. Determine whether the y relationship is a function.

124. Determine the domain of the function g 1 t 2 5 !3 2 t and

express it in interval notation.

x

Solution: y Apply the horizontal line test. Because the horizontal line intersects the graph in two places, this is not a function. x This is incorrect. What mistake was made? 122. Given the function H 1 x2 5 3x 2 2, evaluate the quantity H 132 2 H 1212. Solution: H 132 2 H 1212 5 H 132 1 H 112 5 7 1 1 5 8 This is incorrect. What mistake was made? 123. Given the function ƒ1x2 5 x2 2 x, evaluate the quantity ƒ1x 1 12. This is incorrect. What mistake was made? Solution: ƒ1x 1 12 5 ƒ1x2 1 ƒ112 5 x2 2 x 1 0 ƒ1x 1 12 5 x2 2 x This is incorrect. What mistake was made?

Solution: What can t be? Any nonnegative real number.   3 2 t . 0   3 . t    or    t , 3   Domain:  12q, 32 This is incorrect. What mistake was made? 125. Given the function G 1 x2 5 x2, evaluate G 1 21 1 h 2 2 G 1 21 2 . h Solution: G 1 21 1 h 2 2 G 1 21 2 G 1 21 2 1 G 1 h 2 2 G 1 21 2 5 h h G1h2 h2 5 5 5h h h

This is incorrect. What mistake was made? 126. Given the functions ƒ1x2 5 0 x 2 A 0 2 1 and ƒ112 5 21, find A.

Solution: Since ƒ112 5 21, the point  121, 12 must satisfy the function.      21 5 0 21 2 A 0 21 Add 1 to both sides of the equation.         0 21 2 A 0 5 0 The absolute value of zero is zero, so there is no need for the absolute value signs: 21 2 A 5 0 1 A 5 21. This is incorrect. What mistake was made?



• CONCEPTUAL In Exercises 127–130, determine whether each statement is true or false. 127. If a vertical line does not intersect the graph of an equation, then that equation does not represent a function. 128. If a horizontal line intersects a graph of an equation more than once, the equation does not represent a function.

• CHALLENGE 133. If F 1 x 2 5

C and D.

C2x , F 122 2 is undefined, and F 1212 5 4, find D2x

134. Construct a function that is undefined at x 5 5 and whose

129. If ƒ12a2 5 ƒ1a2, then ƒ does not represent a function. 130. If ƒ12a2 5 ƒ1a2, then ƒ may or may not represent a function. 131. If ƒ1x2 5 Ax2 2 3x and ƒ112 5 21, find A. 132. If g 1 x 2 5

1 and g 132 is undefined, find b. b2x

In Exercises 135 and 136, find the domain of each function, where a is any positive real number. 2100 135. ƒ 1 x 2 5 2 x 2 a2

• TECHNOLOGY

136. ƒ 1 x 2 5 25"x 2 2 a2

137. Using a graphing utility, graph the temperature function in

140. The makers of malted milk balls are considering increasing the

­ xercise 103. What time of day is it the warmest? What is the E temperature? Looking at this function, explain why this model for Tampa, Florida, is valid only from sunrise to sunset (6 to 18). 138. Using a graphing utility, graph the height of the firecracker in Exercise 104. How long after liftoff is the firecracker airborne? What is the maximum height that the firecracker attains? Explain why this height model is valid only for the first 8 seconds. 139. Using a graphing utility, graph the price function in Exercise 105. What are the lowest and highest prices of the cards? Does this agree with what you found in Exercise 106?

size of the spherical treats. The thin chocolate coating on a malted milk ball can be approximated by the surface area, S 1 r2 5 4pr 2. If the radius is increased 3 mm, what is the resulting increase in required chocolate for the thin outer coating? 141. Let ƒ 1 x 2 5 x 2 1 1. Graph y1 5 ƒ 1 x 2 and y2 5 ƒ 1 x 2 2 2 in the same viewing window. Describe how the graph of y2 can be obtained from the graph of y1. 142. Let ƒ 1 x 2 5 4 2 x 2. Graph y1 5 ƒ 1 x 2 and y2 5 ƒ 1 x 1 2 2 in the same viewing window. Describe how the graph of y2 can be obtained from the graph of y1.

graph passes through the point  11, 212.

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255

3.2 G  RAPHS OF FUNCTIONS; PIECEWISE-DEFINED FUNCTIONS; INCREASING AND DECREASING FUNCTIONS; AVERAGE RATE OF CHANGE

SKILLS OBJECTIVES ■■ Classify functions as even, odd, or neither. ■■ Determine whether functions are increasing, decreasing, or constant. ■■ Calculate the average rate of change of a function. ■■ Graph piecewise-defined functions.

CONCEPTUAL OBJECTIVES ■■ Identify common functions and understand that even functions have graphs that are symmetric about the y-axis and odd functions have graphs that are symmetric about the origin. ■■ Understand that intervals of increasing, decreasing, and constant correspond to the x-coordinates. ■■ Understand that the difference quotient is just another form of the average rate of change. ■■ Understand points of discontinutity and domain and range of piecewise-defined functions.

3.2.1  Recognizing and Classifying Functions Common Functions Point-plotting techniques were introduced in Section 2.2, and we noted there that we would explore some more efficient ways of graphing functions in Chapter 3. The nine main functions you will read about in this section will constitute a “library” of functions that you should commit to memory. We will draw on this library of functions in the next section when graphing transformations are discussed. Several of these functions have been shown previously in this chapter, but now we will classify them specifically by name and identify properties that each function exhibits. In Section 2.3, we discussed equations and graphs of lines. All lines (with the exception of vertical lines) pass the vertical line test and hence are classified as functions. Instead of the traditional notation of a line, y 5 mx 1 b, we use function notation and classify a function whose graph is a line as a linear function.

3.2.1 S K I L L

Classify functions as even, odd, or neither. 3.2.1 C O N C E P T U A L

Identify common functions and understand that even functions have graphs that are symmetric about the y-axis and odd functions have graphs that are symmetric about the origin.

LINEAR FUNCTION

ƒ 1 x 2 5 mx 1 b        m and b are real numbers. The domain of a linear function ƒ1x2 5 mx 1 b is the set of all real numbers R. The graph of this function has slope m and y-intercept b. LINEAR FUNCTION: f (x ) 5 mx 1 b

SLOPE: m

y-INTERCEPT: b

ƒ1x2 5 2x 2 7

m52

b 5 27

ƒ1x2 5 2x 1 3

m 5 21

b53

ƒ1x2 5 x

m51

b50

ƒ1x2 5 5

m50

b55

One special case of the linear function is the constant function  1 m 5 02. CONSTANT FUNCTION

ƒ 1 x 2 5 b        b is any real number. The graph of a constant function ƒ1x2 5 b is a horizontal line. The y-intercept corresponds to the point  1 0, b2. The domain of a constant function is the set of all

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CHAPTER 3  Functions and Their Graphs

real numbers R. The range, however, is a single value b. In other words, all x-values correspond to a single y-value.

Identity Function Domain: (–∞, ∞) Range: (–∞, ∞) y (3, 3) (2, 2)

x

Points that lie on the graph of a constant function ƒ1x2 5 b are  125, b2  121, b2  1 0, b2  1 2, b2  14, b2 ...  1 x, b2

y

(–5, b)

(0, b)

(4, b) x

Another specific example of a linear function is the function having a slope of one  1m 5 12 and a y-intercept of zero  1b 5 02. This special case is called the identity function. IDENTITY FUNCTION

ƒ1x2 5 x

(0, 0) (–2, –2) (–3, –3)

Domain:  12q, q2 Range: 3b, b4 or 5b6

The graph of the identity function has the following properties: It passes through the origin, and every point that lies on the line has equal x- and y-coordinates. Both the domain and the range of the identity function are the set of all real numbers R. A function that squares the input is called the square function.

Square Function Domain: (–∞, ∞) Range: [0, ∞) y

SQUARE FUNCTION

ƒ 1 x 2 5 x2 (–2, 4)

(2, 4)

(–1, 1)

x

(1, 1)

The graph of the square function is called a parabola and will be discussed in further detail in Chapters 4 and 8. The domain of the square function is the set of all real numbers R. Because squaring a real number always yields a positive number or zero, the range of the square function is the set of all nonnegative numbers. Note that the intercept is the origin and the square function is symmetric about the y-axis. This graph is contained in quadrants I and II. A function that cubes the input is called the cube function. CUBE FUNCTION

Cube Function Domain: (–∞, ∞) Range: (–∞, ∞) y 10 (2, 8) x –5

(–2, –8)

5

–10

Young_AT_6160_ch03_pp236-297.indd 256

ƒ 1 x 2 5 x3 The domain of the cube function is the set of all real numbers R. Because cubing a negative number yields a negative number, cubing a positive number yields a positive number, and cubing 0 yields 0, the range of the cube function is also the set of all real numbers R. Note that the only intercept is the origin and the cube function is symmetric about the origin. This graph extends only into quadrants I and III. The next two functions are counterparts of the previous two functions: square root and cube root. When a function takes the square root of the input or the cube root of the input, the function is called the square root function or the cube root function, respectively.

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3.2  Graphs of Functions 

SQUARE ROOT FUNCTION

ƒ 1 x 2 5 !x or ƒ 1 x 2 5 x 1/2

Square Root Function Domain: [0, ∞) Range: [0, ∞) y 5

In Section 3.1, we found the domain to be 30, q2. The output of the function will be all real numbers greater than or equal to zero. Therefore, the range of the square root function is 30, q2. The graph of this function will be contained in quadrant I.

(9, 3) (4, 2)

CUBE ROOT FUNCTION

ƒ 1 x 2 5 !3 x or ƒ 1 x 2 5 x 1/3

In Section 3.1, we stated the domain of the cube root function to be  12q, q2. We see by the graph that the range is also 12q, q2. This graph is contained in quadrants I and III and passes through the origin. This function is symmetric about the origin. In Section 1.7, you read about absolute value equations and inequalities. Now we shift our focus to the graph of the absolute value function.

257

x 10 Cube Root Function Domain: (–∞, ∞) Range: (–∞, ∞) y 5 (8, 2) x 10

ABSOLUTE VALUE FUNCTION (–8, –2)

ƒ1 x2 5 0 x 0

Some points that are on the graph of the absolute value function are  121, 12 ,  10, 02 , and  11, 12. The domain of the absolute value function is the set of all real numbers R, yet the range is the set of nonnegative real numbers. The graph of this function is symmetric with respect to the y-axis and is contained in quadrants I and II. A function whose output is the reciprocal of its input is called the reciprocal function. RECIPROCAL FUNCTION

Absolute Value Function Domain: (–∞, ∞) Range: [0, ∞) y

(–2, 2)

1 ƒ1 x2 5 x

(2, 2)

x

x20

The only restriction on the domain of the reciprocal function is that x 2 0. Therefore, we say the domain is the set of all real numbers excluding zero. The graph of the reciprocal function illustrates that its range is also the set of all real numbers except zero. Note that the reciprocal function is symmetric with respect to the origin and is contained in quadrants I and III.

Even and Odd Functions Of the nine functions discussed above, several have similar properties of symmetry. The constant function, square function, and absolute value function are all symmetric with respect to the y-axis. The identity function, cube function, cube root function, and reciprocal function are all symmetric with respect to the origin. The term even is used to describe functions that are symmetric with respect to the y-axis, or vertical axis, and the term odd is used to describe functions that are symmetric with respect to the origin. Recall from Section 2.2 that symmetry can be determined both graphically and algebraically. The following box summarizes the graphic and algebraic characteristics of even and odd functions.

Reciprocal Function Domain: (–∞, 0)  (0, ∞) Range: (–∞, 0)  (0, ∞) y f(x) = x1 (1, 1)

x

(–1, –1)

EVEN AND ODD FUNCTIONS

Function Even Odd

Young_AT_6160_ch03_pp236-297.indd 257

Symmetric with Respect to y-axis, or vertical axis origin

On Replacing x with 2x ƒ12x2 5 ƒ1x2 ƒ12x2 5 2ƒ1x2

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[ CONCEPT CHECK ] Classify the functions f (x) 5 x 2n and g (x) 5 x (2n 1 1) where n is a positive integer (1, 2, 3, …) as Even, Odd, or Neither.

▼ ANSWER f (x) is even; g (x) is odd.

The algebraic method for determining symmetry with respect to the y-axis, or vertical axis, is to substitute 2x for x. If the result is an equivalent equation, the function is ­symmetric with respect to the y-axis. Some examples of even functions are ƒ1x2 5 b, ƒ1x2 5 x2, ƒ1x2 5 x4, and ƒ1x2 5 ) x ). In any of these equations, if 2x is substituted for x, the result is the same; that is, ƒ12x2 5 ƒ1x2. Also note that, with the exception of the absolute value function, these examples are all even-degree polynomial equations. All constant functions are degree zero and are even functions. The algebraic method for determining symmetry with respect to the origin is to substitute 2x for x. If the result is the negative of the original function, that is, if ƒ12x2 5 2ƒ1x2 , then the function is symmetric with respect to the origin and, hence, classified as an odd function. Examples of odd functions are ƒ1x2 5 x, ƒ1x2 5 x3, ƒ1x2 5 x5, and ƒ1x2 5 x1/3. In any of these functions, if 2x is substituted for x, the result is the negative of the original function. Note that with the exception of the cube root function, these equations are odd-degree polynomials. Be careful, though, because functions that are combinations of even- and odd-degree polynomials can turn out to be neither even nor odd, as we will see in Example 1.

EXAMPLE 1   Determining Whether a Function Is Even, Odd, or Neither

Determine whether the functions are even, odd, or neither. a. ƒ1x2 5 x2 2 3  b. g1x2 5 x5 1 x3  c. h1x2 5 x2 2 x Solution (a):

Original function. Replace x with 2x.

ƒ1x2 5 x2 2 3 ƒ12x2 5  12x22 2 3

Simplify. ƒ12x2 5 x2 2 3 5 ƒ1x2 Because ƒ12x2 5 ƒ1x2, we say that ƒ1x2 is an even function . Solution (b):

Original function.

  g1x2 5 x5 1 x3

Replace x with 2x.  g 12x2 5  12x25 1  12x23

Simplify.  g12x2 5 2x5 2 x3 5 21x5 1 x32 5 2g1x2

Because g12x2 5 2g1x2, we say that g 1 x2 is an odd function . Solution (c):

Original function.

h1x2 5 x2 2 x

Replace x with 2x.  h 12x2 5  12x22 2  12x2 Simplify.  h12x2 5 x2 1 x

h12x2 is neither 2h1x2 nor h1x2; therefore the function h 1 x2 is neither even nor odd . In parts (a), (b), and (c), we classified these functions as even, odd, or neither, using the algebraic test. Look back at them now and reflect on whether these classifications agree with your intuition. In part (a), we combined two functions: the square function and the constant function. Both of these functions are even, and adding even functions yields another even function. In part (b), we combined two odd functions: the fifth-power function and the cube function. Both of these functions are odd, and adding two odd functions yields another odd function. In part (c), we combined two functions: the square function and the identity function. The square function is even, and the identity function is

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3.2  Graphs of Functions 

259

odd. In this part, combining an even function with an odd function yields a function that is neither even nor odd and, hence, has no symmetry with respect to the vertical axis or the origin.

▼

▼ ANSWER

Y O U R T U R N   Classify the functions as even, odd, or neither. a. ƒ1x2 5 0 x 0 1 4  b. ƒ1x2 5 x3 2 1

a. even  b. neither

3.2.2  Increasing and Decreasing Functions Look at the figure in the margin. Graphs are read from left to right. If we start at the left side of the graph and trace the red curve with our pen, we see that the function values ­(values in the vertical direction) are decreasing until arriving at the point  122, 222. Then, the function values increase until arriving at the point  121, 12. The values then remain ­constant  1 y 5 12 between the points  121, 12 and  1 0, 12. Proceeding beyond the point  1 0,  12, the function values decrease again until the point  1 2, 222. Beyond the point  1 2, 222, the function values increase again until the point  1 6, 42. Finally, the function values decrease and continue to do so. When specifying a function as increasing, decreasing, or constant, the intervals are classified according to the x-coordinate. For instance, in this graph, we say the function is increasing when x is between x 5 22 and x 5 21 and again when x is between x 5 2 and x 5 6. The graph is classified as decreasing when x is less than 22 and again when x is between 0 and 2 and again when x is greater than 6. The graph is classified as constant when x is between 21 and 0. In interval notation, this is summarized as

Decreasing

12q, 22 2 ∪ 1 0, 2 2 ∪ 1 6, q 2

Increasing Constant 122, 21 2 ∪ 1 2, 6 2

1 21, 0 2

An algebraic test for determining whether a function is increasing, decreasing, or constant is to compare the value ƒ1x2 of the function for particular points in the intervals.

INCREASING, DECREASING, AND CONSTANT FUNCTIONS

1. A function ƒ is increasing on an open interval I if for any x1 and x2 in I, where x1 , x2, then ƒ1x12 , ƒ1x22. 2. A function ƒ is decreasing on an open interval I if for any x1 and x2 in I, where x1 , x2, then ƒ1x12 . ƒ1x22. 3. A function ƒ is constant on an open interval I if for any x1 and x2 in I, then ƒ1x12 5 ƒ1x22.

3.2.2 S K I L L

Determine whether functions are increasing, decreasing, or constant. 3.2.2 C O N C E P T U A L

Understand that intervals of increasing, decreasing, and constant correspond to the x-coordinates. y (6, 4) (–1, 1)

(–2, –2)

(0, 1)

x

(2, –2)

STUD Y T I P • Graphs are read from left to right. • Intervals correspond to the x-coordinates.

In addition to classifying a function as increasing, decreasing, or constant, we can determine the domain and range of a function by inspecting its graph from left to right: The domain is the set of all x-values (from left to right) where the function is defined. The range is the set of all y-values (from bottom to top) that the graph of the function corresponds to. ■■ A solid dot on the left or right end of a graph indicates that the graph terminates there and the point is included in the graph. ■■ An open dot indicates that the graph terminates there and the point is not included in the graph. ■■ Unless a dot is present, it is assumed that a graph continues indefinitely in the same direction. (An arrow is used in some books to indicate direction.) ■■ ■■

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CHAPTER 3  Functions and Their Graphs

[ CONCEPT CHECK ] TRUE OR FALSE  An even function has both increasing and decreasing intervals, but an odd function only has one or the other.

▼ ANSWER True

EXAMPLE 2   Finding Intervals When a Function Is Increasing or Decreasing

Given the graph of a function: a. State the domain and range of the ­ function. b. Find the intervals when the function is increasing, decreasing, or constant.

y 10 (–5, 7) (0, 4) (–2, 4)

Solution (a):

 Domain: 325, q2    Range: 30, q2

x –5

Solution (b):

5

(2, 0)

Reading the graph from left to right, we see that the graph decreases from the point  125, 72 to the point  122, 42. ■■ is constant from the point  122, 42 to the point  1 0, 42. ■■ decreases from the point  1 0, 42 to the point  1 2, 02. ■■

■■

increases from the point  1 2, 02 on.

 he intervals of increasing and decreasing T correspond to the x-coordinates.

y 10 (–5, 7) Decreasing (–2, 4)

(0, 4)

Constant

Increasing x

Decreasing –5

(2, 0)

We say that this function is

5

y

increasing on the interval A2, q B. ■■ decreasing on the interval 125, 22 2 ∪ 1 0, 2 2 . ■■ constant on the interval  122, 02.

10

■■

(–5, 7) (0, 4) (–2, 4) x –5

(2, 0)

5

Constant Increasing Decreasing Decreasing

3.2.3 S K I L L

Note: The intervals of increasing or decreasing are defined on open intervals. This should not be confused with the domain. For example, the point x 5 25 is included in the domain of the function but not in the interval where the function is classified as decreasing.

Calculate the average rate of change of a function. 3.2.3 C O N C E P T U A L

Understand that the difference quotient is just another form of the average rate of change.

3.2.3  Average Rate of Change

y f (x 2, y 2) t

an

sec

(x 1, y1) x x1

Young_AT_6160_ch03_pp236-297.indd 260

x2

How do we know how much a function is increasing or decreasing? For example, is the price of a stock slightly increasing or is it doubling every week? One way we determine how much a function is increasing or decreasing is by calculating its average rate of change. Let  1 x1, y12 and  1 x2, y22 be two points that lie on the graph of a function ƒ. Draw the line that passes through these two points  1 x1, y12 and  1 x2, y22. This line is called a secant line. y2 2 y1 Note that the slope of the secant line is given by m 5 , and recall that the x2 2 x1 slope of a line is the rate of change of that line. The slope of the secant line is used to represent the average rate of change of the function.

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3.2  Graphs of Functions 

261

AVERAGE RATE OF CHANGE

Let  1 x1, ƒ1x122 and  1 x2, ƒ1x222 be two distinct points, 1 x1 2 x2 2 , on the graph of the function ƒ. The average rate of change of ƒ between x1 and x2 is given by ƒ 1 x2 2 2 ƒ 1 x1 2 Average rate of change 5 x2 2 x1

y

(x 2, f (x 2 )) t

an

sec

(x 1, f (x 1 ))

f (x2 ) – f (x 1 )

x2 – x 1 x x1

x2

EXAMPLE 3  Average Rate of Change

Find the average rate of change of ƒ1x2 5 x4 from: a. x 5 21 to x 5 0  b. x 5 0 to x 5 1  c. x 5 1 to x 5 2 Solution (a):

Write the average rate of change formula.

ƒ 1 x2 2 2 ƒ 1 x1 2 x2 2 x1

Let x1 5 21 and x2 5 0.

5

Substitute ƒ1212 5  12124 5 1 and ƒ102 5 04 5 0.

5

ƒ 1 0 2 2 ƒ 121 2 0 2 121 2

021 0 2 121 2

Simplify. 5 21 Solution (b):

Write the average rate of change formula.

ƒ 1 x2 2 2 ƒ 1 x1 2 x2 2 x1

Let x1 5 0 and x2 5 1.

5

ƒ112 2 ƒ102 120

Substitute ƒ102 5 04 5 0 and ƒ112 5  1124 5 1.

5

120 120

Simplify. 5 1 Solution (c):

Write the average rate of change formula.

ƒ 1 x2 2 2 ƒ 1 x1 2 x2 2 x1

Let x1 5 1 and x2 5 2.

5

ƒ122 2 ƒ112 221

Substitute ƒ112 5 14 5 1 and ƒ122 5  1 224 5 16.

5

16 2 1 221

Simplify. 5 15

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CHAPTER 3  Functions and Their Graphs

Graphical Interpretation: Slope of the Secant Line a. Average rate of change of ƒ from x 5 21 to x 5 0:

y

Decreasing at a rate of 1

3

(–1, 1)

x (0, 0)

–2

2

–2

b. Average rate of change of ƒ from x 5 0 to x 5 1:

y

Increasing at a rate of 1

3

(1, 1) (0, 0)

–2

x

2

–2

c. Average rate of change of ƒ from x 5 1 to x 5 2:

y

Increasing at a rate of 15

20 (2, 16)

(1, 1) (0, 0)

–2

▼ ANSWER a. 22  b. 2

x

2

▼ Y O U R T U R N   Find the average rate of change of ƒ1x2 5 x2 from: a. x 5 22 to x 5 0  b. x 5 0 to x 5 2

The average rate of change can also be written in terms of the difference quotient. WORDS

Let the difference between x1 and x2 be h. Solve for x2.

x2 2 x1 5 h x2 5 x1 1 h

Substitute x2 2 x1 5 h into the denominator and

Average rate of change 5

x2 5 x1 1 h into the numerator of the average rate of change.



Let x1 5 x.

Young_AT_6160_ch03_pp236-297.indd 262

MATH

ƒ 1 x2 2 2 ƒ 1 x1 2 x2 2 x1

5

ƒ 1 x1 1 h 2 2 ƒ 1 x1 2 h

5

ƒ1x 1 h2 2 ƒ1x2 h

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3.2  Graphs of Functions 

When written in this form, the average rate of change is called the difference quotient. DEFINITION

The expression

y

f (x + h)

Difference Quotient

ƒ1 x 1 h2 2 ƒ1x2 , where h 2 0, is called the difference quotient. h

f (x) x x+h

x

The difference quotient is more meaningful when h is small. In calculus the difference quotient is used to define a derivative.

h

STUD Y T I P

EXAMPLE 4   Calculating the Difference Quotient

Use brackets or parentheses around ƒ(x) to avoid forgetting to distribute the negative sign:

Calculate the difference quotient for the function ƒ1x2 5 2x2 1 1. Solution:

f 1 x 1 h 2 2 3 f 1 x 24 h

ƒ1x 1 h2 5 21x 1 h22 1 1

Find ƒ1x 1 h2.

5 2 1 x 2 1 2xh 1 h2 2 1 1



263

5 2x 2 1 4xh 1 2h2 1 1



g

Find the difference quotient.

d

ƒ 1 x 1 h 2            ƒ 1 x 2



ƒ1x 1 h2 2 ƒ1x2 2x 2 1 4xh 1 2h2 1 1 2 1 2x 2 1 1 2 5 h h ƒ1x 1 h2 2 ƒ1x2 2x 2 1 4xh 1 2h2 1 1 2 2x 2 2 1 5 h h

Simplify.

ƒ1x 1 h2 2 ƒ1x2 4xh 1 2h2 5 h h

Factor the numerator. Cancel (divide out) the common h.

[ CONCEPT CHECK ] Find the difference quotient for the line f (x) 5 mx 1 b.

▼ ANSWER m

ƒ1x 1 h2 2 ƒ1x2 h 1 4x 1 2h 2 5 h h

ƒ1x 1 h2 2 ƒ1x2 5 4x 1 2h      h 2 0 h

▼ Y O U R T U R N   Calculate the difference quotient for the function ƒ1x2 5 2x2 1 2.

ƒ1x 1 h2 2 ƒ1x2 5 22x 2 h h

3.2.4  Piecewise-Defined Functions Most of the functions that we have seen in this text are functions defined by polynomials. Sometimes the need arises to define functions in terms of pieces. For example, most plumbers charge a flat fee for a house call and then an additional hourly rate for the job. For instance, if a particular plumber charges $100 to drive out to your house and work for 1 hour and then an additional $25 an hour for every additional hour he or she works on your job, we would define this function in pieces. If we let h be the number of hours worked, then the charge is defined as Plumbing charge 5 b

100 100 1 25 1 h 2 1 2

▼ ANSWER

h#1 h.1

3.2.4 S K I L L

Graph piecewise-defined functions. 3.2.4 C O N C E P T U A L

Understand points of discontinutity and domain and range of piecewise-defined functions.

If we were to graph this function, we would see that there is 1 hour that is constant and after that the function continually increases. Another piecewise-defined function is the absolute value function. The absolute value function can be thought of as two pieces: the line y 5 2x (when x is negative) and the line y 5 x (when x is nonnegative). We start by graphing these two lines on the same graph.

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CHAPTER 3  Functions and Their Graphs

200

The absolute value function behaves like the line y 5 2x when x is negative (erase the blue graph in quadrant IV) and like the line y 5 x when x is positive (erase the red graph in quadrant III).

150

Absolute value function

Plumber Charge

$250

ƒ1x2 5 0 x 0 5 b

100 50 1 2 3 4 5 6 7 8 9 10 Time (hours) f (x) y = –x

2x x

x*0 x#0

The next example is a piecewise-defined function given in terms of functions in our “library of functions.” Because the function is defined in terms of pieces of other functions, we draw the graph of each individual function, and then for each function, we darken the piece corresponding to its part of the domain. This is like the procedure above for the absolute value function.

y=x

x

EXAMPLE 5   Graphing Piecewise-Defined Functions

Graph the piecewise-defined function, and state the domain, range, and intervals when the function is increasing, decreasing, or constant. x2 G 1 x 2 5 c1 x

f (x) f (x) = | x |

x , 21 21 # x # 1 x.1

Solution:

f (x) = x2

Graph each of the functions on the same plane. x

Square function:

ƒ 1 x 2 5 x2

Constant function:

ƒ 1x2 5 1

y f (x) = x

2 f (x) = 1

1 (–1, 1) –2

–1

(1, 1) 1

x

2

Identity function:

ƒ 1x2 5 x

The points to focus on in particular are the x-values y where the pieces change over, that is, x 5 21 and 5 x 5 1. Let’s now investigate each piece. When x , 21, this function is defined by the square function, ƒ 1x2 5 x2, so darken that particular function to the left of x 5 21. x (–1, 1) (1, 1) When 21 # x # 1, the function is defined by the –5 5 constant function, ƒ 1x2 5 1, so darken that particular function between the x values of 21 and 1. When x . 1, the function is defined by the identity function, ƒ 1x2 5 x, –5 so darken that function to the right of x 5 1. Erase everything that is not darkened, and the resulting graph of the piecewise-defined function is given on the right. This function is defined for all real values of x, so the domain of this function is the set of all real numbers. The values that this function yields in the vertical direction are all real numbers greater than or equal to 1. Hence, the range of this function is 31, q2. The intervals of increasing, decreasing, and constant are as follows: 12q, 21 2 Decreasing: 

121, 1 2 Constant:  1 1, q 2 Increasing: 

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3.2  Graphs of Functions 

265

The term continuous implies that there are no holes or jumps and that the graph can be drawn without picking up your pencil. A function that does have holes or jumps and cannot be drawn in one motion without picking up your pencil is classified as discontinuous, and the points where the holes or jumps occur are called points of discontinuity. The previous example illustrates a continuous piecewise-defined function. At the x 5 21 junction, the square function and constant function both pass through the point  121, 12. At the x 5 1 junction, the constant function and the identity function both pass through the point  11, 12. Since the graph of this piecewise-defined function has no holes or jumps, we classify it as a continuous function. The next example illustrates a discontinuous piecewise-defined function. EXAMPLE 6 

Graphing a Discontinuous Piecewise-Defined Function

Graph the piecewise-defined function, and state the intervals where the function is increasing, decreasing, or constant, along with the domain and range. 12x ƒ 1 x 2 5 cx 21

x,0 0#x,2 x.2

Solution:

y

Graph these functions on the same plane. Linear function: ƒ 1x2 5 1 2 x

Identity function: ƒ 1x2 5 x

x

Constant function: ƒ 1x2 5 21

Darken the piecewise-defined function on the y graph. For all values less than zero 1x * 02 the function is defined by the linear function. Note the use of an open circle, indicating up to but not including x 5 0. For values 0 " x * 2, the function is defined by the x identity function. The circle is filled in at the left endpoint, x 5 0. An open circle is used at x 5 2. For all values greater than 2, x + 2, the function is defined by the constant function. Because this interval does not include the point x 5 2, an open circle is used. At what intervals is the function increasing, decreasing, or constant? Remember that the intervals correspond to the x-values. Decreasing: 12H, 02

Increasing: 10, 22 Constant: 12, H2

The function is defined for all values of x except x 5 2. Domain: 12q, 2 2 ∪ 1 2, q 2

The output of this function (vertical direction) takes on the y-values y $ 0 and the additional single value y 5 21. Range: 3 21, 21 4 ∪ 3 0, q 2  or  5 21 6 ∪ 3 0, q 2

We mentioned earlier that a discontinuous function has a graph that exhibits holes or jumps. In this example, the point x 5 0 corresponds to a jump because you would have to pick up your pencil to continue drawing the graph. The point x 5 2 corresponds to both a hole and a jump. The hole indicates that the function is not

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266 

CHAPTER 3  Functions and Their Graphs

defined at that point, and there is still a jump because the identity function and the constant function do not meet at the same y-value at x 5 2.

▼

▼ ANSWER

Y O U R T U R N   Graph the piecewise-defined function, and state the intervals

where the function is increasing, decreasing, or constant, along with the domain and range.

Increasing:  11, q2 Decreasing:  12q, 212 Constant:  121, 12 Domain: 1 2q, 1 2 ∪ 1 1, q 2 Range: 31, q2

x # 21 21 , x , 1 x.1

2x ƒ 1 x 2 5 c2 x



y

Piecewise-defined functions whose “pieces” are constants are called step functions. The reason for this name is that the graph of a step function looks like steps of a staircase. A common step function used in engineering is the Heaviside step function (also called the unit step function):

x

H(t)

[ CONCEPT CHECK ]

H1t2 5 b

State the domain, range, and any points of discontinuity for the Heaviside function.

0 1

1

t,0    t$0

t

▼

This function is used in signal processing to represent a signal that turns on at some time and stays on indefinitely. A common step function used in business applications is the greatest integer function.

ANSWER Domain: (2q, q) Range: 304 ∪ 314 Point of Discontinuity: x = 0

GREATEST INTEGER FUNCTION

ƒ 1 x 2 5 3 3 x 4 4 5 greatest integer less than or equal to x.

x ƒ1x2 5 3 3 x4 4

1.0

1.3

1.5

1.7

1.9

2.0

1

1

1

1

1

2

f(x) = x 5

x –5

5

–5

[ S E C T I O N 3 . 2 ]     S U M M A R Y NAME

FUNCTION

DOMAIN

RANGE

Linear

ƒ1x2 5 mx 1 b, m 2 0

 12q, q2

 12q, q2

y

 12q, q2

3c, c4 or 5c6

y

Constant

Young_AT_6160_ch03_pp236-297.indd 266

ƒ1x2 5 c

GRAPH

EVEN/ODD

x

Neither  1 unless y 5 x2 Even

x

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3.2  Graphs of Functions 

NAME

FUNCTION

DOMAIN

RANGE

Identity

ƒ1x2 5 x

 12q, q2

 12q, q2

y

 12q, q2

30, q2

y

 12q, q2

 12q, q2

y

ƒ 1 x 2 5 !x

30, q2

30, q2

y

3 ƒ 1 x 2 5 !x

 12q, q2

 12q, q2

y

ƒ1x2 5 0 x 0

 12q, q2

30, q2

y

12q, 0 2 ∪ 1 0, q 2

12q, 0 2 ∪ 1 0, q 2

y

ƒ1x2 5 x2

Square

ƒ1x2 5 x3

Cube

Square Root

Cube Root

Absolute Value

Reciprocal

ƒ1x2 5

1 x

Domain and Range of a Function

• Implied domain: Exclude any values that lead to the

Difference Quotient 

the set of all outputs (range).

Finding Intervals Where a Function Is Increasing, Decreasing, or Constant

• Increasing: Graph of function rises from left to right. • Decreasing: Graph of function falls from left to right. • Constant: Graph of function does not change height from left

EVEN/ODD

Odd x

Even x

Odd x

Neither x

Odd x

Even x

Odd x

Average Rate of Change 

function being undefined (dividing by zero) or imaginary outputs (square root of a negative real number).

• Inspect the graph to determine the set of all inputs (domain) and

GRAPH

267

ƒ 1 x2 2 2 ƒ 1 x1 2 x2 2 x1

ƒ1x 1 h2 2 ƒ1x2 h

x1 2 x2

h20

Piecewise-Defined Functions

• Continuous: You can draw the graph of a function without picking up the pencil.

• Discontinuous: Graph has holes and/or jumps.

to right.

[SEC TION 3. 2]  E X E R C I S E S • SKILLS In Exercises 1–24, determine whether the function is even, odd, or neither. 1. G 1 x2 5 x 1 4 3

5. g1t2 5 5t 2 3t

Young_AT_6160_ch03_pp236-297.indd 267

3. ƒ1x2 5 3x2 1 1

2. h1x2 5 3 2 x 5

3

6. ƒ1x2 5 3x 1 4x

2

7. h1x2 5 x 1 2x

4. F1x2 5 x4 1 2x2 8. G1x2 5 2x4 1 3x3

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CHAPTER 3  Functions and Their Graphs

9. h 1 x2 5 x1/3 2 x

10. g 1 x2 5 x21 1 x

13. ƒ1x2 5 0 x 0

14. ƒ1x2 5 0 x3 0

11. ƒ1x2 5 0 x 0 1 5

24. y

12. ƒ1x2 5 0 x 0 1 x

15. G 1 t2 5 0 t 2 3 0

y

10

9

16. g 1 t2 5 0 t 1 2 0

17. G 1 t 2 5 !t 2 3

18. ƒ 1 x 2 5 !2 2 x

19. g 1 x 2 5 "x 2 1 x 21. h 1 x 2 5

23.

2

20. ƒ 1 x 2 5 "x 2 1 2

1 13 x

22. h 1 x 2 5

x x

1 2 2x x

–5

–5

5

5

In Exercises 25–36, state the (a) domain, (b) range, and (c) x-interval(s) where the function is increasing, decreasing, or constant. Find the values of (d) ƒ 102 , (e) ƒ 1222 , and (f ) ƒ 122 . 25.

26.

27.

y

5

f (x)

(–3, 3)

y

5

f (x)

(–3, 3)

(2, 1)

(–2, 2)

x (–2, –1)

28.

y

x

(0, –1)

x

(1, –1)

2

–5

30.

–5

(3, –4)

32.

y

5

(6, 0) x 10

(0, 0)

(2, –5)

31.

y

(–6, 0) –10

(–4, –2)

–5

29.

(0, 4)

(–7, 1) –8

(–1, –1)

y

(–3, 4) 5

y

5

y

5

5

(–3, 2) (4, 2) –5

x

5

x –5

–5

x –5 (–2, 0)

5

34. 5

(–2, 3) x –10

10

y

–5

(–4, 0)

(0, 4)

x

–10 (–8, 0)

5

–5

(4, 3)

x

(–2, 3) 10

(5, 0) x 5

–5 (–5, 0)

Young_AT_6160_ch03_pp236-297.indd 268

(–5, 3)

(0, 5)

(2, –3)

–10

y 10

8 (0, 7) (4, 2)

5

36.

y

10

(3, 0)

–5

35.

y

x –5 (–3, 0)

–5 (0, –4)

–5

33.

(2, 0) 5

–2

(0, –4) –10

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3.2  Graphs of Functions 

In Exercises 37–44, find the difference quotient 37. ƒ1x2 5 x2 2 x

ƒ 1x 1 h2 2 ƒ 1x2 for each function. h

38. ƒ1x2 5 x2 1 2x

2

39. ƒ1x2 5 3x 1 x2

2

41. ƒ1x2 5 x 2 3x 1 2

40. ƒ1x2 5 5x 2 x2

2

42. ƒ1x2 5 x 2 2x 1 5

43. ƒ1x2 5 23x 1 5x 2 4

In Exercises 45–52, find the average rate of change of the function from x 5 1 to x 5 3. 1 45. ƒ1x2 5 x3 46. ƒ 1 x 2 5 47. ƒ1x2 5 0 x 0 x 50. ƒ1x2 5 9 2 x2

49. ƒ1x2 5 1 2 2x

51. ƒ1x2 5 0 5 2 2x 0

44. ƒ1x2 5 24x2 1 2x 2 3

48. ƒ1x2 5 2x 52. ƒ 1 x 2 5 "x 2 2 1

In Exercises 53–78, graph the piecewise-defined functions. State the domain and range in interval notation. Determine the intervals where the function is increasing, decreasing, or constant. 53. ƒ 1 x 2 5 b 56. ƒ 1 x 2 5 b

x 2

x,2 x$2

x2 4

54. ƒ 1 x 2 5 b

x,2 x$2

2x 1 2 x2

59. ƒ 1 x 2 5 b

x,1 x $1

1 x 2 62. ƒ 1 x 2 5 d 3 41 x 2 32

1 65. G 1 t 2 5 c t 2

4

2x 2 1 2x 1 1

Young_AT_6160_ch03_pp236-297.indd 269

x . 22

x # 22 22 , x , 1 x.1

60. ƒ 1 x 2 5 b

x20 x , 21 21 # x , 1 x.1

x # 21 21 , x , 1 x.1

x,0 x$0

x x2

21x x2

1 66. G 1 t 2 5 c t 2

4

69. G 1 x 2 5 b 72. G 1 x 2 5 c

x # 21 x . 21 x , 21 21 # x # 3 x.3

21 1 2 63. G x 5 c x 3

x50

3 2! x 74. G 1 x 2 5 c x !x

x 77. ƒ 1 x 2 5 c x 3 x2

x , 22

t,1 1#t#2 t.2

68. ƒ 1 x 2 5 c x 1 1

0 71. G 1 x 2 5 c 1 x

57. ƒ 1 x 2 5 b

x , 21 x $ 21

2x 21

t,1 1,t,2 t.2

0 !x

x,0 x$0

0

x50

1 2 x

x20

x13 75. ƒ 1 x 2 5 c 0 x 0 x2 x2 78. ƒ 1 x 2 5 cx 3 x

x # 22 22 , x , 2 x $ 2

x # 21 21 , x , 1 x $ 1

1 x2

x , 21 x $ 21

2x x2

x#0 x.0

55. ƒ 1 x 2 5 b 58. ƒ 1 x 2 5 b 61. ƒ 1 x 2 5 b

5 2 2x 3x 2 2

x , 21 21 , x , 3 x.3

21 1 2 64. G x 5 c x 3

2x 2 1 67. ƒ 1 x 2 5 c x 1 1

2x 1 1 1 !x

2 !x 73. G 1 x 2 5 c x 2 !x

0x0

76. ƒ 1 x 2 5 c 1

0x0

x , 22 22 , x , 1 x$1

x,1 x.1

70. G 1 x 2 5 b 3

3

x,2 x.2

x # 21 21 , x , 1 x.1

x , 21 21 , x , 1 x.1

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CHAPTER 3  Functions and Their Graphs

• A P P L I C AT I O N S For Exercises 79 and 80, refer to the following: A manufacturer determines that his profit and cost functions over one year are represented by the following graphs. P (profit in millions of dollars) $20 18 16 14 12 10 8 6 4 2

C (cost in millions of dollars) $5 4 3 2 1

2

4

t (time in months) 6 8 10 12



2

4

6

t (time in months) 8 10 12

79.  Business. Find the intervals on which profit is increasing,

decreasing, and constant. 80.  Business. Find the intervals on which cost is increasing,

decreasing, and constant. 81.  Budget: Costs. The Kappa Kappa Gamma sorority decides to order custom-made T-shirts for its Kappa Krush mixer with the Sigma Alpha Epsilon fraternity. If the sorority orders 50 or fewer T-shirts, the cost is $10 per shirt. If it orders more than 50 but less than or equal to 100, the cost is $9 per shirt. If it orders more than 100, the cost is $8 per shirt. Find the cost function C 1 x2 as a function of the number of T-shirts x ordered. 82.  Budget: Costs. The marching band at a university is ordering some additional uniforms to replace existing uniforms that are worn out. If the band orders 50 or fewer, the cost is $176.12 per uniform. If it orders more than 50 but less than or equal to 100, the cost is $159.73 per uniform. Find the cost function C 1 x2 as a function of the number of new uniforms x ordered. 83. Budget: Costs. The Richmond rowing club is planning to enter the Head of the Charles race in Boston and is trying to ­figure out how much money to raise. The entry fee is $250 per boat for the first 10 boats and $175 for each additional boat. Find the cost function C 1 x2 as a function of the number of boats x the club enters. 84.  Phone Cost: Long-Distance Calling. A phone company charges $0.39 per minute for the first 10 minutes of an international long-distance phone call and $0.12 per minute every minute after that. Find the cost function C 1 x2 as a function of the length of the phone call x in minutes. 85.  Event Planning. A young couple are planning their wedding reception at a yacht club. The yacht club charges a flat rate of $1000 to reserve the dining room for a private party. The cost of food is $35 per person for the first 100 people and $25 per person for every additional person beyond the first 100. Write the cost function C 1 x2 as a function of the number of people x attending the reception. 86.  Home Improvement. An irrigation company gives you an ­estimate for an eight-zone sprinkler system. The parts are $1400, and the labor is $25 per hour. Write a function C 1 x2

Young_AT_6160_ch03_pp236-297.indd 270

that determines the cost of a new sprinkler system if you choose this irrigation company. 87.  Sales. A famous author negotiates with her publisher the monies she will receive for her next suspense novel. She will receive $50,000 up front and a 15% royalty rate on the first 100,000 books sold, and 20% on any books sold beyond that. If the book sells for $20 and royalties are based on the selling price, write a royalties function R 1 x2 as a function of total number x of books sold. 88.  Sales. Rework Exercise 87 if the author receives $35,000 up front, 15% for the first 100,000 books sold, and 25% on any books sold beyond that. 89.  Profit. Some artists are trying to decide whether they will make a profit if they set up a Web-based business to market and sell stained glass that they make. The costs associated with this business are $100 per month for the Web site and $700 per month for the studio they rent. The materials cost $35 for each work in stained glass, and the artists charge $100 for each unit they sell.  Write the monthly profit as a function of the number of stained-glass units they sell. 90.  Profit. Philip decides to host a shrimp boil at his house as a fundraiser for his daughter’s AAU basketball team. He orders Gulf shrimp to be flown in from New Orleans. The shrimp costs $5 per pound. The shipping costs $30. If he charges $10 per person, write a function F 1 x2 that represents either his loss or profit as a function of the number of people x that attend. Assume that each person will eat 1 pound of shrimp. 91.  Postage Rates. The following table corresponds to first-class postage rates for the U.S. Postal Service. Write a piecewise-defined function in terms of the greatest integer function that ­ models this cost of mailing flat envelopes first class. WEIGHT LESS THAN (OUNCES)





FIRST-CLASS RATE (FLAT ENVELOPES)

1

$0.98

2

1.20

3

1.42

4

1.64

5

1.86

6

2.08

7

2.30

8

2.52

9

2.74

10

2.96

11

3.18

12

3.40

13

3.62

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3.2  Graphs of Functions 

92.  Postage

Rates. The following table corresponds to first-class postage rates for the U.S. Postal Service. Write a piecewise-defined function in terms of the greatest integer function that models this cost of mailing parcels first class. WEIGHT LESS THAN (OUNCES)



1

1000

$2.54

2

2.54

3

2.54

4

2.74

5

2.94

For Exercises 95 and 96, refer to the following table:

6

3.14

Global Carbon Emissions from Fossil Fuel Burning

7

3.34

8

3.54

YEAR

MILLIONS OF TONS OF CARBON

9

3.74

1990

500

10

3.94

1925

1000

11

4.14

1950

1500

12

4.34

1975

5000

2000

7000





4.54





–1

Climate Change: Global Warming. What is the average 95. 

sciencephotos/Alamy

the following square wave.

An analysis of sales indicates that demand for a product during a calendar year (no leap year) is modeled by

5

t 5

Young_AT_6160_ch03_pp236-297.indd 271

–5

The height (in feet) of a falling object with an initial velocity of 48 feet per second launched straight upward from the ground is given by h 1 t2 5 216t2 1 48t, where t is time (in seconds). 97.  Falling Objects. What is the average rate of change of the height as a function of time from t 5 1 to t 5 2? 98.  Falling Objects. What is the average rate of change of the height as a function of time from t 5 1 to t 5 3? For Exercises 99 and 100, refer to the following:

f (t)



rate of change in global carbon emissions from fossil fuel burning from a. 1900 to 1950? b. 1950 to 2000? 96.  Climate Change: Global Warming. What is the average rate of change in global carbon emissions from fossil fuel burning from a. 1950 to 1975? b. 1975 to 2000? For Exercises 97 and 98, use the following information:

93.  Electronics: Signals. Write a step function ƒ1t2 that represents



the following square wave, where x represents ­frequency in Hz.

PARCELS

A square wave is a waveform used in electronic circuit testing and  signal processing. A square wave alternates regularly and instantaneously between two levels.



94.  Electronics: Signals. Write a step function ƒ1x2 that represents

1

13



271

d 1 t 2 5 3"t 2 1 1 2 2.75t

where d is demand in thousands of units and t is the day of the year and t 5 1 represents January 1. 99.  Economics. Find the average rate of change of the demand of the product over the first quarter. 100. Economics. Find the average rate of change of the demand of the product over the fourth quarter.

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• C AT C H T H E M I S TA K E In Exercises 101–104, explain the mistake that is made. 101. Graph the piecewise-defined function. State the domain and range. ƒ1x2 5 b

2x x

x,0 x.0

y

x

The resulting graph is as shown.

Solution: Draw the graphs of ƒ1x2 5 2x and ƒ1x2 5 x.

Darken the function ƒ1x2 5 2x when x , 1 and the function  ƒ1x2 5 x when x . 1.

Domain:  12q, q2 or R Range:  121, q2

y

This is incorrect. What mistake was made? x

103. The cost of airport Internet access is $15 for the first

30 minutes and $1 per minute for each additional minute. Write a function describing the cost of the service as a function of minutes used online. y

Darken the function ƒ1x2 5 2x when x , 0 and the function ƒ1x2 5 x when x . 0. This gives us the familiar absolute value graph.



15 15 1 x

x # 30 x . 30

This is incorrect. What mistake was made?

x

Domain:  12q, q2 or R Range: 30, q2

104. Most money market accounts pay a higher interest with

a higher principal. If the credit union is offering 2% on accounts with less than or equal to $10,000 and 4% on the additional money over $10,000, write the interest function I 1 x2 that represents the interest earned on an account as a function of dollars in the account.

This is incorrect. What mistake was made?

102. Graph the piecewise-defined function. State the domain and

range. ƒ1x2 5 b

Solution:  C 1 x 2 5 b

2x x

x#1 x.1



Solution:  I 1 x 2 5 b

0.02x 0.02 1 10,000 2 1 0.04x

x # 10,000 x . 10,000

This is incorrect. What mistake was made?

Solution: Draw the graphs of ƒ1x2 5 2x and ƒ1x2 5 x.

y

x

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3.2  Graphs of Functions 

273

• CONCEPTUAL In Exercises 105–108, determine whether each statement is true or false. 105. The identity function is a special case of the linear function. 106. The constant function is a special case of the linear function.

107. If an odd function has an interval where the function is

increasing, then it also has to have an interval where the function is decreasing. 108. If an even function has an interval where the function is increasing, then it also has to have an interval where the function is decreasing.

• CHALLENGE In Exercises 109 and 110, for a and b real numbers, can the function given ever be a continuous function? If so, specify the value for a and b that would make it so. 109. ƒ 1 x 2 5 b

ax bx 2

2 110. ƒ 1 x 2 5 d

x#2 x.2

1 x

x,a

1 x

x$a

• TECHNOLOGY 111. In trigonometry you will

learn about the sine function, sin x. Plot the function ƒ1x2 5 sin x, using a graphing utility. It should look like the graph on the right. Is the sine function even, odd, or neither?

113. In trigonometry you will learn about the tangent function,

y

tan x. Plot the function ƒ1x2 5 tan x, using a graphing utility. If

1

x –10

10

p p , x , , the graph 2 2



you restrict the values of x so that 2



should resemble the graph below. Is the tangent function even, odd, or neither? y 15

–1

y

112. In trigonometry you will

learn about the cosine function, cos x. Plot the function ƒ1x2 5 cos x, using a graphing utility. It should look like the graph on the right. Is the cosine function even, odd, or neither?

x

1

–1.5

x –10

10

–1



1.5

–15

114. Plot the function ƒ 1 x 2 5

sin x . What function is this? cos x

115. Graph the function ƒ 1 x 2 5 3 3 3x 4 4 using a graphing utility.

State the domain and range. 1 116. Graph the function ƒ 1 x 2 5 CC 3 x DD using a graphing utility. State the domain and range.

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CHAPTER 3  Functions and Their Graphs

3.3 GRAPHING TECHNIQUES: TRANSFORMATIONS SKILLS OBJECTIVES ■■ Sketch the graph of a function using horizontal and vertical shifting of common functions. ■■ Sketch the graph of a function by reflecting a common function about the x-axis or y-axis. ■■ Sketch the graph of a function by stretching or compressing a common function.

CONCEPTUAL OBJECTIVES ■■ Understand why a shift in the argument inside the function corresponds to a horizontal shift and a shift outside the function corresponds to a vertical shift. ■■ Understand why a negative argument inside the function corresponds to a reflection about the y-axis and a negative outside the function corresponds to a reflection about the x-axis. ■■ Understand the difference between rigid and nonrigid transformations.

3.3.1  Horizontal and Vertical Shifts 3.3.1 S K I L L

Sketch the graph of a function using horizontal and vertical shifting of common functions. 3.3.1 C O N C E P T U A L

Understand why a shift in the argument inside the function corresponds to a horizontal shift and a shift outside the function corresponds to a vertical shift.

The focus of the previous section was to learn the graphs that correspond to particular functions such as identity, square, cube, square root, cube root, absolute value, and reciprocal. Therefore, at this point, you should be able to recognize and generate the graphs of 1 y 5 x, y 5 x 2, y 5 x 3, y 5 !x, y 5 !3 x, y 5 0 x 0 , and y 5 . In this section, we x will d­ iscuss how to sketch the graphs of functions that are very simple modifications of these functions. For instance, a common function may be shifted (horizontally or ­vertically), reflected, or stretched (or compressed). Collectively, these techniques are called transformations. Let’s take the absolute value function as an example. The graph of ƒ 1x2 5 0  x 0 was given in the last section. Now look at two examples that are much like this function: g1x2 5 0  x 0  1 2 and h1x2 5 0  x 2 1 0 . Graphing these functions by point-plotting yields y g (x) h(x) f (x)

x

x

f (x )

x

g (x )

x

h (x )

22

2

22

4

22

3

21

1

21

3

21

2

0

0

0

2

0

1

1

1

1

3

1

0

2

2

2

4

2

1

Instead of point-plotting the function g1x2 5 0  x 0 1 2, we could have started with the function ƒ 1x2 5 0 x 0 and shifted the entire graph up two units. Similarly, we could have ­generated the graph of the function h1x2 5 0  x 2 1 0 by shifting the function ƒ  1x2 5 0  x 0 to the right one unit. In both cases, the base or starting function is ƒ  1x2 5 0  x 0 . Why did we go up for g 1 x2 and to the right for h 1 x2? Note that we could rewrite the functions g 1 x2 and h 1 x2 in terms of ƒ 1 x2: g 1 x 2 5 0 x0 1 2 5 ƒ 1 x 2 1 2

h 1x2 5 0 x 2 1 0 5 ƒ1x 2 12

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3.3  Graphing Techniques: Transformations 

y

275

y

g (x) f (x)

x

f (x)

f (x) = |x| g(x) = |x| + 2

h(x)

x

f (x) = |x| h(x) = |x – 1|

In the case of g 1 x2, the shift  1122 occurs “outside” the function—that is, outside the ­parentheses showing the argument. Therefore, the output for g1x2 is two more than the typical output for ƒ1x2. Because the output corresponds to the vertical axis, this results in a shift upward of two units. In general, shifts that occur outside the function correspond to a vertical shift corresponding to the sign of the shift. For instance, had the function been G 1 x2 5 0  x 0 2 2, this graph would have started with the graph of the function ƒ1x2 and shifted down two units. In the case of h 1 x2, the shift occurs “inside” the function—that is, inside the parentheses showing the argument. Note that the point  1 0, 02 that lies on the graph of ƒ1x2 was shifted to the point  11, 02 on the graph of the function h 1 x2. The y-value remained the same, but the x-value shifted to the right one unit. Similarly, the points  121, 12 and  11, 12 were shifted to the points  1 0, 12 and  1 2, 12, respectively. In general, shifts that occur inside the function correspond to a horizontal shift opposite the sign. In this case, the graph of the function h 1 x2 5 0  x 2 1 0 shifted the graph of the function ƒ1x2 to the right one unit. If, instead, we had the function H 1 x2 5 0  x 1 1 0 , this graph would have started with the graph of the function ƒ1x2 and shifted to the left one unit.

VERTICAL SHIFTS

Assuming that c is a positive constant, To Graph Shift the Graph of ƒ1x2 ƒ1x2 1 c c units upward ƒ1x2 2 c c units downward

STUD Y T I P Shifts outside the function are vertical shifts with the sign. Up (1) Down (2)

Adding or subtracting a constant outside the function corresponds to a vertical shift that goes with the sign.

HORIZONTAL SHIFTS

STUD Y T I P

Assuming that c is a positive constant,

Shifts inside the function are horizontal shifts opposite the sign.

To Graph ƒ1x 1 c2 ƒ1x 2 c2

Shift the Graph of ƒ1x2 c units to the left c units to the right

Left (1) Right (2)

Adding or subtracting a constant inside the function corresponds to a horizontal shift that goes opposite the sign.

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CHAPTER 3  Functions and Their Graphs

y

EXAMPLE 1   Horizontal and Vertical Shifts

(2, 4)

4

Sketch the graphs of the given functions using horizontal and vertical shifts: a. g 1 x2 5 x2 2 1  b. H 1 x2 5  1 x 1 122

f (x)

(1, 1) (0, 0)

x

–2

2

Solution:

In both cases, the function to start with is ƒ 1x2 5 x2.

y 4

a.  g1x2 5 x2 2 1 can be rewritten as

f (x)

g 1 x2 5 ƒ1x2 2 1.

1.  The shift (one unit) occurs outside of the function. Therefore, we expect a vertical shift that goes with the sign. 2. Since the sign is negative, this corresponds to a downward shift. 3. Shifting the graph of the function ƒ 1x2 5 x2 down one unit yields the graph of g 1x2 5 x2 2 1.

▼ ANSWER a. y 10

(1, 1)

g (x)

(0, 0) –1

b. H1x2 5 1x 1 122 can be rewritten as H 1 x2 5 ƒ1x 1 12. x –5

5

b.

y 10

1. The shift (one unit) occurs inside of the function. Therefore, we expect a horizontal shift that goes opposite the sign. 2. Since the sign is positive, this corresponds to a shift to the left. 3. Shifting the graph of the function ƒ 1x2 5 x2 to the left one unit yields the graph of H1x2 5 1x 1 122.

x

(1, 0) 2

–2 (0, –1)

y 4 f (x)

H(x) x –2

2 –1

▼

Y O U R T U R N   Sketch the graphs of the given functions using horizontal and

x –5

5

vertical shifts. a. g 1 x2 5 x 1 1  b. H 1 x2 5  1 x 2 122 2

It is important to note that the domain and range of the resulting function can be thought of as also being shifted. Shifts in the domain correspond to horizontal shifts, and shifts in the range correspond to vertical shifts. EXAMPLE 2 

[ CONCEPT CHECK ] For the functions F 1 x 2 5 !x 2 a and G 1 x 2 5 !x 2 a where a . 0 explain the shifts on y 5 !x.

▼

ANSWER F (x) is the graph of y 5 !x shifted a units to the right. G (x) is the graph of y 5 !x shifted a units down.

 orizontal and Vertical Shifts and Changes H in the Domain and Range

Graph the functions using translations and state the domain and range of each function. a. g 1 x 2 5 !x 1 1   b. G 1 x 2 5 !x 2 2

f (x) = √x

(9, 3)

Solution:

In both cases, the function to start with is ƒ 1 x 2 5 !x. Domain: 30, H2

Range: 30, H2

Young_AT_6160_ch03_pp236-297.indd 276

y 5

(4, 2) (1, 1) (0, 0)

x 10

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3.3  Graphing Techniques: Transformations 

a. g 1 x 2 5 !x 1 1 can be rewritten as

g 1 x2 5 ƒ1x 1 12. 1. The shift (one unit) is inside the function, which corresponds to a horizontal shift opposite the sign. 2. Shifting the graph of ƒ 1 x 2 5 !x to the left one unit yields the graph of g 1 x 2 5 !x 1 1. Notice that the point 10, 02, which lies on the graph of ƒ1x2, gets shifted to the point 121, 02 on the graph of g 1 x2.

277

y 5

(8, 3) (3, 2) (0, 1) x 9

(1, 0)

Although the original function ƒ 1 x 2 5 !x had an implicit restriction on the domain: 30, H2, the function g 1 x 2 5 !x 1 1 has the implicit restriction that x $ 21. We see that the output or range of g 1 x2 is the same as the output of the original function ƒ1x2.

Domain:  321, H2 Range:  30, H2

b. G 1 x 2 5 !x 2 2 can be rewritten as

G 1 x 2 5 ƒ 1 x 2 2 2. 1. The shift (two units) is outside the function, which corresponds to a vertical shift with the sign. 2. The graph of G 1 x 2 5 !x 2 2 is found by shifting ƒ 1 x 2 5 !x down two units. Note that the point 10, 02, which lies on the graph of ƒ1x2, gets shifted to the point 10, 222 on the graph of G 1 x2.

y

(9, 3)

3 (4, 2)

(0, 0)

x

(4, 0) 10

–2 (0, 2)

The original function ƒ 1 x 2 5 !x has an implicit restriction on the domain: 30, H2. The function G 1 x 2 5 !x 2 2 also has the implicit restriction that x $ 0. The output or range of G 1 x2 is always two units less than the output of the original function ƒ1x2.

Domain:  [0, H2

Range:  [22, H2

▼

▼ ANSWER a. G 1 x 2 5 !x 2 2 y

5

YOUR TURN  S  ketch the graph of the functions using shifts and state the domain

and range. a. G 1 x 2 5 !x 2 2  b. h 1 x 2 5 0 x 0 1 1

The previous examples have involved graphing functions by shifting a known function either in the horizontal or vertical direction. Let us now look at combinations of horizontal and vertical shifts. EXAMPLE 3   Combining Horizontal and Vertical Shifts

Sketch the graph of the function F 1 x2 5  1 x 1 122 2 2. State the domain and range of F. Solution:

The base function is y 5 x2. 1. The shift (one unit) is inside the function, so it represents a horizontal shift opposite the sign.

Young_AT_6160_ch03_pp236-297.indd 277

x –2

8

–5

Domain: 32, q2    Range: 30, q2 b. h 1 x 2 5 0 x 0 1 1 y

5

x –5

5

–5

Domain: 32q, q2    Range: 31, q2

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CHAPTER 3  Functions and Their Graphs

2. The 22 shift is outside the function, which represents a vertical shift with the sign. 3. Therefore, we shift the graph of y 5 x2 to the left one unit and down two units. For instance, the point 10, 02 on the graph of y 5 x2 shifts to the point 121, 222 on the graph of F1x2 5 1x 1 122 2 2.

▼ ANSWER ƒ1x2 5 0  x 2 2 0 1 1

ƒ1x2 5 0  x 0 Domain:  12q, q2 Range: 31, q2

Domain: 12H, H2

x –5

Range: 322, H2

5 (–1, –2)

▼

y

YOUR TURN  S  ketch the graph of the function ƒ 1 x 2 5 0 x 2 2 0 1 1. State the

10

domain and range of ƒ.

x –5

y 8

5

All of the previous transformation examples involve starting with a common function and shifting the function in either the horizontal or the vertical direction (or a combination of both). Now, let’s investigate reflections of functions about the x-axis or y-axis.

3.3.2  Reflection about the Axes 3.3.2 S K I L L

Sketch the graph of a function by reflecting a common function about the x-axis or y-axis. 3.3.2 C O N C E P T U A L

Understand why a negative argument inside the function corresponds to a reflection about the y-axis and a negative outside the function corresponds to a reflection about the x-axis.

To sketch the graphs of ƒ 1x2 5 x2 and g1x2 5 2x2, start by first listing points that are on each of the graphs and then connecting the points with smooth curves. x

f (x )

x

22

4

22

24

21

1

21

21

0

0

0

0

y

g (x )

1

1

1

21

2

4

2

24

5

x –5

5

–5

Note that if the graph of ƒ 1x2 5 x2 is reflected about the x-axis, the result is the graph of g 1x2 5 2x2. Also note that the function g 1 x2 can be written as the negative of the function ƒ1x2; that is, g 1 x2 5 2ƒ1x2. In general, reflection about the x-axis is produced by multiplying a function by 21. Let’s now investigate reflection about the y-axis. To sketch the graphs of ƒ 1 x 2 5 !x and g 1 x 2 5 !2x start by listing points that are on each of the graphs and then connecting the points with smooth curves. x

f (x )

x

g (x )

0

0

29

3

1

1

24

2

4

2

21

1

9

3

0

0

y 5

x –5

Young_AT_6160_ch03_pp236-297.indd 278

5

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3.3  Graphing Techniques: Transformations 

279

Note that if the graph of ƒ 1 x 2 5 !x is reflected about the y-axis, the result is the graph of g 1 x 2 5 !2x. Also note that the function g 1 x2 can be written as g 1 x2 5 ƒ12x2. In general, reflection about the y-axis is produced by replacing x with –x in the function. Notice that the domain of ƒ is 30, q2, whereas the domain of g is  12q, 04. REFLECTION ABOUT THE AXES

The graph of 2ƒ 1x2 is obtained by reflecting the graph of ƒ1x2 about the x-axis. The graph of ƒ 12x2 is obtained by reflecting the graph of ƒ1x2 about the y-axis. EXAMPLE 4   Sketching the Graph of a Function Using Both Shifts and Reflections

Sketch the graph of the function G 1 x 2 5 2 !x 1 1. Solution:

Start with the square root function.

y

ƒ 1 x 2 5 !x

5

Shift the graph of ƒ1x2 to the left one unit to arrive at the graph of ƒ1x 1 12.       ƒ 1 x 1 1 2 5 !x 1 1 Reflect the graph of ƒ1x 1 12 about the x-axis to arrive at the graph of 2ƒ1x 1 12. EXAMPLE 5 

x 9

–5

2ƒ 1 x 1 1 2 5 2 !x 1 1

 ketching the Graph of a Function Using Both S Shifts and Reflections

[ CONCEPT CHECK ] For any even function f (x), describe the graph of f (2x).

Sketch the graph of the function ƒ 1 x 2 5 !2 2 x 1 1.

Solution:

Start with the square root function.

Shift the graph of g 1 x2 to the left two units to arrive at the graph of g 1 x 1 22.

Reflect the graph of g 1 x 1 22 about the

y-axis to arrive at the graph of g 12x 1 22. Shift the graph g 12x 1 22 up one unit to arrive at the graph of g 12x 1 22 1 1.

▼

g 1 x 2 5 !x

ANSWER The graph of f (2x) is identical to the graph of f (x) because for even functions f (2x) 5 f (x).

g 1 x 1 2 2 5 !x 1 2

g 12x 1 2 2 5 !2x 1 2

g 12x 1 2 2 1 1 5 !2 2 x 1 1 y

5

x –5

5

▼ YOUR TURN  U  se shifts and reflections to sketch the graph of the function ƒ 1 x 2 5 2"x 2 1 1 2. State the domain and range of ƒ1x2.

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CHAPTER 3  Functions and Their Graphs

Look back at the order in which transformations were performed in Example 5: ­horizontal shift, reflection, and then vertical shift. Let us consider an alternate order of transformations.

▼ ANSWER

Domain: 31, q2 Range:  12q, 24 y

5

WORDS x 10

–5

MATH

g 1 x 2 5 !x

Start with the square root function. Shift the graph of g 1 x2 up one unit to arrive at the graph of g 1 x2 1 1.

g 1 x 2 1 1 5 !x 1 1

Reflect the graph of g 1 x2 1 1 about the y-axis to arrive at the graph of g 12x2 1 1. Replace x with x 2 2, which corresponds to a shift of the graph of g 12x2 1 1 to the right two units to arrive at the graph of g32 1 x 2 224 1 1.

In the last step, we replaced x with x 2 2, which required us to think ahead knowing the desired result was 2 2 x inside the radical. To avoid any possible confusion, follow this order of transformations:

g 1 2x 2 1 1 5 !2x 1 1 g 12x 1 2 2 1 1 5 "2 2 x 1 1 y

5

1. Horizontal shifts: ƒ1x 6 c2 2. Reflection: ƒ12x2 and/or 2ƒ1x2 x 3. Vertical shifts: ƒ1x2 6 c –5

5

3.3.3  Stretching and Compressing 3.3.3 S K I L L

Sketch the graph of a function by stretching or compressing a common function. 3.3.3 C O N C E P T U A L

Understand the difference between rigid and nonrigid transformations.

Horizontal shifts, vertical shifts, and reflections change only the position of the graph in the Cartesian plane, leaving the basic shape of the graph unchanged. These transformations (shifts and reflections) are called rigid transformations because they alter only the position. Nonrigid transformations, on the other hand, distort the shape of the original graph. We now consider stretching and compressing of graphs in both the vertical and the horizontal direction. A vertical stretch or compression of a graph occurs when the function is multiplied by a positive constant. For example, the graphs of the functions ƒ 1x2 5 x2, g 1x2 5 2 ƒ  1x2 5 2x2, and h 1 x 2 5 12 ƒ 1 x 2 5 12 x2 are illustrated below. Depending on if the constant is larger than 1 or smaller than 1 will determine whether it corresponds to a stretch (expansion) or compression (contraction) in the vertical direction.

x

f (x )

x

g (x )

x

y

h(x )

22

4

22

8

22

2

21

1

21

2

21

1 2

0

0

0

0

0

0

1

1

1

2

1

2

4

2

8

1 2

2

2

20

x –5

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5

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3.3  Graphing Techniques: Transformations 

281

Note that when the function ƒ  1x2 5 x2 is multiplied by 2, so that g 1x2 5 2 ƒ  1x2 5 2x2, the result is a graph stretched in the vertical direction. When the function ƒ  1x2 5 x2 is multiplied by 12, so that h 1 x 2 5 12 ƒ 1 x 2 5 12 x2, the result is a graph that is compressed in the vertical direction. VERTICAL STRETCHING AND VERTICAL COMPRESSING OF GRAPHS

The graph of c ƒ1x2 is found by: ■■ ■■

Vertically stretching the graph of ƒ1x2    if c . 1 Vertically compressing the graph of ƒ1x2   if 0 , c , 1

Note: c is any positive real number.

EXAMPLE 6  Vertically Stretching and Compressing Graphs

Graph the function h 1 x 2 5 14 x 3. Solution:

1. Start with the cube function. 2. Vertical compression is expected because 14 is less than 1. 3. Determine a few points that lie on the graph of h.   10, 02  12, 22  122, 222

ƒ  1x2 5 x3 1 h 1 x 2 5 x3 4

y 10

(2, 2) x –2 (–2, –2)

2

–10

Conversely, if the argument x of a function ƒ is multiplied by a positive real number c, then the result is a horizontal stretch of the graph of ƒ if 0 , c , 1. If c . 1, then the result is a horizontal compression of the graph of ƒ. HORIZONTAL STRETCHING AND HORIZONTAL COMPRESSING OF GRAPHS

The graph of ƒ1cx2 is found by: ■■ ■■

Horizontally stretching the graph of ƒ1x2    if 0 , c , 1 Horizontally compressing the graph of ƒ1x2   if c . 1

Note: c is any positive real number.

[ CONCEPT CHECK ]

EXAMPLE 7  V  ertically Stretching and Horizontally Compressing Graphs

Given the graph of ƒ1x2, graph: a. 2ƒ1x2  b. ƒ12x2

Describe where the graphs of f (x) and a ⋅ f (x) intersect.

y 2 1

▼

( 2 , 1)

ANSWER Only at the points when f (x) 5 0 (x-intercepts)

x

(, 0) –1 –2

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5 2

( 32 , –1)

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CHAPTER 3  Functions and Their Graphs

Solution (a):

y

Since the function is multiplied (on the outside) by 2, the result is that each y-value of ƒ1x2 is multiplied by 2, which corresponds to vertical stretching.

2

( 2 , 2)

1 x 5 2

–1 –2

Solution (b):

( 32 , –2) y

Since the argument of the function is multiplied (on the inside) by 2, the result is that each x-value of ƒ1x2 is divided by 2, which corresponds to horizontal compression.

2 1

( 4 , 1) x

(, 0)

(

–1 –2

)

 ,0 2

5 2

( 34 , –1)

▼

▼ ANSWER

Stretching of the graph ƒ 1x2 5 x3. y

40 g(x) x f (x)

–2

2

Y O U R T U R N   Graph the function g 1 x2 5 4x3.

EXAMPLE 8   Sketching the Graph of a Function Using Multiple Transformations

Sketch the graph of the function H 1 x2 5 22 1 x 2 322. Solution:

–40

ƒ  1x2 5 x2

Start with the square function. Shift the graph of ƒ1x2 to the right three units to arrive at the graph of ƒ1x 2 32.

ƒ 1x 2 32 5 1x 2 322

Vertically stretch the graph of ƒ1x 2 32 by a factor of 2 to arrive at the graph of 2ƒ1x 2 32.

2 ƒ 1x 2 32 5 21x 2 322

Reflect the graph 2 ƒ1x 2 32 about the x-axis to arrive at the graph of 22ƒ1x 2 32.

22 ƒ 1x 2 32 5 221x 2 32 2

y 5

x –4

6

–5

In Example 8, we followed the same “inside out” approach with the functions to determine the order for the transformations: horizontal shift, vertical stretch, and reflection.

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3.3  Graphing Techniques: Transformations 

[ S E C T I O N 3 . 3 ]     S U M M A R Y TRANSFORMATION

TO GRAPH THE FUNCTION…

DRAW THE GRAPH OF f AND THEN…

DESCRIPTION

Horizontal shifts

ƒ1x 1 c2 ƒ1x 2 c2

Shift the graph of ƒ to the left c units. Shift the graph of ƒ to the right c units.

Replace x by x 1 c. Replace x by x 2 c.

Vertical shifts

ƒ1x2 1 c ƒ1x2 2 c

Shift the graph of ƒ up c units. Shift the graph of ƒ down c units.

Add c to ƒ1x2. Subtract c from ƒ1x2.

Reflection about the x-axis

2 ƒ1x2

Reflect the graph of ƒ about the x-axis.

Multiply ƒ1x2 by 21.

Reflection about the y-axis

ƒ12x2

Reflect the graph of ƒ about the y-axis.

Replace x by 2x.

Vertical stretch

c ƒ1x2, where c . 1

Vertically stretch the graph of ƒ.

Multiply ƒ1x2 by c.

Vertical compression

c ƒ1x2, where 0 , c , 1

Vertically compress the graph of ƒ.

Multiply ƒ1x2 by c.

Horizontal stretch

ƒ1cx2, where 0 , c , 1

Horizontally stretch the graph of ƒ.

Replace x by cx.

Horizontal compression

ƒ1cx2, where c . 1

Horizontally compress the graph of ƒ.

Replace x by cx.

[SEC TI O N 3. 3]   E X E R C I S E S • SKILLS In Exercises 1–12, match the function to the graph. 1. ƒ1x2 5 x2 1 1 5. ƒ1x2 5 2 1 x 1 122

2. ƒ1x2 5  1 x 2 122

3. ƒ1x2 5 2 11 2 x22

4. ƒ1x2 5 2x2 2 1

10. ƒ 1 x 2 5 !2x 1 1

11. ƒ 1 x 2 5 2 !2x 1 1

12. ƒ 1 x 2 5 2 !1 2 x 2 1

7. ƒ 1 x 2 5 !x 2 1 1 1

6. ƒ1x2 5 2 11 2 x22 1 1

9. ƒ 1 x 2 5 !1 2 x 2 1 a.

b.

y

c.

y

x

e.

x

Young_AT_6160_ch03_pp236-297.indd 283

y

x

g. y

x

d.

y

f. y

8. ƒ 1 x 2 5 2 !x 2 1

h. y

x

x

y

x

x

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i.

j. y

k. y

x

l. y

x

y

x

x

In Exercises 13–18, write the function whose graph is the graph of y 5 0  x 0 but is transformed accordingly. 13. Shifted up three units

14. Shifted to the left four units

15. Reflected about the y-axis

16. Reflected about the x-axis

17. Vertically stretched by a factor of 3

18. Vertically compressed by a factor of 3

In Exercises 19–24, write the function whose graph is the graph of y 5 x3 but is transformed accordingly. 19. Shifted down four units

20. Shifted to the right three units

21. Shifted up three units and to the left one unit

22. Reflected about the x-axis

23. Reflected about the y-axis

24. Reflected about both the x-axis and the y-axis

In Exercises 25–48, use the given graph to sketch the graph of the indicated functions. 25.

26. y

27. y

x

x

29.

30.

y

a.  y 5 2ƒ 1 x 2

b.  y 5 ƒ 1 2x 2

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31.

y

b.  y 5 ƒ 1 2x 2

a.  y 5 ƒ 1 x 2 1 3

b.  y 5 ƒ 1 x 2 3 2

32.

y

x

a.  y 5 2ƒ 1 x 2

x

a.  y 5 ƒ 1 x 2 2 3

b.  y 5 ƒ 1 x 2 1 2

x

y

x

a.  y 5 ƒ 1 x 1 2 2

a.  y 5 ƒ 1 x 2 2 2

b.  y 5 ƒ 1 x 2 2 2

28. y

b.  y 5 ƒ 1 x 1 3 2 y

x

a.  y 5 2 ƒ 1 x 2 b.  y 5 ƒ 1 2x 2

x

a.  y 5 2 ƒ 1 x 2 b.  y 5 ƒ 1 2x 2

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3.3  Graphing Techniques: Transformations 

33. y 5 ƒ1x 2 22 2 3 34. y 5 ƒ1x 1 12 2 2

1

37. y 5 22 g 1 x 2

y

36. y 5 22 ƒ1x2 1 1

x

1 1

42. y 5 2 F 1 2x 2

g(x)

38. y 5 4 g 1 2x 2

35. y 5 2ƒ1x 2 12 1 2

41. y 5 2 F 1 x 2 1 2 1 2

y

1

f (x)

44. y 5 2F 1 x 2 2 2 2 1

1

x

40. y 5 gA 2 xB

46. y 5 2G 1 2x 2 1 1

F(x)

43. y 5 2F 1 1 2 x 2

39. y 5 2g 1 2x2

45. y 5 2G 1 x 1 1 2 2 4

y

x

285

y G(x)

47. y 5 22G 1 x 2 1 2 1 3 48. y 5 2G 1 x 2 2 2 2 1

x

In Exercises 49–74, graph the function using transformations. 49.  y 5 x2 2 2 53.   y 5  1 x 2 322 1 2 57.   y 5 0 2x 0

61.   y 5 2x2 1 1

65. y 5 2 !2 1 x 2 1

69. y 5

1 12 x13

73. y 5 5!2x

50. y 5 x2 1 3 54. y 5  1 x 1 222 1 1 58. y 5 2 0 x 0

62. y 5 2 0 x 0 1 1

66. y 5 !2 2 x 1 3

70. y 5

1 32x

51. y 5  1 x 1 122

55. y 5 2 11 2 x2

2

59. y 52 0 x 1 2 0 2 1 63. y 5 2 !x 2 2 3

67. y 5 !x 2 1 1 2

71. y 5 2 2

1 x12

52. y 5  1 x 2 222

56. y 5 2 1 x 1 222

60. y 5 0 1 2 x 0 1 2 64. y 5 !2 2 x

3 68. y 5 !x 1221

72. y 5 2 2

1 12x

1

74. y 5 25 !x

In Exercises 75–80, transform the function into the form ƒ1x2 5 c1x 2 h2 2 1 k, where c, k, and h are constants, by completing the square. Use graph-shifting techniques to graph the function. 75. y 5 x2 2 6x 1 11

76. ƒ1x2 5 x2 1 2x 2 2

79. ƒ1x2 5 2x2 2 8x 1 3

80. ƒ1x2 5 3x2 2 6x 1 5

77. ƒ1x2 5 2x2 2 2x

78. ƒ1x2 5 2x2 1 6x 2 7

• A P P L I C AT I O N S 81.  Salary. A manager hires an employee at a rate of $10 per

hour. Write the function that describes the current salary of the employee as a function of the number of hours worked per week, x. After a year, the manager decides to award the employee a raise equivalent to paying him for an additional 5 hours per week. Write a function that describes the salary of the employee after the raise. 82. Profit. The profit associated with St. Augustine sod in Florida is typically P 1 x2 5 2x2 1 14,000x 2 48,700,000, where x is the number of pallets sold per year in a normal year. In rainy years Sod King gives away 10 free pallets per year. Write the function that describes the profit of x pallets of sod in rainy years. 83. Taxes. Every year in the United States each working American typically pays in taxes a percentage of his or her earnings

Young_AT_6160_ch03_pp236-297.indd 285

(minus the standard deduction). Karen’s 2015 taxes were calculated based on the formula T 1 x2 5 0.22 1 x 2 63002. That year the standard deduction was $6300 and her tax bracket paid 22% in taxes. Write the function that will determine her 2016 taxes, assuming she receives the raise that places her in the 33% bracket. 84.  Medication. The amount of medication that an infant

requires is typically a function of the baby’s weight. The number of milliliters of an antiseizure medication A is given by A 1 x 2 5 !x 1 2, where x is the weight of the infant in ounces. In emergencies there is often not enough time to weigh the infant, so nurses have to estimate the baby’s weight. What is the function that represents the actual amount of medication the infant is given if his weight is overestimated by 3 ounces?

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For Exercises 85 and 86, refer to the following: Body Surface Area (BSA) is used in physiology and medicine for many clinical purposes. BSA can be modeled by the function wh BSA 5 Å 3600

where w is weight in kilograms and h is height in centimeters. Since BSA depends on weight and height, it is often thought of as a function

of both weight and height. However, for an individual adult height is generally considered constant; thus BSA can be thought of as a function of weight alone. 85.  Health/Medicine. (a) If an adult female is 162 centimeters

tall, find her BSA as a function of weight. (b) If she loses 3 kilograms, find a function that represents her new BSA. 86.  Health/Medicine. (a) If an adult male is 180 centimeters tall, find his BSA as a function of weight. (b) If he gains 5 kilograms, find a function that represents his new BSA.

• C AT C H T H E M I S TA K E In Exercises 87–90, explain the mistake that is made. 87. Describe a procedure for graphing the function ƒ 1 x 2 5 !x 2 3 1 2. Solution: a. Start with the function ƒ 1 x 2 5 !x. b.  Shift the function to the left three units. c.  Shift the function up two units. This is incorrect. What mistake was made? 88.  Describe a procedure for graphing ƒ 1 x 2 5 2 !x 1 2 2 3.

the

89.  Describe

a procedure ƒ 1 x 2 5 0 3 2 x 0 1 1.

function

Solution: a. Start with the function ƒ 1 x 2 5 !x. b.  Shift the function to the left two units. c.  Reflect the function about the y-axis. d.  Shift the function down three units. This is incorrect. What mistake was made?

for

graphing

Solution: a.  Start with the function ƒ1x2 5 0 x 0 . b.  Reflect the function about the y-axis. c.  Shift the function to the left three units. d.  Shift the function up one unit. This is incorrect. What mistake was made? 90.  Describe a procedure for graphing ƒ1x2 5 22x2 1 1.

the

function

the

function

Solution: a. b. c. d.

Start with the function ƒ1x2 5 x2. Reflect the function about the y-axis. Shift the function up one unit. Expand in the vertical direction by a factor of 2.

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 91–94, determine whether each statement is true or false. 91. The graph of y 5 0 2x 0 is the same as the graph of y 5 0 x 0 . 92. The graph of y 5 !2x is the same as the graph of y 5 !x.

93. If the graph of an odd function is reflected about the x-axis

95. The point  1 a, b2 lies on the graph of the function y 5 ƒ1x2.

96. The point  1 a, b2 lies on the graph of the function y 5 ƒ1x2.

and then the y-axis, the result is the graph of the original odd function. 1 94. If the graph of y 5 x is reflected about the x-axis, it produces the same graph as if it had been reflected about the y-axis.

• CHALLENGE

What point is guaranteed to lie on the graph of ƒ1x 2 32 1 2?

What point is guaranteed to lie on the graph of 2ƒ12x2 1 1?

• TECHNOLOGY 97. Use a graphing utility to graph:

100. Use a graphing utility to graph:

a. y 5 x 2 2 and y 5 0 x 2 2 0

a. y 5 !x and y 5 0.1!x

2

2

b. y 5 x3 2 1 and y 5 0 x3 1 1 0

What is the relationship between ƒ1x2 and 0 ƒ1x2 0 ? 98. Use a graphing utility to graph: a. y 5 x 2 2 and y 5 0 x 0 2 2 2

2

b. y 5 x3 1 1 and y 5 0 x 0 3 1 1

What is the relationship between ƒ1x2 and ƒ1 0 x 0 2?

99. Use a graphing utility to graph: a. y 5 !x and y 5 !0.1x b. y 5 !x and y 5 !10x

What is the relationship between ƒ1x2 and ƒ1ax2, assuming

b. y 5 !x and y 5 10!x

What is the relationship between ƒ1x2 and aƒ1x2, assuming

that a is positive?

101. Use a graphing utility to graph y 5 ƒ 1 x 2 5 3 3 0.5x 4 4 11.

Use transforms to describe the relationship between ƒ1x2 and y 5 3 3 x 4 4. 102. Use a graphing utility to graph y 5 g 1 x 2 5 0.5 3 3 x 4 4 1 1. Use transforms to describe the relationship between g 1 x2 and y 5 33 x44.

that a is positive?

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3.4  Operations on Functions and Composition of Functions 

287

3.4 O  PERATIONS ON FUNCTIONS AND COMPOSITION OF FUNCTIONS SKILLS OBJECTIVES ■■ Add, subtract, multiply, and divide functions. ■■ Evaluate composite functions and determine the corresponding domains.

CONCEPTUAL OBJECTIVES ■■ Understand domain restrictions when dividing functions. ■■ Realize that the domain of a composition of functions excludes values that are not in the domain of the inside function.

Two different functions can be combined using mathematical operations such as addition, subtraction, multiplication, and division. Also, there is an operation on functions called composition, which can be thought of as a function of a function. When we combine f­ unctions, we do so algebraically. Special attention must be paid to the domain and range of the combined functions.

3.4.1 Adding, Subtracting, Multiplying, and Dividing Functions Consider the two functions ƒ1x2 5 x2 1 2x 2 3 and g 1 x2 5 x 1 1. The domain of both of these functions is the set of all real numbers. Therefore, we can add, subtract, or multiply these functions for any real number x. Addition: ƒ 1x2 1 g1x2 5 x2 1 2x 2 3 1 x 1 1 5 x2 1 3x 2 2

The result is in fact a new function, which we denote:

 1ƒ 1 g2  1 x2 5 x2 1 3x 2 2    This is the sum function.

3.4.1 S K I L L

Add, subtract, multiply, and divide functions. 3.4.1 C O N C E P T U A L

Understand domain restrictions when dividing functions.

Subtraction: ƒ 1x2 2 g1x2 5 x2 1 2x 2 3 2 1x 1 12 5 x2 1 x 2 4

The result is in fact a new function, which we denote:

 1ƒ 2 g2  1 x2 5 x2 1 x 2 4    This is the difference function.

Multiplication: ƒ 1x2 • g1x2 5 1x2 1 2x 2 32 1x 1 12 5 x3 1 3x2 2 x 2 3

The result is in fact a new function, which we denote:

 1ƒ • g2  1 x2 5 x3 1 3x2 2 x 2 3    This is the product function.

Although both ƒ and g are defined for all real numbers x, we must restrict x so that x 2 21 ƒ to form the quotient g . Division:

ƒ1x2 x 2 1 2x 2 3 5 , x11 g1x2

x 2 21

The result is in fact a new function, which we denote: ƒ x 2 1 2x 2 3 ,  x 2 21    This is called the quotient function. a g b 1x2 5 x11

Two functions can be added, subtracted, and multiplied. The resulting function domain is therefore the intersection of the domains of the two functions. However, for division, any value of x (input) that makes the denominator equal to zero must be eliminated from the domain. The previous examples involved polynomials. The domain of any polynomial is the set of all real numbers. Adding, subtracting, and multiplying polynomials result in other polynomials, which have domains of all real numbers. Let’s now investigate operations applied to functions that have a restricted domain. The domain of the sum function, difference function, or product function is the intersection of the individual domains of the two functions. The quotient function has

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CHAPTER 3  Functions and Their Graphs

a similar domain in that it is the intersection of the two domains. However, any values that make the denominator zero must also be eliminated. FUNCTION NOTATION

Sum Difference Product Quotient

DOMAIN

 1ƒ 1 g2 1 x2 5 ƒ1x2 1 g 1 x2

 1ƒ 2 g2 1 x2 5 ƒ1x2 2 g 1 x2  1ƒ • g2 1 x2 5 ƒ1x2 • g 1 x2

5domain of ƒ6  " 5domain of g6

5domain of ƒ6  "  5domain of g6

5domain of ƒ6  "  5domain of g6

ƒ1x2 ƒ  5domain of ƒ6  "  5domain of g6  " a b 1x2 5 g g1x2 5g1x2 2 06

We can think of this in the following way: Any number that is in the domain of both the functions is in the domain of the combined function. The exception to this is the quotient function, which also eliminates values that make the denominator equal to zero.

[ CONCEPT CHECK ] Let ƒ 1 x 2 5 !x 1 1 and g 1 x 2 5

g 1x2 find the domain of . ƒ 1x2

▼

ANSWER 121, 02 ∪ 10, q 2

1 x

EXAMPLE 1   Operations on Functions: Determining Domains of New Functions

For the functions ƒ 1 x 2 5 "x 2 1 and g 1 x 2 5 "4 2 x, determine the sum function, ­difference function, product function, and quotient function. State the domain of these four new functions. Solution:

Sum function:  ƒ 1 x 2 1 g 1 x 2 5 "x 2 1 1 "4 2 x Difference function:  ƒ 1 x 2 2 g 1 x 2 5 "x 2 1 2 "4 2 x Product function:  ƒ 1 x 2 ⋅ g 1 x 2 5 !x 2 1⋅ !4 2 x

Quotient function:

5 " 1 x 2 1 2 1 4 2 x 2 5 "2x 2 1 5x 2 4

ƒ1x2 "x 2 1 x21 5 5 g1x2 "4 2 x Å 4 2 x

The domain of the square root function is determined by setting the argument under the radical greater than or equal to zero. Domain of ƒ1x2:  3 1, q 2

Domain of g 1 x2:  12 q, 4 4

The domain of the sum, difference, and product functions is 3 1, q 2 ∩ 12 q, 4 4 5 3 1, 4 4

▼ ANSWER

1 ƒ 1 g 2 1 x 2 5 "x 1 3 1"1 2 x Domain: 3 23, 1 4

Young_AT_6160_ch03_pp236-297.indd 288

The quotient function has the additional constraint that the denominator cannot be zero. This implies that x 2 4, so the domain of the quotient function is 31, 42.

▼

Y O U R T U R N   Given the function ƒ 1 x 2 5 !x 1 3 and g 1 x 2 5 !1 2 x, find

 1ƒ 1 g2 1 x2 and state its domain.

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3.4  Operations on Functions and Composition of Functions 

289

EXAMPLE 2   Quotient Function and Domain Restrictions

Given the functions F 1 x 2 5 !x and G 1 x2 5 0 x 2 3 0 , find the quotient function, F a b 1x 2 , and state its domain. G

Solution:

The quotient function is written as F1x2 !x F 5 a b 1x2 5 G G1x2 0x 2 30

Domain of F 1 x 2 : 3 0, q 2     Domain of G 1 x 2 : 1 2 q, q 2

The real numbers that are in both the domain for F 1 x2 and the domain for G 1 x2 are represented by the intersection 3 0, q 2 ∩ 1 2q, q 2 5 3 0, q 2 . Also, the denominator of the quotient function is equal to zero when x 5 3, so we must eliminate this value from the domain. F Domain of a b 1 x 2 : 3 0, 3 2 ∪ 1 3, q 2 G

▼ Y O U R T U R N   For the functions given in Example 2, determine the quotient

▼ ANSWER

0 x 2 30 G1x2 G 5 a b 1x2 5 F F1x2 !x Domain: 1 0, q 2

G function a b 1 x 2 , and state its domain. F

3.4.2  Composition of Functions Recall that a function maps every element in the domain to exactly one corresponding element in the range as shown in the figure in the margin. Suppose there is a sales rack of clothes in a department store. Let x correspond to the original price of each item on the rack. These clothes have recently been marked down 20%. Therefore, the function g 1 x2 5 0.80x represents the current sale price of each item. You have been invited to a special sale that lets you take 10% off the current sale price and an additional $5 off every item at checkout. The function ƒ1g 1 x22 5 0.90g 1 x2 2 5 determines the checkout price. Note that the output of the function g is the input of the function ƒ as shown in the figure below. Domain of f x

g(x) = 0.80 x

g(x)

Range of f f(x) = 0.90 g (x) − 5

Domain of g

Range of g

Original price

Sale price 20% off original price

3.4.2 S K I L L

Evaluate composite functions and determine the corresponding domains. 3.4.2 C O N C E P T U A L

Realize that the domain of a ­composition of functions excludes values that are not in the domain of the inside.

f(g (x)) x

Additional 10% off sale price and $5 off at checkout

Domain

f

f(x) Range

This is an example of a composition of functions, when the output of one function is the input of another function. It is commonly referred to as a function of a function. An algebraic example of this is the function y 5 "x 2 2 2. Suppose we let g 1 x2 5 x2 2 2 and ƒ 1 x 2 5 "x. Recall that the independent variable in function notation is a placeholder. Since ƒ 1 u 2 5 " 1 u 2 , then ƒ 1 g 1 x 2 2 5 " 1 g 1 x 2 2 . Substituting the expression for g 1 x2, we find ƒ 1 g 1 x 2 2 5 "x 2 2 2. The function y 5 "x 2 2 2 is said to be a composite function, y 5 ƒ1g 1 x22.

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CHAPTER 3  Functions and Their Graphs

▼ CAUTION

ƒ + g 2 ƒ⋅g

Note that the domain of g 1 x2 is the set of all real numbers, and the domain of ƒ1x2 is the set of all nonnegative numbers. The domain of a composite function is the set of all x such that g 1 x2 is in the domain of ƒ. For instance, in the composite function y 5 ƒ1g 1 x22, we know that the allowable inputs into ƒ are all numbers greater than or equal to zero. Therefore, we restrict the outputs of g 1 x2 $ 0 and find the corresponding x-values. Those x-values are the only allowable inputs and constitute the domain of the composite function y 5 ƒ1g 1 x22. The symbol that represents composition of functions is a small open circle; thus 1 ƒ + g 2 1 x 2 5 ƒ 1 g 1 x 2 2 , which is read aloud as “ƒ of g.” It is important not to confuse this with the multiplication sign:  1ƒ • g2 1 x2 5 ƒ1x2g 1 x2. COMPOSITION OF FUNCTIONS

Given two functions ƒ and g, there are two composite functions that can be formed. NOTATION

S TU DY T IP Order is important: 1 f + g 2 1 x 2 5 f 1 g 1 x 22 1 g + f 2 1 x 2 5 g 1 f 1 x 22

ST U DY TIP The domain of f + g is always a subset of the domain of g, and the range of f + g is always a subset of the range of f.

WORDS

DEFINITION

ƒ +g

ƒ composed with g

ƒ1g 1 x22

g +ƒ

g composed with ƒ

g 1ƒ  1 x22

DOMAIN

The set of all real numbers x in the domain of g such that g 1 x2 is also in the domain of ƒ.

The set of all real numbers x in the domain of ƒ such that ƒ1x2 is also in the domain of g.

x

It is important to realize that there are two “filters” that allow certain values of x into the domain of 1 ƒ + g 2 1 x 2 5 ƒ 1 g 1 x 2 2 . The first filter is g 1 x2. If x is not in the domain of g 1 x2, it cannot be in the domain of 1 ƒ + g 2 1 x 2 5 ƒ 1 g 1 x 2 2 . Of those values for x that are in the domain of g 1 x2, only some pass through because we restrict the output of g 1 x2 to values that are allowable as input into ƒ. This adds an additional filter. The domain of ƒ + g is always a subset of the domain of g, and the range of ƒ + g is always a subset of the range of ƒ.

g(x)

f (g(x))

( f g)(x) = f(g(x))

EXAMPLE 3   Finding a Composite Function

Given the functions ƒ1x2 5 x2 1 1 and g 1 x2 5 x 2 3, find 1 ƒ + g 2 1 x 2 . Solution:

Write ƒ1x2 using placeholder notation.

ƒ1u2 5 1u22 1 1

Express the composite function ƒ + g. ƒ1g 1 x22 5  1 g 1 x222 1 1

Substitute g 1 x2 5 x 2 3 into ƒ.

ƒ1g 1 x22 5  1 x 2 322 1 1

Eliminate the parentheses on the right side. ƒ1g 1 x22 5 x2 2 6x 1 10

▼ ANSWER

g + ƒ 5 g 1 ƒ 1 x 22 5 x 2 2 2

Young_AT_6160_ch03_pp236-297.indd 290

▼

1 ƒ + g 2 1 x 2 5 ƒ 1 g 1 x 2 2 5 x 2 2 6x 1 10

Y O U R T U R N   Given the functions in Example 3, find 1 g + ƒ 2 1 x 2 .

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EXAMPLE 4   Determining the Domain of a Composite Function

Given the functions ƒ 1 x 2 5 Solution:

1 1 and g 1 x 2 5 , determine ƒ + g, and state its domain. x x21

Write ƒ1x2 using placeholder notation.  ƒ 1 u 2 5 Express the composite function ƒ + g.   ƒ1g1x2 2 5 Substitute g 1 x 2 5

1 1u2 2 1

1 g1x2 2 1

1 1 into ƒ.   ƒ1g1x2 2 5 x 1 21 x

x 1 x x Multiply the right side by .   ƒ1g1x2 2 5 ⋅ 5 x x 1 12x 21 x x 1ƒ + g2 5 ƒ1g1x2 2 5 12x

What is the domain of  1ƒ + g2 1 x2 5 ƒ1g 1 x22? By inspecting the final result of ƒ1g 1 x22, we see that the denominator is zero when x 5 1. Therefore, x 2 1. Are there any other values for x that are not allowed? The function g 1 x2 has the domain x 2 0; therefore we must also exclude zero. The domain of  1ƒ + g2 1 x2 5 ƒ1g 1 x22 excludes x 5 0 and x 5 1 or, in interval notation,

▼ CAUTION

The domain of the composite function cannot always be determined by examining the final form of ƒ + g.

12 q, 0 2 ∪ 1 0, 1 2 ∪ 1 1, q 2

▼

Y O U R T U R N   For the functions ƒ and g given in Example 4, determine the

composite function g + ƒ and state its domain. The domain of the composite function cannot always be determined by examining the final form of ƒ + g.

▼ ANSWER

g 1 ƒ 1 x 22 5 x 2 1. Domain of g + ƒ is x 2 1, or in interval notation, 1 2q, 1 2 ∪ 1 1, q 2 .

EXAMPLE 5   Determining the Domain of a Composite Function (Without Finding the Composite Function)

[ CONCEPT CHECK ]

1 and g 1 x 2 5 !x 1 3. Find the domain of ƒ1g 1 x22. Do not find the x22 composite function.

For the functions f 1 x 2 5 x 2 a and 1 g 1x2 5 find g(f (x)) and x1a state its domain.

Let ƒ 1 x 2 5 Solution:

▼

3 23, q 2 Find the domain of g. 3 0, q 2 Find the range of g. In ƒ1g 1 x22, the output of g becomes the input for ƒ. Since the domain of ƒ is the set of all real numbers except 2, we eliminate any values of x in the domain of g that correspond to g 1 x2 5 2. !x 1 3 5 2 Let g 1 x2 5 2. Square both sides. x1354 Solve for x. x51 Eliminate x 5 1 from the domain of g, 323, q 2 .

State the domain of ƒ1g 1 x22.

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1 and domain is all real x numbers except x = 0.

ANSWER g 1f 1x22 5

323, 1 2 ∪ 1 1, q 2

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EXAMPLE 6   Evaluating a Composite Function

Given the functions ƒ1x2 5 x2 2 7 and g 1 x2 5 5 2 x2, evaluate: a. ƒ1g 1122  b. ƒ1g 12222  c. g 1ƒ1322  d. g 1ƒ12422

Solution:

One way of evaluating these composite functions is to calculate the two individual composites in terms of x: ƒ1g 1 x22 and g 1ƒ1x22. Once those functions are known, the values can be substituted for x and evaluated.

Another way of proceeding is as follows: a. Write the desired quantity. ƒ1g 1122 Find the value of the inner function g. g 112 5 5 2 12 5 4 Substitute g 112 5 4 into ƒ. ƒ1g 1122 5 ƒ142 Evaluate ƒ142. ƒ142 5 42 2 7 5 9 ƒ1 g112 2 5 9

ƒ1g 12222 Find the value of the inner function g. g 1222 5 5 2  12222 5 1 Substitute g 1222 5 1 into ƒ. ƒ1g 12222 5 ƒ112 Evaluate ƒ112. ƒ112 5 12 2 7 5 26 b. Write the desired quantity.

ƒ 1 g 122 2 2 5 26

g 1ƒ1322 Find the value of the inner function ƒ. ƒ132 5 32 2 7 5 2 Substitute ƒ132 5 2 into g. g 1ƒ1322 5 g 1 22 Evaluate g 1 22. g 1 22 5 5 2 22 5 1 c. Write the desired quantity.

g 1ƒ 132 2 5 1

g 1ƒ12422 Find the value of the inner function ƒ. ƒ1242 5  12422 2 7 5 9 Substitute ƒ1242 5 9 into g. g 1ƒ12422 5 g 1 92 Evaluate g 1 92. g 1 92 5 5 2 92 5 276 d. Write the desired quantity.

▼ ANSWER

ƒ1g 1122 5 5 and g 1ƒ1122 5 27

g 1 ƒ 124 2 2 5 276

▼

Y O U R T U R N   Given the functions ƒ1x2 5 x3 2 3 and g 1 x2 5 1 1 x3, evaluate

ƒ1g 1122 and g 1ƒ1122.

Application Problems

Recall the example at the beginning of this chapter regarding the clothes that are on sale. Often, real-world applications are modeled with composite functions. In the clothes example, x is the original price of each item. The first function maps its input (original price) to an output (sale price). The second function maps its input (sale price) to an output (checkout price). Example 7 is another real-world application of composite functions. Three temperature scales are commonly used: ■■

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The degree Celsius (8C) scale ■■ This scale was devised by dividing the range between the freezing (08C) and boiling (1008C) points of pure water at sea level into 100 equal parts. This scale is used in science and is one of the standards of the “metric” (SI) system of measurements.

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■■

The Kelvin (K) temperature scale ■■ This scale shifts the Celsius scale down so that the zero point is equal to absolute zero (about 2273.158C), a hypothetical temperature at which there is a complete absence of heat energy. ■■ Temperatures on this scale are called kelvins, not degrees kelvin, and kelvin is not capitalized. The symbol for the kelvin is K.

■■

The degree Fahrenheit (8F) scale ■■ This scale evolved over time and is still widely used mainly in the United States, although Celsius is the preferred “metric” scale. ■■ With respect to pure water at sea level, the degrees Fahrenheit are gauged by the spread from 328F (freezing) to 2128F (boiling).

293

The equations that relate these temperature scales are F5

9 C 1 32 5

C 5 K 2 273.15

EXAMPLE 7   Applications Involving Composite Functions

Determine degrees Fahrenheit as a function of kelvins. Solution:

Degrees Fahrenheit is a function of degrees Celsius.

F5

Now substitute C 5 K 2 273.15 into the equation for F.

F5

Simplify.

F5



F5

9 C 1 32 5 9 1 K 2 273.15 2 1 32 5 9 K 2 491.67 1 32 5 9 K 2 459.67 5

[ S E C T I O N 3 . 4 ]     S U M M A R Y Operations on Functions Function Notation Sum  1ƒ 1 g2 1 x2 5 ƒ1x2 1 g 1 x2 Difference Product Quotient

 1ƒ 2 g2 1 x2 5 ƒ1x2 2 g 1 x2  1ƒ⋅g2 1 x2 5 ƒ1x2⋅g 1 x2 ƒ1x2 ƒ a b 1x2 5 g g 1x2

g1x2 2 0

Composition of Functions

1ƒ ° g2 1x2 5 ƒ1 g 1 x2 2

The domain restrictions cannot always be determined simply by inspecting the final form of ƒ1g 1 x22. Rather, the domain of the composite function is a subset of the domain of g 1 x2. Values of x must be eliminated if their corresponding values of g 1 x2 are not in the domain of ƒ.

The domain of the sum, difference, and product functions is the intersection of the domains, or common domain shared by both ƒ and g. The domain of the quotient function is also the intersection of the domain shared by both ƒ and g with an additional restriction that g 1 x 2 2 0.

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[SEC T I O N 3. 4]   E X E R C I S E S • SKILLS ƒ In Exercises 1–10, given the functions  ƒ and g, find  ƒ 1 g,  ƒ 2 g, ƒ . g, and , and state the domain of each. g 1. ƒ1x2 5 2x 1 1

g 1 x2 5 1 2 x



2x 1 3 x24 x24 g 1x2 5 3x 1 2

g 1 x2 5 2x 2 4



g 1 x 2 5 2!x



7. ƒ 1 x 2 5 !x

6. ƒ 1 x 2 5

3. ƒ1x2 5 2x2 2 x

2. ƒ1x2 5 3x 1 2



g 1 x2 5 x2 2 4



g 1 x2 5 2x2



8. ƒ 1 x 2 5 !x 2 1

1 x g 1 x2 5 x

5. ƒ 1 x 2 5

4. ƒ1x2 5 3x 1 2

g 1 x2 5 x2 225



9. ƒ 1 x 2 5 !4 2 x

10. ƒ 1 x 2 5 !1 2 2x

g 1 x 2 5 !x 1 3



g1x2 5

1 x

In Exercises 11–20, for the given functions ƒ and g, find the composite functions ƒ + g and g + ƒ, and state their domains. 12. ƒ1x2 5 x2 2 1

11. ƒ1x2 5 2x 1 1

g 1 x2 5 x2 2 3

16. ƒ1x2 5 0 x 2 1 0

g1x2 5

g 1 x2 5 2 2 x



1 x

g 1 x2 5 x 1 5



2 x23 g 1 x2 5 2 1 x

14. ƒ 1 x 2 5

15. ƒ1x2 5 0 x 0

18. ƒ 1 x 2 5 !2 2 x

19. ƒ1x2 5 x3 1 4

20. ƒ 1 x 2 5 "x 2 2 1



17. ƒ 1 x 2 5 !x 2 1

1 x21 g 1 x2 5 x 1 2

13. ƒ 1 x 2 5





g 1 x2 5 x2 1 2



In Exercises 21–38, evaluate the functions for the specified values, if possible.

ƒ 1 x 2 5 x 2 1 10



g 1 x2 5  1 x 2 421/3

g 1 x 2 5 !x 2 1



g1x2 5

1 x21 3

g 1 x2 5 x2/3 1 1

21.  1ƒ 1 g2 1 22

22.  1ƒ 1 g2 1102

23.  1ƒ 2 g2 1 22

24.  1ƒ 2 g2 1 52

25.  1ƒ ? g2 142

26.  1ƒ ? g2 1 52

33. ƒ1g 1 022

34. g 1ƒ1022

35. ƒ1g 12322

36. gAƒA !7B B

37. 1 ƒ + g 2 1 4 2

38. 1 g + ƒ 2 1 23 2

ƒ g

27. a b 1 10 2

ƒ g

28. a b 12 2

29. ƒ1g 1 222

30. ƒ1g 1122

In Exercises 39–50, evaluate ƒ1g112 2 and g1ƒ122 2 , if possible. 1 1 39. ƒ 1 x 2 5 , g 1 x 2 5 2x 1 1 40. ƒ 1 x 2 5 x 2 1 1, g 1 x 2 5 x 22x 1 43. ƒ 1 x 2 5 , g1x2 5 x 1 3 42. ƒ 1 x 2 5 !3 2 x, g 1 x 2 5 x 2 1 1 0 x 2 10 45. ƒ 1 x 2 5 !x 2 1,

48. ƒ 1 x 2 5

g 1 x 2 5 x2 1 5

x , g 1 x 2 5 4 2 x2 22x

3 46. ƒ 1 x 2 5 !x 2 3,

53. ƒ 1 x 2 5 !x 2 1,

g1x2 5

x21 2

g 1 x 2 5 x 2 1 1 for x $ 1

55. ƒ 1 x 2 5

1 1 , g 1 x 2 5 for x 2 0 x x

59. ƒ 1 x 2 5

1 x11 , g1x2 5  for x 2 0, x 2 1 x x21

57. ƒ 1 x 2 5 4x 2 2 9,

g1x2 5

Young_AT_6160_ch03_pp236-297.indd 294

1 x23

41. ƒ 1 x 2 5 !1 2 x, 44. ƒ 1 x 2 5 47. ƒ 1 x 2 5

32. g 1ƒ1422

g 1 x 2 5 x2 1 2

1 , g 1 x 2 5 0 2x 2 3 0 x

1 , g 1 x 2 5 !x 2 3 x 23 2

49. ƒ1x2 5  1 x 2 121/3,  g 1 x2 5 x2 1 2x 1 1 50. ƒ1x2 5  11 2 x221/2,  g 1 x2 5  1 x 2 321/3

In Exercises 51–60, show that ƒ1g1x2 2 5 x and g1ƒ1x2 2 5 x.

51. ƒ 1 x 2 5 2x 1 1,

g1x2 5

31. g 1ƒ12322

!x 1 9 for x $ 0 2

52. ƒ 1 x 2 5

x22 , g 1 x 2 5 3x 1 2 3

54. ƒ 1 x 2 5 2 2 x 2,

g 1 x 2 5 !2 2 x for x # 2

56. ƒ1x2 5  1 5 2 x21/3,  g 1 x2 5 5 2 x3 3 2 1, 58. ƒ 1 x 2 5 !8x

60. ƒ 1 x 2 5 "25 2 x 2,

g1x2 5

x3 1 1 8

g 1 x 2 5 "25 2 x 2 for 0 # x # 5

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In Exercises 61–66, write the function as a composite of two functions ƒ and g. (There is more than one correct answer.) 61. ƒ1g 1 x2   2 5 2 13x 2 122 1 5 13x 2 12 63. ƒ 1 g 1 x 2 2 5 65. ƒ 1 g 1 x 2 2 5

2 0x 2 30

3 !x 1 1 2 2

62. ƒ 1 g 1 x 2 2 5

1 1 1 x2

64. ƒ 1 g 1 x 2 2 5 "1 2 x 2 66. ƒ 1 g 1 x 2 2 5

!x 3!x 1 2

• A P P L I C AT I O N S Exercises 67 and 68 depend on the relationship between degrees Fahrenheit, degrees Celsius, and kelvins: F5

9 C 1 32 5

72.  Market Price. Typical supply and demand relationships state

that as the number of units for sale increases, the ­market price decreases. Assume that the market price p and the ­number of units for sale x are related by the demand equation:

C 5 K 2 273.15

1 p 5 10,000 2 x 4

67.  Temperature. Write a composite function that converts

kelvins into degrees Fahrenheit. 68.  Temperature. Convert the following degrees Fahrenheit to kelvins: 328F and 2128F. 69.  Dog Run. Suppose that you want to build a square fenced-in area for your dog. Fencing is purchased in linear feet. a. Write a composite function that determines the area of your dog pen as a function of how many linear feet are purchased. b. If you purchase 100 linear feet, what is the area of your dog pen? c. If you purchase 200 linear feet, what is the area of your dog pen? 70.  Dog Run. Suppose that you want to build a circular fenced-in area for your dog. Fencing is purchased in linear feet. a. Write a composite function that determines the area of your dog pen as a function of how many linear feet are purchased. b. If you purchase 100 linear feet, what is the area of your dog pen? c. If you purchase 200 linear feet, what is the area of your dog pen? 71.  Market Price. Typical supply and demand relationships state that as the number of units for sale increases, the m ­ arket price decreases. Assume that the market price p and the ­number of units for sale x are related by the demand equation: 1 p 5 3000 2 x 2





Assume that the cost C 1 x2 of producing x items is governed by the equation C 1 x 2 5 2000 1 10x

and the revenue R 1 x2 generated by selling x units is g­ overned by R 1 x 2 5 100x

a. Write the cost as a function of price p. b. Write the revenue as a function of price p. c. Write the profit as a function of price p.

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Assume that the cost C 1 x2 of producing x items is governed by the equation C 1 x 2 5 30,000 1 5x

and the revenue R 1 x2 generated by selling x units is g­ overned by R 1 x 2 5 1000x

a. Write the cost as a function of price p. b. Write the revenue as a function of price p. c. Write the profit as a function of price p.

In Exercises 73 and 74, refer to the following: The cost of manufacturing a product is a function of the number of hours t the assembly line is running per day. The number of products manufactured n is a function of the number of hours t the assembly line is operating and is given by the function n 1 t2. The cost of manufacturing the product C measured in thousands of dollars is a function of the quantity manufactured, that is, the function C 1 n2. 73.  Business. If the quantity of a product manufactured during a day is given by



n 1 t 2 5 50t 2 t 2

and the cost of manufacturing the product is given by



C 1 n 2 5 10n 1 1375



n 1 t 2 5 100t 2 4t 2

a. Find a function that gives the cost of manufacturing the product in terms of the number of hours t the assembly line was functioning, C 1 n 1 t 2 2 . b. Find the cost of production on a day when the assembly line was running for 16 hours. Interpret your answer. 74.  Business. If the quantity of a product manufactured during a day is given by and the cost of manufacturing the product is given by C 1 n 2 5 8n 1 2375

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a. Find a function that gives the cost of manufacturing the product in terms of the number of hours t the assembly line was functioning, C 1 n 1 t 2 2 . b. Find the cost of production on a day when the assembly line was running for 24 hours. Interpret your answer.

In Exercises 75 and 76, refer to the following: Surveys performed immediately following an accidental oil spill at sea indicate the oil moved outward from the source of the spill in a nearly circular pattern. The radius of the oil spill r measured in miles is a function of time t measured in days from the start of the spill, while the area of the oil spill is a function of radius, that is, the function A 1 r2. 75.  Environment: Oil Spill. If the radius of the oil spill is given by

r 1 t 2 5 10t 2 0.2t 2

and the area of the oil spill is given by



A 1 r 2 5 pr 2



r 1 t 2 5 8t 2 0.1t 2

a. Find a function that gives the area of the oil spill in terms of the number of days since the start of the spill, A 1 r 1 t 2 2 . b. Find the area of the oil spill to the nearest square mile 7 days after the start of the spill. 76.  Environment: Oil Spill. If the radius of the oil spill is given by

a. Find a function that gives the area of the oil spill in terms of the number of days since the start of the spill, A 1 r 1 t 2 2 . b. Find the area of the oil spill to the nearest square mile 5 days after the start of the spill. 77.  Environment: Oil Spill. An oil spill makes a circular pattern around a ship such that the radius in feet grows as a function of time in hours r 1 t 2 5 150!t. Find the area of the spill as a function of time. 78.  Pool Volume. A 20 foot by 10 foot rectangular pool has been built. If 50 cubic feet of water is pumped into the pool per hour, write the water-level height (feet) as a function of time (hours). 79.  Fireworks. A family is watching a fireworks display. If the family is 2 miles from where the fireworks are being launched and the fireworks travel vertically, what is the distance between the family and the fireworks as a function of height above ground? 80.  Real Estate. A couple are about to put their house up for sale. They bought the house for $172,000 a few years ago, and when they list it with a realtor they will pay a 6% commission. Write a function that represents the amount of money they will make on their home as a function of the asking price p.

and the area of the oil spill is given by A 1 r 2 5 pr 2

• C AT C H T H E M I S TA K E In Exercises 81–86, for the functions ƒ1x2 5 x 1 2 and g1x2 5 x2 2 4, find the indicated function and state its domain. Explain the mistake that is made in each problem. 81.

g ƒ

Solution: 

g1x2 x2 2 4 5 x12 ƒ1x2 5

1x 2 22 1x 1 22 x12

5x22

ƒ g

Solution: 

ƒ1x2 x12 5 2 g 1x2 x 24 5

5

1 x12 5 1x 2 22 1x 1 22 x22

1 x22

Domain: 1 2q, 2 2 ∪ 1 2, q 2 This is incorrect. What mistake was made?

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Solution: 

Domain:  12q, q2 This is incorrect. What mistake was made? 82.

83. ƒ + g

ƒ + g 5 ƒ1x2g1x2

5 1 x 1 2 2 1 x2 2 4 2

5 x 3 1 2x 2 2 4x 2 8 Domain:  12q, q2 This is incorrect. What mistake was made? 84. Given the function ƒ1x2 5 x2 1 7 and g 1 x 2 5 !x 2 3, find ƒ + g, and state the domain. Solution: 

2

ƒ + g 5 ƒ 1 g 1 x 2 2 5 A !x 2 3B 1 7 5 ƒ1g1x2 2 5 x 2 3 1 7

5x24 Domain:  12q, q2 This is incorrect. What mistake was made?

85. 1 ƒ 1 g 2 1 2 2 5 1 x 1 2 1 x 2 2 4 2 1 2 2

5 1 x2 1 x 2 2 2 1 2 2 5 2x 2 1 2x 2 4 Domain:  12q, q2 This is incorrect. What mistake was made? 86. ƒ 1 x 2 2 g 1 x 2 5 x 1 2 2 x 2 2 4 5 2x 2 1 x 2 2 Domain:  12q, q2 This is incorrect. What mistake was made?

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• CONCEPTUAL In Exercises 87–90, determine whether each statement is true or false. 87.  When adding, subtracting, multiplying, or dividing two 89. For any functions ƒ and g,  1ƒ + g2 1 x2 exists for all values of functions, the domain of the resulting function is the union of x that are in the domain of g 1 x2, provided the range of g is a the domains of the individual functions. subset of the domain of ƒ. 88. For any functions ƒ and g, ƒ1g 1 x22 5 g 1ƒ1x22 for all values of x 90.  The domain of a composite function can be found by that are in the domain of both ƒ and g. inspection, without knowledge of the domain of the individual functions.

• CHALLENGE 91. For the functions ƒ1x2 5 x 1 a and g 1 x 2 5

g + ƒ and state its domain.

1 , find x2a

92. For the functions ƒ1x2 5 ax2 1 bx 1 c and g 1 x 2 5

find g + ƒ and state its domain.

1 , x2c

93. For the functions ƒ 1 x 2 5 !x 1 a and g 1 x2 5 x2 2 a, find

g + ƒ and state its domain. 1 1 94. For the functions ƒ 1 x 2 5 a and g 1 x 2 5 b , find g + ƒ and x x state its domain. Assume a . 1 and b . 1.

• TECHNOLOGY 95. Using a graphing utility, plot y1 5 !x 1 7 and y2 5 !9 2 x.

Plot y3 5 y1 1 y2. What is the domain of y3?

3 96. Using a graphing utility, plot y1 5 !x 1 5, y2 5



and y3 5

y1 . What is the domain of y3? y2

1 , !3 2 x

98. Using a graphing utility, plot y1 5 !1 2 x, y2 5 x 2 1 2,

and y3 5 y 21 1 2. If y1 represents a function ƒ and y2 represents a function g, then y3 represents the composite function g + ƒ. The graph of y3 is only defined for the domain of g + ƒ. State the domain of g + ƒ.

97. Using a graphing utility, plot y1 5 "x 2 2 3x 2 4,

1 1 , and y3 5 2 . If y1 represents a function x 2 14 y 1 2 14



y2 5



ƒ and y2 represents a function g, then y3 represents the composite function g + ƒ. The graph of y3 is only defined for the domain of g + ƒ. State the domain of g + ƒ.

2

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3.5 ONE-TO-ONE FUNCTIONS AND INVERSE FUNCTIONS SKILLS OBJECTIVES ■■ Determine whether a function is a one-to-one function. ■■ Verify that two functions are inverses of one another. ■■ Graph the inverse function given the graph of the function. ■■ Find the inverse of a function.

CONCEPTUAL OBJECTIVES ■■ Understand why a function that passes the horizontal line test is one-to-one. ■■ Visualize the relationships between the domain and range of a function and the domain and range of its inverse. ■■ Understand why functions and their inverses are symmetric about y 5 x. ■■ Understand why a function has to be one-to-one in order for its inverse to exist.

Every human being has a blood type, and every human being has a DNA sequence. These are examples of functions, where a person is the input and the output is blood type or DNA sequence. These relationships are classified as functions because each person can have one and only one blood type or DNA strand. The difference between these functions is that many people have the same blood type, but DNA is unique to each individual. Can we map backwards? For instance, if you know the blood type, do you know specifically which ­person it came from? No, but, if you know the DNA sequence, you know exactly to which person it corresponds. When a function has a one-to-one correspondence, like the DNA example, then mapping backwards is possible. The map back is called the inverse function.

3.5.1  Determine Whether a Function Is One-to-One 3.5.1 S K I L L

Determine whether a function is a one-to-one function. 3.5.1 C O N C E P T U A L

Understand why a function that passes the horizontal line test is one-to-one.

In Section 3.1, we defined a function as a relationship that maps an input (contained in the domain) to exactly one output (found in the range). Algebraically, each value for x can ­correspond to only a single value for y. Recall the square, identity, absolute value, and ­reciprocal functions from our library of functions in Section 3.3. All of the graphs of these functions satisfy the vertical line test. Although the square function and the absolute value function map each value of x to exactly one value for y, these two functions map two values of x to the same value for y. For example, 121, 12 and  11, 12 lie on both graphs. The identity and reciprocal functions, on the other hand, map each x to a single value for y, and no two x-values map to the same y-value. These two functions are examples of one-to-one functions. DEFINITION

One-to-One Function

A function ƒ1x2 is one-to-one if no two elements in the domain correspond to the same element in the range; that is, if x1 2 x2, then ƒ 1 x1 2 2 ƒ 1 x2 2 . In other words, it is one-to-one if no two inputs map to the same output. EXAMPLE 1   Determining Whether a Function Defined as a Set of Points Is a One-to-One Function

For each of the three relations, determine whether the relation is a function. If it is a function, determine whether it is a one-to-one function. ƒ 5 51 0, 0 2 , 1 1, 1 2 , 1 1, 21 2 6 g 5 51 21, 1 2 , 1 0, 0 2 , 1 1, 1 2 6

h 5 51 21, 21 2 , 1 0, 0 2 , 1 1, 1 2 6

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Solution:

g is a function but not one-to-one.

f is not a function.

h is a one-to-one function.

Just as there is a graphical test for functions, the vertical line test, there is a graphical test for one-to-one functions, the horizontal line test. Note that a horizontal line can be drawn on the square and absolute value functions so that it intersects the graph of each function at two points. The identity and reciprocal functions, however, will intersect a ­horizontal line in at most only one point. This leads us to the horizontal line test for one-to-one functions. Horizontal Line Test

DEFINITION

If every horizontal line intersects the graph of a function in at most one point, then the function is classified as a one-to-one function.

EXAMPLE 2  U  sing the Horizontal Line Test to Determine Whether a Function Is One-to-One

[ CONCEPT CHECK ]

For each of the three relations, determine whether the relation is a function. If it is a function, determine whether it is a one-to-one function. Assume that x is the independent variable and y is the dependent variable. x 5 y2

y 5 x2

Draw a horizontal line at y 5 4 on the graph of y 5 x 2. What are the two points of intersection? Explain why the equation y 5 x 2 cannot be a one-to-one function.

y 5 x3

▼

Solution: 2



2

x5y

y5x

y

y5x

y

5

ANSWER (22, 4) and (2, 4): Two different x values map to the same y value, so this graph fails the horizontal line test.

3

y

10

10

x

x

10

–5

x

–5

–5

5

5

–10

Not a function

Function but not one-to-one

One-to-one function

(fails vertical line test)

(passes vertical line test but fails horizontal line test)

(passes both horizontal and vertical line tests)

▼ Y O U R T U R N   Determine whether each of the functions is a one-to-one function. a. ƒ1x2 5 x 1 2  b. ƒ1x2 5 x2 1 1

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▼ ANSWER a. yes b. no

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CHAPTER 3  Functions and Their Graphs

Another way of writing the definition of a one-to-one function is: If ƒ1x12 5 ƒ1x22, then x1 5 x2.

In the Your Turn following Example 2, we found (using the horizontal line test) that ƒ1x2 5 x 1 2 is a one-to-one function but that ƒ1x2 5 x2 1 1 is not a one-to-one function. We can also use this alternative definition to determine algebraically whether a function is one-to-one. WORDS MATH

State the function. Let there be two real numbers, x1 and x2, such that ƒ1x12 5 ƒ1x22.

Subtract 2 from both sides of the equation.

ƒ1x2 5 x 1 2 x1 1 2 5 x2 1 2 x1 5 x2

ƒ1x2 5 x 1 2 is a one-to-one function.

WORDS MATH

State the function. Let there be two real numbers, x1 and x2, such that ƒ1x12 5 ƒ1x22.

ƒ1x2 5 x2 1 1 x 21 1 1 5 x 22 1 1

Subtract 1 from both sides of the equation.

x 21 5 x 22

Solve for x1.

x1 5 ± x2

ƒ1x2 5 x2 1 2 is not a one-to-one function. EXAMPLE 3  D  etermining Algebraically Whether a Function Is One-to-One

Determine algebraically whether the following functions are one-to-one: a. ƒ1x2 5 5x3 2 2  b. ƒ1x2 5 0 x 1 1 0 Solution (a):

Find ƒ1x12 and ƒ1x22. ƒ1x12 5 5x 31 2 2 and ƒ1x22 5 5x 32 2 2

Let ƒ1x12 5 ƒ1x22. 5x 31 2 2 5 5x 32 2 2 Add 2 to both sides of the equation.

Divide both sides of the equation by 5. Take the cube root of both sides of the equation. Simplify.

5x 31 5 5x 32 x 31 5 x 32

Ax 31 B

1/3

5 Ax 32 B

1/3

x1 5 x2

ƒ1x2 5 5x3 2 2 is a one-to-one function. Solution (b):

Find ƒ1x12 and ƒ1x22. ƒ1x12 5 0 x1 1 1 0 and ƒ1x22 5 0 x2 1 1 0 Let ƒ1x12 5 ƒ1x22.

Solve the absolute value equation.

0 x1 1 1 0 5 0 x2 1 1 0

1x1 1 12 5 1x2 1 12 or 1x1 1 12 5 21x2 1 12

x1 5 x2 or x1 5 2x2 2 2 ƒ1x2 5 0 x 1 1 0 is not a one-to-one function.

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3.5.2  Inverse Functions If a function is one-to-one, then the function maps each x to exactly one y, and no two x-values map to the same y-value. This implies that there is a one-to-one correspondence between the inputs (domain) and outputs (range) of a one-to-one function ƒ 1x2. In the special case of a one-to-one function, it would be possible to map from the output (range of ƒ ) back to the input (domain of ƒ ), and this mapping would also be a function. The function that maps the output back to the input of a function ƒ is called the inverse function and is denoted ƒ21 1x2. A one-to-one function ƒ maps every x in the domain to a unique and distinct ­corresponding y in the range. Therefore, the inverse function ƒ21 maps every y back to a unique and ­distinct x. The function notations ƒ 1x2 5 y and ƒ 21 1 y2 5 x indicate that if the point  1x, y2 satisfies the function, then the point  1 y, x2 satisfies the inverse function. For example, let the function h  1x2 5 5 121, 02,  11, 22,  13, 426.

3.5.2 S K I L L

Verify that two functions are inverses of one another. 3.5.2 C O N C E P T U A L

Visualize the relationships between the domain and range of a function and the domain and range of its inverse. Domain of f f

x Range of f 1

The inverse function undoes whatever the function does. For example, if ƒ1x2 5 5x, then the function ƒ maps any value x in the domain to a value 5x in the range. If we want to map backwards or undo the 5x, we develop a function called the inverse function that takes 5x as input and maps back to x as output. The inverse function is ƒ21 1 x 2 5 15 x. Note that if we input 5x into the inverse function, the output is x: ƒ21 1 5x 2 5 15 1 5x 2 5 x. DEFINITION

Range of f

f

y

1

Domain of f 1

Domain of f

Range of f f (x) = 5x

x

5x

?

f (x) = 5x

x

f 1(5x)

5x

=x

Inverse Function

If ƒ and g denote two one-to-one functions such that ƒ1g 1x22 5 x for every x in the domain of g and

g 1 ƒ 1x22 5 x for every x in the domain of  ƒ ,

then g is the inverse of the function ƒ. The function g is denoted by ƒ21 (read “f-inverse”). Note: ƒ21 is used to denote the inverse of ƒ. The 21 is not used as an exponent and, 1 therefore, does not represent the reciprocal of ƒ: . f Two properties hold true relating one-to-one functions to their inverses: (1) the range of the function is the domain of the inverse, and the range of the inverse is the domain of the function, and (2) the composite function that results with a function and its inverse (and vice versa) is the identity function x.

▼ CAUTION

f 21 2

1 f

Domain of ƒ 5 range of ƒ21 and range of ƒ 5 domain of ƒ21 ƒ21 1ƒ 1x22 5 x  and ƒ 1ƒ21 1x22 5 x

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EXAMPLE 4   Verifying Inverse Functions

Verify that ƒ 21 1 x 2 5 12 x 2 2 is the inverse of ƒ 1x2 5 2x 1 4. Solution:

Show that ƒ 21 1 ƒ 1x22 5 x and ƒ 1  ƒ 21 1x22 5 x.

ƒ 21 1 u 2 5

Write ƒ21 using placeholder notation.

Substitute ƒ1x2 5 2x 1 4 into ƒ21. ƒ 21 1 ƒ 1x22 5

1 1u2 2 2 2

1  12x 1 42 2 2 2

 ƒ 21 1  ƒ 1x22 5 x 1 2 2 2 5 x

Simplify.

Write ƒ using placeholder notation.

 ƒ 21 1 ƒ 1x22 5 x

Substitute ƒ 21 1 x 2 5 12 x 2 2 into ƒ.

1 ƒ 1  ƒ 21 1x22 5 2 a x 2 2b 1 4 2



ƒ 1 u 2 5 21 u 2 1 4

 ƒ 1  ƒ21 1x22 5 x 2 4 1 4 5 x

Simplify.

ƒ 1  ƒ21 1x22 5 x



Note the relationship between the domain and range of ƒ and ƒ21. DOMAIN

ƒ 1x2 5 2x 1 4 f

[ CONCEPT CHECK ] If a one-to-one function f has domain [a, q) and range [b, q) then what are the domain and range of its inverse, f 21?

▼ ANSWER The inverse function, f 21, has domain [b, q) and range [a, q).

21

1x2 5

1 2x

22

RANGE

12 q, q 2

12 q, q 2

12 q, q 2

12 q, q 2

EXAMPLE 5  Verifying Inverse Functions with Domain Restrictions

Verify that ƒ21 1x2 5 x2, for x $ 0, is the inverse of ƒ 1 x 2 5 !x. Solution:

Show that ƒ21 1  ƒ 1x22 5 x and ƒ 1  ƒ21 1x22 5 x.

Write ƒ21 using placeholder notation. Substitute ƒ 1 x 2 5 !x into ƒ21.

ƒ 21 1 ƒ 1 x 2 2 5 1 !x 2 2 5 x

ƒ21 1  ƒ 1x22 5 x for x # 0



ƒ 1u2 5 " 1u2

Write ƒ using placeholder notation. Substitute ƒ21 1x2 5 x2, x $ 0 into ƒ.



ƒ 1 ƒ21 1x22 5 !x2 5 x,  x # 0

ƒ 1 ƒ21 1x22 5 x for x # 0

DOMAIN

ƒ 1 x 2 5 !x

ƒ21 1x2 5 x2, x $ 0

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ƒ21 1u2 5  1u22

30, q2

30, q2

RANGE

30, q2

30, q2

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3.5.3  Graphical Interpretation of Inverse Functions In Example 4, we showed that ƒ 21 1 x 2 5 12 x 2 2 is the inverse of ƒ 1x2 5 2x 1 4. Let’s now investigate the graphs that correspond to the function ƒ and its inverse ƒ 21. f (x)

y

f 21(x)

x

y

x

y

23

22

22

23

22

0

0

22

21

2

2

21

0

4

4

0

(0, 4)

Graph the inverse function given the graph of the function. 3.5.3 C O N C E P T U A L

(–1, 2)

Understand why functions and their inverses are symmetric about y 5 x.

x

(4, 0)

(–2, 0) (–3, –2)

3.5.3 S K I L L

(2, –1) (0, –2) (–2, –3)

Note that the point  123, 222 lies on the function and the point  122, 232 lies on the inverse. In fact, every point  1a, b2 that lies on the function corresponds to a point  1b, a2 that lies on the inverse. Draw the line y 5 x on the graph. In general, the point  1b, a2 on the inverse ƒ21 1x2 is the reflection  1about y 5 x2 of the point  1a, b2 on the function ƒ  1x2. In general, if the point  1a, b2 is on the graph of a function, then the point  1b, a2 is on the graph of its inverse.

STUD Y T I P If the point (a, b) is on the function, then the point (b, a) is on the inverse. Notice the interchanging of the x- and y-coordinates.

EXAMPLE 6  Graphing the Inverse Function

Given the graph of the function ƒ1x2, plot the graph of its inverse ƒ21 1x2.

y

[ CONCEPT CHECK ]

f (x)

x

If the point (b, a ) lies on the graph of a one-to-one function f, then what point must lie on the graph of its inverse, f 21?

▼ ANSWER (a, b)

Solution:

Because the points 123, 222, 122, 02, 10, 22, and 12, 42 lie on the graph of ƒ, then the points  122, 232,  10, 222,  12, 02, and  14, 22 lie on the graph of ƒ21.

y (2, 4) (0, 2) (4, 2) x

(–2, 0) (2, 0) (–3, –2) (0, –2) (–2, –3)

▼ Y O U R T U R N   Given the graph of a function

▼ ANSWER

y

ƒ, plot the inverse function.

y

x

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x

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We have developed the definition of an inverse function and described properties of inverses. At this point, you should be able to determine whether two functions are inverses of one another. Let’s turn our attention to another problem: How do you find the inverse of a function?

3.5.4  Finding the Inverse Function 3.5.4 S K I L L

Find the inverse of a function. 3.5.4 C O N C E P T U A L

Understand why a function has to be one-to-one in order for its inverse to exist. Domain of f x f 1(y) Range of f 1

f

Range of f f (x) y

f 1

Domain of f 1

If the point  1a, b2 lies on the graph of a function, then the point  1b, a2 lies on the graph of the inverse function. The symmetry about the line y 5 x tells us that the roles of x and y interchange. Therefore, if we start with every point  1x, y2 that lies on the graph of a function, then every point  1y, x2 lies on the graph of its inverse. Algebraically, this corresponds to interchanging x and y. Finding the inverse of a finite set of ordered pairs is easy: simply interchange the x- and y-coordinates. Earlier, we found that if h  1x2 5 5 121, 02,  11, 22,  13, 426, then h21  1x2 5 5 10, 212,  12, 12,  14, 326. But how do we find the inverse of a function defined by an equation? Recall the mapping relationship if ƒ is a one-to-one function. This relationship implies that  ƒ 1x2 5 y and ƒ21 1 y2 5 x. Let’s use these two identities to find the inverse. Now consider the function defined by ƒ1x2 5 3x 2 1. To find ƒ21, we let ƒ 1x2 5 y, which yields y 5 3x 2 1. Solve for the variable x:x 5 13 y 1 13. Recall that ƒ21 1 y2 5 x, so we have found the inverse to be ƒ 21 1 y 2 5 13 y 1 13 . It is c­ ustomary to write the independent variable as x, so we write the inverse as ƒ 21 1 x 2 5 13 x 1 13 . Now that we have found the inverse, let’s confirm that the properties ƒ21 1ƒ 1x22 5 x and ƒ 1ƒ 21 1x22 5 x hold. 1 1 ƒAƒ 21 1 x 2 B 5 3 a x 1 b 2 1 5 x 1 1 2 1 5 x 3 3 1 1 1 1 ƒ21 1 ƒ 1 x 2 2 5 1 3x 2 1 2 1 5 x 2 1 5 x 3 3 3 3

FINDING THE INVERSE OF A FUNCTION

Let  ƒ  be a one-to-one function. Then the following procedure can be used to find the inverse function ƒ 21 if the inverse exists. STEP

1

2

3 4

PROCEDURE

Let y 5 ƒ 1x2. Solve the resulting equation for x in terms of y (if possible). Let x 5 ƒ21 1 y2.

Let y 5 x 1interchange x and y2.

EXAMPLE

 

ƒ 1 x 2 5 23x 1 5

   y 5 23x 1 5    3x 5 2y 1 5    x 5 213 y 1

5 3

  ƒ 21 1 y 2 5 213 y 1

5 3

  ƒ 21 1 x 2 5 213 x 1

5 3

The same result is found if we first interchange x and y and then solve for y in terms of x. STEP

EXAMPLE

1

Let y 5 ƒ 1x2.

   y 5 23x 1 5

2

Interchange x and y.

   x 5 23y 1 5

3

Solve for y in terms of x.

   3y 5 2x 1 5

4

Young_AT_6160_ch03_pp298-316.indd 304

PROCEDURE

Let y 5 ƒ21 1x2

 

ƒ 1 x 2 5 23x 1 5

   y 5 213 x 1

5 3

ƒ 21 1 x 2 5 213 x 1

5 3

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305

Note the following: Verify first that a function is one-to-one prior to finding an inverse (if it is not one-to-one, then the inverse does not exist). ■■ State the domain restrictions on the inverse function. The domain of ƒ is the range of ƒ21 and vice versa. ■■ To verify that you have found the inverse, show that ƒ 1 ƒ21 1x22 5 x for all x in the domain of ƒ21 and ƒ21 1 ƒ 1x22 5 x for all x in the domain of ƒ. ■■

EXAMPLE 7   The Inverse of a Square Root Function

Find the inverse of the function ƒ 1 x 2 5 !x 1 2. State the domain and range of both ƒ and ƒ21. Solution:

ƒ 1x2 is a one-to-one function because it passes the horizontal line test.

y 5

x –5

5

–5

STEP 1  Let

y 5 ƒ 1x2.

STEP 2  Interchange STEP 3  Solve

for y.

  Square

both sides of the equation.

  Subtract STEP 4  Let

x and y.

2 from both sides.

y 5 ƒ21 1x2.

y 5 !x 1 2 x 5 !y 1 2 x2 5 y 1 2 x2 2 2 5 y or y 5 x2 2 2 ƒ21 1x2 5 x2 2 2

Note any domain restrictions. (State the domain and range of both ƒ and ƒ21.)

ƒ: Domain: 322, q2 Range: 30, q2 Range: 322, q2 ƒ : Domain: 30, q2  21

The inverse of ƒ 1 x 2 5 !x 1 2 is   ƒ21 1 x 2 5 x 2 2 2 for x $ 0 .

STUDY T I P

Check:

2

ƒ21 1ƒ 1x22 5 x for all x in the ƒ 21 1 ƒ 1 x 2 2 5 A!x 1 2B 2 2 domain of ƒ.

ƒ 1ƒ21 1x22 5 x for all x in the domain of ƒ 21.

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5 x 1 2 2 2 for x $ 22 5x

ƒ 1 ƒ 21 1 x 2 2 5 "Ax 2 2 2B 1 2 5 "x 2 for x $ 0 5x

Had we ignored the domain and range in Example 7, we would have found the inverse function to be the square function f (x ) 5 x2 2 2, which is not a one-to-one function. It is only when we restrict the domain of the square function that we get a one-to-one function.

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Note that the function ƒ 1 x 2 5 !x 1 2 and its inverse ƒ21 1x2 5 x2 2 2 for x $ 0 are symmetric about the line y 5 x.

y

f –1(x)

5

f (x) –5

x 5

–5

▼ ANSWER

x13 a. ƒ 1 x 2 5 7 Domain:  12q, q2, Range:  12q, q2 21

b. g21 1x2 5 x2 1 1

Domain: 30, q2, Range: 31, q2

▼ YOUR TURN  F  ind the inverse of the given function. State the domain and range

of the inverse function. a. ƒ 1x2 5 7x 2 3  b. g 1 x 2 5 !x 2 1 EXAMPLE 8   A Function That Does Not Have an Inverse Function

Find the inverse of the function ƒ1x2 5 0 x 0 if it exists. Solution:

y

The function ƒ1x2 5 0 x 0 fails the horizontal line test and therefore is not a one-to-one function. Because ƒ is not a one-to-one function, its inverse function does not exist.

f (x) = | x |

x

EXAMPLE 9   Finding the Inverse Function

The function ƒ 1 x 2 5

2 , x 2 23, is a one-to-one function. Find its inverse. x13

Solution:

STEP 1  Let

y 5 ƒ1x2.

STEP 2  Interchange

x and y.

y5

2 x13

x5

2 y13

STEP 3  Solve

for y. Multiply the equation by  1y 1 32.   xA y 1 3B 5 2 Eliminate the parentheses.

xy 1 3x 5 2

Subtract 3x from both sides. Divide the equation by x.

S TU DY T IP The range of a function is equal to the domain of its inverse function.

STEP 4  Let

y5

23x 1 2 2 5 23 1 x x

2 y 5 ƒ21 AxB.   ƒ21 1 x 2 5 23 1 x

Note any domain restrictions on ƒ21 1 x2 . The inverse of the function ƒ 1 x 2 5

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xy 5 23x 1 2

x20

2 2 , x 2 23, is ƒ 21 1 x 2 5 23 1 , x 2 0 . x x13

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3.5  One-to-One Functions and Inverse Functions 

2 5 23 1 1 x 1 3 2 5 x, x 2 23 2 a b x13 2 2 5     ƒAƒ21 1 x 2 B 5 5 x, x 2 0 2 2 a23 1 b 1 3 a b x x ▼ 4 Y O U R T U R N   The function ƒ 1 x 2 5 , x 2 1, is a one-to-one function. Find x 2 1 its inverse. Check:

307

ƒ 21 1 ƒ 1x 2 2 5 23 1

▼ ANSWER

4 ƒ21 1 x 2 5 1 1 , x 2 0 x

Note in Example 9 that the domain of ƒ is 12q , 23 2 ∪ 123, q2 and the domain of ƒ 21 is 12q, 0 2 ∪ 1 0, q 2 . Therefore, we know that the range of ƒ is 12q, 0 2 ∪ 1 0, q 2 , and the range of ƒ 21 is 12q , 23 2 ∪ 123, q2 . EXAMPLE 10   Finding the Inverse of a Piecewise-Defined Function

The function ƒ 1 x 2 5 b

3x x2

[ CONCEPT CHECK ]

x,0 is a one-to-one function. Find its inverse. x$0

Explain why you cannot find the inverse of f (x) 5 x 2 without restricting the domain?

y

Solution:

25

From the graph of ƒ we can make a table with corresponding domain and range values.

f(x) = x2

▼ x

(0, 0)

DOMAIN OF f

12 q, 0 2 3 0, q 2

–5

RANGE OF f

12 q, 0 2 3 0, q 2

ANSWER f (x) = x 2 is not a one-to-one function.

5 f(x) = 3x –25

From this information we can also list domain and range values for ƒ 21. DOMAIN OF f 21

RANGE OF f 21

12 q, 0 2 3 0, q 2

12 q, 0 2 3 0, q 2

ƒ 1 x 2 5 3x on 1 2q, 0 2 ; find ƒ21 1x2 on 1 2q, 0 2 .

y 5 3x

STEP 2  Solve

for x in terms of y.

x 5 3y

STEP 3  Solve

for y.

y 5 13 x

STEP 1  Let

STEP 4  Let

y 5 ƒ 1x2.

y 5 ƒ21 1x2.   ƒ21 1 x 2 5 13 x on 1 2q, 0 2

ƒ 1 x 2 5 x 2 on 3 0, q 2 ; find ƒ21 1x2 on 3 0, q 2 .

STEP 1  Let

y 5 ƒ 1x2.   y 5 x2

STEP 2  Solve

for x in terms of y.   x 5 y2

STEP 3  Solve

for y.

STEP 4  Let

y 5 6 !x

y 5 ƒ21 1x2.   ƒ 1 x 2 5 6 !x 21

range of ƒ21 is 3 0, q 2   ƒ21 1 x 2 5 !x Combining the two pieces yields a piecewise-defined inverse function. 1 x x,0 3 21 c 1 2 ƒ x 5 !x x $ 0 STEP 5  The

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[ S E C T I O N 3 . 5 ]     S U M M A R Y One-to-One Functions Properties of Inverse Functions Each input in the domain corresponds to exactly one output in 1. If ƒ is a one-to-one function, then ƒ21 exists. the range, and no two inputs map to the same output. There are 2. Domain and range three ways to test a function to determine whether it is a one-to-one • Domain of ƒ 5 range of ƒ21 function. • Domain of ƒ21 5 range of ƒ 1. Discrete points: For the set of all points  1a, b2 verify that no 3. Composition of inverse functions y-values are repeated. • ƒ21 1ƒ 1x22 5 x for all x in the domain of ƒ. 2. Algebraic equations: Let ƒ 1x12 5 ƒ 1x22; if it can be shown that • ƒ 1ƒ21 1x22 5 x for all x in the domain of ƒ21. x1 5 x2, then the function is one-to-one. 4. The graphs of ƒ and ƒ21 are symmetric with respect to the 3. Graphs: Use the horizontal line test; if any horizontal line line y 5 x. intersects the graph of the function in more than one point, then the function is not one-to-one. Procedure for Finding the Inverse of a Function 1. Let y 5 ƒ 1x2. 2. Interchange x and y. 3. Solve for y. 4. Let y 5 ƒ21 1x2.

[SEC TION 3.5]  E X E R C I S E S • SKILLS In Exercises 1–16, determine whether the given relation is a function. If it is a function, determine whether it is a one-to-one function. 1. 2.

3.

Domain

Range

Domain

Range

MONTH October January April

AVERAGE TEMPERATURE 78°F 68°F

PERSON Mary

10-DIGIT PHONE # (202) 555-1212 (307) 123-4567 (878) 799-6504

Jason Chester

4.

Domain

Range COURSE GRADE

Chris

PERSON Carrie

Alex

Michael

Morgan

Jennifer

Domain

Range

PERSON Jordan

SPOUSE

Pat Tim

A B

Sean

5. 5 10, 12,  11, 22,  12, 32,  13, 426

7. 5 10, 02,  19, 232,  14, 222,  14, 22,  19, 326

9. 5 10, 12,  11, 02,  12, 12,  122, 12,  15, 42,  123, 426 11.

12.

y

6. 5 10, 222,  12, 02,  15, 32,  125, 2726 8. 5 10, 12,  11, 12,  12, 12,  13, 126

10. 5 10, 02,  121, 212,  122, 282,  11, 12,  12, 826 13.

y

y

(6, 4) (–1, 1)

(–2, –2)

(0, 1)

x

x

x

(2, –2)

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3.5  One-to-One Functions and Inverse Functions 

14.

15.

y

16.

y 10

y 5

(–2, 3) x (2, –1)

x –5

x

5

10

In Exercises 17–24, determine algebraically and graphically whether the function is one-to-one. 1 3 17. ƒ1x2 5 0 x 2 3 0 20. ƒ 1 x 2 5 !x 18. ƒ1x2 5  1x 2 222 1 1 19. ƒ 1 x 2 5 x21 1 21. ƒ1x2 5 x2 2 4 23. ƒ1x2 5 x3 2 1 24. ƒ 1 x 2 5 22. ƒ 1 x 2 5 !x 1 1 x12

In Exercises 25–34, verify that the function ƒ21 1x2 is the inverse of  ƒ 1x2 by showing that  ƒ 1  ƒ21 1x22 5 x and  ƒ21 1 ƒ 1x22 5 x. Graph  ƒ 1x2 and  ƒ21 1x2 on the same axes to show the symmetry about the line y 5 x. x21 x22 25. ƒ 1 x 2 5 2x 1 1; ƒ 21 1 x 2 5 26. ƒ 1 x 2 5 ; ƒ 21 1 x 2 5 3x 1 2 2 3 27. ƒ 1 x 2 5 !x 2 1, x $ 1; ƒ 21 1 x 2 5 x 2 1 1 , x $ 0

28. ƒ 1 x 2 5 2 2 x 2 , x $ 0 ; ƒ 21 1 x 2 5 !2 2 x, x # 2

1 21 1 ; ƒ 1x2 5 , x 2 0 x x 1 1 31. ƒ 1 x 2 5 , x 2 23 ; ƒ 21 1 x 2 5 2 3, x 2 0 2x 1 6 2x 29. ƒ 1 x 2 5 33. ƒ 1 x 2 5

30. ƒ  1x2 5  15 2 x21/3; ƒ21 1x2 5 5 2 x3 32. ƒ 1 x 2 5

x13 3 2 4x , x 2 24 ; ƒ 21 1 x 2 5 ,x21 x14 x21

34. ƒ 1 x 2 5

3 3 , x 2 4; ƒ 21 1 x 2 5 4 2 , x 2 0 x 42x

x25 3x 1 5 , x 2 3 ; ƒ21 1 x 2 5 , x 2 21 32x x11

In Exercises 35–42, graph the inverse of the one-to-one function that is given. 35.

36.

37.

y

38. y

y

y

10

5

(–3, 3) (3, 1)

x

(1, 1)

x

x 10

(–1, –3) x –2

39.

40.

41.

y

–5

42. y

y

5

8

y

10

10

x 5

x –5

x –8

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2

5

–10

x 5

–10

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In Exercises 43–60, the function  ƒ is one-to-one. Find its inverse, and check your answer. State the domain and range of both  ƒ and ƒ21. 43. ƒ 1 x 2 5 x 2 1 44. ƒ 1 x 2 5 7x 45. ƒ 1 x 2 5 23x 1 2 46. ƒ 1 x 2 5 2x 1 3

47. ƒ 1 x 2 5 x 3 1 1

48. ƒ 1 x 2 5 x 3 2 1

52. ƒ 1 x 2 5 2x 2 1 1, x $ 0

53. ƒ 1 x 2 5 1 x 1 2 2 2 2 3, x $ 22

54. ƒ 1 x 2 5 1 x 2 3 2 2 2 2, x $ 3

49. ƒ 1 x 2 5 !x 2 3 55. ƒ 1 x 2 5

58. ƒ 1 x 2 5

50. ƒ 1 x 2 5 !3 2 x

2 x

56. ƒ 1 x 2 5 2

7 x12

59. ƒ 1 x 2 5

3 x

7x 1 1 52x

51. ƒ 1 x 2 5 x 2 2 1, x $ 0 57. ƒ 1 x 2 5

60. ƒ 1 x 2 5

2 32x

2x 1 5 71x

In Exercises 61–64, graph the piecewise-defined function to determine whether it is a one-to-one function. If it is a one-to-one function, find its inverse. 61. G 1 x 2 5 e

0 !x

x,0 x$0

1 62. G 1 x 2 5 • x

• A P P L I C AT I O N S

!x

x,0 x$0

x 63. ƒ 1 x 2 5 • x 3

x

x # 21 x13 21 , x , 1 64. ƒ 1 x 2 5 • 0 x 0 x$1 x2

x # 22 22 , x , 2 x$2

Security, write a function E 1x2 that expresses the student’s take-home pay each week. Find the inverse function E21 1x2. What does the inverse function tell you? 70.  Salary. A grocery store pays you $8 per hour for the first 40 hours per week and time and a half for overtime. Write a piecewise-defined function that represents your weekly earnings E 1x2 as a function of the number of hours worked x. Find the inverse function E 21 1x2. What does the inverse function tell you?

65.  Temperature. The equation used to convert from degrees

Celsius to degrees Fahrenheit is ƒ 1 x 2 5 95 x 1 32. Determine the inverse function ƒ 21 1x2. What does the inverse function represent? 66.  Temperature. The equation used to convert from degrees Fahrenheit to degrees Celsius is C 1 x 2 5 59 1 x 2 32 2 . Determine the inverse function C 21 1x2. What does the inverse function represent? 67.  Budget. The Richmond rowing club is planning to enter the Head of the Charles race in Boston and is trying to figure out how much money to raise. The entry fee is $250 per boat for the first 10 boats and $175 for each additional boat. Find the cost function C 1x2 as a function of the number of boats the club enters x. Find the inverse function that will yield how many boats the club can enter as a function of how much money it will raise. 68.  Long-Distance Calling Plans. A phone company charges $0.39 per minute for the first 10 minutes of a long-distance phone call and $0.12 per minute every minute after that. Find the cost function C 1x2 as a function of the length of the phone call in minutes x. Suppose you buy a “prepaid” phone card that is planned for a single call. Find the inverse function that determines how many minutes you can talk as a function of how much you prepaid. 69.  Salary. A student works at Target making $10 per hour, and the weekly number of hours worked per week x varies. If Target withholds 25% of his earnings for taxes and Social

In Exercises 71–74, refer to the following:

By analyzing available empirical data it was determined that during an illness a patient’s body temperature fluctuated during one 24-hour period according to the function T 1 t 2 5 0.0003 1 t 2 24 2 3 1 101.70

where T represents that patient’s temperature in degrees Fahrenheit and t represents the time of day in hours measured from 12:00 a.m. (midnight). 71.  Health/Medicine. Find the domain and range of the function T  1t2. 72.  Health/Medicine. Find time as a function of temperature, that is, the inverse function t  1T2. 73.  Health/Medicine. Find the domain and range of the function t 1T2 found in Exercise 72. 74.  Health/Medicine. At what time, to the nearest hour, was the patient’s temperature 99.5°F?

• C AT C H T H E M I S TA K E In Exercises 75–78, explain the mistake that is made. 75. Is x 5 y2 a one-to-one function? Solution: Yes, this graph represents a one-to-one function because it passes the horizontal line test.

y

x

This is incorrect. What mistake was made?

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3.5  One-to-One Functions and Inverse Functions 

76.  A linear one-to-one function is graphed below. Draw its



inverse. Solution:  Note that the points  13, 32 and  10, 242 lie on the graph of the function.

y

By symmetry, the points  123, 232 and  10, 42 lie on the graph of the inverse.

311

y (0, 4) (3, 3) x

(3, 3) x

(–3, –3) (0, –4)

(0, –4)

77. Given the function ƒ 1x2 5 x2, find the inverse function ƒ21 1x2.

Solution:

STEP 2:  Solve for x. x 5 !y STEP 3:  Interchange x and y. y 5 !x

STEP 4:  Let y 5 ƒ21 1x2. ƒ 21 1 x 2 5 !x



This is incorrect. What mistake was made?

78. Given the function ƒ 1 x 2 5 !x 2 2, find the inverse function

ƒ21 1x2, and state the domain restrictions on ƒ21 1x2.

Solution:

STEP 1:  Let y 5 ƒ 1x2. y 5 x2





2

Check: ƒAƒ 21 1 x 2 B 5 A !x B 5 x and ƒ 21 1 ƒ 1 x 2 2 5 "x 2 5 x. The inverse of ƒ 1x2 5 x2 is ƒ 21 1 x 2 5 !x. This is incorrect. What mistake was made?

STEP 1:  Let y 5 ƒ 1x2.

STEP 2:  Interchange x and y.

y 5 !x 2 2

STEP 3:  Solve for y.

y 5 x2 1 2

x 5 !y 2 2

STEP 4:  Let ƒ 21 1x2 5 y. ƒ21 1x2 5 x2 1 2 STEP 5:  Domain restrictions: ƒ 1 x 2 5 !x 2 2 has the





domain restriction that x $ 2. The inverse of ƒ 1 x 2 5 !x 2 2 is ƒ21 1x2 5 x2 1 2.

The domain of ƒ21 1x2 is x $ 2.

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 79–82, determine whether each statement is true or false. 79. Every even function is a one-to-one function. 80. Every odd function is a one-to-one function. 81. It is not possible that ƒ 5 ƒ21.

• CHALLENGE 85. The unit circle is not a function. If we restrict ourselves to the

semicircle that lies in quadrants I and II, the graph represents a function, but it is not a one-to-one function. If we further restrict ourselves to the quarter circle lying in quadrant I, the graph does represent a one-to-one function. Determine the equations of both the one-to-one function and its inverse. State the domain and range of both.

82. A function ƒ has an inverse. If the function lies in quadrant II,

then its inverse lies in quadrant IV. 83. If  10, b2 is the y-intercept of a one-to-one function ƒ, what is

the x-intercept of the inverse ƒ21? 84. If  1a, 02 is the x-intercept of a one-to-one function ƒ, what is the y-intercept of the inverse ƒ21? 86. Find the inverse of ƒ 1 x 2 5

c , c 2 0. x

87. Under what conditions is the linear function ƒ 1x2 5 mx 1 b a

one-to-one function?

88. Assuming that the conditions found in Exercise 87 are met,

determine the inverse of the linear function.

• TECHNOLOGY In Exercises 89–92, graph the following functions and determine whether they are one-to-one. 89. ƒ 1x2 5 0 4 2 x2 0 91. ƒ 1x2 5 x1/3 2 x5

90.  ƒ 1 x 2 5 92.  ƒ 1 x 2 5

3 x3 1 2 1 x 1/2

In Exercises 93–96, graph the functions ƒ and g and the line y 5 x in the same screen. Do the two functions appear to be inverses of each other? x2 5 93.  ƒ 1 x 2 5 !3x 2 5; g 1 x 2 5 1 3 3 4 x2 94.  ƒ 1 x 2 5 !4 2 3x; g 1 x 2 5 2 , x $ 0 3 3 95. ƒ 1x2 5  1x 2 721/3 1 2;  g 1x2 5 x3 2 6x2 1 12x 2 1 3

96.  ƒ 1 x 2 5 ! x 1 3 2 2;

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g 1 x 2 5 x 3 1 6x 2 1 12x 1 6

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3.6 MODELING FUNCTIONS USING VARIATION SKILLS OBJECTIVES ■■ Develop mathematical models using direct variation. ■■ Develop mathematical models using inverse variation. ■■ Develop mathematical models using joint variation and combined variation.

CONCEPTUAL OBJECTIVES ■■ Understand that direct variation implies that two things grow with one another (as one increases, the other increases). ■■ Understand that inverse variation implies that two things grow opposite of one another (as one increases, the other decreases). ■■ Understand the difference between combined variation and joint variation.

In this section, we discuss mathematical models for different applications. Two quantities in the real world often vary with respect to one another. Sometimes they vary directly. For example, the more money we make, the more total dollars of federal income tax we expect to pay. Sometimes quantities vary inversely. For example, when interest rates on mortgages decrease, we expect the number of homes purchased to increase because a buyer can afford “more house” with the same mortgage payment when rates are lower. In this section, we discuss quantities varying directly, inversely, and jointly.

3.6.1  Direct Variation 3.6.1 S K I L L

Develop mathematical models using direct variation.

When one quantity is a constant multiple of another quantity, we say that the quantities are directly proportional to one another.

3.6.1 C O N C E P T U A L

DIRECT VARIATION

Understand that direct variation implies that two things grow with one another (as one increases, the other increases).

Let x and y represent two quantities. The following are equivalent statements: y 5 kx, where k is a nonzero constant. ■■ y varies directly with x. ■■ y is directly proportional to x. ■■

The constant k is called the constant of variation or the constant of proportionality.

In 2015, the national average cost of residential electricity was 13.06 ¢/kWh (cents per kilowatt-hour). For example, if a residence used 3400 kWh, then the bill would be $444.04, and if a residence used 2500 kWh, then the bill would be $326.50.

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3.6  Modeling Functions Using Variation 

313

EXAMPLE 1   Finding the Constant of Variation

In the United States, the cost of electricity is directly proportional to the number of kilowatt-hours (kWh) used. If a household in Tennessee on average used 3098 kWh per month and had an average monthly electric bill of $320.02, find a mathematical model that gives the cost of electricity in Tennessee in terms of the number of kilowatt-hours used. Solution:

Write the direct variation model.

y 5 kx

Label the variables and constant.

x 5 number of kWh y 5 cost (dollars) k 5 cost per kWh

Substitute the given data x 5 3098 kWh and y 5 $320.02 into y 5 kx.       320.02 5 3098k 320.02 5 0.1033 3098

Solve for k.

k5



y 5 0.1033x

In Tennessee the cost of electricity is  10.33 ¢/kWh .

▼ Y O U R T U R N   Find a mathematical model that describes the cost of electricity in

California if the cost is directly proportional to the number of kWh used and a residence that consumes 4000 kWh is billed $735.20.

▼ ANSWER

y 5 18.38x; the cost of electricity in California is 18.38 ¢/kWh.

Not all variation we see in nature is direct variation. Isometric growth, where the various parts of an organism grow in direct proportion to each other, is rare in living organisms. If organisms grew isometrically, young children would look just like adults, only smaller. In contrast, most organisms grow nonisometrically; the various parts of organisms do not increase in size in a one-to-one ratio. The relative proportions of a human body change dramatically as the human grows. Children have proportionately larger heads and shorter legs than adults. Allometric growth is the pattern of growth whereby different parts of the body grow at different rates with respect to each other. Some human body characteristics vary directly, and others can be mathematically modeled by direct variation with powers. DIRECT VARIATION WITH POWERS

Let x and y represent two quantities. The following are equivalent statements: y 5 kx n, where k is a nonzero constant. ■■ y varies directly with the nth power of x. ■■

■■

y is directly proportional to the nth power of x.

One example of direct variation with powers is height and weight of humans. Weight (in pounds) is directly proportional to the cube of height (feet). W 5 kH 3 EXAMPLE 2   Direct Variation with Powers

The following is a personal ad: Single professional male (6 ft/194 lb) seeks single professional female for long-term relationship. Must be athletic, smart, like the movies and dogs, and have height and weight similarly proportioned to mine.

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[ CONCEPT CHECK ] Students’ final grades in College Algebra are directly proportional to: (A) their financial aid amount or (B) how many hours they spend doing homework and in their instructor’s office hours.

▼ ANSWER B

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Find a mathematical equation that describes the height and weight of the male who wrote the ad. How much would a 5'6" woman weigh who has the same proportionality as the male? Solution:

Write the direct variation (cube) model for height versus weight.

W 5 kH  3

Substitute the given data W 5 194 and H 5 6 into W 5 kH 3. 194 5 k 162 3

Solve for k.

k5

194 5 0.898148 < 0.90 216

W 5 0.9H 3



W 5 0.9 15.523 5 149.73

Let H 5 5.5 ft.

A woman 5960 tall with the same height and weight proportionality as the male would weigh 150 lb . ▼ ANSWER

200 pounds

▼ Y O U R T U R N   A brother and sister both have weight (pounds) that varies as the

cube of height (feet) and they share the same proportionality constant. The sister is 6 feet tall and weighs 170 pounds. Her brother is 6 feet 4 inches. How much does he weigh?

3.6.2  Inverse Variation 3.6.2 S K I L L

Develop mathematical models using inverse variation. 3.6.2 C O N C E P T U A L

Understand that inverse variation implies that two things grow opposite of one another (as one increases, the other decreases).

Two fundamental topics covered in economics are supply and demand. Supply is the quantity that producers are willing to sell at a given price. For example, an artist may be willing to paint and sell five portraits if each sells for $50, but that same artist may be willing to sell 100 portraits if each sells for $10,000. Demand is the quantity of a good that consumers are not only willing to purchase but also have the capacity to buy at a given price. For example, consumers may purchase 1 billion Big Macs from McDonald’s every year, but perhaps only 1 million filet mignons are sold at Outback. There may be 1 billion people who want to buy the filet mignon but don’t have the financial means to do so. Economists study the equilibrium between supply and demand. Demand can be modeled with an inverse variation of price: when the price increases, demand decreases, and vice versa.

INVERSE VARIATION

Let x and y represent two quantities. The following are equivalent statements: k ■■ y 5 , where k is a nonzero constant. x ■■ y varies inversely with x. ■■ y is inversely proportional to x. The constant k is called the constant of variation or the constant of ­proportionality.

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3.6  Modeling Functions Using Variation 

315

[ CONCEPT CHECK ]

EXAMPLE 3   Inverse Variation 1000 Demand (number of potential buyers)

The number of potential buyers of a house decreases as the price of the house increases (see graph on the right). If the number of potential buyers of a house in a particular city is inversely proportional to the price of the house, find a mathematical equation that describes the demand for houses as it relates to price. How many potential buyers will there be for a $2 million house?

(100, 1000)

800

Given a fixed distance, the time it takes you to drive that distance varies inversely with ________?

▼

600

(200, 500)

400

ANSWER rate (or speed)

(400, 250) (600, 167)

200

200 400 600 800 Price of the house (in thousands of dollars)

Solution:

Write the inverse variation model. Label the variables and constant. Select any point that lies on the curve. Substitute the given data x 5 200 k and y 5 500 into y 5 . x

k x x 5 price of house in thousands of dollars y5

y 5 number of buyers  1200, 5002 500 5

k 200

Solve for k.   k 5 200 ? 500 5 100,000

y5

100,000 x

Let x 5 2000.

y5

100,000 5 50 2000

There are only 50 potential buyers for a $2 million house in this city.

▼ Y O U R T U R N   In New York City, the number of potential buyers in the housing

market is inversely proportional to the price of a house. If there are 12,500 potential buyers for a $2 million condominium, how many potential buyers are there for a $5 million condominium?

▼ ANSWER

5000

Two quantities can vary inversely with the nth power of x.

k , then we say that y varies inversely xn with the nth power of x, or y is inversely proportional to the nth power of x.

If x and y are related by the equation y 5

3.6.3  Joint Variation and Combined Variation We now discuss combinations of variations. When one quantity is proportional to the product of two or more other quantities, the variation is called joint variation. When direct variation and inverse variation occur at the same time, the variation is called combined variation.

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3.6.3 S K I L L

Develop mathematical models using joint variation and combined variation. 3.6.3 C O N C E P T U A L

Understand the difference between combined variation and joint variation.

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An example of a joint variation is simple interest (Section 1.2), which is defined as I 5 Prt where I is the interest in dollars P is the principal (initial) dollars ■■ r is the interest rate (expressed in decimal form) ■■ t is time in years ■■ ■■

The interest earned is proportional to the product of three quantities (principal, interest rate, and time). Note that if the interest rate increases, then the interest earned also increases. Similarly, if either the initial investment (principal) or the time the money is invested increases, then the interest earned also increases. An example of combined variation is the combined gas law in chemistry, P5k where

T V

P is pressure T is temperature (kelvins) ■■ V is volume ■■ k is a gas constant ■■ ■■

This relation states that the pressure of a gas is directly proportional to the temperature and inversely proportional to the volume containing the gas. For example, as the temperature increases, the pressure increases, but when the volume decreases, pressure increases. As an example, the gas in the headspace of a soda bottle has a fixed volume. Therefore, as temperature increases, the pressure increases. Compare the different pressures of opening a twist-off cap on a bottle of soda that is cold versus one that is hot. The hot one feels as though it “releases more pressure.”

[ CONCEPT CHECK ] A 5 12 bh,

The area of a triangle, is an example of what type of variation? (A) Combined; (B) Joint

▼ ANSWER B

EXAMPLE 4   Combined Variation

The gas in the headspace of a soda bottle has a volume of 9.0 ml, pressure of 2 atm (atmospheres), and a temperature of 298 K (standard room temperature of 778F). If the soda bottle is stored in a refrigerator, the temperature drops to approximately 279 K (428F). What is the pressure of the gas in the headspace once the bottle is chilled? Solution:

Write the combined gas law.

P5k

T V

Let P 5 2 atm, T 5 298 K, and V 5 9.0 ml.

25k

298 9

Solve for k.

k5

18 298

P5

18 279 ⋅ < 1.87 298 9

Let k 5

18 T , T 5 279, and V 5 9.0 in P 5 k . 298 V

Since we used the same physical units for both the chilled and room-temperature soda bottles, the pressure is in atmospheres.

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P 5 1.87 atm

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3.6  Modeling Functions Using Variation 

317

[ S E C T I O N 3 . 6]     S U M M A R Y Direct, inverse, joint, and combined variation can be used to Joint variation occurs when one quantity is directly proportional model the relationship between two quantities. For two quantities to two or more quantities. Combined variation occurs when one x and y, we say that quantity is directly proportional to one or more quantities and inversely proportional to one or more other quantities. • y is directly proportional to x if y 5 kx. k • y is inversely proportional to x if y 5 . x

[ S E C T I O N 3 . 6]   E X E R C I S E S • SKILLS In Exercises 1–16, write an equation that describes each variation. Use k as the constant of variation. 1. y varies directly with x.  2. s varies directly with t. 3. V varies directly with x3.

 4. A varies directly with x2.

5. z varies directly with m.

 6. h varies directly with !t.

7. ƒ varies inversely with l.

 8. P varies inversely with r2.

9. F varies directly with w and inversely with L.

10. V varies directly with T and inversely with P.

11. v varies directly with both g and t.

12. S varies directly with both t and d.

13. R varies inversely with both P and T.

14. y varies inversely with both x and z.

15. y is directly proportional to the square root of x.

16. y is inversely proportional to the cube of t.

In Exercises 17–36, write an equation that describes each variation. 17. d is directly proportional to t. d 5 r when t 5 1. 18. F is directly proportional to m. F 5 a when m 5 1. 19. V is directly proportional to both l and w. V 5 6h when w 5 3 and l 5 2. 20. A is directly proportional to both b and h. A 5 10 when b 5 5 and h 5 4. 21. A varies directly with the square of r. A 5 9p when r 5 3. 22. V varies directly with the cube of r. V 5 36p when r 5 3.

4 . p 24. W is directly proportional to both R and the square of I. W 5 4 when R 5 100 and I 5 0.25. 23. V varies directly with both h and r 2. V 5 1 when r 5 2 and h 5 25. V varies inversely with P. V 5 1000 when P 5 400. 26. I varies inversely with the square of d. I 5 42 when d 5 16. 27. F varies inversely with both l and L. F 5 20p/m2 when l 5 1 mm and L 5 100 kilometers. 28. y varies inversely with both x and z. y 5 32 when x 5 4 and z 5 0.05. 29. t varies inversely with s. t 5 2.4 when s 5 8. 30. W varies inversely with the square of d. W 5 180 when d 5 0.2. 31. R varies inversely with the square of I. R 5 0.4 when I 5 3.5. 32. y varies inversely with both x and the square root of z. y 5 12 when x 5 0.2 and z 5 4. 33. R varies directly with L and inversely with A. R 5 0.5 when L 5 20 and A 5 0.4. 34. F varies directly with m and inversely with d. F 5 32 when m 5 20 and d 5 8. 35. F varies directly with both m1 and m2 and inversely with the square of d. F 5 20 when m1 5 8, m2 5 16, and d 5 0.4. 36. w varies directly with the square root of g and inversely with the square of t. w 5 20 when g 5 16 and t 5 0.5.

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CHAPTER 3  Functions and Their Graphs

• A P P L I C AT I O N S 37.  Wages. Jason and Valerie both work at Panera Bread and

by about the golden (Fibonacci) ratio. Find an equation that represents the ratio of each section of your finger related to the previous one if one section is eight units long and the next section is five units long.

have the following paycheck information for a certain week. Find an equation that shows their wages W varying directly with the number of hours worked H. EMPLOYEE

HOURS WORKED

WAGES

Jason

23

$172.50

Valerie

32

$240.00

38.  Sales Tax. The sales tax in Orange and Seminole counties in

Florida differs by only 0.5%. A new resident knows this but  doesn’t know which of the counties has the higher tax. The resident lives near the border of the counties and is in the market  for a new plasma television and wants to purchase it in the county with the lower tax. If the tax on a pair of $40 sneakers is $2.60 in Orange County and the tax on a $12 ­T-shirt is $0.84 in Seminole County, write two equations: one for each county that describes the tax T, which is directly proportional to the purchase price P. For Exercises 39 and 40, refer to the following: The ratio of the speed of an object to the speed of sound determines the Mach number. Aircraft traveling at a subsonic speed (less than the speed of sound) have a Mach number less than 1. In other words, the speed of an aircraft is directly proportional to its Mach number. Aircraft traveling at a supersonic speed (greater than the speed of sound) have a Mach number greater than 1. The speed of sound at sea level is approximately 760 miles per hour. 39. Military. The U.S. Navy Blue Angels fly F-18 Hornets that are capable of Mach 1.7. How fast can F-18 Hornets fly at sea level? 40. Military. The U.S. Air Force’s newest fighter aircraft is the F-35, which is capable of Mach 1.9. How fast can an F-35 fly at sea level?

Kim Steele/Getty Images, Inc.

Exercises 41 and 42 are examples of the golden ratio, or phi, a proportionality constant that appears in nature. The numerical approximate value of phi is 1.618. From www.goldenratio.net. 41. Human Anatomy. The length of your forearm F (wrist to elbow) is directly proportional to the length of your hand H (length from wrist to tip of middle finger). Write the equation that describes this relationship if the length of your forearm is 11 inches and the length of your hand is 6.8 inches.

42. Human Anatomy. Each section of your index finger, from

the tip to the base of the wrist, is larger than the preceding one

Young_AT_6160_ch03_pp317-329.indd 318

1

2

2

3

4 3

5

6

7

8 5

9

10 11 12 13 14 15 16 17 18 8

For Exercises 43 and 44, refer to the following: Hooke’s law in physics states that if a spring at rest (equilibrium position) has a weight attached to it, then the distance the spring stretches is directly proportional to the force (weight), according to the formula: F 5 kx where F is the force in Newtons (N), x is the distance stretched in meters (m), and k is the spring constant (N/m).

Equilibrium position

2x

x F

2F

43. Physics. A force of 30 N will stretch the spring 10 centimeters.

How far will a force of 72 N stretch the spring? 44. Physics. A force of 30 N will stretch the spring

10 centimeters. How much force is required to stretch the spring 18 centimeters? 45. Business. A cell phone company develops a pay-as-you-go cell phone plan in which the monthly cost varies directly as the number of minutes used. If the company charges $17.70 in a month when 236 minutes are used, what should the company charge for a month in which 500 minutes are used? 46. Economics. Demand for a product varies inversely with the price per unit of the product. Demand for the product is 10,000 units when the price is $5.75 per unit. Find the demand for the product (to the nearest hundred units) when the price is $6.50. 47. Sales. Levi’s makes jeans in a variety of price ranges for juniors. The Flare 519 jeans sell for about $20, whereas the 646 Vintage Flare jeans sell for $300. The demand for Levi’s jeans is inversely proportional to the price. If 300,000 pairs of the 519 jeans were bought, approximately how many of the Vintage Flare jeans were bought? 48. Sales. Levi’s makes jeans in a variety of price ranges for men. The Silver Tab Baggy jeans sell for about $30, whereas the Offender jeans sell for about $160. The demand for Levi’s jeans is inversely proportional to the price. If 400,000 pairs of the Silver Tab Baggy jeans were bought, approximately how many of the Offender jeans were bought?

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3.6  Modeling Functions Using Variation 

For Exercises 49 and 50, refer to the following: In physics, the inverse square law states that any physical quantity or strength is inversely proportional to the square of the distance from the source of that physical quantity. In particular, the i­ ntensity of light radiating from a point source is inversely ­proportional to the square of the distance from the source. Below is a table of average distances from the Sun:

PLANET

DISTANCE TO THE SUN

Mercury

58,000 km

Earth

150,000 km

Mars

228,000 km

49. Solar Radiation. The solar radiation on the Earth is

approximately 1400 watts per square meter 1 w/m2 2 . How much solar radiation is there on Mars? Round to the nearest ­hundred watts per square meter. 50. Solar Radiation. The solar radiation on the Earth is ­approximately 1400 watts per square meter. How much solar ­radiation is there on Mercury? Round to the nearest hundred watts per square meter.

319

51. Investments. Marilyn receives a $25,000 bonus from her

company and decides to put the money toward a new car that she will need in two years. Simple interest is directly proportional to the principal and the time invested. She ­ compares two different banks’ rates on money market ­ accounts. If she goes  with Bank of America, she will earn $750 in interest, but if she goes with the Navy Federal Credit Union, she will earn $1500. What is the interest rate on money market accounts at both banks? 52. Investments. Connie and Alvaro sell their house and buy a fixer-upper house. They made $130,000 on the sale of their previous home. They know it will take 6 months before the general contractor will start their renovation, and they want to take advantage of a 6-month CD that pays simple interest. What is the rate of the 6-month CD if they will make $3250 in interest? 53. Chemistry. A gas contained in a 4-milliliter container at a ­temperature of 300 K has a pressure of 1 atmosphere. If the ­temperature decreases to 275 K, what is the resulting pressure? 54. Chemistry. A gas contained in a 4-milliliter container at a temperature of 300 K has a pressure of 1 atmosphere. If the container changes to a volume of 3 millileters, what is the resulting pressure?

• C AT C H T H E M I S TA K E In Exercises 55 and 56, explain the mistake that is made. 55.  y varies directly with t and indirectly with x. When x 5 4 and t 5 2, then y 5 1. Find an equation that describes this variation.

56.  y varies directly with t and the square of x. When x 5 4

and t 5 1, then y 5 8. Find an equation that describes this variation. Solution: Write the variation equation.

Solution: Write the variation equation.  y 5 ktx Let x 5 4, t 5 2, and y 5 1. 1 5 k 122 142 1 Solve for k. k5 8 1 1 Substitute k 5 8 into y 5 ktx. y 5 tx 8 This is incorrect. What mistake was made?

y 5 kt!x

Let x 5 4, t 5 1, and y 5 8.  8 5 k 1 1 2 !4 Solve for k.

k54

Substitute k 5 4 into y 5 kt!x.

y 5 4t!x

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 57 and 58, determine whether each statement is true or false. 57. The area of a triangle is directly proportional to both the base and the height of the triangle (joint variation). 58. Average speed is directly proportional to both distance and time (joint variation).

In Exercises 59 and 60, match the variation with the graph. 59. Inverse variation 60. Direct variation

a.

b.

y 10

y 10

x 10

Young_AT_6160_ch03_pp317-329.indd 319

x 10

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CHAPTER 3  Functions and Their Graphs

• CHALLENGE Exercises 61 and 62 involve the theory governing laser propagation through the Earth’s atmosphere. The three parameters that help classify the strength of optical ­turbulence are: C 2n, index of refraction structure parameter ■■ k, wave number of the laser, which is inversely proportional to the wavelength l of the laser: ■■

k5 ■■

61. When C 2n 5 1.0 3 10213 m22/3, L 5 2 km, and l 5 1.55 mm,

the variance of irradiance for a plane wave s2pl is 7.1. Find the equation that describes this variation. 62. When C 2n 5 1.0 3 10213 m22/3, L 5 2 km, and l 5 1.55 mm, the variance of irradiance for a spherical wave s2sp is 2.3. Find the equation that describes this variation.

2p l

L, propagation distance

The variance of the irradiance of a laser s2 is directly proportional to C 2n, k7/6, and L11/16.

• TECHNOLOGY For Exercises 63–66, refer to the following: Data from 1995 to 2006 for oil prices in dollars per barrel, the U.S. Dow Jones Utilities Stock Index, New Privately Owned Housing, and 5-year Treasury Constant Maturity Rate are given in the table. (Data are from Forecast Center’s Historical Economic and Market Home Page at www.neatideas.com/djutil.htm.)

JANUARY OF EACH YEAR

OIL PRICE, $ PER BARREL

  Use the calculator stat edit commands to enter the table with L1 as the oil price, L2 as the utilities stock index, L3 as number ­ aturity rate. of housing units, and L4 as the 5-year m

U.S. DOW JONES UTILITIES STOCK INDEX

NEW, PRIVATELY OWNED HOUSING UNITS

5-YEAR TREASURY CONSTANT MATURITY RATE

1995

17.99

193.12

1407

7.76

1996

18.88

230.85

1467

5.36

1997

25.17

232.53

1355

6.33

1998

16.71

263.29

1525

5.42

1999

12.47

302.80

1748

4.60

2000

27.18

315.14

1636

6.58

2001

29.58

372.32

1600

4.86

2002

19.67

285.71

1698

4.34

2003

32.94

207.75

1853

3.05

2004

32.27

271.94

1911

3.12

2005

46.84

343.46

2137

3.71

2006

65.51

413.84

2265

4.35

63. An increase in oil price in dollars per barrel will drive the

U.S. Dow Jones Utilities Stock Index to soar. a.  Use the calculator commands stat , linReg  1ax 1 b2, and statplot to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the oil price in dollars per barrel. b. If the U.S. Dow Jones Utilities Stock Index varies directly as the oil price in dollars per barrel, then use the calculator commands stat, PwrReg , and statplot to model the data using the power function. Find the variation constant and equation of variation using x as the oil price in dollars per barrel.

Young_AT_6160_ch03_pp317-329.indd 320

c.  Use the equations you found in (a) and (b) to predict the

stock index when the oil price hits $72.70 per barrel in September 2006. Which answer is closer to the actual stock index of 417? Round all answers to the nearest whole number. 64. An increase in oil price in dollars per barrel will affect the interest rates across the board—in particular, the 5-year Treasury constant maturity rate. a.  Use the calculator commands stat , linReg  1ax 1 b2, and statplot to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the oil price in dollars per barrel.

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3.6  Modeling Functions Using Variation 

b. If the 5-year Treasury constant maturity rate varies

321

c.  Use the equations you found in (a) and (b) to predict the

inversely as the oil price in dollars per barrel, then use the calculator commands stat , PwrReg , and statplot to model the data using the power function. Find the variation constant and equation of variation using x as the oil price in dollars per barrel. c.  Use the equations you found in (a) and (b) to predict the maturity rate when the oil price hits $72.70 per barrel in September 2006. Which answer is closer to the actual maturity rate at 5.02%? Round all answers to two decimal places. 65.  An increase in interest rates—in particular, the 5-year Treasury constant maturity rate—will affect the number of new, privately owned housing units. a.  Use the calculator commands stat, linReg  1ax 1 b2, and statplot to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the 5-year rate. b. If the number of new, privately owned housing units varies inversely as the 5-year Treasury constant maturity rate, then use the calculator commands stat , PwrReg , and statplot to model the data using the power function. Find the variation constant and equation of variation using x as the 5-year rate.

number of housing units when the maturity rate is 5.02% in September 2006. Which answer is closer to the actual number of new, privately owned housing units of 1861? Round all answers to the nearest unit. 66. An increase in the number of new, privately owned housing units will affect the U.S. Dow Jones Utilities Stock Index. a.  Use the calculator commands stat, linReg  1ax 1 b2, and statplot to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the number of housing units. b. If the U.S. Dow Jones Utilities Stock Index varies directly as the number of new, privately owned housing units, then use the calculator commands stat, PwrReg , and statplot to model the data using the power function. Find the variation constant and equation of variation using x as the number of housing units. c.  Use the equations you found in (a) and (b) to predict the utilities stock index if there are 1861 new, privately owned housing units in September 2006. Which answer is closer to the actual stock index of 417? Round all answers to the nearest whole number.

For Exercises 67 and 68, refer to the following: Data for retail gasoline price in dollars per gallon for the period March 2006 to March 2015 are given in the following table. (Data are from Energy Information Administration, Official Energy Statistics from the U.S. government at http://tonto.eia.doe. gov/oog/info/gdu/gaspump.html.) Use the calculator stat edit command to enter the table below with L1 as the year  1x 5 1 for year 20062 and L2 as the gasoline price in dollars per gallon. MARCH OF EACH YEAR

2006

2007

2008

2009

2010

2011

2012

2013

2014

2015

RETAIL GASOLINE PRICE $ PER GALLON

2.353

2.599

3.216

1.984

2.768

3.617

3.954

3.888

3.601

2.710

67. a. Use the calculator commands stat LinReg to model the

data using the least-squares regression. Find the equation of the least-squares regression line using x as the year 1x 5 1 for year 20062 and y as the gasoline price in dollars per gallon. Round all answers to three decimal places. b. Use the equation to predict the gasoline price in March 2018. Round all answers to three decimal places. Is the answer close to the actual price? c.  Use the equation to predict the gasoline price in March 2020. Round all answers to three decimal places.

Young_AT_6160_ch03_pp317-329.indd 321

68. a. Use the calculator commands stat PwrReg to model the

data using the power function. Find the v­ ariation constant and equation of variation using x as the year  1x 5 1 for year 20062 and y as the gasoline price in dollars per gallon. Round all answers to three decimal places. b. Use the equation to predict the gasoline price in March 2018. Round all answers to three decimal places. Is the answer close to the actual price? c.  Use the equation to predict the gasoline price in March 2020. Round all answers to three decimal places.

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CHAPTER 3  Functions and Their Graphs

[CHAPTER 3 REVIEW] SECTION

CONCEPT

3.1

Functions

KEY IDEAS/FORMULAS

Relations and functions

All functions are relations, but not all relations are functions.

Functions defined by equations

A vertical line can intersect a function in at most one point.

Function notation

Placeholder notation: ƒ 1 x 2 5 3x 2 2 6x 1 2

CHAPTER 3 REVIEW

Difference quotient:

Domain of a function

3.2

ƒ 1u2 5 3 1u2 2 2 6 1u2 1 2

ƒ1x 1 h2 2 ƒ1x2 ; h20 h Are there any restrictions on x?

Graphs of functions; piecewise-defined functions; increasing and decreasing functions; average rate of change Recognizing and classifying functions

Common functions ƒ1x2 5 mx 1 b, ƒ1x2 5 x, ƒ1x2 5 x2, 3 ƒ 1 x 2 5 x 3, ƒ 1 x 2 5 !x, ƒ 1 x 2 5 ! x,

ƒ1x2 5 0 x 0 , ƒ1x2 5

1 x

Even and odd functions Even: Symmetry about y-axis: ƒ12x2 5 ƒ1x2 Odd: Symmetry about origin: ƒ12x2 5 2ƒ1x2 Increasing and decreasing functions

• Increasing: rises (left to right) • Decreasing: falls (left to right)

3.3

Average rate of change

ƒ 1 x2 2 2 ƒ 1 x1 2    x1 2 x2 x2 2 x1

Piecewise-defined functions

Points of discontinuity

Graphing techniques: Transformations

Shift the graph of ƒ1x2.

Horizontal and vertical shifts

ƒ1x 1 c2

c units to the left, c . 0

ƒ1x 2 c2

c units to the right, c . 0

ƒ1x2 1 c

c units upward, c . 0

ƒ1x2 2 c

c units downward, c . 0

2ƒ1x2

Reflection about the x-axis

ƒ12x2

Reflection about the y-axis

Reflection about the axes Stretching and compressing

cƒ1x2 if c . 1; stretch vertically cƒ1x2 if 0 , c , 1; compress vertically ƒ1cx2 if c . 1; compress horizontally ƒ1cx2 if 0 , c , 1; stretch horizontally

3.4

Operations on functions and composition of functions Adding, subtracting, multiplying, and dividing functions

1 ƒ 1 g2 1x2 5 ƒ1x2 1 g 1x2 1 ƒ 2 g2 1x2 5 ƒ1x2 2 g 1x2 1 ƒ ? g2 1x2 5 ƒ1x2 ? g 1x2

Composition of functions

The domain of the resulting function is the intersection of the individual domains. ƒ1x2 ƒ ,   g1x2 2 0 a b 1x2 5 g g 1x2 The domain of the quotient is the intersection of the domains of ƒ and g, and any points when g 1x2 5 0 must be eliminated. 1 ƒ + g 21 x 2 5 ƒ 1 g 1 x 22

The domain of the composite function is a subset of the domain of g 1x2. Values for x must be eliminated if their corresponding values g 1x2 are not in the domain of ƒ.

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09/12/16 3:00 PM

Chapter Review 

SECTION

CONCEPT

3.5

One-to-one functions and inverse functions Determine whether a function is one-to-one

323

KEY IDEAS/FORMULAS

• No two x-values map to the same y-value. If ƒ 1 x1 2 5 ƒ 1 x2 2 , then x1 5 x2.

• A horizontal line may intersect a one-to-one function in at most one point. Inverse functions

• Only one-to-one functions have inverses. • ƒ21 1ƒ1x22 5 x and ƒ1ƒ21  1x22 5 x.

• Domain of ƒ 5 range of ƒ21. Graphical interpretation of inverse functions

Finding the inverse function

• The graph of a function and its inverse are symmetric about the line y 5 x. • If the point  1a, b2 lies on the graph of a function, then the point  1b, a2 lies on the graph of its inverse. 1. Let y 5 ƒ1x2.

2. Interchange x and y. 3. Solve for y.

3.6

Modeling functions using variation Direct variation Inverse variation Joint variation and combined variation

4. Let y 5 ƒ21 1x2.

CHAPTER 3 REVIEW

Range of ƒ 5 domain of ƒ21.

y 5 kx y5

k x

Joint: One quantity is directly proportional to the product of two or more other quantities. Combined: Direct variation and inverse variation occur at the same time.

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324 

CHAPTER 3  Functions and Their Graphs

[CHAPTER 3 REVIEW EXERCISES] Evaluate the given quantities using the following three ­functions.

3.1  Functions

Determine whether each relation is a function.  1.

Domain

Range

15. ƒ132

NAMES

AGES

Allie

27

Hannah

10

17. ƒ1272 . g 132

Danny

21.

21

Vickie

REVIEW EXERCISES

19.

4

Ethan

F 1t2 5 t2 1 4t 2 3 16.  F 142

ƒ1x2 5 4x 2 7

18. 

ƒ122 2 F122 g102

F102 g102

g 1x2 5 0 x2 1 2x 1 4 0

20.  ƒ 1 3 1 h 2

ƒ13 1 h2 2 ƒ132 h

22. 

F1t 1 h2 2 F1t2 h

 2. 5 11, 22,  13, 42,  12, 42,  13, 726

Find the domain of the given function. Express the domain in interval notation.

 5. x2 1 y2 5 36

25. h 1 x 2 5

 3. 5 122, 32,  11, 232,  10, 42,  12, 626  4. 5 14, 72,  12, 62,  13, 82,  11, 726

23. ƒ1x2 5 23x 2 4

 7. y 5 0 x 1 2 0

27. G 1 x 2 5 !x 2 4

  6. x 5 4

  8. y 5 !x   9.

10. 

y

24.  g 1x2 5 x2 2 2x 1 6

7 x 13 1 28.  H 1 x 2 5 !2x 2 6

1 x14

26.  F 1 x 2 5

Challenge

y

x

x

29. If ƒ 1 x 2 5

2

D , ƒ142 and ƒ1242 are undefined, and x 2 2 16

ƒ 1 5 2 5 2, find D. 30. Construct a function that is undefined at x 5 23 and x 5 2 such that the point  10, 242 lies on the graph of the function. 3.2  Graphs of Functions

Determine whether the function is even, odd, or neither. Use the graphs of the functions to find: 11.

12. 

y

32.  g 1x2 5 7x5 1 4x3 2 2x 34.  ƒ1x2 5 x4 1 3x2 36.  ƒ 1 x 2 5 !x 1 4 1 38.  ƒ 1 x 2 5 2 1 3x 4 1 0 x 0 x

31. ƒ1x2 5 2x 2 7 33. h 1x2 5 x3 2 7x

y

35. ƒ1x2 5 x1/4 1 x

x

x

37. ƒ 1 x 2 5

1 1 3x x3

Use the graph of the functions to find: a. Domain b. Range

c. Intervals on which the function is increasing, decreasing, or

constant.

a. ƒ1242  b. ƒ102 c. x, where ƒ1x2 5 0

a. ƒ1212  b. ƒ112 c.  x, where ƒ1x2 5 0

39.

40. 

y

y

10

13.

14. 

y

10

y

x –10

10

x –10

10

x x –10

–10

41. Find the average rate of change of ƒ1x2 5 4 2 x2 from a. ƒ1222  b. ƒ142 c. x, where ƒ1x2 5 0

Young_AT_6160_ch03_pp317-329.indd 324

a. ƒ1252  b. ƒ102 c. x, where ƒ1x2 5 0

x 5 0 to x 5 2. 42. Find the average rate of change of ƒ1x2 5 0 2x 2 1 0 from x 5 1 to x 5 5.

30/11/16 11:08 AM

Review Exercises 

Graph the piecewise-defined function. State the domain and range in interval notation. 43. F 1 x 2 5 b

x2 2

Write the function whose graph is the graph of y 5 !x, but is transformed accordingly, and state the domain of the ­resulting function.

x,0 x$0

22x 2 3 44. ƒ 1 x 2 5 • 4 x2 1 4

59. Shifted to the left three units 60. Shifted down four units

x#0 0,x#1 x.1

61. Shifted to the right two units and up three units 62. Reflected about the y-axis 63. Stretched by a factor of 5 and shifted down six units

x#0 0,x#1 x.1

x2 46. F 1 x 2 5 • x 3 20 x 0 2 1

64. Compressed by a factor of 2 and shifted up three units

Transform the function into the form ƒ  1x2 5 c 1x 2 h22 1 k by completing the square and graph the resulting function using transformations.

x,0 0,x,1 x$1

65. y 5 x2 1 4x 2 8

3.4  Operations on Functions and Composition

Applications

of Functions

47. Tutoring Costs. A tutoring company charges $25.00 for the

first hour of tutoring and $10.50 for every 30-minute period after that. Find the cost function C 1x2 as a function of the length of the tutoring session. Let x 5 number of 30-minute periods. 48. Salary. An employee who makes $30.00 per hour also earns time and a half for overtime (any hours worked above the normal 40-hour work week). Write a function E 1x2 that describes her weekly earnings as a function of the number of hours worked x. 3.3  Graphing Techniques: Transformations

3 51. y 5 !x 2 3 1 2

1

53. y 5 22 x 3

67. g 1x2 5 23x 2 4 h 1x2 5 x 2 3

68. g 1x2 5 2x 1 3

h 1x2 5 x2 1 6

1 x2

69. g 1 x 2 5

h 1 x 2 5 !x

x13 2x 2 4

50.  y 5 0 2x 1 5 0 2 7

70. g 1 x 2 5

54. y 5 2x2 1 3

71. g 1 x 2 5 !x 2 4

52.  y 5

1 24 x22



Use the given graph to graph the following: 55.

g Given the functions g and h, find g 1 h, g 2 h, g ? h, and , h and state the domain.



Graph the following functions using graphing aids. 49. y 5 2 1x 2 222 1 4

66. y 5 2x2 1 6x 2 5

56. 

y



h1x2 5

3x 2 1 x22

h 1 x 2 5 !2x 1 1

72. g 1 x 2 5 x 2 2 4

y

x

x

REVIEW EXERCISES

x2 45. ƒ 1 x 2 5 • 2"x 0x 1 20

325

h1x2 5 x 1 2

For the given functions ƒ and g, find the composite functions ƒ + g and g + ƒ, and state the domains. 73. ƒ1x2 5 3x 2 4

g 1x2 5 2x 1 1

74. ƒ1x2 5 x3 1 2x 2 1

y 5 ƒ1x 2 22

57.



y 5 3ƒ1x2 58. 

y

y

x

76. ƒ 1 x 2 5 "2x 2 2 5





y 5 22ƒ1x2

Young_AT_6160_ch03_pp317-329.indd 325

y 5 ƒ1x2 1 3

2 x13 1 g1x2 5 42x

75. ƒ 1 x 2 5

x

g 1x2 5 x 1 3

g 1 x 2 5 !x 1 6

77. ƒ 1 x 2 5 !x 2 5



g 1 x 2 5 x2 2 4

30/11/16 11:08 AM

326 

CHAPTER 3  Functions and Their Graphs

78. ƒ 1 x 2 5



g1x2 5

1

 92.

"x

1 x 24 2

Evaluate ƒ  1 g 1322 and g 1ƒ  12122, if possible. 79. ƒ1x2 5 4x2 2 3x 1 2 g 1x2 5 6x 2 3

80. ƒ 1 x 2 5 !4 2 x

g 1x2 5 x2 1 5 x 81. ƒ 1 x 2 5 0 2x 2 3 0

REVIEW EXERCISES





4 x 22 1 g 1x2 5 2 x 29 2

Write the function as a composite ƒ 1  g 1x22 of two functions ƒ and g. 85. h 1x2 5 3 1x 2 222 1 4 1x 2 22 1 7 87. h 1 x 2 5

3 "x

3 1 2 "x 1

"x 2 1 7

88. h 1 x 2 5 " 0 3x 1 4 0

89. Rain. A rain drop hitting a lake makes a circular ripple. If

the radius, in inches, grows as a function of time, in minutes, r 1 t 2 5 25"t 1 2, find the area of the ripple as a function of time. 90. Geometry. Let the area of a rectangle be given by 42 5 l . w, and let the perimeter be 36 5 2 . l 1 2 . w. Express the perimeter in terms of w. 3.5  One-to-One Functions and Inverse Functions

Determine whether the given function is a one-to-one function. Domain

Range

BROTHER Chris

SISTER Paula

Harold

Vickie

Tom

Renee

Danny

Gabriel

Young_AT_6160_ch03_pp317-329.indd 326

1 x2

Verify that the function ƒ21 1x2 is the inverse of ƒ  1x2 by showing that ƒ  1  ƒ 21 1x22 5 x. Graph ƒ  1x2 and ƒ 21 1x2 on the same graph and show the symmetry about the line y 5 x. x24 101. ƒ1x2 5 3x 1 4; ƒ 21 1 x 2 5 3 1 1 1 7x 102. ƒ 1 x 2 5 ; ƒ 21 1 x 2 5 4x 2 7 4x 103. ƒ 1 x 2 5 "x 1 4; ƒ 21 1 x 2 5 x 2 2 4   x $ 0 104. ƒ 1 x 2 5

x 1 2 21 7x 1 2 ; ƒ 1x2 5 x27 x21

The function ƒ is one-to-one. Find its inverse and check your answer. State the domain and range of both ƒ and ƒ21. 105. ƒ1x2 5 2x 1 1

106. ƒ1x2 5 x5 1 2

107. ƒ 1 x 2 5 "x 1 4

108. ƒ1x2 5  1x 1 422 1 3 x > 24

109. ƒ 1 x 2 5

Applications

91.

F

  97. y 5 "x  98.  y 5 x2  99. ƒ1x2 5 x3    100.  ƒ 1 x 2 5

3 g1x2 5 ! x24

86. h 1 x 2 5

C D

 96. 5 128, 262,  124, 22,  10, 32,  12, 282,  17, 426

g 1 x 2 5 x2 2 1 83. ƒ 1 x 2 5 x 2 2 x 1 10



Bill Tracey Tonja Troy Maria Martin

GRADE IN PRECALCULUS COURSE A

 95. 5 122, 02,  14, 52,  13, 726

1 x21

84. ƒ 1 x 2 5

STUDENTS IN PRECALC

Range

 94. 5 123, 92,  15, 252,  12, 42,  13, 926



Function

 93. 5 12, 32,  121, 22,  13, 32,  123, 242,  122, 126

g 1 x 2 5 0 5x 1 2 0

82. ƒ 1 x 2 5

Domain

x16 x13

Applications

3 110.  ƒ 1 x 2 5 2 "x 2528

111. Salary. A pharmaceutical salesperson makes $22,000 base

salary a year plus 8% of the total products sold. Write a function S 1x2 that represents her yearly salary as a function of the total dollars worth of products sold x. Find S21 1x2. What does this inverse function tell you? 112. Volume. Express the volume V of a rectangular box that has a square base of length s and is 3 feet high as a function of the square length. Find V21. If a certain volume is desired, what does the inverse tell you? 3.6  Modeling Functions Using Variation

Write an equation that describes each variation. 113. C is directly proportional to r. C 5 2p when r 5 1. 114. V is directly proportional to both l and w. V 5 12h when

w 5 6 and l 5 2. 115. A varies directly with the square of r. A 5 25p when r 5 5. 116. F varies inversely with both l and L. F 5 20p when

l 5 10 mm and L 5 10 km.

30/11/16 11:08 AM

Review Exercises 

Applications 117. Wages. Cole and Dickson both work at the same museum

and have the following paycheck information for a certain week. Find an equation that shows their wages (W) varying directly with the number of hours (H) worked.

327

126. Using a graphing utility, plot y1 5 "x 2 2 4, y2 5 x 2 2 5,

and y3 5 y21 2 5. If y1 represents a function ƒ and y2 represents a function g, then y3 represents the composite function g + ƒ. The graph of y3 is only defined for the domain of g + ƒ. State the domain of g + ƒ.

HOURS WORKED

WAGE

Section 3.5

Cole

27

$229.50

127. Use a graphing utility to graph the function and determine

Dickson

30

$255.00

EMPLOYEE

118. Sales Tax. The sales tax in two neighboring counties

Technology Exercises

Section 3.1 119. Use a graphing utility to graph the function and find the

domain. Express the domain in interval notation. 1 ƒ1x2 5 2 "x 2 2x 2 3 120. Use a graphing utility to graph the function and find the domain. Express the domain in interval notation. ƒ1x2 5

Section 3.2

x 2 2 4x 2 5 x2 2 9

121. Use a graphing utility to graph the function. State the

(a) domain, (b) range, and (c) x intervals where the function is increasing, decreasing, and constant. 12x 1 2 ƒ x 5 • 3 3x4 4 x11

x , 21 21 # x , 2 x.2

122. Use a graphing utility to graph the function. State the

(a) domain, (b) range, and (c) x intervals where the function is increasing, decreasing, and constant.

Section 3.3

ƒ1x2 5 •

0 x2 2 1 0

!x 2 2 1 4

22 , x , 2 x.2

123. Use a graphing utility to graph ƒ 1 x 2 5 x 2 2 x 2 6 and

g 1 x 2 5 x 2 2 5x. Use transforms to describe the relationship between ƒ 1 x 2 and g 1 x 2 ? 124. Use a graphing utility to graph ƒ 1 x 2 5 2x 2 2 3x 2 5 and g 1 x 2 5 22x 2 2 x 1 6. Use transforms to describe the relationship between ƒ 1 x 2 and g 1 x 2 ? Section 3.4

125. Using a graphing utility, plot y1 5 !2x 1 3, y2 5 !4 2 x,

and y3 5

y1 . What is the domain of y3? y2

Young_AT_6160_ch03_pp317-329.indd 327

ƒ1x2 5

6 5

"x 3 2 1 128. Use a graphing utility to graph the functions ƒ and g and the line y 5 x in the same screen. Are the two functions inverses of each other? 4 ƒ1x2 5 ! x 2 3 1 1, g 1 x 2 5 x 4 2 4x 3 1 6x 2 2 4x 1 3 Section 3.6

From December 2006 to December 2015, data for gold price in dollars per ounce are given in the table below (SOURCE: www .kitco.com/charts/historicalgold.html). Use the calculator stat edit commands to enter the table below with L1 as the year  1x 5 1 for year 20062 and L2 as the gold price in dollars per ounce. DECEMBER OF EACH YEAR

GOLD PRICE PER OUNCE (AVERAGE)

2006

629.79

2007

803.20

2008

816.09

2009

1134.72

2010

1390.55

2011

1652.31

2012

1688.53

2013

1225.40

2014

1202.29

2015

1063.88

REVIEW EXERCISES

differs by 1%. A new resident knows the difference but doesn’t know which county has the higher tax rate. The resident lives near the border of the two counties and wants to buy a new car. If the tax on a $50.00 jacket is $3.50 in County A and the tax on a $20.00 calculator is $1.60 in County B, write two equations (one for each county) that describe the tax (T), which is directly proportional to the purchase price (P).

whether it is one-to-one.

(Source: www.kitco.com/charts/historicalgold.html)

129. a. Use the calculator commands stat LinRege to model

the data using the least-squares regression. Find the equation of the least-squares regression line using x as the year  1x 5 1 for year 20062 and y as the gold price in dollars per ounce. Round all answers to two decimal places. b. Use the equation to predict the gold price in December 2016. Round all answers to two decimal places. Is the answer close to the actual price? c. Use the equation to predict the gold price in December 2017. Round all answers to two decimal places. 130. a. Use the calculator commands stat PwrRegs to model the data using the power function. Find the varia-

tion ­constant and equation of variation using x as the year  1x 5 1 for year 20062 and y as the gold price in dollars per ounce. Round all answers to two decimal places. b. Use the equation to predict the gold price in December 2016. Round all answers to two decimal places. Is the answer close to the actual price? c. Use the equation to predict the gold price in December 2017. Round all answers to two decimal places.

30/11/16 11:08 AM

328 

CHAPTER 3  Functions and Their Graphs

[CHAPTER 3 PRACTICE TEST] Assuming that x represents the independent variable and y represents the dependent variable, classify the relationships as:  a. not a function

3

 1. ƒ1x2 5 0 2x 1 3 0    2.  x 5 y2 1 2    3.  y 5 !x 1 1

Use ƒ 1 x 2 5 !x 2 2 and g 1x2 5 x2 1 11, and determine the desired quantity or expression. In the case of an expression, state the domain. g ƒ  4. ƒ1112 22g 1212   5.  a b 1 x 2      6.  a b 1 x 2 g ƒ   7.  g 1 ƒ 1 x 2 2      8.   1ƒ 1 g2 162   9.  ƒ 1 gA"7B B

Determine whether the function is odd, even, or neither. 10. ƒ1x2 5 0 x 0 2 x2 2 11. ƒ1x2 5 9x3 1 5x 2 3     12. ƒ 1 x 2 5 x Graph the functions. State the domain and range of each ­function.

PRACTICE TEST

13. ƒ 1 x 2 5 2 !x 2 3 1 2 14. ƒ1x2 5 22 1x 2 122

16.

y

Use the graphs of the function to find: a. ƒ132 b.  ƒ102

y  f(x)

c.  ƒ1242 x

17.

d. x, where ƒ1x2 5 3 e.  x, where ƒ1x2 5 0

a. g 132

y

b. g 102

c. g 1242

y  g(x) x

20. ƒ 1 x 2 5 5 2 7x

Find the average rate of change of the given functions.

 c. a one-to-one function

x , 21 21 , x , 2 x$2

ƒ 1x 1 h2 2 ƒ 1x2 for: h

19. ƒ 1 x 2 5 3x 2 2 4x 1 1

 b. a function but not one-to-one

2x 15. ƒ 1 x 2 5 • 1 x2

Find

d. x, where g 1x2 5 0

21. ƒ 1 x 2 5 64 2 16x 2 for x 5 0 to x 5 2 22. ƒ 1 x 2 5 !x 2 1 for x 5 2 to x 5 10

Given the function ƒ , find the inverse if it exists. State the domain and range of both ƒ and ƒ 21. 23. ƒ 1 x 2 5 !x 2 5 24. ƒ  1x2 5 x2 1 5

2x 1 1 52x 2x x#0 26. ƒ 1 x 2 5 e 2x 2 x . 0 25. ƒ 1 x 2 5

27. What domain restriction can be made so that ƒ1x2 5 x2 has an

inverse? 28. If the point  122, 52 lies on the graph of a function, what point lies on the graph of its inverse function? 29. Discount. Suppose a suit has been marked down 40% off the original price. An advertisement in the newspaper has an “additional 30% off the sale price” coupon. Write a function that determines the “checkout” price of the suit. 30. Temperature. Degrees Fahrenheit (8F), degrees Celsius (8C), and kelvins (K) are related by the two equations: F 5 95C 1 32 and K 5 C 1 273.15. Write a function whose input is kelvins and output is degrees Fahrenheit. 31. Circles. If a quarter circle is drawn by tracing the unit circle in quadrant III, what does the inverse of that function look like? Where is it located? 32. Sprinkler. A sprinkler head malfunctions at midfield in a football field. The puddle of water forms a circular pattern around the sprinkler head with a radius in yards that grows as a function of time, in hours: r 1 t 2 5 10!t. When will the puddle reach the sidelines? (A football field is 30 yards from sideline to sideline.) 33. Internet. The cost of airport Internet access is $15 for the first 30 minutes and $1 per minute for each minute after that. Write a function describing the cost of the service as a function of minutes used. Use variation to find a model for the given problem. 34. y varies directly with the square of x. y 5 8 when x 5 5. 35. F varies directly with m and inversely with p. F 5 20 when

a. p 102

y

18.

b. x, where p 1x2 5 0

y  p(x) x

c. p 112 d. p 132

m 5 2 and p 5 3. 36. Use a graphing utility to graph the function. State the

(a) domain, (b) range, and (c) x intervals where the function is increasing, decreasing, and constant. ƒ1x2 5 e

5 0 3 1 2x 2 x 2 0

24 # x , 22 22 , x # 4

37. Use a graphing utility to graph the function and determine

whether it is one-to-one. y 5 x 3 2 12x 2 1 48x 2 65

Young_AT_6160_ch03_pp317-329.indd 328

30/11/16 11:08 AM

Cumulative Test 

329

[CHAPTERS 1–3 CUMULATIVE TEST]  1. Simplify

2

17. Write an equation of a line that passes through the

.

points  11.2, 232 and  120.2, 232.

3 2 "5  2. Factor completely: 10x2 2 29x 2 21.

18. Transform the equation into standard form by completing

x 3 2 4x  3. Simplify and state the domain: . x12 1 1  4. Solve for x: x 5 2 x 1 11. 6 5  5. Perform the operation, simplify, and express in standard

form:  18 2 9i2 18 1 9i2.

  6. Solve for x, and give any excluded values:

5 10 2 10 5 . x 3x

the square, and state the center and radius of the circle: x2 1 y2 1 12x 2 18y 2 4 5 0. 19. Find the equation of a circle with center  122, 212 and passing through the point  124, 32. 20. If a cellular phone tower has a reception radius of 100 miles and you live 85 miles north and 23 miles east of the tower, can you use your cell phone at home? Explain. 21. Use interval notation to express the domain of the function g1x2 5

 7. The original price of a hiking stick is $59.50. The sale price

is $35.70. Find the percent of the markdown.  8. Solve by factoring: x 16x 1 12 5 12. x2 1   9. Solve by completing the square: 2x5 . 2 5 3 10. Solve and check: "x 1 2 5 23.

11. Solve using substitution: x 2 x 2 12 5 0. 4

14. 0 2.7 2 3.2x 0 # 1.3

15. Calculate the distance and midpoint between the segment

23. 24. 25. 26.

x 5 4. Evaluate g 1ƒ12122 for ƒ1x2 5 0 6 2 x 0 and g 1x2 5 x2 2 3. Find the inverse of the function ƒ1x2 5 x2 1 3 for x $ 0. Write an equation that describes the variation: r is inversely proportional to t. r 5 45 when t 5 3. Use a graphing utility to graph the function. State the (a) domain, (b) range, and (c) x intervals where the function is increasing, decreasing, and constant. ƒ1x2 5 e

12 0x 0 1 2 0 x 2 20

21 # x , 1 1,x#3

27. Use a graphing utility to graph the function ƒ 1 x 2 5 x 2 2 3x

and g 1 x 2 5 x 2 1 x 2 2 in the same screen. Find the function h such that g + h 5 ƒ.

CUMULATIVE TEST

joining the points  122.7, 21.42 and  15.2, 6.32. 16. Find the slope of the line passing through the points  10.3, 21.42 and  12.7, 4.32.

Young_AT_6160_ch03_pp317-329.indd 329

22. Find the average rate of change for ƒ1x2 5 5x2, from x 5 2 to

2

Solve and express the solution in interval notation. 12. 27 , 3 2 2 x # 5 x 13. ,0 x25

1 . x21

30/11/16 11:09 AM

[4[ CHAPTER

Polynomial and Rational Functions Indoor football stadiums are designed so that football punters

Focus On Sport/Getty Images, Inc.

will not hit the roof with the football. One of the greatest NFL punters of all time was Ray Guy, who played 14 seasons from 1973 to 1986. “One of his punts hit the giant TV screen hanging from the rafters in the Louisiana Superdome. Not only did Guy punt high and far—‘hang time’ came into the NFL lexicon during his tenure—once he even had an opponent take a ball he punted and test for helium!” (rayguy.net/fact-sheet; Ray Guy Fact Sheet) The path that punts typically follow is called a parabola and is classified as a quadratic function. The distance of a punt is measured in the horizontal direction. The yard line where the punt is kicked from and the yard line where the punt either hits the field or is caught are the zeros of the quadratic function. Zeros are the points where the function value is equal to zero.

Ray Guy

LEARNING OBJECTIVES ■■ Find

the vertex (maximum or minimum) of the graph of a quadratic function. ■■ Graph polynomial functions.

Young_AT_6160_ch04_pp330-387.indd 330

■■ Divide

polynomials using long division and synthetic division. ■■ Develop strategies for searching for zeros of a polynomial function.

■■ Understand

that complex zeros come in conjugate pairs. ■■ Graph rational functions.

25/11/16 3:04 PM

[IN THIS CHAPTER] We will start by discussing quadratic functions (polynomial functions of degree 2) whose graphs are parabolas. We will find the vertex, which is the maximum or minimum point on the graph. Then we will expand our discussion to higher-degree polynomial functions. We will discuss techniques to find zeros of polynomial functions and strategies for graphing polynomial functions. Last we will discuss rational functions, which are ratios of polynomial functions.

POLY N OMIA L AN D R AT I O N AL F UN C T I O N S 4.1

4.2

4.3

4.4

4.5

4.6

QUADRATIC FUNCTIONS

POLYNOMIAL FUNCTIONS OF HIGHER DEGREE

DIVIDING POLYNOMIALS: LONG DIVISION AND SYNTHETIC DIVISION

THE REAL ZEROS OF A POLYNOMIAL FUNCTION

COMPLEX ZEROS: THE FUNDAMENTAL THEOREM OF ALGEBRA

RATIONAL FUNCTIONS

• Graphs of Quadratic Functions: Parabolas • Finding the Equation of a Parabola

• Identifying Polynomial Functions • Graphing Polynomial Functions Using Transformations of Power Functions • Real Zeros of a Polynomial Function • Graphing General Polynomial Functions

• Long Division of Polynomials • Synthetic Division of Polynomials

• The Remainder Theorem and the Factor Theorem • The Rational Zero Theorem and Descartes’ Rule of Signs • Factoring Polynomials • The Intermediate Value Theorem • Graphing Polynomial Functions

• Complex Zeros • Factoring Polynomials

• Domain of Rational Functions • Vertical, Horizontal, and Slant Asymptotes • Graphing Rational Functions

331

Young_AT_6160_ch04_pp330-387.indd 331

25/11/16 3:04 PM

332 

CHAPTER 4  Polynomial and Rational Functions

4.1 QUADRATIC FUNCTIONS SKILLS OBJECTIVES ■■ Graph a quadratic function given in either standard or general form. ■■ Find the equation of a parabola.

CONCEPTUAL OBJECTIVES ■■ Recognize characteristics of graphs of quadratic functions (parabolas): whether the parabola opens up or down; whether the vertex is a maximum or minimum; the axis of symmetry ■■ Understand that as long as you know the vertex and a point that lies along the graph of a parabola, you can determine the equation of the parabola.

4.1.1  Graphs of Quadratic Functions: Parabolas 4.1.1 S K I L L

Graph a quadratic function given in either standard or general form. 4.1.1 C O N C E P T U A L

Recognize characteristics of graphs of quadratic functions (parabolas): whether the parabola opens up or down; whether the vertex is a maximum or minimum; the axis of symmetry.

In Chapter 3, we studied functions in general. In this chapter, we will learn about a special group of functions called polynomial functions. Polynomial functions are simple functions; often, more complicated functions are approximated by polynomial functions. Polynomial functions model many real-world applications such as the stock market, football punts, business costs, revenues and profits, and the flight path of NASA’s “vomit comet.” Let’s start by defining a polynomial function DEFINITION

Polynomial Function

Let n be a nonnegative integer, and let an, an21, . . . , a2, a1, a0 be real numbers with a n 2 0 . The function ƒ 1 x 2 5 anx n 1 an21x n21 1 c1 a2x 2 1 a1x 1 a0

is called a polynomial function of x with degree n. The coefficient an is called the leading coefficient, and a0 is the constant.

Polynomials of particular degrees have special names. In Chapter 3, the library of functions included the constant function ƒ1x2 5 b, which is a horizontal line; the linear function ƒ1x2 5 mx 1 b, which is a line with slope m and y-intercept 10, b2; the square function ƒ1x2 5 x2; and the cube function ƒ1x2 5 x3. These are all special cases of a polynomial function. Here are more examples of polynomial functions of particular degree together with their names:

y 10

POLYNOMIAL f (x) = x2 x –5

5

y 8 f (x) = x 2

x –5

5 F(x) = (x + 1) 2 – 2

Young_AT_6160_ch04_pp330-387.indd 332

ƒ1x2 5 3 ƒ1x2 5 22x 1 1 ƒ1x2 5 7x2 2 5x 1 19 ƒ1x2 5 4x3 1 2x 2 7

DEGREE

SPECIAL NAME

0 1 2 3

Constant function Linear function Quadratic function Cubic function

The leading coefficients of these functions are 3, 22, 7, and 4, respectively. The constants of these functions are 3, 1, 19, and 27, respectively. In Section 2.3, we discussed graphs of linear functions, which are first-degree polynomial functions. In this section, we will discuss graphs of quadratic functions, which are second-degree polynomial functions. In Section 3.2, the library of functions that we compiled included the square function ƒ  1x2 5 x2, whose graph is a parabola. See the graph on the left. In Section 3.3, we graphed functions using transformation techniques such as F1x2 5 1x 1 12 2 2 2, which can be graphed by starting with the square function y 5 x2 and shifting one unit to the left and down two units. See the graph on the left.

25/11/16 3:04 PM

4.1  Quadratic Functions 

333

Note that if we eliminate the parentheses in F1x2 5 1x 1 122 2 2 to get F 1 x 2 5 x 2 1 2x 1 1 2 2 5 x 2 1 2x 2 1



the result is a function defined by a second-degree polynomial (a polynomial with x 2 as the highest-degree term), which is also called a quadratic function. DEFINITION

Quadratic Function

Let a, b, and c be real numbers with a 2 0 . The function ƒ1x2 5 ax2 1 bx 1 c is called a quadratic function.

The graph of any quadratic function is a parabola. If the leading coefficient a is positive, then the parabola opens up. If the leading coefficient a is negative, then the parabola opens down. The vertex (or turning point) is the minimum point, or low point, on the graph if the parabola opens up, whereas it is the maximum point, or high point, on the graph if the parabola opens down. The vertical line that intersects the parabola at the vertex is called the axis of symmetry. The axis of symmetry is the line x 5 h, and the vertex is located at the point 1h, k2, as shown in the following two figures.

x=h Axis of Symmetry

y

y (h, k) Vertex f (x) = ax 2 + bx + c a>0

k f (x) = ax 2 + bx + c a 0 y

an < 0 y

an < 0 y

an > 0 y

x x

x

x

EXAMPLE 7  Graphing a Polynomial Function

Sketch a graph of the polynomial function ƒ1x2 5 2x4 2 8x2. Solution:

the y-intercept: 1x 5 02. ƒ102 5 0 The y-intercept corresponds to the point 10, 02. STEP 1  Determine

STEP 2  Find

ƒ1x2 5 2x4 2 8x2 5 2x21x2 2 42 5 2x21x 2 22 1x 1 22 5 2x21x 2 22 1x 1 22 5 0



the zeros of the polynomial. Factor out the common 2x2. Factor the quadratic binomial. Set ƒ1x2 5 0.



0 is a zero of multiplicity 2. The graph will touch the x-axis. 2 is a zero of multiplicity 1. The graph will cross the x-axis. 22 is a zero of multiplicity 1. The graph will cross the x-axis.

the end behavior. ƒ1x2 5 2x4 2 8x2 behaves like y 5 2x4. y 5 2x4 is of even degree, and the leading coefficient is positive, so the graph rises without bound as x gets large in both the positive and negative directions.

STEP 3  Determine

STEP 4  Sketch

the intercepts and end behavior.

y

x

(0, 0) (–2, 0)

(2, 0)

[ CONCEPT CHECK ] The graph of the function f (x) 5 (x 2 a)3(x 2 b)2 crosses or touches at (a, 0) and (b, 0)?

▼ ANSWER Crosses at (a, 0) and touches at (b, 0).

Young_AT_6160_ch04_pp330-387.indd 358

STEP 5  Find

additional points.

x

21

212

1 2

1

f (x)

26

215 8

215 8

26

25/11/16 3:05 PM

4.2  Polynomial Functions of Higher Degree  STEP 6  Sketch ■■ ■■

the graph.

359

y 10

Estimate additional points. Connect with a smooth curve.

Note the symmetry about the y-axis. This function is an even function: ƒ12x2 5 ƒ1x2.

(–2, 0)

x

(2, 0)

▼ ANSWER

–10

y 10

It is important to note that the local minimums occur at x 5 6 !2 < 6 1.14 but at this time can only be illustrated using a graphing utility.

x –5

5

▼ Y O U R T U R N   Sketch a graph of the polynomial function ƒ1x2 5 x5 2 4x3.

–10

[SEC TION 4.2]  S U M M A RY In general, polynomials can be graphed in one of two ways: n

Use graph-shifting techniques with power functions. General polynomial function.

n

1. Identify intercepts. 2. Determine each real zero and its multiplicity, and ascertain whether the graph crosses or touches the x-axis there.

3. x-intercepts (real zeros) divide the x-axis into intervals. Test points in the intervals to determine whether the graph is above or below the x-axis. 4. Determine the end behavior by investigating the end behavior of the highest-degree monomial. 5. Sketch the graph with a smooth curve.

[SEC TION 4. 2]  E X E R C I S E S • SKILLS In Exercises 1–10, determine which functions are polynomials, and for those that are, state their degree. 1. ƒ1x2 5 23x2 1 15x 2 7

2. ƒ1x2 5 2x5 2 x2 1 13

5. h 1 x 2 5 !x 1 1

6. h  1x2 5 1x 2 121/2 1 5x

9. G 1 x 2 5

x11 x2

10. H 1 x 2 5

3 3. g 1 x 2 5 1 x 1 2 2 3 Ax 2 5 B 2

4. g  1x2 5 x41x 2 1221x 1 2.523

8. F 1 x 2 5 3x 2 1 7x 2

7. F 1x2 5 x1/3 1 7x2 2 2

x2 1 1 2

2 3x

In Exercises 11–18, match the polynomial function with its graph. 11. ƒ1x2 5 23x 1 1

12. ƒ1x2 5 23x2 2 x

13. ƒ1x2 5 x2 1 x

14. ƒ1x2 5 22x3 1 4x2 2 6x

15. ƒ1x2 5 x3 2 x2

16. ƒ1x2 5 2x4 2 18x2

17. ƒ1x2 5 2x4 1 5x3

18. ƒ1x2 5 x5 2 5x3 1 4x

a.

b.

c.

y

d. y

y

5

y

90

4

50

x –5

x

5

–10 x –3

–5

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2 –1

10

x –10

10

–50

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CHAPTER 4  Polynomial and Rational Functions

e.

f.

g.

y

h.

y

5

y

10

x –5

5

y

5

5

x –5

5

–5

x –5

–10

5

x –5

–5

5

–5

In Exercises 19–26, graph each function by transforming a power function y 5 xn. 19. ƒ1x2 5 2x5

20. ƒ1x2 5 2x4

5

21. ƒ1x2 5 1x 2 224

4

23. ƒ1x2 5 x 1 3

25. ƒ1x2 5 3 2 1x 1 12

24. ƒ1x2 5 2x 2 3

22. ƒ1x2 5 1x 1 225

4

26. ƒ1x2 5 1x 2 325 2 2

In Exercises 27–38, find all the real zeros (and state their multiplicities) of each polynomial function. 27. ƒ1x2 5 21x 2 32 1x 1 423

28. ƒ1x2 5 231x 1 2231x 2 122

29. ƒ1x2 5 4x21x 2 7221x 1 42 2

2

30. ƒ1x2 5 5x31x 1 1241x 2 62

2

32. ƒ1x2 5 4x21x2 2 12 1x2 1 92

31. ƒ1x2 5 4x 1x 2 12 1x 1 42 33. ƒ1x2 5 8x3 1 6x2 2 27x

34. ƒ1x2 5 2x4 1 5x3 2 3x2

35. ƒ1x2 5 22.7x3 2 8.1x2 1

36. ƒ1x2 5 1.2x6 2 4.6x4

2

3

2

37. ƒ1x2 5 3 x 6 1 5 x 4

1

38. ƒ1x2 5 7 x 5 2 4 x 4 1 2 x 3

In Exercises 39–52, find a polynomial (there are many) of minimum degree that has the given zeros. 39. 23, 0, 1, 2

40. 22, 0, 2

1 2 3 43. 22 , 3 , 4

3 44. 24 ,

213,

1 2

0,

47. 22 1multiplicity 32, 0 1multiplicity 22

41. 25, 23, 0, 2, 6

42. 0, 1, 3, 5, 10

45. 1 2 !2, 1 1 !2

46. 1 2 !3, 1 1 !3

48. 24 1multiplicity 22, 5 1multiplicity 32 50. 0 1multiplicity 12, 10 1multiplicity 32

49. 23 1multiplicity 22, 7 1multiplicity 52

51. 2 !3 1multiplicity 22, 21 1multiplicity 12, 0 1multiplicity 22, !3 1multiplicity 22 52. 2 !5 1multiplicity 22, 0 1multiplicity 12, 1 1multiplicity 22, !5 1multiplicity 22

In Exercises 53–72, for each polynomial function given: (a) list each real zero and its multiplicity; (b) determine whether the graph touches or crosses at each x-intercept; (c) find the y-intercept and a few points on the graph; (d) determine the end behavior; and (e) sketch the graph. 53. ƒ1x2 5 2x2 2 6x 2 9

54. ƒ1x2 5 x2 1 4x 1 4

3

3

57. ƒ1x2 5 x 2 9x

58. ƒ1x2 5 2x 1 4x

61. ƒ1x2 5 2x4 2 3x3

62. ƒ1x2 5 x5 2 x3

65. ƒ1x2 5 2x5 2 6x4 2 8x3 2

69. ƒ1x2 5 21x 1 22 1x 2 12

55. ƒ1x2 5 1x 2 223

2

3

3

60. ƒ1x2 5 x3 2 6x2 1 9x

59. ƒ1x2 5 2x 1 x 1 2x

63. ƒ1x2 5 12x6 2 36x5 2 48x4 64. ƒ1x2 5 7x5 2 14x4 2 21x3

66. ƒ1x2 5 25x4 1 10x3 2 5x2 2

56. ƒ1x2 5 21x 1 323 2

70. ƒ1x2 5 1x 2 22 1x 1 12

3

67. ƒ1x2 5 x3 2 x2 2 4x 1 4 2

3

71. ƒ1x2 5 x 1x 2 22 1x 1 32

2

68. ƒ1x2 5 x3 2 x2 2 x 1 1 72. ƒ1x2 5 2x31x 2 4221x 1 222

In Exercises 73–76, for each graph given: (a) list each real zero and its smallest possible multiplicity; (b) determine whether the degree of the polynomial is even or odd; (c) determine whether the leading coefficient of the polynomial is positive or negative; (d) find the y-intercept; and (e) write an equation for the polynomial function (assume the least degree possible). 73.

74.

75.

76.

y

y

y

10

16

y

10

8 x –5

x –5

5

5

x –5

5

x –5

5 –4

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–10

–10

–32

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• A P P L I C AT I O N S For Exercises 77 and 78, refer to the following:

Revenue (in millions of dollars)

The relationship between a company’s total revenue R (in millions of dollars) is related to its advertising costs x (in thousands of ­dollars). The relationship between revenue R and advertising costs x is illustrated in the graph. R 50 45 40 35 30 25 20 15 10 5

82. Stock Value. A day trader checks the stock price of

Coca-Cola during a 4-hour period (given below). The price of Coca-Cola stock during this 4-hour period can be modeled as a polynomial function. Plot these data. How many turning points are there? Assuming these are the minimum number of turning points, what is the lowest-degree polynomial that can represent the Coca-Cola stock price? PERIOD WATCHING STOCK MARKET

x 100 200 300 400 500 600 Advertising Costs (in thousands of dollars)

77. Business. Analyze the graph of the revenue function.

a.  Determine the intervals on which revenue is increasing



b.  Identify the zeros of the function. Interpret the meaning

and decreasing. of zeros for this function. 78. Business. Use the graph to identify the maximum revenue for the company and the corresponding advertising costs that produce maximum revenue.

2

where r is the radius of the trachea (in centimeters) during the cough. 79. Health/Medicine. Graph the velocity function and estimate the intervals on which the velocity of air in the trachea is increasing and decreasing. 80. Health/Medicine. Estimate the radius of the trachea that ­produces the maximum velocity of air in the trachea. Use this radius to estimate the maximum velocity of air in the trachea. 81. Weight. Jennifer has joined a gym to lose weight and feel better. She still likes to cheat a little and will enjoy the ­occasional bad meal with an ice cream dream dessert and then miss the gym for a couple of days. Given in the table is Jennifer’s weight for a period of 8 months. Her weight can be modeled as a polynomial. Plot these data. How many turning points are there? Assuming these are the minimum number of turning points, what is the lowest-degree polynomial that can represent Jennifer’s weight? MONTH

Young_AT_6160_ch04_pp330-387.indd 361

WEIGHT

1

169

2

158­

3

150

4

161

5

154

6

159

7

148

8

153

$53.00

2

$56.00

3

$52.70

4

$51.50

4-hour period is given in the following table. If a third-degree polynomial models this stock, do you expect the stock to go up or down in the fifth period? PERIOD WATCHING STOCK MARKET

PRICE

1

$15.10

2

$14.76

3

$15.50

4

$14.85

84. Stock Value. The stock prices for Coca-Cola during a

4-hour period on another day yield the following results. If a third-degree polynomial models this stock, do you expect the stock to go up or down in the fifth period? PERIOD WATCHING STOCK MARKET

PRICE

1

$52.80

2

$53.00

3

$56.00

4

$52.70

 or Exercises 85 and 86, the following graph illustrates the F average annual federal funds rate in the month of January (2006–2014). Annual Average Federal Funds Rate

v 1 r 2 5 2120r 1 80r 3

1

83. Stock Value. The price of Tommy Hilfiger stock during a

For Exercises 79 and 80, refer to the following: During a cough, the velocity v (in meters per second) of air in the trachea may be modeled by the function

PRICE

10.00% 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 2006 2008 2010 2012 2014 Year

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CHAPTER 4  Polynomial and Rational Functions

85. Finance. If a polynomial function is used to model the

f­ederal funds rate data shown in the graph, determine the degree of the lowest-degree polynomial that can be used to model those data.

86. Finance. Should the leading coefficient in the polynomial

found in Exercise 85 be positive or negative? Explain.

• C AT C H T H E M I S TA K E In Exercises 87–90, explain the mistake that is made. 87. Find a fourth-degree polynomial function with zeros 22, 21,

3, 4.

Solution:

Solution:

The zeros are 21 and 1, so the x-intercepts are 121, 02 and 11, 02.

ƒ1x2 5 1x 2 22 1x 2 12 1x 1 32 1x 1 42

This is incorrect. What mistake was made? 88. Determine the end behavior of the polynomial function ƒ1x2 5 x  1x 2 223.

The y-intercept is 10, 12.

Plotting these points and connecting with a smooth curve yield the graph on the right.

Solution:

This polynomial has similar end behavior to the graph of y 5 x3.

90. Graph the polynomial function ƒ1x2 5 1x 1 1221x 2 122. y 5

x –5

5

–5

End behavior falls to the left and rises to the right.

This graph is incorrect. What did we forget to do?



This is incorrect. What mistake was made? 89. Graph the polynomial function ƒ1x2 5 1x 2 1221x 1 223.

Solution:

The zeros are 22 and 1, and therefore, the x-intercepts are 122, 02 and 11, 02. The y-intercept is 10, 82.

Plotting these points and connecting with a smooth curve yield the graph on the right.

y 10

x –5

5

–10

This graph is incorrect. What did we forget to do?

• CONCEPTUAL In Exercises 91–94, determine if each statement is true or false. 91.  The graph of a polynomial function might not have any

y-intercepts. 92.  The graph of a polynomial function might not have any x-intercepts. 93. The domain of all polynomial functions is 12 q , q 2.

94. The range of all polynomial functions is 12 q , q 2.

95. What is the maximum number of zeros that a polynomial of

degree n can have?

96­. What is the maximum number of turning points a graph of an

nth-degree polynomial can have?

• CHALLENGE 97. Find a seventh-degree polynomial that has the following

graph characteristics: The graph touches the x-axis at x 5 21, and the graph crosses the x-axis at x 5 3. Plot this polynomial function. 98. Find a fifth-degree polynomial that has the following graph characteristics: The graph touches the x-axis at x 5 0 and crosses the x-axis at x 5 4. Plot the polynomial function.

Young_AT_6160_ch04_pp330-387.indd 362

 99. Determine the zeros of the polynomial

ƒ1x2 5 x3 1 1b 2 a2x2 2 abx for the positive real numbers a and b. 100. Graph the function ƒ1x2 5 x21x 2 a221x 2 b22 for the p ­ ositive real numbers a, b, where b . a.

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363

• TECHNOLOGY In Exercises 101 and 102, use a graphing calculator or ­computer to graph each polynomial. From that graph, ­estimate the x-intercepts (if any). Set the function equal to zero, and solve for the zeros of the polynomial. Compare the zeros with the x-intercepts. 4

2

101. ƒ1x2 5 x 1 2x 1 1

3

2

102. ƒ1x2 5 1.1x 2 2.4x 1 5.2x

In Exercises 105 and 106, use a graphing calculator or a computer to graph each polynomial. From the graph, ­estimate the x-intercepts and state the zeros of the function and their multiplicities. 105. ƒ1x2 5 x4 2 15.9x3 1 1.31x2 1 292.905x 1 445.7025 106. ƒ1x2 5 2x5 1 2.2x4 1 18.49x3 2 29.878x2 2 76.5x 1 100.8

For each polynomial in Exercises 103 and 104, determine the power function that has similar end behavior. Plot this power function and the polynomial. Do they have similar end behavior?

In Exercises 107 and 108, use a graphing calculator or a computer to graph each polynomial. From the graph, estimate the coordinates of the relative maximum and minimum points. Round your answers to two decimal places.

103. ƒ1x2 5 22x5 2 5x4 2 3x3

107. ƒ 1 x 2 5 2x 4 1 5x 3 2 10x 2 2 15x 1 8

104. ƒ1x2 5 x4 2 6x2 1 9

108. ƒ 1 x 2 5 2x 5 2 4x 4 2 12x 3 1 18x 2 1 16x 2 7

4.3 DIVIDING POLYNOMIALS: LONG DIVISION AND SYNTHETIC DIVISION SKILLS OBJECTIVES ■■ Divide polynomials with long division. ■■ Divide polynomials with synthetic division.

CONCEPTUAL OBJECTIVES ■■ Extend long division of real numbers to polynomials. ■■ Understand that synthetic division can only be used when dividing a polynomial by a linear factor.

To divide polynomials, we rely on the technique we use for dividing real numbers. For example, if you were asked to divide 6542 by 21, the long division method used is illustrated in the margin. This solution can be written two ways: 311 R11 or 311 1

11 . 21

In this example, the dividend is 6542, the divisor is 21, the quotient is 311, and the remainder is 11. We employ a similar technique (dividing the leading terms) when dividing polynomials.

311 21q6542 263 24 221 32 221 11

4.3.1  Long Division of Polynomials Let’s start with an example whose answer we already know. We know that a quadratic ­expression can be factored into the product of two linear factors: x  2 1 4x 2 5 5 1x 1 52 1x 2 12. Therefore, if we divide both sides of the equation by 1x 2 12, we get x 2 1 4x 2 5 5x15 x21

We can state this by saying x2 1 4x 2 5 divided by x 2 1 is equal to x 1 5. Confirm this statement by long division:

4.3.1 S K I L L

Divide polynomials with long division. 4.3.1 C O N C E P T U A L

Extend long division of real numbers to polynomials.

x 2 1 q x 2 1 4x 2 5 Note that although this is standard division notation, the dividend and the divisor are both polynomials that consist of multiple terms. The leading terms of each algebraic expression will guide us.

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CHAPTER 4  Polynomial and Rational Functions

WORDS

MATH 2

Q: x times what quantity gives x ? x 2 1 q x2 1 4x 2 5 A: x        x x 2 1qx 2 1 4x 2 5

.

       x . 2 x 2 1 q x 2 1 4x 2 5 Multiply x  1x 2 12 5 x 2 x.       x2 2 x Subtract 1x2 2 x2 from x2 1 4x 2 5. x x 2 1q x 2 1 4x 2 5 Note: 21x2 2 x2 5 2x2 1 x.       2x 2 1 x Bring down the 25.      5x 2 5

.

  x15 . Q: x times what quantity is 5x? x 2 1 q x 2 1 4x 2 5 2x 2 1 x A: 5  5x 2 5     Multiply 51x 2 12 5 5x 2 5. 5x 2 5 x15   x 2 1 qx 2 1 4x 2 5       2x 2 1 x  5x 2 5     25x 15 Subtract 15x 2 52. Note: 215x 2 52 5 25x 1 5.      0

As expected, the remainder is 0. By long division we have shown that x 2 1 4x 2 5 5x15 x21

Check: Multiplying the equation by x 2 1 yields x2 1 4x 2 5 5 1x 1 52 1x 2 12, which we know to be true.

EXAMPLE 1  D  ividing Polynomials Using Long Division; Zero Remainder

Divide 2x3 2 9x 2 1 7x 1 6 by 2x 1 1. Solution:

Multiply: x 2 12x 1 12.

Subtract: Bring down the 7x. Multiply: 25x  12x 1 12.

Subtract: Bring down the 6. Multiply: 6 12x 1 12. Subtract.

Quotient:

Young_AT_6160_ch04_pp330-387.indd 364

x2 2 5x 1 6 2x 1 1q2x 3 2 9x 2 1 7x 1 6 2 1 2 x3 1 x2 2 210x 2 1 7x 2 1 210x2 2 5x 2 12x  1 6 1 2 12 x 1 6 2 0

x 2 2 5x 1 6

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365

Check: 12x 1 12 1x2 2 5x 1 62 5 2x3 2 9x2 1 7x 1 6.

Note: Since the divisor cannot be equal to zero, 2x 1 1 Z 0, then we say x Z 212.

▼

▼ ANSWER

Y O U R T U R N   Divide 4x3 1 13x2 2 2x 2 15 by 4x 1 5.

x2 1 2x 2 3, remainder 0.

Why are we interested in dividing polynomials? Because it helps us find zeros of polynomials. In Example 1, using long division, we found that 2x3 2 9x2 1 7x 1 6 5 12x 1 12 1x2 2 5x 1 62

Factoring the quadratic expression enables us to write the cubic polynomial as a product of three linear factors: 2x3 2 9x2 1 7x 1 6 5 12x 1 12 1x2 2 5x 1 62 5 12x 1 12 1x 2 32 1x 2 22

Set the value of the polynomial equal to zero, 12x 1 12 1x 2 32 1x 2 22 5 0, and solve for x. The zeros of the polynomial are 212, 2, and 3. In Example 1 and in the Your Turn, the remainder was 0. Sometimes there is a nonzero remainder (Example 2).

EXAMPLE 2  D  ividing Polynomials Using Long Division; Nonzero Remainder

Divide 6x2 2 x 2 2 by x 1 1. Solution:

Multiply 6x  1x 1 12. Subtract and bring down 22. Multiply 271x 1 12. Subtract and identify the remainder.

6x 2 7 x 1 1q6x 2 2 x 2 2 2 1 6x 2 1 6x 2

27x 2 2 2 1 27x 2 7 2 15



Dividend 



6x2 2 x 2 2 5 5 6x 2 7 1    x 2 21 x11 x11



Divisor         Divisor

Check: Multiply the quotient and       remainder by x 1 1. The result is the dividend.

  Quotient Remainder

16x 2 72 1x 1 12 1

5 ⋅ 1x 1 12 1x 1 12

5 6x2 2 x 2 7 1 5 5 6x2 2 x 2 2 ✓

▼ Y O U R T U R N   Divide 2x3 1 x2 2 4x 2 3 by x 2 1.

▼ ANSWER

2x2 1 3x 2 1 R: 24 or

In general, when a polynomial is divided by another polynomial, we express the result in the following form:

2x 2 1 3x 2 1 2

4 x21

r 1x2 P1x2 5 Q1x2 1 d 1x2 d1x2

where P 1x2 is the dividend, d 1 x 2 2 0 is the divisor, Q 1x2 is the quotient, and r  1x2 is the remainder. Multiplying this equation by the divisor, d  1x2, leads us to the division algorithm.

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[ CONCEPT CHECK ] To use long division to divide two polynomials, the degree of the numerator must be (greater/less) than the degree of the denominator.

▼ ANSWER greater than

THE DIVISION ALGORITHM

If P  1x2 and d  1x2 are polynomials with d 1 x 2 2 0, and if the degree of P  1x2 is greater than or equal to the degree of d  1x2, then unique polynomials Q1x2 and r  1x2 exist such that P1x2 5 d 1x2 ⋅ Q1x2 1 r 1x2

If the remainder r  1x2 5 0, then we say that d  1x2 divides P  1x2 and that d  1x2 and Q1x2 are factors of P  1x2. EXAMPLE 3  Long Division of Polynomials with “Missing” Terms

Divide x3 2 8 by x 2 2. Solution:

x2 1 2x 1 4 x 2 2qx 3 1 0x2 1 0x 2 8 2 1 x 3 2 2x 2 2

2

Insert 0x 1 0x for placeholders. Multiply x21x 2 22 5 x3 2 2x2. Subtract and bring down 0x. Multiply 2x  1x 2 22 5 2x2 2 4x. Subtract and bring down 28. Multiply 41x 2 22 5 4x 2 8. Subtract and get remainder 0. Since the remainder is 0, x 2 2 is a factor of x3 2 8.

2x 2 1 0x 2 1 2x 2 2 4x 2 4x 2 8 2 1 4x 2 8 2 0

x3 2 8 5 x 2 1 2x 1 4, x 2 2 x22

▼ ANSWER

Check: x3 2 8 5 1x2 1 2x 1 42 1x 2 22 5 x3 1 2x2 1 4x 2 2x2 2 4x 2 8 5 x3 2 8    ✓

▼

Y O U R T U R N   Divide x3 2 1 by x 2 1.

x2 1 x 1 1

EXAMPLE 4  Long Division of Polynomials

Divide 3x4 1 2x3 1 x2 1 4 by x2 1 1. Solution:

Insert 0x as a placeholder in both 3x2 1 2x 2 2 2 the divisor and the dividend. x 1 0x 1 1q3x 4 1 2x 3 1 x 2 1 0x 1 4 Multiply 3x21x2 1 0x 1 12. 2 1 3x 4 1 0x 3 1 3x 2 2 Subtract and bring down 0x. 2x 3 2 2x 2 1 0x Multiply 2x  1x2 1 0x 1 12. 2 1 2x 3 1 0x 2 1 2x 2 Subtract and bring down 4. 22x2 2 2x 1 4 2 2 1 22x 2 1 0x 2 2 2 Multiply 221x 2 2x 1 12. 22x 1 6 Subtract and get remainder 22x 1 6. 3x 4 1 2x 3 1 x 2 1 4 22x 1 6 5 3x 2 1 2x 2 2 1 2 2 x 11 x 11 ▼ ANSWER

11x 2 1 18x 1 36 2x 1 6 1 x 3 2 3x 2 4 2

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▼ Y O U R T U R N   Divide 2x5 1 3x2 1 12 by x3 2 3x 2 4.

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In Examples 1 through 4 the dividends, divisors, and quotients were all polynomials with integer coefficients. In Example 5, however, the resulting quotient has rational (noninteger) coefficients EXAMPLE 5  L  ong Division of Polynomials Resulting in Quotients with Rational Coefficients

Divide 8x 4 2 5x 3 1 7x 2 2 by 2x 2 1 1. Solution:

4x2 2 52x 2 2 Insert 0x 2 as a placeholder in the dividend and 0x as a placeholder in the divisor. 2x 2 1 0x 1 1q8x 4 2 5x 3 1 0x 2 1 7x 2 2 Multiply 4x 2 1 2x 2 1 0x 1 1 2 . 2 1 8x 4 1 0x 3 1 4x 2 2 Subtract and bring down remaining terms. Multiply 252 x 1 2x 2 1 0x 1 1 2 .

Subtract and bring down remaining terms. Multiply 22 1 2x 2 1 0x 1 1 2 . Subtract and bring down the remainder 19 2 x.

25x 3 2 4x 2 1 7x 5 2 1 25x 3 1 0x 2 2 2 x 2

24x 2 1 19 2x 2 2 2 1 2 24x 1 0x 2 2 2 19 2x

19 x 5 8x 2 5x 1 7x 2 2 2 2 5 4x 2 x 2 2 1 2 2 2x 2 1 1 2x 1 1 4

3

▼

▼ ANSWER

Y O U R T U R N   Divide 10x 4 2 3x 3 1 5x 2 4 by 2x 2 2 1.

7 3 3 5 2x 2 2 5x 2 2 x 1 1 2 2 2 2x 2 1

4.3.2  Synthetic Division of Polynomials In the special case when the divisor is a linear factor of the form x 2 a or x 1 a, there is another, more efficient way to divide polynomials. This method is called synthetic division. It is called synthetic because it is a contrived shorthand way of dividing a polynomial by a linear factor. A detailed step-by-step procedure is given below for ­synthetic division. Let’s divide x4 2 x3 2 2x 1 2 by x 1 1 using synthetic division. STEP 1  Write

the division in synthetic form. ■■ List the coefficients of the dividend. Remember to use 0 for a placeholder. ■■ The divisor is x 1 1. Since x 1 1 5 0 x 5 21 is used. down the first term 112 in the dividend.

STEP 2  Bring

  

            Coefficients of Dividend

21  1  21 0 22 2

4.3.2 S K I L L

Divide polynomials with synthetic division. 4.3.2 C O N C E P T U A L

Understand that synthetic division can only be used when dividing a polynomial by a linear factor.

21  1  21 0 22 2    Bring down the 1 1

the 21 by this leading coefficient 112, and place the product up and to the right in the second column.

STEP 3  Multiply

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21  1  21 0 22 2         21 1

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CHAPTER 4  Polynomial and Rational Functions

ST U DY TIP

STEP 4  Add

the values in the second column.

If (x 2 a) is the divisor, then a is the number used in synthetic division.

21  1  21  0  22  2                21 ADD 1    22

STEP 5  Repeat

Steps 3 and 4 until all columns are filled.

21  1 21  0  22 2     21 2 22 4           1  22 2  24  6 21  1  21  0  22  2

the quotient by assigning powers of x in descending order, beginning with xn21 5 x421 5 x3. The last term is the remainder.

STEP 6  Identify

      21 2 22 4 f

          1  22 2 24  6 Quotient Coefficient x3 2 2x2 1 2x 2 4

Remainder

We know that the degree of the first term of the quotient is 3 because a fourth-degree polynomial was divided by a first-degree polynomial. Let’s compare dividing x4 2 x3 2 2x 1 2 by x 1 1 using both long division and synthetic division. Long Division 3

ST U DY TIP Synthetic division can only be used when the divisor is of the form x 2 a or x 1 a.

2

x 2 2x 1 2x 2 4 x 1 1qx4 2 x3 1 0x2 2 2x 1 2 x4 1 x3   22x 3 1 0x 2   3 2 1 2 2 22x 2 2x   2x 2 2 2x    2 1 2x 2 1 2x 2    24x 1 2 2 1 24x 2 42 16

Synthetic Division 21  1  21  0  22  2       21 2 22 4           1  22  2  24  6 f



x3 2 2x2 1 2x 2 4

Both long division and synthetic division yield the same answer. x 4 2 x 3 2 2x 1 2 6 5 x 3 2 2x 2 1 2x 2 4 1 x11 x11

EXAMPLE 6  Synthetic Division

Use synthetic division to divide 3x5 2 2x3 1 x2 2 7 by x 1 2. Solution:

[ CONCEPT CHECK ] Can synthetic division be used to divide f (x) 5 x4 2 1 by g(x) 5 x3 1 1?

▼ ANSWER No. The degree of the denominator has to be one.

Young_AT_6160_ch04_pp330-387.indd 368

STEP 1  Write

the division in synthetic form.

22 3 0 22 1 0  27

 ist the coefficients of the L dividend. Remember to use 0 for a placeholder. ■■ The divisor of the original problem is x 1 2. If we set x 1 2 5 0, we find that x 5 22, so 22 is the divisor for synthetic division. ■■

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4.3  Dividing Polynomials: Long Division and Synthetic Division  STEP 2  Perform

the synthetic division steps.

the quotient and remainder.

22

3

0 26

22 12

1 220

0 38

27 276

3

26

10

219

38

283

3

0 26

22 12

1 220

0 38

27 276

3

26

10

219

38

283

f

STEP 3  Identify

22

369

3x 4 2 6x 3 1 10x 2 2 19x 1 38 3x 5 2 2x 3 1 x 2 2 7 83 5 3x 4 2 6x 3 1 10x 2 2 19x 1 38 2 x12 x12

▼ Y O U R T U R N   Use synthetic division to divide 2x3 2 x 1 3 by x 2 1.

▼ ANSWER

2x 2 1 2x 1 1 1

4 x21

[SEC TION 4.3]  S U M M A RY D I V I S I O N O F P O LY N O M I A L S n n

Long division can always be used. Synthetic division is restricted to when the divisor is of the form x 2 a or x 1 a.

E X P R E S S I N G R E S U LT S n

Dividend remainder 5 quotient 1 Divisor divisor

n Dividend

5 1quotient2 1divisor2 1 remainder

WHEN REMAINDER IS ZERO n Dividend n

5 1quotient2 1divisor2

Quotient and divisor are factors of the dividend.

[SEC TI O N 4 . 3]   E X E R C I S E S • SKILLS In Exercises 1–30, divide the polynomials using long division. Use exact values. Express the answer in the form Q 1x2 5 ?, r 1x2 5 ?. 1. 12x2 1 5x 2 32 4 1x 1 32 2. 12x2 1 5x 2 32 4 1x 2 32 2 3. 1x 2 5x 1 62 4 1x 2 22 4. 12x2 1 3x 1 12 4 1x 1 12 5. 13x2 2 9x 2 52 4 1x 2 22 6. 1x2 1 4x 2 32 4 1x 2 12 2 7. 13x 2 13x 2 102 4 1x 1 52 8. 13x2 2 13x 2 102 4 1x 2 52 9. 1x2 2 42 4 1x 1 42 10. 1x2 2 92 4 1x 2 22 2 11. 19x 2 252 4 13x 2 52 12. 15x2 2 32 4 1x 1 12 13. 14x2 2 92 4 12x 1 32 14. 18x3 1 272 4 12x 1 32 2 3 15. 111x 1 20x 1 12x 1 22 4 13x 1 22 16. 112x3 1 2 1 11x 1 20x22 4 12x 1 12 17. 14x3 2 2x 1 72 4 12x 1 12 18. 16x4 2 2x2 1 52 4 123x 1 22 1 3 2 1 12x 3 1 1 1 7x 1 16x 2 2 4 Ax 1 31 B 19. 1 4x 2 12x 2 x 1 3 2 4 Ax 2 2 B 20.

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CHAPTER 4  Polynomial and Rational Functions

21. 122x5 1 3x4 2 2x22 4 1x3 2 3x2 1 12 22. 129x6 1 7x4 2 2x3 1 52 4 13x4 2 2x 1 12 23.

213x 2 1 4x 4 1 9 x4 2 1 x4 2 9 40 2 22x 1 7x 3 1 6x 4 24. 25. 26. 2 2 2 x 21 x 13 6x 1 x 2 2 4x 2 2 9

27.

23x 4 1 7x 3 2 2x 1 1 2x 5 2 4x 3 1 3x 2 1 5 28. x 2 0.6 x 2 0.9

29. 1x4 1 0.8x3 2 0.26x2 2 0.168x 1 0.04412 4 1x2 1 1.4x 1 0.492

30. 1x5 1 2.8x4 1 1.34x3 2 0.688x2 2 0.2919x 1 0.08822 4 1x2 2 0.6x 1 0.092

In Exercises 31–50, divide the polynomial by the linear factor with synthetic division. Indicate the quotient Q 1x2 and the ­remainder r 1x2  . 31. 13x2 1 7x 1 22 4 1x 1 22 32. 12x2 1 7x 2 152 4 1x 1 52 33. 17x2 2 3x 1 52 4 1x 1 12 34. 14x2 1 x 1 12 4 1x 2 22 2 4 3 35. 13x 1 4x 2 x 2 2x 2 42 4 1x 1 22 36. 13x2 2 4 1 x32 4 1x 2 12 37. 1x4 1 12 4 1x 1 12 38. 1x4 1 92 4 1x 1 32 4 39. 1x 2 162 4 1x 1 22 40. 1x4 2 812 4 1x 2 32 1 1 3x 3 2 8x 2 1 1 2 4 Ax 1 13 B 41. 1 2x 3 2 5x 2 2 x 1 1 2 4 Ax 1 2 B 42. 2 4 3 2 1 3x 4 1 x 3 1 2x 2 3 2 4 Ax 2 34 B 43. 1 2x 2 3x 1 7x 2 4 2 4 Ax 2 3 B 44. 45. 12x4 1 9x3 2 9x2 2 81x 2 812 4 1x 1 1.52 46. 15x3 2 x2 1 6x 1 82 4 1x 1 0.82 47.

x 7 2 8x 4 1 3x 2 1 1 x 6 1 4x 5 2 2x 3 1 7 48. x21 x11

49. 1 x 6 2 49x 4 2 25x 2 1 1225 2 4 Ax 2 !5B

50. 1 x 6 2 4x 4 2 9x 2 1 36 2 4 Ax 2 !3B

In Exercises 51–60, divide the polynomials by either long division or synthetic division. 51. 16x2 2 23x 1 72 4 13x 2 12 52. 16x2 1 x 2 22 4 12x 2 12 3 2 53. 1x 2 x 2 9x 1 92 4 1x 2 12 54. 1x3 1 2x2 2 6x 2 122 4 1x 1 22 55. 1x5 1 4x3 1 2x2 2 12 4 1x 2 22 56. 1x4 2 x2 1 3x 2 102 4 1x 1 52 4 2 57. 1x 2 252 4 1x 2 12 58. 1x3 2 82 4 1x2 2 22 59. 1x7 2 12 4 1x 2 12 60. 1x6 2 272 4 1x 2 32

• A P P L I C AT I O N S 61. Geometry. The area of a rectangle is 6x4 1 4x3 2 x2 2 2x 2 1

square feet. If the length of the rectangle is 2x2 2 1 feet, what is the width of the rectangle? 62. Geometry. If the rectangle in Exercise 61 is the base of a ­rectangular box with volume 18x5 1 18x4 1 x3 2 7x2 2 5x 2 1 cubic feet, what is the height of the box?

63. Travel. If a car travels a distance of x3 1 60x2 1 x 1 60 miles

at an average speed of x 1 60 miles per hour, how long does the trip take? 64. Sports. If a quarterback throws a ball 2x2 2 5x 1 50 yards in 5 2 x seconds, how fast is the football traveling?

• C AT C H T H E M I S TA K E In Exercises 65–68, explain the mistake that is made. 65. Divide x3 2 4x2 1 x 1 6 by x2 1 x 1 1.

66. Divide x4 2 3x2 1 5x 1 2 by x 2 2.

Solution:

Solution: 16 16 23 13

This is incorrect. What mistake was made?

Young_AT_6160_ch04_pp330-387.indd 370

22

1 23

5

2

22

10 230

1 25

15 228

2

f

x23 x 2 1 x 1 1q x 3 2 4x 2 1 x x3 1 x2 1 x 23x 2 1 2x 23x 2 2 3x 2x

x 2 5x 1 15 This is incorrect. What mistake was made?

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4.3  Dividing Polynomials: Long Division and Synthetic Division 

67. Divide x3 1 4x 2 12 by x 2 3.

68. Divide x3 1 3x2 2 2x 1 1 by x2 1 1.

Solution:

Solution:

1

4 212 3

21

7

9

21

1

1

3 22

1

21 22

4

2 24

5

f

1

f

3

371

x17

x2 2 2x 2 4

This is incorrect. What mistake was made?

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 69–72, determine whether each statement is true or false. 69. A fifth-degree polynomial divided by a third-degree polynomial

71. Synthetic division can be used whenever the degree of the

will yield a quadratic quotient. 70. A third-degree polynomial divided by a linear polynomial will yield a linear quotient.

dividend is exactly one more than the degree of the divisor. 72. When the remainder is zero, the divisor is a factor of the dividend.

• CHALLENGE 73. Is x 1 b a factor of x3 1 12b 2 a2x2 1 1b2 2 2ab2x 2 ab2? 4

2

2

2

2 2

74. Is x 1 b a factor of x 1 1b 2 a 2x 2 a b ?

75. Divide x3n 1 x2n 2 xn 2 1 by xn 2 1. 76. Divide x3n 1 5x2n 1 8xn 1 4 by xn 1 1.

• TECHNOLOGY 2x 3 2 x 2 1 10x 2 5 . What type of function is it? x2 1 5 Perform this division using long division, and confirm that the graph corresponds to the quotient.

7 7. P lot

78. Plot

x 3 2 3x 2 1 4x 2 12 . What type of function is it? x23

Perform this division using synthetic division, and confirm that the graph corresponds to the quotient.

x 4 1 2x 3 2 x 2 2 79. Plot . What type of function is it? Perform x12 this division using synthetic division, and confirm that the graph corresponds to the quotient.

Young_AT_6160_ch04_pp330-387.indd 371

x 5 2 9x 4 1 18x 3 1 2x 2 2 5x 2 3 . What type of function x 4 2 6x 3 1 2x 1 1 is it? Perform this division using long division, and confirm that the graph corresponds to the quotient.

80. Plot

26x 3 1 7x 2 1 14x 2 15 . What type of function is it? 2x 1 3 Perform this division using long division, and confirm that the graph corresponds to the quotient.

81. Plot

23x 5 2 4x 4 1 29x 3 1 36x 2 2 18x . What type of function 3x 2 1 4x 2 2 is it? Perform this division using long division, and confirm that the graph corresponds to the quotient.

82. Plot

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CHAPTER 4  Polynomial and Rational Functions

4.4 THE REAL ZEROS OF A POLYNOMIAL FUNCTION SKILLS OBJECTIVES ■■ Apply the remainder theorem to evaluate a polynomial function and use the factor theorem to factor polynomials. ■■ Use the rational zero (root) theorem to list possible rational zeros and Descartes’ rule of signs to determine the possible combination of positive and negative real zeros. ■■ Express polynomials as a product of linear or irreducible quadratic factors. ■■ Employ the intermediate value theorem to approximate an irrational zero. ■■ Graph any polynomial function.

CONCEPTUAL OBJECTIVES ■■ Understand that a polynomial degree of n has at most n real zeros. ■■ Understand that a real zero can be either rational or irrational and that irrational zeros will not be listed as possible zeros through the rational zero test. ■■ Irreducible factors are quadratic expressions that are not factorable over the real numbers. ■■ Realize that rational zeros can be found exactly, whereas irrational zeros must be approximated. ■■ Understand that the rational root test and the intermediate value theorem make it possible for us to graph any polynomial function and that Descartes’ rule of signs and upper and lower bound rules help us determine the graphs more efficiently.

4.4.1 The Remainder Theorem and the Factor Theorem 4.4.1 S K I L L

Apply the remainder theorem to evaluate a polynomial function and use the factor theorem to factor polynomials. 4.4.1 C O N C E P T U A L

Understand that a polynomial degree of n has at most n real zeros.

The zeros of a polynomial function assist us in finding the x-intercepts of the graph of a polynomial function. How do we find the zeros of a polynomial function? For polynomial functions of degree 2, we have the quadratic formula, which allows us to find the two zeros. For polynomial functions whose degree is greater than 2, much more work is required.*1 In this section, we focus our attention on finding the real zeros of a polynomial function. Later, in Section 4.5, we expand our discussion to complex zeros of polynomial functions. In this section, we start by listing possible rational zeros. As you will see, there are sometimes many possibilities. We can then narrow the search using Descartes’ rule of signs, which tells us possible combinations of positive and negative real zeros. We can narrow the search even further with the upper and lower bound rules. Once we have tested possible values and determined a zero, we will employ synthetic division to divide the polynomial by the linear factor associated with the zero. We will continue the process until we have factored the polynomial function into a product of either linear factors or irreducible quadratic factors. Last, we will discuss how to find irrational real zeros using the intermediate value theorem. If we divide the polynomial function ƒ1x2 5 x3 2 2x2 1 x 2 3 by x 2 2 using synthetic division, we find the remainder is 21. 2

1

1

22

1

23

2

0

2

0

1

21

Notice that if we evaluate the function at x 5 2, the result is 21. ƒ122 5 21 This leads us to the remainder theorem.

*There are complicated formulas for finding the zeros of polynomial functions of degree 3 and 4, but there are no such formulas for degree 5 and higher polynomials (according to the Abel–Ruffini theorem)

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4.4  The Real Zeros of a Polynomial Function 

WORDS

MATH

Recall the division algorithm.

P1x2 5 d1x2 ⋅ Q1x2 1 r 1x2

The degree of the remainder is always less than the degree of the divisor: therefore, the remainder must be a constant 1Call it r, r 1 x 2 5 r 2 . Let x 5 a.

P1x2 5 1x 2 a2 ⋅ Q1x2 1 r 1x2 P1x2 5 1x 2 a2 ⋅ Q1x2 1 r

P1a2 5 1a 2 a2 ⋅ Q1x2 1 r f

Let d 1 x 2 5 x 2 a for any real number a.

373

0

P 1 a 2 5 r

Simplify.

REMAINDER THEOREM

If a polynomial P  1x2 is divided by x 2 a, then the remainder is r 5 P  1a2. The remainder theorem tells you that polynomial division can be used to evaluate a polynomial function at a particular point. EXAMPLE 1  Two Methods for Evaluating Polynomials

Let P  1x2 5 4x5 2 3x4 1 2x3 2 7x2 1 9x 2 5 and evaluate P  122 by a. Evaluating P  122 directly b. The remainder theorem and synthetic division Solution:

a. P  122 5 41225 2 31224 1 21223 2 71222 1 9122 2 5

5 41322 2 31162 1 2182 2 7142 1 9122 2 5 5 128 2 48 1 16 2 28 1 18 2 5 5 81

b. 2

4

23

2

27

9

25

8

10

24

34

86

5

12

17

43

81

4

▼ Y O U R T U R N   Let P  1x2 5 2x3 1 2x2 2 5x 1 2 and evaluate P  1222 using the

remainder theorem and synthetic division.

▼ ANSWER

P  12 22 5 28

Recall that when a polynomial is divided by x 2 a, if the remainder is zero, we say that x 2 a is a factor of the polynomial. Through the remainder theorem, we now know that the remainder is related to evaluation of the polynomial at the point x 5 a. We are then led to the factor theorem. FACTOR THEOREM

If P  1a2 5 0, then x 2 a is a factor of P  1x2. Conversely, if x 2 a is a factor of P  1x2, then P  1a2 5 0.

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CHAPTER 4  Polynomial and Rational Functions

EXAMPLE 2  Using the Factor Theorem to Factor a Polynomial

Determine whether x 1 2 is a factor of P  1x2 5 x3 2 2x2 2 5x 1 6. If so, factor P  1x2 completely. Solution:

By the factor theorem, x 1 2 is a factor of P  1x2 5 x3 2 2x2 2 5x 1 6 if P  1222 5 0. By the remainder theorem, if we divide P  1x2 5 x3 2 2x2 2 5x 1 6 by x 1 2, then the remainder is equal to P  1222. P  1x2 5 x3 2 2x2 2 5x 1 6 by x 1 2 using synthetic division.

STEP 1  Divide

1

22

25

6

22

8

26

24

3

0

f

1

22

x2 2 4x 1 3 Since the remainder is zero, P  1222 5 0, x 1 2 is a factor of P  1x2 5 x3 2 2x2 2 5x 1 6. P  1x2 as a product. P  1x2 5 1x 1 22 1x2 2 4x 1 32

STEP 2  Write

STEP 3 

▼ ANSWER

1x 2 12 is a factor; P  1x2 5 1x 2 52 1x 2 12 1x 1 22

▼

 actor the quadratic polynomial. F P  1x2 5 1x 1 22 1x 2 32 1x 2 12

Y O U R T U R N   Determine whether x 2 1 is a factor of P  1x2 5 x3 2 4x2 2 7x 1 10.

If so, factor P  1x2 completely.

EXAMPLE 3  Using the Factor Theorem to Factor a Polynomial

If a fourth-degree polynomial can be factored into four distinct linear factors P(x) 5 (x 2 a)(x 2 b)(x 2 c)(x 2 d ), what are the x-intercepts of the graph of this polynomial?

▼

Determine whether x 2 3 and x 1 2 are factors of P  1x2 5 x4 2 13x2 1 36. If so, factor P  1x2 completely. Solution:

STEP 1  With

synthetic division divide P  1x2 5 x4 2 13x2 1 36 by x 2 3.

3

1

1

ANSWER (a, 0) (b, 0) (c, 0) (d, 0)

0

213

0

36

3

9

212

236

24

212

0

3

f

[ CONCEPT CHECK ]

3

2

x 1 3x 2 4x 2 12 Because the remainder is 0,  x 2 3 is a factor , and we can write the polynomial as

synthetic division divide the remaining cubic polynomial 1x3 1 3x2 2 4x 2 122 by x 1 2.

22

1

1 2

3

24

212

22

22

12

1

26

0

f

STEP 2  With

P 1 x 2 5 1 x 2 3 2 1 x 3 1 3x 2 2 4x 2 12 2

x 1x26

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4.4  The Real Zeros of a Polynomial Function 

375

 ecause the remainder is 0,  x 1 2 is a factor , and we can now write the B polynomial as P 1 x 2 5 1 x 2 3 2 1 x 1 2 2 1 x2 1 x 2 6 2 STEP 3  Factor

the quadratic polynomial: x2 1 x 2 6 5 1x 1 32 1x 2 22.

P  1x2 as a product of linear factors:

STEP 4  Write

P1x2 5 1x 2 32 1x 2 22 1x 1 22 1x 1 32

▼ Y O U R T U R N   Determine whether x 2 3 and x 1 2 are factors of

P  1x2 5 x4 2 x3 2 7x2 1 x 1 6. If so, factor P  1x2 completely.

The Search for Real Zeros

▼ ANSWER

1 x 2 3 2 and 1 x 1 2 2 are factors; P  1x2 5 1x 2 32 1x 1 22 1x 2 12 1x 1 12

In all of the examples thus far, the polynomial function and one or more real zeros (or linear factors) were given. Now, we will not be given any real zeros to start with. Instead, we will develop methods to search for them. Each real zero corresponds to a linear factor, and each linear factor is of degree 1. Therefore, the largest number of real zeros a polynomial function can have is equal to the degree of the polynomial.

STUDY T I P THE NUMBER OF REAL ZEROS

A polynomial function cannot have more real zeros than its degree.

The largest number of zeros a polynomial can have is equal to the degree of the polynomial.

The following functions illustrate that a polynomial function of degree n can have at most n real zeros. POLYNOMIAL FUNCTION

REAL ZEROS

COMMENTS

2

x 5 63

Two real zeros

ƒ1x2 5 x 1 4

2

None

No real zeros

ƒ1x2 5 x3 2 1

3

x51

One real zero

3

x 5 22, 0, 3

Three real zeros

2 2

ƒ1x2 5 x 2 9

3

2

ƒ1x2 5 x 2 x 2 6x

DEGREE

Now that we know the maximum number of real zeros a polynomial function can have, let us discuss how to find these zeros.

4.4.2 The Rational Zero Theorem and Descartes’ Rule of Signs When the coefficients of a polynomial are integers, then the rational zero theorem (rational root test) gives us a list of possible rational zeros. We can then test these possible values to determine whether they really do correspond to actual zeros. Descartes’ rule of signs tells us the possible combinations of positive real zeros and negative real zeros. Using Descartes’ rule of signs will help us narrow down the large list of possible zeros ­generated through the rational zero theorem to a (hopefully) shorter list of possible zeros. First, let’s look at the rational zero theorem; then we’ll turn to Descartes’ rule of signs.

Young_AT_6160_ch04_pp330-387.indd 375

4.4.2 S K I L L

Use the rational zero (root) theorem to list possible rational zeros and Descartes’ rule of signs to determine the possible combination of positive and negative real zeros. 4.4.2 C O N C E P T U A L

Understand that a real zero can be either rational or irrational and that irrational zeros will not be listed as possible zeros through the rational zero test.

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CHAPTER 4  Polynomial and Rational Functions

THE RATIONAL ZERO THEOREM (RATIONAL ROOT TEST)

If the polynomial function P 1 x 2 5 anx n 1 an21x n21 1 c1 a2x 2 1 a1x 1 a0 has integer coefficients, then every rational zero of P  1x2 has the form: Rational zero 5

integer factors of a0 integer factors of constant term 5 integer factors of an integer factors of leading coefficient

positive integer factors of constant term   56 positive integer factors of leading coefficient

To use this theorem, simply list all combinations of integer factors of both the constant term a0 and the leading coefficient term an and take all appropriate combinations of ratios. This procedure is illustrated in Example 4. Notice that when the leading coefficient is 1, then the possible rational zeros will simply be the possible integer factors of the constant term.

EXAMPLE 4  Using the Rational Zero Theorem

Determine possible rational zeros for the polynomial P  1x2 5 x4 2 x3 2 5x2 2 x 2 6 by the rational zero theorem. Test each one to find all rational zeros Solution:

STEP 1  List

factors of the constant a0 5 26 61, 62, 63, 66 and leading coefficient terms       an 5 1   61

STEP 2  List

possible

a0 rational zeros .    an

There are three ways to test whether any of these are zeros: Substitute these values into the polynomial to see which ones yield zero; use either polynomial division or synthetic division to divide the polynomial by these possible zeros; and look for a zero remainder.

S TU DY TIP The remainder can be found by ­evaluating the function or synthetic division. For simple values like x 5 61, it is easier to evaluate the polynomial function. For other ­values, it is often easier to use synthetic division.

61 62 63 66 , , , 5 61, 62, 63, 66 61 61 61 61

STEP 3  Test



possible zeros by looking for zero remainders.

1 is not a zero:

P  112 5 1124 2 1123 2 51122 2 112 2 6 5 212

21 is not a zero: P  1212 5 12124 2 12123 2 512122 2 1212 2 6 5 28 We could continue testing with direct substitution, but let us now use synthetic division as an alternative. 2

1

21

25

21

26

2

2

26

214

1

1

23

27

220

1

21

25

21

26

22

6

22

6

23

1

23

0

2 is not a zero:

22 is a zero:

22

1

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4.4  The Real Zeros of a Polynomial Function 

377

Since 22 is a zero, then x 1 2 is a factor of P  1x2, and the remaining quotient is x3 2 3x2 1 x 2 3. Therefore, if there are any other real roots remaining, we can now use the simpler x3 2 3x2 1 x 2 3 for the dividend. Also note that the rational zero theorem can be applied to the new dividend and possibly shorten the list of possible rational zeros. In this case, the possible rational zeros of F1x2 5 x3 23x2 1 x 23 are 61 and 63.

3

3 is a zero:

1

23

1

23

3

0

3

0

1

0

1 We now know that 22 and 3 are confirmed zeros. If we continue testing, we will find that the other possible zeros fail. This is a fourth-degree polynomial, and we have found two rational real zeros. We see in the graph on the right that these two real zeros correspond to the x-intercepts.

y

STUD Y T I P Notice in Step 3 that the polynomial F 1 x 2 5 x 3 2 3x 2 1 x 2 3 can be facto­red by grouping: F 1 x 2 5 1 x 2 3 21 x 2 1 1 2 .

20

x –5

5

–20

▼ Y O U R T U R N   List the possible rational zeros of the polynomial

P  1x2 5 x4 1 2x3 2 2x2 1 2x 2 3, and determine rational real zeros.

Notice in Example 4 that the polynomial function P  1x2 5 x4 2 x3 2 5x2 2 x 2 6 had two rational real zeros, x 5 22 and x 5 3. This implies that x 1 2 and x 2 3 are factors of P  1x2. Also note in the last step that when we divided by the zero x 5 3, the quotient was x2 1 1. Therefore, we can write the polynomial in factored form as

▼ ANSWER

Possible rational zeros: 61 and 63. Rational real zeros: 1 and 23.

P  1x2 5 1x 1 22 1x 2 32 1x2 1 12 ()* ()* ()* linear factor

linear factor

irreducible quadratic factor

Notice that the first two factors are of degree 1, so we call them linear factors. The third expression, x2 1 1, is of degree 2 and cannot be factored in terms of real numbers. We will ­discuss complex zeros in the next section. For now, we say that a quadratic expression, ax2 1 bx 1 c, is called irreducible if it cannot be factored over the real numbers.

EXAMPLE 5  Factoring a Polynomial Function

Write the following polynomial function as a product of linear and/or irreducible quadratic factors:

Solution:

P  1x2 5 x4 2 4x3 1 4x2 2 36x 2 45.

Use the rational zero theorem to list possible rational roots.

x 5 61, 63, 65, 69, 615, 645

Test possible zeros by evaluating the function or by utilizing synthetic division. x 5 1 is not a zero.       P  112 5 280 x 5 21 is a zero.       P  1212 5 0

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CHAPTER 4  Polynomial and Rational Functions

Divide P  1x2 by x 1 1.

21

1

1 5

x 5 5 is a zero.

24

4

236

245

21

5

29

45

25

9

245

0

1

25

9

245

5

0

45

1 0 9 (++)'+* x2 1 9

0

The factor x2 1 9 is irreducible. Write the polynomial as a product of linear and/or irreducible quadratic factors.

P  1x2 5 1x 2 52 1x 1 12 1x2 1 92

Notice that the graph of this polynomial function has x-intercepts at x 5 21 and x 5 5.

▼

▼ ANSWER

Y O U R T U R N   Write the following polynomial function as a product of linear 2

P  1x2 5 1x 1 12 1x 2 32 1x 1 22

and/or irreducible quadratic factors. P  1x2 5 x4 2 2x3 2 x2 2 4x 2 6

The rational zero theorem lists possible zeros. It would be helpful if we could narrow that list. Descartes’ rule of signs determines the possible combinations of positive real zeros and negative real zeros through variations of sign. A variation in sign is a sign difference seen between consecutive coefficients. Sign Change 2 to 1

P 1 x 2 5 2x 6 2 5x 5 2 3x 4 1 2x 3 2 x 2 2 x 2 1 Sign Change 1 to 2

Sign Change 1 to 2

This polynomial experiences three sign changes or variations in sign.

DESCARTES’ RULE OF SIGNS

If the polynomial function P 1 x 2 5 a n x n 1 a n21x n21 1 c1 a2 x 2 1 a1 x 1 a 0 has real coefficients and a0 2 0, then: ■ The number of positive real zeros of the polynomial is either equal to the number of variations of sign of P  1x2 or less than that number by an even integer. ■ The number of negative real zeros of the polynomial is either equal to the number of variations of sign of P  12x2 or less than that number by an even integer. Descartes’ rule of signs narrows our search for real zeros because we don’t have to test all of the possible rational zeros. For example, if we know there is one positive real zero, then if we find a positive rational zero we no longer need to continue to test possible positive zeros

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4.4  The Real Zeros of a Polynomial Function 

379

EXAMPLE 6  Using Descartes’ Rule of Signs

Determine the possible combinations of zeros for P  1x2 5 x3 2 2x2 2 5x 1 6. Solution:

Determine the number of variations of sign in P  1x2.

Sign Change

P  1x2 5 x3 2 2x2 2 5x 1 6 Sign Change

P  1x2 has 2 variations in sign.

Apply Descartes’ rule of signs.     P  1x2 has either 2 or 0 positive real zeros.

Determine the number of variations of sign in P  12x2.

12x23 2 212x22 2 512x2 1 6 2x3 2 2x2 1 5x 1 6

P  12x2 has 1 variation in sign.

P  1x2 5 2x3 2 2x2 1 5x 1 6

Apply Descartes’ rule of signs.

Sign Change

P  1x2 must have 1 negative real zero.

Since P  1x2 5 x3 2 2x2 2 5x 1 6 is a third-degree polynomial, there are at most 3 real zeros. One zero is a negative real number, and there can be either 2 positive real zeros or 0 positive real zeros. Now look back at Example 2 and see that in fact there were 1 negative real zero and 2 positive real zeros.

EXAMPLE 7  U  sing Descartes’ Rule of Signs to Find Possible Combinations of Real Zeros

Determine the possible combinations of real zeros for P  1x2 5 x4 2 2x3 1 x2 1 2x 2 2. Sign Change

Solution:

P  1x2 has 3 variations in sign.

4

3

2

P  1x2 5 x 2 2x 1 x 1 2x 2 2 Sign Change

[ CONCEPT CHECK ] TRUE OR FALSE  The rational zeros are a subset of the real zeros.

▼ ANSWER True

Sign Change

Apply Descartes’ rule of signs.     P  1x2 has either 3 or 1 positive real zero. Find P  12x2.         P  12x2 5 12x24 2 212x23 1 12x22 1 212x2 2 2 5 x4 1 2x3 1 x2 2 2x 2 2 P  12x2 has 1 variation in sign. Apply Descartes’ rule of signs.

P  12x2 5 x4 1 2x3 1 x2 2 2x 2 2 Sign Change

P  1x2 has 1 negative real zero.

Since P  1x2 5 x4 2 2x3 1 x2 1 2x 2 2 is a fourth-degree polynomial, there are at most 4 real zeros. P  1x2 has 1 negative real zero and could have 3 or 1 positive real zeros.

▼

Y O U R T U R N   Determine the possible combinations of zeros for

P  1x2 5 x4 1 2x3 1 x2 1 8x 2 12

Young_AT_6160_ch04_pp330-387.indd 379

▼ ANSWER

Positive real zeros: 1 Negative real zeros: 3 or 1

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CHAPTER 4  Polynomial and Rational Functions

4.4.3  Factoring Polynomials 4.4.3 S K I L L

Express polynomials as a product of linear or irreducible quadratic factors. 4.4.3 C O N C E P T U A L

Irreducible factors are quadratic expressions that are not factorable over the real numbers.

Now let’s draw on the tests discussed in this section thus far to help us in finding all real zeros of a polynomial function. Doing so will enable us to factor polynomials. EXAMPLE 8  Factoring a Polynomial

Write the polynomial P  1x2 5 x5 1 2x4 2 x 2 2 as a product of linear and/or irreducible quadratic factors. Solution: STEP 1  Determine

variations in sign.

P  1x2 has 1 sign change.       P  1x2 5 x5 1 2x4 2 x 2 2 P  12x2 has 2 sign changes.      P  12x2 5 2x5 1 2x4 1 x 2 2

STEP 2  Apply

Positive Real Zeros:   1 Negative Real Zeros:    2 or 0

Descartes’ rule of signs.

STEP 3  Use

the rational zero theorem to determine the possible rational zeros.  61, 62

We know (Step 2) that there is one positive real zero, so test the possible positive rational zeros first. STEP 4  Test



[ CONCEPT CHECK ] TRUE OR FALSE  An irreducible quadratic factor corresponds to two additional x-intercepts.

▼ ANSWER False

possible rational zeros.

1

1

1 is a zero: 1

Now that we have found the positive zero, we can 21 test the other two possible negative zeros—because either they both are zeros or neither is a zero.

2

0

0

21

22

1

3

3

3

2

3

3

3

2

0

1

3

3

3

2

21

22

21

22

1

2

1

2

0

1

2

1

2

22

0

22

0

1

0

21 is a zero:



22 is a zero:

STEP 5  Three

1

f

At this point, from Descartes’ 22 rule of signs we know that 22 must also be a zero, since there are either 2 or 0 negative zeros. Let’s confirm this.

2

x 11

of the five zeros have been found to be zeros: 21, 22, and 1.

STEP 6  Write

the fifth-degree polynomial as a product of 3 linear factors and an irreducible quadratic factor.

▼ ▼ ANSWER

P  1x2 5 1x 2 22 1x 1 12 1x 2 12 1x2 1 22

Young_AT_6160_ch04_pp330-387.indd 380

P 1 x 2 5 1 x 2 1 2 1 x 1 1 2 1 x 1 2 2 Ax 2 1 1B

Y O U R T U R N   Write the polynomial P  1x2 5 x5 2 2x4 1 x3 2 2x2 2 2x 1 4 as a

product of linear and/or irreducible quadratic factors.

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4.4  The Real Zeros of a Polynomial Function 

The rational zero theorem gives us possible rational zeros of a polynomial, and Descartes’ rule of signs gives us possible combinations of positive and negative real zeros. Additional aids that help eliminate possible zeros are the upper and lower bound rules. These rules can give you an upper and lower bound on the real zeros of a polynomial function. If ƒ1x2 has a common monomial factor, you should factor it out first, and then follow the upper and lower bound rules.

381

STUDY T I P If f (x) has a common monomial factor, it should be factored out before applying the bound rules.

UPPER AND LOWER BOUND RULES

Let ƒ1x2 be a polynomial with real coefficients and a positive leading coefficient. Suppose ƒ1x2 is divided by x 2 c using synthetic division. 1. If c . 0 and each number in the bottom row is either positive or zero, c is an upper bound for the real zeros of ƒ. 2. If c , 0 and the numbers in the bottom row are alternately positive and negative (zero entries count as either positive or negative), c is a lower bound for the real zeros of ƒ. EXAMPLE 9  Using  Upper and Lower Bounds to Eliminate Possible Zeros

Find the real zeros of ƒ1x2 5 4x3 2 x2 1 36x 2 9. Solution: STEP 1  The

rational zero theorem gives possible rational zeros. Factors of 9 61, 63, 69 5 Factors of 4 61, 62, 64



1 1 3 3 9 9 5 61, 6 , 6 , 6 , 6 , 6 , 63, 6 , 69 2 4 4 2 4 2

STEP 2  Apply



Descartes’ rule of signs: ƒ1x2 has 3 sign variations. ƒ12x2 has no sign variations.

STEP 3  Try

x 5 1.

3 or 1 positive real zeros no negative real zeros 1

4

4

21

36

29

4

3

39

3

39

30

 x 5 1 is not a zero, but because the last row contains all positive entries, x 5 1 is an upper bound. Since we know there are no negative real zeros, we restrict our search to between 0 and 1. STEP 4  Try

x 5 14.

1 4

4

4

21

36

29

1

0

9

0

36

0

1 2  4 is a zero and the quotient 4x 1 36 has all positive coefficients; therefore,



1 4

is the only real zero .

 Note: If ƒ1x2 has a common monomial factor, it should be factored out first before applying the bound rules.

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CHAPTER 4  Polynomial and Rational Functions

4.4.4  The Intermediate Value Theorem 4.4.4 S K I L L

Employ the intermediate value theorem to approximate an irrational zero. 4.4.4 C O N C E P T U A L

Realize that rational zeros can be found exactly, whereas irrational zeros must be approximated.

In our search for zeros, we sometimes encounter irrational zeros, as in, for example, the polynomial ƒ 1 x 2 5 x5 2 x4 2 1

Descartes’ rule of signs tells us there is exactly one real positive zero. However, the rational zero test yields only x 5 61, neither of which is a zero. So if we know there is a real positive zero and we know it’s not rational, it must be irrational. Notice that ƒ112 5 21 and ƒ122 5 15. Since polynomial functions are continuous and the function goes from negative to positive between x 5 1 and x 5 2, we expect a zero somewhere in that interval. Generating a graph with a graphing utility, we find that there is a zero around x 5 1.3.

The intermediate value theorem is based on the fact that polynomial functions are continuous. INTERMEDIATE VALUE THEOREM

y

Let a and b be real numbers such that a , b and ƒ1x2 be a polynomial function. If ƒ1a2 and ƒ1b2 have opposite signs, then there is at least one real zero between a and b.

f (x) f (b)

x

a b

f (a)

zero

If the intermediate value theorem tells us that there is a real zero in the interval 1a, b2, how do we approximate that zero? The bisection method* is a root-finding algorithm that approximates the solution to the equation ƒ1x2 5 0. In the bisection method, the interval is divided in half, and then the subinterval that contains the zero is selected. This is repeated until the bisection method converges to an approximate root of ƒ. EXAMPLE 10  Approximating Real Zeros of a Polynomial Function

Approximate the real zero of ƒ1x2 5 x5 2 x4 2 1. Note: Descartes’ rule of signs tells us that there are no real negative zeros and there is exactly one real positive zero. Solution:

Find two consecutive integer values for x that have corresponding function values opposite in sign.

x

f (x)

1

21

2

15

Note that a graphing utility would have shown an x-intercept between x 5 1 and x 5 2. * In calculus you will learn Newton’s method, which is a more efficient approximation technique for finding zeros.

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4.4  The Real Zeros of a Polynomial Function 

Apply the bisection method, with a 5 1 and b 5 2. Evaluate the function at x 5 c. Compare the values of ƒ at the endpoints and midpoint. Select the subinterval corresponding to the opposite signs of ƒ. Apply the bisection method again (repeat the algorithm). Evaluate the function at x 5 1.25.

c5

a1b 112 3 5 5 2 2 2

ƒ 1 1.5 2 < 1.53

ƒ112 5 21, ƒ 1 1.5 2 < 1.53, ƒ122 5 15

383

[ CONCEPT CHECK ] TRUE OR FALSE  If we know there is one positive zero and all of the potential rational roots have been found not to be zeros, then the real zero must be irrational.

▼ ANSWER True

11, 1.52

1 1 1.5 5 1.25 2

ƒ 1 1.25 2 < 20.38965

Compare the values of ƒ at the endpoints and midpoint. ƒ112 5 21, ƒ 1 1.25 2 < 20.38965, ƒ 1 1.5 2 < 1.53 Select the subinterval corresponding to the opposite signs of ƒ. 11.25, 1.52 Apply the bisection method again (repeat the algorithm).

1.25 1 1.5 5 1.375 2

Evaluate the function at x 5 1.375. ƒ 1 1.375 2 < 0.3404 Compare the values of ƒ at the endpoints and midpoint. ƒ 1 1.25 2 < 20.38965, ƒ 1 1.375 2 < 0.3404, ƒ 1 1.5 2 < 1.53 Select the subinterval corresponding to the opposite signs of ƒ. 11.25, 1.3752 We can continue this procedure (applying the bisection method ) to find that the zero is somewhere between  ƒ 1 1.32 2 < 20.285 and ƒ 1 1.33 2 < 0.0326.

1.32 is an approximation to the real zero. We find that to three significant digits   

4.4.5  Graphing Polynomial Functions In Section 4.2, we graphed simple polynomial functions that were easily factored. Now that we have procedures for finding real zeros of polynomial functions (rational zero theorem, Descartes’ rule of signs, and upper and lower bound rules for rational zeros, and the intermediate value theorem and the bisection method for irrational zeros), let us return to the topic of graphing polynomial functions. Since a real zero of a polynomial function corresponds to an x-intercept of its graph, we now have methods for finding (or estimating) any x-intercepts of the graph of any polynomial function. EXAMPLE 11  Graphing a Polynomial Function

Graph the function ƒ1x2 5 2x4 2 2x3 1 5x2 1 17x 2 22. Solution: STEP 1  Find

4.4.5 S K I L L

Graph any polynomial function. 4.4.5 C O N C E P T U A L

Understand that the rational root test and the intermediate value theorem make it possible for us to graph any polynomial function and Descartes’ rule of signs and that upper and lower bound rules help us determine the graphs more efficiently.

the y-intercept. ƒ102 5 222

STEP 2  Find

any x-intercepts (real zeros). Apply Descartes’ rule of signs.  3 sign changes correspond to 3 or 1 positive real zeros. ƒ1x2 5 2x4 2 2x3 1 5x2 1 17x 2 22

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CHAPTER 4  Polynomial and Rational Functions



1 sign change corresponds to 1 negative real zero. Apply the rational zero theorem.



Let a0 5 222 and an 5 2.

ƒ12x2 5 2x4 1 2x3 1 5x2 2 17x 2 22

Factors of a0 1 11 5 6 , 61, 62, 6 , 611, 622 Factors of an 2 2

Test the possible zeros. x 5 1 is a zero. There are no other rational zeros. Apply the upper bound rule.

ƒ112 5 0 1

2

22

5

17

222

2

0

5

22

0

5

22

0

2

Since x 5 1 is positive and all of the numbers in the bottom row are positive (or zero), x 5 1 is an upper bound for the real zeros. We know there is exactly one negative real zero, but none of the possible zeros from the rational zero theorem is a zero. Therefore, the negative real zero is irrational. Apply the intermediate value theorem and the bisection method. ƒ is positive at x 5 22. ƒ1222 5 12 ƒ is negative at x 5 21. ƒ1212 5 230 Use the bisection method to find the negative x < 21.85 real zero between 22 and 21.

[ CONCEPT CHECK ]

STEP 3  Determine

y 5 2x4

the end behavior.

y

TRUE OR FALSE  We now have the tools to find all x-intercepts (exactly or approximated).

▼

x

ANSWER True

STEP 4  Find

additional points.

x

22

21.85

21

0

1

2

f (x)

12

0

230

222

0

48

122, 122

121.85, 02

121, 2302

10, 2222

11, 02

12, 482

Point

STEP 5  Sketch

y

the graph.

48

(2, 48)

(–2, 12) x

(–1.85, 0) (1, 0) –2

2 (0, –22)

(–1, –30)

Young_AT_6160_ch04_pp330-387.indd 384

–32

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4.4  The Real Zeros of a Polynomial Function 

385

[SEC TION 4.4]  S U M M A RY In this section, we discussed how to find the real zeros of a P R O C E D U R E F O R FA C T O R I N G A polynomial function. Once real zeros are known, it is possible P O LY N O M I A L F U N C T I O N to write the polynomial function as a product of linear and/or n List possible rational zeros (rational zero theorem). irreducible quadratic factors. n List possible combinations of positive and negative real zeros (Descartes’ rule of signs). THE NUMBER OF ZEROS n Test possible values until a zero is found.* n Once a real zero is found, repeat testing on the quotient until n A polynomial of degree n has at most n real zeros. linear and/or irreducible quadratic factors remain. n Descartes’ rule of signs determines the possible combinan If there is a real zero but all possible rational roots have tions of positive and negative real zeros. failed, then approximate the zero using the intermediate n Upper and lower bounds help narrow the search for zeros. value theorem and the bisection method. HOW TO FIND ZEROS n

Rational zero theorem: List possible rational zeros: Factors of constant, a0 Factors of leading coefficient, an

n

Irrational zeros: Approximate zeros by determining when the polynomial function changes sign (intermediate value theorem).

*

Depending on the form of the quotient, upper and lower bounds may eliminate possible zeros.

[SEC T I O N 4 . 4]   E X E R C I S E S • SKILLS In Exercises 1–6, find the following values by using synthetic division. Check by substituting the value into the function

1. ƒ112

2. ƒ1212

ƒ1x2 5 3x4 2 2x3 1 7x2 2 8    g  1x2 5 2x3 1 x2 1 1 3. g  112

4. g  1212

5. ƒ1222

In Exercises 7–10, determine whether the number given is a zero of the polynomial.

6. g  122

7. 27, P  1x2 5 x3 1 2x2 2 29x 1 42 8. 2, P  1x2 5 x3 1 2x2 2 29x 1 42 9. 23, P  1x2 5 x3 2 x2 2 8x 1 12 10. 1, P  1x2 5 x3 2 x2 2 8x 1 12

In Exercises 11–20, given a real zero of the polynomial, determine all other real zeros, and write the polynomial in terms of a product of linear and/or irreducible quadratic factors.   Polynomial 11. P  1x2 5 x3 2 13x 1 12 13. P  1x2 5 2x3 1 x2 2 13x 1 6 15. P  1x2 5 x4 2 2x3 2 11x2 2 8x 2 60 17. P  1x2 5 x4 2 5x2 1 10x 2 6

19. P  1x2 5 x4 1 6x3 1 13x2 1 12x 1 4

Zero Polynomial 1 12. P  1x2 5 x3 1 3x2 2 10x 2 24 1 P  1x2 5 3x3 2 14x2 1 7x 1 4 2 14. 23, 5 16. P  1x2 5 x4 2 x3 1 7x2 2 9x 2 18 1, 23 18. P  1x2 5 x4 2 4x3 1 x2 1 6x 2 40 22 1multiplicity 22 20. P  1x2 5 x4 1 4x3 2 2x2 2 12x 1 9

In Exercises 21–28, use the rational zero theorem to list the possible rational zeros. 21. P  1x2 5 x4 1 3x2 2 8x 1 4 22. P  1x2 5 2x4 1 2x3 2 5x 1 4 23. P  1x2 5 x5 2 14x3 1 x2 2 15x 1 12 24. P  1x2 5 x5 2 x3 2 x2 1 4x 1 9 6 4 3 25. P  1x2 5 2x 2 7x 1 x 2 2x 1 8 26. P  1x2 5 3x5 1 2x4 2 5x3 1 x 2 10 27. P  1x2 5 5x5 1 3x4 1 x3 2 x 2 20 28. P  1x2 5 4x6 2 7x4 1 4x3 1 x 2 21

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Zero 3 231 21, 2 4, 22 1 1multiplicity 22

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CHAPTER 4  Polynomial and Rational Functions

In Exercises 29–32, list the possible rational zeros, and test to determine all rational zeros. 29. P  1x2 5 x4 1 2x3 2 9x2 2 2x 1 8 30. P  1x2 5 x4 1 2x3 2 4x2 2 2x 1 3 31. P  1x2 5 2x3 2 9x2 1 10x 2 3 32. P  1x2 5 3x3 2 5x2 2 26x 2 8

In Exercises 33–44, use Descartes’ rule of signs to determine the possible number of positive real zeros and negative real zeros. 33. P  1x2 5 x4 2 32 34. P  1x2 5 x4 1 32

35. P  1x2 5 x5 2 1 36. P  1x2 5 x5 1 1 37. P  1x2 5 x5 2 3x3 2 x 1 2 38. P  1x2 5 x4 1 2x2 2 9 39. P  1x2 5 9x7 1 2x5 2 x3 2 x 40. P  1x2 5 16x7 2 3x4 1 2x 2 1

41. P  1x2 5 x6 2 16x4 1 2x2 1 7 42. P  1x2 5 27x6 2 5x4 2 x2 1 2x 1 1 43. P  1x2 5 23x4 1 2x3 2 4x2 1 x 2 11 44. P  1x2 5 2x4 2 3x3 1 7x2 1 3x 1 2

For each polynomial in Exercises 45–62: (a) use Descartes’ rule of signs to determine the possible combinations of positive real zeros and negative real zeros; (b) use the rational zero test to determine possible rational zeros; (c) test for rational zeros; and (d) factor as a product of linear and/or irreducible quadratic factors. 45. P(x) 5 x3 1 6x2 1 11x 1 6 46. P  1x2 5 x3 2 6x2 1 11x 2 6

47. P  1x2 5 x3 2 7x2 2 x 1 7 48. P  1x2 5 x3 2 5x2 2 4x 1 20 49. P  1x2 5 x4 1 6x3 1 3x2 2 10x 50. P  1x2 5 x4 2 x3 2 14x2 1 24x

51. P  1x2 5 x4 2 7x3 1 27x2 2 47x 1 26 52. P  1x2 5 x4 2 5x3 1 5x2 1 25x 2 26 53. P  1x2 5 10x3 2 7x2 2 4x 1 1 54. P  1x2 5 12x3 2 13x2 1 2x 2 1 55. P  1x2 5 6x3 1 17x2 1 x 2 10 56. P  1x2 5 6x3 1 x2 2 5x 2 2

57. P  1x2 5 x4 2 2x3 1 5x2 2 8x 1 4 58. P  1x2 5 x4 1 2x3 1 10x2 1 18x 1 9 59. P  1x2 5 x6 1 12x4 1 23x2 2 36 60. P  1x2 5 x4 2 x2 2 16x2 1 16

61. P  1x2 5 4x4 2 20x3 1 37x2 2 24x 1 5 62. P  1x2 5 4x4 2 8x3 1 7x2 1 30x 1 50

In Exercises 63–66, use the information found in Exercises 47, 51, 55, and 61 to assist in sketching a graph of each polynomial function. 63. See Exercise 47.

64. See Exercise 51.

65. See Exercise 55.

66. See Exercise 61.

In Exercises 67–72, use the intermediate value theorem and the bisection method to approximate the real zero in the indicated interval. Approximate to two decimal places. 67. ƒ1x2 5 x4 2 3x3 1 4   31, 24 68. ƒ1x2 5 x5 2 3x3 1 1   30, 14

69. ƒ1x2 5 7x5 2 2x2 1 5x 2 1   30, 14 70. ƒ1x2 5 22x3 1 3x2 1 6x 2 7   322, 214 71. ƒ1x2 5 x3 2 2x2 2 8x 2 3   321, 04 72. ƒ1x2 5 x4 1 4x2 2 7x 2 13   322, 214

• A P P L I C AT I O N S

73. Profit. A mathematics honorary society wants to sell

­ agazine subscriptions to Math Weekly. If there are x hundred m subscribers, its monthly revenue and cost are given by: R 1 x 2 5 46 2 3x 2 and C 1 x 2 5 20 1 2x



a.  Determine the profit function. Hint: P 5 R 2 C.



b.  Determine the number of subscribers needed in order to

break even. 74. Profit. Using the profit equation P  1x2 5 x3 2 5x2 1 3x 1 6,

when will the company break even if x represents the units sold?

For Exercises 75 and 76, refer to the following: The demand function for a product is p 1 x 2 5 28 2 0.0002x

where p is the unit price (in dollars) of the product and x is the number of units produced and sold. The cost function for the ­product is C 1 x 2 5 20x 1 1500

Young_AT_6160_ch04_pp330-387.indd 386

where C is the total cost (in dollars) and x is the number of units produced. The total profit obtained by producing and selling x units is P 1 x 2 5 xp 1 x 2 2 C 1 x 2

75. Business. Find the total profit function when x units are

p­ roduced and sold. Use Descartes’ rule of signs to d­ etermine possible combinations of positive zeros for the profit ­function 76. Business. Find the break-even point(s) for the product to the nearest unit. Discuss the significance of the break-even point(s) for the product. 77. Health/Medicine. During the course of treatment of an ­illness, the concentration of a dose of a drug (in mcg/mL) in the bloodstream fluctuates according to the mode



C 1 t 2 5 15.4 2 0.05t 2

where t 5 0 is when the drug was administered. Assuming a single dose of the drug is administered, in how many hours (to the nearest hour) after being administered will the drug be eliminated from the bloodstream?

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4.4  The Real Zeros of a Polynomial Function 

78. Health/Medicine. During the course of treatment of an



387

where t 5 0 is when the drug was administered. Assuming a single dose of the drug is administered, in how many hours (to the nearest hour) after being administered will the drug be eliminated from the bloodstream?

illness, the concentration of a dose of a drug (in mcg/mL) in the bloodstream fluctuates according to the mode C 1 t 2 5 60 2 0.75t 2

• C AT C H T H E M I S TA K E In Exercises 79 and 80, explain the mistake that is made. 79.  Use Descartes’ rule of signs to determine the possible

80. Determine whether x 2 2 is a factor of

P  1x2 5 x3 2 2x2 2 5x 1 6.

combinations of zeros of

Solution:

P 1 x 2 5 2x 5 1 7x 4 1 9x 3 1 9x 2 1 7x 1 2



22

1

No sign changes, so no positive real zeros. P 1 x 2 5 2x 1 7x 1 9x 1 9x 1 7x 1 2 5



Solution:

4

3

22

25

6

22

8

26

24

3

0

2

1

Five sign changes, so five negative real zeros

P 1 2x 2 5 22x 5 1 7x 4 2 9x 3 1 9x 2 2 7x 1 2

This is incorrect. What mistake was made?

Yes, x 2 2 is a factor of P  1x2.

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 81–84 determine whether each statement is true or false. 81. All real zeros of a polynomial correspond to x-intercepts.

83. A polynomial of degree n, n . 0, can be written as a product

82. A polynomial of degree n, n . 0, must have at least one zero.

of n linear factors over real numbers. 84. The number of sign changes in a polynomial is equal to the number of positive real zeros of that polynomial.

• CHALLENGE 85. Given that x 5 a is a zero of

P  1x2 5 x3 2 1a 1 b 1 c2x2 1 1ab 1 ac 1 bc2x 2 abc, find the other two zeros, given that a, b, c are real numbers and a . b . c.

86. Given that x 5 a is a zero of

p  1x2 5 x3 1 12a 1 b 2 c2x2 2 1ab 1 bc 2 ac2x 1 abc, find the other two real zeros, given that a, b, c are real positive numbers.

• TECHNOLOGY In Exercises 87 and 88, determine all possible rational zeros of the polynomial. There are many possibilities. Instead of ­trying them all, use a graphing calculator or software to graph P 1x2 to help find a zero to test. 87. P  1x2 5 x3 2 2x2 1 16x 2 32 88. P  1x2 5 x3 2 3x2 1 16x 2 48

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In Exercises 89 and 90: (a) determine all possible rational zeros of the polynomial, use a graphing calculator or software to graph P 1x2 to help find the zeros; and (b) factor as a product of linear and/or irreducible quadratic factors. 89. P 1 x 2 5 12x 4 1 25x 3 1 56x 2 2 7x 2 30 90. P 1 x 2 5 – 3x 3 2 x 2 2 7x 2 49

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4.5 COMPLEX ZEROS: THE FUNDAMENTAL THEOREM OF ALGEBRA SKILLS OBJECTIVES ■■ Factor a polynomial function given certain zeros. ■■ Factor a polynomial function of degree n into n linear factors.

CONCEPTUAL OBJECTIVES ■■ Understand why complex zeros occur in conjugate pairs. ■■ Understand why an odd-degree polynomial must have at least one real zero.

4.5.1  Complex Zeros 4.5.1 S K I L L

Factor a polynomial function given certain zeros. 4.5.1 C O N C E P T U A L

Understand why complex zeros occur in conjugate pairs.

[

[

S TU DY TIP

The zeros of a polynomial can be complex numbers. Only when the zeros are real numbers do we interpret zeros as x-intercepts. y 8 (0, 4) (–2, 0)

x (2, 0) (0, –4)

8

S TU DY TIP The largest number of zeros a polynomial can have is equal to the degree of the polynomial.

[ CONCEPT CHECK ] TRUE OR FALSE  Complex zeros correspond to x-intercepts.

▼ ANSWER False

In Section 4.4, we found the real zeros of a polynomial function. In this section we find the complex zeros of a polynomial function. In this chapter, we assume the coefficients of polynomial functions are real numbers. The domain of polynomial functions thus far has been the set of all real numbers. Now, we consider a more general case. In this section, the coefficients of a polynomial function and the domain of a polynomial function are complex numbers. Note that the set of real numbers is a subset of the complex numbers. (Choose the imaginary part to be zero.) It is important to note, however, that when we are discussing graphs of polynomial functions, we restrict the domain to the set of real numbers. A zero of a polynomial P  1x2 is the solution or root of the equation P  1x2 5 0. The zeros of a polynomial can be complex numbers. However, since the xy-plane represents real numbers, we interpret zeros as x-intercepts only when the zeros are real numbers. We can illustrate the relationship between real and complex zeros of polynomial functions and their graphs with two similar examples. Let’s take the two quadratic functions ƒ 1x2 5 x2 2 4 and g  1x2 5 x2 1 4. The graphs of these two functions are parabolas that open upward with ƒ 1x2 shifted down four units and g  1x2 shifted up four units as shown on the left. Setting each function equal to zero and solving for x, we find that the zeros for ƒ 1x2 are 22 and 2 and the zeros for g  1x2 are 22i and 2i. Notice that the x-intercepts for ƒ 1x2 are  122, 02 and  12, 02 and g 1x2 has no x-intercepts.

The Fundamental Theorem of Algebra

In Section 4.4, we were able to write a polynomial function as a product of linear and/or irreducible quadratic factors. Now, we consider factors over complex numbers. Therefore, what were irreducible quadratic factors over real numbers will now be a product of two linear factors over the complex numbers. What are the minimum and maximum number of zeros a polynomial can have? Every polynomial has at least one zero (provided the degree is greater than zero). The largest number of zeros a polynomial can have is equal to the degree of the polynomial. THE FUNDAMENTAL THEOREM OF ALGEBRA

Every polynomial P  1x2 of degree n . 0 has at least one zero in the complex ­number system. The fundamental theorem of algebra and the factor theorem are used to prove the ­following n zeros theorem. n ZEROS THEOREM

Every polynomial P  1x2 of degree n . 0 can be expressed as the product of n linear factors. Hence, P  1x2 has exactly n zeros, not necessarily distinct.

These two theorems are illustrated with five polynomials below.

 a. The first-degree polynomial ƒ1x2 5 x 1 3 has exactly one zero: x 5 23.  b. The second-degree polynomial ƒ1x2 5 x2 1 10x 1 25 5 1x 1 52 1x 1 52 has exactly two zeros: x 5 25 and x 5 25. It is customary to write this as a single zero of multiplicity 2 or refer to it as a repeated root.

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389

 c. The third-degree polynomial ƒ1x2 5 x3 1 16x 5 x  1x2 1 162 5 x  1x 1 4i2 1x 2 4i2 has exactly three zeros: x 5 0, x 5 24i, and x 5 4i.  d. The fourth-degree polynomial ƒ1x2 5 x4 2 1 5 1x2 2 12 1x2 1 12 5 1x 2 12 1x 1 12 1x 2 i2 1x 1 i2 has exactly four zeros: x 5 1, x 5 21, x 5 i, and x 5 2i.  e. The fifth-degree polynomial ƒ1x2 5 x5 5 x ⋅ x⋅ x⋅ x⋅ x has exactly five zeros: x 5 0, which has multiplicity 5. The fundamental theorem of algebra and the n zeros theorem only tell you that the zeros exist—not how to find them. We must rely on techniques discussed in Section 4.4 and additional strategies discussed in this section to determine the zeros.

Complex Conjugate Pairs Often, at a grocery store or a drugstore, we see signs for special offers—“buy one, get one free.” A similar phenomenon occurs for complex zeros of a polynomial function with real coefficients. If we restrict the coefficients of a polynomial to real numbers, complex zeros always come in conjugate pairs. In other words, if a zero of a polynomial function is a complex number, then another zero will always be its complex conjugate. Look at the third-degree polynomial in the above illustration, part (c), where two of the zeros were 24i and 4i, and in part (d), where two of the zeros were i and 2i. In general, if we restrict the coefficients of a polynomial to real numbers, complex zeros always come in conjugate pairs.

STUD Y T I P If we restrict the coefficients of a polynomial to real numbers, complex zeros always come in conjugate pairs.

COMPLEX CONJUGATE ZEROS THEOREM

If a polynomial P  1x2 has real coefficients, and if a 1 bi is a zero of P  1x2, then its ­complex conjugate a 2 bi is also a zero of P  1x2. EXAMPLE 1  Z  eros That Appear as Complex Conjugates

Find the zeros of the polynomial P  1x2 5 x2 2 4x 1 13. Solution:

Set the polynomial equal to zero.   P  1x2 5 x2 2 4x 1 13 5 0

Use the quadratic formula to solve for x.

x5

2 1 24 2 6 " 1 24 2 2 2 4 1 1 2 1 13 2 2112

x 5 2 6 3i Simplify. The zeros are the complex conjugates 2 2 3i and 2 1 3i. Check: This is a second-degree polynomial, so we expect two zeros. EXAMPLE 2  F  inding a Polynomial Given Its Zeros

Find a polynomial of minimum degree that has the zeros: 22, 1 2 i, 1 1 i. Solution:

Write the factors corresponding to each zero: 22: 1x 1 22 1 1 i: 3x 2 11 1 i24 1 2 i: 3x 2 11 2 i24 Express the polynomial as the product P1x2 5 1x 1 22 3x 2 11 1 i2 4 3x 2 11 2 i2 4 of the three factors. Regroup inner parentheses.

P1x2 5 1x 1 22 3 1x 2 12 2 i4 3 1x 2 12 1 i4

g

Use the difference of two squares formula P1x2 5 1x 1 22 3 1x 2 12 2 i4 3 1x 2 12 1 i4 1a 2 b2 1a 1 b2 5 a2 2 b2 for the product of the latter two factors.

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1x 2 1222i2

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CHAPTER 4  Polynomial and Rational Functions

P1x2 5 1x 1 22 3 1x 2 122 2 i24 d

Simplify.

x2 2 2x + 1 11

P 1 x 2 5 1 x 1 2 2 1 x 2 2 2x 1 2 2 P 1 x 2 5 x 3 2 2x 2 1 2x 1 2x 2 2 4x 1 4 P 1 x 2 5 x 3 2 2x 1 4

▼ ANSWER

d

390 

▼

Y O U R T U R N   Find a polynomial of minimum degree that has the zeros:

1, 2 2 i, 2 1 i.

P 1 x 2 5 x 3 2 5x 2 1 9x 2 5

EXAMPLE 3  Factoring a Polynomial with Complex Zeros

Factor the polynomial P  1x2 5 x4 2 x3 2 5x2 2 x 2 6 given that i is a zero of P  1x2. Since P  1x2 is a fourth-degree polynomial, we expect four zeros. The goal in this problem is to write P  1x2 as a product of four linear factors: P  1x2 5 1x 2 a2 1x 2 b2 1x 2 c2 1x 2 d2, where a, b, c, and d are complex numbers and represent the zeros of the polynomial.

Solution:

Write known zeros and linear factors. Since i is a zero, we know that 2i is a zero.

x 5 i and x 5 2i

We now know two linear factors of P  1x2.

1x 2 i2 and 1x 1 i2

Write P  1x2 as a product of four factors.

P  1x2 5 1x 2 i2 1x 1 i2 1x 2 c2 1x 2 d2

Multiply the two known factors.    1x 1 i2 1x 2 i2 5 x2 2 i2 5 x2 2 1212 5 x2 1 1 Rewrite the polynomial.  P  1x2 5 1x2 1 12 1x 2 c2 1x 2 d2 Divide both sides of the equation by x2 1 1.

P1x2 5 1x 2 c2 1x 2 d 2 x 11 2

Divide P  1x2 by x2 1 1 using long division. x2 2 x 2 6 2 x 1 0x 11 qx 4 2 x 3 2 5x 2 2 x 2 6 4 3 2 2 1 x 1 0x 1 x 2 2x 3 2 6x 2 2 x 2 1 2x 3 1 0x 2 2 x 2



26x 2 1 0x 2 6 2 2 126x 1 0x 2 6 2 0 2 Since the remainder is 0, x 2 x 2 6 is a factor. P  1x2 5 1x2 1 12 1x2 2 x 2 62 x2 2 x 2 6 5 1x 2 32 1x 1 22 Factor the quotient x2 2 x 2 6. Write P  1x2 as a product of four linear factors. P  1x2 5 1x 2 i2 1x 1 i2 1x 2 32 1x 1 22 ▼ A N S W E R 

P  1x2 5 1x 2 2i2 1x 1 2i2 1x 2 12 1x 2 22 Note: The zeros of P  1x2 are 1, 2, 2i, and 22i.

Young_AT_6160_ch04_pp388-423.indd 390

 heck: P  1x2 is a fourth-degree polynomial and we found four zeros, two of which C are complex conjugates.

▼ Y O U R T U R N   Factor the polynomial P  1x2 5 x4 2 3x3 1 6x2 2 12x 1 8 given

that x 2 2i is a factor.

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391

EXAMPLE 4  Factoring a Polynomial with Complex Zeros

Factor the polynomial P  1x2 5 x4 2 2x3 1 x2 1 2x 2 2 given that 1 1 i is a zero of P  1x2.

Since P  1x2 is a fourth-degree polynomial, we expect four zeros. The goal in this problem is to write P  1x2 as a product of four linear factors: P  1x2 5 1x 2 a2 1x 2 b2 1x 2 c2 1x 2 d2, where a, b, c, and d are complex numbers and represent the zeros of the polynomial. Solution:

STEP 1  Write

known zeros and linear factors. Since 1 1 i is a zero, we know that 1 2 i is a zero.

x 5 1 1 i and x 5 1 2 i

We now know two linear factors of P  1x2.

3x 2 11 1 i24 and 3x 2 11 2 i24

P  1x2 as a product of four factors.

STEP 2  Write

STEP 3  Multiply

P  1x2 5 3x 2 11 1 i24 3x 2 11 2 i24 1x 2 c2 1x 2 d  2 3x 2 11 1 i24 3x 2 11 2 i24

the first two terms. First, group the real parts together in each bracket. Use the special product 1a 2 b2 1a 1 b2 5 a2 2 b2, where a is 1x 2 12 and b is i. STEP 4  Rewrite

the polynomial.

STEP 5  Divide

31x 2 12 2 i4 31x 2 12 1 i4

1x 2 122 2 i2 1x2 2 2x 1 12 2 1212 x2 2 2x 1 2

P  1x2 5 1x2 2 2x 1 22 1x 2 c2 1x 2 d 2

both sides of the equation by x2 2 2x 1 2, x 4 2 2x 3 1 x 2 1 2x 2 2 and substitute in the original 5 1x 2 c2 1x 2 d 2 x 2 2 2x 1 2 polynomial P  1x2 5 x4 2 2x3 1 x2 1 2x 2 2.

STEP 6  Divide

the left side of the equation using long division.

STEP 7  Factor STEP 8  Write

x 4 2 2x 3 1 x 2 1 2x 2 2 5 x2 2 1 x 2 2 2x 1 2

x2 2 1.

P  1x2 as a product of four linear factors.

1x 2 12 1x 1 12

P  1x2 5 3x 2 11 1 i24 3x 2 11 2 i24 3x 2 14 3x 1 14



▼

or P  1x2 5 1x 2 1 2 i2 1x 2 1 1 i2 1x 2 12 1x 1 12

Y O U R T U R N   Factor the polynomial P  1x2 5 x4 2 2x2 1 16x 2 15 given that

1 1 2i is a zero.

Because an n-degree polynomial function has exactly n zeros and since complex zeros always come in conjugate pairs, if the degree of the polynomial is odd, there is guaranteed to be at least one zero that is a real number. If the degree of the ­polynomial is even, there is no guarantee that a zero will be real—all the zeros could be complex.

Young_AT_6160_ch04_pp388-423.indd 391

▼ ANSWER

P  1x2 5 3x 2 11 1 2i24 # 3x 2 11 2 2i24 1x 2 12 1x 1 32  Note: The zeros of P  1x2 are 1, 23, 1 1 2i, and 1 2 2i.

[

STUD Y T I P Odd-degree polynomials have at least one real zero.

[

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EXAMPLE 5  Finding  Possible Combinations of Real and Complex Zeros

List the possible combinations of real and complex zeros for the given polynomials. a. 17x5 1 2x4 2 3x3 1 x2 2 5 b. 5x4 1 2x3 2 x 1 2 Solution: a. Since this is a fifth-degree polynomial,

REAL ZEROS

there are five zeros. Because complex zeros come in conjugate pairs, the table describes the possible five zeros.

Applying Descartes’ rule of signs, we find that there are 3 or 1 positive real zeros and 2 or 0 negative real zeros.

POSITIVE REAL ZEROS

REAL ZEROS

6

2

4

4

2

6

0

2

5

0

NEGATIVE REAL ZEROS

COMPLEX ZEROS

4

3

0

2

1

2

2

3

2

0

REAL ZEROS

POSITIVE REAL ZEROS

COMPLEX ZEROS

0

4

3

0

polynomial, there are four zeros. Since complex zeros come in conjugate pairs, the table describes the possible four zeros.

▼ ANSWER

1

1

b. Because this is a fourth-degree

Applying Descartes’ rule of signs, we find that there are 2 or 0 positive real zeros and 2 or 0 negative real zeros.

COMPLEX ZEROS

COMPLEX ZEROS

0

4

2

2

4

0

NEGATIVE REAL ZEROS

COMPLEX ZEROS

0

0

4

2

0

2

0

2

2

2

2

0

▼ Y O U R T U R N   List the possible combinations of real and complex zeros for

P  1x2 5 x6 2 7x5 1 8x3 2 2x 1 1

4.5.2  Factoring Polynomials 4.5.2 S K I L L

Factor a polynomial function of degree n into n linear factors.

Now let’s draw on the tests discussed in this chapter to help us find all the zeros of a polynomial. Doing so will enable us to write polynomials as a product of linear factors. Before reading Example 6, reread Section 4.4, Example 8.

4.5.2 C O N C E P T U A L

Understand why an odd-degree polynomial must have at least one real zero.

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393

EXAMPLE 6  Factoring a Polynomial

STUDY T I P

Factor the polynomial P  1x2 5 x5 1 2x4 2 x 2 2.

From Step 2 we know there is one positive real zero, so test the positive possible rational zeros first in Step 4.

Solution:

STEP 1  Determine

variations in sign. P  1x2 has 1 sign change. P  12x2 has 2 sign changes. STEP 2  Apply

P  1x2 5 x5 1 2x4 2 x 2 2 P  12x2 5 2x5 1 2x4 1 x 2 2

Descartes’ rule of signs and summarize the results in a table. POSITIVE REAL ZEROS

NEGATIVE REAL ZEROS

COMPLEX ZEROS

1

2

2

1

0

4

rational zero theorem to determine the possible rational zeros.

[ CONCEPT CHECK ] A polynomial of the form P(x) 5 (x 2 1 a2)(x 2 2 b2) where a and b are real numbers has how many real and how many ­imaginary zeros?

STEP 3  the

STEP 4  Test

1

1

0

0

21

22

1

3

3

3

2

3

3

3

2

0

3

3

3

2

21

22

21

22

1

2

1

2

0

1

2

1

2

22

0

22

0

1

0

1



21 is a zero:

22 is a zero:

21

22

1

1

STEP 5  Write

▼

2

P  1x2 as a product of linear factors.

d



possible rational zeros. 1 is a zero:

61, 62

x 2 1 1  5 1x 2 i2 1x 1 i2

ANSWER Two real ( 6 b) and two ­imaginary ( 6 ai)

[

STUD Y T I P P(x) is a fifth-degree polynomial, so we expect five zeros.

P  1x2 5 1x 2 12 1x 1 12 1x 1 22 1x 2 i2 1x 1 i2

[

[SEC TION 4.5]  S U M M A RY In this section, we discussed complex zeros of polynomial functions. A polynomial function, P  1x2, of degree n with real coefficients has the following properties:

Young_AT_6160_ch04_pp388-423.indd 393

n

P  1x2 has at least one zero and no more than n zeros. a 1 bi is a zero, then a 2 bi is also a zero.

n If n

The polynomial can be written as a product of linear factors, not necessarily distinct.

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[SEC TION 4.5]  E X E R C I S E S • SKILLS In Exercises 1–8, find all zeros (real and complex). Factor the polynomial as a product of linear factors. 1. P  1x2 5 x2 1 4 2. P  1x2 5 x2 1 9 3. P  1x2 5 x2 2 2x 1 2 4. P  1x2 5 x2 2 4x 1 5 5. P  1x2 5 x4 2 16 6. P  1x2 5 x4 2 81 7. P  1x2 5 x4 2 25 8. P  1x2 5 x4 2 9

In Exercises 9–16, a polynomial function with real coefficients is described. Find all remaining zeros. 9. Degree: 3 Zeros: 21, i 10. Degree: 3 Zeros: 1, 2i 11. Degree: 4 Zeros: 2i, 3 2 i 12. Degree: 4 Zeros: 3i, 2 1 i 13. Degree: 6 Zeros: 2 1multiplicity 22, 1 2 3i, 2 1 5i 14. Degree: 6 Zeros: 22 1multiplicity 22, 1 2 5i, 2 1 3i 15. Degree: 6 Zeros: 2i, 1 2 i 1multiplicity 22 16. Degree: 6 Zeros: 2i, 1 1 i 1multiplicity 22 In Exercises 17–22, find a polynomial of minimum degree that has the given zeros.

17. 0, 1 2 2i, 1 1 2i 18. 0, 2 2 i, 2 1 i 19. 1, 1 2 5i, 1 1 5i 20. 2, 4 2 i, 4 1 i 21. 1 2 i, 1 1 i, 23i, 3i 22. 2i, i, 1 2 2i, 1 1 2i

In Exercises 23–34, given a zero of the polynomial, determine all other zeros (real and complex), and write the polynomial in terms of a product of linear factors.

Polynomial 23. P  1x2 5 x4 2 2x3 2 11x2 2 8x 2 60 4

3

2

25. P  1x2 5 x 2 4x 1 4x 2 4x 1 3

27. P  1x2 5 x4 2 2x3 1 10x2 2 18x 1 9 29. P  1x2 5 x4 2 9x2 1 18x 2 14

31. P  1x2 5 x4 2 6x3 1 6x2 1 24x 2 40

33. P  1x2 5 x4 2 9x3 1 29x2 2 41x 1 20

Zero

Polynomial

22i 24. P  1x2 5 x4 2 x3 1 7x2 2 9x 2 18 i 26. P  1x2 5 x4 2 x3 1 2x2 2 4x 2 8 23i 28. P  1x2 5 x4 2 3x3 1 21x2 2 75x 2 100 1 1 i 30. P  1x2 5 x4 2 4x3 1 x2 1 6x 2 40 3 2 i 32. P  1x2 5 x4 2 4x3 1 4x2 1 4x 2 5 2 2 i 34. P  1x2 5 x4 2 7x3 1 14x2 1 2x 2 20

Zero 3i 22i 5i 1 2 2i 21i 31i

In Exercises 35–58, factor each polynomial as a product of linear factors. 35. P  1x2 5 x3 2 x2 1 9x 2 9 36. P  1x2 5 x3 2 2x2 1 4x 2 8 3 2 37. P  1x2 5 x 2 5x 1 x 2 5 38. P  1x2 5 x3 2 7x2 1 x 2 7 39. P  1x2 5 x3 1 x2 1 4x 1 4 40. P  1x2 5 x3 1 x2 2 2 3 2 41. P  1x2 5 x 2 x 2 18 42. P  1x2 5 x4 2 2x3 2 2x2 2 2x 2 3 43. P  1x2 5 x4 2 2x3 2 11x2 2 8x 2 60 44. P  1x2 5 x4 2 x3 1 7x2 2 9x 2 18 4 3 2 45. P  1x2 5 x 2 4x 2 x 2 16x 2 20 46. P  1x2 5 x4 2 3x3 1 11x2 2 27x 1 18 47. P  1x2 5 x4 2 7x3 1 27x2 2 47x 1 26 47. P  1x2 5 x4 2 5x3 1 5x2 1 25x 2 26 4 3 2 49. P  1x2 5 2x 2 3x 1 x 1 13x 1 10 50. P  1x2 5 2x4 2 x3 1 12x2 1 26x 1 24 51. P  1x2 5 x4 2 2x3 1 5x2 2 8x 1 4 52. P  1x2 5 x4 1 2x3 1 10x2 1 18x 1 9 6 4 2 53. P  1x2 5 x 1 12x 1 23x 2 36 54. P  1x2 5 x6 2 2x5 1 9x4 2 16x3 1 24x2 2 32x 1 16 55. P  1x2 5 4x4 2 20x3 1 37x2 2 24x 1 5 56. P  1x2 5 4x4 2 44x3 1 145x2 2 114x 1 26 5 4 3 2 57. P  1x2 5 3x 2 2x 1 9x 2 6x 2 12x 1 8 58. P  1x2 5 2x5 2 5x4 1 4x3 2 26x2 1 50x 2 25

• A P P L I C AT I O N S In Exercises 59–62, assume the profit model is given by a polynomial function P 1x2 where x is the number of units sold by the company per year. 59. Profit. If the profit function of a given company has all 61. Profit. If the profit function of a company is modeled by a imaginary zeros and the leading coefficient is positive, third-degree polynomial with a negative leading coefficient would you invest in this company? Explain. and this polynomial has two complex conjugates as zeros and one positive real zero, would you invest in this company? 60. Profit. If the profit function of a given company has all Explain. imaginary zeros and the leading coefficient is negative, would you invest in this company? Explain.

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395

62. Profit. If the profit function of a company is modeled by a

64. Business. If the profit function pictured is a fourth-degree

third-degree polynomial with a positive leading coefficient and this polynomial has two complex conjugates as zeros and one positive real zero, would you invest in this company? Explain.

polynomial with a negative leading coefficient, how many real and how many complex zeros does the function have? Discuss the implications of these zeros.

For Exercises 63 and 64, refer to the following:  he following graph models the profit P of a company where t is T months and t $ 0.

For Exercises 65 and 66, refer to the following: The following graph models the concentration, C 1in mg/mL2 of a drug in the bloodstream; t is time in hours after the drug is administered where t $ 0. C (mcg/mL)

P (profit in millions of dollars)

50 45 40 35 30 25 20 15 10 5

15 12 9 6 3 1

2

3

4

t (months) 5 6

63. Business. If the profit function pictured is a third-degree

polynomial, how many real and how many complex zeros does the function have? Discuss the implications of these zeros.

1

2

3

4 5 6

7

t (hours) 8

65. Health/Medicine. If the concentration function pictured

is a third-degree polynomial, how many real and how many c­omplex zeros does the function have? Discuss the ­implications of these zeros. 66. Health/Medicine. If the concentration function pictured is a fourth-degree polynomial with a negative leading coefficient, how many real and how many complex zeros does the ­function have? Discuss the implications of these zeros.

• C AT C H T H E M I S TA K E In Exercises 67 and 68, explain the mistake that is made. 67. Given that 1 is a zero of P  1x2 5 x3 2 2x2 1 7x 2 6, find all other zeros. Solution: Step 1:  P  1x2 is a third-degree polynomial, so we expect

three zeros.

68. Factor the polynomial P  1x2 5 2x3 1 x2 1 2x 1 1.

Solution:

Step 1:  Since P  1x2 is an odd-degree polynomial, we are guaranteed

one real zero (since complex zeros come in conjugate pairs).

Step 2:  Because 1 is a zero, 21 is a zero, so two linear

Step 2:  Apply the rational zero test to develop a list of potential

Step 3:  Write the polynomial as a product of three linear



factors are 1x 2 12 and 1x 1 12.

factors.

P  1x2 5 1x 2 12 1x 1 12 1x 2 c2 P  1x2 5 1x2 2 12 1x 2 c2

Step 4:  To find the remaining linear factor, we divide P  1x2

by x2 2 1.



x 3 2 2x 2 1 7x 2 6 6x 2 8 5x221 2 x2 2 1 x 21

rational zeros. Possible zeros: 61

Step 3:  Test possible zeros.



1 is not a zero:

P  1x2 5 21123 1 1122 1 2112 1 1 5 6

21 is not a zero: P  1x2 5 212123 1 12122 1 21212 1 1 5 22

Note: 212 is the real zero. Why did we not find it?

Which has a nonzero remainder? What went wrong?

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• CONCEPTUAL In Exercises 69–72, determine whether each statement is true or false. 69. If x 5 1 is a zero of a polynomial function, then x 5 21 is

also a zero of the polynomial function. 70. All zeros of a polynomial function correspond to x-intercepts. 71. A polynomial function of degree n, n . 0 must have at least one zero.

72. A polynomial function of degree n, n . 0 can be written as a

product of n linear factors. 73.  Is it possible for an odd-degree polynomial to have all

imaginary complex zeros? Explain. 74. Is it possible for an even-degree polynomial to have all

imaginary zeros? Explain.

• CHALLENGE In Exercises 75 and 76, assume a and b are nonzero real numbers. 75. Find a polynomial function that has degree 6 and for which bi 76. F  ind a polynomial function that has degree 4 and for which is a zero of multiplicity 3. a 1 bi is a zero of multiplicity 2.

• TECHNOLOGY For Exercises 77 and 78, determine possible combinations of real and complex zeros. Plot P 1x2 and identify any real zeros with a graphing calculator or software. Does this agree with your list? 77. P  1x2 5 x4 1 13x2 1 36 78. P  1x2 5 x6 1 2x4 1 7x2 2 130x 2 288

For Exercises 79 and 80, find all zeros (real and complex). Factor the polynomial as a product of linear factors. 79. P  1x2 5 25x5 1 3x4 2 25x3 1 15x2 2 20x 1 12 80. P  1x2 5 x5 1 2.1x4 2 5x3 2 5.592x2 1 9.792x 2 3.456

4.6 RATIONAL FUNCTIONS SKILLS OBJECTIVES ■■ Find the domain of a rational function. ■■ Determine vertical, horizontal, and slant asymptotes of rational functions. ■■ Graph rational functions.

CONCEPTUAL OBJECTIVES ■■ Understand that the domain of a rational function is the set of all real numbers except those that correspond to the denominator being equal to zero. ■■ Understand that the graph of a rational function can have either a horizontal asymptote or a slant asymptote but not both. ■■ Any real number excluded from the domain of a rational function corresponds to either a vertical asymptote or a hole in the graph.

4.6.1  Domain of Rational Functions 4.6.1 S K I L L

Find the domain of a rational function. 4.6.1 C O N C E P T U A L

Understand that the domain of a rational function is the set of all real numbers except those that correspond to the denominator being equal to zero.

Young_AT_6160_ch04_pp388-423.indd 396

So far in this chapter we have discussed polynomial functions. We now turn our ­attention to rational functions, which are ratios of polynomial functions. Ratios of ­integers are called rational numbers. Similarly, ratios of polynomial functions are called r­ ational functions. DEFINITION

Rational Function

A function ƒ1x2 is a rational function if n1x2 d1x2 2 0 ƒ1x2 5 d1x2 where the numerator, n  1x2, and the denominator, d  1x2, are polynomial functions. The domain of ƒ1x2 is the set of all real numbers x such that d  1x2 Þ 0. Note: If d  1x2 is a constant, then ƒ1x2 is a polynomial function.

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4.6  Rational Functions 

397

The domain of any polynomial function is the set of all real numbers. When we divide two polynomial functions, the result is a rational function, and we must exclude any values of x that make the denominator equal to zero. EXAMPLE 1  Finding the Domain of a Rational Function

x11 Find the domain of the rational function ƒ 1 x 2 5 2 . Express the domain x 2x26 in interval notation. Solution:

x2 2 x 2 6 5 0 1x 1 22 1x 2 32 5 0 x 5 22 or x 5 3 x 2 22 or x 2 3

Set the denominator equal to zero. Factor. Solve for x. Eliminate these values from the domain. State the domain in interval notation.

12q, 22 2 ∪ 122, 3 2 ∪ 1 3, q 2

▼

Y O U R T U R N   Find the domain of the rational function ƒ 1 x 2 5

Express the domain in interval notation.

x22 . x 2 3x 2 4 2

It is important to note that there are not always restrictions on the domain. For e­ xample, if the denominator is never equal to zero, the domain is the set of all real numbers.

▼ ANSWER

The domain is the set of all real numbers such that x 2 21 or x 2 4. Interval notation: 12q, 21 2 ∪ 121, 4 2 ∪ 1 4, q 2

EXAMPLE 2  W  hen the Domain of a Rational Function Is the Set of All Real Numbers

3x Find the domain of the rational function g 1 x 2 5 2 . Express the domain in x 19 interval notation. Solution:

Set the denominator equal to zero. Subtract 9 from both sides. Solve for x. There are no real solutions; therefore, the domain has no restrictions. State the domain in interval notation.

▼

x2 1 9 5 0 x2 5 29 x 5 23i or x 5 3i R, the set of all real numbers 12q, q 2

Y O U R T U R N   Find the domain of the rational function g 1 x 2 5

the domain in interval notation.

5x . Express x 14 2

▼ ANSWER

The domain is the set of all real numbers. Interval notation: 1 2q, q 2 y

x2 2 4 , where x 2 22 and g 1 x 2 5 x 2 2 are not x12 the same function. Although ƒ1x2 can be written in the factored form 1x 2 22 1x 1 22 ƒ1x2 5 5 x 2 2, its domain is different. The domain of g  1x2 is the set x12 of all real numbers, whereas the domain of ƒ1x2 is the set of all real numbers such that x 2 22. If we were to plot ƒ1x2 and g  1x2, they would both look like the line y 5 x 2 2. However, ƒ1x2 would have a hole, or discontinuity, at the point x 5 22.

5

It is important to note that ƒ 1 x 2 5

Young_AT_6160_ch04_pp388-423.indd 397

x –5

5 hole –5

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CHAPTER 4  Polynomial and Rational Functions

4.6.2  Vertical, Horizontal, and Slant Asymptotes

4.6.2 C O N C E P T U A L

Understand that the graph of a rational function can have either a horizontal asymptote or a slant asymptote but not both.

[ CONCEPT CHECK ]

If a function is not defined at a point, then it is still useful to know how the ­function behaves near that point. Let’s start with a simple rational function, the reciprocal 1 ­function ƒ 1 x 2 5 . This function is defined everywhere except at x 5 0. x 1 10

x

2

1 f (x) 5 – x

210

2

1 1 2 100 1000

2100

21000

0

1 1000

1 100

1 10

undefined

1000

100

10

x approaching 0 from the left

g

Determine vertical, horizontal, and slant asymptotes of rational functions.

g

4.6.2 S K I L L

x approaching 0 from the right

x

1 f (x) 5 – x

210

1 210

21

21

1

1

10

1 10

TRUE OR FALSE  The domain of 1 and the domain of x 1 a2 1 g1x2 5 2 are equal. x 1 b2 f 1x2 5

2

▼

ANSWER True

y f (x) = 1x (1, 1)

WORDS MATH x

(–1, –1)

f (x) = 1x

We cannot let x 5 0 because that point is not in the domain of the function. We should, however, ask the question, “how does ƒ1x2 behave as x approaches zero?” Let 1 1 1 us take values that get closer and closer to x 5 0, such as 10 , 100 , 1000 , . . . (See the table above.) We use an arrow to represent the word approach, a positive superscript to represent from the right, and a negative superscript to represent from the left. A plot of this f­unction can be generated using point-plotting techniques. The following are 1 ­observations of the graph ƒ1x2 5 . x

As x approaches zero from the right, the function ƒ 1x2 increases without bound.

x S 01 1 S` x

 s x approaches zero from the left, the A function ƒ 1x2 decreases without bound.

x S 02 1 S 2` x xS` 1 S 01 x

As x approaches negative infinity (decreases without bound), the function ƒ 1x2 approaches zero from below.

x S 2` 1 S 02 x

As x approaches infinity (increases without bound), the function ƒ 1x2 approaches zero from above.

y

(1, 1) (–1, –1)

x

The symbol q does not represent an actual real number. This symbol represents growing without bound. 1. Notice that the function is not defined at x 5 0. The y-axis, or the vertical line x 5 0, represents the vertical asymptote. 2. Notice that the value of the function is never equal to zero. The x-axis is never touched by the function. The x-axis, or y 5 0, is a horizontal asymptote. Asymptotes are lines that the graph of a function approaches. Suppose a football team’s defense is its own 8-yard line and the team gets an “offsides” penalty that results in loss of “half the distance to the goal.” Then the offense would get the ball on the 4-yard line. Suppose the defense gets another penalty on the next play that results in “half the distance to the goal.” The offense would then get the ball on the 2-yard line. If the defense received 10 more penalties, all resulting in “half the distance to the goal,” would the referees give the offense a touchdown? No, because although the offense may appear

Young_AT_6160_ch04_pp388-423.indd 398

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4.6  Rational Functions 

399

to be snapping the ball from the goal line, technically it has not actually reached the goal line. Asymptotes utilize the same concept. 1 We will start with vertical asymptotes. Although the function ƒ1x2 5 had one x vertical asymptote, in general, rational functions can have none, one, or several ­vertical asymptotes. We will first formally define a vertical asymptote and then discuss how to find it. DEFINITION

Vertical Asymptotes

The line x 5 a is a vertical asymptote for the graph of a function if ƒ1x2 either increases or decreases without bound as x approaches a from either the left or the right. y

y

x=a

y x=a

x=a

y f (x)

f (x)

x=a

f (x) x

x

x

x

f (x)

Vertical asymptotes assist us in graphing rational functions since they essentially “steer” the function in the vertical direction. How do we locate the vertical asymptotes of a rational function? Set the denominator equal to zero. If the numerator and denominator have no common factors, then any numbers that are excluded from the domain of a rational function locate vertical asymptotes. n1x2 is said to be in lowest terms if the numerator n  1x2 A rational function ƒ 1 x 2 5 d1x2 n1x2 and denominator d  1x2 have no common factors. Let ƒ 1 x 2 5 be a rational function d1x2 in lowest terms; then any zeros of the numerator n  1x2 correspond to x-intercepts of the graph of ƒ, and any zeros of the denominator d  1x2 correspond to vertical asymptotes of the graph of ƒ. If a rational function does have a common factor (is not in ­lowest terms), then the common factor(s) should be canceled, resulting in an equivalent ­rational ­function R  1x2 in lowest terms. If 1x 2 a2p is a factor of the numerator and 1x 2 a2q is a factor of the denominator, then there is a hole in the graph at x 5 a provided p $ q and x 5 a is a vertical asymptote if p , q.

LOCATING VERTICAL ASYMPTOTES

n1x2 Let ƒ 1 x 2 5 be a rational function in lowest terms (that is, assume n  1x2 and d  1x2 d1x2 are polynomials with no common factors); then the graph of ƒ has a vertical asymptote at any real zero of the denominator d  1x2. That is, if d  1a2 5 0, then x 5 a corresponds to a vertical asymptote on the graph of ƒ.

STUDY T I P The vertical asymptotes of a rational function in lowest terms occur at x-values that make the denominator equal to zero.

Note: If ƒ is a rational function that is not in lowest terms, then divide out the common factors, resulting in a rational function R that is in lowest terms. Any common factor x 2 a of the function ƒ corresponds to a hole in the graph of ƒ at x 5 a provided the multiplicity of a in the numerator is greater than or equal to the multiplicity of a in the denominator.

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CHAPTER 4  Polynomial and Rational Functions

EXAMPLE 3  Determining Vertical Asymptotes

Locate any vertical asymptotes of the rational function ƒ 1 x 2 5

Solution:

ƒ1x2 5

Factor the denominator.

5x 1 2 . 6x 2 2 x 2 2

5x 1 2 1 2x 1 1 2 1 3x 2 2 2

The numerator and denominator have no common factors. 2x 1 1 5 0 and  3x 2 2 5 0 Set the denominator equal to zero. 1 x 5 2  and 2

Solve for x.

x5

2 3

x 5 212 and    x 5 23 . The vertical asymptotes are     ▼ ANSWER

x5

2 25

▼ Y O U R T U R N   Locate any vertical asymptotes of the following rational function:

and x 5 3

ƒ1x2 5

3x 2 1 2x 2 2 x 2 15

EXAMPLE 4  Determining  Vertical Asymptotes When the Rational Function Is Not in Lowest Terms

Locate any vertical asymptotes of the rational function ƒ 1 x 2 5 Solution:

x12 . x 2 3x 2 2 10x 3

x3 2 3x2 2 10x 5 x  1x2 2 3x 2 102 5 x  1x 2 52 1x 1 22

Factor the denominator.

Write the rational function in factored form.   ƒ 1 x 2 5 Cancel (divide out) the common factor 1x 1 22.

 ind the values when the denominator of R is F equal to zero. x55. 0 The vertical asymptotes are  x 5   and    

R1x2 5

1x 1 22 x1x 2 52 1x 1 22

1   x 2 22 x1x 2 52

x 5 0 and x 5 5

Note: x 5 22 is not in the domain of ƒ1x2, even though there is no vertical asymptote there. There is a “hole” in the graph at x 5 22. Graphing calculators do not always show such “holes.” ▼ ANSWER

▼ Y O U R T U R N   Locate­any vertical asymptotes of the following rational function:

x53

ƒ1x2 5

x 2 2 4x x 2 7x 1 12 2

We now turn our attention to horizontal asymptotes. As we have seen, rational functions can have several vertical asymptotes. However, rational functions can have at most one horizontal asymptote. Horizontal asymptotes imply that a function approaches a constant value as x becomes large in the positive or negative direction. Another difference between vertical and horizontal asymptotes is that the graph of a function never touches a vertical asymptote but, as you will see in the next box, the graph of a function may cross a horizontal asymptote, just not at the “ends” 1 x S 6q 2 . Young_AT_6160_ch04_pp388-423.indd 400

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4.6  Rational Functions 

DEFINITION

401

Horizontal Asymptote

The line y 5 b is a horizontal asymptote of the graph of a function if ƒ1x2 approaches b as x increases or decreases without bound. The following are three examples: As x S q, ƒ 1 x 2 S b

y

y

f (x)

y f (x)

y=b

x

y=b

x

y=b

x

f (x)

Note: A horizontal asymptote steers a function as x gets large. Therefore, when x is not large, the function may cross the asymptote. How do we determine whether a horizontal asymptote exists? And, if it does, how do we locate it? We investigate the value of the rational function as x S q or as x S 2q. One of two things will happen: either the rational function will increase or decrease without bound or the rational function will approach a constant value. We say that a rational function is proper if the degree of the numerator is less 1 than the degree of the denominator. Proper rational functions, like ƒ 1 x 2 5 , approach x zero as x gets large. Therefore, all proper rational functions have the specific horizontal asymptote, y 5 0 (see Example 5a). We say that a rational function is improper if the degree of the numerator is greater than or equal to the degree of the denominator. In this case, we can divide the numerator by the denominator and determine how the quotient behaves as x increases without bound. If the quotient is a constant (resulting when the degrees of the numerator and denominator are equal), then as x S q or as x S 2q, the rational function approaches the constant quotient (see Example 5b). ■■ If the q ­ uotient is a polynomial function of degree 1 or higher, then the quotient depends on x and does not approach a constant value as x increases (see Example 5c). In this case, we say that there is no horizontal asymptote. ■■

We find horizontal asymptotes by comparing the degree of the numerator and the degree of the denominator. There are three cases to consider: 1. The degree of the numerator is less than the degree of the denominator. 2. The degree of the numerator is equal to the degree of the denominator. 3. The degree of the numerator is greater than the degree of the denominator. LOCATING HORIZONTAL ASYMPTOTES

Let ƒ be a rational function given by an x n 1 an21x n21 1 c1 a1x 1 a0 n1x2 5 ƒ1x2 5 d1x2 bm x m 1 bm21x m21 1 c1 b1x 1 b0 where n  1x2 and d  1x2 are polynomials.

1. When n , m, the x-axis 1y 5 02 is the horizontal asymptote. an 2. When n 5 m, the line y 5 (ratio of leading ­coefficients) is the horizontal bm asymptote. 3. When n . m, there is no horizontal asymptote.

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CHAPTER 4  Polynomial and Rational Functions

In other words, 1.  When the degree of the numerator is less than the degree of the denominator, then y 5 0 is the horizontal asymptote. 2.  When the degree of the numerator is the same as the degree of the denominator, then the horizontal asymptote is the ratio of the leading coefficients. 3.  If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote. EXAMPLE 5  Finding Horizontal Asymptotes

 etermine whether a horizontal asymptote exists for the graph of each of the given D rational functions. If it does, locate the horizontal asymptote. a.  ƒ 1 x 2 5

8x 1 3 4x 2 1 1

Solution (a):

b.  g 1 x 2 5

8x 2 1 3 4x 2 1 1

c.  h 1 x 2 5

The degree of the numerator 8x 1 3 is 1.

8x 3 1 3 4x 2 1 1

n51

2

The degree of the denominator 4x 1 1 is 2.

m52

 he degree of the numerator is less than the T degree of the denominator.

n,m

The x-axis is the horizontal asymptote for the graph of ƒ1x2. The line      y 5 0  is the horizontal asymptote for the graph of ƒ1x2.

y50

Solution (b):

The degree of the numerator 8x2 1 3 is 2. 2

n52

The degree of the denominator 4x 1 1 is 2.

m52

The degree of the numerator is equal to the degree of the denominator.

n5m

 he ratio of the leading coefficients is the T horizontal asymptote for the graph of g  1x2.

y5

8 52 4

y 5 2  is the horizontal asymptote for the graph of g  1x2. The line      If we divide the numerator by the denominator, 8x 2 1 3 1 g1x2 5 521 2 the resulting quotient is the constant 2. 2 4x 1 1 4x 1 1 Solution (c):

The degree of the numerator 8x3 1 3 is 3. 2

n53

The degree of the denominator 4x 1 1 is 2.

m52

The degree of the numerator is greater than the degree of the denominator.

n . m

no horizontal asymptote   The graph of the rational function h  1x2 has           .

▼ ANSWER

y5

274

is the horizontal asymptote.

S TU DY TIP There are three types of linear asymptotes: horizontal, vertical, and slant.

Young_AT_6160_ch04_pp388-423.indd 402

If we divide the numerator by the denominator, 8x 3 1 3 22x 1 3 h1x2 5 2 5 2x 1 the resulting quotient is a linear function. 4x 1 1 4x 2 1 1

▼

Y O U R T U R N   Find the horizontal asymptote (if one exists) for the graph of the

rational function ƒ 1 x 2 5

7x 3 1 x 2 2 . 24x 3 1 1

Thus far we have discussed linear asymptotes: vertical and horizontal. There are three types of lines: horizontal (slope is zero), vertical (slope is undefined), and slant (nonzero slope). Similarly, there are three types of linear asymptotes: horizontal, v­ ertical, and slant.

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4.6  Rational Functions 

403

Recall that when dividing polynomials the degree of the quotient is always the d­ ifference between the degree of the numerator and the degree of the denominator. For example, a cubic (third-degree) polynomial divided by a quadratic (second-degree) polynomial results in a linear (first-degree) polynomial. A fifth-degree polynomial divided by a fourth-degree polynomial results in a first-degree (linear) polynomial. When the degree of the numerator is exactly one more than the degree of the denominator, the quotient is linear and represents a slant asymptote.

SLANT ASYMPTOTES

n1x2 , where n  1x2 and d  1x2 are d1x2 ­polynomials and the degree of n  1x2 is one more than the degree of d  1x2. On ­dividing n  1x2 by d  1x2, the rational function can be expressed as Let ƒ be a rational function given by ƒ 1 x 2 5

ƒ 1 x 2 5 mx 1 b 1

[ CONCEPT CHECK ]

r 1x2 d1x2

where the degree of the remainder r  1x2 is less than the degree of d  1x2 and the line y 5 mx 1 b is a slant asymptote for the graph of ƒ. Note that as x S 2q or x S q, ƒ 1 x 2 S mx 1 b.

TRUE OR FALSE  You can have both a horizontal and a slant asymptote.

▼ ANSWER False

EXAMPLE 6  Finding Slant Asymptotes

Determine the slant asymptote of the rational function ƒ 1 x 2 5

4x 3 1 x 2 1 3 . x2 2 x 1 1

Solution:

4x 1 5 q4x x 2 x 1 1 3 1 x 2 1 0x 1 3 2 1 4x 3 2 4x 2 1 4x 2 5x 2 2 4x 1 3 2 1 5x 2 2 5x 1 5 2 x22 x22 Note that as x S 6q, the rational ƒ 1 x 2 5 4x 1 5 1 2 x 2x11 expression approaches 0. S 0 as x S 6q The quotient is the slant asymptote. y 5 4x 1 5 2

d

Divide the numerator by the denominator with long division.

▼ YOUR TURN  F  ind the slant asymptote of the rational function

ƒ1x2 5

Young_AT_6160_ch04_pp388-423.indd 403

x 2 1 3x 1 2 . x22

▼ ANSWER

y5x15

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CHAPTER 4  Polynomial and Rational Functions

ST U DY TIP Common factors need to be divided out first; then the ­remaining x-values ­corresponding to a denominator value of 0 are vertical asymptotes.

4.6.3 S K I L L

Graph rational functions. 4.6.3 C O N C E P T U A L

Any real number excluded from the domain of a rational function corresponds to either a vertical asymptote or a hole in the graph.

4.6.3  Graphing Rational Functions We can now graph rational functions using asymptotes as graphing aids. The following box summarizes the five-step procedure for graphing rational functions. Graphing Rational Functions Let ƒ be a rational function given by ƒ 1 x 2 5

Step 1: Step 2: Step 3:

n1x2 . d1x2

Find the domain of the rational function ƒ. Find the intercept(s). n y-intercept: evaluate ƒ102. n x-intercept: solve the equation n  1x2 5 0 for x in the domain of ƒ. Find any holes. n Factor the numerator and denominator. n Divide out common factors. n A common factor x 2 a corresponds to a hole in the graph of ƒ at x 5 a if the multiplicity of a in the numerator is greater than or equal to the multiplicity of a in the denominator. p1x2 n The result is an equivalent rational function R 1 x 2 5 in lowest terms. q1x2 Step 4: Find any asymptotes. n Vertical asymptotes: solve q 1x2 5 0. n Compare the degree of the numerator and the degree of the denominator to determine whether either a horizontal or slant asymptote exists. If one exists, find it. Step 5: Find additional points on the graph of ƒ—particularly near asymptotes. Step 6: Sketch the graph; draw the asymptotes, label the intercept(s) and additional points, and complete the graph with a smooth curve between and beyond the vertical asymptotes.

It is important to note that any real number eliminated from the domain of a r­ ational function corresponds to either a vertical asymptote or a hole on its graph. EXAMPLE 7  Graphing a Rational Function

Graph the rational function ƒ 1 x 2 5

x . x 24 2

Solution:

the domain. Set the denominator equal to zero. Solve for x. State the domain.

STEP 1  Find





the intercepts. y-intercept:



x-intercepts:



10, 02  . The only intercept is at the point   

STEP 2  Find

S TU D Y T IP Any real number excluded from the domain of a rational function corresponds to either a vertical asymptote or a hole on its graph.

Young_AT_6160_ch04_pp388-423.indd 404

x2 2 4 5 0 x 5 62 1 2q, 22 2 ∪ 1 22, 2 2 ∪ 1 2, q 2 0 5 0 24 x ƒ1x2 5 2 5 0 x 24

ƒ102 5

y50 x50

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4.6  Rational Functions 

STEP 3  Find



ƒ1x2 5

any holes.

405

x 1x 1 22 1x 2 22

There are no common factors, so ƒ is in lowest terms. Since there are no common factors, there are no holes on the graph of ƒ. any asymptotes. Vertical asymptotes:

STEP 4  Find



d  1x2 5 1x 1 22 1x 2 22 5 0 x 5 22



Degree of numerator 5 1 Degree of denominator 5 2



Horizontal asymptote:



Degree of numerator , Degree of denominator y 5 0

STEP 5  Find

x52

and

additional points on the graph.

x

23

21

1

3

f (x)

235

1 3

213

235

STEP 6  Sketch

the graph; label the intercepts, asymptotes, and additional points and complete with a smooth curve approaching the asymptotes.

y 5

x –5

5

–5

▼ Y O U R T U R N   Graph the rational function ƒ 1 x 2 5

x . x 21 2

▼ ANSWER y 5

EXAMPLE 8  Graphing  a Rational Function with No Horizontal or Slant Asymptotes

x –5

5

State the asymptotes (if there are any) and graph the rational function ƒ 1 x 2 5

x 4 2 x 3 2 6x 2 . x2 2 1

–5

Solution:

STEP 1  Find



the domain. Set the denominator equal to zero. Solve for x. State the domain.

STEP 2  Find the intercepts. y-intercept:

x2 2 1 5 0 x 5 61 1 2q, 21 2 ∪ 1 21, 1 2 ∪ 1 1, q 2 ƒ102 5

0 50 21

x-intercepts: n  1x2 5 x4 2 x3 2 6x2 5 0 Factor. x 2  1x 2 32 1x 1 22 5 0 Solve. x 5 0, x 5 3, and x 5 22 The intercepts are the points  10, 02 ,  13, 02 , and  122, 02 .

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CHAPTER 4  Polynomial and Rational Functions

STEP 3  Find



ƒ1x2 5

any holes.

x2 1 x 2 3 2 1 x 1 2 2 1x 2 12 1x 1 12

There are no common factors, so ƒ is in lowest terms. Since there are no common factors, there are no holes on the graph of ƒ.

STEP 4  Find

the asymptotes.

Factor.

d  1x2 5 x2 2 1 5 0

Solve. No horizontal asymptote: degree of n  1x2 . degree of d  1x2 No slant asymptote: degree of n  1x2 2 degree of d  1x2 . 1 The asymptotes are x 5 21 and x 5 1.

x 5 21 and x 5 1 34 . 24 34 2 2 5 2 . 14



Vertical asymptote:

STEP 5  Find

additional points on the graph.

1x 1 12 1 x 2 12 5 0

x

23

20.5

0.5

2

4

f (x)

6.75

1.75

2.08

25.33

6.4

STEP 6  Sketch

the graph; label the intercepts and asymptotes, and complete with a smooth curve between and beyond the vertical asymptote.

▼ ANSWER

y 10 x=1

Vertical asymptotes: x 5 22. No horizontal or slant asymptotes.

(–2, 0)

y

x

(3, 0)

–5

5 x = –1

50

–10 x = –2 (–1, 0) –5

▼ x (3, 0)

10

Y O U R T U R N   State the asymptotes (if there are any) and graph the rational

­function ƒ 1 x 2 5

–25

x 3 2 2x 2 2 3x . x12

EXAMPLE 9  Graphing  a Rational Function with a Horizontal Asymptote

State the asymptotes (if there are any) and graph the rational function

Solution:

ƒ1x2 5

4x 3 1 10x 2 2 6x 8 2 x3

the domain. Set the denominator equal to zero. Solve for x. State the domain.

STEP 1  Find



Young_AT_6160_ch04_pp388-423.indd 406

8 2 x3 5 0 x52 1 2q, 2 2 ∪ 1 2, q 2

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407

4.6  Rational Functions  STEP 2  Find

the intercepts. ƒ102 5

y-intercept:

0 50 8

n  1x2 5 4x3 1 10x2 2 6x 5 0

x-intercepts:

2x  12x 2 12 1x 1 32 5 0

Factor.

1 x 5 0, x 5 , and x 5 23 2

Solve.

The intercepts are the points  10, 02, A 12, 0B, and  123, 02. STEP 3  Find



ƒ1x2 5

the holes.

2x 1 2x 2 1 2 1 x 1 3 2 1 2 2 x 2 1 x 2 1 2x 1 4 2

There are no common factors, so ƒ is in lowest terms (no holes). the asymptotes. Vertical asymptote:

STEP 4  Find



d  1x2 5 8 2 x3 5 0

Solve.

x52



Horizontal asymptote:



Use leading coefficients.

degree of n  1x2 5 degree of d  1x2 y5

4 5 24 21

The asymptotes are x 5 2 and y 5 24. additional points on the graph.

STEP 5  Find

x

24

21

1 4

1

3

f (x)

21

1.33

20.10

1.14

29.47

y

STEP 6  Sketch

the graph; label the intercepts and asymptotes and complete with a smooth curve.

▼ ANSWER

10

x

x=2

(–3, 0) –10

(

1 2

, 0)

10

y = –4

Vertical asymptotes: x 5 4, x 5 21 Horizontal asymptote: y 5 2 Intercepts: A0, 2 32 B, A 32 , 0B, 1 2, 0 2 y

5

–10

y=2 x

▼ 2x 2 2 7x 1 6 . Give ­equations x 2 2 3x 2 4 of the vertical and horizontal asymptotes and state the intercepts.

Y O U R T U R N   Graph the rational function ƒ 1 x 2 5

Young_AT_6160_ch04_pp388-423.indd 407

–4

6 x = –1

x=4

–5

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CHAPTER 4  Polynomial and Rational Functions

EXAMPLE 10  Graphing a Rational Function with a Slant Asymptote

Graph the rational function ƒ 1 x 2 5 Solution:

x 2 2 3x 2 4 . x12

the domain. Set the denominator equal to zero. x 1 2 5 0 Solve for x. x 5 22 1 2q,22 2 ∪ 1 22, q 2 State the domain.

STEP 1  Find



STEP 2  Find

the intercepts.

4 ƒ 1 0 2 5 2 5 22 2

y-intercept:

x-intercepts: n  1x2 5 x2 2 3x 2 4 5 0 Factor. 1x 1 12 1x 2 42 5 0 Solve. x 5 21 and x 5 4 The intercepts are the points 10, 222,  121, 02, and  14, 02. STEP 3  Find

ƒ1x2 5

any holes.

1x 2 42 1x 1 12 1x 1 22

There are no common factors, so ƒ is in lowest terms. Since there are no common factors, there are no holes on the graph of ƒ. the asymptotes. Vertical asymptote: Solve. Slant asymptote: STEP 4  Find

d  1x2 5 x 1 2 5 0 x 5 22 degree of n  1x2 2 degree of d  1x2 5 1

ƒ1x2 5

Divide n  1x2 by d  1x2.

x 2 2 3x 2 4 6 5x251 x12 x12

Write the equation of the asymptote. y 5 x 2 5 The asymptotes are x 5 22 and y 5 x 2 5. additional points on the graph.

STEP 5  Find

x

26

25

23

5

6

f (x)

212.5

212

214

0.86

1.75

y

STEP 6  Sketch

the graph; label the intercepts and asymptotes, and complete with a smooth curve between and beyond the vertical asymptote.

▼ ANSWER

Horizontal asymptote: x 5 3 Slant asymptote: y 5 x 1 4

20 x = –2

–20

(–1, 0) (0, –2)

x (4, 0)

20

y=x–5

y 20

–20

(0, 23 )

y=x+4 –10 (–2, 0)

x

(1, 0) 10 x=3

–10

Young_AT_6160_ch04_pp388-423.indd 408

▼

x2 1 x 2 2 , state the asymptotes x23 (if any exist) and graph the function.

Y O U R T U R N   For the function ƒ 1 x 2 5

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409

EXAMPLE 11  Graphing a Rational Function with a Hole in the Graph

Graph the rational function ƒ 1 x 2 5 Solution:

x2 1 x 2 6 . x2 2 x 2 2

the domain. Set the denominator equal to zero. Solve for x. State the domain. STEP 1  Find

x2 2 x 2 2 5 0 1x 2 22 1x 1 12 5 0 x 5 21 or x 5 2 1 2q, 21 2 ∪ 1 21, 2 2 ∪ 1 2, q 2

the intercepts. y-intercept: STEP 2  Find

26 5 3  y 5 3 22 x-intercepts: n  1x2 5 x2 1 x 2 6 5 0 1x 1 32 1x 2 22 5 0 x 5 23 or x 5 2 123, 02 . The point 12, 02 The intercepts correspond to the points 10,   32 and      appears to be an x-intercept; however, x 5 2 is not in the domain of the ­function. 1x 2 22 1x 1 32 STEP 3  Find any holes. ƒ1x2 5 1x 2 22 1x 1 12 Since x 2 2 is a common factor, there is a 1x 1 32 hole in the graph of ƒ at x 5 2. R1x2 5 1x 1 12 Dividing out the common factor generates an equivalent rational function in lowest terms. ƒ102 5

the asymptotes. Vertical asymptotes:

STEP 4  Find



x1150 x 5 21

Horizontal Degree of numerator of ƒ 5 Degree of denominator of ƒ 5 2 asymptote: and Degree of numerator of R 5 Degree of denominator of R 5 1 Since the degree of the numerator 1 equals the degree of the denominator, y 5 5 1 1 use the leading coefficients. STEP 5  Find



additional points on the graph.

TRUE OR FALSE  The only time a hole will exist on the graph of a rational function is when the numerator and the denominator have a common factor.

▼ ANSWER True

x

24

22

212

1

3

f (x) or R(x)

1 3

21

5

2

3 2

y

STEP 6  Sketch

the graph; label the intercepts, asymptotes, and additional points and complete with a smooth curve approaching asymptotes. Recall the hole at x 5 2. Note that R 1 2 2 5 53, so the open “hole” is located at the point 12, 5/32.

5

x –5

▼ Y O U R T U R N   Graph the rational function ƒ 1 x 2 5

▼ ANSWER y

5

–5

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[ CONCEPT CHECK ]

x2 2 x 2 2 . x2 1 x 2 6

5

x –7

3

–5

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CHAPTER 4  Polynomial and Rational Functions

[SEC TION 4 .6]  S U M M A RY 1. If degree of the numerator 2 degree of the denominator 5 1, then there is a slant asymptote. 2. Divide the numerator by the denominator. The quotient ­corresponds to the equation of the line (slant a­ symptote).

In this section, rational functions were discussed. ƒ1x2 5

n1x2 d1x2

Domain: All real numbers except the x-values that make the denominator equal to zero, d  1x2 5 0. n Vertical Asymptotes: Vertical lines, x 5 a, where d  1a2 5 a, after all common factors have been divided out. Vertical asymptotes steer the graph are never touched. n Horizontal Asymptotes: Horizontal lines, y 5 b, that steer the graph as x S 6q. n

1. If degree of the numerator , degree of the denominator, then y 5 0 is a horizontal asymptote. 2. If degree of the numerator 5 degree of the denominator, then y 5 c is a horizontal asymptote where c is the ratio of the leading coefficients of the numerator and denominator, respectively. 3. If degree of the numerator . degree of the denominator, then there is no horizontal asymptote. n

Slant Asymptotes: Slant lines, y 5 mx 1 b, that steer the graph as x S 6q.

P R O C E D U R E F O R G R A P H I N G R AT I O N A L FUNCTIONS

1. Find the domain of the function. 2. Find the intercept(s). n n

y-intercept x-intercepts (if any)

3. Find any holes. n If x 2 a is a common factor of the numerator and ­denominator, then x 5 a corresponds to a hole in the graph of the rational function if the multiplicity of a in the numerator is greater than or equal to the multiplicity of a in the denominator. The result after the common factor is ­canceled is an equivalent rational function in lowest terms (no common factor). 4. Find any asymptotes. n Vertical asymptotes n Horizontal/slant asymptotes 5. Find additional points on the graph. 6. Sketch the graph: Draw the asymptotes and label the ­intercepts and points and connect with a smooth curve.

[ S E C T I O N 4 . 6]   E X E R C I S E S • SKILLS In Exercises 1–10, find the domain of each rational function 1 x13 x14 5. ƒ 1 x 2 5 2 x 1 x 2 12 3 1 x2 1 x 2 2 2 9. ƒ 1 x 2 5 2 2 1 x2 2 x 2 6 2 1. ƒ 1 x 2 5

3 42x x21 6. ƒ 1 x 2 5 2 x 1 2x 2 3 5 1 x 2 2 2x 2 3 2 10. ƒ 1 x 2 5 1 x2 2 x 2 6 2 2. ƒ 1 x 2 5

2x 1 1 5 2 3x 4. ƒ 1 x 2 5 1 3x 1 1 2 1 2x 2 1 2 1 2 2 3x 2 1 x 2 7 2 7x 2x 7. ƒ 1 x 2 5 2 8. ƒ 1 x 2 5 2 2 x 1 16 x 19 3. ƒ 1 x 2 5

In Exercises 11–20, find all vertical asymptotes and horizontal asymptotes (if there are any). 11. ƒ 1 x 2 5 15. ƒ 1 x 2 5 19. ƒ 1 x 2 5

1 x12 6x 5 2 4x 2 1 5 6x 2 1 5x 2 4 1 0.2x 2 3.1 2 1 1.2x 1 4.5 2 0.7 1 x 2 0.5 2 1 0.2x 1 0.3 2

Young_AT_6160_ch04_pp388-423.indd 410

12. ƒ 1 x 2 5 16. ƒ 1 x 2 5 20. ƒ 1 x 2 5

1 52x 6x 2 1 3x 1 1 3x 2 2 5x 2 2 4

0.8x 2 1 x 2 2 0.25

13. ƒ 1 x 2 5 17. ƒ 1 x 2 5

7x 3 1 1 x15 1 2 3x

1 13x 2 x2 1

1 9

1 4

14. ƒ 1 x 2 5

18. ƒ 1 x 2 5

2 2 x3 2x 2 7 1 2 10 Ax

2 2x 1 2x 2 1

3 10 B

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411

In Exercises 21–26, find the slant asymptote corresponding to the graph of each rational function. 21. ƒ 1 x 2 5 24. ƒ 1 x 2 5

x 2 1 10x 1 25 x 2 1 9x 1 20 2x 2 1 14x 1 7 22. ƒ1x2 5 23. ƒ1x2 5 x14 x23 x25 8x 4 1 7x 3 1 2x 2 5 3x 3 1 4x 2 2 6x 1 1 2x 6 1 1 1 2 1 2 25. ƒ x 5 26. ƒ x 5 x 2 2 x 2 30 2x 3 2 x 2 1 3x 2 1 x5 2 1

In Exercises 27–32, match the function to the graph. 27. ƒ 1 x 2 5

3 x24

30. ƒ 1 x 2 5 2

28. ƒ 1 x 2 5

3x 2 x 14

31. ƒ 1 x 2 5

2

a.

3x x24

29. ƒ 1 x 2 5

3x 2 4 2 x2

32. ƒ 1 x 2 5

b.

3x 2 x2 2 4 3x 2 x14

c.

y

y

y

10

10

150

x

x –10

–10

10

x

10

–10

–10

–10

d.

–150

e.

f.

y

y

10

y

5

5

x –10

10

x

10

–5

–10

5

x –5

–5

5

–5

In Exercises 33–58, use the graphing strategy outlined in this section to graph the rational functions. 33. ƒ 1 x 2 5 39. ƒ 1 x 2 5 43. ƒ 1 x 2 5 47. ƒ 1 x 2 5

2 4 2x 4x x21 21x   34.  ƒ 1 x 2 5   35.  ƒ 1 x 2 5   36.  ƒ 1 x 2 5   37.  ƒ 1 x 2 5   38.  ƒ 1 x 2 5 x x11 x22 x21 x12 x21 2 1 x 2 2 2x 2 3 2 x 2 1 2x 2x 3 2 x 2 2 x x2 2 4 7x 2 1 2x 1 1 2 2

51. ƒ 1 x 2 5 3x 1 55. ƒ 1 x 2 5

4 x

1 x 2 1 2 1 x2 2 4 2 1 x 2 2 2 1 x2 1 1 2

Young_AT_6160_ch04_pp388-423.indd 411

40. ƒ 1 x 2 5 44. ƒ 1 x 2 5 48. ƒ 1 x 2 5

3 1 x2 2 1 2 x 2 2 3x

3x 3 1 5x 2 2 2x x2 1 4 12x 4 1 3x 1 1 2 4

52. ƒ 1 x 2 5 x 2 56. ƒ 1 x 2 5

4 x

1 x 2 1 2 1 x2 2 9 2 1 x 2 3 2 1 x2 1 1 2

41. ƒ 1 x 2 5 45. ƒ 1 x 2 5 49. ƒ 1 x 2 5 53. ƒ 1 x 2 5 57. ƒ 1 x 2 5

x2 x11

x2 1 1 x2 2 1 1 2 9x 2 1 1 2 4x 2 2 3 1x 2 122 1 x2 2 1 2

3x 1 x 2 1 2 x 1 x2 2 4 2

42. ƒ 1 x 2 5 46. ƒ 1 x 2 5

50. ƒ 1 x 2 5 54. ƒ 1 x 2 5 58. ƒ 1 x 2 5

x2 2 9 x12 1 2 x2 x2 1 1 25x 2 2 1 1 16x 2 2 1 2 2

1x 1 122 1 x2 2 1 2

22x 1 x 2 3 2 x 1 x2 1 1 2

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In Exercises 59–62, for each graph of the rational function given determine: (a) all intercepts, (b) all asymptotes, and (c) equation of the rational function. 59.

60.

61.

y

62.

y

5

y

10

x –5

y

10

10

x

5

–10

–5

10

–10

x –10

10

x –10

–10

10

–10

• A P P L I C AT I O N S 63. Medicine. The concentration C of a particular drug in a

­person’s bloodstream t minutes after injection is given by C1t2 5

2t t 1 100 2

66. Memorization. A professor teaching a large lecture course

tries to learn students’ names. The number of names she can remember N  1t2 increases with each week in the semester t and is given by the rational function: N1t2 5

a  W  hat is the concentration in the bloodstream after 1 minute? b.  What is the concentration in the bloodstream after 1 hour? c.  What is the concentration in the bloodstream after 5 hours?

d.  Find the horizontal asymptote of C1t2. What do you expect

the concentration to be after several days? 64. Medicine. The concentration C of aspirin in the bloodstream

t hours after consumption is given by C 1 t 2 5

t . t 1 40 2

1

a.  What is the concentration in the bloodstream after 2 hour? b.  What is the concentration in the bloodstream after 1 hour? c.  What is the concentration in the bloodstream after 4 hours? d.   Find the horizontal asymptote for C  1t2. What do you

How many students’ names does she know by the third week in the semester? How many students’ names should she know by the end of the semester (16 weeks)? According to this function, what are the most names she can remember? 67. Food. The amount of food that cats typically eat increases as their weight increases. A rational function that describes this 10x 2 is F 1 x 2 5 2 , where the amount of food F1x2 is given in x 14 ounces and the weight of the cat x is given in pounds. ­Calculate the ­horizontal asymptote. How many ounces of food will most adult cats eat?

expect the concentration to be after several days?

65. Typing. An administrative assistant is hired after graduating

from high school and learns to type on the job. The number of words he can type per minute is given by N1t2 5

130t 1 260 t15

Young_AT_6160_ch04_pp388-423.indd 412

y 10

t$0

where t is the number of months he has been on the job. a.  How many words per minute can he type the day he starts? b.  How many words per minute can he type after 12 months? c.  How many words per minute can he type after 3 years? d.  How many words per minute would you expect him to type if he worked there until he retired?

600t t 1 20

x 10

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413

68. Memorization. The Guinness Book of World Records,

71. Business. Find the number of units that must be sold to

2004 states that Dominic O’Brien (England) memorized on a single sighting a random sequence of 54 separate packs of cards all shuffled together (2808 cards in total) at Simpson’s-in-the-Strand, London, England, on May 1, 2002. He memorized the cards in 11 hours 42 minutes and then recited them in exact sequence in a time of 3 hours 30 minutes. With only a 0.5% m ­ argin of error allowed (no more than 14 errors), he broke the record with just 8 errors. If we let x represent the time (hours) it takes to memorize the cards and y represent the number of cards memorized, then a rational

produce an average profit of $16,000 per thousand units. Convert the answer to average profit in dollars per unit. 72. Business. Find the number of units that must be sold to produce an average profit of $25,000 per thousand units. Convert the answer to average profit in dollars per unit.

2800x 2 1 x . x2 1 2 According to this model, how many cards could be memorized in an hour? What is the greatest number of cards that can be memorized? 69. Gardening. A 500-square-foot rectangular garden will be enclosed with fencing. Write a rational function that describes how many linear feet of fence will be needed to enclose the garden as a function of the width of the garden w. 70. Geometry. A rectangular picture has an area of 414 square inches. A border (matting) is used when framing. If the top and bottom borders are each 4 inches and the side borders are 3.5 inches, write a function that represents the area A1l2 of the entire frame as a function of the length of the picture l.

function that models this event is given by y 5

For Exercises 71 and 72, refer to the following: The monthly profit function for a product is given by P 1 x 2 5 2x 3 1 10x 2

where x is the number of units sold measured in thousands and P is profit measured in thousands of dollars. The average profit, which represents the profit per thousand units sold, for this product is given by P1x2 5

2x 3 1 10x 2 x

For Exercises 73 and 74, refer to the following: Some medications, such as Synthroid, are prescribed as a maintenance drug because they are taken regularly for an ongoing condition, such as hypothyroidism. Maintenance drugs function by maintaining a therapeutic drug level in the bloodstream over time. The concentration of a maintenance drug over a 24-hour period is modeled by the function C1t2 5

22 1 t 2 1 2 1 24 t2 1 1

where t is time in hours after the dose was administered and C is the concentration of the drug in the bloodstream measured in mg/mL. This medication is designed to maintain a consistent concentration in the bloodstream of approximately 25 mg/mL. Note: This drug will become inert; that is, the concentration will drop to 0 mg/mL, during the 25th hour after taking the medication. 73. Health/Medicine. Find the concentration of the drug, to the nearest tenth of mg/mL, in the bloodstream 15 hours after the dose is administered. Is this the only time this concentration of the drug is found in the bloodstream? At what other times is this concentration reached? Round to the nearest hour. Discuss the significance of this answer. 74. Health/Medicine. Find the time, after the first hour and a half, at which the concentration of the drug in the bloodstream has dropped to 25 mg/mL. Find the concentration of the drug 24 hours after taking a dose to the nearest tenth of a mgmL. Discuss the importance of taking the medication every 24 hours rather than every day.

where x is units sold measured in thousands and P is profit measured in thousands of dollars.

• C AT C H T H E M I S TA K E In Exercises 75–78, explain the mistake that is made. 75. Determine the vertical asymptotes of the function



ƒ1x2 5

The following is a correct graph of the function. Note that only x 5 21 is an asymptote. What went wrong? y

x21 . x2 2 1

10

Solution:

Set the denominator equal to zero.

x2 2 1 5 0



Solve for x.

x 5 61



The vertical asymptotes are x 5 21 and x 5 1.

x –10

10

–10

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CHAPTER 4  Polynomial and Rational Functions

76. Determine the vertical asymptotes of ƒ 1 x 2 5

2x . x2 1 1

78. Determine whether a horizontal or a slant asymptote exists

for the function ƒ 1 x 2 5

Solution:

x 2 1 2x 2 1 . If one does, find it. 3x 3 2 2x 2 2 1



Set the denominator equal to zero.

x2 1 1 5 0

Solution:



Solve for x.

x 5 61

Step 1:  The degree of the denominator is exactly one more



The vertical asymptotes are x 5 21 and x 5 1.

than the degree of the numerator, so there is a slant asymptote.

The following is a correct graph of the function. Note that there are no vertical asymptotes. What went wrong? y

Step 2:  Divide.  



2

x –10

10



The slant asymptote is y 5 3x 2 8.

The following is the c­ orrect graph of the function. Note that y 5 3x 2 8 is not an asymptote. What went wrong?

–2

y 10

77. Determine whether a horizontal or a slant asymptote exists

for the function ƒ 1 x 2 5

3x 2 8 x 2 1 2x 2 1 q 3x 3 2 2x 2 1 0x 2 1 3x 3 1 6x 2 2 3x 2 8x 2 1 3x 2 1 8x 2 2 16x 1 8 19x 2 9

9 2 x2 . If one does, find it. x2 2 1 x

Solution:

–5

5

Step 1:  The degree of the numerator equals the degree of the

denominator, so there is a horizontal asymptote. –10

Step 2:  The horizontal asymptote is the ratio of the lead coef-



ficients: y 5 91 5 9. The horizontal asymptote is y 5 9.

The following is a correct graph of the function. Note that there is no horizontal asymptote at y 5 9. What went wrong? y 20

x –10

10

–30

• CONCEPTUAL For Exercises 79–82, determine whether each statement is true or false. 79. A rational function can have either a horizontal asymptote or

81. A rational function can cross a vertical asymptote.

a slant asymptote but not both. 80. A rational function can have at most one vertical asymptote.

82. A rational function can cross a horizontal or a slant asymptote.

Young_AT_6160_ch04_pp388-423.indd 414

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4.6  Rational Functions 

415

• CHALLENGE 83. Determine the asymptotes of the rational function

1x 2 a2 1x 1 b2 ƒ1x2 5 . 1x 2 c2 1x 1 d2

84. Determine the asymptotes of the rational function

ƒ1x2 5

2

2

3x 1 b . x 2 1 a2

85.  Write a rational function that has vertical asymptotes at

x 5 23 and x 5 1 and a horizontal asymptote at y 5 4. 86.  Write a rational function that has no vertical asymptotes,

approaches the x-axis as a horizontal asymptote, and has an x-intercept of 13, 02.

• TECHNOLOGY x24 . x 2 2 2x 2 8 Graph this function utilizing a graphing utility. Does the graph confirm the asymptotes? 2x 1 1 88. Determine the vertical asymptotes of ƒ 1 x 2 5 2 . 6x 1 x 2 1  Graph this function utilizing a graphing utility. Does the graph confirm the asymptotes? 89. Find the asymptotes and intercepts of the rational function 1 2 ƒ1x2 5 2 . Note: Combine the two expressions into x 3x 1 1 a single rational expression. Graph this function utilizing a graphing utility. Does the graph confirm what you found? 90. Find the asymptotes and intercepts of the rational function 1 1 ƒ1x2 5 2 2 1 . Note: Combine the two expressions x x 11 into a single rational expression. Graph this function utilizing a graphing utility. Does the graph confirm what you found? 87. Determine the vertical asymptotes of ƒ 1 x 2 5

Young_AT_6160_ch04_pp388-423.indd 415

For Exercises 91 and 92: (a) Identify all asymptotes for each function. (b) Plot ƒ 1x2 and g  1x2 in the same window. How does the end behavior of the function ƒ differ from g? (c) Plot g  1x2 and h 1x2 in the same window. How does the end ­behavior of g differ from h? (d) Combine the two expressions into a single rational expression for the functions g and h. Does the strategy of finding horizontal and slant asymptotes agree with your findings in (b) and (c)? 91. ƒ 1 x 2 5 92. ƒ 1 x 2 5

1 1 1 , g1x2 5 2 1 , h 1 x 2 5 23 1 x23 x23 x23

2x 2x 2x , g1x2 5 x 1 2 , h1x2 5 x 2 3 1 2 x 21 x 21 x 21 2

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CHAPTER 4  Polynomial and Rational Functions 

[CHAPTER 4 REVIEW] SECTION

CONCEPT

4.1

Quadratic functions Graphs of quadratic functions: Parabolas

KEY IDEAS/FORMULAS

Parabolas Graphing quadratic functions in standard form ƒ1x2 5 a 1x 2 h22 1 k

CHAPTER 4 REVIEW

n Vertex:

Finding the equation of a parabola

4.2

Graphing polynomial functions using transformations of power functions

Real zeros of a polynomial function

4.4

Opens Up: a . 0

n 

Opens Down: a , 0

Graphing quadratic functions in general form b b ƒ1x2 5 ax2 1 bx 1 c, vertex is 1 h, k 2 5 a2 , f a2 bb 2a 2a

Polynomial functions of higher degree Identifying polynomial functions

4.3

1h, k2

n 

P 1 x 2 5 anx n 1 an21x n21 1 c1 a2x 2 1 a1x 1 a0 is a polynomial of degree n. Shift power functions

y 5 xn behave similar to: n 

y 5 x2, when n is even.

n 

y 5 x3, when n is odd.

P  1x2 5 1x 2 a2 1x 2 b2n 5 0 n 

a is a zero of multiplicity 1.

n 

b is a zero of multiplicity n.

Graphing general polynomial functions

Intercepts; zeros and multiplicities; end behavior

Dividing polynomials: Long division and synthetic division

Use zero placeholders for missing terms.

Long division of polynomials

Can be used for all polynomial division.

Synthetic division of polynomials

Can only be used when dividing by 1x 6 a2.

The real zeros of a polynomial function The remainder theorem and the factor theorem The rational zero theorem and Descartes’ rule of signs

Factoring polynomials

P 1 x 2 5 anx n 1 an21x n21 1 c1 a2x 2 1 a1x 1 a0 If P  1c2 5 0, then c is a zero of P  1x2.

If P  1x2 is divided by x 2 a, then the remainder r is r 5 P  1a2. Possible zeros 5

Factors of a0

Factors of an  umber of positive or negative real zeros is related to the N number of sign variations in P  1x2 or P  12x2. 1.  List possible rational zeros (rational zero theorem).

2.  List possible combinations of positive and negative real zeros (Descartes’ rule of signs). 3.  Test possible values until a zero is found. 4.  Once a real zero is found, use synthetic division. Then repeat testing on quotient until linear and/or irreducible quadratic factors are reached. 5.  If there is a real zero but all possible rational roots have failed, then approximate the real zero using the intermediate value theorem/bisection method. The intermediate value theorem

The intermediate value theorem and the bisection method are used to approximate irrational zeros.

Graphing polynomial functions

1.  Find the intercepts.       2.  Determine end behavior. 3.  Find additional points.     4.  Sketch a smooth curve.

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Chapter Review  CONCEPT

4.5

Complex zeros: The fundamental theorem of algebra

KEY IDEAS/FORMULAS

P 1 x 2 5 anx n 1 an21x n21 1 c1 a2x 2 1 a1x 1 a0 P 1 x 2 5 1 x 2 c1 21 x 2 c2 2 c1 x 2 cn 2

d

SECTION

417

n factors  here the c’s represent complex (not necessarily distinct) zeros. w P  1x2 of degree n has at least one zero and at most n zeros.

n 

Complex conjugate pairs:

Factoring polynomials

 he polynomial can be written as a product of linear factors, not T necessarily distinct.

n If

4.6

Rational functions Domain of rational functions

Horizontal, vertical, and slant asymptotes

a 1 bi is a zero of P  1x2, then a 2 bi is also a zero.

ƒ1x2 5

n1x2

d 1x2

d 1x2 2 0

Domain: All real numbers except x-values that make the denominator equal to zero; that is, d  1x2 5 0. n1x2 A rational function ƒ 1 x 2 5 is said to be in lowest terms if n  1x2 d 1x2 and d  1x2 have no common factors.

 rational function that has a common factor x 2 a in both the numerator and A ­denominator has a hole at x 5 a in its graph if the multiplicity of a in the numerator is greater than or equal to the multiplicity of a in the denominator.

CHAPTER 4 REVIEW

The fundamental theorem of algebra Complex zeros

Vertical Asymptotes A rational function in lowest terms has a vertical asymptote corresponding to any x-values that make the denominator equal to zero. Horizontal Asymptotes n  n 

y 5 0 if degree of n  1x2 , degree of d  1x2.

No horizontal asymptote if degree of n  1x2 . degree of d  1x2.

n  y

5

Leading coefficient of n 1 x 2

Leading coefficient of d 1 x 2

  if degree of n  1x2 5 degree of d  1x2.

Graphing rational functions

Slant Asymptotes If degree of n  1x2 2 degree of d  1x2 5 1. Divide n  1x2 by d  1x2 and the quotient determines the slant asymptote; that is, y 5 quotient. 1.  Find the domain of the function. 2.  Find the intercept(s). 3.  Find any holes. 4.  Find any asymptotes. 5.  Find additional points on the graph. 6.  Sketch the graph: Draw the asymptotes and label the intercepts and points and connect with a smooth curve.

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418 

CHAPTER 4  P olynomial and Rational Functions

[CHAPTER 4 REVIEW EXERCISES] 4.1  Quadratic Functions

19. ƒ1x2 5 20.45x2 2 0.12x 1 3.6

Match the quadratic function with its graph.

20. ƒ 1 x 2 5 2 4 x 2 1 5 x 1 4

 1. ƒ1x2 5 221x 1 622 1 3  3. ƒ1x2 5 x2 1 x 2 6

3

2. ƒ 1 x 2 5 14 1 x 2 4 2 2 1 2 4. ƒ1x2 5 23x2 2 10x 1 8

a.

Find the quadratic function that has the given vertex and goes through the given point. 21. vertex: 122, 32  point: 11, 42

b. y

22. vertex: 14, 72  point: 123, 12

y

10

2

23. vertex: 12.7, 3.42  point: 13.2, 4.82

10

5 7

REVIEW EXERCISES

x –10

10

x –10

–10

Applications

25. Profit. The revenue and the cost of a local business are given

10

below as functions of the number of units x in thousands produced and sold. Use the cost and the revenue to answer the questions that follow.

–10

c.

d. y

1 C 1 x 2 5 x 1 2 and R 1 x 2 5 22x 2 1 12x 2 12 3 a. Determine the profit function. b. State the break-even points.

y

10

1 3

24. vertex: A2 2 , 4 B  point: A 2 , 5 B

c. Graph the profit function.

16

d. What is the range of units to make and sell that will x –10

10 x –10 –10

10 –4

Graph the quadratic function given in standard form.  5. ƒ1x2 5 21x 2 722 1 4 6.  ƒ1x2 5 1x 1 322 2 5 1

1 2

2

 7. ƒ 1 x 2 5 2 2 Ax 2 3 B 1 5 8.  ƒ1x2 5 0.61x 2 0.7522 1 0.5

Rewrite the quadratic function in standard form by completing the square.  9. ƒ1x2 5 x2 2 3x 2 10 11. ƒ1x2 5 4x2 1 8x 2 7

10. ƒ1x2 5 x2 2 2x 2 24 12. ƒ 1 x 2 5 2 41 x 2 1 2x 2 4

Graph the quadratic function given in general form. 13. ƒ1x2 5 x2 2 3x 1 5

correspond to a profit? 26. Geometry. Given the length of a rectangle is 2x 2 4 and the width is x 1 7, find the area of the rectangle. What dimensions correspond to the largest area? 27. Geometry. A triangle has a base of x 1 2 units and a height of 4 2 x units. Determine the area of the triangle. What dimensions correspond to the largest area? 28. Geometry. A person standing at a ridge in the Grand Canyon throws a penny upward and toward the pit of the canyon. The height of the penny is given by the function: h 1 t 2 5 212t 2 1 80t

a. What is the maximum height that the penny will reach? b. How many seconds will it take the penny to hit the

ground below? 4.2  Polynomial Functions of Higher Degree

Determine which functions are polynomials, and for those, state their degree. 29. ƒ1x2 5 x6 2 2x5 1 3x2 1 9x 2 42

15. ƒ1x2 5 24x2 1 2x 1 3

30. ƒ1x2 5 13x 2 4231x 1 622

16. ƒ1x2 5 20.75x2 1 2.5

32. ƒ1x2 5 5x3 2 2x2 1

14. ƒ1x2 5 2x2 1 4x 1 2

4

31. ƒ1x2 5 3x4 2 x31 x2 1 ! x 1 5

4x 23 7

Find the vertex of the parabola associated with each ­quadratic function. 17. ƒ1x2 5 13x2 2 5x 1 12 2

18. ƒ 1 x 2 5 5 x 2 2 4x 1 3

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Review Exercises 

Match the polynomial function with its graph.

53. ƒ1x2 5 6x7 1 3x5 2 x2 1 x 2 4

33. ƒ1x2 5 2x 2 5 34. ƒ1x2 5 23x2 1 x 2 4

54. ƒ1x2 5 2x413x 1 6231x 2 723

35. ƒ1x2 5 x4 2 2x3 1 x2 2 6 36. ƒ1x2 5 x7 2 x5 1 3x4 1 3x 1 7 a.

Applications

55. Salary. Tiffany has started tutoring students x hours per

b. y

y

10

419

week. The tutoring job corresponds to the following additional income:

x

–5

5

ƒ 1 x 2 5 1 x 2 1 21 x 2 3 21 x 2 7 2

a. Graph the polynomial function. b. Give any real zeros that occur.

x –5

5

c.  How many hours of tutoring are financially beneficial to

–10

c.



d. y



y

40

10

P 1 x 2 5 3 1 x 2 2 2 2 1 x 2 5 2 2 1 x 2 10 2 2    1 # x # 12

Graph the polynomial. When are the peak seasons?

4.3  Dividing Polynomials: Long Division and Synthetic

Division

x –10

10

x –5

57. 1x2 1 2x 2 62 4 1x 2 22

5 –10

58. 12x2 2 5x 2 12 4 12x 2 32

–10

Graph each function by transforming a power function y 5 xn. 7

37. ƒ1x2 5 2x 38.  ƒ1x2 5 1x 2 32 39. ƒ1x2 5 x4 2 2

3

40.  ƒ1x2 5 26 2 1x 1 725

Find all the real zeros of each polynomial function, and state their multiplicities. 2

41. ƒ1x2 5 31x 1 42 1x 2 62

5

3

42.  ƒ1x2 5 7x  12x 2 42 1x 1 52

43. ƒ1x2 5 x5 2 13x3 1 36x 44.  ƒ1x2 5 4.2x4 2 2.6x2

Find a polynomial of minimum degree that has the given zeros. 45. 23, 0, 4

46. 2, 4, 6, 28

47. 2 52, 34, 0

48. 2 2 !5, 2 1 !5

49. 22 1multiplicity of 22, 3 1multiplicity of 22

50. 3 1multiplicity of 22, 21 1multiplicity of 22, 0 1multiplicity of 32

For each polynomial function given: (a) list each real zero and its multiplicity; (b) determine whether the graph touches or crosses at each x-intercept; (c) find the y-intercept and a few points on the graph; (d) determine the end behavior; and (e) sketch the graph. 51. ƒ1x2 5 x2 2 5x 2 14 52. ƒ1x2 5 21x 2 525

Young_AT_6160_ch04_pp388-423.indd 419

Divide the polynomials with long division. If you choose to use a calculator, do not round off. Keep the exact values instead. Express the answer in the form Q  1x2 5 ?,   r 1x2 5 ?.

REVIEW EXERCISES

–10

Tiffany? 56. Profit. The following function is the profit for Walt Disney World, where P  1x2 represents profit in millions of dollars and x represents the month 1x 5 1 corresponds to January2:

59. 14x4 2 16x3 1 x 2 9 1 12x22 4 12x 2 42

60. 16x2 1 2x3 2 4x4 1 2 2 x2 4 12x2 1 x 2 42

Use synthetic division to divide the polynomial by the linear factor. Indicate the quotient Q  1x2 and the remainder r 1x2. 61. 1x4 1 4x3 1 5x2 2 2x 2 82 4 1x 1 22 62. 1x3 2 10x 1 32 4 12 1 x2 63. 1x6 2 642 4 1x 1 82

3 64. 1 2x 5 1 4x 4 2 2x 3 1 7x 1 5 2 4 Ax 2 4 B

Divide the polynomials with either long division or synthetic division. 65. 15x3 1 8x2 2 22x 1 12 4 15x2 2 7x 1 32 66. 1x4 1 2x3 2 5x2 1 4x 1 22 4 1x 2 32 67. 1x3 2 4x2 1 2x 2 82 4 1x 1 12

68. 1x3 2 5x2 1 4x 2 202 4 1x2 1 42

Applications

69. Geometry. The area of a rectangle is given by the

polynomial 6x4 2 8x3 2 10x2 1 12x 2 16. If the width is 2x 2 4, what is the length of the rectangle? 70. Volume. A 10-inch by 15-inch rectangular piece of cardboard is used to make a box. Square pieces x inches on a side are cut out from the corners of the cardboard, and then the sides are folded up. Find the volume of the box.

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420 

CHAPTER 4  P olynomial and Rational Functions

4.4  The Real Zeros of a Polynomial Function

Find the following values by applying synthetic division. Check by substituting the value into the function. ƒ 1 x 2 5 6x 5 1 x 4 2 7x 2 1 x 2 1 g 1 x 2 5 x 3 1 2x 2 2 3 71. ƒ1222  72. ƒ112  73.  g  112  74.  g  1212 Determine whether the number given is a zero of the polynomial. 75. 23, P  1x2 5 x3 2 5x2 1 4x 1 2 76. 2 and 22, P  1x2 5 x4 2 16

REVIEW EXERCISES

77. 1, P  1x2 5 2x4 2 2x

78. 4, P  1x2 5 x4 2 2x3 2 8x

Given a zero of the polynomial, determine all other real zeros, and write the polynomial in terms of a product of linear or irreducible factors. Zero P  1x2 5 x 2 6x 1 32x 22 3 2 P  1x2 5 x 2 7x 1 36 3 P  1x2 5 x5 2 x4 2 8x3 1 12x2 0 P  1x2 5 x4 2 32x2 2 144 6

79. 80. 81. 82.

Polynomial 4

3

Use Descartes’ rule of signs to determine the possible number of positive real zeros and negative real zeros. 4

3

5

3

9

7

84. P  1x2 5 x 1 6x 2 4x 2 2 4

 96. P  1x2 5 x3 1 3x2 2 6x 2 8

 97. P  1x2 5 x3 2 9x2 1 20x 2 12

 98. P  1x2 5 x4 2 x3 2 7x2 1 x 1 6

 99. P  1x2 5 x4 2 5x3 2 10x2 1 20x 1 24 100. P  1x2 5 x5 2 3x3 2 6x2 1 8x

4.5  Complex Zeros: The Fundamental Theorem

of Algebra

Find all zeros. Factor the polynomial as a product of linear factors. 101. P  1x2 5 x2 1 25

102. P  1x2 5 x2 1 16 104. P  1x2 5 x2 1 4x 1 5

2

103. P  1x2 5 x 2 2x 1 5

A polynomial function is described. Find all remaining zeros. 105. Degree: 4

Zeros: 22i, 3 1 i Zeros: 3i, 2 2 i Zeros: i, 2 2 i 1multiplicity 22 Zeros: 2i, 1 2 i 1multiplicity 22

107. Degree: 6 3

85. P  1x2 5 x 2 2x 1 x 2 3x 1 2x 2 1 3

 95. P  1x2 5 x3 1 3x 2 5

106. Degree: 4

83. P  1x2 5 x 1 3x 2 16 5

For each polynomial: (a) use Descartes’ rule of signs to determine the possible combinations of positive real zeros and negative real zeros; (b) use the rational zero test to determine possible rational zeros; (c) use the upper and lower bound rules to eliminate possible zeros; (d) test for rational zeros; (e) factor as a product of linear and/or ­irreducible quadratic factors; and (f) graph the polynomial function.

108. Degree: 6

2

86. P  1x2 5 2x 2 4x 1 2x 2 7

Use the rational zero theorem to list the possible rational zeros.

87. P  1x2 5 x3 2 2x2 1 4x 1 6

88. P  1x2 5 x5 2 4x3 1 2x2 2 4x 2 8

89. P  1x2 5 2x4 1 2x3 2 36x2 2 32x 1 64 90. P  1x2 5 24x5 2 5x3 1 4x 1 2

Given a zero of the polynomial, determine all other zeros (real and complex) and write the polynomial in terms of a product of linear factors. Polynomial 4

3

2

109. P  1x2 5 x 2 3x 2 3x 2 3x 2 4

110. P  1x2 5 x4 2 4x3 1 x2 1 16x 2 20

111. P  1x2 5 x4 2 2x3 1 11x2 2 18x 1 18 112. P  1x2 5 x4 2 5x2 1 10x 2 6

Zero i 22i 23i 11i

List the possible rational zeros, and test to determine all rational zeros.

Factor each polynomial as a product of linear factors.

91. P  1x2 5 2x3 2 5x2 1 1

113. P  1x2 5 x4 2 81

94. P  1x2 5 24x4 2 4x3 2 10x2 1 3x 2 2

116. P  1x2 5 x4 2 5x3 1 12x2 2 2x 2 20

92. P  1x2 5 12x3 1 8x2 2 13x 1 3 93. P  1x2 5 x4 2 5x3 1 20x 2 16

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114. P  1x2 5 x3 2 6x2 1 12x

115. P  1x2 5 x3 2 x2 1 4x 2 4

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Review Exercises 

4.6  Rational Functions

Section 4.3

Determine the vertical, horizontal, or slant asymptotes (if they exist) for the following rational functions.

133. Plot

117. ƒ 1 x 2 5 119. ƒ 1 x 2 5 121. ƒ 1 x 2 5

72x x12

118. ƒ 1 x 2 5

4x 2 x11 2x 2 2 3x 1 1 x2 1 4

120. ƒ 1 x 2 5 122. ƒ 1 x 2 5

2 2 x2 1x 2 123

3x 2 2 x 19

22x 2 1 3x 1 5 x15

Graph the rational functions.



421

15x 3 2 47x 2 1 38x 2 8 . What type of function is it? 3x 2 2 7x 1 2 Perform this division using long division, and confirm that the graph corresponds to the quotient. 24x 3 1 14x 2 2 x 2 15 . What type of function is it? x23 Perform this division using synthetic division, and confirm that the graph corresponds to the quotient.

134. Plot

Section 4.4

5 124. ƒ 1 x 2 5 x11

In Exercises 135 and 136: (a) From the list of all possible rational zeros of the polynomial, use a graphing calculator or software to graph P 1x2 to find the rational zeros. (b) Factor as a product of linear and/or irreducible quadratic factors.

125. ƒ 1 x 2 5

126. ƒ 1 x 2 5

135. P 1 x 2 5 x 4 2 3x 3 2 12x 2 1 20x 1 48

127. ƒ 1 x 2 5

x2 x2 1 4

x 2 2 49 x17

128. ƒ 1 x 2 5

x 2 2 36 x 2 1 25 2x 2 2 3x 2 2 2x 2 2 5x 2 3

Technology Section 4.1 129. On a graphing calculator, plot the quadratic function: ƒ 1 x 2 5 0.005x 2 2 4.8x 2 59

a. Identify the vertex of this parabola. b. Identify the y-intercept. c.  Identify the x-intercepts (if any). d. What is the axis of symmetry?

130. Determine the quadratic function whose vertex is

12.4, 23.12 and passes through the point 10, 5.542. a. Write the quadratic function in general form. b. Plot this quadratic function with a graphing calculator. c.  Zoom in on the vertex and y-intercept. Do they agree with the given values?

136. P 1 x 2 5 25x 5 2 18x 4 2 32x 3 2 24x 2 1 x 1 6

Section 4.5

Find all zeros (real and complex). Factor the polynomial as a product of linear factors. 137. P 1 x 2 5 2x 3 1 x 2 2 2x 2 91

138. P 1 x 2 5 22x 4 1 5x 3 1 37x 2 2 160x 1 150

Section 4.6

REVIEW EXERCISES

2 123. ƒ 1 x 2 5 2 x23

In Exercises 139 and 140: (a) graph the function ƒ 1x2 utilizing a graphing utility to determine if it is a one-to-one function; (b) if it is, find its inverse; and (c) graph both functions in the same viewing window. 139. ƒ 1 x 2 5 140. ƒ 1 x 2 5

2x 2 3 x11 4x 1 7 x22

Section 4.2

In Exercises 131 and 132, use a graphing calculator or a computer to graph each polynomial. From the graph, estimate the x-intercepts and state the zeros of the function and their multiplicities. 131. ƒ 1 x 2 5 5x 3 2 11x 2 2 10.4x 1 5.6

132. ƒ 1 x 2 5 2x 3 2 0.9x 2 1 2.16x 2 2.16

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422 

CHAPTER 4  Polynomial and Rational Functions

[CHAPTER 4 PRACTICE TEST]  1. Graph the parabola y 5 21x 2 422 1 1.

20. Profit. The profit of a company is governed by the polynomial

 2. Write the parabola in standard form y 5 2x2 1 4x 2 1. 1

 3. Find the vertex of the parabola ƒ1x2 5 22 x 2 1 3x 2 4.

PRACTICE TEST

 4. Find a quadratic function whose vertex is 123, 212 and

whose graph passes through the point 124, 12.  5. Find a sixth-degree polynomial function with the given zeros: 2 of multiplicity 3  1 of multiplicity 2  0 of multiplicity 1  6. For the polynomial function ƒ1x2 5 x4 1 6x3 2 7x: a. List each real zero and its multiplicity. b. Determine whether the graph touches or crosses at each x-intercept. c. Find the y-intercept and a few points on the graph. d. Determine the end behavior. e. Sketch the graph.  7. Divide 24x4 1 2x3 2 7x2 1 5x 2 2 by 2x2 2 3x 1 1.  8. Divide 17x5 2 4x3 1 2x 2 10 by x 1 2.  9. Is x 2 3 a factor of x4 1 x3 2 13x2 2 x 1 12? 10. Determine whether 21 is a zero of P  1x2 5 x21 2 2x  18 1 5x12 1 7x3 1 3x2 1 2. 11. Given that x 2 7 is a factor of P  1x2 5 x3 2 6x2 2 9x 1 14, factor the polynomial in terms of linear factors. 12. Given that 3i is a zero of P  1x2 5 x4 2 3x3 1 19x2 2 27x 1 90, find all other zeros. 13. Can a polynomial have zeros that are not x-intercepts? Explain. 14. Apply Descartes’ rule of signs to determine the possible combinations of positive real zeros, negative real zeros, and complex zeros of P  1x2 5 3x5 1 2x4 2 3x3 1 2x2 2 x 1 1. 15. From the rational zero test, list all possible rational zeros of P  1x2 5 3x4 2 7x2 1 3x 1 12. In Exercises 16–18, determine all zeros of the polynomial function and graph.

16. P  1x2 5 2x3 1 4x

17. P  1x2 5 2x3 2 3x2 1 8x 2 12 4

3

In Exercises 22–25, determine (if any) the: a. x- and y-intercepts b. vertical asymptotes c. horizontal asymptotes d. slant asymptotes e. graph 22. ƒ 1 x 2 5 24. h 1 x 2 5

2x 2 9 x13

23. g 1 x 2 5

3x 3 2 3 x2 2 4

25. F 1 x 2 5

x x2 2 4 x23 x 2 2 2x 2 8

26. Food. On eating a sugary snack, the average body almost

doubles its glucose level. The percentage increase in glucose level y can be approximated by the rational function 25x y5 2 , where x represents the number of minutes after x 1 50 eating the snack. Graph the function. 27.  a. Use the calculator commands STAT QuadReg to model the data using a quadratic function. b. Write the quadratic function in standard form and identify the vertex. c.  Find the x-intercepts. d. Plot this quadratic function with a graphing calculator. Do they agree with the given values? x

23

2.2

7.5

y

10.01

29.75

25.76

2

18. P  1x2 5 x 2 6x 1 10x 2 6x 1 9

19. Sports. A football player shows up in August at 300 pounds.

After 2 weeks of practice in the hot sun, he is down to 285 pounds. Ten weeks into the season he is up to 315 pounds because of weight training. In the spring he does not work out, and he is back to 300 pounds by the next August. Plot these points on a graph. What degree polynomial could this be?

Young_AT_6160_ch04_pp388-423.indd 422

P  1x2 5 x3 2 13x2 1 47x 2 35, where x is the number of units sold in thousands. How many units does the company have to sell to break even? 21. Interest Rate. The interest rate for a 30-year fixed mortgage fluctuates with the economy. In 1970, the mortgage interest rate was 8%, and in 1988 it peaked at 13%. In 2002, it dipped down to 4%, and in 2005, it was up to 6%. In 2015 it dipped back down to 4.25%. What is the lowest-degree polynomial that can represent this function?

28. Find the asymptotes and intercepts of the rational function

x 1 2x 2 3 2 1 1. Note: Combine the two expressions x 2 2 3x into a single rational expression. Graph this function utilizing a graphing utility. Does the graph confirm what you found? ƒ1x2 5

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Cumulative Test 

423

[CHAPTERS 1–4 CUMULATIVE TEST]   1. Simplify

1 5x 21y 22 2 3 and express in terms of positive 1210x 2y 2 2 2

exponents.

  2. Factor 2xy 2 2x 1 3y 2 3.   3. Multiply and simplify

4x 2 2 36 x 3 1 6x 2 ⋅ . x2 x 2 1 9x 1 18

  4. Solve for x: 0 2x 2 5 0 1 3 . 10.

  5. Austin can mow a lawn in 75 minutes. The next week Stacey

  6.   7.   8.

  9. 10. 11. 12. 13. 14.

16. Find the composite function f + g and state the domain for

ƒ1x2 5 x2 2 3 and g 1 x 2 5 !x 1 2.

Young_AT_6160_ch04_pp388-423.indd 423

18. Find the inverse of the function ƒ1x2 5 1x 2 422 1 2, 19.

20. 21. 22. 23. 24.

25. 26.

where x $ 4. Find a quadratic function that has the vertex 122, 32 and point 121, 42. Find all of the real zeros and state their multiplicities of the function ƒ1x2 5 23.7x4 2 14.8x3. Use long division to find the quotient Q 1x2 and the remainder r  1x2 of 1220x3 2 8x2 1 7x 2 52 4 125x 1 32. Use synthetic division to find the quotient Q 1x2 and the remainder r  1x2 of 12x3 1 3x2 2 11x 1 62 4 1x 2 32. List the possible rational zeros, and test to determine all rational zeros for P  1x2 5 12x3 1 29x2 1 7x 2 6. Given the real zero x 5 5 of the polynomial P  1x2 5 2x3 2 3x2 2 32x 2 15, determine all the other zeros and write the polynomial in terms of a product of linear factors. 3x 2 5 Find all vertical and horizontal asymptotes of ƒ 1 x 2 5 2 . x 24 2x 3 2 x 2 2 x Graph the function ƒ 1 x 2 5 . x2 2 1

27. Find the asymptotes and intercepts of the rational function 5 1 ƒ1x2 5 2 . Note: Combine the two expressions into 2x 2 3 x a single rational expression. Graph this function utilizing a

graphing utility. Does the graph confirm what you found? 28. Find the asymptotes and intercepts of the rational function 6x 6x ƒ1x2 5 2 . Note: Combine the two ­expressions 3x 1 1 4x 2 1 into a single rational expression. Graph this function utilizing

a graphing utility. Does the graph confirm what you found?

CUMULATIVE TEST

15.

mows the same lawn in 60 minutes. How long would it take them to mow the lawn working together? Use the discriminant to determine the number and type of roots: 24x2 1 3x 1 15 5 0. Solve and check "16 1 x 2 5 x 1 2. Apply algebraic tests to determine whether the equation’s graph is symmetric with respect to the x-axis, y-axis, or origin for y 5 0 x 0 2 3. Write an equation of the line that is parallel to the line x 2 3y 5 8 and has the point 14, 12. Find the x-intercept and y-intercept and sketch the graph for 2y 2 6 5 0. Write the equation of a circle with center 10, 62 and that passes through the point 11, 52. Express the domain of the function ƒ 1 x 2 5 !6x 2 7 with interval notation. Determine whether the function g 1 x 2 5 !x 1 10 is even, odd, or neither. For the function y 5 21x 1 122 1 2, identify all of the transformations of y 5 x2. Sketch the graph of y 5 !x 2 1 1 3 and identify all transformations.

17. Evaluate g  1 ƒ 1212 2 for ƒ1x2 5 7 2 2x2 and g  1x2 5 2x 2 10.

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[5[ CHAPTER

Exponential and Logarithmic Functions

Francois Gohier/Science Source

How do archaeologists and anthropologists date fossils? One method is carbon testing. The ratio of carbon 12 (the more stable kind of carbon) to carbon 14 at the moment of death is the same as the ratio in all living things while they are still alive, but the carbon 14 decays and is not replaced. By looking at the ratio of carbon 12 to carbon 14 in the sample and comparing it with the ratio in a living organism, it is possible to determine the age of a formerly living thing fairly precisely. The amount of carbon in a fossil is a function of how many years the organism has been dead. The amount is modeled by an exponential function, and the inverse of the exponential function, a logarithmic function, is used to determine the age of the fossil.

LEARNING OBJECTIVES ■■ Graph

exponential functions. ■■ Graph logarithmic functions.

Young_AT_6160_ch05_pp424-455.indd 424

■■ Apply

properties of logarithms. ■■ Solve exponential and ­logarithmic equations.

■■ Use

exponential and logarithmic models to represent a variety of real-world phenomena.

25/11/16 3:17 PM

[IN THIS CHAPTER] We will discuss exponential functions and their inverses, logarithmic functions. We will graph these functions and use their properties to solve exponential and logarithmic equations. Last, we will discuss particular exponential and logarithmic models that represent phenomena such as compound interest, world populations, conservation biology models, carbon dating, pH values in chemistry, and the bell curve that is fundamental in statistics for describing how quantities vary in the real world.

EX PONEN T I AL AN D L O G AR I T H M I C F UN C T I O NS 5.1

5.2

5.3

5.4

5.5

EXPONENTIAL FUNCTIONS AND THEIR GRAPHS

LOGARITHMIC FUNCTIONS AND THEIR GRAPHS

PROPERTIES OF LOGARITHMS

EXPONENTIAL AND LOGARITHMIC EQUATIONS

EXPONENTIAL AND LOGARITHMIC MODELS

• Evaluating Exponential Functions • Graphs of Exponential Functions • The Natural Base e • Applications of Exponential Functions

• Evaluating Logarithms • Common and Natural Logarithms • Graphs of Logarithmic Functions • Applications of Logarithms

• Properties of Logarithmic Functions • Change-of-Base Formula

• Exponential Equations • Solving Logarithmic Equations

• Exponential Growth Models • Exponential Decay Models • Gaussian (Normal) Distribution Models • Logistic Growth Models • Logarithmic Models

425

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CHAPTER 5  Exponential and Logarithmic Functions

5.1 EXPONENTIAL FUNCTIONS AND THEIR GRAPHS SKILLS OBJECTIVES ■■ Evaluate exponential functions. ■■ Graph exponential functions. ■■ Evaluate exponential functions of base e. ■■ Apply exponential functions to economics and the natural sciences.

CONCEPTUAL OBJECTIVES ■■ Understand that irrational exponents lead to approximations. ■■ Understand characteristics of exponential functions (implying domain, range, asymptotes, intercepts, etc.). ■■ Understand that e is irrational and why it is the “natural” base. ■■ Understand why compounding continuously results in higher interest than compounding daily.

5.1.1  Evaluating Exponential Functions 5.1.1 S K I L L

Evaluate exponential functions. 5.1.1 C O N C E P T U A L

Understand that i­ rrational ­exponents lead to approximations.

Most of the functions (polynomial, rational, radical, etc.) we have studied thus far have been algebraic functions. Algebraic functions involve basic operations, powers, and roots. In this chapter, we discuss exponential functions and logarithmic functions, which are called transcendental functions because they transcend our ability to define them with a finite number of algebraic expressions. The following table illustrates the difference between algebraic functions and exponential functions. FUNCTION

VARIABLE IS IN THE

CONSTANT IS IN THE

EXAMPLE

EXAMPLE

Algebraic

Base

Exponent

ƒ1x2 5 x2

g 1x2 5 x1/3

Exponential

Exponent

Base

F 1x2 5 2x

DEFINITION

1 x G1x2 5 a b 3

Exponential Function

An exponential function with base b is denoted by ƒ1x2 5 bx

where b and x are any real numbers such that b . 0 and b 2 1. Note: We eliminate b 5 1 as a value for the base because it merely yields the constant function ƒ1x2 5 1x 5 1. ■■ We eliminate negative values for b because they would give non–real-number values such as 129 2 1/2 5 !29 5 3i. ■■ We eliminate b 5 0 because 0x corresponds to an undefined value when x is negative. ■■

Sometimes the value of an exponential function for a specific argument can be found by inspection as an exact number. x

F1x2 5 2x

23

223 5

1 1 5 8 23

21

221 5

1 1 5 2 21

0

1

3

20 5 1

21 5 2

23 5 8

If an exponential function cannot be evaluated exactly, then we find the decimal approximation using a calculator. Most calculators have either a base to a power­ button x y or a caret ` button for working with exponents. x

F1x2 5 2x

Young_AT_6160_ch05_pp424-455.indd 426

22.7

2 54

5 7

2.7

222.7 < 0.154

224/5 < 0.574

25/7 < 1.641

22.7 < 6.498

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5.1  Exponential Functions and Their Graphs 

The domain of exponential functions, ƒ1x2 5 bx, is the set of all real numbers. It is important to note that in Chapter 0 we discussed exponents of the form bx, where x is an integer or a rational number. What happens if x is irrational? We can approximate the irrational number with a decimal approximation such as bp < b3.14 or b!2 < b1.41. Consider 7!3, and realize that the irrational number !3 is a decimal that never terminates or repeats: !3 5 1.7320508 . . . . We can show in advanced mathematics that there is a number 7!3, and although we cannot write it exactly, we can approximate the number. In fact, the closer the exponent is to !3, the closer the approximation is to 7!3. It is important to note that the properties of exponents (Chapter 0) hold when the exponent is any real number (rational or irrational).

427

71.7 < 27.3317 71.73 < 28.9747 71.732 < 29.0877 ... !3 < 29.0906 7

EXAMPLE 1  Evaluating Exponential Functions x

Let ƒ1x2 5 3x, g 1 x 2 5 A 14 B , and h1x2 5 10 x22. Find the following values. If an approximation is required, approximate to four decimal places. 3 a. ƒ122  b. ƒ1p2  c. gA2 2 B  d. h12.32  e. ƒ102  f. g102 Solution:

a. ƒ122 5 32 5 9 b. ƒ 1 p 2 5 3p < 3 c. gA22 B

5

23/2 A 14 B

31.5443 * 5 43/2 5 A!4B 3 5 23 5 8

d. h 1 2.3 2 5 102.322 5 100.3 < 1.9953 e. ƒ102 5 30 5 1 1 0

f. g 1 0 2 5 A 4 B 5 1

Notice that parts (a) and (c) were evaluated exactly, whereas parts (b) and (d) required approximation using a calculator.

▼

1 x

Y O U R T U R N   Let ƒ1x2 5 2x and g 1 x 2 5 A 9 B and h1x2 5 5x22. Find the following

values. Evaluate exactly when possible, and round to four decimal places when a calculator is needed. 3 a. ƒ142  b. ƒ1p2  c. gA2 2 B  d. h12.92

*In part (b), the p button on the calculator is selected. If we instead approximate p by 3.14, we get a slightly ­different approximation for the function value: ƒ1p2 5 3 < 3 p

3.14

< 31.4891

▼ ANSWER a. 16   b. 8.8250 c. 27   d. 4.2567

[ CONCEPT CHECK ] Find the approximation to f (x) 5 px for f (22.3); round to four decimal places.

▼ ANSWER 0.0719

5.1.2  Graphs of Exponential Functions

x

Let’s graph two exponential functions, y 5 2x and y 5 22x 5 A 12 B , by plotting points. y 5 2x

x

1 1 5 2 5 2 4

22

2

21

221 5

0

0

22

1 1 5 21 2

2 51 1

1

2 52

2

22 5 4

3

23 5 8

(x, y )

1 a22, b 4 1 a21, b 2 10, 12 11, 22 12, 42 13, 82

x

y 5 22x

(x, y )

23

221232 5 23 5 8

22

221222 5 22 5 4

123, 82

21

221212 5 21 5 2

0

0

2 51 1 1 5 21 2

1

221 5

2

1 1 222 5 2 5 2 4

122, 42 121, 22

10, 12

1 a1, b 2

y y=

2–x

y = 2x

(–3, 8)

(–2, 4) (–1, 2)

(3, 8)

(2, 4) (1, 2) x

1 a2, b 4

Notice that both graphs’ y-intercepts are 10, 12 (as shown to the right) and neither graph has an x-intercept. The x-axis is a horizontal asymptote for both graphs. The following box summarizes general characteristics of exponential functions.

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CHAPTER 5  Exponential and Logarithmic Functions

5.1.2 S K I L L

CHARACTERISTICS OF GRAPHS OF EXPONENTIAL FUNCTIONS

Graph exponential functions.

ƒ 1 x 2 5 bx,

b . 0,

b21

Domain: 12q, q2 ■■ Range: 10, q2 ■■ x-intercepts: none ■■ y-intercept: 10, 12 ■■ Horizontal asymptote: x-axis ■■

5.1.2 C O N C E P T U A L

Understand characteristics of exponential functions ­(including domain, range, asymptotes, ­intercepts, etc.).

1 The graph passes through 11, b2 and a21, b . b ■■ As x increases, ƒ1x2 increases if b + 1 and decreases if 0 * b * 1. ■■ The function ƒ is one-to-one. ■■

y b>1

0 2x + 1 x

Step 2:  Draw the line that corresponds to the resulting equation in Step 1. •  If the inequality is strict, , or ., use a dashed line. •  If the inequality is not strict, # or $, use a solid line.

y < 2x + 1

Step 3:  Test a point. •  Select a point in one half-plane and test to see whether it satisfies the inequality. If it does, then so do all the points in that region (half-plane). If not, then none of the points in that half-plane satisfy the inequality. •  Repeat this step for the other half-plane.

S TU DY TIP A dashed line means that the points that lie on the line are not included in the solution of the linear inequality.

[ CONCEPT CHECK ] TRUE OR FALSE:  A dashed line indicates that the points along the line satisfy the inequality.

▼

Step 4:  Shade the half-plane that satisfies the inequality. EXAMPLE 1  Graphing a Strict Linear Inequality in Two Variables

Graph the inequality 3x 1 y , 2. Solution: STEP 1  Change

the inequality sign to an equal sign.

ANSWER False

the line. Convert from standard form to slope–intercept form. Since the inequality , is a strict inequality, use a dashed line.

3x 1 y 5 2

STEP 2  Draw

y 5 23x 1 2 y

y = –3x + 2 x

STEP 3  Test points in each half-plane. Substitute 13, 02 into 3x 1 y , 2. 3132 1 0 , 2 The point 13, 02 does not satisfy the inequality. 9,2 Substitute 122, 02 into 3x 1 y , 2. 31222 1 0 , 2 The point 122, 02 does satisfy the inequality. 26 , 2 STEP 4  Shade

▼ ANSWER

the region containing the point 122, 02.

y

3x + y < 2 or y < –3x + 2 x

y

(–2, 0)

(3, 0)

–x + y > –1 or y>x–1 x

▼ Y O U R T U R N   Graph the inequality 2x 1 y . 21.

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9.4  Systems of Linear Inequalities in Two Variables 

905

EXAMPLE 2  G  raphing a Nonstrict Linear Inequality in Two Variables

Graph the inequality 2x 2 3y $ 6. Solution: STEP 1  Change

the inequality sign to an equal sign.

the line. Convert from standard form to slope–intercept form.

2x 2 3y 5 6

STEP 2  Draw



Since the inequality $ is not a strict inequality, use a solid line.

2 y5 x22 3 y

x y=

2 x 3

–2

STEP 3  Test points in each half-plane. Substitute 15, 02 into 2x 2 3y $ 6. 2152 2 3102 $ 6 The point 15, 02 satisfies the inequality. 10 $ 6 Substitute 10, 02 into 2x 2 3y $ 6. 2102 2 3102 $ 6 0$6 The point 10, 02 does not satisfy the inequality. STEP 4  Shade



the region containing the point 15, 02.

y

▼ ANSWER

x (5, 0)

y x – 2y ≤ 6 or 1 y≥ 2x–3

2x –3y ≥ 6 or y≤ 2x–2 3

x

▼ Y O U R T U R N   Graph the inequality x 2 2y # 6.

9.4.2  Systems of Linear Inequalities in Two Variables Systems of linear inequalities are similar to systems of linear equations. In systems of linear equations, we sought the points that satisfied all of the equations. The solution set of a system of inequalities contains the points that satisfy all of the inequalities. The graph of a system of inequalities can be obtained by simultaneously graphing each individual inequality and finding where the shaded regions intersect (or overlap), if at all.

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9.4.2 S K I L L

Graph a system of linear inequalities in two variables. 9.4.2 C O N C E P T U A L

Interpret an overlapped shaded region as a solution.

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CHAPTER 9  Systems of Linear Equations and Inequalities

EXAMPLE 3  Graphing a System of Two Linear Inequalities

Graph the system of inequalities: x 1 y $ 22 x 1 y #      2 Solution: STEP 1  Change

the inequality signs to equal signs.

STEP 2  Draw



x 1 y 5 22 x 1 y 5    2

the two lines.

Because the inequality signs are not strict, use solid lines.

y

y = –x + 2 x y = –x – 2

STEP 3  Test points for each inequality. x 1 y # 22 Substitute 124, 02 into x 1 y $ 22. 24 $ 22 The point 124, 02 does not satisfy the inequality. Substitute 10, 02 into x 1 y $ 22. 0 $ 22 The point 10, 02 does satisfy the inequality. x1y"2 Substitute 10, 02 into x 1 y # 2. 0#2 The point 10, 02 does satisfy the inequality. Substitute 14, 02 into x 1 y # 2. 4#2 The point 14, 02 does not satisfy the inequality.

x 1 y # 22, shade the region above that includes 10, 02. y

STEP 4  For



For x 1 y " 2, shade the region below that includes 10, 02. y

y ≥ –x – 2

x

x

y ≤ –x + 2

STEP 5  The overlapping region is the solution. Notice that the points 10, 02, 121, 12, and 11, 212 all lie in the shaded region and all three satisfy both inequalities.

y

y ≥ –x – 2 x

y ≤ –x + 2

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9.4  Systems of Linear Inequalities in Two Variables 

[ CONCEPT CHECK ]

EXAMPLE 4  G  raphing a System of Two Linear Inequalities with No Solution

Graph the system of inequalities:

TRUE OR FALSE  Only the points that lie in the shaded region satisfy both linear inequalities.

x 1 y # 22 x 1 y $    2

▼

Solution: STEP 1  Change

ANSWER True

the inequality signs to equal signs.



x 1 y 5 22 x 1 y 5    2

STEP 2  Draw



907

the two lines. Because the inequality signs are not strict, use solid lines.

y

y = –x + 2 x y = –x – 2

STEP 3  Test

points for each inequality. x 1 y " 22 Substitute 124, 02 into x 1 y # 22. 24 # 22 The point 124, 02 does satisfy the inequality. Substitute 10, 02 into x 1 y # 22. 0 # 22 The point 10, 02 does not satisfy the inequality. x1y#2 Substitute 10, 02 into x 1 y $ 2. 0$2 The point 10, 02 does not satisfy the inequality. Substitute 14, 02 into x 1 y $ 2. 4$2 The point 14, 02 does satisfy the inequality. x 1 y " 22, shade the region below that includes 124, 02. For x 1 y " 2, shade the region above that includes 14, 02. STEP 4  For

STEP 5  There is no overlapping region. Therefore, no points satisfy both inequalities.  We say there is no solution .

y

y ≥ –x + 2 x (–4, 0) y ≤ –x – 2

(4, 0)

▼ ANSWER a. No solution. b.

▼ Y O U R T U R N   Graph the solution to the system of inequalities.



a.  y . x 1 1

b.  y , x 1 1

y , x 2 1

y.x21

y

yx–1

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908 

CHAPTER 9  Systems of Linear Equations and Inequalities

Thus far we have addressed only systems of two linear inequalities. Systems with more than two inequalities are treated in a similar manner. The solution is the set of all points that satisfy all of the inequalities. When there are more than two linear inequalities, the solution may be a bounded region. We can algebraically determine where the lines intersect by ­setting the y-values equal to each other.

EXAMPLE 5  Graphing a System of Multiple Linear Inequalities

Solve the system of inequalities:

y #    x y $ 2x y ,    3

Solution: STEP 1  Change

the inequalities to equal signs.

y 5    x y 5 2x y 5    3

STEP 2  Draw

the three lines. To determine the points of intersection, set the y-values equal. Point where y 5 x and y 5 2x intersect: x 5 2x x50 Substitute x 5 0 into y 5 x. 10, 02 Point where y 5 2x and 2x 5 3 y y 5 3 intersect: x 5 23 y=3 123, 32 (3, 3) Point where y 5 3 and x53 y 5 x intersect: 13, 32 y=x

STEP 3  Test

points to determine the shaded half-planes corresponding to y " x, y # 2x, and y * 3.

(3, 3) x y = –x

y

x y ≥ –x

y 4

• CONCEPTUAL In Exercises 55 and 56, determine whether each statement is true or false. 55. A nonlinear inequality always represents a bounded region.

56. A system of inequalities always has a solution.

• CHALLENGE x 2 1 y 2 $ a2 , what x 2 1 y 2 # b2 restriction must be placed on the values of a and b for this system to have a solution? Assume a and b are real numbers.

57. For the system of nonlinear inequalities

58. Can x 2 1 y 2 , 21 ever have a real solution? What types

of numbers would x and/or y have to be to satisfy this inequality?

• TECHNOLOGY Use a graphing utility to graph the following inequalities. 59. x 2 1 y 2 2 2x 1 4y 1 4 $ 0 60. x 2 1 y 2 1 2x 2 2y 2 2 # 0 x

61. y $ e

62. y # ln x

64. y , 10x

y . log x    x . 0 65. x 2 2 4y 2 1 5x 2 6y 1 18 $ 0 66. x 2 2 2xy 1 4y 2 1 10x 2 25 # 0

63. y , ex

y . ln x    x . 0

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11.7  Rotation of Axes 

1057

11.7 ROTATION OF AXES SKILLS OBJECTIVES ■■ Transform general second-degree equations into recognizable equations of conics by analyzing rotations of axes. ■■ Determine the angle of rotation that will transform a general second-degree equation into a familiar equation of a conic section.

CONCEPTUAL OBJECTIVES ■■ Understand how the equation of a conic section is altered by rotation of axes. ■■ Understand how the angle of rotation formula is derived.

11.7.1  Rotation of Axes Formulas In Sections 11.1 through 11.4 we learned the general equations of parabolas, ellipses, and hyperbolas that were centered at any point in the Cartesian plane and whose vertices and foci were aligned either along or parallel to either the x-axis or the y-axis. We learned, for example, that the general equation of an ellipse centered at the origin is given by y2 x2 2 1 2 5 1 a b

11.7.1  S K I L L

Transform general second-degree equations into recognizable equations of conics by analyzing rotations of axes. 11.7.1  C O N C E P T U A L

where the major and minor axes are, respectively, either the x- or the y-axis, depending on whether a is greater than or less than b. Now let us look at an equation of a conic section whose graph is not aligned with the x- or y-axis: the equation 5x2 2 8xy 1 5y2 2 9 5 0.

Understand how the equation of a conic section is altered by rotation of axes.

y Y

–5

5 4 3 2

–3

X

45º

x

2 3 4 5 –2 –3 –4 –5

This graph can be thought of as an ellipse that started with the major axis along the x-axis and the minor axis along the y-axis and then was rotated counterclockwise 45°. A new XY-coordinate system can be introduced such that this system has the same origin, but the XY-coordinate system is rotated by a certain amount from the standard xy-coordinate system. In this example, the major axis of the ellipse lies along the new X-axis, and the minor axis lies along the new Y-axis. We will see that we can write the equation of this ellipse as X2 Y2 2 51 9 1 We will now develop the rotation of axes formulas, which allow us to transform the generalized second-degree equation in xy, that is, Ax 2 1 Bxy 1 Cy 2 1 Dx 1 Ey 1 F 5 0, into an equation in XY of a conic that is familiar to us.

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CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

Let the new XY-coordinate system be displaced from the xy-coordinate system by rotation through an angle u. Let P represent some point a distance r from the origin.

Y

y

P r

X

α x

θ

We can represent the point P as the point 1x, y2 or the point 1X, Y 2.

Y

y

Y

r

We define the angle a as the angle r makes with the X-axis and a 1 u as the angle r makes with the x-axis.

We can represent the point P in polar coordinates using the following relationships:

P y

α X

X

θ

x

x

x 5 r cos 1 a 1 u 2 y 5 r sin 1 a 1 u 2 X 5 r cos a Y 5 r sin a

WORDS MATH

Start with the x-term and write the cosine identity for a sum. Eliminate the parentheses and group r with the a-terms. Substitute according to the relationships X 5 r cos a and Y 5 r sin a. Start with the y-term and write the sine identity for a sum. Eliminate the parentheses and group r with the a-terms. Substitute according to the relationships X 5 r cos a and Y 5 r sin a.

x 5 rcos 1 a 1 u 2 5 r 1 cos a cos u 2 sin a sin u 2

x 5 1 rcos a 2 cos u 2 1 rsin a 2 sin u x 5 X cos u 2 Ysin u y 5 rsin 1 a 1 u 2 5 r 1 sin a cos u 1 cos a sin u 2

y 5 1 rsin a 2 cos u 1 1 rcos a 2 sin u

y 5 Ycos u 1 Xsin u

By treating the highlighted equations for x and y as a system of linear equations in X and Y, we can then solve for X and Y in terms of x and y. The results are summarized in the following box:

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11.7  Rotation of Axes 

1059

ROTATION OF AXES FORMULAS

Suppose that the x- and y-axes in the rectangular coordinate plane are rotated through an acute angle u to produce the X- and Y-axes. Then, the coordinates (x, y) and (X, Y) are related according to the following equations: x 5 X cos u 2 Y sin u   X 5 x cos u 1 y sin u or y 5 X sin u 1 Y cos u     Y 5 2x sin u 1 y cos u

EXAMPLE 1  Rotating the Axes

If the xy-coordinate axes are rotated 60°, find the XY-coordinates of the point 1 x, y 2 5 1 23, 4 2 . Solution:

Start with the rotation formulas.

X 5 x cos u 1 y sinu

  Y 5 2x sinu 1 y cosu Let x 5 23, y 5 4, and u 5 60°. X 5 23 cos 1 60° 2 1 4 sin 1 60° 2

•



•

  Y 5 2 1 23 2 sin 1 60° 2 1 4 cos 1 60° 2 Simplify. X 5 23 cos 1 60° 2 1 4 sin 1 60° 2 1 2

!3 2



!3 2

•

•

  Y 5 3 sin 1 60° 2 1 4 cos 1 60° 2 1 2

3 X 5 2 1 2!3 2



  Y5

3 !3 12 2

3 3 !3 The XY-coordinates are a2 1 2!3, 1 2b  . 2 2

▼

Y O U R T U R N   If the xy-coordinate axes are rotated 30°, find the XY-coordinates

of the point 1 x, y 2 5 1 3, 24 2 .

▼ ANSWER

a

3 !3 3 2 2, 2 2 2 !3b 2 2

EXAMPLE 2  Rotating an Ellipse

Show that the graph of the equation 5x 2 2 8xy 1 5y 2 2 9 5 0 is an ellipse aligning with coordinate axes that are rotated by 45°. Solution:



y 5 X sin 1 45° 2 1 Y cos 1 45° 2



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!2 2

•



!2 2

•

•

x 5 X cos u 2 Y sinu y 5 X sinu 1 Y cosu x 5 X cos 1 45° 2 2 Y sin 1 45° 2 •

Start with the rotation formulas. Let u 5 45°.

!2 2

!2 2

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1060 

CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

Simplify.

x5



y5

Substitute x 5 5c

!2 1X 2 Y2 2 !2 1X 1 Y2 2

!2 !2 1 X 2 Y 2 and y 5 1 X 1 Y 2 into 5x 2 2 8xy 1 5y 2 2 9 5 0. 2 2

2 2 !2 !2 !2 !2 1X 2 Y2 d 2 8 c 1X 2 Y2 d c 1X 1 Y2 d 1 5 c 1X 1 Y2 d 2 9 5 0 2 2 2 2

Simplify.

5 2 5 1 X 2 2XY 1 Y 2 2 2 4 1 X 2 2 Y 2 2 1 1 X 2 1 2XY 1 Y 2 2 2 9 5 0 2 2 5 2 5 5 5 X 2 5XY 1 Y 2 2 4X 2 1 4Y 2 1 X 2 1 5XY 1 Y 2 5 9 2 2 2 2

X 2 1 9Y 2 5 9

Combine like terms.

X2 Y2 1 51 9 1

Divide by 9. This (as discussed earlier) is an ellipse whose major axis is along the X-axis. The vertices are at the points 1 X, Y 2 5 1 6 3, 0 2 .

y Y

–5

5 4 3 2

–3

X

45º

x

2 3 4 5 –2 –3 –4 –5

11.7.2  The Angle of Rotation Necessary to Transform a General Second-Degree Equation into an Equation of a Conic 11.7.2  S K I L L

Determine the angle of rotation that will transform a general second-degree equation into a familiar equation of a conic section. 11.7.2  C O N C E P T U A L

Understand how the angle of rotation formula is derived.

In Section 11.1 we stated that the general second-degree equation Ax 2 1 Bxy 1 Cy 2 1 Dx 1 Ey 1 F 5 0 corresponds to a graph of a conic. Which type of conic it is depends on the value of the ­discriminant, B2 2 4AC. In Sections 11.2–11.4 we discussed graphs of parabolas, ellipses, and hyperbolas with vertices along either the axes or lines parallel (or perpendicular) to the axes. In all cases the value of B was taken to be zero. When the value of B is nonzero, the result is a conic with vertices along the new XY-axes (or, respectively, parallel and perpendicular to them), which are the original xy-axes rotated through an angle u. If given u, we can determine the rotation equations as illustrated in Example 2, but how do we find the angle u that ­represents the angle of rotation? To find the angle of rotation, let us start with a general second-degree polynomial equation: Ax 2 1 Bxy 1 Cy 2 1 Dx 1 Ey 1 F 5 0

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11.7  Rotation of Axes 

1061

We want to transform this equation into an equation in X and Y that does not contain an XY-term. Suppose we rotate our coordinates by an angle u and then use the rotation equations x 5 X cos u 2 Y sin u    y 5 X sin u 1 Y cos u in the general second-degree polynomial equation; then the result is A 1 X cos u 2 Y sin u 2 2 1 B 1 X cos u 2 Y sin u 2 1 X sin u 1 Y cos u 2

1 C 1 X cos u 1 Y sin u 2 2 1 D 1 X cos u 2 Y sin u 2 1 E 1 X sin u 1 Y cos u 2 1 F 5 0

If we expand these expressions and collect like terms, the result is an equation of the form aX 2 1 bXY 1 cY 2 1 dX 1 eY 1 f 5 0 where a 5 A cos2 u 1 B sin u cos u 1 C sin2 u b 5 B 1 cos2 u 2 sin2 u 2 1 2 1 C 2 A 2 sin u cos u c 5 A sin2 u 2 B sin u cos u 1 C cos2 u

d 5 D cos u 1 E sin u e 5 2D sin u 1 E cos u ƒ5F

WORDS MATH

We do not want this new equation to have an XY-term, so we set b 5 0.

B 1 cos2 u 2 sin2 u 2 1 2 1 C 2 A 2 sin u cos u 5 0 f

µ

B 1 cos2 u 2 sin2 u 2 1 1 C 2 A 2 2 sin u cos u 5 0

We can use the double-angle formulas to simplify.

cos 2u

sin 2u

Subtract the sine term.  B cos 2u 5 1 A 2 C 2 sin 2u

1 A 2 C 2 sin 2u B cos 2u 5 B sin 2u B sin 2u

Divide by B sin 2u. Simplify.

cot 2u 5

A2C B

ANGLE OF ROTATION FORMULA

To transform the equation of a conic Ax 2 1 Bxy 1 Cy 2 1 Dx 1 Ey 1 F 5 0 into an equation in X and Y without an XY-term, rotate the xy-axes by an acute angle u that satisfies the equation cot 2u 5

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A2C B

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CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

A2C can be solved exactly for B some values of u (Example 3) and will have to be approximated with a calculator for other values of u (Example 4). Notice that the trigonometric equation cot 2u 5

EXAMPLE 3  D  etermining the Angle of Rotation I: The Value of the Cotangent Function Is That of a Known (Special) Angle

Determine the angle of rotation necessary to transform the following equation into an equation in X and Y with no XY-term.

cot 2u 5

Let A 5 3, B 5 2 !3, and C 5 1.

cot 2u 5

Simplify.

•

Write the rotation formula.

A B

cot 2u 5

e

3x 2 1 2 !3 xy 1 1y 2 1 2x 2 2 !3y 5 0

Identify the A, B, and C parameters in the equation.

{

Solution:

3x 2 1 2 !3xy 1 y 2 1 2x 2 2!3y 5 0 C

A2C B 321 2 !3 1 !3

Apply the reciprocal identity. tan 2u 5 !3 From our knowledge of trigonometric exact values, we know that 2u 5 60° or u 5 30°  .

EXAMPLE 4  D  etermining the Angle of Rotation II: The Argument of the Cotangent Function Needs to Be Approximated with a Calculator

Determine the angle of rotation necessary to transform the following equation into an equation in X and Y with no XY-term. Round to the nearest tenth of a degree. 4x 2 1 2xy 2 6y 2 2 5x 1 y 2 2 5 0 Solution:

Young_AT_6160_ch11_pp1057-1077.indd 1062

{

e

4x 2 1 2 xy 2 6y 2 2 5x 1 y 2 2 5 0

{

Identify the A, B, and C parameters in the equation.

A B C

A2C B 4 2 1 26 2 5 2 55 1 5 5 0.2 5 5 tan 21 0.2

Write the rotation formula.

cot 2u 5

Let A 5 4, B 5 2, and C 5 26.

cot 2u

Simplify.

cot 2u

Apply the reciprocal identity.

tan 2u

Write the result as an inverse tangent function.

2u

With a calculator evaluate the right side of the equation.

2u < 11.31°

Solve for u and round to the nearest tenth of a degree.

u 5 5.7°

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1063

Special attention must be given when evaluating the inverse tangent function on a calculator, as the result is always in quadrant I or IV. If 2u turns out to be negative, then 180° must be added so that 2u is in quadrant II (as opposed to quadrant IV). Then u will be an acute angle lying in quadrant I. Recall that we stated (without proof) in Section 11.1 that we can identify a general equation of the form Ax 2 1 Bxy 1 Cy 2 1 Dx 1 Ey 1 F 5 0 as that of a particular conic depending on the discriminant.

Parabola

B2 2 4AC 5 0

Ellipse

B2 2 4AC , 0

Hyperbola

B2 2 4AC . 0

EXAMPLE 5  Graphing a Rotated Conic

For the equation x 2 1 2xy 1 y 2 2 !2x 2 3!2y 1 6 5 0: a. Determine which conic the equation represents. b. Find the rotation angle required to eliminate the XY-term in the new coordinate system. c. Transform the equation in x and y into an equation in X and Y. d. Graph the resulting conic. Solution (a):

A 5 1, B 5 2, C 5 1

Compute the discriminant.

{



{

{

1x 2 1 2 xy 1 1y 2 2 !2 x 2 3!2 y 1 6 5 0

Identify A, B, and C.

A B C

B2 2 4AC 5 22 2 4 1 1 2 1 1 2 5 0

Since the discriminant equals zero, the equation represents a parabola. Solution (b):

Write the rotation formula.

cot 2u 5

A2C B

Let A 5 1, B 5 2, and C 5 1.

cot 2u 5

121 2

Simplify.

cot 2u 5 0

Write the cotangent function in terms of the sine and cosine functions.

cos 2u 50 sin 2u

The numerator must equal zero. cos 2u 5 0 From our knowledge of trigonometric exact values, we know that 2u 5 90° or u 5 45°  .

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Solution (c):

!2 1X 2 Y2 Start with the equation x 5 X cos 1 45° 2 2 Y sin 1 45° 2 5 2 2 2 x 1 2xy 1 y 2 !2x 2 3 !2y 1 6 5 0, and use the rotation formulas !2 1X 1 Y2 y 5 X sin 1 45° 2 1 Y cos 1 45° 2 5 with u 5 45°. 2 Find x 2, xy, and y 2. Substitute the values for x, y, x 2, xy, and y 2 into the original equation.

x2 5 c

2 !2 1 1 X 2 Y 2 d 5 1 X 2 2 2XY 1 Y 2 2 2 2

y2 5 c

2 !2 1 1 X 1 Y 2 d 5 1 X 2 1 2XY 1 Y 2 2 2 2

xy 5 c

!2 1 !2 1X 2 Y 2 d c 1X 1 Y 2 d 5 1X 2 2 Y 22 2 2 2

x 2 1 2xy 1 y 2 2 !2x 2 3 !2y 1 6 5 0 1 2 1 1 X 2 2XY 1 Y 2 2 1 2 1 X 2 2 Y 2 2 2 2



1

Eliminate the parentheses and combine like terms. Divide by 2. Add Y. Complete the square on X.

1 2 !2 1 X 1 2XY 1 Y 2 2 2 !2 c 1X 2 Y 2 d 2 2

2 3!2 c

!2 1X 1 Y 2 d 1 6 5 0 2

Solution (d):

2X 2 2 4X 2 2Y 1 6 5 0 X 2 2 2X 2 Y 1 3 5 0 Y 5 AX 2 2 2XB 1 3 Y 5 1X 2 122 1 2

This is a parabola opening upward in the XY-coordinate system shifted to the right one unit and up two units.

y Y

2 –5

X

5 4

–3

45º

x

2 3 4 5 –2 –3 –4 –5

[ S E C T I O N 11 . 7 ]     S U M M A R Y In this section, we found that the graph of the general seconddegree equation

The following are the rotation formulas relating the xy-coordinate system to a rotated coordinate system with axes X and Y

Ax 2 1 Bxy 1 Cy 2 1 Dx 1 Ey 1 F 5 0

x 5 X cos u 2 Y sin u y 5 X sin u 1 Y cos u

can represent conics in a system of rotated axes.

where the rotation angle u is found from the equation cot 2u 5

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A2C B

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1065

[ S E C T I O N 11 . 7 ]   E X E R C I S E S • SKILLS In Exercises 1–8, the coordinates of a point in the xy-coordinate system are given. Assuming that the XY-axes are found by rotating the xy-axes by an angle u, find the corresponding coordinates for the point in the XY-system. 2. 1 5, 1 2 , u 5 60° 3. 1 23, 2 2 , u 5 30° 4. 1 24, 6 2 , u 5 45° 1. 1 2, 4 2 , u 5 45° 5. 1 21, 23 2 , u 5 60°

6. 1 4, 24 2 , u 5 45°

7. 1 0, 3 2 , u 5 60°

8. 1 22, 0 2 , u 5 30°

In Exercises 9–24, (a) identify the type of conic from the discriminant, (b) transform the equation in x and y into an equation in X and Y (without an XY-term) by rotating the x- and y-axes by an angle u to arrive at the new X- and Y-axes, and (c) graph the resulting equation (showing both sets of axes). 9. xy 2 1 5 0, u 5 45° 2

10. xy 2 4 5 0, u 5 45°

2

11. x 1 2xy 1 y 1 !2x 2 !2y 2 1 5 0, u 5 45°

12. 2x 2 2 4xy 1 2y 2 2 !2x 1 1 5 0, u 5 45°

15. 7x 2 2 2!3xy 1 5y 2 2 8 5 0, u 5 60°

14. x 2 2 !3xy 2 3 5 0, u 5 60°

17. 3x 2 1 2!3xy 1 y 2 1 2x 2 2!3y 2 2 5 0, u 5 30°

16. 4x 2 1 !3xy 1 3y 2 2 45 5 0, u 5 30°

18. x 2 1 2!3xy 1 3y 2 2 2!3x 1 2y 2 4 5 0, u 5 60°

13. y 2 2 !3xy 1 3 5 0, u 5 30°

19. 7x 2 1 4!3xy 1 3y 2 2 9 5 0, u 5

p 6

21. 7x 2 2 10!3xy 2 3y 2 1 24 5 0, u 5

20. 37x 2 1 42!3xy 1 79y 2 2 400 5 0, u 5

p 3

22. 9x 2 1 14!3xy 2 5y 2 1 48 5 0, u 5

p 3

p 6

p p 24. x 2 1 2xy 1 y 2 1 3!2x 1 !2y 5 0, u 5 4 4 In Exercises 25–38, determine the angle of rotation necessary to transform the equation in x and y into an equation in X and Y with no XY-term. 23. x 2 2 2xy 1 y 2 2 !2x 2 !2y 2 8 5 0, u 5 25. x 2 1 4xy 1 y 2 2 4 5 0

26. 3x 2 1 5xy 1 3y 2 2 2 5 0

27. 2x 2 1 !3xy 1 3y 2 2 1 5 0

28. 4x 2 1 !3xy 1 3y 2 2 1 5 0

31. !2x 2 1 xy 1 !2y 2 2 1 5 0

32. x 2 1 10xy 1 y 2 1 2 5 0

33. 12!3x 2 1 4xy 1 8!3y 2 2 1 5 0

34. 4x 2 1 2xy 1 2y 2 2 7 5 0

35. 5x 2 1 6xy 1 4y 2 2 1 5 0

36. x 2 1 2xy 1 12y 2 1 3 5 0

37. 3x 2 1 10xy 1 5y 2 2 1 5 0

38. 10x 2 1 3xy 1 2y 2 1 3 5 0

29. 2x 2 1 !3xy 1 y 2 2 5 5 0

30. 2!3x 2 1 xy 1 3!3y 2 1 1 5 0

In Exercises 39–48, graph the second-degree equation. (Hint: Transform the equation into an equation that contains no xy-term.) 39. 21x 2 1 10!3xy 1 31y 2 2 144 5 0

40. 5x 2 1 6xy 1 5y 2 2 8 5 0

41. 8x 2 2 20xy 1 8y 2 1 18 5 0

42. 3y 2 2 26!3xy 2 23x 2 2 144 5 0

43. 3x 2 1 2!3xy 1 y 2 1 2x 2 2!3y 2 12 5 0

44. 3x 2 2 2!3xy 1 y 2 2 2x 2 2!3y 2 4 5 0

45. 37x 2 2 42!3xy 1 79y 2 2 400 5 0

46. 71x 2 2 58!3xy 1 13y 2 1 400 5 0

47. x 2 1 2xy 1 y 2 1 5!2x 1 3!2y 5 0

48. 7x 2 2 4!3xy 1 3y 2 2 9 5 0

• CONCEPTUAL In Exercises 49–52, determine whether each statement is true or false. 49. The graph of the equation x 2 1 kxy 1 9y 2 5 5, where k is 51. The reciprocal function is a rotated hyperbola. any positive constant less than 6, is an ellipse. 52. The equation !x 1 !y 5 3 can be transformed into the 50. The graph of the equation x 2 1 kxy 1 9y 2 5 5, where k is equation X 2 1 Y 2 5 9. any constant greater than 6, is a parabola.

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• CHALLENGE 53. Determine the equation in X and Y that corresponds to 2

2

y x 1 2 5 1 when the axes are rotated through 2 a b a. 90°    b. 180° 54. Determine the equation in X and Y that corresponds to









55. Identify the conic section with equation y 2 1 ax 2 5 x for

various values of a. 56. Identify the conic section with equation x 2 2 ay 2 5 y for various values of a.

y2 x2 2 2 5 1 when the axes are rotated through 2 a b a. 90°    b. 180°

• TECHNOLOGY For Exercises 57–62, refer to the following: To use a TI-83 or TI-83 Plus (function-driven software or graphing utility) to graph a general second-degree equation, you need to solve for y. Let us consider a general second-degree equation Ax 2 1 Bxy 1 Cy 2 1 Dx 1 Ey 1 F 5 0. Group y 2 terms together, y terms together, and the remaining terms together. Ax 2 1 Bxy 1 Cy 2 1 Dx 1 Ey 1 F 5 0 Cy 2 1 1 Bxy 1 Ey 2 1 1 Ax 2 1 Dx 1 F 2 5 0

Factor out the common y in the first set of parentheses.

Cy 2 1 y 1 Bx 1 E 2 1 1 Ax 2 1 Dx 1 F 2 5 0

Now this is a quadratic equation in y: ay 2 1 by 1 c 5 0. Use the quadratic formula to solve for y.

Cy 2 1 y 1 Bx 1 E 2 1 1 Ax 2 1 Dx 1 F 2 5 0 a 5 C, b 5 Bx 1 E, c 5 Ax 2 1 Dx 1 F



y5



y5



y5

2 1 Bx 1 E 2 6 " 1 Bx 1 E 2 2 2 4 1 C 2 1 Ax 2 1 Dx 1 F 2 2b 6 "b2 2 4ac    y 5 2a 21C2 2 1 Bx 1 E 2 6 "B2x 2 1 2BEx 1 E 2 2 4ACx 2 2 4CDx 2 4CF 2C

2 1 Bx 1 E 2 6 " 1 B2 2 4AC 2 x 2 1 1 2BE 2 4CD 2 x 1 1 E 2 2 4CF 2 2C

Case I: B2 2 4AC 5 0 S The second-degree equation Ax 2 1 Bxy 1 Cy 2 1 Dx 1 Ey 1 F 5 0 is a parabola. y5

2 1 Bx 1 E 2 6 " 1 2BE 2 4CD 2 x 1 1 E 2 2 4CF 2 2C

Case II: B2 2 4AC , 0 S The second-degree equation Ax 2 1 Bxy 1 Cy 2 1 Dx 1 Ey 1 F 5 0 is an ellipse. y5

2 1 Bx 1 E 2 6 " 1 B2 2 4AC 2 x 2 1 1 2BE 2 4CD 2 x 1 1 E 2 2 4CF 2 2C

Case III: B2 2 4AC . 0 S The second-degree equation Ax 2 1 Bxy 1 Cy 2 1 Dx 1 Ey 1 F 5 0 is a hyperbola. y5

2 1 Bx 1 E 2 6 " 1 B2 2 4AC 2 x 2 1 1 2BE 2 4CD 2 x 1 1 E 2 2 4CF 2 2C

57. Use a graphing utility to explore the second-degree equation

58. Use a graphing utility to explore the second-degree equation







3x 2 1 2!3xy 1 y 2 1 Dx 1 Ey 1 F 5 0 for the following values of D, E, and F: a. D 5 1, E 5 3, F 5 2 b. D 5 21, E 5 23, F 5 2 Show the angle of rotation to the nearest degree. Explain the differences.

Young_AT_6160_ch11_pp1057-1077.indd 1066



x 2 1 3xy 1 3y 2 1 Dx 1 Ey 1 F 5 0 for the following values of D, E, and F: a. D 5 2, E 5 6, F 5 21 b. D 5 6, E 5 2, F 5 21 Show the angle of rotation to the nearest degree. Explain the differences.

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11.8  Polar Equations of Conics 

59. Use a graphing utility to explore the second-degree equation





60.





2x 2 1 3xy 1 y 2 1 Dx 1 Ey 1 F 5 0 for the following values of D, E, and F: a. D 5 2, E 5 1, F 5 22 b. D 5 2, E 5 1, F 5 2 Show the angle of rotation to the nearest degree. Explain the differences. Use a graphing utility to explore the second-degree equation 2!3x 2 1 xy 1 !3y 2 1 Dx 1 Ey 1 F 5 0 for the following values of D, E, and F: a. D 5 2, E 5 1, F 5 21 b. D 5 2, E 5 6, F 5 21 Show the angle of rotation to the nearest degree. Explain the differences.

1067

61. Use a graphing utility to explore the second-degree equation





62.





Ax 2 1 Bxy 1 Cy 2 1 2x 1 y 2 1 5 0 for the following values of A, B, and C: a. A 5 4, B 5 24, C 5 1 b. A 5 4, B 5 4, C 5 21 c. A 5 1, B 5 24, C 5 4 Show the angle of rotation to the nearest degree. Explain the differences. Use a graphing utility to explore the second-degree equation Ax 2 1 Bxy 1 Cy 2 1 3x 1 5y 2 2 5 0 for the following values of A, B, and C: a. A 5 1, B 5 24, C 5 4 b. A 5 1, B 5 4, D 5 24 Show the angle of rotation to the nearest degree. Explain the differences.

11.8 POLAR EQUATIONS OF CONICS SKILLS OBJECTIVE ■■

Express equations of conics in polar form and graph.

CONCEPTUAL OBJECTIVE ■■

Define all conics in terms of a focus and a directrix.

11.8.1  Equations of Conics in Polar Coordinates In Section 11.1 we discussed parabolas, ellipses, and hyperbolas in terms of geometric definitions. Then in Sections 11.2–11.4 we examined the rectangular equations of these conics. The equations for the conics are simpler when their centers are at the origin than when they are not (when conics are shifted). In Section 8.8, we discussed polar coordinates and graphing of polar equations. In this section, we develop a more unified definition of the three conics in terms of a single focus and a directrix. You will see in this section that if the focus is located at the origin, then equations of conics are simpler when written in polar coordinates.

11.8.1  S K I L L

Express equations of conics in polar form and graph. 11.8.1  C O N C E P T U A L

Define all conics in terms of a focus and a directrix.

Alternative Definition of Conics Recall that when we work with rectangular coordinates we define a parabola (Sections 11.1 and 11.2) in terms of a fixed point (focus) and a line (directrix), whereas we define an ellipse and hyperbola (Sections 11.1, 11.3, and 11.4) in terms of two fixed points (the foci). However, it is possible to define all three conics in terms of a single focus and a directrix. The following alternative representation of conics depends on a parameter called eccentricity. ALTERNATIVE DESCRIPTION OF CONICS

Let D be a fixed line (the directrix), F be a fixed point (a focus) not on D, and e be a fixed positive number (eccentricity). The set of all points P such that the ratio of the distance from P to F to the distance from P to D equals the constant e defines a conic section. d 1 P, F 2 5e d 1 P, D 2 If e 5 1, the conic is a parabola. If e , 1, the conic is an ellipse. ■■ If e . 1, the conic is a hyperbola. ■■ ■■

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When e 5 1, the result is a parabola, described by the same definition we used previously in Section 11.1. When e 2 1, the result is either an ellipse or a hyperbola. The major axis of an ellipse passes through the focus and is perpendicular to the directrix. The transverse axis of a hyperbola also passes through the focus and is perpendicular to the directrix. If we let c represent the distance from the focus to the center and a represent the distance from the vertex to the center, then eccentricity is given by e5

c a

In polar coordinates, if we locate the focus of a conic at the pole and the directrix is either perpendicular or parallel to the polar axis, then we have four possible scenarios: The directrix is perpendicular to the polar axis and p units to the right of the pole. ■■ The directrix is perpendicular to the polar axis and p units to the left of the pole. ■■ The directrix is parallel to the polar axis and p units above the pole. ■■ The directrix is parallel to the polar axis and p units below the pole. ■■

Let us take the case in which the directrix is perpendicular to the polar axis and p units to the right of the pole. y

D P x

F

Polar axis

In polar coordinates 1 r, u 2 , we see that the distance from the focus to a point P is equal to r, that is, d 1 P, F 2 5 r, and the distance from P to the closest point on the directrix is d 1 P, D 2 5 p 2 r cos u. y

D

P d(P, D) r F θ r cos θ p

x Polar axis

x=p

WORDS MATH

Substitute d 1 P, F 2 5 r and d 1 P, D 2 5 p 2 r cos u d 1 P, F 2 5 e. into the formula for eccentricity, d 1 P, D 2 Multiply the result by p 2 r cos u. Eliminate the parentheses. Add er cos u. Factor out the common r. Divide by 1 1 e cos u.

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r 5e p 2 r cos u r 5 e 1 p 2 r cos u 2 r 5 ep 2 er cos u r 1 er cos u 5 ep 1 r 1 1 e cos u 2 5 ep r5

ep 1 1 e cos u

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We need not derive the other three cases here, but note that if the directrix is perpendicular to the polar axis and p units to the left of the pole, the resulting polar equation is r5

ep 1 2 e cos u

If the directrix is parallel to the polar axis, the directrix is either above 1 y 5 p 2 or below 1 y 5 2p 2 the polar axis and we get the sine function instead of the cosine function, as summarized in the following box: POLAR EQUATIONS OF CONICS

The following polar equations represent conics with one focus at the origin and with eccentricity e. It is assumed that the positive x-axis represents the polar axis. EQUATION

DESCRIPTION

r5

ep 1 1 e cos u

The directrix is vertical and p units to the right of the pole.

r5

ep 1 2 e cos u

The directrix is vertical and p units to the left of the pole.

r5

ep 1 1 e sin u

The directrix is horizontal and p units above the pole.

r5

ep 1 2 e sin u

The directrix is horizontal and p units below the pole.

ECCENTRICITY

THE CONIC IS A ___________

THE _____ IS PERPENDICULAR TO THE DIRECTRIX

e51

Parabola

Axis of symmetry

e,1

Ellipse

Major axis

e.1

Hyperbola

Transverse axis

EXAMPLE 1  Finding the Polar Equation of a Conic

Find a polar equation for a parabola that has its focus at the origin and whose directrix is the line y 5 3. Solution:

The directrix is horizontal and above the pole.

r5

ep 1 1 e sin u

A parabola has eccentricity e 5 1 and p 5 3.

r5

3 1 1 sin u

▼ Y O U R T U R N   Find a polar equation for a parabola that has its focus at the origin

and whose directrix is the line x 5 23.

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▼ ANSWER

r5

3 12 6 cos u

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CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

EXAMPLE 2  Identifying a Conic from Its Equation

Identify the type of conic represented by the equation

10 . 3 1 2 cos u

Solution:

To identify the type of conic, we need to rewrite the equation in the form:

Divide the numerator and denominator by 3.

r5

ep 1 6 e cos u

10 3 2 a1 1 cos ub 3

10 3 5 2 a1 1 cos ub 3

Identify e in the denominator.

{

e

e

{

p

{

2 5⋅ 3 5 2 a1 1 cos ub 3

The numerator is equal to ep.

{

e

2 , 1, the conic is an ellipse  . The directrix is x 5 5, so the major axis 3 is along the x-axis (perpendicular to the directrix).

Since e 5

▼ ANSWER

hyperbola, e 5 5, with transverse axis along the y-axis

▼ Y O U R T U R N   Identify the type of conic represented by the equation

r5

10 2 2 10 sin u

10 is an ellipse with 3 1 2 cos u major axis along the x-axis. We will graph this ellipse in Example 3. In Example 2 we found that the polar equation r 5

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EXAMPLE 3  Graphing a Conic from Its Equation

The graph of the polar equation r 5

10 is an ellipse. 3 1 2 cos u

a.  Find the vertices.                                                b.  Find the center of the ellipse. c.  Find the lengths of the major and minor axes.   d.  Graph the ellipse. Solution (a):

From Example 2 we see that e 5 23, which corresponds to an ellipse, and x 5 5 is the directrix. The major axis is perpendicular to the directrix. Therefore, the major axis lies along the polar axis. To find the vertices (which lie along the major axis), let u 5 0 and u 5 p. u 5 0:

r5

10 10 5 52 3 1 2 cos u 5

u 5 p:

r5

10 10 5 5 10 3 1 2 cos p 1

The vertices are the points V1 5 1 2, 0 2 and V2 5 1 10, p 2  . Solution (b):

The vertices in rectangular coordinates are V1 5 1 2, 0 2 and V2 5 1 210, 0 2 .

The midpoint (in rectangular coordinates) between the two vertices is the point 1 24, 0 2 , which corresponds to the point 1 4, p 2 in polar coordinates. Solution (c):

The length of the major axis, 2a, 2a 5 12 is the distance between the vertices. The length a 5 6 corresponds to the distance from the center to a vertex. Apply the formula e 5 and e 5 23 to find c.

c with a 5 6 a

2 c 5 ae 5 6 5 4 3

b2 5 62 2 42 5 20 Let a 5 6 and c 5 4 in b2 5 a2 2 c2. Solve for b. b 5 2 !5 The length of the minor axis is 2b 5 4 !5 . Solution (d):

Graph the ellipse. 2√5 V2 –10

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–8

–6

y 5 4 3 2 1

–4

–2

–1 –2 –3 –4 –5

V1

x

1 2 Polar axis

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EXAMPLE 4  Identifying and Graphing a Conic from Its Equation

2 . 2 1 3 sin u

Solution:

e

e

p

The conic is a hyperbola since e 5

3 2

2 3 a ba b 3 2 2 r5 5 2 1 3 sin u 3 1 1 a b sin u 2 e

Rewrite the equation in ep the form r 5 . 1 1 e sin u

e

Identify and graph the conic defined by the equation r 5

. 1.

e

The directrix is horizontal and 23 unit above the pole (origin). To find the vertices, let u 5 u5

p : 2

u5

3p : 2

p 3p and u 5 . 2 2 2

2 5 p 5 2 1 3 sin a b 2 2 2 r5 5 5 22 3p 21 2 1 3 sin a b 2 r5

3p 2 p The vertices in polar coordinates are a , b and a22, b. 5 2 2

The vertices in rectangular coordinates are V1 5 A0, 25 B and V2 5 1 0, 2 2 . The center is the midpoint between the vertices: A0, 65 B. The distance from the center to a focus is c 5 65.

Apply the formula e 5 e 5 32 to find a.

c with c 5 65 and a

Let a 5 45 and c 5 65 in b2 5 c2 2 a2. Solve for b.

6 5 c 4 a5 5 5 e 3 5 2 6 2 4 2 20 b2 5 a b 2 a b 5 5 5 25 b5

2 !5 5

The asymptotes are given by a y 5 6 1 x 2 h 2 1 k, b

4 2 !5 6 , and 1 h, k 2 5 a0, b . where a 5 , b 5 5 5 5 2 6 y56 x1 5 !5

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y 4 b= 3

2√5 5

2 1 x –2

–1

1 –1

2

Polar axis

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1073

It is important to note that although we relate specific points (vertices, foci, etc.) to rectangular coordinates, another approach to finding a rough sketch is to simply pointplot the equation in polar coordinates.

EXAMPLE 5  Graphing a Conic by Point-Plotting in Polar Coordinates

4 . 1 2 sin u

Sketch a graph of the conic r 5 Solution: STEP 1  The

conic is a parabola because the equation is in the form r5



142 112 1 2 1 1 2 sin u

Make a table with key values for u and r.

u

r5

4 1 2 sin u

0

r5

4 4 5 54 1 2 sin 0 1

p 2

r5

p

r5

3p 2

r5

2p

r5

4 p 1 2 sin 2

(r, u)

5

1 4, 0 2

4 4 5 121 0

undefined

4 4 5 54 1 2 sin p 1 4 3p 1 2 sin 2

5

1 4, p 2

4 4 5 52 2 1 2 1 21 2

a2,

4 4 5 54 1 2 sin 2p 1

the points on a polar graph and connect them with a smooth parabolic curve.

3p b 2

1 4, 2p 2

STEP 2  Plot

5π 6 11π 12

3π 4

2π 3

7π 12

π 2 5

5π 12

π 3

π 4

3

π 12

1

π

0

13π 12 7π 6

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π 6

5π 4

4π 3

17π 12

3π 2

19π 12

5π 3

7π 4

23π 12 11π 6

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CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

[ S E C T I O N 11 . 8 ]     S U M M A R Y In this section, we found that we could graph polar equations of conics by identifying a single focus and the directrix. There are four ­possible equations in terms of eccentricity e: EQUATION

r5

DESCRIPTION

ep 1 1 e cos u

The directrix is vertical and p units to the right of the pole.

ep 1 2 e cos u ep r5 1 1 e sin u ep r5 1 2 e sin u

The directrix is vertical and p units to the left of the pole.

r5

The directrix is horizontal and p units above the pole. The directrix is horizontal and p units below the pole.

[ S E C T I O N 11 . 8 ]   E X E R C I S E S • SKILLS In Exercises 1–14, find the polar equation that represents the conic described (assume that a focus is at the origin).   Conic

Eccentricity

Directrix

  Conic

Eccentricity

1. Ellipse

e5

y 5 25

2. Ellipse

e5

3. Hyperbola

e52

y54

4. Hyperbola

e53

y 5 22

5. Parabola

e51

x51

6. Parabola

e51

x 5 21

x52

8. Ellipse

e5

2 3

x 5 24

1 2

Directrix y53

1 3

7. Ellipse

e5

3 4

9. Hyperbola

e5

4 3

x 5 23

10. Hyperbola

e5

3 2

x55

11. Parabola

e51

y 5 23

12. Parabola

e51

y54

13. Ellipse

e5

y56

14. Hyperbola

e5

8 5

y55

3 5

In Exercises 15–26, identify the conic (parabola, ellipse, or hyperbola) that each polar equation represents. 15. r 5

4 1 1 cos u

16. r 5

3 2 2 3 sin u

17. r 5

2 3 1 2 sin u

18. r 5

3 2 2 2 cos u

19. r 5

2 4 1 8 cos u

20. r 5

1 4 2 cos u

21. r 5

7 3 1 cos u

22. r 5

4 5 1 6 sin u

23. r 5

40 5 1 5 sin u

24. r 5

5 5 2 4 sin u

25. r 5

1 1 2 6 cos u

26. r 5

5 3 2 3 sin u

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11.8  Polar Equations of Conics 

1075

In Exercises 27–40, for the given polar equations: (a) identify the conic as a parabola, an ellipse, or a hyperbola; (b) find the eccentricity and vertex (or vertices); and (c) graph. 27. r 5

2 1 1 sin u

28. r 5

4 1 2 cos u

29. r 5

4 1 2 2 sin u

30. r 5

3 3 1 8cos u

31. r 5

2 2 1 sin u

32. r 5

1 3 2 sin u

33. r 5

1 2 2 2sin u

34. r 5

1 1 2 2sin u

35. r 5

4 3 1 cos u

36. r 5

2 5 1 4 sin u

37. r 5

6 2 1 3sin u

38. r 5

6 1 1 cos u

39. r 5

2 5 1 5 cos u

40. r 5

10 6 2 3cos u

• A P P L I C AT I O N S For Exercises 41 and 42, refer to the following: Planets travel in elliptical orbits around a single focus, the Sun. Pluto (orange), the dwarf planet furthest from the Sun, has a ­pronounced elliptical orbit, whereas Earth (royal blue) has an almost circular orbit. The polar equation of a planet’s orbit can be expressed as r5

41.  Planetary Orbits. Mercury’s orbit is summarized in the

picture below. Find the eccentricity of Mercury’s orbit. Find the polar equation that governs Mercury’s orbit. Mercury’s Orbit

a 1 1 2 e2 2 1 1 2 e cosu 2 Perihelion 46,000,000 km

Aphelion 69,820,000 km

Mercury has a very elliptical orbit, which is highly inclined with respect to the plane of the ecliptic. The Sun and Mercury are not to scale in this drawing.

42.  Planetary Orbits. Earth’s orbit is summarized in the picture

below. Find the eccentricity of Earth’s orbit. Find the polar equation that governs Earth’s orbit. Earth's Orbit

where e is the eccentricity and 2a is the length of the major axis. It can also be shown that the perihelion distance (minimum distance from the Sun to a planet) and the aphelion distance (maximum distance from the Sun to the planet) can be represented by r 5 a 1 1 2 e 2 and r 5 a 1 1 1 e 2 , respectively.

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Perihelion 147,100,000 km

Aphelion 152,600,000 km

The Sun and Earth are not to scale in this drawing.

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CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

For Exercises 43 and 44, refer to the following:

43.  Asteroids. The asteroid 433 or Eros is the second largest

Asteroids, meteors, and comets all orbit the Sun in elliptical patterns and often cross paths with Earth’s orbit, making life a little tense now and again. Asteroids are large rocks (bodies under 1000 kilometers across), meteors range from sand particles to rocks, and comets are masses of debris. A few asteroids have orbits that cross the Earth’s orbits—called Apollos or Earth-crossing asteroids. In recent years, asteroids have passed within 100,000 kilometers of Earth!

near-Earth asteroid. The semimajor axis of its orbit is 150 million kilometers, and the eccentricity is 0.223. Find the polar equation of Eros’s orbit. 44.  Asteroids. The asteroid Toutatis is the largest near-Earth asteroid. The semimajor axis of its orbit is 350 million kilometers, and the eccentricity is 0.634. On September 29, 2004, it missed Earth by 961,000 miles. Find the polar equation of Toutatis’s orbit.

• CONCEPTUAL 45. When 0 , e , 1, the conic is an ellipse. Does the conic

become more elongated or elliptical as e approaches 1 or as e approaches 0? ep 46. Show that r 5 is the polar equation of a conic 1 2 e sin u with a horizontal directrix that is p units below the pole.

47. Convert from rectangular to polar coordinates to show that

y2 x2 the equation of a hyperbola, 2 2 2 5 1, in polar form is a b b2 r2 5 2 . 1 2 e2 cos2 u 48. Convert from rectangular to polar coordinates to show that y2 x2 the equation of an ellipse, 2 1 2 5 1, in polar form is a b b2 r2 5 . 1 2 e2 cos2 u

• CHALLENGE 49. Find the major diameter of the ellipse with polar equation

ep r5 in terms of e and p. 1 1 e cos u 50. Find the minor diameter of the ellipse with polar equation ep r5 in terms of e and p. 1 1 e cos u

51. Find the center of the ellipse with polar equation

ep in terms of e and p. 1 1 e cos u 52. Find the length of the latus rectum of the parabola with polar p equation r 5 . Assume that the focus is at the origin. 1 1 cos u

r5

• TECHNOLOGY ep and 1 1 e cos u with eccentricity e 5 1. With a graphing

53. Let us consider the polar equations r 5

ep 1 2 e cos u utility, explore the equations with p 5 1, 2, and 6. Describe the behavior of the graphs as p S q and also the difference between the two equations. ep 54. Let us consider the polar equations r 5 and 1 1 e sin u ep r5 with eccentricity e 5 1. With a graphing 1 2 e sin u utility, explore the equations with p 5 1, 2, and 6. Describe the behavior of the graphs as p S q and also the difference between the two equations. ep 55. Let us consider the polar equations r 5 and 1 1 e cos u ep r5 with p 5 1. With a graphing utility, explore 1 2 e cos u the equations with e 5 1.5, 3, and 6. Describe the behavior of the graphs as e S q and also the difference between the two equations.

r5

Young_AT_6160_ch11_pp1057-1077.indd 1076

ep and 1 1 e sin u with p 5 1. With a graphing utility, explore

56. Let us consider the polar equations r 5

ep 1 2 e sin u the equations with e 5 1.5, 3, and 6. Describe the behavior of the graphs as e S q and also the difference between the two equations. ep 57. Let us consider the polar equations r 5 and 1 1 e cos u ep r5 with p 5 1. With a graphing utility, explore 1 2 e cos u the equations with e 5 0.001, 0.5, 0.9, and 0.99. Describe the behavior of the graphs as e S 1 and also the difference between the two equations. Be sure to set the window parameters properly. ep 58. Let us consider the polar equations r 5 and 1 1 e sin u ep r5 with p 5 1. With a graphing utility, explore 1 2 e sin u the equations with e 5 0.001, 0.5, 0.9, and 0.99. Describe the behavior of the graphs as e S 1 and also the difference between the two equations. Be sure to set the window parameters properly.

r5

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11.8  Polar Equations of Conics 

5 . 5 1 2 sin u Explain why the graphing utility gives the following graphs with the specified window parameters: p a.  3 22, 2 4 by 3 22, 2 4 with u step 5 2 59. Let us consider the polar equation r 5

b.  3 22, 2 4 by 3 22, 2 4 with u step 5

p 3

2 . Explain 1 1 cos u why a graphing utility gives the following graphs with the specified window parameters: p a.  3 22, 2 4 by 3 24, 4 4 with u step 5 2 60. Let us consider the polar equation r 5

b.  3 22, 2 4 by 3 24, 4 4 with u step 5

Young_AT_6160_ch11_pp1057-1077.indd 1077

p 3

1077

6 . 1 1 3 sin u Explain why a graphing utility gives the following graphs with the specified window parameters: p a.  3 28, 8 4 by 3 22, 4 4 with u step 5 2 61. Let us consider the polar equation r 5

b.  3 24, 8 4 by 3 22, 6 4 with u step 5 0.4p

2 . Explain 1 2 sin u why a graphing utility gives the following graphs with the specified window parameters: p a.  3 24, 4 4 by 3 22, 4 4 with u step 5 3 62. Let us consider the polar equation r 5

b.  3 24, 4 4 by 3 22, 6 4 with u step 5 0.8p

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CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

11.9 PARAMETRIC EQUATIONS AND GRAPHS SKILLS OBJECTIVES ■■ Graph parametric equations. ■■ Graph cycloids and the curves representing projectile motion.

CONCEPTUAL OBJECTIVES ■■ Understand that the result of increasing the value of the parameter reveals the orientation of a curve or the direction of motion along it. ■■ Use time as a parameter in parametric equations such as cycloids and projectile motion.

11.9.1  Parametric Equations of a Curve 11.9.1  S K I L L

Graph parametric equations. 11.9.1  C O N C E P T U A L

Understand that the result of increasing the value of the parameter reveals the orientation of a curve or the direction of motion along it.

Thus far we have talked about graphs in planes. For example, the equation x 2 1 y 2 5 1 when graphed in a plane is the unit circle. Similarly, the function ƒ 1 x 2 5 sin x when graphed in a plane is a sinusoidal curve. Now, we consider the path along a curve. For example, if a car is being driven on a circular racetrack, we want to see the movement along the circle. We can determine where (position) along the circle the car is at some time t using parametric equations. Before we define parametric equations in general, let us start with a simple example. Let x 5 cos t and y 5 sin t and t $ 0. We then can make a table of some corresponding values. t SECONDS

x 5 cos t

y 5 sin t

(x, y )

0

x 5 cos 0 5 1

y 5 sin 0 5 0

p 2

p x 5 cos a b 5 0 2 x 5 cos p 5 21

p y 5 sin a b 5 1 2

11, 02

3p x 5 cos a b 5 0 2

3p y 5 sin a b 5 21 2

p 3p 2 2p

10, 212 11, 02

y 5 sin 1 2p2 5 0

If we plot these points and note the correspondence to time ( by converting all numbers to decimals), we will be tracing a path counterclockwise along the unit circle.

t = 1.57 (0, 1)

(–1, 0)

(1, 0)

t = 3.14

t = 0 t = 6.28

(0, –1) t = 4.71

121, 02

y 5 sin p 5 0

x 5 cos 1 2p 2 5 1

y

10, 12

x

TIME (SECONDS)

t50

t 5 1.57

t 5 3.14

t 5 4.71

POSITION

11, 02

10, 12

121, 02

10, 212

Notice that at time t 5 6.28 seconds we are back to the point 11, 02. We can see that the path represents the unit circle, since x 2 1 y 2 5 cos2 t 1 sin2 t 5 1. DEFINITION

Parametric Equations

Let x 5 ƒ 1 t 2 and y 5 g 1 t 2 be functions defined for t on some interval. The set of points 1 x, y 2 5 1 ƒ 1 t 2 , g 1 t 2 2 represents a plane curve. The equations x 5 ƒ 1 t 2  and y 5 g 1 t 2

are called parametric equations of the curve. The variable t is called the parameter.

Parametric equations are useful for showing movement along a curve. We insert arrows in the graph to show direction, or orientation, along the curve as t increases.

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11.9  Parametric Equations and Graphs 

1079

EXAMPLE 1  G  raphing a Curve Defined by Parametric Equations

Graph the curve defined by the parametric equations x 5 t 2  y 5 1 t 2 1 2   t in 3 22, 2 4

Indicate the orientation with arrows. Solution: STEP 1  Make

a table and find values for t, x, and y.

t

x 5 t   2

y 5 (t 2 1)

(x, y  )

t 5 22

x 5 1 22 2 2 5 4

y 5 1 22 2 1 2 5 23

14, 232

t 5 21

x 5 1 21 2 2 5 1

t50

x50 50

t51

2

2

x51 51

t52

STEP 2  Plot

2

x52 54

y 5 1 21 2 1 2 5 22

11, 222

y 5 11 2 12 5 0

11, 02

10, 212

y 5 1 0 2 1 2 5 21

14, 12

y 5 12 2 12 5 1

the points in the xy-plane.

y 2 (4, 1)

(1, 0) t=1

1

t=2

–1

1

(0, –1) –1

t=0 t = –1

–2

2

3

(1, –2)

t = –2

–3

STEP 3  Connect

the points with a s mooth curve and use arrows to indicate direction.

x

4

(4, –3) y

2 1

(4, 1)

(1, 0) t=1

t=2

–1

1

(0, –1) –1

t=0 t = –1

–2

(1, –2)

–3

2

3

x

4

t = –2 (4, –3)

The shape of the graph appears to be a parabola. The parametric equations are x 5 t 2 and y 5 1 t 2 1 2 . If we solve the second equation for t, getting t 5 y 1 1, and substi­tute this expression into x 5 t 2, the result is x 5 1 y 1 1 2 2. The graph of x 5 1 y 1 1 2 2 is a parabola with vertex at the point 1 0, 21 2 and opening to the right.

▼

Y O U R T U R N   Graph the curve defined by the parametric equations



x 5 t 1 1  y 5 t 2  t in 3 22, 2 4 Indicate the orientation with arrows.

▼ ANSWER t = –2 (–1, 4)

y

t=2

4

(3, 4)

3 2 1 (0, 1)

Sometimes it is easier to show the rectangular equivalent of the curve and eliminate the parameter.

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–1 –1

t = –1 t=0 1 2 (1, 0)

t=1 (2, 1) 3

x 4

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CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

EXAMPLE 2  G  raphing a Curve Defined by Parametric Equations by First Finding an Equivalent Rectangular Equation

Graph the curve defined by the parametric equations x 5 4 cos t  y 5 3sin t  t is any real number Indicate the orientation with arrows. Solution:

One approach is to point-plot as in Example 1. A second approach is to find the equivalent rectangular equation that represents the curve. We apply the Pythagorean identity.

sin2 t 1 cos2 t 5 1

Find sin2 t from the parametric equation for y.

y 5 3sin t y 2 5 9sin2 t

Square both sides.

Similarly, find cos2 t.

y2 9 x 5 4cos t

Square both sides.

x 2 5 16cos2 t

sin2 t 5

Divide by 9.

cos2 t 5

Divide by 16. y2 x2 and cos2 t 5 9 16 into sin2 t 1 cos2 t 5 1. Substitute sin2 t 5

S TU DY TIP For open curves the orientation can be determined from two values of t. However, for closed curves three points should be chosen to ensure clockwise or counterclockwise orientation.

The curve is an ellipse centered at the origin and elongated horizontally.

x2 16

y2 x2 1 51 9 16 y 5 π 4 (0, 3) 3 t = 2 2 t = 2π (4, 0) t = 0 x (–4, 0) t = π 1 –5

–3

–1

–2 (0, –3)–3 –4 –5

1 2 3 4 5 t = 3π 2

The orientation is counterclockwise. For example, when t 5 0, the position is 14, 02, p when t 5 , the position is 10, 32, and when t 5 p, the position is 1 2 4, 0 2 . 2

11.9.2  Applications of Parametric Equations 11.9.2  S K I L L

Graph cycloids and the curves representing projectile motion. 11.9.2  C O N C E P T U A L

Use time as a parameter in parametric equations such as cycloids and projectile motion.

Young_AT_6160_ch11_pp1078-1095.indd 1080

Parametric equations can be used to describe motion in many applications. Two that we will discuss are the cycloid and a projectile. Suppose that you paint a red X on a bicycle tire. As the bicycle moves in a straight line, if you watch the motion of the red X, you will see that it follows the path of a cycloid. X

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11.9  Parametric Equations and Graphs 

The parametric equations that define a cycloid are

1081

STUD Y T I P

x 5 a 1 t 2 sin t 2   and  y 5 a 1 1 2 cos t 2

A cycloid is a curve that does not have a simple rectangular equation. The only convenient way to describe its path is with parametric equations.

where t is any real number.

EXAMPLE 3  Graphing a Cycloid

Graph the cycloid given by x 5 2 1 t 2 sin t 2 and y 5 2 1 1 2 cos t 2 for t in 3 0, 4p 4 . Solution:

STEP 1  Make

a table and find key values for t, x, and y.

t

x 5 2(t 2 sin t)

y 5 2(1 2 cos t)

(x, y  )

t50

x 5 210 2 02 5 0

y 5 211 2 12 5 0

10, 02

x 5 2 1 2p 2 0 2 5 4p

y 5 211 2 12 5 0

14p, 02

t5p t 5 2p t 5 3p t 5 4p

x 5 2 1 p 2 0 2 5 2p

x 5 2 1 3p 2 0 2 5 6p x 5 2 1 4p 2 0 2 5 8p

points in a plane and connect them with a smooth curve.

y 5 2 3 1 2 1 21 24 5 4

12p, 42

y 5 2 3 1 2 1 21 24 5 4

16p, 42

y 5 211 2 12 5 0

STEP 2  Plot

y 4 3 2 1 t=0

18p, 02

t=π

t = 3π

t = 2π 2π

–1 –2 –3 –4

t = 4π 4π

6π

x 8π

Fitzer/Getty Images

Another example of parametric equations describing real-world phenomena is ­projectile motion. The accompanying photo of a golfer hitting a golf ball illustrates an example of a projectile.

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CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

Let v0 be the initial velocity of an object, u be the initial angle of inclination with the horizontal, and h be the initial height above the ground. Then the parametric equations describing the projectile motion (which will be developed in calculus) are 1 2 x 5 1 v0 cos u 2 t  and    y 5 22 gt 1 1 v0 sin u 2 t 1 h

where t is the time and g is the constant acceleration due to gravity (9.8 meters per square second or 32 feet per square second).

EXAMPLE 4  Graphing Projectile Motion

Suppose a golfer hits his golf ball with an initial velocity of 160 feet per second at an angle of 30° with the ground. How far is his drive, assuming the length of the drive is from the tee to where the ball first hits the ground? Graph the curve ­representing the path of the golf ball. Assume that he hits the ball straight off the tee and down the fairway. Solution: STEP 1  Find

the parametric equations that describe the golf ball that the golfer drove. First, write the parametric equations for projectile motion.



x 5 1 v0 cos u 2 t and y 5 212gt 2 1 1 v0 sin u 2 t 1 h



Let g 5 32 ft/sec2, v0 5 160 ft/sec, h 5 0, and u 5 30°.

x 5 1 160 # cos 30° 2 t and y 5 216t 2 1 1 160 # sin 30° 2 t



Evaluate the sine and cosine functions and simplify.

STEP 2  Graph t

t50 t51 t52 t53 t54 t55

x 5 80 !3 t and y 5 216t 2 1 80t

the projectile motion. x 5 80 !3 t

x 5 80!3 1 0 2 5 0

y 5 216t 2 1 80t

y 5 216 1 0 2 2 1 80 1 0 2 5 0

1x, y2

1 0, 0 2

x 5 80!3 1 1 2 < 139

y 5 216 1 1 2 1 80 1 1 2 5 64

x 5 80!3 1 3 2 < 416

y 5 216 1 3 2 1 80 1 3 2 5 96

1 416, 96 2

x 5 80!3 1 5 2 < 693

y 5 216 1 5 2 1 80 1 5 2 5 0

1 693, 0 2

x 5 80!3 1 2 2 < 277

x 5 80!3 1 4 2 < 554

2

y 5 216 1 2 2 1 80 1 2 2 5 96 2

2

y 5 216 1 4 2 1 80 1 4 2 5 64 2 2

1 139, 64 2

1 277, 96 2

1 554, 64 2

y 700 500 300 100 t = 1 t= 0 100

t= 2 300

t= 3 t =4 t= 5x 500

700

 e can see that we selected our time increments well (the last point, 1 693, 0 2 , W ­corresponds to the ball hitting the ground 693 feet from the tee).

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11.9  Parametric Equations and Graphs  STEP 3  Identify

1083

the horizontal distance from the tee to where the ball first hits the

ground. Algebraically, we can determine the distance of the tee shot by setting the height y equal to zero.

Factor (divide) the common, 216t.



Solve for t.



The ball hits the ground after 5 seconds.

y 5 216t 2 1 80t 5 0 216t 1 t 2 5 2 5 0 t 5 0 or t 5 5

Let t 5 5 in the horizontal distance, x 5 80 !3 t.

x 5 80 !3 1 5 2 < 693

The ball hits the ground 693 feet from the tee.

With parametric equations, we can also determine when the ball lands (5 seconds).

[ S E C T I O N 11 .9 ]     S U M M A R Y Parametric equations are a way of describing the path an object takes along a curve in the xy-plane. Parametric equations have equivalent ­ rectangular equations. Typically, the method of graphing a set of parametric equations is to eliminate t and graph the corresponding rectangular equation. Once the curve is found,

orientation along the curve can be determined by finding points corresponding to different t-values. Two important applications are cycloids and projectiles, whose paths we can trace using parametric equations.

[ S E C T I O N 11 .9 ]   E X E R C I S E S • SKILLS In Exercises 1–30, graph the curve defined by the parametric equations. 1. x 5 t 1 1, y 5 !t, t $ 0 2. x 5 3t, y 5 t 2 2 1, t in 3 0, 4 4 3. x 5 23t, y 5 t 2 1 1, t in 3 0, 4 4

4. x 5 t 2 2 1, y 5 t 2 1 1, t in 3 23, 3 4

7. x 5 !t, y 5 t, t in 3 0, 10 4

8. x 5 t, y 5 "t 2 1 1, t in 3 0, 10 4

5. x 5 t 2, y 5 t 3, t in 3 22, 2 4

6. x 5 t 3 1 1, y 5 t 3 2 1, t in 3 22, 2 4

9. x 5 1 t 1 1 2 2, y 5 1 t 1 2 2 3, t in 3 0, 1 4 11. x 5 e t, y 5 e2t, 2ln 3 # t # ln 3

10. x 5 1 t 2 1 2 3, y 5 1 t 2 2 2 2, t in 3 0, 4 4

13. x 5 2t 4 2 1, y 5 t 8 1 1, 0 # t # 4

14. x 5 3t 6 2 1, y 5 2t 3, 21 # t # 1

15. x 5 t 1 t 2 2 2 3, y 5 t 1 t 2 2 2 3, 0 # t # 4

16. x 5 2t!t, y 5 2 5t 8 2 2, 23 # t # 3

19. x 5 sin t 1 1, y 5 cos t 2 2, t in 3 0, 2p 4

20. x 5 tan t, y 5 1, t in c 2

12. x 5 e22t, y 5 e2t 1 4, 2ln 2 # t # ln 3 3

17. x 5 3 sin t, y 5 2 cos t, t in 3 0, 2p 4

18. x 5 cos 1 2t 2 , y 5 sin t, t in 3 0, 2p 4

21. x 5 1, y 5 sin t, t in 3 22p, 2p 4

22. x 5 sin t, y 5 2, t in 3 0, 2p 4

23. x 5 sin t, y 5 cos t, t in 3 0, 2p 4 2

2

25. x 5 2 sin 1 3t 2 , y 5 3 cos 1 2t 2 , t in 3 0, 2p 4

t 2

t 2

27. x 5 cos a b 2 1, y 5 sin a b 1 1, 22p # t # 2p 29. x 5 2 sin at 1

p p p 7p b, y 5 22 cos at 1 b, 2 # t # 4 4 4 4

Young_AT_6160_ch11_pp1078-1095.indd 1083

p p , d 4 4

24. x 5 2 sin2 t, y 5 2 cos2 t, t in 3 0, 2p 4 26. x 5 4 cos 1 2t 2 , y 5 t, t in 3 0, 2p 4

t 3

t 3

28. x 5 sin a b 1 3, y 5 cos a b 2 1, 0 # t # 6p 30. x 5 23 cos2 1 3t 2 , y 5 2 cos 1 3t 2 , 2

p p #t# 3 3

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CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

In Exercises 31–40, the given parametric equations define a plane curve. Find an equation in rectangular form that also corresponds to the plane curve. 1 31. x 5 , y 5 t 2 32. x 5 t 2 2 1, y 5 t 2 1 1 t 33. x 5 t 3 1 1, y 5 t 3 2 1 34. x 5 3t, y 5 t 2 2 1 35. x 5 t, y 5 "t 2 1 1

36. x 5 sin2 t, y 5 cos2 t

37. x 5 2 sin2 t, y 5 2 cos2 t

38. x 5 sec2 t, y 5 tan2 t

39. x 5 4 1 t 2 1 1 2 , y 5 1 2 t 2

40. x 5 !t 2 1, y 5 !t

• A P P L I C AT I O N S For Exercises 41–50, recall that the flight of a projectile can be modeled with the parametric equations x 5 1v0 cos u2 t   y 5 216t 2 1 1v0 sin u2 t 1 h

where t is in seconds, v0 is the initial velocity, u is the angle with the horizontal, and x and y are in feet. 41.  Flight of a Projectile. A projectile is launched from the ground at a speed of 400 feet per second at an angle of 45° with the horizontal. After how many seconds does the projectile hit the ground? 42.  Flight of a Projectile. A projectile is launched from the ground at a speed of 400 feet per second at an angle of 45° with the horizontal. How far does the projectile travel (what is the horizontal distance), and what is its maximum altitude? 43.  Flight of a Baseball. A baseball is hit at an initial speed of 105 mph and an angle of 20° at a height of 3 feet above the ground. If home plate is 420 feet from the back fence, which is 15 feet tall, will the baseball clear the back fence for a home run? 44.  Flight of a Baseball. A baseball is hit at an initial speed of 105 mph and an angle of 20° at a height of 3 feet above the ground. If there is no back fence or other obstruction, how far does the baseball travel (horizontal distance), and what is its maximum height? 45.  Bullet Fired. A gun is fired from the ground at an angle of 60°, and the bullet has an initial speed of 700 feet per second. How high does the bullet go? What is the horizontal (ground) distance between the point where the gun is fired and the point where the bullet hits the ground? 46.  Bullet Fired. A gun is fired from the ground at an angle of 60°, and the bullet has an initial speed of 2000 feet per second. How high does the bullet go? What is the horizontal (ground) distance between the point where the gun is fired and the point where the bullet hits the ground? 47.  Missile Fired. A missile is fired from a ship at an angle of 30°, an initial height of 20 feet above the water’s surface, and a speed of 4000 feet per second. How long will it be before the missile hits the water? 48.  Missile Fired. A missile is fired from a ship at an angle of 40°, an initial height of 20 feet above the water’s surface, and a speed of 5000 feet per second. Will the missile be able to hit a target that is 2 miles away? 49.  Path of a Projectile. A projectile is launched at a speed of 100 feet per second at an angle of 35° with the horizontal. Plot the path of the projectile on a graph. Assume that h 5 0.

Young_AT_6160_ch11_pp1078-1095.indd 1084

50.  Path of a Projectile. A projectile is launched at a speed of

150 feet per second at an angle of 55° with the horizontal. Plot the path of the projectile on a graph. Assume that h 5 0. For Exercises 51 and 52, refer to the following: Modern amusement park rides are often designed to push the envelope in terms of speed, angle, and ultimately gs, and usually take the form of gargantuan roller coasters or skyscraping towers. However, even just a couple of decades ago, such creations were depicted only in fantasy-type drawings, with their creators never truly believing their construction would become a reality. Nevertheless, thrill rides capable of nauseating any would-be rider were still able to be constructed; one example is the Calypso. This ride is a not-too-distant cousin of the more well-known Scrambler. It consists of four rotating arms (instead of three like the Scrambler), and on each of these arms, four cars (equally spaced around the circumference of a circular frame) are attached. Once in motion, the main piston to which the four arms are connected rotates clockwise, while each of the four arms themselves rotates counterclockwise. The combined motion appears as a blur to any onlooker from the crowd, but the motion of a single rider is much less chaotic. In fact, a single rider’s path can be modeled by the following graph: y 10 8 6 4 2 –10

–6

–2

x 2 4 6 8 10

–4 –6 –8 –10

The equation of this graph is defined parametrically by x 1 t 2 5 A cos t 1 B cos 1 23t 2 y 1 t 2 5 A sin t 1 B sin 1 23t 2 0 # t # 2p 51.  Amusement Rides. What is the location of the rider at p 3p t 5 0, t 5 , t 5 p, t 5 , and t 5 2p? 2 2

52.  Amusement Rides. Suppose that the ride conductor was

rather sinister and speeded up the ride to twice the speed. How would you modify the parametric equations to model such a change? Now vary the values of A and B. What do you think these parameters are modeling in this problem?

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11.9  Parametric Equations and Graphs 

1085

• C AT C H T H E M I S TA K E In Exercises 53 and 54, explain the mistake that is made. 53. Find the rectangular equation that corresponds to the plane

curve defined by the parametric equations x 5 t 1 1 and y 5 !t. Describe the plane curve.

Solution:

Square y 5 !t.

Substitute t 5 y 2 into x 5 t 1 1.

y2 5 t x 5 y2 1 1

The graph of x 5 y 2 1 1 is a parabola opening to the right with vertex at 11, 02.



This is incorrect. What mistake was made?

54. Find the rectangular equations that correspond to the plane

curve defined by the parametric equations x 5 !t and y 5 t 2 1. Describe the plane curve.

Solution: Square x 5 !t.

x2 5 t

Substitute t 5 x 2 into y 5 t 2 1. y 5 x 2 2 1 The graph of y 5 x 2 2 1 is a parabola opening up with ­vertex at 1 0, 21 2 .

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 55 and 56, determine whether each statement is true or false. 55. Curves given by equations in rectangular form have

57. Determine what type of curve the parametric equations

orientation. 56. Curves given by parametric equations have orientation.

58. Determine what type of curve the parametric equations

x 5 !t and y 5 !1 2 t define. x 5 ln t and y 5 t define.

• CHALLENGE p 3p , 2 2 are parametric equations for a hyperbola. Assume that a and b are nonzero constants. t t Prove that x 5 a csc a b, y 5 b cot a b, 0 # t # 4p, 2 2 t 2 p, 3p are parametric equations for a hyperbola. Assume that a and b are nonzero constants. Consider the parametric curve x 5 a sin2 t 2 b cos2 t, p y 5 b cos2 t 1 a sin2 t, 0 # t # . 2 Assume that a and b are nonzero constants. Find the Cartesian equation for this curve.

59. Prove that x 5 a tan t, y 5 b sec t, 0 # t # 2p, t 2



60.



61.







62. Consider the parametric curve x 5 a sin t 1 a cos t,

y 5 a cos t 2 a sin t, 0 # t # 2p.

Assume that a is not zero. Find the Cartesian equation for this curve. 63. Consider the parametric curve x 5 eat, y 5 bet, t . 0. Assume that a is a positive integer and b is a positive real number. Determine the Cartesian equation. 64. Consider the parametric curve x 5 a ln t, y 5 ln 1 bt 2 , t . 0. Assume that b is a positive integer and a is a positive real number. Determine the Cartesian equation.

• TECHNOLOGY 65. Consider the parametric equations: x 5 a sin t 2 sin 1 at 2 and

y 5 a cos t 1 cos 1 at 2 . With a graphing utility, explore the graphs for a 5 2, 3, and 4. 66. Consider the parametric equations: x 5 a cos t 2 b cos 1 at 2 and y 5 a sin t 1 sin 1 at 2 . With a graphing utility, explore the graphs for a 5 3 and b 5 1, a 5 4 and b 5 2, and a 5 6 and b 5 2. Find the t-interval that gives one cycle of the curve. 67. Consider the parametric equations: x 5 cos 1at2 and y 5 sin(bt). With a graphing utility, explore the graphs for a 5 2 and b 5 4, a 5 4 and b 5 2, a 5 1 and b 5 3, and a 5 3 and b 5 1. Find the t-interval that gives one cycle of the curve.

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68. Consider the parametric equations: x 5 a sin 1 at 2 2 sin t and

y 5 a cos 1 at 2 2 cos t. With a graphing utility, explore the graphs for a 5 2 and 3. Describe the t-interval for each case. 69. Consider the parametric equations x 5 a cos 1 at 2 2 sin t and y 5 a sin 1 at 2 2 cos t. With a graphing utility, explore the graphs for a 5 2 and 3. Describe the t-interval for each case. 70. Consider the parametric equations x 5 a sin 1 at 2 2 cos t and y 5 a cos 1 at 2 2 sin t. With a graphing utility, explore the graphs for a 5 2 and 3. Describe the t-interval for each case.

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CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

[CHAPTER 11 REVIEW] SECTION

CONCEPT

11.1

Conic basics

KEY IDEAS/FORMULAS

CHAPTER 11 REVIEW

Three types of conics

11.2

Parabola: Distance to a reference point (focus) and a reference line (directrix) is constant. Ellipse: Sum of the distances between the point and two reference points (foci) is constant. Hyperbola: Difference of the distances between the point and two reference points (foci) is a c­ onstant.

The parabola Parabola with a vertex at the origin

y

y

y 2 = 4px

x 2 = 4py Focus (p, 0)

Focus (0, p) p p

Directrix y = –p

Up: p > 0

Right: p > 0

Left: p < 0

1 y 2 k 2 2 5 4p 1 x 2 h 2

1 x 2 h 2 2 5 4p 1 y 2 k 2

1 p 1 h, k 2

x 5 2p 1 h

1 h, p 1 k 2

y 5 2p 1 k

y5k

x5h

p+0

Opens to the right

Opens upward

p*0

Opens to the left

Opens downward

Equation

1h, k2

Vertex Focus Directrix Axis of symmetry

11.3

p

Directrix x = –p

Down: p < 0

Parabola with a vertex at the point 1h, k2

Applications

p

x

x

1h, k2

Satellite Dish

The ellipse Ellipse centered at the origin

y

y (0, a) (0, b) (–c, 0)

(0, c) (c, 0)

(–a, 0)

x

(–b, 0)

(b, 0)

x

(a, 0) (0, –b)

(0, –c)

(0, –a)

x2 + y2 = 1 a2 b2 c2 = a2 – b2

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x2 + y2 = 1 b2 a2 c2 = a2 – b2

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Chapter Review  SECTION

CONCEPT

1087

KEY IDEAS/FORMULAS

Ellipse centered at the point 1h, k2

Horizontal (Parallel to the x-axis)

Orientation of Major Axis

1x 2 h2

Equation

2

1

a2

1y 2 k2

Vertical (Parallel to the y-axis) 1x 2 h22

2

b2

51

b2

1

1y 2 k22 a2

51

Graph

Applications

11.4

1 h, k 2 c 2    1 h, k 1 c 2

1 h 2 a, k 2    1 h 1 a, k 2

Vertices

CHAPTER 11 REVIEW

1 h 2 c, k 2    1 h 1 c, k 2

Foci

1 h, k 2 a 2    1 h, k 1 a 2

Blimps, football, and orbits

The hyperbola Hyperbola centered at the origin

y y = – ba x

y y = ba x

(0, b)

(–a, 0) (–c, 0)

x (–b, 0) y= a x b

(b, 0)

(0, –a)

x2 – y2 = 1 a2 b2 c2 = a2 + b2 Hyperbola centered at the point 1h, k2 Orientation of Transverse Axis

Horizontal Parallel to the x-axis

Equation

1x 2 h22 a

2

2

1 y 2 k22 2

b

51

(0, c)

(a, 0) x (c, 0)

(0, –b)

(0, a)

y= –ax b

(0, –c)

y2 x2 – =1 a2 b2 c2 = a2 + b2

Vertical Parallel to the y-axis 1 y 2 k22 a

2

2

1x 2 h22 b2

51

Graph

Vertices Foci Applications

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1 h 2 a, k 2    1 h 1 a, k 2 1 h 2 c, k 2    1 h 1 c, k 2 Loran

1 h, k 2 a 2    1 h, k 1 a 2 1 h, k 2 c 2    1 h, k 1 c 2

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CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

SECTION

CONCEPT

KEY IDEAS/FORMULAS

11.5

Systems of nonlinear equations

There is no procedure guaranteed to solve nonlinear equations.

Solving a system of nonlinear equations

Elimination: Eliminate a variable by either adding one equation to or subtracting one equation from the other. Substitution: Solve for one variable in terms of the other and substitute into the second equation.

CHAPTER 11 REVIEW

11.6

11.7

Systems of nonlinear inequalities

# or $ use solid curves

■

, or . use dashed curves

Step 1: Rewrite the inequality as an equation. Step 2: Graph the equation. Step 3: Test points. Step 4: Shade.

Systems of nonlinear inequalities

Graph the individual inequalities and the solution in the common (overlapping) shaded region.

Rotation of axes

The angle of rotation necessary to transform a general second-degree equation into an equation of a conic

11.9

■

Nonlinear inequalities in two variables

Rotation of axes formulas

11.8

Solutions are determined graphically by finding the common shaded regions.

x 5 X cos u 2 Y sin u x 5 X sin u 1 Y cos u cot 2u 5

A2C B

Polar equations of conics

All three conics (parabolas, ellipses, and hyperbolas) are defined in terms of a single focus and a directrix.

Equations of conics in polar coordinates

The directrix is vertical and p units to the right of the pole. ep r5 1 1 e cos u The directrix is vertical and p units to the left of the pole. ep r5 1 2 e cos u The directrix is horizontal and p units above the pole. ep r5 1 1 e sin u The directrix is horizontal and p units below the pole. ep r5 1 2 e sin u

Parametric equations and graphs Parametric equations of a curve Applications of parametric equations

Parametric equations: x 5 ƒ 1 t 2 and y 5 g 1 t 2 Plane curve: 1 x, y 2 5 1 ƒ 1 t 2 , g 1 t 22 Cycloid: x 5 a 1 t 2 sin t 2   y 5 a 1 1 2 cos t 2

Projectile motion: x 5 1 n0 cos u2 t   y 5 2 21 gt 2 1 1 n0 sin u 2 t 1 h

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Review Exercises 

1089

[CHAPTER 11 REVIEW EXERCISES] 31. Major axis vertical with length of 16, minor axis length of

11.1  Conic Basics

Determine whether each statement is true or false.  1. The focus is a point on the graph of the parabola.

6 and centered at 10, 02.

32. Major axis horizontal with length of 30, minor axis length of

20 and centered at 10, 02.

 2. The graph of y 2 5 8x is a parabola that opens upward.

Graph each ellipse. 33.

1 y 1 522 1 x 2 722 1 51 100 36

34. 20 1 x 1 3 2 2 1 1 y 2 4 2 2 5 120

35. 4x 2 2 16x 1 12y 2 1 72y 1 123 5 0 36. 4x 2 2 8x 1 9y 2 2 72y 1 147 5 0

11.2  The Parabola

Find an equation for the parabola described.  5. vertex at 1 0, 0 2 ; focus at 13, 02  6. vertex at 10, 02; focus at 1 0, 2 2

 7. vertex at 10, 02; directrix at x 5 5  8. vertex at 10, 02; directrix at y 5 4  9. 10. 11. 12.

vertex at 1 2, 3 2 ; focus at 1 2, 5 2 vertex at 121, 22 2 ; focus at 1 1, 22 2 focus at 1 1, 5 2 ; directrix at y 5 7 focus at 1 2, 2 2 ; directrix at x 5 0

Find the standard form of an equation of the ellipse with the given characteristics. 37. foci: 121, 3 2 and 1 7, 3 2 vertices: 122, 3 2 and 1 8, 3 2

38. foci: 1 1, 23 2 and 1 1, 21 2 vertices: 1 1, 24 2 and 1 1, 0 2

Applications

39. Planetary Orbits. Jupiter’s orbit is summarized in the

picture. Utilize the fact that the Sun is a focus to determine an equation for Jupiter’s elliptical orbit around the Sun. Round to the nearest hundred thousand kilometers.

REVIEW EXERCISES

y2 x2 2 5 1 is the graph of a hyperbola that has a horizontal 9 1 transverse axis. 1 y 2 322 1 x 1 122  4. 1 5 1 is a graph of an ellipse whose 9 16 center is 11, 32.  3.

Jupiter's Orbit

Find the focus, vertex, directrix, and length of the latus rectum, and graph the parabola. 13. x 2 5 212y 2

14. x 2 5 8y 16. y 2 5 26x

15. y 5 x 17. 1 y 1 2 2 2 5 4 1 x 2 2 2

19. 1 x 1 3 2 5 28 1 y 2 1 2 2

21. x 2 1 5x 1 2y 1 25 5 0

18. 1 y 2 2 2 2 5 24 1 x 1 1 2

Perihelion 740,900,000 km

20. 1 x 2 3 2 2 5 28 1 y 1 2 2

22. y 2 1 2y 2 16x 1 1 5 0

Applications 23. Satellite Dish. A satellite dish measures 10 feet across its

The Sun and Jupiter are not to scale in this drawing.

40. Planetary Orbits. Mars’s orbit is summarized in the picture

below. Utilize the fact that the Sun is a focus to determine an equation for Mars’s elliptical orbit around the Sun. Round to the nearest million kilometers.

opening and 2 feet deep at its center. The receiver should be placed at the focus of the parabolic dish. Where should the receiver be placed? 24. Clearance under a Bridge. A bridge with a parabolic shape reaches a height of 40 feet in the center of the road, and the width of the bridge opening at ground level is 30 feet combined (both lanes). If an RV is 14 feet tall and 8 feet wide, will it make it through the tunnel?

Mars' Orbit

11.3  The Ellipse

Perihelion 207,000,000 km

Graph each ellipse. 25.

y2 x2 1 5 1 9 64

27. 25x 2 1 y 2 5 25

26.

Find the standard form of an equation of the ellipse with the given characteristics. 29. foci: 123, 0 2 and 1 3, 0 2 vertices: 125, 0 2 and 1 5, 0 2

30. foci: 1 0, 22 2 and 1 0, 2 2 vertices: 1 0, 23 2 and 1 0, 3 2

Young_AT_6160_ch11_pp1078-1095.indd 1089

Aphelion 249,000,000 km

The Sun and Mars are not to scale in this drawing.

y2 x2 1 51 81 49

28. 4x 2 1 8y 2 5 64

Aphelion 815,700,000 km

11.4  The Hyperbola

Graph each hyperbola. 41.

y2 x2 2 5 1 9 64

43. x 2 2 25y 2 5 25

42.

y2 x2 2 51 81 49

44. 8y 2 2 4x 2 5 64

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1090 

CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

Find the standard form of an equation of the hyperbola with the given characteristics. 45. vertices: 123, 0 2 and 13, 02 foci: 125, 0 2 and 15, 02

46. vertices: 1 0, 21 2 and 10, 12 foci: 1 0, 23 2 and 10, 32

47. center: 10, 02; transverse: y-axis; asymptotes: y 5 3x and

y 5 23x 1 48. center: 10, 02; transverse axis: y-axis; asymptotes: y 5 2 x 1 and y 5 22x Graph each hyperbola. 1 y 2 12 1 x 2 22 2 51 36 9 2

REVIEW EXERCISES

49.

2



2 1 1 2 5 15 68. x 2 1 y 2 5 2 x2 y x2 1 y2 5 4 1 1 2 2 5 23 x2 y

11.6  Systems of Nonlinear Inequalities

Graph the nonlinear inequality. 69. y $ x 2 1 3

70. x 2 1 y 2 . 16

71. y # ex

72. y , 2x 3 1 2

73. y $ ln 1 x 2 1 2

74. 9x 2 1 4y 2 # 36

Solve each system of inequalities and shade the region on a graph or indicate that the system has no solution.

50. 3 1 x 1 3 2 2 12 1 y 2 4 2 5 72 2

67.

2

x 2 2 2 76. x 2 1 y 2 # 4 2 y # 2x 1 2 y #x

51. 8x 2 2 32x 2 10y 2 2 60y 2 138 5 0

75. y $

52. 2x 2 1 12x 2 8y 2 1 16y 1 6 5 0



Find the standard form of an equation of the hyperbola with the given characteristics. 53. vertices: 10, 32 and 18, 32 foci: 121, 3 2 and 19, 32

54. vertices: 1 4, 22 2 and 14, 02 foci: 1 4, 23 2 and 14, 12

Applications

55. Ship Navigation. Two loran stations are located 220 miles

apart along a coast. If a ship records a time difference of 0.00048 second and continues on the hyperbolic path corresponding to that difference, where would it reach shore? Assume the speed of radio signals is 186,000 miles per second. 56. Ship Navigation. Two loran stations are located 400 miles apart along a coast. If a ship records a time difference of 0.0008 second and continues on the hyperbolic path corresponding to that difference, where would it reach shore? 11.5  Systems of Nonlinear Equations

Solve the system of equations with the elimination method. 57. x2 1 y 5 23

x   2 y 5                 5

59.    x2 1 y2 5 5

58. x2 1 y2 5 4

x2 1 y        5 2

60.        x2 1 y2 5 16

2x2 2 y  5 0

6x2 1 y2 5 16

77. y $ 1 x 1 1 2 2 2 2

y # 10 2 x 2

2

79. 4y 2 9x # 36

78. 3x 2 1 3y 2 # 27

y $ x 1 1

y $x21 2

80. 9x 1 16y 2 # 144

y $ 1 2 x2



11.7  Rotation of Axes

The coordinates of a point in the xy-coordinate system are given. Assuming the X- and Y-axes are found by rotating the x- and y-axes by an angle u, find the corresponding coordinates for the point in the XY-system. 81. 123, 22,  u 5 60°

82. 14, 232,  u 5 45°

Transform the equation of the conic into an equation in X and Y (without an XY-term) by rotating the x- and y-axes through an angle u. Then graph the resulting equation. 83. 2x 2 1 4!3xy 2 2y 2 2 16 5 0,

u 5 30° p 84. 25x 2 1 14xy 1 25y 2 2 288 5 0, u 5 4 Determine the angle of rotation necessary to transform the equation in x and y into an equation in X and Y with no XY-term. 85. 4x 2 1 2!3xy 1 6y 2 2 9 5 0 86. 4x 2 1 5xy 1 4y 2 2 11 5 0

Solve the system of equations with the substitution method.

Graph the second-degree equation.

61. x 1 y         5 3

87. x 2 1 2xy 1 y 2 1 !2x 2 !2y 1 8 5 0



x2 1 y2 5 4

63. x2 1 xy 1 y2 5 212

62.                       xy 5  4

x2 1 y2 5 16

64. 3x 1 y  5                  3

                                           x 2 y 5                       2

 x 2 y2 5 29

Solve the system of equations by applying any method. 65. x 3 2 y 3 5 219

66. 2x 2 1 4xy 5 9

x 2 y 5      21

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x 2 2 2xy 5 0

88. 76x 2 1 48!3xy 1 28y 2 2 100 5 0 11.8  Polar Equations of Conics

Find the polar equation that represents the conic described. 3

89. An ellipse with eccentricity e 5 7 and directrix y 5 27 90. A parabola with directrix x 5 2

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Review Exercises 

1091

Identify the conic (parabola, ellipse, or hyperbola) that each polar equation represents.

Section 11.4

6 91. r 5 4 2 5 cos u

4 1 2x 2 2 2 y 2 5 1, and 4 1 3x 2 2 2 y 2 5 1. What can be said to happen to hyperbola 4 1 cx 2 2 2 y 2 5 1 as c increases? 110. Graph the following three hyperbolas: x 2 2 4y 2 5 1, x 2 2 4 1 2y 2 2 5 1, and x 2 2 4 1 3y 2 2 5 1. What can be said to happen to hyperbola x 2 2 4 1 cy 2 2 5 1 as c increases?

92. r 5

2 5 1 3 sin u

For the given polar equations, find the eccentricity and vertex (or vertices), and graph the curve. 93. r 5

4 2 1 cos u

94. r 5

6 1 2 sin u

11.9  Parametric Equations and Graphs

Graph the curve defined by the parametric equations.  96. x 5 5 sin t, y 5 2 cos t for t in 32p, p 4 2

2

Section 11.5

With a graphing utility, solve the following systems of ­equations. 111. 7.5x 2 1 1.5y 2 5 12.25

x 2y 5 1 

112. 4x 2 1 2xy 1 3y 2 5 12 x 3y 5 3 2 3x 3

 97. x 5 4 2 t , y 5 t for t in 3 23, 3 4

Section 11.6

The given parametric equations define a plane curve. Find an equation in rectangular form that also corresponds to the plane curve.

113. y $ 10x 2 1

2

2

 98. x 5 t 1 3, y 5 4 for t in 324, 4 4

 99. x 5 4 2 t 2, y 5 t

100. x 5 5 sin2 t, y 5 2 cos2 t

101. x 5 2 tan2 t, y 5 4 sec2 t

102. x 5 3t 2 1 4, y 5 3t 2 2 5

Technology Exercises

Section 11.2 103. In your mind, picture the parabola given by

1 x 2 0.6 2 2 5 24 1 y 1 1.2 2 . Where is the vertex? Which way does this parabola open? Now plot the parabola with a graphing utility. 104. In your mind, picture the parabola given by 1 y 2 0.2 2 2 5 3 1 x 2 2.8 2 . Where is the vertex? Which way does this parabola open? Now plot the parabola with a graphing utility. 105. Given is the parabola y 2 1 2.8y 1 3x 2 6.85 5 0. a.  Solve the equation for y, and use a graphing utility to plot the parabola. b. Transform the equation into the form 1 y 2 k 2 2 5 4p 1 x 2 h 2 . Find the vertex. Which way does the parabola open? c. Do (a) and (b) agree with each other? 106. Given is the parabola x 2 2 10.2x 2 y 1 24.8 5 0. a.  Solve the equation for y, and use a graphing utility to

plot the parabola. b. Transform the equation into the form

1 x 2 h 2 2 5 4p 1 y 2 k 2 . Find the vertex. Which way does the parabola open? c. Do (a) and (b) agree with each other?

With a graphing utility, graph the following systems of nonlinear inequalities.

2

y#12x

114. x 2 1 4y 2 # 36 y $ ex

Section 11.7 115. With a graphing utility, explore the second-degree equation

REVIEW EXERCISES

 95. x 5 sin t, y 5 4 cos t for t in 3 2p, p 4

109. Graph the following three hyperbolas: 4x 2 2 y 2 5 1,

Ax 2 1 Bxy 1 Cy 2 1 10x 2 8y 2 5 5 0 for the following values of A, B, and C: a. A 5 2, B 5 23, C 5 5 b. A 5 2, B 5 3, C 5 25

Show the angle of rotation to one decimal place. Explain the differences. 116. With a graphing utility, explore the second-degree equation

Ax 2 1 Bxy 1 Cy 2 1 2x 2 y 5 0 for the following values of A, B, and C: a. A 5 1, B 5 22, C 5 21 b. A 5 1, B 5 2, D 5 1

Show the angle of rotation to the nearest degree. Explain the differences. Section 11.8

8 . Explain 4 1 5 sin u why a graphing utility gives the following graph with the specified window parameters: p 3 26, 6 4 by 3 23, 9 4 with u step 5 4

117. Let us consider the polar equation r 5

Section 11.3 107. Graph the following three ellipses: 4x 2 1 y 2 5 1,

4 1 2x 2 2 1 y 2 5 1, and 4 1 3x 2 2 1 y 2 5 1. What can be said to happen to ellipse 4 1 cx 2 2 1 y 2 5 1 as c increases?

108. Graph the following three ellipses: x 2 1 4y 2 5 1,

x 2 1 4 1 2y 2 2 5 1, and x 2 1 4 1 3y 2 2 5 1. What can be said to happen to ellipse x 2 1 4 1 cy 2 2 5 1 as c increases?

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CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

9 . Explain 3 2 2 sin u why a graphing utility gives the following graph with the specified window parameters: p 3 26, 6 4 by 3 23, 9 4 with u step 5 2

Section 11.9 119. Consider the parametric equations x 5 a cos at 1 b sin bt

and y 5 a sin at 1 b cos bt. Use a graphing utility to explore the graphs for 1 a, b 2 5 1 2, 3 2 and 1 a, b 2 5 1 3, 2 2 . Describe the t-interval for each case.

120. Consider the parametric equations x 5 a sin at 2 b cos bt

and y 5 a cos at 2 b sin bt. Use a graphing utility to explore the graphs for 1 a, b 2 5 1 1, 2 2 and 1 a, b 2 5 1 2, 1 2 . Describe the t-interval for each case.

REVIEW EXERCISES

118. Let us consider the polar equation r 5

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07/12/16 10:59 AM

Practice Test 

1093

[CHAPTER 11 PRACTICE TEST] Match the equation to the graph. 2

 1. x 5 16y

2. y 5 16x

 3. x 2 1 16y 2 5 1

4. x 2 2 16y 2 5 1

2

2

foci: 126, 23 2 , 125, 23 2

15. Hyperbola vertices: 121, 0 2 and 1 1, 0 2

asymptotes: y 5 22x and y 5 2x

6. 16y 2 2 x 2 5 1

 5. 16x 1 y 5 1 a.

14. Ellipse vertices: 127, 23 2 , 124, 23 2

2

16. Hyperbola vertices: 1 0, 21 2 and 1 0, 1 2

b. y

y

5

17. Hyperbola foci: 1 2, 26 2 , 1 2, 6 2

vertices: 1 2, 24 2 , 1 2, 4 2

18. Hyperbola foci: 127, 23 2 , 124, 23 2

x –5

asymptotes: y 5 231 x and y 5 31 x



10

vertices: 126, 23 2 , 125, 23 2

5

Graph the following equations. x

–5

c.

–1

1

20. 4x 2 2 8x 1 y 2 1 10y 1 28 5 0

d. y

21. y 2 1 4y 2 16x 1 20 5 0

y

1

19. 9x 2 1 18x 2 4y 2 1 16y 2 43 5 0

22. x 2 2 4x 1 y 1 1 5 0

1

23. Eyeglass Lens. Eyeglass lenses can be thought of as very x –1

1

24. Planetary Orbits. The planet Uranus’s orbit is described in –1

e.

the following picture, with the Sun as a focus of the elliptical orbit. Write an equation for the orbit.

–1

f. y

Uranus' Orbit

y

5

1

x –5

5

PRACTICE TEST

10

wide parabolic curves. If the focus occurs 1.5 centimeters from the center of the lens and the lens at its opening is 4 centimeters across, find an equation that governs the shape of the lens.

x

x –1

1

Perihelion 2,739,000,000 km

Aphelion 3,003,000,000 km

The Sun and Uranus are not to scale in this drawing. –5

–1

Graph the following nonlinear inequalities.

Find the equation of the conic with the given characteristics.

25. y , x 3 1 1

 7. Parabola vertex: 1 0, 0 2 focus: 124, 0 2

Graph the following systems of nonlinear inequalities.

 8. Parabola vertex: 1 0, 0 2 directrix: y 5 2  9. Parabola vertex: 121, 5 2 focus: 121, 2 2

10. Parabola vertex: 1 2, 23 2 directrix: x 5 0 11. Ellipse

c enter: 1 0, 0 2 vertices: 1 0, 24 2 , 1 0, 4 2 foci: 1 0, 23 2 , 1 0, 3 2

12. Ellipse center: 1 0, 0 2

vertices: 123, 0 2 , 1 3, 0 2 foci: 121, 0 2 , 1 1, 0 2

13. Ellipse vertices: 1 2, 26 2 , 1 2, 6 2

foci: 1 2, 24 2 , 1 2, 4 2

Young_AT_6160_ch11_pp1078-1095.indd 1093

26. y 2 $ 16x

y # 4 2 x 2 28. y # e2x 16x 1 25y 2 # 400 y $ x2 2 4

27.

2

29. Identify the conic represented by the equation



r5

12 . State the eccentricity. 3 1 2 sin u

30. Use rotation of axes to transform the equation in x and y into

an equation in X and Y that has no XY-term:

6!3x 2 1 6xy 1 4!3y 2 5 21!3. State the rotation angle.

31. A golf ball is hit with an initial speed of 120 feet per second at

an angle of 458 with the ground. How long will the ball stay in the air? How far will the ball travel (horizontal ­distance) before it hits the ground?

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1094 

CHAPTER 11  Analytic Geometry and Systems of Nonlinear Equations and Inequalities

32. Describe (classify) the plane curve defined by the parametric

35. Given is the parabola x 2 1 4.2x 2 y 1 5.61 5 0.

equations x 5 !1 2 t and y 5 !t for t in 3 0, 1 4 .

a.  Solve the equation for y and use a graphing utility to plot the

parabola.

33. Use a graphing utility to graph the following nonlinear

b. Transform the equation into the form

inequality:

2

1x 2 h22 5 4p 1 y 2 k2. Find the vertex. Which way does

2

x 1 4xy 2 9y 2 6x 1 8y 1 28 # 0 34. Use a graphing utility to solve the following systems of

­equations: 2

the parabola open?

c. Do (a) and (b) agree with each other?

36. With a graphing utility, explore the second-degree equation

Ax 2 1 Bx y 1 Cy2 1 10x 2 8y 2 5 5 0 for the following ­values of A, B, and C:

2

0.1225x 1 0.0289y 5 1 y 3 5 11x Round your answers to three decimal places.



b. A 5 2, B 5 !3, C 5 21

Show the angle of rotation to one d­ ecimal place. Explain the differences.

PRACTICE TEST



a. A 5 2, B 5 2!3, C 5 1

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Cumulative Test 

1095

[CHAPTERS 1–11 CUMULATIVE TEST]  1. Solve for x: 1 x 1 2 2 2 2 1 x 1 2 2 2 20 5 0.

 2. Find an equation of a circle centered at 1 5, 1 2 and passing

through the point 1 6, 22 2 .

 3. Evaluate the difference quotient  4.

 5.  6.

 7.  8.

 9. 10.

11.

ƒ1x 1 h2 2 ƒ1x2 for the h

function ƒ 1 x 2 5 8 2 7x. Write an equation that describes the following variation: I is directly proportional to both P and t, and I 5 90 when P 5 1500 and t 5 2. Find the quadratic function that has vertex 17, 72 and goes through the point 110, 102. Compound Interest. How much money should you put in a savings account now that earns 4.7% interest a year compounded weekly if you want to have $65,000 in 17 years? Solve the logarithmic equation exactly: log x 2 2 log 16 5 0. In a 30°-60°-90° triangle, if the shortest leg has length 8 inches, what are the lengths of the other leg and the hypotenuse? Use a calculator to evaluate cot 1 227° 2 . Round your answer to four decimal places. Sound Waves. If a sound wave is represented by y 5 0.007 sin 850pt cm, what are its amplitude and ­frequency? For the trigonometric expression tan u 1 csc u 1 cos u 2 , ­perform the operations and simplify. Write the answer in terms of sin u and cos u.

12. Find the exact value of cos a2

11p b. 12

17. At a food court, 3 medium sodas and 2 soft pretzels cost

$6.77. A second order of 5 medium sodas and 4 soft pretzels costs $12.25. Find the cost of a soda and the cost of a soft pretzel. 18. Find the partial fraction decomposition for the rational 3x 1 5 expression . 1 x 2 3 2 1 x2 1 5 2 19. Graph the system of inequalities or indicate that the system has no solution. y $ 3x 2 2 y # 3x 1 2 20. Solve the system using Gauss–Jordan elimination.



21. Given A 5 c

C5 c

9 1

3 0

4 1

27 8 d, B 5 c 5 9

22 0

0 d , find 2B 2 3A. 2

6 d , and 21

22. Use Cramer’s rule to solve the system of equations.

25x 1 40y 5 212 75x 2 105y 5 69 23. Find the standard form of the equation of an ellipse with

foci 1 6, 2 2 and 1 6, 26 2 and vertices 1 6, 3 2 and 1 6, 27 2 .

24. Find the standard form of the equation of a hyperbola with

­vertices 1 5, 22 2 and 1 5, 0 2 and foci 1 5, 23 2 and 1 5, 1 2 .

25. Solve the system of equations.

exactly over the interval 0 # u # 2p. 14. Airplane Speed. A plane flew due north at 450 mph for 2 hours. A second plane, starting at the same point and at the same time, flew southeast at an angle of 1358 clockwise from due north at 375 mph for 2 hours. At the end of 2 hours, how far apart were the two planes? Round to the nearest mile. 15. Find the vector with magnitude  u  5 15 and direction angle u 5 110°. 16. Given z1 5 5 3 cos 1 15° 2 1 i sin 1 15° 2 4 and z2 5 2 3 cos 1 75° 2 1 i sin 1 75° 2 4 , find the product z1z2 and express it in rectangular form.

26. Use a graphing utility to graph the following equation:

x 2 2 3xy 1 10y 2 2 1 5 0 27. Use a graphing utility to graph the following system of

nonlinear inequalities:



y $ e20.3x 2 3.5 y # 4 2 x2

CUMULATIVE TEST

x 1y 5 6 x 2 1 y 2 5 20

2

13. Solve the trigonometric equation 4 cos x 1 4 cos 2x 1 1 5 0

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x 2 2y 1 z 5 7 23x 1 y 1 2z 5 211

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Sequences, Series, and Probability

Have you ever been to a casino and played blackjack? It is the

Dealer’s Up Card

Your Hand

only game in the casino that you can win based on the law of large numbers. In the early 1990s, a group of math and science majors from the Massachusetts Institute of Technology (MIT) devised a foolproof scheme to win at blackjack. A professor at MIT developed a basic strategy outlined in the figure on the right that is based on the probability of combinations of particular cards being dealt, given certain cards already showing. To play blackjack (also called 21), each person is dealt two cards with the option of taking additional cards. The goal is to get a combination of cards that is worth 21 points (or less) without going over (called a bust). You have to avoid going over 21 or staying too far below 21. All face cards (jacks, queens, and kings) are worth 10 points, and an ace in blackjack is worth either 1 or 11 points. The students used the professor’s strategy along with a card-counting technique to place higher bets when there were more high-value cards left in the deck. It is reported that in 1992 the team won $4 million from Las Vegas casinos. The casinos caught on, and the students were all banned within 2 years. The 2008 movie 21 was based on this event.

A BASIC STRATEGY FOR BLACKJACK

17+ 16 15 I 14 13 12 11 10 II 9 5-8 A, 8 - 10 A, 7 A, 6 A, 5 III A, 4 A, 3 A, 2 A, A; 8, 8 10, 10 9, 9 IV 7, 7 6, 6 5, 5 4, 4 3, 3 2, 2 HIT

2 S S S S S H D D H H S S H H H H H SP S SP SP H D H H H

3 S S S S S H D D D H S D D H H H H SP S SP SP SP D H H H

4 S S S S S S D D D H S D D D D H H SP S SP SP SP D H SP SP

5 S S S S S S D D D H S D D D D D D SP S SP SP SP D H SP SP

6 S S S S S S D D D H S D D D D D D SP S SP SP SP D H SP SP

7 S H H H H H D D H H S S H H H H H SP S S SP H D H SP SP

8 S H H H H H D D H H S S H H H H H SP S SP H H D H H H

9 S H H H H H D D H H S H H H H H H SP S SP H H D H H H

10 S H H H H H D H H H S H H H H H H SP S S H H H H H H

A S H H H H H H H H H S H H H H H H SP S S H H H H H H

When surrender is allowed, surrender 9, 7 or 10, 6 vs 9, 10, A; 9, 6 or 10, 5 vs 10 When doubling down after splitting is allowed, split: 2’s, 3’s, 7’s vs 2-7; 4’s vs 5 or 6; 6’s vs 2-6

[12[ CHAPTER

STAND DOUBLE DOWN SPLIT

LEARNING OBJECTIVES ■■ Find

the general nth term of a sequence or series. ■■ Evaluate a finite arithmetic series. ■■ Determine if an infinite geometric series converges or diverges.

Young_AT_6160_ch12_pp1096-1132.indd 1096

■■ Prove

a mathematical statement using induction. ■■ Use the binomial theorem to expand a binomial raised to a positive integer power.

■■ Understand

the difference between permutations and combinations. ■■ Calculate the probability of an event.

07/12/16 11:12 AM

[IN THIS CHAPTER] We will discuss counting and probability in addition to three other topics: sequences and series, mathematical induction, and the binomial theorem.

S EQ U EN C E S , S E R I E S , AN D P R O B AB I L I T Y 12.1

12.2

12.3

12.4

12.5

12.6

12.7

SEQUENCES AND SERIES

ARITHMETIC SEQUENCES AND SERIES

GEOMETRIC SEQUENCES AND SERIES

MATHEMATICAL INDUCTION

THE BINOMIAL THEOREM

COUNTING, PERMUTATIONS, AND COMBINATIONS

PROBABILITY

• Sequences • Factorial Notation • Recursion Formulas • Sums and Series

• Arithmetic Sequences • The General (nth) Term of an Arithmetic Sequence • The Sum of an Arithmetic Sequence

• Geometric Sequences • The General (nth) Term of a Geometric Sequence • Geometric Series

• Mathematical Induction

• Binomial Coefficients • Binomial Expansion • Pascal’s Triangle • Finding a Particular Term of a Binomial Expansion

• The Fundamental Counting Principle • Permutations • Combinations • Permutations with Repetition

• Sample Space • Probability of an Event • Probability of an Event Not Occurring • Mutually Exclusive Events • Independent Events

1097

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1098 

CHAPTER 12  Sequences, Series, and Probability

12.1 SEQUENCES AND SERIES SKILLS OBJECTIVES ■■ Find terms of a sequence given the general term. ■■ Apply factorial notation. ■■ Apply recursion formulas. ■■ Use summation (sigma) notation to represent a series and evaluate finite series and infinite series (if possible).

CONCEPTUAL OBJECTIVES ■■ Recognize patterns in a sequence in order to identify the general term. ■■ Understand why (mn)! Is not equal to m!n!. ■■ Understand why the Fibonacci sequence is a recursion formula. ■■ Understand why all finite series sum and why infinite series may or may not sum.

12.1.1  Sequences 12.1.1 S K I L L

Find terms of a sequence given the general term.

The word sequence means an order in which one thing follows another in succession. In mathematics, it means the same thing. For example, if we write x, 2x 2, 3x 3, 4x 4, 5x 5, ?, what would the next term in the sequence be, the one where the question mark now stands? The answer is 6x 6.

12.1.1 C O N C E P T U A L

Recognize patterns in a sequence in order to identify the general term.

DEFINITION

A Sequence

A sequence is a function whose domain is a set of positive integers. The function values, or terms, of the sequence are written as a1, a2, a3, c, an, c Rather than using function notation, sequences are usually written with subscript (or index) notation, asubscript.

A finite sequence has the domain 5 1, 2, 3, c, n 6 for some positive integer n. An infinite sequence has the domain of all positive integers 5 1, 2, 3, c 6 . There are times when it is convenient to start the indexing at 0 instead of 1: a0, a1, a2, a3, c, an, c

Sometimes a pattern in the sequence can be obtained and the sequence can be written using a general term. In the previous example, x, 2x 2, 3x 3, 4x 4, 5x 5, 6x 6, c, each term has the same exponent and coefficient. We can write this sequence as an 5 nx n, n 5 1, 2, 3, 4, 3, 6, c, where an is called the general term.

EXAMPLE 1  Finding the Sequence, Given the General Term

Find the first four 1 n 5 1, 2, 3, 4 2 terms of the sequences, given the general term. a. an 5 2n 2 1 121 2 n b. bn 5 n11

Young_AT_6160_ch12_pp1096-1132.indd 1098

Solution (a):

an 5 2n 2 1

Find the first term, n 5 1. Find the second term, n 5 2. Find the third term, n 5 3. Find the fourth term, n 5 4. The first four terms of the sequence are  1, 3, 5, 7  .

a1 5 a2 5 a3 5 a4 5

2112 2122 2132 2142

2 2 2 2

1 1 1 1

5 5 5 5

1 3 5 7

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12.1  Sequences and Series 

Solution (b):

bn 5

Find the first term, n 5 1.

b1 5

121 2 n n11

121 2 1 111 121 2 2 b2 5 211 121 2 3 b3 5 311 121 2 4 b4 5 411

Find the second term, n 5 2. Find the third term, n 5 3. Find the fourth term, n 5 4.

52 5

1 2

1 3

52 5

1099

1 4

1 5

The first four terms of the sequence are  2 21, 13, 2 41, 15  .

▼ Y O U R T U R N   Find the first four terms of the sequence an 5

121 2 n . n2

EXAMPLE 2  F  inding the General Term, Given Several Terms of the Sequence

Find the general term of the sequence, given the first five terms. 1 1 1 1 a. 1, 4 , 9 , 16 , 25 , c        b. 21, 4, 29, 16, 225, c

▼ ANSWER 1 21, 14 , 219 , 16

[ CONCEPT CHECK ] What is the next term in the sequence: 1, 23, 9, 227, ?

▼ ANSWER 81

Solution (a):

Write 1 as 11. Notice that each denominator is an integer squared. Identify the general term.

 an 5

1 1 1 1 1 , , , , ,c 1 4 9 16 25 1 1 1 1 1 , , , , ,c 12 22 32 42 52

STUDY T I P (21)n or (21)n11 is a way to represent an alternating sequence.

1     n 5 1, 2, 3, 4, 5, c n2

Solution (b):

Notice that each term includes an integer squared. 212, 22, 232, 42, 252, c Identify the general term.

▼

bn 5 121 2 n n2    n 5 1, 2, 3, 4, 5, c

Y O U R T U R N   Find the general term of the sequence, given the first five terms. 1 1 1 1 1 1 1 1 1 1 a. 2 2 , 4 , 2 6 , 8 , 210 , c      b. 2 , 4 , 8 , 16 , 32 , c

▼ ANSWER a. an 5

121 2 n 1   b. an 5 n 2n 2

Parts (b) in both Example 1 and Example 2 are called alternating sequences because the terms alternate signs (positive and negative). If the odd terms, a1, a3, a5, c are negative and the even terms, a2, a4, a6, c, are positive, we include 121 2 n in the general term. If the opposite is true, and the odd terms are positive and the even terms are negative, we include 121 2 n11 in the general term.

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CHAPTER 12  Sequences, Series, and Probability

12.1.2  Factorial Notation 12.1.2 S K I L L

Apply factorial notation.

Many important sequences that arise in mathematics involve terms that are defined with products of consecutive positive integers. The products are expressed in factorial notation.

12.1.2 C O N C E P T U A L

Understand why (mn)! Is not equal to m!n!. DEFINITION

Factorial

If n is a positive integer, then n! (stated as “n factorial”) is the product of all positive integers from n down to 1. n! 5 n 1 n 2 1 2 1 n 2 2 2 c3⋅ 2 ⋅ 1  n $ 2 and 0! 5 1 and 1! 5 1.

The values of n! for the first six nonnegative integers are 0! 1! 2! 3! 4! 5!

5 5 5 5 5 5

1 1 2⋅1 5 2 3⋅2⋅1 5 6 4 ⋅ 3 ⋅ 2⋅ 1 5 24 5 ⋅ 4 ⋅ 3⋅ 2 ⋅ 1 5 120

Notice that 4! 5 4 ⋅ 3 ⋅ 2⋅ 1 5 4 ⋅ 3!. In general, we can apply the formula n! 5 n 3 1 n 2 1 2 ! 4 . Often the brackets are not used, and the notation n! 5 n 1 n 2 1 2 ! implies calculating the factorial 1 n 2 1 2 ! and then multiplying that quantity by n. For example, to find 6!, we employ the relationship n! 5 n 1 n 2 1 2 ! and set n 5 6: 6! 5 6 ⋅ 5! 5 6 ⋅ 120 5 720

EXAMPLE 3  Finding the Terms of a Sequence Involving Factorials

Find the first four terms of the sequence, given the general term an 5 Solution:

Find the first term, n 5 1.

a1 5

x1 5x 1!

Find the second term, n 5 2.

a2 5

x2 x2 x2 5 5 2! 2⋅1 2

Find the third term, n 5 3.

a3 5

x3 x3 x3 5 5 3! 3 ⋅ 2⋅ 1 6

Find the fourth term, n 5 4.

a4 5

x4 x4 x4 5 5 4! 4 ⋅ 3⋅ 2 ⋅ 1 24

The first four terms of the sequence are x,

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xn . n!

x2 x3 x4 , ,  . 2 6 24

07/12/16 11:12 AM

12.1  Sequences and Series 

1101

EXAMPLE 4  Evaluating Expressions with Factorials

[ CONCEPT CHECK ]

Evaluate each factorial expression.

TRUE OR FALSE  (mn)! 5 m!n!

a.

▼

1n 1 12! 6!         b.  1n 2 12! 2! ⋅ 3!

ANSWER False

Solution (a):

Expand each factorial in the numerator and denominator.

6! 6⋅ 5 ⋅ 4 ⋅ 3⋅ 2 ⋅ 1 5 2! ⋅ 3! 2⋅ 1 ⋅ 3⋅ 2 ⋅ 1

Cancel the 3 ⋅ 2 ⋅ 1 in both the numerator and denominator.

5

6⋅5⋅4 2⋅1

Simplify.

5

6 ⋅ 5⋅ 2 5 60 1

6! 5 60 2! ⋅ 3!

Solution (b):

Expand each factorial in the numerator and denominator.

1 n 1 1 2 1 n 2 1 n 2 1 2 1 n 2 2 2 c3 ⋅ 2 ⋅ 1 1n 1 12! 5 1 n 2 1 2 1 n 2 2 2 c3 ⋅ 2 ⋅ 1 1n 2 12!

Cancel the 1 n 2 1 2 1 n 2 2 2 c3⋅ 2 ⋅ 1 in the numerator and denominator. Alternatively,

common mistake

1n 1 12! 5 1n 1 12 1n2 1n 2 12!

1n 1 12 1n2 1n 2 12! 1n 1 12! 5 1n 2 12! 1n 2 12!

6! 5 60. It is important to note that 2! ⋅ 3! 2 6! 2! ⋅ 3!

▼ Y O U R T U R N   Evaluate the factorial expressions. a. 

In general, m!n! 2 (mn)!

1n 1 12! 5 n2 1 n 1n 2 12!



In Example 4 we found

STUDY T I P

1n 1 22! 3! ⋅ 4! .      b.  2! ⋅ 6! n!

▼ ANSWER 1

a. 10    b. n2 1 3n 1 2

12.1.3  Recursion Formulas Another way to define a sequence is recursively, or using a recursion formula. The first few terms are listed, and the recursion formula determines the remaining terms based on previous terms. For example, the famous Fibonacci sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, c. Each term in the Fibonacci sequence is found by adding the previous two terms. 1 1 1 5 2    1 1 2 5 3    2 1 3 5 5 3 1 5 5 8    5 1 8 5 13  8 1 13 5 21 13 1 21 5 34  21 1 34 5 55  34 1 55 5 89

12.1.3 S K I L L

Apply recursion formulas. 12.1.3 C O N C E P T U A L

Understand why the Fibonacci sequence is a recursion formula.

We can define the Fibonacci sequence using a general term: a1 5 1, a2 5 1, and an 5 an22 1 an21  n $ 3

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1102 

CHAPTER 12  Sequences, Series, and Probability

The Fibonacci sequence is found in places we least expect them (for example, p­ ineapples, broccoli, and flowers). The number of petals in a flower is a Fibonacci number. For example, a wild rose has 5 petals, lilies and irises have 3 petals, and daisies have 34, 55, or even 89 petals. The number of spirals in an Italian broccoli is a Fibonacci number (13). EXAMPLE 5  Using a Recursion Formula to Find a Sequence

Find the first four terms of the sequence: a1 5 2 and an 5 2an21 2 1, n $ 2. Solution:

ST U DY TIP If an 5 an21 1 an22, then a100 5 a99 1 a98.

Write the first term, n 5 1.

a1 5 2

Find the second term, n 5 2.

a2 5 2a1 2 1 5 2 1 2 2 2 1 5 3 a3 5 2a2 2 1 5 2 1 3 2 2 1 5 5 a4 5 2a3 2 1 5 2 1 5 2 2 1 5 9

Find the third term, n 5 3.

[ CONCEPT CHECK ]

Find the fourth term, n 5 4.

What is the next term in the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ?

The first four terms of the sequence are 2, 3, 5, 9  .

▼ Y O U R T U R N   Find the first four terms of the sequence:

▼ ANSWER 144



a1 5 1    and    an 5

an21     n $ 2 n!

▼ ANSWER 1 1 1, 12 , 12 , 288

12.1.4  Sums and Series When we add the terms in a sequence, the result is a series.

12.1.4 S K I L L

Use summation (sigma) notation to represent a series and evaluate finite series and infinite series (if possible). 12.1.4 C O N C E P T U A L

Understand why all finite series sum and why infinite series may or may not sum.

DEFINITION

Series

Given the infinite sequence a1, a2, a3, . . . , an, . . . , the sum of all of the terms in the infinite sequence is called an infinite series and is denoted by a1 1 a2 1 a3 1 ? ? ? 1 an 1 ? ? ? and the sum of only the first n terms is called a finite series, or nth partial sum, and is denoted by Sn 5 a1 1 a2 1 a3 1 ? ? ? 1 an

ST U DY TIP S is a regular Greek letter, but when used to represent the mathematical sum operation, we oversize it.

The capital Greek letter g (sigma) corresponds to the capital S in our alphabet. Therefore, we use g as a shorthand way to represent a sum (series). For example, the sum of the first five terms of the sequence 1, 4, 9, 16, 25, . . . , n2, . . . can be represented using sigma (or summation) notation: 2 2 2 2 2 2 a n 5 112 1 122 1 132 1 142 1 152 5

S TU DY TIP When seen in running text, we often use the following notation: a 5 a n51  5

5

n51

S TU DY TIP A series can start at any integer (not just 1).

n51

5 1 1 4 1 9 1 16 1 25

This is read “the sum as n goes from 1 to 5 of n2.” The letter n is called the index of summation, and often other letters are used instead of n. It is important to note that the sum can start at other numbers besides 1. If we wanted the sum of all of the terms in the sequence, we would represent that infinite series using summation notation as 2 c a n 5 1 1 4 1 9 1 16 1 25 1 q

n51

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12.1  Sequences and Series 

1103

EXAMPLE 6  Writing a Series Using Sigma Notation

Write the following series using sigma notation. 1 1 1 1 a. 1 1 1 1 2 1 6 1 24 1 120   b. 8 1 27 1 64 1 125 1 c

Solution (a):

1 1 1 1 1 1 1 1 1 1 1 1 1 2 6 24 120 1 1 1 1 1 1 5 1 1 1 1 1 1 1 2! 3! 4! 5! 1 1 1 1 1 1 5 1 1 1 1 1 0! 1! 2! 3! 4! 5! 1 an 5   n 5 0, 1, 2, 3, 4, 5 n!

Write 1 as 11. Notice that we can write the denominators using factorials. Recall that 0! 5 1 and 1! 5 1. Identify the general term.

1 a n! 5

Write the finite series using sigma notation.

n50

Solution (b):

8 1 27 1 64 1 125 1 c 5 23 1 33 1 43 1 53 1 c an 5 n3  n $ 2

Write the infinite series as a sum of terms cubed. Identify the general term of the series.

Write the infinite series using sigma notation. a n3 q

n52

▼ Y O U R T U R N   Write the following series using sigma notation. a. 1 2 2 1 6 2 24 1 120 2 c b. 4 1 8 1 16 1 32 1 64 1 c 1

1

1

1

▼ ANSWER a.  a

q 121 2 n11   b.  a 2n n! n51 n52 q

Now that we are comfortable with sigma (summation) notation, let’s turn our attention to evaluating a series (calculating the sum). You can always evaluate a finite series. However, you cannot always evaluate an infinite series.

EXAMPLE 7  Evaluating a Finite Series

Evaluate the series a i50 1 2i 1 1 2 . Solution: Write out the partial sum. 4

1i 5 12 1i 5 32 T T 4 1 2 a 2i 1 1 5 1 1 3 1 5 1 7 1 9 i50 c c c 1i 5 02 1i 5 22 1i 5 42 5 25 a 1 2i 1 1 2 5 25 4

Simplify.

i50

Y O U R T U R N   Evaluate the series a n51 121 2 n n.

▼

Young_AT_6160_ch12_pp1096-1132.indd 1103

5

▼ ANSWER

23

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1104 

CHAPTER 12  Sequences, Series, and Probability

S TU DY TIP The sum of a finite series always exists. The sum of an infinite series may or may not exist.

Infinite series may or may not have a finite sum. For example, if we keep adding 1 1 1 1 1 1 1 1 c, then there is no single real number that the series sums to because the sum continues to grow without bound. However, if we add 0.9 1 0.09 1 0.009 1 0.0009 1 c, this sum is 0.9999 c 5 0.9, which is a rational number, and it can be proven that 0.9 5 1. EXAMPLE 8  Evaluating an Infinite Series, If Possible

Evaluate the following infinite series, if possible. q q 3 a. a n    b.  a n2 n51 n51 10 Solution (a):

3 3 3 3 3 c a 10n 5 10 1 100 1 1000 1 10,000 1 n51 q

Expand the series. Write in decimal form.

q 3 c a 10n 5 0.3 1 0.03 1 0.003 1 0.0003 1

n51

3 1 a 10n 5 0.3333333 5 3 n51 q

Calculate the sum.

3 1 a 10n 5 3 n51 q

Solution (b):

2 c a n 5 1 1 4 1 9 1 16 1 25 1 36 1 q

Expand the series.

n51

This sum is infinite since it continues to grow without any bound. In part (a) we say that the series converges to 13 , and in part (b) we say that the series diverges.

▼

▼ ANSWER

Y O U R T U R N   Evaluate the following infinite series, if possible.

a. a 2n   b.  a 6A 10 B

a. Series diverges. b. Series converges to

2 3.

q

q

n51

n51

1 n

[ CONCEPT CHECK ]

Applications

TRUE OR FALSE  An infinite series may or may not converge.

The annual sales at an electronics retailer from 2015 to 2017 can be approximated by the model an 5 45.7 1 9.5n 2 1.6n2, where an is the yearly sales in billions of dollars and n 5 0, 1, 2.

▼

Annual Sales (in billions)

ANSWER True

$90.0 70.0 50.0

$45.7

$53.6

$58.3

30.0 10.0 2015

2016 Year

2017

Young_AT_6160_ch12_pp1096-1132.indd 1104

TOTAL SALES IN BILLIONS

YEAR

n

an 5 45.7 1 9.5n 2 1.6n2

2015

0

$45.7

2016

1

a0 5 45.7 1 9.5 1 0 2 2 1.6 1 0 2 2 2

$53.6

2017

2

a2 5 45.7 1 9.5 1 2 2 2 1.6 1 2 2

2

$58.3

a1 5 45.7 1 9.5 1 1 2 2 1.6 1 1 2

What does the finite series 13 a n50 an tell us? It tells us the average yearly sales over 3 years. 2

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12.1  Sequences and Series 

1105

[ S E C T I O N 12 .1]     S U M M A R Y In this section, we discussed finite and infinite sequences and The sum of a finite series is always finite. series. When the terms of a sequence are added together, the The sum of an infinite series is either result is a series. ■■ convergent or Finite sequence: a1, a2, a3, c, an ■■ divergent. Infinite sequence: a , a , a , c, a , c 1

Finite series: Infinite series:

2

3

n

a1 1 a2 1 a3 1 c 1 an a1 1 a2 1 a3 1 c 1 an 1 c

Sigma notation is used to express a series.

Finite series: a ai 5 a1 1 a2 1 c 1 an n

■■

Factorial notation was also introduced:

i51 q

n! 5 n⋅ 1 n 2 1 2 ⋅ c ⋅3⋅2⋅1  n $ 2

■■

Infinite series: a an 5 a1 1 a2 1 a3 1 c n51

and 0! 5 1 and 1! 5 1.

[ S E C T I O N 12 .1]   E X E R C I S E S • SKILLS In Exercises 1–12, write the first four terms of the sequence. Assume n starts at 1. 1. an 5 n 5. an 5

n 1n 1 12

6. an 5

9. an 5 121 2 nx n11

1n 1 12 n

7. an 5

10. an 5 121 2 n11 n2

11. an 5

In Exercises 13–20, find the indicated term of the sequence. 1 2

n

14. an 5

13. an 5 a b     a9 5 ? 15. an 5

121 2 n n!     a19 5 ? 1n 1 22!

17. an 5 a1 1

16. an 5

1 2 b     a100 5 ? n

19. an 5 log 10n    a23 5 ?

4. an 5 x n

3. an 5 2n 2 1

2. an 5 n2

2n n!

8. an 5

121 2 n 1n 1 12 1n 1 22

12. an 5

n! 1n 1 12!

1n 2 122 1n 1 122

n     a15 5 ? 1n 1 122

121 2 n11 1 n 2 1 2 1 n 1 2 2     a13 5 ? n

18. an 5 1 2

1     a10 5 ? n2

20. an 5 elnn    a49 5 ?

In Exercises 21–28, write an expression for the nth term of the given sequence. 21. 2, 4, 6, 8, 10, . . .

2 4 3 9

25. 2 , , 2

22. 3, 6, 9, 12, 15, . . .

8 16 , ,c 27 81

26.

1 3 9 27 81 , , , , ,c 2 4 8 16 32

23.

1 1 1 1 1 , , , , ,c 2⋅1 3⋅2 4⋅3 5⋅4 6⋅5

27. 1, 21, 1, 21, 1, c

24.

1 1 1 1 1 , , , , ,c 2 4 8 16 32

28.

1 2 3 4 5 ,2 , ,2 , ,c 3 4 5 6 7

In Exercises 29–40, simplify the ratio of factorials. 29.

9! 7!

30.

4! 6!

31.

35.

97! 93!

36.

101! 98!

37.

Young_AT_6160_ch12_pp1096-1132.indd 1105

29! 27! 1n 2 12! 1n 1 12!

32.

32! 30!

33.

38.

1n 1 22! n!

39.

75! 77! 1 2n 1 3 2 ! 1 2n 1 1 2 !

34.

40.

100! 103! 1 2n 1 2 2 ! 1 2n 2 1 2 !

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1106 

CHAPTER 12  Sequences, Series, and Probability

In Exercises 41–50, write the first four terms of the sequence defined by the recursion formula. Assume the sequence begins at 1. 41. a1 5 7    an 5 an21 1 3

42. a1 5 2    an 5 an21 1 1

43. a1 5 1    an 5 n⋅an21

44. a1 5 2    an 5 1 n 1 1 2 ⋅an21

an21 45. a1 5 100    an 5 n!

46. a1 5 20    an 5

an21 n2

an22 an21

47. a1 5 1, a2 5 2    an 5 an21 ⋅an22

48. a1 5 1, a2 5 2    an 5

2 2 49. a1 5 1, a2 5 21    an 5 121 2 n C an21 1 an22 D

50. a1 5 1, a2 5 21    an 5 1 n 2 1 2 an21 1 1 n 2 2 2 an22

In Exercises 51–64, evaluate the finite series. 51. a 2

52. a 7

53. a n2

54. a n

57. a 1n

58. a 2n

63. a k50 k!

64. a

59. a 12x 2 n

60. a 12x 2 n11

5

n51

4

n50

x

n50

n51

4

4

4

5

3

n50

n50

121 2 x k! k50

k

4

k k

4

55. a 1 2n 2 1 2 6

1

n51

n51

3

n50

61. a k50 k! 5

k

2

56. a 1 n 1 1 2 6

n51

62. a

121 2 k k50 k! 5

In Exercises 65–68, evaluate the infinite series, if possible. 65. a 2⋅ 1 0.1 2 j

66. a 5⋅ A 10 B

q

q

j50

1

j

j50

67. a n j    n $ 1 q

j50

68. a 1j q

j50

In Exercises 69–76, apply sigma notation to write the sum. 69. 1 2

1 1 1 1 1 2 1 c1 2 4 8 64

71. 1 2 2 1 3 2 4 1 5 2 6 1 c 73.

2⋅1 3⋅2⋅1 4⋅3⋅2⋅1 5⋅4⋅3⋅2⋅1 6⋅5⋅4⋅3⋅2⋅1 1 1 1 1 1 1 2⋅1 3⋅2⋅1 4⋅3⋅2⋅1

75. 1 2 x 1

x2 x3 x4 x5 2 1 2 1c 2 6 24 120

70. 1 1

1 1 1 1 1 1 1 c1 1c 2 4 8 64

72. 1 1 2 1 3 1 4 1 5 1 c1 21 1 22 1 23 74. 1 1

2 22 23 24 1 1 1 1c 1 2⋅1 3⋅2⋅1 4⋅3⋅2⋅1

76. x 1 x 2 1

x3 x4 x5 x6 1 1 1 2 6 24 120

• A P P L I C AT I O N S 77. Money. Upon graduation Jessica receives a commission from

79.  Salary. An attorney is trying to calculate the costs associated

the U.S. Navy to become an officer and a $20,000 signing bonus for selecting aviation. She puts the entire bonus in an account that earns 6% interest compounded monthly. The balance in the account after n months is

with going into private practice. If she hires a paralegal to assist her, she will have to pay the paralegal $20 per hour. To be competitive with most firms, she will have to give her paralegal a $2 per hour raise per year. Find a general term of a sequence an, which represents the hourly salary of a paralegal with n years of experience. What will be the paralegal’s salary with 20 years of experience? 80.  NFL Salaries. A player in the NFL typically has a career that lasts 3 years. The practice squad makes the league minimum of $275,000 (2004) in the first year, with a $75,000 raise per year. Write the general term of a sequence an that represents the salary of an NFL player making the league minimum during his entire career. Assuming n 5 1 corresponds to the 3 first year, what does a n51 an represent?

An 5 20,000 a1 1

0.06 n b   n 5 1, 2, 3, c 12

Her commitment to the Navy is 6 years. Calculate A72. What does A72 represent? 78.  Money. Dylan sells his car in his freshman year and puts $7000 in an account that earns 5% interest compounded quarterly. The balance in the account after n quarters is An 5 7000 a1 1

0.05 n b   n 5 1, 2, 3, c 4

Calculate A12. What does A12 represent?

Young_AT_6160_ch12_pp1096-1132.indd 1106

81.  Salary. Upon graduation Sheldon decides to go to work for a

local police department. His starting salary is $30,000 per year,

07/12/16 11:13 AM

12.1  Sequences and Series 

and he expects to get a 3% raise per year. Write the recursion formula for a sequence that represents his salary n years on the job. Assume n 5 0 represents his first year making $30,000. 82.  Escherichia coli. A single cell of bacteria reproduces through a process called binary fission. Escherichia coli cells divide into two every 20 minutes. Suppose the same rate of division is maintained for 12 hours after the original cell enters the body. How many E. coli bacteria cells would be in the body 12 hours later? Suppose there is an infinite nutrient source so that the E. coli bacteria maintain the same rate of division for 48 hours after the original cell enters the body. How many E. coli bacteria cells would be in the body 48 hours later? 83.  AIDS/HIV. A typical person has 500 to 1500 T cells per drop of blood in the body. HIV destroys the T cell count at a rate of 50–100 cells per drop of blood per year, depending on how aggressive it is in the body. Generally, the onset of AIDS occurs once the body’s T cell count drops below 200. Write a sequence that represents the total number of T cells in a person infected with HIV. Assume that before infection the person has a 1000 T cell count 1 a0 5 1000 2 and the rate at which the infection spreads corresponds to a loss of 75 T cells per drop of blood per year. How much time will elapse until this person has full-blown AIDS? 84.  Company Sales. A national furniture store reported total sales from 2016 through 2017 in the billions. The sequence an 5 3.8 1 1.6n represents the total sales in billions of dollars. Assuming n 5 3 corresponds to 2016, what were 4 the reported sales in 2016 and 2017? What does 12 ⋅ a n53 an represent? 85.  Cost of Eating Out. A college student tries to save money by bringing a bag lunch instead of eating out. He will be able to save $100 per month. He puts the money into his savings account, which draws 1.2% interest and is compounded monthly. The balance in his account after n months of bagging his lunch is



An 5 100,000 3 1 1.001 2 n 2 1 4       n 5 1, 2, c

1107

86.  Cost of Acrylic Nails. A college student tries to save money

by growing her own nails out and not spending $50 per month on acrylic fills. She will be able to save $50 per month. She puts the money into her savings account, which draws 1.2% interest and is compounded monthly. The balance in her account after n months of natural nails is An 5 50,000 3 1 1.001 2 n 2 1 4       n 5 1, 2, c

 alculate the first four terms of this sequence. Calculate the C amount after 4 years (48 months). xn q 87.  Math and Engineering. The formula a n50 can be used n! x to approximate the function y 5 e . Compute the first five terms of this formula to approximate e x. Apply the result to find e2 and compare this result with the calculator value of e2. 88.  Home Prices. If the inflation rate is 3.5% per year and the average price of a home is $195,000, the average price of a home after n years is given by An 5 195,000 1 1.035 2 n. Find the average price of the home after 6 years. 89.  Approximating Functions. Polynomials can be used to approximate transcendental functions such as ln 1x2 and ex, which are found in advanced mathematics and engineering. 1 x 2 1 2 n11 q can be used to For example, a n50 121 2 n n11 approximate  ln 1 x 2 , where x is close to 1. Use the first five terms of the series to approximate ln 1 x 2 . Next, find ln 1 1.1 2 and compare with the value given by your calculator. 90.  Future Value of an Annuity. The future value of an ordinary annuity is given by the formula FV 5 PMT 3 1 1 1 1 i 2 n 2 1 2 /i 4 , where PMT 5 amount paid into the account at the end of each period, i 5 interest rate per period, and n 5 number of compounding periods. If you invest $5000 at the end of each year for 5 years, you will have an accumulated value of FV as given in the above equation at the end of the nth year. Determine how much is in the account at the end of each year for the next 5 years if i 5 0.06.

Calculate the first four terms of this sequence. Calculate the amount after 3 years (36 months).

• C AT C H T H E M I S TA K E In Exercises 91–94, explain the mistake that is made. 1 3! 2 1 5! 2  91. Simplify the ratio of factorials: . 6! Solution:

Express 6! in factored form.

Cancel the 3! in the numerator and denominator.

Write out the factorials.

1 3! 2 1 5! 2 1 3! 2 1 2! 2 1 5! 2 1 2! 2

5⋅4⋅3⋅2⋅1 2⋅1 5⋅4⋅3 5 60

Simplify. 1 3! 2 1 5! 2 2 60. What mistake was made? 1 3! 2 1 2! 2

Young_AT_6160_ch12_pp1096-1132.indd 1107

 92. Simplify the factorial expression:

Solution:



2n 1 2n 2 2 2 ! . 1 2n 1 2 2 !

Express factorials in factored form. 2n 1 2n 2 2 2 1 2n 2 4 2 1 2n 2 6 2 c 1 2n 1 2 2 1 2n 2 1 2n 2 2 2 1 2n 2 4 2 1 2n 2 6 2 c

Cancel common terms.

1 2n 1 2

This is incorrect. What mistake was made?

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1108 

CHAPTER 12  Sequences, Series, and Probability

 94. Evaluate the series a k50 121 2 kk 2.

  93. Find the first four terms of the sequence defined by

3

an 5 121 2 n11n2.

Solution:

Solution: Find the n 5 1 term.

a1 5 21



Find the n 5 2 term.

a2 5 4



Find the n 5 3 term.

a3 5 29



Find the n 5 4 term.

a4 5 16



k 2 a 121 2 k 5 21 1 4 2 9 3

Write out the sum.

k50

k 2 a 121 2 k 5 26 3



Simplify the sum.

k56



This is incorrect. What mistake was made?

The sequence 21, 4, 29, 16, c, is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 95–98, determine whether each statement is true or false. 95. a cx k 5 c a x k n

n

k50

k50

96. a ai 1 bi 5 a ai 1 a bi n

n

n

i51

i51

i51

97. a akbk 5 a ak ⋅ a bk n

n

n

k51

k51

k51

98. 1 a! 2 1 b! 2 5 1 ab 2 !

• CHALLENGE   99. Write the first four terms of the sequence defined by the

recursion formula: a1 5 C    an 5 an21 1 D    D 2 0 100.  Write the first four terms of the sequence defined by the

recursion formula: a1 5 C    an 5 Dan21    D 2 0

101. Fibonacci Sequence. An explicit formula for the nth term of

the Fibonacci sequence is: n

Fn 5

A1 1 !5 B 2 A1 2 !5 B

n

2n !5

Apply algebra (not your calculator) to find the first two terms of this sequence and verify that these are indeed the first two terms of the Fibonacci sequence. 102. Let an 5 !an21 for n $ 2 and a1 5 7. Find the first five terms of this sequence and make a generalization for the nth term.

• TECHNOLOGY 1 n b approaches the n number e as n gets large. Use a graphing calculator to find a100, a1000, a10,000, and keep increasing n until the terms in the sequence approach 2.7183. 104. The Fibonacci sequence is defined by a1 5 1, a2 5 1, and an11 an 5 an22 1 an21 for n $ 3. The ratio is an approximation an of the golden ratio. The ratio approaches a constant f (phi) as n gets large. Find the golden ratio using a graphing utility. 103. The sequence defined by an 5 a1 1

Young_AT_6160_ch12_pp1096-1132.indd 1108

105. Use a graphing calculator “SUM” to sum a k50 . Compare k! 5

it with your answer to Exercise 61.

2k

106. Use a graphing calculator “SUM” to sum a k50 . k! Compare it with your answer to Exercise 62. 5

121 2 k

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12.2  Arithmetic Sequences and Series 

1109

12.2 ARITHMETIC SEQUENCES AND SERIES SKILLS OBJECTIVES ■■ Determine if a sequence is arithmetic. ■■ Find the general nth term of an arithmetic sequence. ■■ Evaluate a finite arithmetic series.

CONCEPTUAL OBJECTIVES ■■ Understand that the common difference of an arithmetic sequence can be either positive or negative. ■■ Derive the formula for the nth term of an arithmetic sequence in terms of n, the first term, and the common difference. ■■ Derive the formula for the sum of a finite arithmetic sequence.

12.2.1  Arithmetic Sequences The word arithmetic (with emphasis on the third syllable) often implies adding or subtracting of numbers. Arithmetic sequences are sequences whose terms are found by adding a constant to each previous term. The sequence 1, 3, 5, 7, 9, . . . is arithmetic because each ­successive term is found by adding 2 to the previous term.

12.2.1 S K I L L

Determine if a sequence is arithmetic. 12.2.1 C O N C E P T U A L

DEFINITION

Arithmetic Sequences

A sequence is arithmetic if each term in the sequence is found by adding a real number d to the previous term, so that an11 5 an 1 d. Because an11 2 an 5 d, the number d is called the common difference.

EXAMPLE 1  I dentifying the Common Difference in Arithmetic Sequences

 etermine whether each sequence is arithmetic. If so, find the common difference D for each of the arithmetic sequences. 1 5 11 a. 5, 9, 13, 17, c  b. 18, 9, 0, 29, c  c. 2 , 4 , 2, 4 , c Solution (a):

Label the terms.

a1 5 5, a2 5 9, a3 5 13, a4 5 17, c

Find the difference d 5 an11 2 an.

d 5 a2 2 a1 5 9 2 5 5 4

Check that the difference for the next two terms is also 4.

d 5 a3 2 a2 5 13 2 9 5 4

Understand that the common difference of an arithmetic sequence can be either positive or negative.

[ CONCEPT CHECK ] TRUE OR FALSE  If the second term is less than the first term in an arithmetic sequence, then the common difference is negative. If the second term is greater than the first term in an arithmetic sequence, then the common difference is positive.

▼ ANSWER True

d 5 a4 2 a3 5 17 2 13 5 4

There is a common difference of 4  . Therefore, this sequence is arithmetic, and each successive term is found by adding 4 to the previous term. Solution (b):

Label the terms.

a1 5 18, a2 5 9, a3 5 0, a4 5 29, c

Find the difference d 5 an11 2 an.

d 5 a2 2 a1 5 9 2 18 5 29

Check that the difference for the next two terms is also 29.

d 5 a3 2 a2 5 0 2 9 5 29 d 5 a4 2 a3 5 29 2 0 5 29

There is a common difference of 29  . Therefore, this sequence is arithmetic, and each successive term is found by subtracting 9 from (that is, adding 29 to) the previous term.

Young_AT_6160_ch12_pp1096-1132.indd 1109

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1110 

CHAPTER 12  Sequences, Series, and Probability

ST U DY TIP We check several terms to confirm that a sequence is arithmetic.

Solution (c):

Label the terms. Find the difference d 5 an11 2 an. Check that the difference for the next two terms is also 34 .

1 5 11 a1 5 , a2 5 , a3 5 2, a4 5 , c 2 4 4 5 1 3 d 5 a2 2 a1 5 2 5   4 2 4 5 3 d 5 a3 2 a2 5 2 2 5 4 4 11 3 d 5 a4 2 a3 5 225 4 4

There is a common difference of 34 . Therefore, this sequence is arithmetic, and each successive term is found by adding 34 to the previous term. ▼ ANSWER 2 a. 25    b. 3

▼ Y O U R T U R N   Find the common difference for each of the arithmetic sequences. a. 7, 2, 23, 28, c  b. 1, 3 , 3 , 3, c 5 7

12.2.2  The General (nth) Term of an Arithmetic Sequence 12.2.2 S K I L L

Find the general nth term of an arithmetic sequence.

To find a formula for the general, or nth, term of an arithmetic sequence, write out the first several terms and look for a pattern.

12.2.2 C O N C E P T U A L

Derive the formula for the nth term of an arithmetic sequence in terms of n, the first term, and the common difference.

First term, n 5 1.

a1

Second term, n 5 2.

a2 5 a1 1 d

Third term, n 5 3.

a3 5 a2 1 d 5 1 a1 1 d 2 1 d 5 a1 1 2d

Fourth term, n 5 4.

a4 5 a3 1 d 5 1 a1 1 2d 2 1 d 5 a1 1 3d

In general, the nth term is given by an 5 a1 1 1 n 2 1 2 d. THE nTH TERM OF AN ARITHMETIC SEQUENCE

The nth term of an arithmetic sequence with common difference d is given by an 5 a1 1 1 n 2 1 2 d      for n $ 1 EXAMPLE 2  Finding the nth Term of an Arithmetic Sequence

Find the 13th term of the arithmetic sequence 2, 5, 8, 11, . . . . Solution:

▼ ANSWER

Identify the common difference. d552253 Identify the first 1 n 5 1 2 term. a1 5 2 Substitute a1 5 2 and d 5 3 into an 5 a1 1 1 n 2 1 2 d. an 5 2 1 3 1 n 2 1 2 Substitute n 5 13 into an 5 2 1 3 1 n 2 1 2 . a13 5 2 1 3 1 13 2 1 2 5 38

▼

Y O U R T U R N   Find the 10th term of the arithmetic sequence 3, 10, 17, 24, . . . .

66

Young_AT_6160_ch12_pp1096-1132.indd 1110

07/12/16 11:13 AM

12.2  Arithmetic Sequences and Series 

1111

EXAMPLE 3  Finding the Arithmetic Sequence

[ CONCEPT CHECK ]

The 4th term of an arithmetic sequence is 16, and the 21st term is 67. Find a1 and d and construct the sequence.

For the arithmetic sequence 3, 10, 17, 24, . . . what is the general term for the nth term?

Solution:

Write the 4th and 21st terms. a4 5 16 and a21 5 67 Adding d 17 times to a4 results in a21.   a21 5 a4 1 17d Substitute a4 5 16 and a21 5 67. 67 5 16 1 17d Solve for d.

▼ ANSWER an 5 3 1 7(n 2 1)

d53

Substitute d 5 3 into an 5 a1 1 1 n 2 1 2 d. an 5 a1 1 3 1 n 2 1 2 Let a4 5 16. 16 5 a1 1 3 1 4 2 1 2 Solve for a1.

a1 5 7

The arithmetic sequence that starts at 7 and has a common difference of 3 is 7, 10, 13, 16, . . . .

▼ Y O U R T U R N   Construct the arithmetic sequence whose 7th term is 26 and

whose 13th term is 50.

▼ ANSWER

2, 6, 10, 14, . . .

12.2.3  The Sum of an Arithmetic Sequence 12.2.3 S K I L L

What is the sum of the first 100 counting numbers 1 1 2 1 3 1 4 1 c1 99 1 100 5 ? If we write this sum twice (one in ascending order and one in descending order) and add, we get 100 pairs of 101. 11

21

31

4 1 ? ? ? 1 99 1 100

100 1 99 1 98 1 97 1 ? ? ? 1

21

Evaluate a finite arithmetic series. 12.2.3 C O N C E P T U A L

Derive the formula for the sum of a finite arithmetic sequence.

1

101 1 101 1 101 1 101 1 ? ? ? 1 101 1 101 5 100 11012

Since we added twice the sum, we divide by 2.

1 101 2 1 100 2 1 1 2 1 3 1 4 1 c1 99 1 100 5 5 5050 2 Now, let us develop the sum of a general arithmetic series. The sum of the first n terms of an arithmetic sequence is called the nth partial sum, or finite arithmetic series, and is denoted by Sn. An arithmetic sequence can be found by ­starting at the first term and adding the common difference to each successive term, and so the nth ­partial sum, or finite series, can be found the same way, but terminating the sum at the nth term: Sn 5 a1 1 a2 1 a3 1 a4 1 c

Sn 5 a1 1 1 a1 1 d 2 1 1 a1 1 2d 2 1 1 a1 1 3d 2 1 c1 1 a n 2

Similarly, we can start with the nth term and find terms going backward by subtracting the common difference until we arrive at the first term: Sn 5 an 1 an21 1 an22 1 an23 1 c

Sn 5 an 1 1 an 2 d 2 1 1 an 2 2d 2 1 1 an 2 3d 2 1 c 1 1 a1 2

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CHAPTER 12  Sequences, Series, and Probability

Add these two representations of the nth partial sum. Notice that the d terms are eliminated: Sn 5 a1 1 1 a1 1 d 2 1 1 a1 1 2d 2 1 1 a1 1 3d 2 1 c1 1 a n 2

Sn 5 an 1 1 an 2 d 2 1 1 an 2 2d 2 1 1 an 2 3d 2 1 c 1 1 a1 2

2Sn 5 1 a1 1 an 2 1 1 a1 1 an 2 1 1 a1 1 an 2 1 c1 1 a1 1 an 2 ('''''''''''')''''''''''''*

n 1 a1 1 an 2



2Sn 5 n 1 a1 1 an 2        or        Sn 5



ST U DY TIP Sn can also be written as n Sn 5 3 2a1 1 1 n 2 1 2 d 4 . 2

Let an 5 a1 1 1 n 2 1 2 d. Sn 5 5

n 1 a 1 an 2 2 1

n 3 a 1 a1 1 1 n 2 1 2 d 4 2 1

n1n 2 12d n 3 2a1 1 1 n 2 1 2 d 4 5 na1 1 2 2

Evaluating a Finite Arithmetic Series

DEFINITION

The sum of the first n terms of an arithmetic sequence, called a finite arithmetic series, is given by the formula Sn 5

n n 3 2a1 1 1 n 2 1 2 d 4       or      Sn 5 1 a1 1 an 2   n $ 2 2 2

EXAMPLE 4  Evaluating a Finite Arithmetic Series

Evaluate the finite arithmetic series a k51 k. Solution: 100

c1 99 1 100 ak 5 1 1 2 1 3 1 100

Expand the arithmetic series.

k51

This is the sum of an arithmetic sequence of numbers with a common difference of 1. Identify the parameters of the arithmetic sequence. Substitute these values into Sn 5 Simplify.

n 1 a 1 an 2 . 2 1

a1 5 1, an 5 100, and n 5 100 S100 5

100 1 1 1 100 2 2

S100 5 5050

The sum of the first 100 natural numbers is 5050. ▼ ANSWER a. 465      b. 440

Young_AT_6160_ch12_pp1096-1132.indd 1112

▼ Y O U R T U R N   Evaluate the following finite arithmetic series. a.  a k        b.  a 1 2k 1 1 2 30

20

k51

k51

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12.2  Arithmetic Sequences and Series 

[ CONCEPT CHECK ]

EXAMPLE 5  Finding the nth Partial Sum of an Arithmetic Sequence

Find the sum of the first 20 terms of the arithmetic sequence 3, 8, 13, 18, 23, . . . . Solution:

Recall the partial-sum formula.

Sn 5

Find the 20th partial sum of this arithmetic sequence. S20 5 Recall that the general nth term of an arithmetic sequence is given by:

1113

n 1 a 1 an 2 2 1

Derive the formula for the sum of an arithmetic sequence given by 2, 6, 10, 14, 18, . . .

▼ ANSWER Sn 5 2n2

20 1 a 1 a20 2 2 1

an 5 a1 1 1 n 2 1 2 d

Note that the first term of the arithmetic sequence is 3. a1 5 3 This is an arithmetic sequence with a common difference of 5.

d55

Substitute a1 5 3 and d 5 5 into an 5 a1 1 1n 2 12 d. an 5 3 1 1 n 2 1 2 5 Substitute n 5 20 to find the 20th term.

a20 5 3 1 1 20 2 1 2 5 5 98

Substitute a1 5 3 and a20 5 98 into the partial sum. S20 5 10 1 3 1 98 2 5 1010 The sum of the first 20 terms of this arithmetic sequence is 1010.

▼ Y O U R T U R N   Find the sum of the first 25 terms of the arithmetic sequence 2, 6,

10, 14, 18, . . . .

▼ ANSWER

1250

Applications EXAMPLE 6  Marching Band Formation

David Young-Wolff/PhotoEdit

Suppose a band has 18 members in the first row, 22 members in the second row, and 26 members in the third row and continues with that pattern for a total of nine rows. How many marchers are there all together?

UC Berkeley marching band

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CHAPTER 12  Sequences, Series, and Probability

Solution:

The number of members in each row forms an arithmetic sequence with a common difference of 4, and the first row has 18 members. Calculate the nth term of the sequence an 5 a1 1 1 n 2 1 2 d. Find the 9th term n 5 9. Calculate the sum Sn 5

a1 5 18

d54

an 5 18 1 1 n 2 1 2 4

a9 5 18 1 1 9 2 1 2 4 5 50

n 9 1 a1 1 an 2 of the nine rows. S9 5 1 a1 1 a9 2 2 2 9 5 1 18 1 50 2 2 9 5 1 68 2 2

There are 306 members in the marching band.

5 306

▼

▼ ANSWER

Y O U R T U R N   Suppose a bed of tulips is arranged in a garden so that there are

20 tulips in the first row, 26 tulips in the second row, and 32 tulips in the third row and the rows continue with that pattern for a total of 8 rows. How many tulips are there all together?

328

[ S E C T I O N 12 . 2 ]     S U M M A R Y In this section, arithmetic sequences were defined as sequences of which each successive term is found by adding the same constant d to the previous term. Formulas were developed for the general, or nth, term of an arithmetic sequence, and for the nth partial sum of an arithmetic sequence, also called a finite arithmetic series.

an 5 a1 1 1 n 2 1 2 d    n $ 1 n n Sn 5 1 a1 1 an 2 5 3 2a1 1 1 n 2 1 2 d 4 2 2

[ S E C T I O N 12 . 2 ]   E X E R C I S E S • SKILLS In Exercises 1–10, determine whether the sequence is arithmetic. If it is, find the common difference. 1. 2, 5, 8, 11, 14, c

2. 9, 6, 3, 0, 23, 26, c

3. 12 1 22 1 32 1 c

4. 1! 1 2! 1 3! 1 c

5. 3.33, 3.30, 3.27, 3.24, c

6. 0.7, 1.2, 1.7, 2.2, c

7. 4, 3 , 3 , 6, c

8. 2, 3 , 3 , 3, c

9. 101, 102, 103, 104, c

10. 120, 60, 30, 15, c

14 16

7 8

In Exercises 11–20, find the first four terms of the sequence. Determine whether the sequence is arithmetic, and if so, find the common difference. 11. an 5 22n 1 5

12. an 5 3n 2 10

13. an 5 n2

15. an 5 5n 2 3

16. an 5 24n 1 5

19. an 5 121 2 n

20. an 5 121 2 n112n

17. an 5 10 1 n 2 1 2

n

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n2 n! 18. an 5 8n 2 4 14. an 5

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12.2  Arithmetic Sequences and Series 

1115

In Exercises 21–28, find the general, or nth, term of the arithmetic sequence given the first term and the common difference. 22. a1 5 5    d 5 11

21. a1 5 11    d 5 5 25. a1 5 0    d 5

2 3

26. a1 5 21    d 5

243

23. a1 5 24    d 5 2

24. a1 5 2    d 5 24

27. a1 5 0    d 5 e

28. a1 5 1.1    d 5 20.3

In Exercises 29–34, find the specified term for each arithmetic sequence. 29. The 10th term of the sequence 7, 20, 33, 46, c

31. The 100th term of the sequence 9, 2, 25, 212, c 33. The 21st term of the sequence

1 7 5 13 3 , 12 , 6 , 12 ,

c

30. The 19th term of the sequence 7, 1, 25, 211, c 32. The 90th term of the sequence 13, 19, 25, 31, c 34. The 33rd term of the sequence 5 , 15 , 15 , 5 , c 1 8 13 6

In Exercises 35–40, for each arithmetic sequence described, find a1 and d and construct the sequence by stating the general, or nth, term. 35. The 5th term is 44 and the 17th term is 152.

36. The 9th term is 219 and the 21st term is 255.

37. The 7th term is 21 and the 17th term is 241.

38. The 8th term is 47 and the 21st term is 112.

39. The 4th term is 3 and the 22nd term is 15.

40. The 11th term is 23 and the 31st term is 213.

In Exercises 41–52, find the sum. 41. a 2k 23

k51

42. a 5k 20

k50

47. 2 1 7 1 12 1 17 1 c 1 62

43. a 1 22n 1 5 2 30

n51

49. 4 1 7 1 10 1 c 1 151 13 1 1 1 51. 6 2 6 2 2 2 c 2 2

44. a 1 3n 2 10 2 17

n50

45. a 0.5j

48. 1 2 3 2 7 2 c 2 75

14

j53

46. a j51 4 33

j

50. 2 1 0 2 2 2 c 2 56

7 17 11 14 52. 12 1 6 1 12 1 c 1 3

In Exercises 53–58, find the partial sum of the arithmetic series. 53. The first 18 terms of 1 1 5 1 9 1 13 1 c. 54. The first 21 terms of 2 1 5 1 8 1 11 1 c. 3 3 1 1 55. The first 43 terms of 1 1 2 1 0 2 2 2 c. 56. The first 37 terms of 3 1 2 1 0 2 2 2 c. 57. The first 18 terms of 29 1 1 1 11 1 21 1 31 1 c. 58. The first 21 terms of 22 1 8 1 18 1 28 1 c.

• A P P L I C AT I O N S 59.  Comparing Salaries. Colin and Camden are twin brothers

graduating with B.S. degrees in biology. Colin takes a job at the San Diego Zoo making $28,000 for his first year with a $1500 raise per year every year after that. Camden accepts a job at Florida Fish and Wildlife making $25,000 with a guaranteed $2000 raise per year. How much will each of the brothers have made in a total of 10 years? 60.  Comparing Salaries. On graduating with a Ph.D. in optical sciences, Jasmine and Megan choose different career paths. Jasmine accepts a faculty position at the University of Arizona making $80,000 with a guaranteed $2000 raise every year. Megan takes a job with the Boeing Corporation making $90,000 with a guaranteed $5000 raise each year. Calculate how many total dollars each woman will have made after 15 years. 61.  Theater Seating. You walk into the premiere of Chris Pine’s new movie, and the theater is packed, with almost every seat filled. You want to estimate the number of people in the theater. You quickly count to find that there are 22 seats in the front row, and there are 25 rows in the theater. Each row appears to have 1 more seat than the row in front of it. How many seats are in that theater? 62.  Field of Tulips. Every spring the Skagit County Tulip Festival plants more than 100,000 bulbs. In honor of the Tri-Delta sorority that has sent 120 sisters from the University of Washington to ­volunteer for the festival, Skagit County has

Young_AT_6160_ch12_pp1096-1132.indd 1115

planted tulips in the shape of DDD. In each of the triangles there are 20 rows of tulips, each row having one less than the row before. How many tulips are planted in each delta if there is 1 tulip in the first row? 63.  World’s Largest Champagne Fountain. From December 28 to 30, 1999, Luuk Broos, director of Maison Luuk-Chalet Fontaine, constructed a 56-story champagne fountain at the Steigenberger Kurhaus Hotel, Scheveningen, Netherlands. The fountain consisted of 30,856 champagne glasses. Assuming there was one glass at the top and the number of glasses in each row forms an arithmetic sequence, how many were on the bottom row (story)? How many fewer glasses did each successive row (story) have? Assume each story is one row. 64.  Stacking of Logs. If 25 logs are laid side by side on the ground, and 24 logs are placed on top of those, and 23 logs are placed on the 3rd row, and the pattern continues until there is a single log on the 25th row, how many logs are in the stack? 65.  Falling Object. When a skydiver jumps out of an airplane, she falls approximately 16 feet in the 1st second, 48 feet during the 2nd second, 80 feet during the 3rd second, 112 feet during the 4th second, and 144 feet during the 5th second, and this pattern continues. If she deploys her parachute after 10 seconds have elapsed, how far will she have fallen during those 10 seconds?

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CHAPTER 12  Sequences, Series, and Probability

66.  Falling Object. If a penny is dropped out of a plane, it falls

69.  Theater Seating. At a theater, seats are arranged in a

approximately 4.9 meters during the 1st second, 14.7 meters during the 2nd second, 24.5 meters during the 3rd second, and 34.3 meters during the 4th second. Assuming this pattern continues, how many meters will the penny have fallen after 10 seconds? 67.  Grocery Store. A grocer has a triangular display of oranges in a window. There are 20 oranges in the bottom row, and the number of oranges decreases by one in each row above this row. How many oranges are in the display? 68.  Salary. Suppose your salary is $45,000 and you receive a $1500 raise for each year you work for 35 years. a. How much will you earn during the 35th year? b. What is the total amount you earned over your 35-year career?

triangular pattern of rows with each succeeding row having one more seat than the previous row. You count the number of seats in the fourth row and determine that there are 26 seats. a. How many seats are in the first row? b. Now, suppose there are 30 rows of seats. How many total seats are there in the theater? 70.  Mathematics. Find the exact sum of 1 3 5 23 1 1 1 c1 e e e e

• C AT C H T H E M I S TA K E In Exercises 71–74, explain the mistake that is made. 71.  Find the general, or nth, term of the arithmetic sequence 3, 4, 5, 6, 7, . . . . Solution: The common difference of this sequence is 1. The first term is 3. The general term is an 5 a1 1 nd.

d51 a1 5 3 an 5 3 1 n

This is incorrect. What mistake was made?

72. Find the general, or nth, term of the arithmetic sequence

10, 8, 6, . . . . Solution: The common difference of this sequence is 2. The first term is 10. The general term is an 5 a1 1 1 n 2 1 2 d.

73. Find the sum a k50 2n 1 1. 10

Solution: n The sum is given by Sn 5 1 a1 1 an 2 , where n 5 10. 2 Identify the 1st and 10th terms. a1 5 1    a10 5 21 Substitute a1 5 1, a10 5 21, and n 5 10 into n 10 1 1 1 21 2 5 110 Sn 5 1 a1 1 an 2 . S10 5 2 2 This is incorrect. What mistake was made? 74. Find the sum 3 1 9 1 15 1 21 1 27 1 33 1 c 1 87.

Solution: d52 a1 5 10 an 5 10 1 2 1 n 2 1 2

This is incorrect. What mistake was made?

This is an arithmetic sequence with common difference of 6. The general term is given by an 5 a1 1 1 n 2 1 2 d. 87 is the 15th term of the series.

d56 an 5 3 1 1 n 2 1 2 6

a15 5 3 1 1 15 2 1 2 6 5 87

The sum of the series is n 15 1 87 2 3 2 5 630 Sn 5 1 an 2 a1 2 . S15 5 2 2 This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 75–78, determine whether each statement is true or false. 75.  An arithmetic sequence and a finite arithmetic series are the

77. An alternating sequence cannot be an arithmetic

same. 76.  The sum of all infinite and finite arithmetic series can always be found.

78.  The common difference of an arithmetic sequence is

Young_AT_6160_ch12_pp1096-1132.indd 1116

sequence. always positive.

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12.3  Geometric Sequences and Series 

1117

• CHALLENGE

79. Find the sum a 1 1 a 1 b 2 1 1 a 1 2b 2 1 c1 1 a 1 nb 2 . 80. Find the sum

30 a k5 229

k

ln e .

• TECHNOLOGY 83.  Use a graphing calculator “SUM” to sum the natural numbers

from 1 to 100. 84.  Use a graphing calculator to sum the even natural numbers from 1 to 100. 85.  Use a graphing calculator to sum the odd natural numbers from 1 to 100.

81. The wave number, v (reciprocal of wave length), of certain

light waves in the spectrum of light emitted by hydrogen is 1 1 given by v 5 R a 2 2 2 b, n . k, where R 5 109,678. A k n series of lines is given by holding k constant and varying the value of n. Suppose k 5 2 and n 5 3, 4, 5, c⋅ Find the limit of the wave number of the series. 82.  In a certain arithmetic sequence a1 5 24 and d 5 6. If Sn 5 570, find the value of n.

86.  Use a graphing calculator to sum a n51 122n 1 5 2 . Compare 30

it with your answer to Exercise 43.

87.  Use a graphing calculator to sum a n51 3259 1 5 1 n 2 1 2 4 . 100

88.  Use a graphing calculator to sum a n51 C218 1 5 1 n 2 1 2 D . 200

4

12.3 GEOMETRIC SEQUENCES AND SERIES SKILLS OBJECTIVES ■■ Determine if a sequence is geometric. ■■ Find the general, nth, term of a geometric sequence. ■■ Evaluate finite geometric series and infinite geometric series (if possible).

CONCEPTUAL OBJECTIVES ■■ Understand that the common ratio of a geometric sequence can be positive or negative. ■■ Derive the formula for the nth term of a geometric sequence in terms of n, the first term of the sequence, and the common ratio. ■■ Understand why it is not possible to evaluate all infinite geometric series.

12.3.1  Geometric Sequences In Section 12.2, we discussed arithmetic sequences, where successive terms had a common difference. In other words, each term was found by adding the same constant to the pre­vious term. In this section we discuss geometric sequences, where successive terms have  a common ratio. In other words, each term is found by multiplying the previous term by the same constant. The sequence 4, 12, 36, 108, c is geometric because each successive term is found by multiplying the previous term by 3.

DEFINITION

12.3.1 S K I L L

Determine if a sequence is geometric. 12.3.1 C O N C E P T U A L

Understand that the common ratio of a geometric sequence can be positive or negative.

Geometric Sequences

A sequence is geometric if each term in the sequence is found by multiplying the an11 previous term by a number r, so that an11 5 r ⋅ an. Because 5 r, the number an r is called the common ratio.

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CHAPTER 12  Sequences, Series, and Probability

[ CONCEPT CHECK ]

EXAMPLE 1  Identifying the Common Ratio in Geometric Sequences

TRUE OR FALSE  If the geometric sequence is alternating, then the common ratio is negative.

Find the common ratio for each of the geometric sequences. 1 1 1 a. 5, 20, 80, 320, c    b. 1, 22 , 4 , 28 , c    c. $5000, $5500, $6050, $6655, c

▼

Solution (a):

ANSWER True

Label the terms. Find the ratio r 5

an11 . an



a1 5 5, a2 5 20, a3 5 80, a4 5 320, c a2 20  r 5 5 54 a1 5 a3 80  r 5 5 54 a2 20 a4 320  r 5 5 54 a3 80 The common ratio is 4.

Solution (b):

Label the terms. Find the ratio r 5

an11 . an



1 1 1 a1 5 1, a2 5 2 , a3 5 , a4 5 2 , c 2 4 8 a2 21/2 1  r 5 5 52 a1 1 2 a3 1/4 1  r 5 5 52 a2 21/2 2 a4 21/8 1  r 5 5 52 a3 1/4 2 The common ratio is 2 21.

Solution (c):

a1 5 $5000, a2 5 $5500, a3 5 $6050, a4 5 $6655, c an11 a2 $5500 Find the ratio r 5 .   r5 5 5 1.1 an a1 $5000 a3 $6050   r5 5 5 1.1 a2 $5500 a4 $6655   r5 5 5 1.1 a3 $6050 Label the terms.



▼

▼ ANSWER 1 a. 23      b. 4

The common ratio is 1.1.

Y O U R T U R N   Find the common ratio of each geometric sequence. or 0.25

12.3.2 S K I L L

Find the general, nth, term of a geometric sequence. 12.3.2 C O N C E P T U A L

Derive the formula for the nth term of a geometric sequence in terms of n, the first term of the sequence, and the common ratio.

Young_AT_6160_ch12_pp1096-1132.indd 1118

a. 1, 23, 9, 227, c      b. 320, 80, 20, 5, c

12.3.2  The General (nth) Term of a Geometric Sequence To find a formula for the general, or nth, term of a geometric sequence, write out the first several terms and look for a pattern. First term, n 5 1. Second term, n 5 2. Third term, n 5 3. Fourth term, n 5 4.

a1 a2 5 a1 ⋅ r a3 5 a2 ⋅ r 5 1 a1 ⋅ r 2 ⋅ r 5 a1 ⋅ r 2 a4 5 a3 ⋅ r 5 1 a1 ⋅ r 2 2 ⋅ r 5 a1 ⋅ r 3

In general, the nth term is given by an 5 a1 ⋅ r n21.

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12.3  Geometric Sequences and Series 

1119

THE nTH TERM OF A GEOMETRIC SEQUENCE

The nth term of a geometric sequence with common ratio r is given by an 5 a1 ⋅ r n21 for n $ 1

EXAMPLE 2  Finding the nth Term of a Geometric Sequence

Find the 7th term of the sequence 2, 10, 50, 250, . . . . Solution:

Identify the common ratio. Identify the first 1 n 5 1 2 term.

10 50 250 5 5 55 2 10 50 a1 5 2 r5

Substitute a1 5 2 and r 5 5 into an 5 a1 ⋅ r n21.

an 5 2 ⋅ 5n21

Substitute n 5 7 into an 5 2 ⋅ 5n21.

a7 5 2 ⋅ 5721 5 2 ⋅ 56 5 31,250

The 7th term of the geometric sequence is 31,250. ▼ ANSWER

▼ Y O U R T U R N   Find the 8th term of the sequence 3, 12, 48, 192, . . . .

49,152

[ CONCEPT CHECK ] EXAMPLE 3  Finding the Geometric Sequence

For the geometric sequence 3, 12, 48, 192, . . . find the general nth, term of the sequence.

Find the geometric sequence whose 5th term is 0.01 and whose common ratio is 0.1.

▼ ANSWER an 5 3 3 4(n21)

Solution:

Label the common ratio and 5th term. Substitute a5 5 0.01, n 5 5, and r 5 0.1 into an 5 a1 ⋅ r n21. Solve for a1.

a5 5 0.01 and r 5 0.1 0.01 5 a1 ⋅ 1 0.1 2 521 0.01 0.01 a1 5 5 5 100 1 0.1 2 4 0.0001

The geometric sequence that starts at 100 and has a common ratio of 0.1 is 100, 10, 1, 0.1, 0.01, . . . .

▼ Y O U R T U R N   Find the geometric sequence whose 4th term is 3 and whose

common ratio is 13.

▼ ANSWER

81, 27, 9, 3, 1, . . .

12.3.3 S K I L L

12.3.3  Geometric Series The sum of the terms of a geometric sequence is called a geometric series. a1 1 a1 ⋅ r 1 a1 ⋅ r 2 1 a1 ⋅ r 3 1 c If we only sum the first n terms of a geometric sequence, the result is a finite geometric series given by Sn 5 a1 1 a1 ⋅ r 1 a1 ⋅ r 2 1 a1 ⋅ r 3 1 c 1 a1 ⋅ r n21

Young_AT_6160_ch12_pp1096-1132.indd 1119

Evaluate finite geometric series and infinite geometric series (if possible). 12.3.3 C O N C E P T U A L

Understand why it is not possible to evaluate all infinite geometric series.

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CHAPTER 12  Sequences, Series, and Probability

To develop a formula for the nth partial sum, we multiply the above equation by r: r ⋅ Sn 5 a1 ⋅ r 1 a1 ⋅ r 2 1 a1 ⋅ r 3 1 c 1 a1 ⋅ r n21 1 a1 ⋅ r n Subtracting the second equation from the first equation, we find that all of the terms on the right side drop out except the first term in the first equation and the last term in the second equation: Sn 5 a1 1 a1 ⋅ r 1 a1 ⋅ r 2 1 c1 a1r n21

2rSn 5



2 a1 ⋅ r 2 a1 ⋅ r 2 2 c2 a1r n21 2 a1r n 2a1r n

Sn 2 rSn 5 a1

Factor the Sn out of the left side and the a1 out of the right side: Sn 1 1 2 r 2 5 a1 1 1 2 r n 2

ST U DY TIP The underscript k 5 1 applies only when the summation starts at the a1 term. It is important to note which term is the starting term.

Divide both sides by 1 1 2 r 2 , assuming r 2 1. The result is a general formula for the sum of a finite geometric series: 11 2 r n2 Sn 5 a1 r21 11 2 r2 EVALUATING A FINITE GEOMETRIC SERIES

The sum of the first n terms of a geometric sequence, called a finite geometric series, is given by the formula 11 2 r n2       r 2 1 Sn 5 a1 11 2 r2

It is important to note that a finite geometric series can also be written in sigma (summation) notation: Sn 5 a a1 ⋅ r k21 5 a1 1 a1 ⋅ r 1 a1 ⋅ r 2 1 a1 ⋅ r 3 1 c1 a1 ⋅ r n21 n

k51

EXAMPLE 4  Evaluating a Finite Geometric Series

Evaluate the finite geometric series. a. a 3⋅ 1 0.4 2 k21 13

k51

b. The first nine terms of the series 1 1 2 1 4 1 8 1 16 1 32 1 64 1 c

Solution (a):

Identify a1, n, and r.  a1 5 3, n 5 13, and r 5 0.4 Substitute a1 5 3, n 5 13, and r 5 0.4 A1 2 0.413 B 11 2 r n2 into sn 5 a1 S13 5 3 . 11 2 r2 1 1 2 0.4 2 S13 < 4.99997

Simplify.

Solution (b):

Identify the first term and common ratio. Substitute a1 5 1 and r 5 2 into Sn 5 a1

 a1 5 1 and r 5 2

11 2 r n2 1 1 2 2n 2 . Sn 5 11 2 r2 11 2 22

A1 2 29 B 11 2 22 Simplify.  S9 5 511 To sum the first nine terms, let n 5 9.  S9 5

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12.3  Geometric Sequences and Series 

1121

The sum of an infinite geometric sequence is called an infinite geometric series. Some infinite geometric series converge (yield a finite sum), and some diverge (do not have a finite sum). For example, 1 1 1 1 1 1 1 1 1 1 1 c1 n 1 c 5 1 1converges2 2 4 8 16 32 2 n c 2 1 4 1 8 1 16 1 32 1 1 2 1 c 1diverges2 For infinite geometric series that converge, the partial sum Sn approaches a single number as n gets large. The formula used to evaluate a finite geometric series Sn 5 a1

11 2 r n2 11 2 r2

can be extended to an infinite geometric series for certain values of r. If 0 r 0 , 1, then when r is raised to a power, it continues to get smaller, approaching 0. For those values of r, the infinite geometric series converges to a finite sum. Let n S q; then a1

11 2 02 11 2 r n2 1 S a1 , if  0 r 0 , 1. 5 a1 11 2 r2 11 2 r2 12r

EVALUATING AN INFINITE GEOMETRIC SERIES

STUD Y T I P

The sum of an infinite geometric series is given by the formula

The formula used to evaluate an infinite geometric series is:

1 n a a1 ⋅ r 5 a1 1 1 2 r 2   0 r 0 , 1 n50 q

1 First term 2 1 2 1 Ratio 2

EXAMPLE 5  D  etermining Whether the Sum of an Infinite Series Exists

Determine whether the sum exists for each of the geometric series. 1 1 1 a. 3 1 15 1 75 1 375 1 c      b. 8 1 4 1 2 1 1 1 2 1 4 1 8 1 c Solution (a):

Identify the common ratio. Since 5 is greater than 1, the sum does not exist  . Solution (b):

Identify the common ratio. Since 12 is less than 1, the sum exists  .

r55 r55.1 1 2 1 r5 ,1 2 r5

▼ Y O U R T U R N  Determine whether the sum exists for the following geometric a. 81, 9, 1,

1 9,

series. c      b. 1, 5, 25, 125, c

▼ ANSWER a. yes      b. no

1 1 1 1 1 1 Do you expect 14 1 12 1 36 1 64 1 c and 14 2 12 1 36 2 64 1 c to sum to the same number? The answer is no, because the second series is an alternating series and terms are both added and subtracted. Hence, we would expect the second series to sum to a smaller number than the first series sums to.

Young_AT_6160_ch12_pp1096-1132.indd 1121

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1122 

CHAPTER 12  Sequences, Series, and Probability

[ CONCEPT CHECK ]

EXAMPLE 6  Evaluating an Infinite Geometric Series

TRUE OR FALSE  All infinite series with a negative common ratio converge.

Evaluate each infinite geometric series. 1 1 1 1 a. 1 1 3 1 9 1 27 1 c      b. 1 2 3 1

▼

Solution (a):

ANSWER False

ST U DY TIP

q First term 1 n a a1 ⋅ r 5 1 2 Ratio 5 1 2 1/3 n50

1 9

1 27

2

1c

Identify the first term and the common ratio.

a1 5 1    r 5

Since 0 r 0 5  13  , 1, the sum of the series exists.

1 3

1 1 q Substitute a1 5 1 and r 5 13 into a n50 a1 ⋅ r n 5 a1 . 11 2 r2 1 2 1/3 1 3 Simplify. 5 5 2/3 2 11

1 1 1 3 1 1 1 c5 3 9 27 2

Solution (b):

Identify the first term and the common ratio.

a1 5 1    r 5 2

Since 0 r 0 5 0 2 31 0 , 1, the sum of the series exists. Substitute a1 5 1 and r 5 2 31

1 3

into a n50 a1 ⋅ r n 5 a1 q

ST U DY TIP

Simplify.

q First term 1 n a a1 ⋅ r 5 1 2 Ratio 5 12 121/3 2 n50

12

1 . 11 2 r2 1 3 1 1 5 5 5 5 4/3 4 1 1 1 1/3 2 1 2 121/3 2

1 1 1 3 1 2 1 c5 3 9 27 4

Notice that the alternating series summed to 34, whereas the positive series summed to 32. ▼ ANSWER 3 3 a. 8       b. 16

▼ Y O U R T U R N   Find the sum of each infinite geometric series. 1 1 1 1 1 1 1 1 a. 4 1 12 1 36 1 108 1 c      b.  4 2 12 1 36 2 108 1 c

It is important to note the restriction on the common ratio r. The absolute value of the common ratio has to be strictly less than 1 for an infinite geometric series to converge. Otherwise the infinite geometric series diverges.

EXAMPLE 7  Evaluating an Infinite Geometric Series

Evaluate the infinite geometric series, if possible. a. a 2 a2 b       b.  a 3 ⋅ 1 2 2 n21 4

1

n

n50

Identify a1 and r.

n51

r 5 2 41

1 n 1 1 1 1 2 1 2c a2 b 522 1 2 a 4 2 8 32 128 n50 a1

e

Solution (a):

q

q

{

q

e

e

r 5 2 41

Young_AT_6160_ch12_pp1096-1132.indd 1122

r 5 2 41

07/12/16 11:14 AM

12.3  Geometric Sequences and Series 

a1 n a a1 ⋅ r 5 1 1 2 r 2 n50 q

Since 0 r 0 5 0 2 14 0 5 14 , 1, the infinite geometric series converges. Let a1 5 2 and r 5 2 41.

5

Simplify. 5

Solution (b):

r52

n21

5 3 1 6 1 12 1 24 1 48 1 c a1

e

a 3 ⋅ 122

n51

2 2 8 5 5 1 1 1/4 5/4 5

e

Identify a1 and r.

q

2 3 1 2 121/4 2 4

q 1 n 8 a 2 a2 4 b 5 5 n50

{

This infinite geometric series converges.

1123

r52

Since r 5 2 . 1, this infinite geometric series diverges.

Applications Suppose you are given a job offer with a guaranteed percentage raise per year. What will your annual salary be 10 years from now? That answer can be obtained using a geometric sequence. Suppose you want to make voluntary contributions to a retirement account directly debited from your paycheck every month. Suppose the account earns a fixed percentage rate: How much will you have in 30 years if you deposit $50 a month? What is the difference in the total you will have in 30 years if you deposit $100 a month instead? These important questions about your personal finances can be answered using geometric sequences and series. EXAMPLE 8  Future Salary: Geometric Sequence

Suppose you are offered a job as an event planner for the PGA Tour. The starting salary is $45,000, and employees are given a 5% raise per year. What will your annual salary be during the 10th year with the PGA Tour?

STUD Y T I P a10 5 45,000 1 1.05 2 9 < 69,809.77

Solution:

Every year the salary is 5% more than the previous year. a1 5 45,000 Label the year 1 salary. Calculate the year 2 salary. a2 5 1.05⋅ a1 Calculate the year 3 salary. a3 5 1.05⋅ a2 5 1.05 1 1.05⋅ a1 2 5 1 1.05 2 2a1 Calculate the year 4 salary. a4 5 1.05⋅ a3 5 1.05 1 1.05 2 2a1 5 1 1.05 2 3a1 Identify the year n salary. an 5 1.05n21a1 Substitute n 5 10 and a1 5 45,000. a10 5 1 1.05 2 9 ⋅ 45,000 Simplify. a10 < 69,809.77 During your 10th year with the company your salary will be $69,809.77.

▼ Y O U R T U R N  Suppose you are offered a job with AT&T at $37,000 per year

with a guaranteed raise of 4% after every year. What will your annual salary be after 15 years with the company?

Young_AT_6160_ch12_pp1096-1132.indd 1123

▼ ANSWER

$64,072.03

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1124 

CHAPTER 12  Sequences, Series, and Probability

EXAMPLE 9  Savings Growth: Geometric Series

Karen has maintained acrylic nails by paying for them with money earned from a part-time job. After hearing a lecture from her economics professor on the importance of investing early in life, she decides to remove the acrylic nails, which cost $50 per month, and do her own manicures. She has that $50 automatically debited from her checking account on the first of every month and put into a money market account that receives 3% interest compounded monthly. What will the balance be in the money market account exactly 2 years from the day of her initial $50 deposit? Solution:

r nt A 5 P a1 1 b n

Recall the compound interest formula. Substitute r 5 0.03 and n 5 12 into the compound interest formula.

A 5 P a1 1

0.03 12t b 12

5 P 1 1.0025 2 12t

n Let t 5 , where n is the number of months 12 of the investment:

An 5 P 1 1.0025 2 n

The first deposit of $50 will gain interest for 24 months. A24 5 50 1 1.0025 2 24 The second deposit of $50 will gain interest for A23 5 50 1 1.0025 2 23 23 months. The third deposit of $50 will gain interest for 22 months. A22 5 50 1 1.0025 2 22 The last deposit of $50 will gain interest for 1 month. Sum the amounts accrued from the 24 deposits.

A1 5 50 1 1.0025 2 1

A1 1 A2 1 c1 A24 5 50 1 1.0025 2 1 50 1 1.0025 2 2 1 50 1 1.0025 2 3 1 c1 50 1 1.0025 2 24

Identify the first term and common ratio.

Sum the first n terms of a geometric series. Substitute n 5 24, a1 5 50 1 1.0025 2 , and r 5 1.0025. Simplify.

a1 5 50 1 1.0025 2 and r 5 1.0025

Sn 5 a1

11 2 r n2 11 2 r2

S24 5 50 1 1.0025 2 S24 < 1238.23

A1 2 1.002524 B 1 1 2 1.0025 2

Karen will have $1238.23 saved in her money market account in 2 years. ▼ ANSWER

$5105.85

▼ Y O U R T U R N  Repeat Example 9 with Karen putting $100 (instead of $50) in the

same money market. Assume she does this for 4 years (instead of 2 years).

[ S E C T I O N 12 . 3 ]     S U M M A R Y In this section, we discussed geometric sequences, in which each successive term is found by multiplying the previous term by a constant, so that an11 5 r ⋅an. That constant, r, is called the common ratio. The nth term of a geometric sequence is given by an 5 a1r n21, n $ 1 or an11 5 a1r n, n $ 0. The sum of the terms of a geometric sequence is called a geometric series. Finite geometric series converge to a number. Infinite geometric series converge to a number if the absolute value of the common ratio is less than 1. If the absolute value of the common ratio is greater

Young_AT_6160_ch12_pp1096-1132.indd 1124

than or equal to 1, the infinite geometric series diverges and the sum does not exist. Many real-world applications involve geometric sequences and series, such as growth of salaries and annuities through percentage increases. n 11 2 r n2 k a r 5 a Finite Geometric Series: 1 1 a 11 2 r2 k50 q 1 Infinite Geometric Series:     a a1r k 5 a1       0 r 0 , 1 1 1 2 r2 k50

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12.3  Geometric Sequences and Series 

1125

[ S E C T I O N 12 . 3 ]   E X E R C I S E S • SKILLS In Exercises 1–8, determine whether the sequence is geometric. If it is, find the common ratio. 1. 1, 3, 9, 27, c

2. 2, 4, 8, 16, c

5. 8, 4, 2, 1, c

6. 8, 24, 2, 21, c

3. 1, 4, 9, 16, 25, c

4. 1, 4 , 9 , 16 , c 7. 800, 1360, 2312, 3930.4, c 8. 7, 15.4, 33.88, 74.536, c 1 1 1

In Exercises 9–16, write the first five terms of the geometric series. 9. a1 5 6    r 5 3

10. a1 5 17    r 5 2

11. a1 5 1    r 5 24

13. a1 5 10,000    r 5 1.06

14. a1 5 10,000    r 5 0.8

15. a1 5 3     r 5 2

2

12. a1 5 23    r 5 22

1

1

1

16. a1 5 10     r 5 25

In Exercises 17–24, write the formula for the nth term of the geometric series. 17. a1 5 5    r 5 2

18. a1 5 12    r 5 3

21. a1 5 1000    r 5 1.07

19. a1 5 1    r 5 23

22. a1 5 1000    r 5 0.5

23. a1 5

16 3     r

5

20. a1 5 24    r 5 22

2 41

1

24. a1 5 200     r 5 5

In Exercises 25–30, find the indicated term of the geometric sequence. 25. 7th term of the sequence 22, 4, 28, 16, c

26. 10th term of the sequence 1, 25, 25, 2225, c

27. 13th term of the sequence 3 , 3 , 3 , 3 , c

28. 9th term of the sequence 100, 20, 4, 0.8, c

1 2 4 8

30. 8th term of the sequence 1000, 2800, 640, 2512, c

29. 15th term of the sequence 1000, 50, 2.5, 0.125, c

In Exercises 31–40, find the sum of the finite geometric series. 1 2 22 c 212 1 1 1 1 3 3 3 3 c 33. 2 1 6 1 18 1 54 1 1 2A39 B

1 1 1 1 1 2 1 3 1 c1 10 3 3 3 3 34. 1 1 4 1 16 1 64 1 c1 49

31.

32. 1 1

36.  a 3 1 0.2 2 n

35.  a 2 1 0.1 2 n

37.  a 2 1 3 2 n21

11

10

n50

38.  a

8

n50

9

2 n21 152 n51 3

n51

39.  a 2k

40.  a a b

13

13

k50

k50

In Exercises 41–54, find the sum of the infinite geometric series, if possible. 41. a a b n50 2 q

42. a a b n51 3

n50

46. a 1.01n n50

q

q

q

1

n

45. a 1n

49. a 10,000 1 0.05 2 n n50

53. a 0.99n q

n50

q

1

n

q

50. a 200 1 0.04 2 n n50

54. a a b n50 4 q

5

43. a a2 b 3 q

1

n51

n

1 n 47. a 29 a b 3 n50 q

51. a 0.4n q

44. a a2 b 2 q

1

1 2

k

n

n50

48. a 28 a2 b 2 q

1

n

n50

52. 0.3 1 0.03 1 0.003 1 0.0003 1 c

n51

n

• A P P L I C AT I O N S 55.  Salary. Jeremy is offered a government job with the

Department of Commerce. He is hired on the “GS” scale at a base rate of $34,000 with 2.5% increases in his salary per year. Calculate what his salary will be after he has been with the Department of Commerce for 12 years. 56.  Salary. Alison is offered a job with a small start-up company that wants to promote loyalty to the company with incentives for employees to stay with the company. The company offers her a starting salary of $22,000 with a guaranteed 15% raise per year. What will her salary be after she has been with the company for 10 years? 57.  Depreciation. Brittany, a graduating senior in high school, receives a laptop computer as a graduation gift from her Aunt

Young_AT_6160_ch12_pp1096-1132.indd 1125

Jeanine so that she can use it when she gets to the University of Alabama. If the laptop costs $2000 new and ­depreciates 50% per year, write a formula for the value of the laptop n years after it was purchased. How much will the laptop be worth when Brittany graduates from college (assuming she will graduate in 4 years)? How much will it be worth when she finishes graduate school? Assume graduate school is another 3 years. 58.  Depreciation. Derek is deciding between a new Honda Accord and the BMW 325 series. The BMW costs $35,000 and the Honda costs $25,000. If the BMW depreciates at 20% per year and the Honda depreciates at 10% per year, find formulas for the value of each car n years after it is purchased. Which car is worth more in 10 years?

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CHAPTER 12  Sequences, Series, and Probability

59.  Bungee Jumping. A bungee jumper rebounds 70% of the

Laurence Fordyce/Eye Ubiquitous/Corbis Images

height jumped. Assuming the bungee jump is made with a cord that stretches to 100 feet, how far will the bungee jumper travel upward on the fifth rebound?

60.  Bungee Jumping. A bungee jumper rebounds 65% of the height

jumped. Assuming the bungee cord stretches 200 feet, how far will the bungee jumper travel upward on the eighth rebound? 61.  Population Growth. One of the fastest-growing universities in the country is the University of Central Florida. The student populations each year starting in 2000 were 36,000, 37,800, 39,690, 41,675, c. If this rate continued, how many students were at UCF in 2010? 62.  Website Hits. The website for a band has noticed that every week the number of hits to its website increases 5%. If there were 20,000 hits this week, how many will there be exactly 52 weeks from now? 63.  Rich Man’s Promise. A rich man promises that he will give you $1000 on January 1, and every day after that, he will pay you 90% of what he paid you the day before. How many days will it take before you are making less than $1? How much will the rich man pay out for the entire month of January? Round answers to the nearest dollar. 64.  Poor Man’s Clever Deal. A poor man promises to work for you for $0.01 the first day, $0.02 on the second day, $0.04 on the third day; his salary will continue to double each day. If he started on January 1, how much would he be paid to work

on January 31? How much total would he make during the month? Round answers to the nearest dollar. 65.  Investing Lunch. A newlywed couple decides to stop going out to lunch every day and instead brings their lunch. They estimate it will save them $100 per month. They invest that $100 on the first of every month into an account that is compounded monthly and pays 5% interest. How much will be in the account at the end of 3 years? 66.  Pizza as an Investment. A college freshman decides to stop ordering late-night pizzas (for both health and cost reasons). He realizes that he has been spending $50 a week on pizzas. Instead, he deposits $50 into an account that compounds weekly and pays 4% interest. How much money will be in the account in 52 weeks? 67.  Tax-Deferred Annuity. Dr. Schober contributes $500 from her paycheck (weekly) to a tax-deferred investment account. Assuming the investment earns 6% and is compounded weekly, how much will be in the account in 26 weeks? 52 weeks? 68.  Saving for a House. If a new graduate decides she wants to save for a house and she is able to put $300 every month into an account that earns 5% compounded monthly, how much will she have in the account in 5 years? 69.  House Values. In 2017, you buy a house for $240,000. The value of the house appreciates 3.1% per year, on the average. How much is the house worth after 15 years? 70.  The Bouncing Ball Problem. A ball is dropped from a height of 9 feet. Assume that on each bounce, the ball rebounds to one-third of its previous height. Find the total distance that the ball travels. 71.  Probability. A fair coin is tossed repeatedly. The probability that the first head noccurs on the nth toss is given by the function p 1 n 2 5 A 12 B , where n $ 1. Show that q 1 n a a 2 b 5 1.0 n51

72.  Salary. Suppose you work for a supervisor who gives you

two different options to choose from for your monthly pay. Option 1: The company pays you $0.01 for the first day of work, $0.02 for the second day, $0.04 for the third day, $0.08 for the fourth day, and so on for 30 days. Option 2: You can receive a check right now for $10 million. Which pay option is better? How much better is it?

• C AT C H T H E M I S TA K E In Exercises 73–76, explain the mistake that is made. 73. Find the nth term of the geometric sequence: 21,

1 3,

1 2 91, 27 ,

Solution: Identify the first term and common ratio. Substitute a1 5 21 and r 5 13 into an 5 a1 ⋅r n21.

74. Find the sum of the first n terms of the finite geometric series:

2, 4, 8, 16, c.

c.

a1 5 21 and r 5

1 3

1 n21 an 5 1 21 2 ⋅ a b 3 21 Simplify. an 5 n21 3 This is incorrect. What mistake was made?

Young_AT_6160_ch12_pp1096-1132.indd 1126

Solution: Write the sum in sigma notation. Identify the first term and common ratio. Substitute a1 5 1 and r 5 2 11 2 r n2 . into Sn 5 a1 11 2 r2

a 122 n

k

k51

a1 5 1 and r 5 2 Sn 5 1

1 1 2 2n 2 11 2 22

Simplify. Sn 5 2n 2 1 This is incorrect. What mistake was made?

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12.3  Geometric Sequences and Series 

75. Find the sum of the finite geometric series a 4 123 2 n. 8

n51

Solution: Identify the first term and common ratio.

a1 5 4 and r 5 23

Substitute a1 5 4 and r 5 23

76. Find the sum of the infinite geometric series a 2⋅3n21. q

n51

Solution: Identify the first term and common ratio.

a1 5 2 and r 5 3

Substitute a1 5 2 and r 5 3

11 2 r n2 into Sn 5 a1 . 11 2 r2

3 1 2 1 23 2 n 4 Sn 5 4 3 1 2 1 23 2 4

Simplify.

Sn 5 3 1 2 1 23 2 n 4



54

3 1 2 1 23 2 n 4 4

into Sq 5 a1

1 . 11 2 r2

Simplify.

Sq 5 2

1 11 2 32

Sq 5 21

This is incorrect. The series does not sum to 21. What mistake was made?

S8 5 3 1 2 1 23 2 8 4 5 2 6,560

Substitute n 5 8.

1127

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 77–80, determine whether each statement is true or false. 77. An alternating sequence cannot be a geometric sequence.

79. The common ratio of a geometric sequence can be

78. All finite and infinite geometric series can always be

positive or negative. 80. An infinite geometric series can be evaluated if the common ratio is less than or equal to 1.

evaluated.

• CHALLENGE 84. Suppose the sum of an infinite geometric series is

81. State the conditions for the sum

a 1 a⋅b 1 a⋅b 1 c1 a⋅b 1 c



to exist. Assuming those conditions are met, find the sum.



2



n

82. Find the sum of a k50 log 102 . 83. Represent the repeating decimal 0.474747. . . as a fraction 20

k



2 , where x is a variable. 12x a. Write out the first five terms of the series. b. For what values of x will the series converge? S5

(ratio of two integers).

• TECHNOLOGY

85. Sum the series: a k51 122 2 k21. Apply a graphing utility to confirm your answer. 50

86. Does the sum of the infinite series

graphing calculator to find it.

q 1 n a n50 A 3 B

exist? Use a

87. Apply a graphing utility to plot y1 5 1 1 x 1 x 2 1 x 3 1 x 4

1 , and let the range of x be 320.5, 0.5 4 . Based 12x on what you see, what do you expect the geometric series q n a n50 x to sum to? and y2 5

Young_AT_6160_ch12_pp1096-1132.indd 1127

88.  Apply a graphing utility to plot y1 5 1 2 x 1 x 2 2 x 3 1 x 4

1 , and let x range from 320.5, 0.5 4 . Based 11x on what you see, what do you expect the geometric series q n n a n50 121 2 x to sum to? and y2 5

89.  Apply a graphing utility to plot

1 , and 1 2 2x let x range from 320.3, 0.3 4 . Based on what you see, what do  y1 5 1 1 2x 1 4x 2 1 8x 3 1 16x 4 and y2 5

you expect the geometric series a n50 1 2x 2 n to sum to? q

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CHAPTER 12  Sequences, Series, and Probability

12.4 MATHEMATICAL INDUCTION SKILLS OBJECTIVE ■■ Prove mathematical statements using mathematical induction.

CONCEPTUAL OBJECTIVE ■■ Understand that just because there appears to be a pattern in a mathematical expression, the pattern is not necessarily true for all values.

12.4.1  Mathematical Induction n

n2 2 n 1 41

PRIME?

1

41

Yes

2

43

Yes

3

47

Yes

4

53

Yes

5

61

Yes

[ CONCEPT CHECK ] TRUE OR FALSE  If you can show that a statement is true for n 5 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10, then we can claim it is true for all n.

▼ ANSWER False

12.4.1 S K I L L

Prove mathematical statements using mathematical induction. 12.4.1 C O N C E P T U A L

Understand that just because there appears to be a pattern in a mathematical expression, the pattern is not necessarily true for all values.

Is the expression n2 2 n 1 41 always a prime number if n is a natural number? Your instinct may lead you to try a few values for n. It appears that the statement might be true for all natural numbers. However, what about when n 5 41? n2 2 n 1 41 5 1 41 2 2 2 41 1 41 5 412

We find that when n 5 41, n2 2 n 1 41 is not prime. The moral of the story is that just because a pattern seems to exist for some values, the pattern is not necessarily true for all values. We must look for a way to show whether a statement is true for all values. In this section we talk about mathematical induction, which is a way to show a statement is true for all values. Mathematics is based on logic and proof (not assumptions or belief). One of the most famous mathematical statements was Fermat’s last theorem. Pierre de Fermat (1601–1665) conjectured that there are no positive integer values for x, y, and z such that x n 1 y n 5 z n, if n $ 3. Although mathematicians believed that this theorem was true, no one was able to prove it until 350 years after the assumption was made. Professor Andrew Wiles at Princeton University received a $50,000 prize for successfully proving Fermat’s last theorem in 1994. Mathematical induction is a technique used in college algebra and even in very advanced mathematics to prove many kinds of mathematical statements. In this section, you will use it to prove statements like “if x . 1, then x n . 1 for all natural numbers n.” The principle of mathematical induction can be illustrated by a row of standing dominos, as in the picture. We make two assumptions: 1.  The first domino is knocked down. If a domino is knocked down, then the domino 2.  immediately following it will also be knocked down. If both of these assumptions are true, then it is also true that all of the dominos will fall.

PRINCIPLE OF MATHEMATICAL INDUCTION

Let Sn be a statement involving the positive integer n. To prove that Sn is true for all positive integers, the following steps are required. Step 1:  Show that S1 is true. Step 2: Assume Sk is true and show that Sk11 is true 1 k 5 positive integer 2 .

Combining Steps 1 and 2 proves the statement is true for all positive integers.

Young_AT_6160_ch12_pp1096-1132.indd 1128

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12.4  Mathematical Induction 

1129

EXAMPLE 1  Using Mathematical Induction

Apply the principle of mathematical induction to prove this statement: If x . 1, then x n . 1 for all natural numbers n. Solution: STEP 1  Show

x 1 . 1 because x . 1

the statement is true for n 5 1.

the statement is true for n 5 k. xk . 1 Show the statement is true for k 1 1.   Multiply both sides by x. x k ⋅ x . 1⋅ x   (Since x . 1, this step does not reverse the inequality sign.) STEP 2  Assume

Simplify. x k11 . x Recall that x . 1. x k11 . x . 1 Therefore, we have shown that x k11 . 1. This completes the induction proof. Thus, the following statement is true. “If x . 1, then x n . 1 for all natural numbers n.”

EXAMPLE 2  Using Mathematical Induction

Use mathematical induction to prove that n2 1 n is divisible by 2 for all natural numbers n. Solution: STEP 1  Show

the statement we are testing is true for n 5 1.



Show it is true for k 1 1 where k $ 1.

1k 1 122 1 1k 1 12 0 an integer 2

k 2 1 2k 1 1 1 k 1 1 0 an integer 2

Regroup terms.



k2 1 k 5 an integer 2

the statement is true for n 5 k.



2 51 2

2 is divisible by 2.

STEP 2  Assume



12 1 1 5 2

Ak 2 1 kB 1 2 1 k 1 1 2 2

Ak 2 1 kB 2

We assumed

k2 1 k 5 an integer. 2

Since k is a natural number,

1

0 an integer

21k 1 12 0 an integer 2

an integer 1 1 k 1 1 2 0 an integer

an integer 1 an integer 5 an integer

This completes the induction proof. The following statement is true: “n2 1 n is divisible by 2 for all natural numbers n.”

Young_AT_6160_ch12_pp1096-1132.indd 1129

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CHAPTER 12  Sequences, Series, and Probability

Mathematical induction is often used to prove formulas for partial sums.

EXAMPLE 3  P  roving a Partial-Sum Formula with Mathematical Induction

Apply mathematical induction to prove the following partial-sum formula: n1n 1 12 for all positive integers n 1 1 2 1 3 1 c1 n 5 2 Solution: STEP 1

Show the formula is true for n 5 1.

15

STEP 2

111 1 12 2 5 51 2 2

k1k 1 12 1 1 2 1 3 1 c1 k 5 2 1k 1 12 1k 1 22 1 1 2 1 3 1 c1 k 1 1 k 1 1 2 0 2

Assume the formula is true for n 5 k. Show it is true for n 5 k 1 1.

1k 1 12 1k 1 22 1 1 2 1 3 1 c1 k 1 1 k 1 1 2 0 2 k 1k 1 12 2



µ

1k 1 12 1k 1 22 k1k 1 12 1 1k 1 12 0 2 2



1k 1 12 1k 1 22 k1k 1 12 1 21k 1 12 0 2 2



1k 1 12 1k 1 22 k 2 1 3k 1 2 0 2 2



1k 1 12 1k 1 22 1k 1 12 1k 1 22 5 2 2

This completes the induction proof. The following statement is true:

n1n 1 12 “1 1 2 1 3 1 c1 n 5 for all positive integers n.” 2

[ S E C T I O N 12 . 4 ]     S U M M A R Y Just because we believe something is true does not mean that it is. In mathematics we rely on proof. In this section, we discussed mathematical induction, a process of proving mathematical statements. The two-step procedure for mathematical induction

Young_AT_6160_ch12_pp1096-1132.indd 1130

is to (1) show the statement is true for n 5 1, then (2) assume the statement is true for n 5 k and show the statement must be true for n 5 k 1 1. The combination of Steps 1 and 2 proves the statement.

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12.4  Mathematical Induction 

1131

[ S E C T I O N 12 . 4 ]   E X E R C I S E S • SKILLS In Exercises 1–24, prove the statements using mathematical induction for all positive integers n. 1. n2 # n3

2. If 0 , x , 1, then 0 , x n , 1.

3. 2n # 2n

4. 5n , 5n11

5. n! . 2n      n $ 4      (Show it is true for n 5 4, instead of n 5 1. 2 6. 1 1 1 c 2 n $ nc    c . 1

8. n3 2 n is divisible by 3.

7. n 1 n 1 1 2 1 n 2 1 2 is divisible by 3.

10. n 1 n 1 1 2 1 n 1 2 2 is divisible by 6.

11. 2 1 4 1 6 1 8 1 c1 2n 5 n 1 n 1 1 2

9. n2 1 3n is divisible by 2.

3n11 2 1 13. 1 1 3 1 32 1 33 1 c1 3n 5

12. 1 1 3 1 5 1 7 1 c1 1 2n 2 1 2 5 n2

2

14. 2 1 4 1 8 1 c1 2n 5 2n11 2 2 16. 13 1 23 1 33 1 c1 n3 5 18.

n2 1 n 1 1 2 2 4

n 1 1 1 5 1 1 c1 1 1n 1 12 1n 1 22 2⋅3 3⋅4 2 n 1 22

19. 1 1⋅2 2 1 1 2⋅3 2 1 1 3⋅4 2 1 c1 n 1 n 1 1 2 5

20. 1 1⋅3 2 1 1 2⋅4 2 1 1 3⋅5 2 1 c1 n 1 n 1 2 2 5

n 1 n 1 1 2 1 2n 1 1 2 15. 12 1 22 1 32 1 c1 n2 5 6 n 1 1 1 1 5 17. 1 1 1 c1 n11 1⋅2 2⋅3 3⋅4 n1n 1 12

n1n 1 12 1n 1 22 3

n 1 n 1 1 2 1 2n 1 7 2 6

1 2 xn 21. 1 1 x 1 x 2 1 x 3 1 c1 x n21 5      x 2 1 12x

22.

1 1 1 1 1 1 1 1 c1 n 5 1 2 n 2 4 8 2 2

n 23. The sum of an arithmetic sequence: a1 1 1 a1 1 d 2 1 1 a1 1 2d 2 1 c1 3 a1 1 1 n 2 1 2 d 4 5 3 2a1 1 1 n 2 1 2 d 4 . 2

n

12r 24. The sum of a geometric sequence: a1 1 a1r 1 a1r 2 1 c1 a1r n21 5 a1 a b. 12r

• A P P L I C AT I O N S

Andy Washnik

The Tower of Hanoi. This is a game with three pegs and n disks (largest on the bottom and smallest on the top). The goal is to move this entire tower of disks to another peg (in the same order). The challenge is that you may move only one disk at a time, and at no time can a larger disk be resting on a smaller disk. You may want to first go online to www.mazeworks.com/hanoi/index/htm and play the game.

Tower of Hanoi

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CHAPTER 12  Sequences, Series, and Probability

25. What is the smallest number of moves needed if there are

three disks? 26. What is the smallest number of moves needed if there are

four disks? 27. What is the smallest number of moves needed if there are

five disks? 28. What is the smallest number of moves needed if there are n

disks? Prove it by mathematical induction. 29.  Telephone Infrastructure. Suppose there are n cities that are

to be connected with telephone wires. Apply mathematical

induction to prove that the number of telephone wires n1n 2 12 . Assume required to connect the n cities is given by 2 each city has to connect directly with any other city. 30.  Geometry. Prove, with mathematical induction, that the sum of the interior angles of a regular polygon of n sides is given by the formula: 1 n 2 2 2 1 180 2 for n $ 3. Hint: Divide a polygon into triangles. For example, a four-sided polygon can be divided into two triangles. A five-sided polygon can be divided into three triangles. A six-sided polygon can be divided into four triangles, and so on.

• CONCEPTUAL In Exercises 31 and 32, determine whether each statement is true or false. 31. Assume Sk is true. If it can be shown that Sk11 is true, then

Sn is true for all n, where n is any positive integer.

32. Assume S1 is true. If it can be shown that S2 and S3 are

true, then Sn is true for all n, where n is any positive integer.

• CHALLENGE 33. Apply mathematical induction to prove: 4 ak 5 n

k51

n 1 n 1 1 2 1 2n 1 1 2 A3n2 1 3n 2 1B

5 ak 5

k51

1 1 1 1 a1 1 b a1 1 b a1 1 b c a1 1 b 5 n 1 1 n 1 2 3

30

34. Apply mathematical induction to prove: n

35. Apply mathematical induction to prove:

n 1 n 1 1 2 2 A2n2 1 2n 2 1B 2

36. Apply mathematical induction to prove that x 1 y is a factor

of x 2n 2 y 2n. 37. Apply mathematical induction to prove:

12

ln 1 c1 ⋅c2 ⋅c3 c cn 2 5 ln c1 1 ln c2 1 c 1 ln cn

• TECHNOLOGY 38. Use a graphing calculator to sum the series



11 • 22 1 12 • 32 1 13 • 42 1 ? ? ? 1 n 1n 1 12 on the left side, and n1n 1 12 1n 1 22 evaluate the expression on the right side for 3  n 5 200. Do they agree with each other? Do your answers confirm the proof for Exercise 19?

Young_AT_6160_ch12_pp1096-1132.indd 1132

39. Use a graphing calculator to sum the series

1 1 1 1 1 1 1 ? ? ? 1 n on the left side, and evaluate the 2 4 8 2 1 expression 1 2 n on the right side for n 5 8. Do they agree 2 with each other? Do your answers confirm the proof for Exercise 22?

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12.5  The Binomial Theorem 

1133

12.5 THE BINOMIAL THEOREM SKILLS OBJECTIVES ■■ Evaluate a binomial coefficient. ■■ Use the binomial theorem to expand a binomial raised to a positive integer power. ■■ Evaluate a binomial coefficient using Pascal’s triangle. ■■ Find a particular term of a binomial expansion.

CONCEPTUAL OBJECTIVES ■■ Understand why n has to be greater than or equal to k in order for “n choose k” to be defined. ■■ Recognize patterns in binomial expansions that lead to the binomial theorem. ■■ Understand why the nth row of Pascal’s triangle has n 1 1 coefficients. ■■ Understand that it is not necessary to perform an entire binomial expansion if you only are looking for a single term.

12.5.1  Binomial Coefficients A binomial is a polynomial that has two terms. The following are all examples of binomials:

12.5.1 S K I L L

Evaluate a binomial coefficient.

x 2 1 2y   a 1 3b  4x 2 1 9 In this section, we will develop a formula for raising a binomial to a power n, where n is a positive integer. 1 x 2 1 2y 2 6   1 a 1 3b 2 4   1 4x 2 1 9 2 5

12.5.1 C O N C E P T U A L

Understand why n has to be greater than or equal to k in order for “n choose k” to be defined.

To begin, let’s start by writing out the expansions of 1 a 1 b 2 n for several values of n. 1a 1 b20 5 1

1a 1 b21 5 a 1 b

1 a 1 b 2 2 5 a2 1 2ab 1 b2

1 a 1 b 2 3 5 a3 1 3a2b 1 3ab2 1 b3

1 a 1 b 2 4 5 a4 1 4a3b 1 6a2b2 1 4ab3 1 b4

1 a 1 b 2 5 5 a5 1 5a4b 1 10a3b2 1 10a2b3 1 5ab4 1 b5

All of the binomial expansions have several patterns. 1. The number of terms in each resulting polynomial is always one more than the power of the binomial n. Thus, there are n 1 1 terms in each expansion. 1 a 1 b 2 3 5 a3 1 3a2b 1 3ab2 1 b3

f

n 5 3:

four terms

2. Each expansion has symmetry. For example, a and b can be interchanged, and you will arrive at the same expansion. Furthermore, the powers of a decrease by 1 in each successive term, and the powers of b increase by 1 in each successive term. 1 a 1 b 2 3 5 a3b0 1 3a2b1 1 3a1b2 1 a0b3

3.  The sum of the powers of each term in the expansion is n. f

11253 01353 f

n 5 3:

21153 f

f

31053

1 a 1 b 2 3 5 a3b0 1 3a2b1 1 3a1b2 1 a0b3

4.  The coefficients increase and decrease in a symmetric manner. 1 a 1 b 2 5 5 1a5 1 5a4b 1 10a3b2 1 10a2b3 1 5ab4 1 1b5

Using these patterns, we can develop a generalized formula for 1 a 1 b 2 n. 1a 1 b2 n 5  an 1  an21b 1  an22b2 1 . . . 1  a2bn22 1  abn21 1 Where

is a placeholder for a constant to be determined.

Young_AT_6160_ch12_pp1133-1150.indd 1133

 bn

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CHAPTER 12  Sequences, Series, and Probability

We know that there are n 1 1 terms in the expansion. We also know that the sum of the powers of each term must equal n. The powers increase and decrease by 1 in each successive term, and if we interchanged a and b, the result would be the same expansion. The question that remains is, what coefficients go in the blanks? We know that the coefficients must increase and then decrease in a symmetric order (similar to walking up and then down a hill). It turns out that the binomial coefficients are represented by a symbol that we will now define. Binomial Coefficients

DEFINITION

n For nonnegative integers n and k, where n $ k, the symbol a b is called the k binomial coefficient and is defined by n n! a b 5 1 n 2 k 2 !k! k

n a b is read “n choose k.” k

You will see in the following sections that “n choose k” comes from combinations.

[ CONCEPT CHECK ] TRUE OR FALSE ”n choose k” is always greater than or equal to 1 when n is greater than or equal to k.

▼ ANSWER True

EXAMPLE 1 

Evaluating a Binomial Coefficient

Evaluate the following binomial coefficients. 6 4

5 5

4 0

a.  a b   b.  a b   c.  a b   d.  a

Solution:

10 b 9

Select the top number as n and the bottom number as k and substitute into the n n! binomial coefficient formula a b 5 . 1 k n 2 k 2 !k! 6 4

a. a b 5

5 5

b. a b 5

4 0

c. a b 5 d. a ▼ ANSWER a. 84  b. 28

▼

6! 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2⋅ 1 6⋅5 6! 5 5 15 5 5 1 2 ⋅ 1 2 1 4 ⋅ 3⋅ 2 ⋅ 1 2 1 6 2 4 2 !4! 2 2!4! 5! 1 1 5! 5 5 5 5 1 1 5 2 5 2 !5! 0!5! 0! 1 4! 1 4! 5 5 5 1 1 4 2 0 2 !0! 4!0! 0!

10! 10 ⋅ 9! 10 10! 5 5 5 10 b 5 1 10 2 9 2 !9! 1!9! 9! 9

Y O U R T U R N   Evaluate the following binomial coefficients.

9 6

8 6

a.  a b   b.  a b

Parts (b) and (c) of Example 1 lead to the general formulas: n a b 51 n

Young_AT_6160_ch12_pp1133-1150.indd 1134

and

n a b 51 0

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12.5  The Binomial Theorem 

1135

12.5.2  Binomial Expansion Let’s return to the question of the binomial expansion and how to determine the coefficients: n

1a 1 b2  5

n

 a 1

n21

 a

b1

b 1…1

n22 2

 a

2 n22

 a b

1

n21

 ab

1

n

 b

n The symbol a b is called a binomial coefficient because the coefficients in the blanks k in the binomial expansion are equivalent to this symbol.

12.5.2 S K I L L

Use the binomial theorem to expand a binomial raised to a positive integer power. 12.5.2 C O N C E P T U A L

Recognize patterns in binomial expansions that lead to the binomial theorem.

THE BINOMIAL THEOREM

Let a and b be real numbers; then for any positive integer n, n n n 1 a 1 b 2 n 5 a b an 1 a b an21b 1 a b an22b2 1 0 1 2

or in sigma (summation) notation as

P

1 a

n n n b a2bn22 1 a b abn21 1 a b bn n2 2 n21 n

n n 1 a 1 b 2 n 5 a a b an2kbk k50 k

EXAMPLE 2  Applying the Binomial Theorem

Expand 1 x 1 2 2 3 with the binomial theorem. Solution:

Substitute a 5 x, b 5 2, n 5 3 into the equation of the binomial theorem.

3 3 1 x 1 2 2 3 5 a a b x 32k2k k50 k

Find the binomial coefficients.

3 3 3 3 5 a b x 3 1 a b x 2 ⋅ 2 1 a b x ⋅ 22 1 a b 23 0 1 2 3

Simplify.

5 x 3 1 6x 2 1 12x 1 8

Expand the summation.

5 x 3 1 3x 2 ⋅ 2 1 3x ⋅ 22 1 23

▼

Y O U R T U R N   Expand 1 x 1 5 2 4 with the binomial theorem.

EXAMPLE 3 

Applying the Binomial Theorem

Expand 1 2x 2 3 2 4 with the binomial theorem. Solution:

Substitute a 5 2x, b 5 23, n 5 4 into the equation of the binomial theorem. Expand the summation.

4 4 1 2x 2 3 2 4 5 a a b 1 2x 2 42k 123 2 k k50 k

▼ ANSWER

x 4 1 20x 3 1 150x 2 1 500x 1 625

[ CONCEPT CHECK ] In the binomial expansion of (3x 2 2)4, what is the first term and last term?

▼ ANSWER 81x 4; 16

4 4 4 4 4 5 a b 1 2x 2 4 1 a b 1 2x 2 3 123 2 1 a b 1 2x 2 2 123 2 2 1 a b 1 2x 2 123 2 3 1 a b 123 2 4 0 1 2 3 4

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CHAPTER 12  Sequences, Series, and Probability

Find the binomial coefficients.

5 1 2x 2 4 1 4 1 2x 2 3 123 2 1 6 1 2x 2 2 123 2 2 1 4 1 2x 2 123 2 3 1 123 2 4 5 16x 4 2 96x 3 1 216x 2 2 216x 1 81

Simplify. ▼ ANSWER

81x 4 2 216x 3 1 216x 2 2 96x 1 16

▼

Y O U R T U R N   Expand 1 3x 2 2 2 4 with the binomial theorem.

12.5.3  Pascal’s Triangle 12.5.3 S K I L L

Evaluate a binomial coefficient using Pascal’s triangle. 12.5.3 C O N C E P T U A L

Understand why the nth row of Pascal’s triangle has n 1 1 coefficients.

1 1    1 1    2    1 1    3    3    1 1    4    6    4    1 1    5    10    10    5    1

[ CONCEPT CHECK ] The top row of Pascal’s triangle corresponds to (a 1 b)0 (zero power or n 5 0 row). The next row of Pascal’s triangle corresponds to (a 1 b)1 (1st power or n 5 1 row). The next row of Pascal's triangle corresponds to (a 1 b)2 (2nd power or n 5 2 row) which is equal to a2 1 2ab 1 b2. So the n 5 0 row has one term, the n 5 1 row has two terms, and the n 5 2 row has three terms. How many terms does the nth row of Pascal's triangle have?

▼ ANSWER n 1 1

Young_AT_6160_ch12_pp1133-1150.indd 1136

Instead of writing out the binomial theorem and calculating the binomial coefficients using factorials every time you want to do a binomial expansion, we now present an alternative, more convenient way of remembering the binomial coefficients, called Pascal’s triangle. Notice that the first and last number in every row is 1. Each of the other numbers is found by adding the two numbers directly above it. For example, 3 5 2 1 1  4 5 1 1 3  10 5 6 1 4 n Let’s arrange values of a b in a triangular pattern. k n 5 0 row n 5 1 row n 5 2 row n 5 3 row n 5 4 row n 5 5 row n 5 6 row

5 a b 0

4 a b 0

3 a b 0

5 a b 1

2 a b 0

4 a b 1

1 a b 0

3 a b 1

5 a b 2

0 a b 0

2 a b 1

4 a b 2

1 a b 1

3 a b 2

5 a b 3

2 a b 2

4 a b 3

3 a b 3

5 a b 4

4 a b 4

5 a b 5

6 6 6 6 6 6 6 a b         a b         a b         a b         a b         a b         a b 0 1 2 3 4 5 6

It turns out that these numbers in Pascal’s triangle are exactly the coefficients in a binomial expansion. n 5 0 row

1

n 5 1 row

1a 1 1b

n 5 2 row

1a2 1 2ab 1 1b2

n 5 3 row

1a3 1 3a2b 1 3ab2 1 1b3

n 5 4 row

1a4 1 4a3b 1 6a2b2 1 4ab3 1 1b4

n 5 5 row

1a5 1 5a4b 1 10a3b2 1 10a2b3 1 5ab4 1 1b5

n 5 6 row

?

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12.5  The Binomial Theorem 

The top row is called the zero row because it corresponds to the binomial raised to the zero power, n 5 0. Since each row in Pascal’s triangle starts and ends with a 1 and all other values are found by adding the two numbers directly above it, we can now easily calculate the sixth row.   1n 5 5 row2

1137

STUDY T I P Since the top row of Pascal’s triangle is called the zero row, the fifth row is the row with 6 coefficients. The nth row is the row with n 1 1 coefficients.

  1n 5 6 row2 EXAMPLE 4  Applying Pascal’s Triangle in a Binomial Expansion

Use Pascal’s triangle to determine the binomial expansion of 1 x 1 2 2 5. Solution:

Write the binomial expansion with blanks for coefficients. 1x 1 22 5 5

 x5 1

 x4 ⋅ 2 1

 x3 ⋅ 22 1

 x2 ⋅ 23 1

 x ⋅ 24 1

Write the binomial coefficients in the fifth row of Pascal’s triangle.

 25

1, 5, 10, 10, 5, 1 Substitute these coefficients into the blanks of the binomial expansion. 1 x 1 2 2 5 5 1x 5 1 5x 4 ⋅ 2 1 10x 3 ⋅ 22 1 10x 2 ⋅ 23 1 5x ⋅ 24 1 1 ⋅ 25

Simplify.   1 x 1 2 2 5 5 x 5 1 10x 4 1 40x 3 1 80x 2 1 80x 1 32

▼

Y O U R T U R N   Apply Pascal’s triangle to determine the binomial expansion of

1 x 1 3 2 4.

▼ ANSWER

x 4 1 12x 3 1 54x 2 1 108x 1 81

EXAMPLE 5  Applying Pascal’s Triangle in a Binomial Expansion

Use Pascal’s triangle to determine the binomial expansion of 1 2x 1 5 2 4. Solution:

Write the binomial expansion with blanks for coefficients. 12x 1 52 4 5

 12x2 4 1

 12x2 3 ⋅ 5 1

 12x2 2 ⋅ 52 1

 12x2 ⋅ 53 1

Write the binomial coefficients in the fourth row of Pascal’s triangle.

 54

1, 4, 6, 4, 1 Substitute these coefficients into the blanks of the binomial expansion. 1 2x 1 5 2 4 5 1 1 2x 2 4 1 4 1 2x 2 3 ⋅ 5 1 6 1 2x 2 2 ⋅ 52 1 4 1 2x 2 ⋅ 53 1 1⋅ 54

Simplify.   1 2x 1 5 2 4 5 16x 4 1 160x 3 1 600x 2 1 1000x 1 625

▼ ANSWER a.  27x 3 1 54x 2 1 36x 1 8 b.  243x 5 2 810x 4 1 1080x 3 2 720x 2 1 240x 2 32

▼

Y O U R T U R N   Use Pascal’s triangle to determine the binomial expansion of: a.  1 3x 1 2 2 3  b.  1 3x 2 2 2 5

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CHAPTER 12   Sequences, Series, and Probability

12.5.4  Finding a Particular Term of a Binomial Expansion What if we don’t want to find the entire expansion, but instead want just a single term? For example, what is the fourth term of 1 a 1 b 2 5?

WORDS MATH

Recall the sigma notation.

n n 1 a 1 b 2 n 5 a a b an2k bk k50 k 5 5 1 a 1 b 2 5 5 a a b a52k bk k50 k

Let n 5 5.

5 5 5 5 5 5 1 a 1 b 2 5 5 a b a5 1 a b a4b 1 a b a3b2 1 a b a2b3 1 a b ab4 1 a b b5 0 1 2 3 4 5

Expand.

f

10a2b3

Simplify the fourth term.

fourth term

12.5.4 S K I L L

Find a particular term of a binomial expansion.

FINDING A PARTICULAR TERM OF A BINOMIAL EXPANSION

n The 1 r 1 1 2 term of the expansion 1 a 1 b 2 n is a b an2rbr. r

12.5.4 C O N C E P T U A L

Understand that it is not necessary to perform an entire binomial expansion if you only are looking for a single term.

EXAMPLE 6  Finding a Particular Term of a Binomial Expansion

Find the 5th term of the binomial expansion of 1 2x 2 7 2 6. Solution:

[ CONCEPT CHECK ] TRUE OR FALSE  To find a particular term of a binomial expansion, you must always first perform the entire binomial expansion.

n Recall that the r 1 1 term of 1 a 1 b 2 n is a b an2rbr. r For the 5th term, let r 5 4.

For this expansion, let a 5 2x, b 5 27, n 5 6.

▼ ANSWER False

6 Note that a b 5 15. 4

Simplify.

n a b an24b4 4

6 a b 1 2x 2 624 127 2 4 4 15 1 2x 2 2 127 2 4

144,060x 2

▼

Y O U R T U R N   What is the third term of the binomial expansion of 1 3x 2 2 2 5?

▼ ANSWER

1080x 3

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12.5  The Binomial Theorem 

1139

[ S E C T I O N 12 . 5 ]    S U M M A R Y In this section, we developed a formula for raising a binomial expression to an integer power, n $ 0. The patterns that surfaced were that the expansion displays symmetry between the two terms; that is, every expansion has n 1 1 terms, the powers sum to n, and the coefficients, called binomial coefficients, are ratios of factorials:

are that every row begins and ends with 1 and all other numbers are found by adding the two numbers above the entry. 1 1    1 1    2    1 1    3    3    1 1    4    6    4    1 1    5    10    10    5    1

n n 1 a 1 b 2 n 5 a a ban2kbk k50 k

n n! a b5 1 n 2 k 2 ! k! k



Also, Pascal’s triangle, a shortcut method for evaluating the binomial coefficients, was discussed. The patterns in the triangle

Last, a formula was given for finding a particular term of a binomial n expansion; the 1 r 1 1 2 term of 1 a 1 b 2 n is a b an2rbr. r

[ S E C T I O N 12 . 5 ]   E X E R C I S E S • SKILLS In Exercises 1–10, evaluate the binomial coefficients. 7 3

8 2

1. a b 5. a 9. a

2. a b

17 b 0

6. a

48 b 45

10. a

3. a

100 b 0

7. a

29 b 26

10 b 8

4. a

99 b 99

8. a

23 b 21 52 b 52

In Exercises 11–32, expand the expression using the binomial theorem. 11. 1 x 1 2 2 4

12. 1 x 1 3 2 5

13. 1 y 2 3 2 5

19. 1 5x 2 2 2 3

20. 1 a 2 7b 2 3

21. a

23. 1 x 2 1 y 2 2 4 27. A !x 1 2B

17. 1 x 1 3y 2 3

16. 1 x 2 y 2 6

15. 1 x 1 y 2 5

24. 1 r 3 2 s3 2 3

6

31. Ax 1/4 1 2!yB

28. A3 1 !yB

4

14. 1 y 2 4 2 4

4 1 1 5yb x 25. 1 ax 1 by 2 5

3 4 b y 26. 1 ax 2 by 2 5 22. a2x 1

4 29. 1 a3/4 1 b1/4 2

4

32. A !x 2 3y 1/4 B

18. 1 2x 2 y 2 3

8

3 30. 1 x 2/3 1 y 1/3 2

In Exercises 33–36, expand the expression using Pascal’s triangle. 7 34. 1 x 2 1 y 2 2

33. 1 r 2 s 2 4

36. 1 x 1 3y 2 4

35. 1 ax 1 by 2 6

In Exercises 37–44, find the coefficient C of the term in the binomial expansion. Binomial 37. 1 x 1 2 2 10

41. 1 2x 1 3y 2 7

Term Cx

6

Cx 3y 4

• A P P L I C AT I O N S

Binomial 38. 1 3 1 y 2 9

42. 1 3x 2 5y 2 9

Term Cy

5

Cx 2y 7

Binomial 39. 1 y 2 3 2 8

8 43. 1 x 2 1 y 2

Term Cy

4

Cx 8y 4

Binomial 40. 1 x 2 1 2 12

10 44. 1 r 2 s2 2

Term Cx 5

Cr 6s8

In later sections, you will learn the “n choose k” notation for combinations. 45. Lottery. In a state lottery in which six numbers are drawn 46. Lottery. In a state lottery in which six numbers are drawn from a possible 40 numbers, the number of possible sixfrom a possible 60 numbers, the number of possible six40 60 number combinations is equal to a b. How many possible number combinations is equal to a b . How many possible 6 6 combinations are there? combinations are there?

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CHAPTER 12  Sequences, Series, and Probability

47. Poker. With a deck of 52 cards, 5 cards are dealt in a game of

48. Canasta. In the card game canasta, two decks of cards

52 poker. There are a total of a b different 5-card poker hands 5 that can be dealt. How many possible hands are there?

including the jokers are used and 11 cards are dealt to each 108 person. There are a total of a b different 11-card canasta 11 hands that can be dealt. How many possible hands are there?

• C AT C H T H E M I S TA K E In Exercises 49 and 50, explain the mistake that is made. 7 5

49. Evaluate the expression a b .

Solution:

Solution:

Write out the binomial coefficient in terms of factorials.

7 7! a b5 5 5!

7 7! 7⋅6⋅5⋅4⋅3⋅2⋅1 5 a b5 5 5! 5⋅4⋅3⋅2⋅1

Write out the factorials.

7 7! 7⋅6 5 5 42 a b5 5 5! 1

Simplify.

50. Expand 1 x 1 2y 2 4.



Write out with blanks. 1x 1 2y2 4 5

x4 1

x3y 1

x2y2 1

x y3 1

y4

Write out the terms from the fifth row of Pascal’s triangle. 1, 4, 6, 4, 1



Substitute these coefficients into the binomial expansion.



1 x 1 2y 2 4 5 x 4 1 4x 3y 1 6x 2y 2 1 4xy 3 1 y 4

This is incorrect. What mistake was made?

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 51–54, determine whether each statement is true or false. 51. The binomial expansion of 1 x 1 y 2 10 has 10 terms.

52. The binomial expansion of 1 x 2 1 y 2 2 15 has 16 terms.

n n

n 2n

53. a b 5 1 54. a b 5 21

• CHALLENGE n k

55. Show that a b 5 a

n b, if 0 # k # n. n2k

56. Show that if n is a positive integer, then:



n n n n a b 1 a b 1 a b 1 c1 a b 5 2n 0 1 2 n

Hint: Let 2n 5 1 1 1 1 2 n and use the binomial theorem to expand.

• TECHNOLOGY 57. With a graphing utility, plot y1 5 1 2 3x 1 3x 2 2 x 3,

y2 5 21 1 3x 2 3x 2 1 x 3, and y3 5 1 1 2 x 2 3 in the same viewing screen. What is the binomial expansion of 1 1 2 x 2 3? 58. With a graphing utility, plot y1 5 1 x 1 3 2 4, y2 5 x 4 1 4x 3 1 6x 2 1 4x 1 1, and y3 5 x 4 1 12x 3 1 54x 2 1 108x 1 81. What is the binomial expansion of 1 x 1 3 2 4? 59. With a graphing utility, plot y1 5 1 2 3x, y2 5 1 2 3x 1 3x 2, y3 5 1 2 3x 1 3x 2 2 x 3, and y4 5 1 1 2 x 2 3 for 21 , x , 1. What do you notice ­happening each time an additional term is added? Now, let 1 , x , 2. Does the same thing happen? 3 3 3 60. With a graphing utility, plot y1 5 1 2 , y2 5 1 2 1 2 , x x x 3 3 1 1 3 y3 5 1 2 1 2 2 3 , and y4 5 a1 2 b for 1  , x , 2. x x x x

Young_AT_6160_ch12_pp1133-1150.indd 1140

What do you notice happening each time an additional term is added? Now, let 0 , x , 1. Does the same thing happen? 3 3 3 , y 5 1 1 1 2, x 2 x x 3 3 1 1 3 y3 5 1 1 1 2 2 3 , and y4 5 a1 1 b for x x x x

61. With a graphing utility, plot y1 5 1 1



 1 , x , 2. What do you notice happening each time an additional term is added? Now, let 0 , x , 1. Does the same thing happen?

62. With a graphing utility, plot y1 5 1 1

x x x2 , y2 5 1 1 1 , 1! 1! 2!

x x2 x3 1 2 , and y4 5 e x for 2  1 , x , 1. 1! 2! 3! What do you notice happening each time an additional term is added? Now, let 1 , x , 2. Does the same thing happen?

y3 5 1 1

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12.6  Counting, Permutations, and Combinations 

1141

12.6 COUNTING, PERMUTATIONS, AND COMBINATIONS SKILLS OBJECTIVES ■■ Apply the fundamental counting principle to solve counting problems. ■■ Apply permutations to solve counting problems. ■■ Apply combinations to solve counting problems. ■■ Apply distinguishable permutations to solve counting problems.

CONCEPTUAL OBJECTIVES ■■ Understand that the number of ways successive things can occur is found by multiplying the number of ways each thing can occur. ■■ Understand that in a permutation order matters. ■■ Understand that in combinations order does not matter. ■■ Understand that permutation with repetition means that some objects are nondistinguishable from each other.

12.6.1  The Fundamental Counting Principle You are traveling through Europe for the summer and decide the best packing option is to select separates that can be mixed and matched. You pack one pair of shorts and one pair of khaki pants. You pack a pair of Teva sport sandals and a pair of sneakers. You have three shirts (red, blue, and white). How many different outfits can be worn using only the clothes mentioned above?

12.6.1 S K I L L

Apply the fundamental counting principle to solve counting problems. 12.6.1 C O N C E P T U A L

Understand that the number of ways successive things can occur is found by multiplying the number of ways each thing can occur.

[ CONCEPT CHECK ] If you packed four shirts, a pair of pants and a pair of shorts, and two types of shoes (sandals and sneakers), how many different outfits can be worn on your vacation?

▼ ANSWER 4⋅2⋅2 5 16

The answer is 12. There are two options for bottoms (pants or shorts), three options for shirts, and two options for shoes. The product of these is 2 ⋅ 3 ⋅ 2 5 12 The general formula for counting possibilities is given by the fundamental counting principle.

FUNDAMENTAL COUNTING PRINCIPLE

Let E1 and E2 be two independent events. The first event E1 can occur in m1 ways. The second event E2 can occur in m2 ways. The number of ways that the combination of the two events can occur is m1 ⋅ m2.

STUDY T I P The fundamental counting principle can be extended to more than two events.

In other words, the number of ways in which successive things can occur is found by multiplying the number of ways each thing can occur.

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CHAPTER 12  Sequences, Series, and Probability

EXAMPLE 1  Possible Meals Served at a Restaurant

A restaurant is rented for a retirement party. The owner offers an appetizer, an entrée, and a dessert for a set price. The following are the choices that people attending the party may choose from. How many possible dinners could be served that night?

Appetizers: calamari, stuffed mushrooms, or caesar salad



Entrées: tortellini alfredo, shrimp scampi, eggplant parmesan, or chicken marsala



Desserts:

tiramisu or flan

Solution:

There are three possible appetizers, four possible entrées, and two possible desserts. Write the product of possible options.

3 ⋅ 4 ⋅ 2 5 24

There are 24 possible dinners for the retirement party. ▼ ANSWER

▼ Y O U R T U R N   In Example 1, the restaurant will lower the cost per person for the

retirement party if the number of appetizers and entrées is reduced. Suppose the appetizers are reduced to either soup or salad and the entrées are reduced to either tortellini or eggplant parmesan. How many possible dinners could be served at the party?

8

EXAMPLE 2  T  elephone Numbers (When to Require 10-Digit Dialing)

In many towns in the United States, residents can call one another using a 7-digit dialing system. In some large cities, 10-digit dialing is required because two or more area codes coexist. Determine how many telephone numbers can be allocated in a 7-digit dialing area. Solution:

 ith 7-digit telephone numbers, the first number cannot be a 0 or a 1, but each of W the following six numbers can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. First number: 2, 3, 4, 5, 6, 7, 8, or 9. 8 possible digits Second number: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9.

10 possible digits

Third number:

10 possible digits

Fourth number:

10 possible digits

Fifth number:

10 possible digits

Sixth number:

10 possible digits

Seventh number:

10 possible digits

Counting principle:

8 ⋅ 10⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10

Possible telephone numbers:

8,000,000

Eight million 7-digit telephone numbers can be allocated within one area code. ▼ ANSWER

8 billion

Young_AT_6160_ch12_pp1133-1150.indd 1142

▼ Y O U R T U R N   If the first digit of an area code cannot be 0 or 1, but the second

and third numbers of an area code can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9, how many 10-digit telephone numbers can be allocated in the United States?

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12.6  Counting, Permutations, and Combinations 

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The fundamental counting principle applies when an event can occur more than once. We now introduce two other concepts, permutations and combinations, which allow individual events to occur only once. For example, in Example 2, the allowable telephone numbers can include the same number in two or more digit places, as in 555-1212. However, in many state lottery games, once a number is selected, it cannot be used again. An important distinction between a permutation and a combination is that in a permutation order matters, but in a combination order does not matter. For example, the Florida winning lotto numbers one week could be 2–3–5–11–19–27. This would be a combination because the order in which they are drawn does not matter. However, if you were betting on a trifecta at the Kentucky Derby, to win you must not only select the first, second, and third place horses, you must select them in the order in which they finished. This would be a permutation.

12.6.2  Permutations

12.6.2 S K I L L

Apply permutations to solve counting problems. DEFINITION

Permutation

A permutation is an ordered arrangement of distinct objects without repetition.

EXAMPLE 3  Finding the Number of Permutations of n Objects

How many permutations are possible for the letters A, B, C, and D? Solution:

ABCD ABDC ACBD ACDB ADCB ADBC BACD BADC BCAD BCDA BDCA BDAC CABD CADB CBAD CBDA CDAB CDBA DABC DACB DBCA DBAC DCAB DCBA

12.6.2 C O N C E P T U A L

Understand that in a permutation order matters.

[ CONCEPT CHECK ] TRUE OR FALSE  In a permutation, an event only occurs once and order matters.

▼ ANSWER True

There are 24 (or 4!) possible permutations of the letters A, B, C, and D.

Notice that in the first row of permutations in Example 3, A was selected for the first space. That left one of the remaining three letters to fill the second space. Once that was selected there remained two letters to choose from for the third space, and then the last space was filled with the unselected letter. In general, there are n! ways to order n objects.

NUMBER OF PERMUTATIONS OF n OBJECTS

The number of permutations of n objects is n! 5 n ⋅ 1 n 2 1 2 ⋅ 1 n 2 2 2 c 2⋅ 1

Young_AT_6160_ch12_pp1133-1150.indd 1143

STUDY T I P In a permutation of objects, order matters. That is, the same objects arranged in a different order are considered to be a distinct permutation.

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CHAPTER 12  Sequences, Series, and Probability

EXAMPLE 4  Running Order of Dogs

Courtesy Cynthia Young, University of Central Florida

In a field trial sponsored by the American Kennel Club (AKC), the dogs compete in random order. If there are nine dogs competing in the trials, how many possible running orders are there?

Solution:

There are nine dogs that will run. The number of possible running orders is n!

n59 n! 5 9! 5 362,880

There are 362,880 different possible running orders of nine dogs. ▼ ANSWER

120

▼ Y O U R T U R N   Five contestants in the Miss America pageant reach the live

television interview round. In how many possible orders can the contestants compete in the interview round?

In Examples 3 and 4, we were interested in all possible permutations. Sometimes, we are interested in only some permutations. For instance, 20 horses usually run in the Kentucky Derby. If you bet on a trifecta, you must pick the top three places in the correct order to win. Therefore, we do not consider all possible permutations of 20 horses finishing (places 1–20). Instead, we only consider the possible permutations of first-, second-, and third-place finishes of the 20 horses. We would call this a permutation of 20 objects taken 3 at a time. In general, this ordering is called a permutation of n objects taken r at a time. If 20 horses are entered in the Kentucky Derby, there are 20 possible first-place ­finishers. We consider the permutations with one horse in first place. That leaves 19 possi­ ble horses for second place and then 18 possible horses for third place. Therefore, there are 20 ⋅ 19 ⋅ 18 5 6840 possible winning scenarios for the trifecta. This 20! 20! 5 . can also be represented as 1 20 2 3 2 ! 17!

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12.6  Counting, Permutations, and Combinations 

1145

NUMBER OF PERMUTATIONS OF n OBJECTS TAKEN r AT A TIME

The number of permutations of n objects taken r at a time is nPr

5

n! 5 n1n 2 12 1n 2 22 c 1n 2 r 1 12 1n 2 r2!

EXAMPLE 5  Starting Lineup for a Volleyball Team

The starters for a six-woman volleyball team have to be listed in a particular order (1–6). If there are 13 women on the team, how many possible starting lineups are there? Solution:

Identify the total number of players. Identify the total number of starters in the lineup. Substitute n 5 13 and r 5 6 into nPr 5 13P6

5

n 5 13 r56

n! . 1n 2 r2!

13! 13! 5 5 13 ⋅ 12 ⋅ 11⋅ 10 ⋅ 9 ⋅ 8 5 1,235,520 1 13 2 6 2 ! 7!

There are 1,235,520 possible combinations.

▼ Y O U R T U R N   A softball team has 12 players, 10 of whom will be in the starting

lineup (batters 1–10). How many possible starting lineups are there for this team?

▼ ANSWER

239,500,800

12.6.3  Combinations The difference between a permutation and a combination is that a permutation has an order associated with it, whereas a combination does not have an order associated with it. DEFINITION

Combination

A combination is an arrangement, without specific order, of distinct objects without repetition.

12.6.3 S K I L L

Apply combinations to solve counting problems. 12.6.3 C O N C E P T U A L

Understand that in combinations order does not matter.

The six winning Florida lotto numbers and the NCAA men’s Final Four basketball tournament are examples of combinations (six numbers and four teams) without regard to order. The number of combinations of n objects taken r at a time is equal to the n binomial coefficient a b . r NUMBER OF COMBINATIONS OF n OBJECTS TAKEN r AT A TIME

The number of combinations of n objects taken r at a time is nCr

Young_AT_6160_ch12_pp1133-1150.indd 1145

n n! 5 a b 5 1 n 2 r 2 !r! r

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CHAPTER 12  Sequences, Series, and Probability

[ CONCEPT CHECK ] In permutations and combinations, an event can only occur once and in permutations order of events matters, whereas in combinations order does not matter, so do we expect a greater number of permutations or combinations given the same n object taken r at a time?

Compare the number of permutations nPr 5

n! and the number of 1n 2 r2!

n! . It makes sense that the number of combinations is 1 n 2 r 2 !r! less than the number of permutations. The denominator is larger because there are no separate orders associated with a combination.

combinations nCr 5

EXAMPLE 6 

Possible Combinations to the Lottery

If there are a possible 59 numbers and the lottery officials draw 6 numbers, how many possible combinations are there?

▼ ANSWER Permutations

Solution:

Identify how many numbers are in the drawing.   n 5 59 Identify how many numbers are chosen. r56 Substitute n 5 59 and r 5 6 into nCr 5 Simplify.

n! . 1 n 2 r 2 !r!

59C6

5 5



5

59! 1 59 2 6 2 !6!

59 ⋅ 58 ⋅ 57 ⋅ 56⋅ 55 ⋅ 54⋅ 1 53 2 ! 53! ⋅ 6! 59 ⋅ 58 ⋅ 57 ⋅ 56 ⋅ 55 ⋅ 54 6 ⋅ 5 ⋅ 4 ⋅ 3⋅ 2

5 45,057,474 There are 45,057,474 possible combinations.­

▼ ▼ ANSWER

Y O U R T U R N   What are the possible combinations for a lottery with 49 possible

­numbers and 6 drawn numbers?

13,983,816

12.6.4  Permutations with Repetition 12.6.4 S K I L L

Apply distinguishable permutations to solve counting problems. 12.6.4 C O N C E P T U A L

Understand that permutation with repetition means that some objects are nondistinguishable from each other.

Permutations and combinations are arrangements of distinct (nonrepeated) objects. A ­permutation in which some of the objects are repeated is called a permutation with ­repetition or a nondistinguishable permutation. For example, if a sack has three red ­marbles, two blue marbles, and one white marble, how many possible permutations would there be when drawing six marbles, one at a time? This is a different problem from writing the numbers 1 through 6 on pieces of paper, putting them in a hat, and drawing them out. The reason the problems are different is that the two blue balls are indistinguishable and the three red balls are also indistinguishable. The possible permutations for drawing numbers out of the hat are 6!, whereas the possible permutations for drawing balls out of the sack are given by 6! 3! ⋅ 2! ⋅ 1! NUMBER OF DISTINGUISHABLE PERMUTATIONS

If a set of n objects has n1 of one kind of object, n2 of another kind of object, n3 of a third kind of object, and so on for k different types of objects so that n 5 n1 1 n2 1 c 1 nk, then the number of distinguishable permutations of the n objects is n! n1! ⋅ n2! ⋅ n3! c nk!

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12.6  Counting, Permutations, and Combinations 

In our sack of marbles, there were six marbles n 5 6. Specifically, there were three red marbles 1 n1 5 3 2 , two blue marbles 1 n2 5 2 2 , and one white marble 1 n3 5 1 2 . Notice that n 5 n1 1 n2 1 n3 and that the number of distinguishable permutations is equal to 6! 5 60 3! ⋅ 2! ⋅ 1!

1147

[ CONCEPT CHECK ] Is the number of permutations of n objects that have some objects that are repeated or nondistinguishable greater than or less than the number of permutations of n distinguishable (nonrepeated) objects?

▼

EXAMPLE 7  Peg Game at Cracker Barrel

ANSWER Less than

Andy Washnik

The peg game on the tables at Cracker Barrel is a triangle with 15 holes drilled in it, in which pegs are placed. There are 5 red pegs, 5 white pegs, 3 blue pegs, and 2 yellow pegs. If all 15 pegs are in the holes, how many different ways can the pegs be aligned?

Solution:

There are four different colors of pegs (red, white, blue, and yellow). n1 5 5 5 red pegs:

5 white pegs:

n2 5 5



3 blue pegs:

n3 5 3



2 yellow pegs:

n4 5 2



There are 15 pegs total: n 5 15

Substitute n 5 15, n1 5 5, n2 5 5, n3 5 3, and n4 5 2 into

n! . n1! ⋅ n2! ⋅ n3!cnk!

Simplify.

15! 5! ⋅ 5! ⋅ 3! ⋅ 2! 5 7,567,560

There are a possible  7,567,560  ways to insert the 15 colored pegs at the Cracker Barrel.

▼ Y O U R T U R N   Suppose a similar game to the peg game at Cracker Barrel is set up

with only 10 holes in a triangle. With 5 red pegs, 2 white pegs, and 3 blue pegs, how many different permutations can fill that board?

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▼ ANSWER

2520

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CHAPTER 12  Sequences, Series, and Probability

[ S E C T I O N 12 . 6]     S U M M A R Y In this section, we discussed the fundamental counting principle, C O M B I N AT I O N S permutations, and combinations. ■■ Objects cannot be repeated. T H E F U N D A M E N TA L C O U N T I N G P R I N C I P L E IS APPLICABLE WHEN ■■ ■■ ■■

Objects can be repeated. The objects can occur in any order. The first event E1 can occur m1 ways, and the second event E2 can occur m2 ways: the number of ways successive events can occur is m1m2 ways.

Order does not matter. Number of combinations of n objects taken r at a time:

■■ ■■

nCr

Some objects are repeated because they are not distinguishable. For n objects with k different types of objects:

P E R M U TAT I O N S ■■

■■ ■■ ■■

Objects cannot be repeated. Order matters. Number of permutations of n objects: n!. Number of permutations of n objects taken r at a time: nPr

5

n! 1 n 2 r 2 !r!

N O N D I S T I N G U I S H A B L E P E R M U TAT I O N S ■■

■■

5

n! n1!n2!n3! c nk!

n! 1n 2 r2!

[ S E C T I O N 12 . 6]   E X E R C I S E S • SKILLS In Exercises 1–8, use the formula for nPr to evaluate each expression. 1. 6P4

2. 7P3

3. 9P5

4. 9P4

5. 8P8

6. 6P6

7.

13P3

8.

20P3

In Exercises 9–18, use the formula for nCr to evaluate each expression. 9.

10C5

10. 9C4

15.

30C4

16.

13C5

11.

50C6

12.

50C10

17.

45C8

18.

30C4

13. 7C7

14. 8C8

• A P P L I C AT I O N S 19. Computers. At the www.dell.com website, a customer can

“build” a system. If there are four monitors to choose from, three different computers, and two different keyboards, how many possible system configurations are there? 20. Houses. In a “new home” community, a person can select from one of four models, five paint colors, three tile selections, and two landscaping options. How many different houses (interior and exterior) are there to choose from? 21. Wedding Invitations. An engaged couple is ordering wedding invitations. The wedding invitations come in white or ivory. The writing can be printed, embossed, or engraved. The envelopes can come with liners or without. How many possible designs of wedding invitations are there to choose from? 22. Dinner. Siblings are planning their father’s 65th birthday dinner and have to select one of four main courses (baked chicken,

Young_AT_6160_ch12_pp1133-1150.indd 1148

grilled mahi-mahi, beef Wellington, or lasagna), one of two starches (rosemary potatoes or rice), one of three vegetables (green beans, carrots, or zucchini), and one of five appetizers (soup, salad, pot stickers, artichoke dip, or calamari). How many possible dinner combinations are there? 23. PIN Number. Most banks require a four-digit ATM PIN code for each customer’s bank card. How many possible four-digit PIN codes are there to choose from? 24. Password. All e-mail accounts require passwords. If a four-character password is required that can contain letters (but no numbers), how many possible passwords can there be? (Assume letters are not case sensitive.) 25. Leadership. There are 15 professors in a department, and there are four leadership positions (chair, assistant chair, undergraduate coordinator, and graduate coordinator). How many possible leadership teams are there?

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1149

26. Fraternity Elections. A fraternity is having elections. There

38. Lotto. If a state lottery picks from 53 numbers and 5 numbers

are three men running for president, two men running for vice president, four men running for secretary, and one man running for treasurer. How many possible outcomes do the elections have? Multiple-Choice Tests. There are 20 questions on a multiple -choice exam, and each question has four possible answers (A, B, C, and D). Assuming no answers are left blank, how many different ways can you answer the questions on the exam? Multiple-Choice Tests. There are 25 questions on a multiple-choice exam, and each question has five possible answers (A, B, C, D, and E). Assuming no answers are left blank, how many different ways can you answer the questions on the exam? Zip Codes. In the United States a five-digit zip code is used to route mail. How many five-digit zip codes are possible? (All numbers can be used.) If 0s were eliminated from the first and last digits, how many possible zip codes would there be? License Plates. In a particular state, there are six characters in a license plate: three letters followed by three numbers. If 0s and 1s are eliminated from possible numbers and Os and Is are eliminated from possible letters, how many different license plates can be made? Class Seating. If there are 30 students in a class and there are exactly 30 seats, how many possible seating charts can be made, assuming all 30 students are present? Season Tickets. Four friends buy four season tickets to the Green Bay Packers. To be fair, they change the seating arrangement every game. How many different seating arrangements are there for the four friends? Combination “Permutation” Lock. A combination lock on most lockers will open when the correct choice of three numbers (1 to 40) is selected and entered in the correct order. Therefore, a combination lock should really be called a permutation lock. How many possible permutations are there, assuming no numbers can be repeated? Safe. A safe will open when the correct choice of three numbers (1 to 50) is selected and entered in the correct order. How many possible permutations are there, assuming no numbers can be repeated? Raffle. A fundraiser raffle is held to benefit cystic fibrosis research, and 1000 raffle tickets are sold. Three prizes are raffled off. First prize is a round-trip ticket on Delta Air Lines, second prize is a round of golf for four people at a local golf course, and third prize is a $50 gift certificate to Chili’s. How many possible winning scenarios are there if all 1000 tickets are sold to different people? Ironman Triathlon. If 100 people compete in an ironman triathlon, how many possible placings are there (first, second, and third place)? Lotto. If a state lottery picks from 53 numbers and 6 numbers are selected, how many possible 6-number combinations are there?

are selected, how many possible 5-number combinations are there? Cards. In a deck of 52 cards, how many different 5-card hands can be dealt? Cards. In a deck of 52 cards, how many different 7-card hands can be dealt? Blackjack. In a single-deck blackjack game (52 cards), how many different 2-card combinations are there? Blackjack. In a single deck, how many two-card combinations are there that equal 21: ace (worth 11) and a 10 or face card—jack, queen, or king? March Madness. Every spring, the NCAA men’s basketball tournament starts with 64 teams. After two rounds, it is down to the Sweet Sixteen, and after two more rounds, it is reduced to the Final Four. Once 64 teams are selected (but not yet put in brackets), how many possible scenarios are there for the Sweet Sixteen? March Madness. Every spring, the NCAA men’s basketball tournament starts with 64 teams. After two rounds, it is down to the Sweet Sixteen, and after two more rounds, it is reduced to the Final Four. Once the 64 teams are identified (but not yet put in brackets), how many possible scenarios are there for the Final Four? NFL Playoffs. There are 32 teams in the National Football League (16 AFC and 16 NFC). How many possible combinations are there for the Superbowl? (Assume one team from the AFC plays one team from the NFC in the Superbowl.) NFL Playoffs. After the regular season in the National Football League, 12 teams make the playoffs (6 from the AFC and 6 from the NFC). How many possible combinations are there for the Superbowl once the 6 teams in each conference are identified? Survivor. On the television show Survivor, one person is voted off the island every week. When it is down to six contestants, how many possible voting combinations remain, if no one will vote him- or herself off the island? Assume that the order (who votes for whom) makes a difference. How many total possible v­ oting outcomes are there? American Idol. On the television show American Idol, a young rising star is eliminated from the competition every week. The first week, each of the 12 contestants sings one song. How many possible ways could the contestants be ordered 1–12? How many possible ways could 6 men and 6 women be ordered to alternate female and male contestants? Dancing with the Stars. In the popular TV show Dancing with the Stars, 12 entertainers (6 men and 6 women) compete in a dancing contest. The first night, the show decides to select 3 men and 3 women. How many ways can this be done? Dancing with the Stars. See Exercise 49. How many ways can six male celebrities line up for a picture alongside six female celebrities?

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.

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39. 40. 41. 42.

43.

44.

45.

46.

47.

48.

49.

50.

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CHAPTER 12  Sequences, Series, and Probability

• C AT C H T H E M I S TA K E In Exercises 51 and 52, explain the mistake that is made. 51. In a lottery that picks from 30 numbers, how many five-

52. A homeowners association has 12 members on the board of

number combinations are there?

directors. How many ways can the board elect a president, vice president, secretary, and treasurer?

Solution:

Solution:

Let n 5 30 and r 5 5. Calculate nPr 5 Simplify.

n! . 1n 2 r2!

30P5 5 30P5

30! 25!

Let n 5 12 and r 5 4. Calculate nCr 5

5 17,100,720

This is incorrect. What mistake was made?

Simplify.

n! 12! .       12C4 5 1 n 2 r 2 !⋅r! 8!⋅4! 12C4

5 495

This is incorrect. What mistake was made?

• CONCEPTUAL In Exercises 53–56, determine whether each statement is true or false. 53. The number of permutations of n objects is always greater

55. The number of four-letter permutations of the letters A, B, C,

than the number of combinations of n objects if the objects are distinct. 54. The number of permutations of n objects is always greater than the number of combinations of n objects even when the objects are indistinguishable.

and D is equal to the number of four-letter permutations of ABBA. 56. The number of possible answers to a true/false question is a permutation problem.

• CHALLENGE 57. What is the relationship between nCr and nCr11?

59. Simplify the expression nCr ⋅r!.

58. What is the relationship between nPr and nPr21?

60. What is the relationship between nCr and nCn2r?

• TECHNOLOGY 61. Employ a graphing utility with a nPr feature and compare it 62. 63.

with answers to Exercises 1–8. Employ a graphing utility with a nCr feature and compare it with answers to Exercises 9–18. Use a graphing calculator to evaluate: a.  10P4   b.  4! 1 10C4 2

c.  Are answers in (a) and (b) the same? d.  Why?

Young_AT_6160_ch12_pp1133-1150.indd 1150

64. Use a graphing calculator to evaluate:



a.  12P5 b.  5! 1 12C5 2

c.  Are answers in (a) and (b) the same? d.  Why?

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12.7 Probability 

1151

12.7 PROBABILITY SKILLS OBJECTIVES ■■ Determine the sample space of an outcome. ■■ Find the probability of an event. ■■ Find the probability that an event will not occur. ■■ Find the probability of mutually exclusive events. ■■ Find the probability of independent events.

CONCEPTUAL OBJECTIVES ■■ Understand that all possible outcomes of an experiment constitute the sample space. ■■ Understand that the probability of an event occurring is the ratio of the number of outcomes in an event divided by the number of outcomes in the sample space. ■■ Understand that the sum of the probability of an event occurring and the probability of an event not occurring is one. ■■ Understand the difference between the probability of: event 1 and event 2; event 1 or event 2. ■■ Understand that the probability of two independent events both occurring is the product of the individual probabilities.

12.7.1  Sample Space You are sitting at a blackjack table at Caesar’s Palace, and the dealer is showing a 7. You have a 9 and a 7; should you hit? Will it rain today? What will the lotto numbers be this week? Will the coin toss at the Superbowl result in a head or a tail? Will Derek Jeter get a hit at his next trip to the plate? These are all questions where probability is used to guide us. Anything that happens for which the result is uncertain is called an experiment. Each trial of an experiment is called an outcome. All of the possible outcomes of an experiment constitute the sample space. The term event is used to describe the kind of possible outcomes. For example, a coin toss is an experiment. Every outcome is either heads or tails. The sample space of a single toss is {heads, tails}. The result of one experiment has no certain outcome. However, if the experiment is performed many times, the results will produce regular patterns. For example, if you toss a fair coin, you don’t know whether it will come up heads or tails. You can toss a coin 10 times and get 10 heads. However, if you made 1,000,000 tosses, you would get about 500,000 heads and 500,000 tails. Therefore, since we assume a head is equally likely as a tail and there are only two possible events (heads or tails), we assign a probability of a head equal to 12 and a probability of a tail equal to 12. EXAMPLE 1  Finding the Sample Space

Find the sample space for each of the following outcomes. a. Tossing a coin once  b.  Tossing a coin twice  c.  Tossing a coin three times

12.7.1 S K I L L

Determine the sample space of an outcome. 12.7.1 C O N C E P T U A L

Understand that all possible ­outcomes of an experiment ­constitute the sample space.

[ CONCEPT CHECK ] TRUE OR FALSE  The ­sample space includes all possible outcomes and order matters (for example, Heads then Tails is not the same as Tails then Heads, even though the result of two rolls is one head and one tail).

▼ ANSWER True

Solution (a):

Tossing a coin one time will result in one of two events: heads (H) or tails (T ). The sample space S is written as  S 5 5 H, T 6 . Solution (b):

Tossing a coin twice can result in one of four possible outcomes. The sample space consists of all possible outcomes. S 5 5 HH, HT, TH, TT 6

Note that TH and HT are two different outcomes.

STUDY T I P

Solution (c):

If the coin is tossed n times, there are 2n possible outcomes.

There are eight possible outcomes when a coin is tossed three times.

▼

S 5 5 HHH, HHT, HTH, HTT, TTT, TTH, THT, THH 6

Y O U R T U R N   Find the sample space associated with having three children

(B boys or G girls).

Young_AT_6160_ch12_pp1151-1168.indd 1151

▼ ANSWER

S 5 5GGG, GGB, GBG, GBB, BBB, BBG, BGB, BGG6

07/12/16 11:10 AM

1152 

CHAPTER 12  Sequences, Series, and Probability

12.7.2  Probability of an Event 12.7.2 S K I L L

Find the probability of an event.

To calculate the probability of an event, start by counting the number of outcomes in the event and the number of outcomes in the sample space. The ratio is equal to the probability if all outcomes are equally likely.

12.7.2 C O N C E P T U A L

Understand that the probability of an event occurring is the ratio of the number of outcomes in an event divided by the number of outcomes in the sample space.

DEFINITION

Probability of an Event

If an event E has n 1 E 2 equally likely outcomes and its sample space S has n 1 S 2 equally likely outcomes, then the probability of event E, denoted P 1 E 2 , is P1E2 5

[ CONCEPT CHECK ]

n1E2 Number of outcomes in event E 5 Number of outcomes in sample space S n1S2

Since the number of outcomes in an event must be less than or equal to the number of outcomes in the sample space, the probability of an event must be a number between 0 and 1 or equal to 0 or 1; to be precise, 0 # P 1 E 2 # 1. If P 1 E 2 5 0, then the event can never happen, and if P 1 E 2 5 1, the event is certain to happen. EXAMPLE 2  Finding the Probability of Two Girls

Would the probability of having a boy and a girl be greater or less than the probability of having two boys?

If two children are born, what is the probability that they are both girls?

▼

The sample space is all four possible outcomes.

ANSWER Greater than

Solution:

The event is both children being girls. The number of outcomes in the event is 1. The number of events in the sample space is 4. n1E2 Compute the probability using P 1 E 2 5 . n1S2

E 5 5 GG 6

S 5 5 BB, BG, GB, GG 6

n1E2 5 1 n1S2 5 4

P1E2 5

1 4

The probability that both children are girls is 14, or 0.25.

EXAMPLE 3  Finding the Probability of Drawing a Face Card

Find the probability of drawing a face card (jack, queen, or king) out of a 52-card deck. Solution:

There are 12 face cards in a deck. There are 52 cards in a deck. Compute the probability using P 1 E 2 5

n1E2 . n1S2

n 1 E 2 5 12 n 1 S 2 5 52

P1E2 5

12 3 5 52 13

3 The probability of drawing a face card out of a 52-card deck is 13 or 22.303 2 75. x 5 2, x 5 3 77. x 5 54 , x 5 73.

3 4

79. y 5 14 , y 5 1

81. x 5 2125 8 ,x 5 1



83. x 5 218 , x 5 21

85. x 5 63i, x 5 62



87. x 5 0, 28, 4

89. p 5 62, 3

91. p 5

212 , 52 ,

  

b 5 7: x 5 22 6 !11   b 5 12: x 5 26, 2

3 93. y 5 69

95. 12q, 244 97. 32, 64 99. x . 26

101. 23 # x # 7

103. x $ 24, 324, q2

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09/12/16 2:23 PM

Answers to Odd-Numbered Exercises  

153. x 5 A21 1 !7B 4 < 7.34

1177

27. 627 # minutes # 722 y1 5 2x

1/4

y2 5 2x 1/2 1 6

29. x 5 7.95

155.  a.  12q, 15.4 2   b.  y1 5 20.61x 1 7.62, y2 5 0.24x 2 5.47

Cumulative Test

x 12 y3 7. 21a 1 102 1a2 2 10a 1 1002

1. 215

3.

 c. agree

5. x4 1 2x3 2 15x2

157. 12q, 25 2 ∪ 1 5.25, q 2

9.

x 2 1 22x x2 2 4 13. x 5 24



11. x 5 56 15. 11.25 hours

17. x 5 26 6 2i

19. x 5 4

21. 323, 42

23. 12q, 23 2 ∪ 3 22, 3 2

25. x 5 21, x 5 217 3 27. Q21, 12 R

159. A 75 , 27 B

y1 5 `



3x ` , y2 5 1 x22

161. 122.19, 20.9 2 ∪ 1 0.9, 2.19 2

CHAPTER 2 Section 2.1 1. 14, 22

Practice Test 1. 23 5 p 5. x 5

212 , 83

9. x 5 4 13. x 5 0, 2, 6 17. 12q, 174

21. 12q, 21 4 ∪ S 43 , qR

3. t 5 24, 7 7. y 5 28 11. y 5 1 P 2 2W 15. L 5 2

3. 123, 02

7. A: quadrant II B:  quadrant I C:  quadrant III D:  quadrant IV E:  on negative y-axis F:  on positive x-axis

5. 10, 232

19. Q232 5 , 26T

23. Q212 , 3T 25. 1000 feet

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1178 

Answers to Odd-Numbered Exercises

9.

59. P1 5 1a, b2, P2 5 1c, d2,

d1 5 distance from P1 to P2, and

d2 5 distance from P2 to P1, d1 5 " 1 a 2 c 2 2 1 1 b 2 d 2 2 d2 5 " 1 c 2 a 2 2 1 1 d 2 b 2 2

since 1c 2 a22 5 121a 2 c222 5 1a 2 c22,

1d 2 b22 5 121b 2 d222 5 1b 2 d22 [ d1 5 d2,

so does not matter what point is labeled “first.”

11. d 5 4, 13, 32 13. d 5 4!2, 1 1, 2 2

7 15. d 5 3!10, A217 2 , 2 B

17. d 5 5, A25, 12 B

63. d > 3.111

10.7, 5.152

12.2, 3.32

21. d 5 5, A 32 , 11 6B

19. d 5 4!2, 124, 262

!4049 5 1 , A224 , 15 B 60

23. d 5

61. d > 6.357

25. d 5 3.9, 10.3, 3.952

27. d 5 44.64, 11.05, 21.22 29. d 5 4!2, A!3, 3!2B

31. d 5 #10 1 2!2 1 4!3, a

1 2 !2 22 1 !3 , b 2 2

33. The perimeter of the triangle rounded to two decimal places is 21.84. 35. right triangle

37. isosceles

39. 128.06 miles 41. 268 miles

Section 2.2 1. a. no      b. yes

3. a. yes   b. no

5. a. yes        b. no

7. a. yes   b. no

9.

43. 12003, 3302; $330 million 45.

Dollars per Gallon (2012)

x

y521x

22

0

0

2

1

3

x

y 5 x2 2 x

21

2

0

0

1 2

214

1

0

2

2

4

3.5 3 2.5 2

1x, y2

122, 02 10, 22 11, 32

1.5 1 0.5

Month 1

2

3

4

5

6

7

8

9

10

11

12

47. x values were subtracted from y values. The correct distance would be d 5 !58.

49. The values were not used in the correct positions. The correct midpoint would be A2, 13 2 B. 51. true 53. true 55. d 5 !2 0 a 2 b 0 , a 57. d 5

5

5 5

Ç

ax1 2

a1b b1a , b 2 2

y1 1 y2 2 x1 1 x2 2 b 1 ay1 2 b 2 2

2y1 2 y1 2 y2 2 2x1 2 x1 2 x2 2 b 1 a b Ç 2 2 a

11.

1x, y2

121, 22 10, 02

A 12 ,

214 B

11, 02 12, 22

y1 2 y2 2 x1 2 x2 2 b 1 a b Ç 2 2 a

1 " 1 x1 2 x2 2 2 1 1 y1 2 y2 2 2 2

Using 1x2, y22 with the midpoint yields the same result.

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Answers to Odd-Numbered Exercises  

13.

x 1

1

5

2

10

3

69.

11, 02

0

2

67.

1x, y2

y 5 !x 2 1

1179

12, 12 15, 22

110, 32 71. 73.

15.

17.

pr 7 5. 0.00014

pt

0.00012 0.00010 0.00008 0.00006

19.

0.00004

21.

0.00002

km 2000

6000

10,000

77. Break even units: 2000 2000 , x , 4000 23. x-intercept: 13, 02, y-intercept: 10, 262

25. x-intercept: 163, 02, y-intercept: 10, 292 27. x-intercept: 14, 02, no y-intercept

29. no x-intercept, y-intercept: A0, 14 B

31. x-intercept: 162, 02, y-intercept: 10, 642 33. d

35. a

39. 121, 232

41. 127, 102

45. x-axis

47. origin

43. 13, 22, 123, 22, 123, 222

37. b

p (price per unit) 3

49. x-axis

51. x-axis, y-axis, origin 53. y-axis

79. x $ 1 or 3 1, q 2 ; the demand model is defined when at least 1000 units per day are demanded.

2

55. y-axis 57. origin

59.

61.

1

x (thousands of units demanded) 2

4

6

8

81. The equation is not linear 2 you need more than two points to plot the graph. 63.

Young_AT_6160_ANS_SE_pg1169-1210.indd 1179

65.

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1180 

Answers to Odd-Numbered Exercises

83. You are checking to see whether the graph is symmetric with respect to the y-axis. Thus the substitution shown is incorrect. The correct substitution would be plugging 2x for x into the equation, not 2y for y. 85. false 87. true 91. y-axis

29. x-intercept: A272 , 0B y-intercept: none

89. origin

93. x-axis, y-axis, and origin

31. y 5 25 x 2 2, m 5 25 , y-intercept: 10, 222

33. y 5 213 x 1 2, m 5 213 , y-intercept: 10, 22

35. y 5 4x 2 3, m 5 4, y-intercept: 10, 232

95. x-axis, y-axis, and origin

37. y 5 22x 1 4, m 5 22, y-intercept: 10, 42 39. y 5 23 x 2 2, m 5 23 , y-intercept: 10, 222

41. y 5 234 x 1 6, m 5 234 , y-intercept: 10, 62

43. y 5 2x 1 3 45. y 5 213 x 47. y 5 2 49. x 5

3 2

51. y 5 5x 1 2 53. y 5 23x 2 4

Section 2.3

55. y 5 34 x 2 74 57. y 5 4

1. m 5 3

3. m 5 22

5. m 5 219 10

7. m < 2.379

59. x 5 21 61. y 5 35 x 1 9. m 5 23

63. y 5 25x 2 16 65. y 5 16 x 2

11. x-intercept: 10.5, 02, y-intercept: 10, 212, m 5 2, rising

67. y 5 23x 1 1 69. y 5 32 x

15. x-intercept: none, y-intercept: 10, 12, m 5 0, horizontal

75. y 5 65 x 1 6

13. x-intercept: 11, 02, y-intercept: 10, 12, m 5 21, falling A 32 ,

0B 17. x-intercept: y-intercept: 10, 232

19.  x-intercept: 14, 02 y-intercept: 10, 22

1 5 121 3

71. x 5 3 73. y 5 7 79. x 5

2 5

77. x 5 26 81. y 5 x 2 1

83. y 5 22x 1 3

85. y 5 212 x 1 1

87. y 5 2x 1 7

89. y 5 32 x

91. y 5 5

93. y 5 2

95. y 5

3 2x

2 4 97. y 5 54 x 1

3 2

99. y 5 37 x 1 52 101. 32-hour job will cost $2,000 103. $375 21. x-intercept: 12, 02 y-intercept: A0, 243 B

23. x -intercept: 122, 02 y-intercept: 10, 222

107. F 5

9 5C

105. 347 units 1 32, 2408C 5 2408F

1 109. The rate of change in inches per year is 50 .

111. 0.06 ounces per year. In 2040 we expect a baby to weigh 6 pounds 12.4 ounces. 113. y-intercept 0.35 is the flat monthly charge of $35. 115. 20.35 in./yr, 1 inch

25. x-intercept: 121, 02 y-intercept: nonev

27.  x-intercept: none y-intercept: 10, 1.52

117. 2.4 plastic bags per year 1in billions2, 418.4 billion 119. a.  11, 31.932  12, 51.182  15, 111.832

b. m 5 25.59. This means that when you buy one bottle of hoisin it costs $31.93 per bottle. c. m 5 31.93. This means that when you buy two bottles of hoisin it costs $25.59 per bottle. d. m 5 22.366. This means that when you buy five bottles of hoisin it costs $22.37 per bottle. 121. The correction that needs to be made is that for the x-intercept, y 5 0, and for the y-intercept, x 5 0.

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Answers to Odd-Numbered Exercises  

123. Values for the numerator and denominator reversed. 125. true

127. false

129. The line perpendicular is vertical and has undefined slope. 131. y 5 2AB x 1 1

b.  This placement of cell phone towers will provide cell phone coverage for the entire 10 mile by 10 mile square.

133. y 5 BA x 1 1 2B 2 1 2

y 10

135. Let y1 5 mx 1 b1 and y2 5 mx 1 b2, assuming that b1 Þ b2. At a point of intersection of these two lines, y1 5 y2. This is equivalent to mx 1 b1 5 mx 1 b2, which implies b1 5 b2, which contradicts our assumption. Hence, there are no points of intersection. 137. perpendicular

1181

8

6

4

139. perpendicular

2

x 2

4

6

8

10

67. The standard form of an equation of a circle is 1x 2 h22 1 1 y 2 k22 5 25, which would lead to a center of 14, 232, not 14, 32.

69. This equation is not a circle. The standard equation of a circle is 1x 2 h22 1 1 y 2 k22 5 r2.

141. neither

71. true 73. true 75. single point 125, 32 77. 1x 2 322 1 1 y 1 222 5 20 79. 4c 5 a2 1 b2

81. C1a, 02, r 5 10

83. no graph (because no solution)

Section 2.4

85. single point 125, 32

1. 1x 2 122 1 1 y 2 222 5 9

b. y 5 21.5 6 "39.69 2 1 x 2 5.5 2 2

87. a.  1 x 2 5.5 2 2 1 1 y 1 1.5 2 2 5 39.69, 1 5.5, 21.5 2 , r 5 6.3 c. 

3. 1x 1 322 1 1 y 1 422 5 100 5. 1x 2 522 1 1 y 2 722 5 81

7. 1x 1 1122 1 1 y 2 1222 5 169

9. x 2 1 y 2 5 4 11. x 2 1 1 y 2 222 5 9

13. x 2 1 y 2 5 2 17. Ax 2

2 2 3B

1 Ay 1

2

3 2 5B

5

1 16

15. 1x 2 522 1 1 y 1 322 5 12

19. 1x 2 1.32 1 1 y 2 2.722 5 10.24 21. C11, 32, r 5 5

23. C12, 252, r 5 7

25. C14, 92, r 5 2!5 29. C11.5, 22.72, r 5 1.3

27. C A 25 , 17 B, r 5

33. C122, 232, r 5 4

35. C123, 242, r 5 10

37. C15, 72, r 5 9

39. C10, 12, r 5 4

2 3

31. C10, 02, r 5 5!2

41. C11, 32, r 5 3 43. C15, 232, r 5 2!3 45. C13, 22, r 5 2! 3 47. CA 12 , 2 12 B, r 5

1 2

49. C11.3, 2.72, r 5 3.2 51. 1x 1 122 1 1 y 1 222 5 8 53. 1x 1 222 1 1 y 2 322 5 41 55. 1x 1 222 1 1 y 1 522 5 25

57. no A"952 1 332 ≅ 100.568B

59. x 2 1 y 2 5 2500 61. x2 1 y2 5 2,250,000

d. The graphs in (a) and (c) are the same.

Section 2.5* 1. Negative linear association because the data closely cluster around what is reasonably described as a linear with negative slope. 3. Although the data seem to comprise two linear segments, the overall data set cannot be described as having a positive or negative direction of association. Moreover, the pattern of the data is not linear, per se; rather, it is nonlinear and conforms to an identifiable curve (an upside down V called the absolute value function). 5. B because the association is positive, thereby eliminating choices A and C. And, since the data are closely clustered around a linear curve, the bigger of the two correlation coefficients, 0.80 and 0.20, is more appropriate. 7. C because the association is negative, thereby eliminating choices B and D. And the data are more loosely clustered around a linear curve than are those pictured in number 6. So, the correlation ­coefficient is the negative choice closer to 0.

63. x 2 1 y 2 5 40,000 65. a.  Tower 1: 1x 2 2.522 1 1 y 2 2.522 5 3.52

Tower 2: 1x 2 2.522 1 1 y 2 7.522 5 3.52 Tower 3: 1x 2 7.522 1 1 y 2 2.522 5 3.52 Tower 4: 1x 2 7.522 1 1 y 2 7.522 5 3.52

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1182 

Answers to Odd-Numbered Exercises

9. a. 

15. a. 

20

10 8

15

6

10

4

5

2 0

0 -4

-2

-5

0

2

4

-5

-10

6

-2 0

5

10

15

-4 -6

-10

-8

-15

-10

b. The data seem to be nearly perfectly aligned to a line with negative slope. So, it is reasonable to guess that the correlation coefficient is very close to 21.



c. The equation of the best fit line is y 5 23x 1 5 with a ­correlation coefficient of r 5 21.

b. The values x 5 0 and x 5 26 are within the range of the data set, so that using the best fit line for predictive purposes is reasonable. This is not the case for the values x 5 12 and x 5 215. The predicted value of y when x 5 0 is ­approximately 20.4392, and the predicted value of y when x 5 26 is 24.971.

d. There is a perfect negative linear association between x and y. 11. a. 

2 0 -20

-10

-2 0

10

20

30

-4 -6 -8 -10

The equation of the best fit line is about y 5 0.7553x 2 0.4392 with a correlation coefficient of about r 5 0.868.

c. Solve the equation 2 5 0.7553x 2 0.4392 for x to obtain: 2.4392 5 0.7553x so that x 5 3.229. So, using the best fit line, you would expect to get a y-value of 2 when x is approximately 3.229. 17. a. 

-12

20

-14

15

-16

10

-18

5

b. The data tend to fall from left to right, so that the correlation coefficient should be negative. Also, the data do not seem to stray too far from a linear curve, so the r value should be reasonably close to 21, but not equal to it. A reasonable guess would be around 20.90. c. The equation of the best fit line is approximately y 5 20.5844x 2 3.801 with a correlation coefficient of about r 5 0.9833. d. There is a strong (but not perfect) negative linear relationship between x and y. 13. a. 

4 2 0 -2

-2

-20

-15

-10

-5

-5 0

5

-10 -15 -20



The equation of the best fit line is about y 5 21.2631x 2 11.979 with a correlation coefficient of about r 5 20.980.

b. The values x 5 215, 26, and 0 are within the range of the data set, so that using the best fit line for predictive purposes is reasonable. This is not the case for the value x 5 12. The predicted value of y when x 5 215 is approximately 6.9675, the predicted value of y when x 5 26 is about 24.4004, and the predicted value of y when x 5 0 is 211.979.

6

-4

0 -25

0

2

4

6

-4 -6 -8

c. Solve the equation 2 5 21.2631x 2 11.979 for x to obtain: 13.979 5 21.2631x so that x 5 211.067. So, using the best fit line, you would expect to get a y-value of 2 when x is ­approximately 211.067. 19. a.  The scatterplot for the entire data set is: 20

b. The data seem to rise from left to right, but it is difficult to be certain about this relationship since the data stray considerably away from an identifiable line. As such, it is reasonable to guess that r is a rather small value close to 0, say around 0.30. c. The equation of the best fit line is approximately y 5 0.5x 1 0.5 with a correlation coefficient of about 0.349. d. There is a very loose (bordering on unidentifiable) positive linear relationship between x and y.

15 10 5 0 -4

-2

-5

0

2

4

6

-10 -15

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Answers to Odd-Numbered Exercises  



The equation of the best fit line is y 5 23x 1 5 with a ­correlation coefficient of r 5 21.

23. a. 

5 0

b. The scatterplot for the data set obtained by removing the starred data point 15, 2102 is:

-10

-20

20

-15 -20

8

-25

6 4



10

-10

12 10

2 0 -2 0 -4 -6

-2

0 -5

16 14

-4

1183

2

4

The equation of the best fit line of this modified data set is y 5 23x 1 5 with a correlation coefficient of r 5 21.

c. Removing the data point did not result in the slightest change in either the equation of the best fit line or the correlation coefficient. This is reasonable since the relationship between x and y in the original data set is perfectly linear, so that all of the points lie ON the same line. As such, removing one of them has no effect on the line itself.

b. The correlation coefficient is approximately r 5 20.980. This is identical to the r-value from Problem 17. This makes sense because simply interchanging the x- and y-values does not change how the points cluster together in the xy-plane. c. The equation of the best fit line for the paired data 1 y, x2 is x 5 20.7607y 2 9.4957.

d. It is not reasonable to use the best fit line in (c) to find the ­predicted value of x when y 5 23 because this value falls outside the range of the given data. However, it is okay to use the best fit line to find the predicted values of x when y 5 2 or y 5 216. Indeed, the predicted value of x when y 5 2 is about 211.0171, and the predicted value of x when y 5 216 is about 2.6755. 25. First, note that the scatterplot is given by

21. a.   The scatterplot for the entire data set is:

20

20 15

15

10

10

5 0 -4

-2

-5

5 0

2

4

6

8

0

-10

0

-20



1

b. The scatterplot for the data set obtained by removing the starred data points 13, 242 and 16, 2162 is: 20

r5

10 5 0 -5

0

2

4

6

-10 -15 -20



4

The paired data all lie identically on the vertical line x 5 3. As such, you might think that the square of the correlation coefficient would be 1 and the best fit line is, in fact, x 5 3. However, since there is absolutely no variation in the x-values for this data set, it turns out that in the formula for the correlation coefficient

15

-2

3

-10

The equation of the best fit line is y 5 23.4776x 1 4.6076 with a correlation coefficient of about r 5 20.993.

-4

2

-5

-15

The equation of the best fit line of this modified data set is y 5 23.6534x 1 4.2614 with a correlation coefficient of r 5 20.995.

c. Removing the data point did change both the best fit line and the correlation coefficient, but only very slightly.

Young_AT_6160_ANS_SE_pg1169-1210.indd 1183

n a xy 2 a a xb a a yb

n x 2 2 a a xb ⋅ n a y 2 2 a a yb Å a Å 2

2

n x 2 2 a a xb turns out to be zero. Å a 2

the quantity

(Check this on Excel for this data set!) As such, there is no ­meaningful r-value for this data set.   Also, the best fit line is definitely the vertical line x 5 3, but the technology cannot provide it because its slope is undefined. 27. The y-intercept 1.257 is mistakenly interpreted as the slope. The correct interpretation is that for every unit increase in x, the y-value increases by about 5.175.

09/12/16 2:23 PM

1184 

Answers to Odd-Numbered Exercises

29. a.   Here is a table listing all of the correlation coefficients between each of the events and the total points: EVENT

100 m

33. a.  A scatterplot illustrating the relationship between average wait times and average rating of enjoyment is shown below.

r

20.567

Long Jump

0.842

Shot Put

0.227

High Jump

0.587

400 m

20.715

110 m hurdle

20.488

Discus

0.173

Pole Vault

0.446

Javelin

0.542

  Long jump has the strongest relationship to the total points. b. The correlation coefficient between long jump and total events is r 5 0.842. c. The equation of the best fit line between the two events in (b) is y 5 939.98x 1 1261.2. d. Evaluate the equation in (c) at x 5 40 to get the total points are about 38.860. 31. a.  A scatterplot illustrating the relationship between % residents immunized and % residents with influenza is shown below.

b. The correlation coefficient between average wait times and average rating of enjoyment is r 5 0.348. c. The correlation coefficient 1r 5 0.3482 indicates a somewhat weak relationship between average wait times and average rating of enjoyment. d. The equation of the best fit line that describes the relationship between average wait times and average rating of enjoyment is y 5 0.31x 1 37.83. e. No, given the somewhat weak correlation coefficient between average wait times and average rating of enjoyment, one could not use the best fit line to produce accurate predictions. 35. a.  A scatterplot illustrating the relationship between average wait times and average rating of enjoyment for Park 2 is shown below.

b. The correlation coefficient between % residents immunized and % residents with influenza is r 5 20.812. c. Based on the correlation coefficient 1r 5 20.8122, we would believe that there is a strong relationship between % residents immunized and % residents with influenza.

b. The correlation coefficient between average wait times and average rating of enjoyment is r 5 20.064.

d. The equation of the best fit line that describes the relationship between % residents immunized and % residents with influenza is y 5 20.27x 1 32.20.

c. The correlation coefficient 1r 5 20.0642 indicates a weak relationship between average wait times and average rating of enjoyment for Park 2.

e. Since r 5 20.812 indicates a strong relationship between % residents immunized and % residents with influenza, we can make a reasonably accurate prediction. However it will not be completely accurate.

d. The equation of the best fit line that describes the relationship between average wait times and average rating of enjoyment is y 5 20.05x 1 52.53.

Young_AT_6160_ANS_SE_pg1169-1210.indd 1184

e. No, given the weak correlation coefficient between average wait times and average rating of enjoyment for Park 2, one could not use the best fit line to produce accurate predictions.

09/12/16 2:23 PM

Answers to Odd-Numbered Exercises  

37. a.  The scatterplot for this data set is given by

9. A 52 , 6B 13. d < 52.20 units

30

11. 13.85, 5.32

20

15. x-intercepts: 162, 02, y-intercepts: 10, 612

15

17. x-intercepts: 163, 02, no y-intercepts 19. y-axis

10

23. 25.

25

1185

21. origin

5 0 0

2

4

6

8

10

b. The equation of the best fit line is y 5 21.9867x 1 27.211 with a correlation coefficient of r 5 20.671. This line does not seem to accurately describe the data because some of the points rise as you move left to right, while others fall as you move left to right; a line cannot capture both types of behavior ­simultaneously. Also, r being negative has no meaning here.

27. 29.

c. The best fit is provided by QuadReg. The associated equation of the best fit quadratic curve, the correlation coefficient, and associated scatterplot are: 31. y 5 23x 1 6  m 5 23  y-intercept: 10, 62

33. y 5 232 x 212   m 5 232   y-intercept: A0, 212 B

35. x-intercepts A 54 , 0B, y-intercepts: 10, 252, m 5 4 39. a.   The scatterplot for this data set is given by 6 5 4

37. x-intercepts: 14, 02, y-intercepts: 10, 42, m 5 21

3 2 1 0 0

2

4

6

8

10

12

14

16

b. The equation of the best fit line is y 5 0.3537x 1 0.5593 with a correlation coefficient of r 5 0.971. This line seems to provide a very good fit for this data, although not perfect. c. The best fit is provided by LnReg. The associated equation of the best fit logarithmic curve, the correlation coefficient, and associated scatterplot are:

Review Exercises 1. quadrant II 3. quadrant III 5. d 5 3!5 7. d 5 !205

Young_AT_6160_ANS_SE_pg1169-1210.indd 1185

39. no x-intercepts, y-intercepts: 10, 22, m 5 0

41. y 5 4x 2 3

43. x 5 23

45. y 5 22x 2 2

47. y 5 6

49. y 5

5 6x

1

4 3 51.

y 5 22x 2 1

y 5 234 x 1 31 53. y 5 23 x 1 13 55. 16 57. y 5 1.2x 1 100 (x is pretest score, y is posttest score) 2

2

59. 1x 1 222 1 1 y 2 322 5 36 61. Ax 2 43 B 1 Ay 2 25 B 5 63. C122, 232, r 5 9

65. C A2 43 , 12 B, r 5

4 25

2 3

09/12/16 2:24 PM

1186 

Answers to Odd-Numbered Exercises

67. not a circle 69. C A 13 , 2 32 B, r 5 3 2

2

71. 1x 2 22 1 1 y 2 72 5 2 75. x-axis, y-axis, origin

27. neither

73. right triangle

77. perpendicular

CHAPTER 3 79. y 5 223 6"9 2 Ax 2 13 B 2

Section 3.1 1. function 3. not a function 5. function 7. not a function 9. not a function 11. function 13. not a function 15. not a function 17. function 19. not a function 21. function 23. not a function 25. a. 5  b. 1  c.  23 27. a. 3  b. 2  c. 5 29. a.  25  b.  25  c.  25 31. a. 2  b.  28  c.  25

Practice Test 1. d 5 !82 3. d 5 !29, A 12 , 5B 5. y 5 1, 9 7. x-axis 9.

33. 1 35. 23 and 1 37. 324, 44

39. 6 41. 27 43. 6 276 45. 21 47. 233 49. 2

51. 3 53. 4 55. 82x2a 57. 2 59. 1 61. 2 63. 1 65. 12q, q2 67. 12q, q2

69. 12q, 52 < 15, q2 71. 12q, 222 < 122, 22 < 12, q2

11. x-intercepts: 16, 02; y-intercepts: 10, 222

13. y 5

8 3x

5 73. 12q, q2 75. 12q, 74 77. C22 , qB

28

15. y 5 x 1 5

17. y 5 22x 1 3

79. 12q, 224 < 32, q2 81. 13, q2 83. 12q, q2 A2q, 32 B 85. 12q, 242 < 124, q2 87.

89. 12q, 222 < 13, q2 91. 12q, 244 < 34, q2 12q, q2 93. A2q, 32 B 95.

97. x 5 22, 4 99. x 5 21, 5, 6

19. y 5 2x 1 2 2

2

23. 1x 2 42 1 1 y 2 92 5 20

21. 1x 2 622 1 1 y 1 722 5 64

25. both

Cumulative Test 1. 12 3. x4 2 32x2 1 256 52x 5. 7. 12 1 30i 51x 9. x 5 1 11. x 5 63 13. discriminant is negative; two complex (conjugate) roots 15. x 5 2

17. 10, 122

21. origin 23. x 5 5 25. C125, 232, r 5 !30

Young_AT_6160_ANS_SE_pg1169-1210.indd 1186

19. 122, 62

101. y 5 45x, 175, q2 103. T 162 5 64.88F, T 1122 5 908F 105. P1102 > $641.66, P11002 > $634.50 107. V1x2 = x 110 2 2x22, 10, 52

109. E 142 > 84 Yen E 172 > 84 Yen E 182 > 83 Yen 111. 229 people

113. a.   A1x2 5 x 1x − 3.3752 5 x2 − 3.375x b. A14.52 5 5.0625 window 1area of window < 5 sq in.2. c. A18.52 5 43.5625 (not possible because the window cannot be larger than the envelope).

115. Yes, because every input (year) corresponds to exactly one output (federal funds rate). 117. 11999, 25002, 12003, 37002, 12007, 47002, 12011, 59002, 12015, 66002

119. a.  F1502 5 0    b.  g1502 5 1000 c. H1502 5 2000

09/12/16 2:24 PM

Answers to Odd-Numbered Exercises  

121. You must apply the vertical line test instead of the horizontal line test. Applying the vertical line test would show that the graph given is actually a function.

53.

domain: 12q, q2 range: 12q, 24

increasing: 12q, 22

123. ƒ1x 112 2 ƒ1x2 1 ƒ112, in general.

decreasing: nowhere

125. G1211 h2 2 G1212 1 G1h2, in general.

constant: 12, q2

127. false 129. false 131. A52 133. C 5 25, D 5 22 135. 12q, 2a2 < 12a, a2 < 1a, q2 137.

55.

domain: 12q, q2 range: 30, q2

Warmest at noon: 908 F. Outside the interval 36, 184 the temperatures are too low.

increasing: 10, q2

decreasing: 121, 02

constant: 12q, 212 domain: 12q, q2

57.

139.

1187

range: 12q, q2

increasing: 12q, q2

lowest price $10, highest $642.38 agrees (if there is only 1 card for sale.)

decreasing: nowhere constant: nowhere

59.

141.

domain: 12q, q2 range: 31, q2

increasing: 11, q2

Graph of y2 can be obtained by shifting the graph of y1 two units to the right.

decreasing: 12q, 12 constant: nowhere 61.

Section 3.2 1. neither 3. even 5. odd

domain: 12q, 22 < 12, q2

10 9 8 7 6 5 4 3 2 1 -3

-2

-1

range: 11, q2 5-2*x 3*x-2 0

1

2

3

4

5

range: 321, 34

increasing: 121, 32

13. even 15. neither 17. neither

decreasing: nowhere

19. neither 21. neither 23. neither

constant: 12q, 212 < 13, q2

65.

domain: 12q, q2 range: 31, 44

27. a. 327, 24  b.  325, 44  c. increasing: 124, 02, decreasing: 127, 242 < 10, 22, constant: nowhere   d. 4   e. 1   f.  25

increasing: 11, 22

decreasing: nowhere

29. a. 12q, q2  b.  12q, q2  c. increasing: 12q, 232 < 14, q2, decreasing: nowhere, constant: 123, 42  d. 2   e. 2   f. 2

31. a. 12q, q2  b.  324, q2  c. increasing: 10, q2, decreasing: 12q, 02, constant: nowhere   d.  24  e. 0   f. 0

33. a. 12q, 02 < 10, q2  b.  12q, 02 < 10, q2 c. increasing: 12q, 02 < 10, q2, decreasing: nowhere, constant: nowhere   d. undefined   e. 3   f.  23

35. a. 12q, 02 < 10, q2  b.  12q, 52 < 374  c. increasing: 12q, 02, decreasing: 15, q2, constant: 10, 52  d. undefined   e. 3 f. 7

37. 2x 1 h 21 39. 2x 1 h 1 3 41. 2x 1 h 2 3

43. 26x 2 3h 1 5 45. 13 47. 1 49. 22 51. 21

Young_AT_6160_ANS_SE_pg1169-1210.indd 1187

decreasing: 12q, 22 domain: 12q, q2

63.

7. neither 9. odd 11. even

25. a. 12q, q2  b.  321, q2  c. increasing: 121, q2, decreasing: 123, 222, constant: 12q, 232 < 122, 212   d. 0   e. 21  f. 2

increasing: 12, q2

constant: 12q, 12 < 12, q2 domain: 12q, 222 < 122, q2

67.

range: 12q, q2

increasing: 122, 12

decreasing: 12q, 222 < 11, q2 constant: nowhere open holes 122,12, 122, 212, 11, 22

closed hole 11, 02

09/12/16 2:24 PM

Answers to Odd-Numbered Exercises

69.

domain: 12q, q2

97. 0 ft/sec

range: 30, q2

99. Demand for product is increasing at an approximate rate of 236 units over the first quarter.

decreasing: nowhere constant: 12q, 02

101. The domain is incorrect. It should be 12q, 02 < 10, q2. The range is also incorrect and should be 10, q2. The graph is also incorrect. It should contain an open circle at 10, 02.

increasing: 10, q2

71.

domain: 12q, q2

15 1   x 2 30

range: 12q, q2

decreasing: 12q, 02 < 10, q2

105. true 107. false 109. yes, if a 5 2b

constant: nowhere

111. odd 113. odd

increasing: nowhere

73.

closed hole 10, 02

103. The portion of C1x2 for x . 30 should be:

domain: 12q, 12 < 11, q2

e

1188 

Number miles beyond first 30

115. domain: all real numbers range: set of integers

range: 12q, 212 < 121, q2 increasing: 121, 12

decreasing: 12q, 212 < 11, q2 constant: nowhere open holes 121, 212, 11, 12, 11, 212

3 graph of 2! x on 12q, 212, closed hole 121, 12

domain: 12q, q2

75.

range: 12q, 22 < 34, q2

increasing: 12q, 222 < 10, 22 < 12, q2

decreasing: 122, 02 constant: nowhere

77.

open holes 122, 22, 12, 22

Section 3.3 1.   l  3.    a  5.    b  7.    i   9.    c  11.    g y 5 0 2x 0 5 0 x 0 13. y 5 0 x 0 1 3 15.

y 5 x3 2 4 17. y 5 3 0 x 0 19. y 5 1 2x 2 3 21. y 5 1 x 1 1 2 3 1 3 23. 25. 27.

closed holes 122, 12, 12, 42

domain: 12q, 12 < 11, q2 range: 12q, 12 < 11, q2

increasing: 12q, 12 < 11, q2 decreasing: nowhere constant: nowhere open hole 11, 12

79. Profit is increasing from October through December and ­decreasing from January through October. 10x, 81. C 1 x 2 5 • 9x, 8x,

87. R 1 x 2 5 e

31.

33.

35.

0 # x # 50 50 , x # 100 x . 100

250x, 83. C 1 x 2 5 e 175x 1 750,

85. C 1 x 2 5 e

29.

0 # x # 10 x . 10

1000 1 35x, 2000 1 25x,

0 # x # 100 x . 100

50,000 1 3x, 4x 2 50,000,

0 # x # 100,000 x . 100,000

89. P1x2 5 65x 2 800 91. ƒ 1 x 2 5 0.98 1 0.22 3 3 x 4 4 , x $ 0 93. ƒ 1 t 2 5 3 1 21 2 33t44, t $ 0

95. a.  20 million tons per year b.  110 million tons per year

Young_AT_6160_ANS_SE_pg1169-1210.indd 1188

09/12/16 2:24 PM

Answers to Odd-Numbered Exercises  

37.

39.

1189

71.

69.

73.

41.

43.

75. ƒ 1 x 2 5 1 x 2 3 2 2 1 2

45.

47.

77. ƒ 1 x 2 5 2 1 x 1 1 2 2 1 1 49.

51.

79. ƒ 1 x 2 5 2 1 x 2 2 2 2 2 5 53.

55.

81. S 1x2 5 10x and S 1x2 5 10x 1 50 83. T 1x2 5 0.331x 2 63002 57.

59.

85. a.  BSA 1 w 2 5

9w ; Å 200

b.  BSA 1 w 2 3 2 5

91w 2 32 Ç 200

87. (b) shifted to the right 3 units

61.

63.

89. (b) should be deleted since 0 3 2 x 0 5 0 x 2 3 0 . The correct sequence of steps would be: (a) S (c)* S (d ), where (c)*: Shift to the right 3. 91. true

65.

Young_AT_6160_ANS_SE_pg1169-1210.indd 1189

67.

95. 1a 1 3, b 1 22

93. true

97. Any part of the graph of ƒ 1x2 that is below the x-axis is reflected above it for 0 ƒ 1 x 2 0 . a.

b.

09/12/16 2:24 PM

1190 

Answers to Odd-Numbered Exercises

99. If 0 , a , 1, you have a horizontal stretch. If a . 1, the graph is a horizontal compression. a.

b.

1     domain: 12q, 21 2 ∪ 121, q 2 x11 1 1 g + ƒ 21 x 2 5 1 2    domain: 12q, 1 2 ∪ 1 1, q 2 x21 1 15. 1 ƒ + g 21 x 2 5     domain: 12q, 1 2 ∪ 1 1, q 2 0x 2 10

13. 1 ƒ + g 21 x 2 5

1

1 g + ƒ 21 x 2 5 101. Each horizontal line in the graph of y 5 Œ x œ is stretched twice in length. There is a vertical shift of one unit up.

0x0 2 1

    domain: 12q, 21 2 ∪ 121, 1 2 ∪ 1 1, q 2

17. 1 ƒ + g 21 x 2 5 !x 1 4    domain: 324, q2

1 g + ƒ 21 x 2 5 !x 2 1 1 5    domain: 31, q2 19. 1 ƒ + g 21 x 2 5 x domain: 12q, q2

1 g + ƒ 21 x 2 5 x domain: 12q, q2

26!3 21. 15 23. 13 25.

27.

110 11 31. 3!2 3 29.

33. undefined 35. undefined 37. 13

1. ƒ1x2 1 g 1x2 5 x 1 2 ƒ 1 x 2 2 g 1 x 2 5 3x s  domain: 12q, q 2 ƒ 1 x 2 ⋅g 1 x 2 5 22x 2 1 x 1 1 ƒ1x2 2x 1 1 5 domain: 12q, 1 2 ∪ 1 1, q 2 12x g 1x2

3. ƒ 1 x 2 1 g 1 x 2 5 3x 2 2 x 2 4 s  domain: 12q, q 2 ƒ 1 x 2 2 g 1 x 2 5 x 2 2 x 1 4 4 3 2 1 2 1 2 ƒ x ⋅ g x 5 2x 2 x 2 8x 1 4x

5.

ƒ1x2 2x 2 2 x   domain: 12q, 22 2 ∪ 122, 2 2 ∪ 1 2, q 2 5 2 g 1x2 x 24

1 1 x2 x 1 2 x2 ƒ1x2 2 g 1x2 5 x w  domain: 12q, 0 2 ∪ 1 0, q 2 ƒ1x2 ⋅ g 1x2 5 1 ƒ1x2 1 5 2 g 1x2 x ƒ1x2 1 g 1x2 5

39. ƒ 1 g 1 1 22 5

1 3

43. ƒ 1 g 1 1 22 5

1 3

41. ƒ1g1122 5 undefined  g 1  ƒ1222 5 undefined

9.



ƒ1x2 1 5    domain: 10, q2 2 g 1x2

ƒ 1 x 2 1 g 1 x 2 5 !4 2 x 1 !x 1 3 ƒ 1 x 2 2 g 1 x 2 5 !4 2 x 2 !x 1 3 s   domain: 323, 44 ƒ 1 x 2 ⋅g 1 x 2 5 !4 2 x⋅ !x 1 3 ƒ1x2 !4 2 x !x 1 3 5  domain: 123, 44 x13 g 1x2

11. 1 ƒ + g 21 x 2 5 2x 2 2 5       domain: 12q, q2

1 g + ƒ 21 x 2 5 4x 2 1 4x 2 2  domain: 12q, q2

Young_AT_6160_ANS_SE_pg1169-1210.indd 1190

g 1 ƒ 1 2 22 5 4

45. ƒ 1 g 1 1 22 5 !5  g1ƒ1222 5 6

47. ƒ1g1122 5 undefined  g1ƒ1222 5 undefined 3 g1ƒ1222 5 4 49. ƒ 1 g 1 1 22 5 "3 

51. ƒ 1 g 1 x 22 5 2a

x21 b 115x21115x 2

1 2x 1 1 2 21 2x 5 5x 2 2 53. ƒ 1 g 1 x 22 5 " 1 x 2 1 1 2 2 1 5 "x 2 5 0 x 0 5 x g 1 ƒ 1 x 22 5

Since x $ 1

g 1 ƒ 1 x 22 5 A !x 2 1B 2 1 1 5 1 x 2 1 2 1 1 5 x 55. ƒ 1 g 1 x 22 5

1 1 x

5x

57. ƒ 1 g 1 x 22 5 4a

5 4a

7.

ƒ 1 x 2 1 g 1 x 2 5 3!x    ƒ 1 x 2 2 g 1 x 2 5 2 !xs domain: 3 0, q 2 ƒ 1 x 2 ⋅ g 1 x 2 5 2x

g 1 ƒ 1 2 22 5 2

e

Section 3.4

g 1 ƒ 1 x 22 5

5

59. ƒ 1 g 1 x 22 5

g 1 ƒ 1 x 22 5

g 1 ƒ 1 x 22 5

1 1 x

5x

!x 1 9 2 b 29 2

x19 b29 5 x 4

" 1 4x 2 2 9 2 1 9 2 "4x 2 2x 5 5x 2 2 1 x11 x

21

1 x2 1 1 1 x2 1

1

5

5

1 x1 1 2x x 1 1x2 1 x2 1 1 x2 1

5

5

1 1 x

5x

x x21 1 x21

5x

61.  ƒ1x2 5 2x2 1 5x g1x2 5 3x 2 1

63. ƒ 1 x 2 5 0 2x 0   g 1 x 2 5 x 2 3 3 65. ƒ 1 x 2 5   g1x2 5 x 1 1 !x 2 2

09/12/16 2:24 PM

Answers to Odd-Numbered Exercises  

67. F 1 C 1 K 22 5 95 1 K 2 273.15 2 1 32

21. not one-to-one function

c.  A12002 5 2500 ft2

25.

27.

29.

31.

33.

35.

37.

39.

69. a.  A 1 x 2 5

A 4x B 2 

b.  A11002 5 625 ft2

1191

23. one-to-one function

71. a.  C1 p2 5 62,000 2 20p b. R1 p2 5 600,000 2 200p c. P1 p2 5 538,000 2 180p 73. a.  C1n1t22 5 210t2 1 500t 1 1375 b. C1n11622 5 6815 The cost of production on a day when the assembly line was running for 16 hours is $6,815,000. 75. a.  A 1 r 1 t 22 5 p 1 10t 2 0.2t 2 2 2

b. 11,385 square miles

77. A 1 t 2 5 p C 150!t D 2 5 22,500pt ft2 79. d 1 h 2 5 "h2 1 4

81. Must exclude 22 from the domain. 83. The operation is composition, not multiplication. 85. Function notation, not multiplication. 87. false

89. true 1 91. 1 g + ƒ 21 x 2 5 domain: x 2 0. x 93. 1 g + ƒ 21 x 2 5 x domain: C 2a, qB 95.

domain: 327, 94

41.

97.

domain: 1 2q, 3 2 ∪ 1 23, 21 4 ∪ 3 4, 6 2 ∪ 1 6, q 2

Section 3.5

1. function, not one-to-one 3. function, one-to-one 5. function, one-to-one 7. not a function 9. function, not one-to-one 11. function, not one-to-one 13. function, one-to-one 15. function, not one-to-one 17. not one-to-one function 19. one-to-one function

Young_AT_6160_ANS_SE_pg1169-1210.indd 1191

43. ƒ211x2 5 x 1 1 domain ƒ: 1 2q, q 2   domain ƒ21: 1 2q, q 2 range ƒ: 1 2q, q 2   range ƒ21: 1 2q, q 2 45. ƒ21 1 x 2 5 213 x 1

2 3

domain ƒ: 1 2q, q 2   domain ƒ21: 1 2q, q 2 range ƒ: 1 2q, q 2   range ƒ21: 1 2q, q 2

3 x21 47. ƒ21 1 x 2 5 ! domain ƒ: 1 2q, q 2   domain ƒ21: 1 2q, q 2 range ƒ: 1 2q, q 2   range ƒ21: 1 2q, q 2

49. ƒ21 1 x 2 5 x 2 1 3 domain ƒ: 3 3, q 2   domain ƒ21: 3 0, q 2 range ƒ: 3 0, q 2   range ƒ21: 3 3, q 2 51. ƒ21 1 x 2 5 !x 1 1

domain ƒ: 3 0, q 2   domain ƒ21: 3 21, q 2 range ƒ: 3 21, q 2   range ƒ21: 3 0, q 2

09/12/16 2:24 PM

1192 

Answers to Odd-Numbered Exercises

53. ƒ21 1 x 2 5 22 1 !x 1 3

domain ƒ: 3 22, q 2   domain ƒ21: 3 23, q 2 range ƒ: 3 23, q 2   range ƒ21: 3 22, q 2 55. ƒ21 1 x 2 5

2 x

domain ƒ: 1 2q, 0 2 ∪ 1 0, q 2 range ƒ: 1 2q, 0 2 ∪ 1 0, q 2 domain ƒ21: 1 2q, 0 2 ∪ 1 0, q 2 range ƒ21: 1 2q, 0 2 ∪ 1 0, q 2 57. ƒ21 1 x 2 5 3x x2 2 5 322x domain ƒ: 1 2q, 3 2 ∪ 1 3, q 2 range ƒ: 1 2q, 0 2 ∪ 1 0, q 2 domain ƒ21: 1 2q, 0 2 ∪ 1 0, q 2 range ƒ21: 1 2q, 3 2 ∪ 1 3, q 2 59. ƒ 1 x 2 5 21

5x 2 1 x1 7

domain ƒ: 1 2q, 5 2 ∪ 1 5, q 2

range ƒ: 1 2q, 27 2 ∪ 1 27, q 2

85. ƒ 1 x 2 5 "1 2 x 2,  0 # x # 1, 0 # y # 1, ƒ 21 1 x 2 5 "1 2 x 2,  both are 30, 14

0 # x # 1, 0 # y # 1, domain and range of

87. m 2 0

89. not one-to-one

91. not one-to-one





93. No, the functions are not 95. Yes, the functions are inverses of each other. Had we inverses of each other. restricted the domain of g1x2, then they would be inverses.

domain ƒ21: 1 2q, 27 2 ∪ 1 27, q 2 range ƒ21: 1 2q, 5 2 ∪ 1 5, q 2 61. not one-to-one

Section 3.6 1. y 5 kx 3. V 5 kx3 5. z 5 km k kw F5 11. v 5 kgt 7. ƒ 5 9. l L k 13. R 5 15. y 5 k !x 17. d 5 rt PT p 2 2 V5 hr 19. V 5 lwh 21. A 5 pr 23. 16 400,000 2p 19.2 25. V 5 27. F5 29. t5 P lL s 0.025m1m2 4.9 0.01L 31. R 5 2 33. R 5 35. F5 A I d2 37. W 5 7.5H 39. 1292 mph 41. F 5 1.618H

63. one-to-one

x 3 ƒ 21 1 x 2 5 c !x x

x # 21 21 , x , 1 x$1

65. ƒ 21 1 x 2 5 59 1 x 2 32 2 ; this now represents degrees Fahrenheit being turned into degrees Celsius. 67. C 1 x 2 5 b C 21 1 x 2 5

250x, 175x 1 750, x 250 , bx 2 750 175 ,

0 # x # 10 x . 10

49. 600 w/m

51. Bank of America: 1.5% Navy Federal Credit Union: 3% 53.

11 12

or 0.92 atm

55. Should be y is inversely proportional to x. 57. true 59. b 61. spl2 5 1.23C 2n k 7/6 L11/6

x . 2500

63. a.   The least-squares regression line is y 5 2.93x 1 201.72.

x 7.5 ,

71. domain: 30, 244 range: 397.5528, 101.704 73. domain: 397.5528, 101.704 range: 30, 244

75. No, it’s not a function because it fails the vertical line test. 77. ƒ is not one-to-one, so it does not have an inverse function. 79. false

45. $37.50 47. 20,000 2

0 # x # 2500

69. E 1 x 2 5 7.5x, E 1 x 2 5 x $ 0 tells you how many hours the student will have to work to bring home x dollars. 21

43. 24 cm

81. false 83. 1b, 02

Young_AT_6160_ANS_SE_pg1169-1210.indd 1192



b.  The variation constant is 120.07, and the equation of the direct variation is y 5 120.074x 0.259.



c. When the oil price is $72.70 per barrel in September 2006, the predicted stock index from the least-squares regression line is 415 and from the equation of direct ­variation is 364. The least-squares regression line gives a closer approximation to the actual value, 417.

65. a.  y 5 2141.73x 1 2,419.35 3217.69 b.  3,217.69, y 5 x 0.41

13/12/16 4:03 PM

Answers to Odd-Numbered Exercises  

c.  When the 5-year maturity rate is 5.02% in September 2006, the predicted number of housing units from the least-squares regression line is 1708, and from the equation of inverse variation is 1661. The equation of the least-squares regression line gives a closer approximation to the actual value, 1861. 67. a. y 5 0.123x 1 2.391   b.  About $3.38 per gallon. Yes. c. $3.74

1193

59. y 5 !x 1 3 domain: 323, q2

61. y 5 !x 2 2 1 3 domain: 32, q2

63. y 5 5!x 2 6 domain: 30, q2 65. y 5 1x 1 222 2 12

Review Exercises 1. yes 3. yes 5. no 7. yes 9. no 11. a. 2   b. 4   c.  x 5 23, 4 13. a. 0   b.  22  c.  x 5 25, 2 15. 5 17. 2665 19. 22 21. 4 1 2q, 24 2 ∪ 1 24, q 2 23. 12q, q2 25.

27. 34, q2 29. D 5 18

31. neither

33. odd 35. neither 37. odd 39. a.  324, 74  b.  322, 44

c. increasing: 13, 72, decreasing: 10, 32, constant: 124, 02 41. 22 43.

domain: 12q, q2 range: 10, q2

open hole 10, 02, closed hole 10, 22

45.

domain: 12q, q2 range: 321, q2 open hole 11, 32

closed hole 11, 212

25,     x # 2, 47. C 1 x 2 5 b 25 1 10.50 1 x 2 2 2 ,    x . 2 49.

51.

67. g1x2 1 h1x2 5 22x 2 7 g1x2 2 h1x2 5 24x 2 1

t

g1x2 ? h1x2 5 23x2 1 5x 1 12 domain: 12q, q2 g 1x2 23x 2 4 5  domain: 1 2q, 3 2 ∪ 1 3, q 2 x23 h1x2 1 69. g 1 x 2 1 h 1 x 2 5 2 1 !x x 1 g 1 x 2 2 h 1 x 2 5 2 2 !x x w domain: 10, q2 1    g 1 x 2 ⋅h 1 x 2 5 3/2       x g1x2 1 5 5/2     h1x2 x 71. g 1 x 2 1 h 1 x 2 5 !x 2 4 1 !2x 1 1

g 1 x 2 2 h 1 x 2 5 !x 2 4 2 !2x 1 1  domain: 34, q2 t

g 1 x 2 ⋅ h 1 x 2 5 !x 2 4 ⋅ !2x 1 1 g1x2 !x 2 4 5 domain: 34, q2 h1x2 !2x 1 1

73. 1 ƒ + g 21 x 2 5 6x 2 1  domain: 12q, q2 1 g + ƒ 21 x 2 5 6x 2 7  domain: 12q, q2

75. 1 ƒ + g 21 x 2 5

8 2 2x 13 2 3x

13   domain: 1 2q, 4 2 ∪ Q4, 13 3 R ∪ Q 3 , qR

1 g + ƒ 21 x 2 5

53.

55.

x13 4x 1 10

  domain: 12q, 23 2 ∪ Q23, 252 R ∪ Q252 , qR 77. 1 ƒ + g 21 x 2 5 "x 2 2 9

  domain: 1 2q, 23 4 ∪ 3 3, q 2

1 g + ƒ 21 x 2 5 x 2 9  domain: 35, q2

79. ƒ1g 1322 5 857,  g 1ƒ12122 5 51 57.

81. ƒ 1 g 1 3 22 5

17 31 , 

g 1ƒ12122 5 1

83. ƒ1g 1322 5 12,  g 1ƒ12122 5 2 85. ƒ1x2 5 3x2 1 4x 1 7,  g 1x2 5 x 2 2 1 87. ƒ 1 x 2 5 ,  g 1x2 5 x2 1 7 !x 89. A 1t2 5 625p 1t 1 22 in2 91. yes 93. no

Young_AT_6160_ANS_SE_pg1169-1210.indd 1193

95. yes 97. yes 99. yes

09/12/16 2:24 PM

1194 

Answers to Odd-Numbered Exercises

101.

103.

127. yes

125. domain: 3 21.5, 4 2





x21 2 domain ƒ: 12q, q2 domain ƒ21: 12q, q2 105. ƒ21 1 x 2 5 12 1 x 2 1 2 5

129. a. y 5 64.67x 1 805

range ƒ: 12q, q2 range ƒ21: 12q, q2

b. No, it is not close to the actual price at $517.20.

domain ƒ: 324, q2 domain ƒ21: 30, q2

Practice Test

107. ƒ211x2 5 x2 2 4

range ƒ: 30, q2 range ƒ : 324, q2 21

6 2 3x x21 domain ƒ: 12q, 23 2 ∪ 123, q 2 109. ƒ21 1 x 2 5

!x 2 2 5. 2  domain: 32, q2 x 1 11 7. x 1 9  domain: 12, q2 13.

domain ƒ 21: 12q, 1 2 ∪ 1 1, q 2

111. S1x2 5 22,000 1 0.08 x,  S 21 1 x 2 5

1. b 3. c

9. 4 11. neither

range ƒ: 12q, 1 2 ∪ 1 1, q 2

range ƒ 21: 12q, 23 2 ∪ 123, q 2

c. $1581.05

domain: 33, q2

range: 12q, 24

x 2 22,000 , 0.08

sales required to earn a desired income 113. C 5 2pr 115. A 5 pr 2 117. W 5 8.5H 119. domain: 12q, 212 ∪ 13, q2

15.

domain: 12q, 21 2 ∪ 121, q 2

y 8 7 6 5 4 3 2

range: 31, q2 x

–5

–3

–1

1 2 3 4 5

–2

121. a.  12q, 22 ∪ 12, q2  b.  521, 0, 16 ∪ 12, q2

c. increasing: 12, q2, decreasing: 12q, 212, constant: 121, 02 ∪ 10, 12 ∪ 11, 22 4 3 2 1

–3 –2 – 1



1 –1 –2 –3

17. a.  22  b. 4   c.  23  d.  x 5 23, 2 19. 6x 1 3h 2 4 21. 232 23. ƒ211x2 5 x2 1 5

domain ƒ: 35, q2

2

3

123. The graph of ƒ can be obtained by shifting the graph of g two units to the left. That is, ƒ1x2 5 g1x 1 22.

range ƒ: 30, q2

domain ƒ21: 30, q2

range ƒ21: 35, q2

5x 2 1 x12 domain ƒ: 12q, 5 2 ∪ 1 5, q 2

25. ƒ21 1 x 2 5

range ƒ: 12q, 22 2 ∪ 122, q 2

domain ƒ21: 12q, 22 2 ∪ 122, q 2 range ƒ21: 12q, 5 2 ∪ 1 5, q 2 27. 30, q2

29. S1x2 5 0.42 x where x is the original price of a suit 31. quadrant III, “quarter of unit circle” 15, 33. C 1 x 2 5 e x 2 15,

35. F 5

Young_AT_6160_ANS_SE_pg1169-1210.indd 1194

0 # x # 30 x . 30

30 m P

09/12/16 2:24 PM

Answers to Odd-Numbered Exercises  

37. yes

1195

23. ƒ1x2 5 1x 1 322 2 12

21.

25. ƒ1x2 5 21x 1 522 1 28 27. ƒ1x2 5 21x 1 222 2 10 29. ƒ1x2 5 241x 2 222 1 9 31. ƒ1x2 5 12 1 x 2 4 2 2 2 5 33.



35.

Cumulative Test 3 1 !5 3. x 1 x 2 2 2 , x 2 22 2 5. 145 1 0i 7. 40% 9. x 5 1 6 1.

11. x 5 62, 6i !3 13. 10, 52

15. d 5 11.03, 1 1.25, 2.45 2 17. y 5 23

!35 5

37.



39.

19. 1 x 1 2 2 1 1 y 1 1 2 5 20 21. 1 2q, 1 2 ∪ 1 1, q 2 2

2

23. 46

25. r 5

27. h 1 x 2 5 x 2 2

135 t

1 494 41. A 33 , 33 B

45. 1275,12.952



43. A7, 2 39 2 B 829 47. A 15 28 , 392 B

2

49. y 5 221x 1 12 1 4

51. y 5 251x 2 222 1 5

53. y 5 59 1 x 1 1 2 2 2 3

3 2 55. y 5 12 A x 2 21 B 24

57. y 5 54 1 x 2 2.5 2 2 2 3.5 59. a.  350,000 units

b. $12,262,500

61. Gaining: January 2010; Losing: February 2010–June 2011

CHAPTER 4

63. a.  120 feet

Section 4.1

65. 2,083,333 sq. ft

1. b

3. a

9.

5. b 11.



7. c

b. 50 yards

67. a. 1 second, 116 feet

b. 3.69 seconds

69. a.  26,000 feet

b. 8944 feet

71. a.  100 boards

b. $24,000

73. 15 to 16 units to break even or 64 to 65 units to break even. 75. a.  ƒ 1 t 2 5

349 2 162 t

1 16

b. 878 million

77. a.  y5 20.011t 2 22522 1 400 b. 425 minutes 13.



15.

79. The corrections that need to be made are vertex 123, 212 and x-intercepts 122, 02 124, 02.

81. ƒ1x2 5 21x 2 122 1 4. The negative must be factored out of the x2 and x terms. 83. true 87. ƒ 1 x 2 5 a Ax 1

85. false b 2 2a B

1

4ac 2 b2 4a

89. a. The maximum area of the rectangular pasture is 62,500 square feet. 17.



19.

b. The maximum area of the circular fence is approximately 79,577 square feet. 91. a.  11425, 4038.252 c.  14.04, 02, 12845.96, 02 d.  x 5 1425

Young_AT_6160_ANS_SE_pg1169-1210.indd 1195

b. 10, 2232

09/12/16 2:24 PM

1196 

Answers to Odd-Numbered Exercises

93. a.  y 5 22x 2 1 12.8x 1 4.32 b. y 5 22 1 x 2 3.2 2 2 1 24.8, 1 3.2, 24.8 2

23.

25.

95. a. 

29. 0 1multiplicity 22, 7 1multiplicity 22, 24 1multiplicity 12

c. Yes

27. 3 1multiplicity 12, 24 1multiplicity 32

Ben’s Data

Vertical Distance (in feet)

45 40

31. 0 1multiplicity 22, 1 1multiplicity 22

35

33. 0 1multiplicity 12, 32 1multiplicity 12, 294 1multiplicity 12

30 25

35. 0 1multiplicity 22, 23 1multiplicity 12

20 15

37. 0 1multiplicity 42

10 5 0 0

2

4 Time (in seconds)

6

8

45. P1x2 5 x2 2 2x 2 1 47. P1x2 5 x2  1x 1 223

49. P1x2 5 1x 1 322  1x 2 725

Ben’s Data

2

51. P 1 x 2 5 x 2 1 x 1 1 2 Ax 1 !3B Ax 2 !3B

45 Vertical Distance (in feet)

41. P1x2 5 x 1x 1 52 1x 1 32 1x 2 2 2 1x 2 62 43. P1x2 5 12x 1 12 13x 2 22 14x 2 32

b. The equation of the best fit parabola is y 5 21.0589x2 1 12.268x 1 4.3042 with r2 5 0.9814. This is shown below on the scatterplot:

39. P1x2 5 x 1x 1 32 1x 2 12 1x 2 22

40

2

53. ƒ1x2 5 21x 1 322  a. 23 1multiplicity 22 b. touches at 23  c. 10, 292  d. falls left and right, without bound  e.

35 30 25 20 15 10 5 0

0

2

4 Time (in seconds)

6

8

c. i. Using the equation from part 1b2, the initial height from which the pumpkin was thrown is about 4.3 feet, and the initial velocity is about 12.3 feet per second.

55. ƒ1x2 5 1x 2 223  a. 2 1multiplicity 32 b. crosses at 2  c. 10, 282  d. falls left, rises right  e. 

ii. The maximum height occurs at the vertex and is ­approximately 39.8 feet; this occurs about 5.85 seconds into the flight. iii. The pumpkin lands approximately 12 seconds after it is launched.

Section 4.2 1. polynomial; degree 2

3. polynomial; degree 5

5. not a polynomial

7. not a polynomial

9. not a polynomial 13. b

57. ƒ1x2 5 x 1x 2 32 1x 1 32  a. 0, 3, 23 1multiplicity 12 b. crosses at each zero  c. 10, 02  d. falls left, rises right  e. 

11. h 15. e

17. c

19. 21. 59. ƒ1x2 5 2x 1x 2 22 1x 1 12  a. 0, 2, –1 1multiplicity 12  b. crosses at each zero c. 10, 02  d. falls right, rises left e. 

Young_AT_6160_ANS_SE_pg1169-1210.indd 1196

09/12/16 2:24 PM

Answers to Odd-Numbered Exercises  

61. ƒ1x2 5 2x31x 1 32  a. 0 1multiplicity 32, 23 1multiplicity 12  b. crosses at both 0 and 23 c. 10, 02  d. falls left and right, without bound e. 

1197

75. a. 0 1multiplicity 22, 22 1multiplicity 22, 32 1multiplicity 12 b. odd  c. positive  d. 10, 02 e. ƒ1x2 5 x212x 2 32 1x 1 222

77. a. Revenue for the company is increasing when advertising costs are less than $400,000. Revenue for the company is decreasing when advertising costs are between $400,000 and $600,000. b. The zeros of the revenue function occur when $0 and $600,000 are spent on advertising. When either $0 or $600,000 is spent on advertising, the company’s revenue is $0.

63. ƒ1x2 5 12x41x 2 42 1x 1 12  a. 0 1multiplicity 42, 4 1multiplicity 12, 21 1multiplicity 12 b. touches at 0 and crosses at 4 and 21 c. 10, 02  d. rises left and right, without bound e. 

79. The velocity of air in the trachea is increasing when the radius of the trachea is between 0 and 0.45 cm and decreasing when the radius of the trachea is between 0.45 cm and 0.65 cm.

v 5

4

3

2

1 r 0.2

65. ƒ1x2 5 2x31x 2 42 1x 1 12  a. 0 1multiplicity 32, 4 1multiplicity 12, 21 1multiplicity 12 b. crosses at each zero  c. 10, 02 d. falls left, rises right  e. 

67. ƒ1x2 51x 2 22 1x 1 22 1x 2 12 a. 1, 2, 221multiplicity 12  b. crosses at each zero c. 10, 42  d. falls left, rises right e. 

0.4

81. Sixth-degree polynomial because there are five known turning points. 83. down

85. 6th degree

87. If h is a zero, then 1x 2 h2 is a factor. So, in this case the function would be: P1x2 5 1x 1 22 1x 1 12 1x 2 32 1x 2 42

89. ƒ1x2 5 1x 2 1221x 1 223 Yes, the zeros are 22 and 1. But you must remember that this is a fifth-degree polynomial. At the 22 zero the graph crosses, but it should be noted that at 1 it only touches at this value. The correct graph would be the following:

91. false 69. ƒ1x2 521x 1 2221x 2 122  a. 22 1multiplicity 22, 1 1multiplicity 22  b. touches at both 22 and 1 c. 10, 242  d. falls left and right, without bound e. 

71. ƒ1x2 5 x21x 2 2231x 1 322  a. 0 1multiplicity 22, 2 1multiplicity 32, 23 1multiplicity 22 b. touches at both 0 and 23, and crosses at 2 c. 10, 02  d. falls left, and rises right e. 

0.6

93. true 2

5

95. n 4

97. ƒ1x2 5 1x 1 12 1x 2 32 , g1x2 5 1x 1 12 1x 2 323

   h1x2 5 1x 1 1261x 2 32

99. The zeros of the polynomial are 0, a, and 2b. 101. no x-intercepts

103. y 5 22x5 , yes

73. a. 23 1multiplicity 12, 21 1multiplicity 22, 2 1multiplicity 12 b. even  c. negative  d. 10, 62 e. ƒ1x2 5 21x 1 1221x 2 22 1x 1 32

Young_AT_6160_ANS_SE_pg1169-1210.indd 1197

09/12/16 2:24 PM

1198 

Answers to Odd-Numbered Exercises

105. x-intercepts: 122.25, 0 2 , 1 6.2, 0 2 , 1 14.2, 0 2 zeros: 22.25 ­1multiplicity 22, 6.2 1multiplicity12, 14.2 1multiplicity 12

61. The width is 3x2 1 2x 1 1 feet. 63. The trip takes x2 1 1 hours. 65. In long division, you must subtract 1not add2 each term. 67. Forgot the “0” placeholder.

107.  122.56, 217.122, 120.58, 12.592, 11.27, 211.732

Section 4.3

69. true

71. false

73. yes

75. Q1x2 5 x2n 1 2x n 1 1, r 1x2 5 0

77. 2x 2 1; linear 79. x3 2 1; cubic

8 1. 26x2 1 16x 2 10; quadratic function

1. Q1x2 5 2x 2 1, r1x2 5 0 3. Q1x2 5 x 2 3, r1x2 5 0 5. Q1x2 5 3x 2 3, r1x2 5 211 7. Q1x2 5 3x 2 28, r1x2 5 130 9. Q1x2 5 x 2 4, r1x2 5 12 11. Q1x2 5 3x 1 5, r1x2 5 0

Section 4.4

13. Q1x2 5 2x 2 3, r1x2 5 0

1. ƒ112 5 0

15. Q1x2 5 4x2 1 4x 1 1, r1x2 5 0 17. Q1x2 5 2x 2 2 x 2 12 , r1x2 5

3. g112 5 4

5. ƒ1222 5 84

7. Yes, the given is a zero of the polynomial.

15 2

9. Yes, the given is a zero of the polynomial.

19. Q1x2 5 4x2 2 10x 2 6, r1x2 5 0

11. 24, 1, 3; P1x2 5 1x 2 12 1x 1 42 1x 2 32

21. Q1x2 5 22x2 2 3x 2 9, r1x2 5 227x2 1 3x 1 9

13. 23, 12 , 2; P1x2 5 12x 2 12 1x 1 32 1x 2 22

23. Q1x2 5 x2 1 1, r1x2 5 0 25. Q1x2 5 x 2 1 x 1 16 , r1x2 5 2 121 6 x1

121 3

27. Q1x2 5 23x3 1 5.2x2 1 3.12x 2 0.128, r1x2 5 0.9232 29. Q1x2 5 x2 2 0.6x 1 0.09, r1x2 5 0

15. 23, 5; P1x2 5 1x2 1 42 1x 2 52 1x 1 32

17. 23, 1; P1x2 5 1x 2 12 1x 1 32 1x2 2 2x 1 22

19. 22, 21 1both multiplicity 22; P1x2 5 1x 1 222 1x 1 122 21. 61, 62, 64 612 ,

23. 61, 62, 63, 64, 66, 612

31. Q1x2 5 3x 1 1, r1x2 5 0

25.

33. Q1x2 5 7x 2 10, r1x2 5 15

27. 61, 62, 64, 65, 610, 620, 615 , 625 , 645

35. Q1x2 5 2x3 1 3x 2 2, r1x2 5 0

29. 61, 62, 64, 68; rational zeros: 24, 21, 2, 1

61, 62, 64, 68

37. Q1x2 5 x 2 x 1 x 2 1, r1x2 5 2

31. 61, 63, 612 , 632 ; rational zeros: 12 , 1, 3

39. Q1x2 5 x3 2 2x2 1 4x 2 8, r1x2 5 0

33.

3

2

2

41. Q1x2 5 2x 2 6x 1 2, r1x2 5 0 43. Q 1 x 2 5 2x 2 3

3

5 2 3x

1

53 9x

1

2

106 27 ,

r1x2 5

POSITIVE REAL ZEROS

2 112 81

1

45. Q1x2 5 2x 1 6x 2 18x 2 54, r1x2 5 0

47. Q1x2 5 x6 1 x5 1 x4 2 7x3 2 7x2 2 4x 2 4, r1x2 5 23

POSITIVE REAL ZEROS

r1x2 5 0

1

51. Q1x2 5 2x 2 7, r1x2 5 0

55. Q1x2 5 x4 1 2x3 1 8x2 1 18x 1 36, r1x2 5 71 57. Q1x2 5 x2 1 1, r1x2 5 224 59. Q1x2 5 x6 1 x5 1 x4 1 x3 1 x2 1 x 1 1, r1x2 5 0

Young_AT_6160_ANS_SE_pg1169-1210.indd 1198

1

35.

49. Q 1 x 2 5 x 5 1 !5 x 4 2 44x 3 2 44!5 x 2 2 245x 2 245!5, 53. Q1x2 5 x2 2 9, r1x2 5 0

NEGATIVE REAL ZEROS

NEGATIVE REAL ZEROS

0

37. POSITIVE REAL ZEROS

NEGATIVE REAL ZEROS

2

1

0

1

09/12/16 2:24 PM

Answers to Odd-Numbered Exercises  

1199

51. a.  Number of sign variations for P1x2: 4

39. POSITIVE REAL ZEROS

NEGATIVE REAL ZEROS

1

POSITIVE REAL ZEROS

1

41. POSITIVE REAL ZEROS

NEGATIVE REAL ZEROS

2

2

0

2

2

0

0

0

   Number of sign variations for P12x2: 0 NEGATIVE REAL ZEROS

4

0

2

0

0

0

b. possible rational zeros: 61, 62, 613, 626 c. rational zeros: 1, 2 d. P1x2 5 1x 2 12 1x 2 22 1x2 2 4x 1 132

43.

53. a.  Number of sign variations for P1x2: 2 POSITIVE REAL ZEROS

NEGATIVE REAL ZEROS

4

0

2

0

0

0

   Number of sign variations for P12x2: 1 POSITIVE REAL ZEROS

NEGATIVE REAL ZEROS

2

1

0

1

45. a.  Number of sign variations for P1x2: 0    Number of sign variations for P12x2: 3 POSITIVE REAL ZEROS

1 b. possible rational zeros: 61, 612 , 615 , 610

NEGATIVE REAL ZEROS

0

3

0

1

b. possible rational zeros: 61, 62, 63, 66 c. rational zeros: 21, 22, 23 d. P1x2 5 1x 1 12 1x 1 22 1x 1 32

47. a.  Number of sign variations for P1x2: 2    Number of sign variations for P12x2: 1 POSITIVE REAL ZEROS

1

0

1

b. possible rational zeros: 61, 67

49. a.  Number of sign variations for P1x2: 1

POSITIVE REAL ZEROS

NEGATIVE REAL ZEROS

1

2

1

0

61, 62, 65, 610, 612 , 6 13 , 6 16 , 623 ,

652 , 653 , 656 , 610 3

c. rational zeros: 21, 252 , 23 5 2 d. P1x2 5 61x 1 12 ax 1 b ax 2 b 2 3

POSITIVE REAL ZEROS

   Number of sign variations for P12x2: 2 NEGATIVE REAL ZEROS

1

2

1

0

b. possible rational zeros: 61, 62, 65, 610

Young_AT_6160_ANS_SE_pg1169-1210.indd 1199

   Number of sign variations for P12x2: 2

Number of sign variations for P12x2: 0

d. P1x2 5 1x 1 12 1x 2 12 1x 2 72

d. P1x2 5 x 1 x 2 12 1x 1 22 1x 1 52

55. a.  Number of sign variations for P1x2: 1

57. a. Number of sign variations for P1x2: 4

c. rational zeros: 21, 1, 7

c. rational zeros: 0, 1, 22, 25

d. P1x2 5 1x 2 12 1 2x 1 12 15x 2 12

b. possible rational zeros:

NEGATIVE REAL ZEROS

2

POSITIVE REAL ZEROS

c. rational zeros: 21, 212 , 15

NEGATIVE REAL ZEROS

4

0

2

0

0

0

b. possible rational zeros: 61, 62, 64 c. rational zeros: 1 d. P1x2 5 1x 2 122 1x2 1 42

09/12/16 2:24 PM

1200 

Answers to Odd-Numbered Exercises

59. a.  Number of sign variations for P1x2: 1

79. It is true that one can get 5 negative zeros here, but there may be just 1 or 3.

Number of sign variations for P12x2: 1 POSITIVE REAL ZEROS

POSITIVE REAL ZEROS

NEGATIVE REAL ZEROS

1

1

b. possible rational zeros: 61, 62, 63, 64, 66, 69, 612, 618, 636 c. rational zeros: 21, 1

NEGATIVE REAL ZEROS

0

5

0

3

0

1

81. true 83. false

85. b, c

87. possible rational zeros: 61, 62, 64, 68, 616, 632 zeros: 2

d. P1x2 5 1x 1 12 1x 2 12 1x2 1 92 1x2 1 42 61. a.  Number of sign variations for P1x2: 4

Number of sign variations for P12x2: 0 POSITIVE REAL ZEROS

NEGATIVE REAL ZEROS

4

0

2

0

0

0

89. a.  2 43 , 23 b. P 1 x 2 5 1 3x 2 2 21 4x 1 3 21 x 2 1 2x 1 5 2

b. possible rational zeros: 61, 65, 612 , 614 , 6 52 , 654 c. rational zeros: 12 d. P 1 x 2 5 4Ax 2 12 B 2 1 x 2 2 4x 1 5 2 63. 65.

Section 4.5 1. x 5 62i; P1x2 51x 1 2i 2 1x 2 2i 2

3. x 5 1 6 i; P1x2 5 1x 2 11 2 i 22 1x 2 11 1 i 22

5. x 5 62, 62i; P1x2 5 1x 2 22 1x 1 22 1x 2 2i2 1x 1 2i 2 7. x 5 6 !5, 6 !5 i ; P 1 x 2 5 Ax 2 !5B Ax 1 !5B Ax 2 !5 iB Ax 1 !5 iB

9. 2i 67. x 5 1.34 69. x 5 0.22

11. 22i, 3 1 i

13. 1 1 3i, 2 2 5i 3

2

17. P1x2 5 x 2 2x 1 5x

15. i, 1 1 i 19. P1x2 5 x3 2 3x2 1 28x 2 26

21. P1x2 5 x4 2 2x3 1 11x2 2 18x 1 18 23. 62i, 23, 5; P1x2 5 1x 2 2i 2 1x 1 2i 2 1x 2 52 1x 1 32 25. 6i, 1, 3; P1x2 51x 2 i2 1x 1 i 2 1x 2 32 1x 2 12

27. 63i, 1 1multiplicity 22; P1x2 51x 2 3i 2 1x 1 3i 2 1x 2 122

71. x 5 20.43

29. 1 6 i, 21 6 2!2; P 1 x 2 5 1 x 2 1 1 1 i 221 x 2 1 12 i 221 x2 A212 2!2BB Ax2 A2112!2BB

31. 3 6 i, 62; P1x2 5 1x 2 13 1 i 22 1x 2 13 2 i 22 1x 2 22 1x 1 22 33. 2 6 i, 1, 4; P1x2 5 1x – 12 1 i 22 1x 2 12 2 i 22 1x 2 12 1x 2 42 35. P1x2 5 1x 1 3i 2 1x 2 3i 2 1x 2 12

2

73. a.  P1x2 5 23x 2 2x 1 26, x $ 0 b. 263 subscribers 2

75. P1x2 5 20.0002x 1 8x 2 1500; 0 or 2 positive real zeros. 77. 18 hours

37. P1x2 5 1x 1 i 2 1x 2 i 2 1x 2 52

39. P1x2 5 1x 1 2i 2 1x 2 2i 2 1x 1 12

41. P 1 x 2 5 1 x 2 3 21 x 2 1 21 1 !5i 221 x 2 1 21 2 !5i 22

43. P1x2 5 1x 1 32 1x 2 52 1x 1 2i2 1x 2 2i2 45. P1x2 5 1x 1 12 1x 2 52 1x 1 2i2 1x 2 2i2

47. P1x2 5 1x 2 12 1x 2 22 1x 2 12 2 3i 22 1x 2 12 1 3i 22

Young_AT_6160_ANS_SE_pg1169-1210.indd 1200

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1201

Answers to Odd-Numbered Exercises  

49. P1x2 5 21x 1 12 1x 2 22 1x 2 12 2 i 22 1x 2 12 1 i 22

15. HA: none VA: x 5 12 , x 5 243

51. P1x2 5 1x 2 1221x 1 2i2 1x 2 2i2

17. HA: y 5 13 VA: none

53. P1x2 5 1x 2 12 1x 1 12 1x 2 2i 2 1x 1 2i 2 1x 2 3i2 1x 1 3i2

19. HA: y 5 12 ­7 VA: x 5 0.5, x 5 21.5

57. P 1 x 2 5 1 x 2 1 21 x 1 1 21 3x 2 2 21 x 2 2i 21 x 1 2i 2

55. P1x2 5 12x 2 1221x 2 12 2 i22 1x 2 12 1 i 22

21. y 5 x 1 6

23. y 5 2x 1 24

25. y 5 4x 1

27. b

29. a

31. e

59. Yes—since there are no real zeros, this function is either always positive or always negative; the end behavior is similar to an even power function with a positive leading coefficient 1rising in both ends2. We expect profit to eventually rise 1without bound!2.

33.

61. No—since there is only one real zero 1and it corresponds to producing a positive number of units2, then that is the “break-even point”; the end behavior is similar to an odd power function with a negative leading coefficient 1falling to the right2. We expect profit to fall 1without bound!2 after the break-even point. 63. Since the profit function is a third-degree polynomial, we know that the function has three zeros and at most two turning points. Looking at the graph, we can see there is one real zero where t # 0. There are no real zeros when t . 0; therefore, the other two zeros must be complex conjugates. Therefore, the company always has a profit greater than approximately $5.1 million and, in fact, the profit will increase towards infinity as t increases.

65. Since the concentration function is a third-degree polynomial, we know the function has three zeros and at most two turning points. Looking at the graph, we can see there will be one real zero where t $ 8. The remaining zeros are a pair of complex conjugates. Therefore, the concentration of the drug in the bloodstream will decrease to zero as the hours go by. Note that the concentration will not approach negative infinity since concentration is a nonnegative quantity.

11 2

35.

3 7. 39.

4 1. 43.

67. Just because 1 is a zero does not mean 21 is a zero 1holds only for complex conjugates2. Should have used synthetic division with 1. 69. false

71. true

73. No—if there is an imaginary zero, then its conjugate is also a zero. Therefore, all imaginary zeros can only correspond to an even-degree polynomial.

45.

47.

75. P1x2 5 x6 1 3b2 x4 1 3b4 x2 1 b6 77. all roots are complex REAL ZEROS

COMPLEX ZEROS

0

4

2

2

4

0

49. 51.

53. 55. 79. 35 , 6i, 62i; P1x2 5 251x 2 0.62 1x 2 2i2 1x 1 2i2 1x 2 i2 1x 1 i2



Section 4.6 1. 12q, 23 2 ∪ 123, q 2

3. A2 q, 2 @3 B ∪ A2 @3, @2 B ∪ A @2, qB

7. 1 2q, q 2

9. 12q, 22 2 ∪ 122, 3 2 ∪ 1 3, q 2

1

1

1

5. 12q, 24 2 ∪ 124, 3 2 ∪ 1 3, q 2

1

y 5 4 3 2 1

–5 –3

–1

x 1 2 3 4 5

–3 –4 –5

11. HA: y 5 0 VA: x 5 22 13. HA: none VA: x 5 25

Young_AT_6160_ANS_SE_pg1169-1210.indd 1201

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1202 

Answers to Odd-Numbered Exercises

57.

89. HA: y 5 0  VA: x 5 0, x 5 213

Intercepts: A225 , 0B, yes 

59. a. x-intercept: 12, 02; y-intercept: 10, 0.52 b. HA: y 5 0  VA: x 5 21, x 5 4 x22 c. f 1 x 2 5 1 x 1 1 21 x 2 4 2

61. a. x-intercept: 10, 02; y-intercept: 10, 02 b. HA: y 5 23 VA: x 5 24, x 5 4 23x 2 c. f 1 x 2 5 1 x 1 4 21 x 2 4 2

91. a. f: HA: y 5 0 VA: x 5 3 g: HA: y 5 2 VA: x 5 3 h: HA: y 5 23 VA: x 5 3 b. graphs of ƒ and g: as x S 6q, f 1 x 2 S 0 and g 1 x 2 S 2

63. a. C112 > 0.0198  b.  C1602 > 0.0324 c. C13002 > 0 d. y 5 0; after several days, C1t2 > 0 65. a. N102 5 52 wpm

b. N1122 > 107 wpm

c. N1362 > 120 wpm

d. y 5 130; 130 wpm

c. graphs of g and h below: as x S 6q, g 1 x 2 S 2 and h 1 x 2 S 23

67. y 5 10, 10 ounces of food 69.

2w2 1 1000 w

71. 2000 or 8000 units; average profit of $16 per unit. 73. The concentration of the drug in the bloodstream 15 hours after taking the dose is approximately 25.4 µg/mL. There are two times, 1 hour and 15 hours, after taking the medication at which the concentration of the drug in the bloodstream is approximately 25.4 µg/mL. The first time, approximately 1 hour, occurs as the concentration of the drug is increasing to a level high enough that the body will be able to maintain a concentration of approximately 25 µg/mL throughout the day. The second time, approximately 15 hours, occurs many hours later in the day as the concentration of the medication in the bloodstream drops.

d. g 1 x 2 5

2x 2 5 23x 1 10 , h1x2 5 x23 x23

Yes, if the degree of the numerator is the same as the degree of the denominator, then the horizontal asymptote is the ratio of the leading coefficients for both g and h.

Review Exercises 1. b

3. a

5.    7.



75. There is a common factor in the numerator and denominator. There is a hold at x 5 1. 77. The horizontal asymptote was found incorrectly. The correct ­horizontal asymptote is y 5 21. 79. true

9. f 1 x 2 5 Ax 2 32 B 2 49 4 2

81. false

83. HA: y 5 1  VA: x 5 c, x 5 2d

11. ƒ1x2 5 41x 1 122 2 11

13. 15.

2

85. Two possibilities: y 5 y5

4x 5 1x 1 3231x 2 122

4x and 1 x 1 3 21 x 2 1 2

87. VA: x 5 22, yes

5 599 2 451 1 7. A 26 , 52 B 19. A215 , 125 B 1 2 21. y 5 9 1 x 1 2 2 1 3 23. y 5 5.61x 2 2.722 1 3.4

25. a. P 1 x 2 5 22x 2 1 35 3 x 2 14 b.  x > 4.1442433, 1.68909 c. d. 11.6891, 4.1442 or 1689 to 4144

Young_AT_6160_ANS_SE_pg1169-1210.indd 1202

09/12/16 2:25 PM

Answers to Odd-Numbered Exercises  

27. A 1 x 2 5 212 1 x 2 1 2 2 1 92 , ­maximum at x 5 1 base: 3 units, height: 3 units 29. yes, 6

31. no

33. d

37.

1203

85. POSITIVE REAL ZEROS

35. a

39.

41. 6 1multiplicity 52, 24 1multiplicity 22

43. 0, 22, 2, 3, 23, all multiplicity 1

NEGATIVE REAL ZEROS

5

2

5

0

3

2

3

0

1

2

1

0

87. possible rational zeros: 61, 62, 63, 66

45. ƒ1x2 5 x 1x 1 32 1x 2 42

89. possible rational zeros: 61, 62, 64, 68, 616, 632, 664, 612

47. ƒ1x2 5 x 15x 1 22 14x 2 32

91. possible rational zeros: 61, 612 ; zeros: 12

49. ƒ1x2 5 x4 2 2x3 2 11x2 1 12x 1 36 51. ƒ1x2 5 1x 2 72 1x 1 22  a.  22, 7 1both multiplicity 12 b. crosses at 22, 7  c.  10, 2142  d.  rises right and left e. 

93. possible rational zeros: 61, 62, 64, 68, 616; zeros: 1, 2, 4, 22 95. a. POSITIVE REAL ZEROS

1

NEGATIVE REAL ZEROS

0

b. 61, 65 7

5

2

53. ƒ1x2 5 6x 1 3x 2 x 1 x 2 4  a.  10.8748, 02 with multiplicity 1 b. crosses at its only real zero  c.  10, 242  d.  falls left and rises right  e.

55. a. 

c. 21 is a lower bound, 5 is an upper bound d. none  e.  not possible f. 

97. a.  POSITIVE REAL ZEROS

b. 1, 3, 7 1all with multiplicity 12 c.  between 1 and 3 hours, and more than 7 hours is financially beneficial 57. Q1x2 5 x 1 4,  r1x2 5 2

NEGATIVE REAL ZEROS

3

0

1

0

b. 61, 62, 63, 64, 66, 612 c. 24 is a lower bound, 12 is upper bound d. 1, 2, 6   e.  P1x2 5 1x 2 12 1x 2 62 1x 2 22

59. Q1x2 5 2x3 2 4x2 2 2x 272 ,  r  1x2 5 223

f.

61. Q1x2 5 x3 1 2x2 1 x 2 4,  r1x2 5 0

63. Q1x2 5 x5 2 8x4 1 64x3 2 512x2 1 4096x 2 32,768, r 1x2 5 262,080 65. Q1x2 5 x 1 3,  r1x2 5 24x 2 8 67. Q1x2 5 x2 2 5x 1 7,  r1x2 5 215 3

2

69. 3x 1 2x 2 x 1 4 feet 73. g112 5 0

99. a. 

71. ƒ1222 5 2207

75. no

77. yes

2

2

79. P1x2 5 x 1x 1 22 1x 2 42 81. P1x2 5 x 1x 1 32 1x 2 222 83.

POSITIVE REAL ZEROS

1

Young_AT_6160_ANS_SE_pg1169-1210.indd 1203

NEGATIVE REAL ZEROS

1

POSITIVE REAL ZEROS

NEGATIVE REAL ZEROS

0

0

0

2

2

2

2

0

b. 61, 62, 63, 64, 66, 68, 612, 624

09/12/16 2:25 PM

1204 

Answers to Odd-Numbered Exercises

c. 24 is a lower bound, 8 is upper bound d. 22, 21, 1, 6

135. a.  22, 1multiplicity 22, 3, 4

b.  P1x2 5 1x 1 2221x 2 32 1x 2 42

e. P1x2 5 1x 2 22 1x 1 12 1x 1 22 1x 2 62 f.

101. P1x2 5 1x 2 5i2 1x 1 5i2

103. P1x2 5 1x 2 11 2 2i 22 1x 2 11 1 2i 22 105. 2i, 3 2 i

7 2,

22 6 3i; P 1 x 2 5 1 2x 2 7 21 x 1 2 2 3i 21 x 1 2 1 3i 2 x13 139. a.  yes, one-to-one  b.   f 21 1 x 2 5 22x c. 137.

107. 2i, 2 1 I

109. 2i, 4, 21; P1x2 5 1x 2 i2 1x 1 i 2 1x 2 42 1x 1 12

111. 3i, 1 6 i; P1x2 5 1x 2 3i2 1x 1 3i2 1x 2 11 1 i22 1x 2 11 2 i22

Practice Test

115. P1x2 51x 2 2i2 1x 1 2i2 1x 2 12

1.

113. P1x2 5 1x 2 32 1x 1 32 1x 2 3i 2 1x 1 3i 2 117. HA: y 5 21 VA: x 5 22

3. 1 3, 12 2

119. HA: none  VA: x 5 21 Slant: y 5 4x 2 4 121. HA: y 5 2  VA: none 123.

125. 5. ƒ1x2 5 x 1x 2 2231x 2 122

19 7. Q1x2 5 22x2 2 2x 211 2 , r1x2 5 2 2 x 1

9. yes, complex zeros. 13. yes, complex zeros. 127.

7 2

11. P1x2 5 1x 2 72 1x 1 22 1x 2 12

15. possible rational zeros: 61, 62, 63, 64, 66, 612, 613 , 623 , 643 17. 32 , 62i

129. a.  1480, 212112  b.  10, 2592  c.  1212.14, 02, 1972.14, 02 d.  x 5 480 19. Given the points 10, 3002, 12, 2852, 110, 3152, 152, 3002, you can have a polynomial of degree 3 because there are 2 turning points.

131. x-intercepts: 121, 02, 10.4, 02, 12.8, 02; zeros: 21, 0.4, 2.8, each with multiplicity 1 21. Given the points 11970, 0.082, 11988, 0.132, 12002, 0.042, 12005, 0.062, the lowest-degree polynomial that can be represented is a third-degree polynomial. 133. linear function

23. a.  x-intercept: 10, 02, y-intercept: 10, 02 b.  x 5 62  c.  y 5 0  d. none

e.

Young_AT_6160_ANS_SE_pg1169-1210.indd 1204

09/12/16 2:25 PM

Answers to Odd-Numbered Exercises  

33. y-intercept: 10, 12 HA: y 5 0 domain: 12q, q2 range: 10, q2 other points: A21, 61 B, 11, 62

25. a.  x-intercept: 13, 02, y-intercept: A0, 38 B

b. x 5 22, x 5 4  c.  y 5 0  d. none e. 

27. a.  y 5 x 2 2 3x 2 7.99 b. y 5 1 x 2 1.5 2 2 2 10.24

35. y-intercept: 10, 12 HA: y 5 0 domain: 12q, q2 range: 10, q2 other points: 11, 0.12, 121, 102

c. 121.7, 0 2 and 1 4.7, 0 2 d. yes

37. y-intercept: 10, 12 HA: y 5 0 domain: 12q, q2 range: 10, q2 other points: A1, 1e B, 121, e2

Cumulative Test 5 4x 7y 10 7. x 5 3 1.

3. 4x 2 12 9. y 5 13 x 2

11. x2 1 1 y 2 622 5 2

1205

5. 3313 min. 1 3

13. neither

15. right 1 unit and then up 3 units

39. y-intercept: 10, 02 HA: y 5 21 domain: 12q, q2 range: 121, q2 other points: 12, 32, 11, 12 19. ƒ1x2 5 1x 1 222 1 3

17. g1ƒ12122 5 0

21. Q1x2 5 4x2 1 4x 1 1, r  1x2 5 28

23. possible rational zeros: 1 61, 62, 63, 64, 66, 6 21 , 6 31 , 6 41 , 6 61 ,612 , 6 32 , 6 23 , 6 43 , 6 41 3 1 zeros: 22, 2 4 , 3 25. HA:  y 5 0  VA:  x 5 62 27. f 1 x 2 5

31x 1 12

x 1 2x 2 3 2

HA:  y 5 0

VA:  x 5 0, x 5

x-intercept: 121, 02

41. y-intercept: 10, 12 HA: y 5 2 domain: 12q, q2 range: 12q, 22 other points: 11, 2 2 e2, A21, 2 2 1e B

3 2

yes 43. y-intercept: 10, e 2 42 HA: y 5 24 domain: 12q, q2 range: 124, q2 other points: 121, 232, 11, e2 242

CHAPTER 5 Section 5.1 1. 16

3.

1 25 5. 4 7. 27 9.

21

11. 5.2780 13. 9.7385 15. 7.3891 17. 0.0432 19. 27 21. 16 23. 4 25. 19.81 27. f 29. e 31. b

Young_AT_6160_ANS_SE_pg1169-1210.indd 1205

09/12/16 2:25 PM

Answers to Odd-Numbered Exercises

87. close on the interval 123, 32

45. y-intercept: 10, 32 HA: y 5 0 domain: 12q, q2 range: 10, q2 other points: 12, 3e2, 11, 3!e2

89. as x increases, ƒ1x2 S e, g1x2 S e2, h1x2 S e4

49. 10.4 million 51. P1302 5 150012 ⁄ 2 > 96,000 30

53. 168 mg 55. $3031

5

57. $3,448.42

59. $13,011.03 61. $4319.55 63. $13,979.42 65. 3.4 mg/L 67.

D 1p2—approximate demand for product in units

p 1price per unit2

1.00

1,955,000

5.00

1,020,500

10.00

452,810

20.00

89,147

40.00

3455

60.00

134

80.00

5

90.00

1

69. The mistake is that 42 @2 2 42. Rather, 42 @2 5 1

1

1 1

4 @2 71. 2.5% needs to be converted to a decimal, 0.025. 73. false 77.

91. a. 

210 Temperature (in degrees Fahrenheit)

47. y-intercept: 10, 52 HA: y 5 1 domain: 12q, q2 range: 11, q2 other points: 10, 52, 12, 22

190 180 170 160 150 0

4

2

6

Time (in minutes)

b.  The best fit exponential curve is y 5 228.34 1 0.9173 2 x with r 2 5 0.9628. This best fit curve is shown on the following scatterplot. The fit is very good, as evidenced by the fact that the square of the correlation coefficient is very close to 1. 220

5 21 .

210 200 190 180 170 160 150 140 0

75. true 79.

200

140

Temperature (in degrees Fahrenheit)

1206 

2

4

6

Time (in minutes)

c. i. Compute the y-value when x 5 6 to obtain about 136 degrees Fahrenheit.   ii. The temperature of the soup the moment it was taken out of the microwave is the y-value at x 5 0, namely, about 228 degrees Fahrenheit.

81. y-intercept: 10, be 2 a2 HA: y 5 2a 83. Domain: 12q, q2

d. The shortcoming of this model for large values of x is that the curve approaches the x-axis, not 72 degrees. As such, it is no longer useful for describing the temperature beyond the x-value at which the temperature is 72 degrees.

Section 5.2 1. 53 5 125 3. 81 ⁄ 5 3 5. 225 5 1

8 5.

4

7. 1022 5 0.01 9. 104 5 10,000 11. A 14 B

23

1 32

5 64

13. e21 5 1e 15. e0 5 1 17. e x 5 5 19. xz 5 y 21. log8 15122 5 3

1 2 213

23. log10.000012 5 25 25. log225 1 15 2 5

8 27. log 2@5 A 125 B 5 3 29. log 1@27 1 3 2 5

Young_AT_6160_ANS_SE_pg1169-1210.indd 1206

09/12/16 2:25 PM

Answers to Odd-Numbered Exercises  

31. ln 6 5 x

33. logy x 5 z 35. 0

37. 5

39. 7 41. 26

43. undefined

45. undefined 47. 1.46

49. 5.94

1207

High Frequency— Shortwave broadcasts, Citizens Band Radio

100 m

3,000,000 Hz

51. undefined 53. 28.11

Very High Frequency— FM Radio, Television

10 m

30,000,000 Hz

55. 125, q2

57. A2q, 52 B 59. A2q, 72 B

Ultra High Frequency— Television, Mobile Phones

0.050 m

6,000,000,000 Hz

67. c

69. d

61. 1 2q, 0 2 ∪ 1 0, q 2 63. R 65. b 71. domain: 11, q2 range: 12q, q2

73. domain: 10, q2 range: 12q, q2

75. domain: 122, q2 range: 12q, q2

77. domain: 10, q2 range: 12q, q2

b.

103. log2 4 5 x is equivalent to 2x 5 4 1 not x 5 24 2 .

105. The domain is the set of all real numbers such that x 1 5 . 0, which is written as 125, q2. 107. false

79. domain: 124, q2 range: 12q, q2

111. domain: 1a, q2 range: 12q, q2 x-intercept: 1a 1 eb, 02 113.

81.  domain: 10, q2 range: 12q, q2

109. true

115. y 5 x

117. x-intercept: 11, 02 VA: x 5 0 119. 10, q2

83. 60 decibels 85. 117 decibels 87. 8.5 89. 6.6 91. 3.3 93. Normal rainwater: 5.6 Acid rain/tomato juice: 4 95. 3.6 97. 13,236 years 99. 25 dB loss 101. a.  USAGE

WAVELENGTH

FREQUENCY

Super Low Frequency— Communication with Submarines

10,000,000 m

30 Hz

Ultra Low Frequency— Communication within Mines

1,000,000 m

300 Hz

Very Low Frequency— Avalanche Beacons

100,000 m

3000 Hz

Low Frequency— Navigation, AM Longwave Broadcasting

10,000 m

30,000 Hz

Medium Frequency—AM Broadcasts, Amateur Radio

1,000 m

300,000 Hz

Young_AT_6160_ANS_SE_pg1169-1210.indd 1207

121. a.

S

b.  A reasonable estimate for Vmax is about 156 µmol/min. c.  Km is the value of [S] that results in the velocity being half of its maximum value, which by 1b2 is about 156. So, we need the value of [S] that corresponds to v 5 78. From the graph, this is very difficult to ascertain because of the very small units. We can simply say that it occurs between 0.0001 and 0.0002. A more accurate estimate can be obtained if a best fit curve is known.

09/12/16 2:25 PM

1208 

Answers to Odd-Numbered Exercises

d. 1i2 v 5 33.70 ln 13 S 4 2 1 395.80 with r 2 5 0.9984. It is shown on the scatterplot below.

71. yes

73. no





75. no

77. yes





     

1ii2 Using the equation, we must solve the following equation for [S]:     100 5 33.70 ln13S 42 1 395.80    2295.80 5 33.70 ln3S 4   28.77745 5 ln 3 S 4

Section 5.4

Section 5.3

7. x 5 24 9. x 5 2 23 11. x 5 21

   e28.77745 5 S

S < 0.000154171

 

1. 0

3. 1

9. 32 17.

1 9

21.

1 2

1. x 5 4 3. x 5 22 5. x 5 62

5. 8 7. 23

11. 5 13. x 1 5 15. 8

logb 1 x 2 1

1 3

19. 3 logb 1x2 1 5 logb 1 y2

logb 1 y 2 23.

25. logb 1 x 2 2 logb 1 y 2 2 logb 1 z 2 27. 2 log x 1

1 2 log 1 x

1 3

logb 1 r 2 2 12 logb 1 s 2

1 52

29. 3 ln x 1 2 ln 1 x 2 2 2 2 12 ln 1 x 2 1 5 2

31. 2 log 1x 2 12 2 log 1x 2 32 2 log 1x 1 32 33. logb 1x3y52 35. logb a 37. logb 1 x 41. ln a

1/2

u5 b v2

u2 y 2 39. log a 3 2 b vz 2/3

1 x 1 3 2 1/2 x2 2 1 b 43. ln a b 2 2 1x 1 32 x 1 x 1 2 2 1/3

13. x 5 3, 4 19. x 5

15. x 5 0, 6 17. x 5 1, 4

3 1 log 1 81 2 2

< 2.454

21. x 5 log3 1 5 2 2 1 < 0.465 log2 1 27 2 1 1

< 1.918 3 25. x 5 ln 5 < 1.609 27. x 5 10 ln4 < 13.863 23. x 5

29. x 5 log3 1 10 2 < 2.096 31. x 5 33. x 5

ln 1 22 2 2 4 < 20.303 3

ln 6 27 1 !61 < 0.896 35. x 5 ln a b < 20.904 2 2

37. x 5 0 39. x 5 ln 7 < 1.946 log10 1 9 2 41. x 5 0 43. x 5 < 0.477 2 9 45. x 5 50 47. x 5 40 49. x 5 32 51. x 5 63 53. x 5 5 55. x 5 6

45. 1.2091 47. 22.3219 49. 1.6599

57. x 5 21 59. no solution 61. x 5

51. 2.0115 53. 3.7856 55. 110 decibels

63. x 5 6"e5 < 612.182 65. x 5 47.5

57. 5.5

67. x 5 6"e4 2 1 < 67.321

59. 3 log 5 2 log 52 5 3 log 5 2 2 log 5 5 log 5 61. Cannot apply the product and quotient properties to logarithms with different bases. So, you cannot reduce the given expression further without using the change of base formula. 63. true 65. false 67. Proof: Let u 5 logb M, v 5 logb N. Then, u

v

b 5 M, b 5 N.  Observe that u

b logb A M N B 5 logb A bv B

5 logb Abu2v B 5 u 2 v 5 logb M 2 logb N

69. 6 logb x 2 9 logb y 1 15 logb z

25 8

1 1 23 1 e22 2 < 21.432 2 71. x < 21.25 69. x 5

73. x 5

2 1 "4 1 4e4 < 8.456 2

23 1 "13 < 0.303 2 77. x 5 1 1 !7 < 3.646 75. x 5

79. a. 151 beats per minute b. 7 minutes c. 66 beats per minute 81. 31.9 years

83. 19.74 years

15

85. 3.16 3 10 joules 87. 1 W@m2 89. 4.61 hours 91. 15.89 years 93. 6.2

Young_AT_6160_ANS_SE_pg1169-1210.indd 1208

09/12/16 2:25 PM

Answers to Odd-Numbered Exercises  

95. ln 1 4ex 2 2 4x. Should first divide both sides by 4, then take the natural log:

41. a. 

1209

  b. 75  c. 4  d. 4

4ex 5 9 ex 5

9 4

ln 1 e x 2 5 lnA 94 B

x 5 ln A 94 B

97. The correction that needs to be made is that x 5 25 needs to be removed from the list of solutions. The domain of the logs cannot include a negative solution.

43. a. 18 years b. 10 years

45. a. 30 years  b. $328, 120

47. r 5 0.07, not 7 49. true 51. false 53. less time

99. true 101. false 1 1 "1 1 4b2 3000 2 y 103. x 5 105. t 5 25 lna 2y b 2

55. a. For the same periodic payment, it will take Wing Shan fewer years to pay off the loan if she can afford to pay biweekly.

107. f 21 1 x 2 5 lnAx 1 "x 2 2 1B; x $ 1 109. x 5

36 !5 2    

b. 11.58 years  c. 10.33 years  d. 8.54 years, 7.69 years, respectively

Review Exercises 1. 17,559.94 3. 5.52

111.

y 16 14 12 10 8 6 4 2 –5

–3

–1 –4

7. 5.89 9. 73.52

x = 2.40

13. b

5. 24.53 11. 6.25

15. c

17. y-intercept: 10, 212. 

x

HA:  y 5 0

2 3 4 5 x = –0.25

113. domain: 12q, q2  y-axis symmetry

19. y-intercept: 10, 22 HA:  y 5 1

21. y-intercept: 10, 12 HA:  y 5 0







Section 5.5 1. c 1iv2 3. a 1iii2

5. f 1i2

7. 119 million 9. 7.7 years, 2010

23. y-intercept: 10, 3.22 HA:  y 5 0



11. 2151.9 subscribers 13. $6,770,673 15. 332 million 17. 1.53 million 19. 13.53 ml 8 b > 0.6286 15 b.   636,000 mp3 players

21. a.  k 5 2ln a

25. $6144.68 27. $23,080.29 29. 43 5 64

23. 7575 years 25. 131,158,556 years old

37. 0 39. 24 41. 1.51

27. 1058F 29. 3.8 hours before 7 am

43. 22.08 45. 122, q2

31. $19,100

33. a. 84,520 b. 100,000 c. 100,000

31. 1022 5

49. b

1 100

4 33. log6 216 5 3 35. log 2@13 A 169 B52

51. d

47. 12q, q2

35. 202.422 cases 37. 1.89 years 39. r 5 0

Young_AT_6160_ANS_SE_pg1169-1210.indd 1209

09/12/16 2:25 PM

1210 

Answers to Odd-Numbered Exercises

53.

55.





17. x 5 lnA 12 B 5 2ln 1 2 2 < 20.693 19. 121, 0 2 ∪ 1 1, q 2

21. x-intercept: none y-intercept: 10, 22 HA: y 5 1 57. 6.5 59. 50 decibels

61. 1

63. 6 65. a logc 1x2 1 b logc 1 y  2

67. logj 1r2 1 logj 1s2 2 3 logj t

69. 12 log 1 a 2 2 32 log 1 b 2 2 25 log 1 c 2

71. 0.5283 73. 0.2939

75. x 5 24

77. x 5 43

81. x < 20.218

79. x 5 26

83. no solution

85. x 5 0

87. x 5

3 1 1@e

2 y-intercept: none VA: x 5

, 0b

3 2

100 3

89. x 5 128!2 91. x < 63.004

93. x < 0.449

95. $28,536.88 97. 16.6 years

99. 4.59 million

101. 6250 bacteria 103. 56 years

23. x-intercept: a

105. 16 fish

25. $8051.62 27. 90 decibels

107. 343 mice

29. 7.9 3 1011 , E , 2.5 3 1013 joules

109. HA:  y 5 e!2 < 4.11 111. 12.376, 2.0712

31. 7800 bacteria 33. 3 days 35. domain: 12q, q2 symmetric origin

113. 10, q2

Cumulative Test 1. x ⁄ y2 3. x 5 5

115. domain: 12q, q2, symmetric origin HA: y 5 21 1 as x S 2q 2 , y 5 1 1 as x S q 2

6

5. 12q, 219 D

2 6 !19 5

7. y 5 34 x 1

3 4

9. a. 1   b. 5   c. 1   d. undefined e. domain: 122, q2  range: 10, q2

f. increasing: 14, q2, decreasing: 10, 42, constant: 122, 02 11. yes

13. 11, 212

15. Q1x2 5 2x3 1 3x 2 5, r1x2 5 0

17. HA: none VA: x 5 3 Slant:  y 5 x 1 3



19. 125 21. 5

117. a.  N 5 4e20.038508t < 4 1 0.9622 2 t b.  N 5 4 1 0.9622 2 t c. yes

23. x 5 0.5 25. 8.62 years

1. x3

CHAPTER 6

27. a.  N 5 6e20.247553t < 6 1 0.9755486421 2 t

b. 2.72 grams

Practice Test

3. 24

5. x 5 6 !1 1 ln 42 < 62.177

7. x 5

21 1 ln A 300 27 B 0.2

Section 6.1 1. 180°

> 7.04

5. 300°

9. x 5 4 1 e2 < 11.389 11. x 5 ee < 15.154 13. x 5 9

15. x 5

Young_AT_6160_ANS_SE_pg1169-1210.indd 1210

23 6 !9 1 4e < 0.729 2

3. 2120° 7. 2288°

9. 270°

11. a. 72°  b. 162°

13. a. 48°  b. 138°

15. a. 1°  b. 91°

17. 54°/ 36°

19. 120°/ 60°

09/12/16 2:25 PM

Answers to Odd-Numbered Exercises  

21. g 5 30°

45.

23. a 5 120°, b 5 30°, g 5 30°

2!3 3

47.

!3 3

27. c 5 5

29. b 5 8

49. !2

53. 0.1392

55. 1.3764

31. c 5 !89

33. c 5 25

57. 1.0098

59. 1.0002

37. 2 cm

61. 0.7002

63. 0.4337

39. Other leg: 5!3 < 8.66 m;  Hypotenuse: 10 m

65. 30°

67. a. 2/5  b. 5/2

41. Other leg: 4!3 < 6.93 yd;  Hypotenuse: 8!3 < 13.9 yd

69. 10 mi

71. 2.405

43. Short leg: 5 in.;

73. 1.335

75. 0 points

Long leg: 5!3 < 8.66 in.

77. 50 points (a bullseye)

25. a 5 18°, b 5 108°, g 5 54°

35. 10!2 < 14.14 in.

45. ƒ 5 3

47. a 5 15 1 3

51. 0.6018

79. 1.09 angstroms

49. c 5 11.55 km

51. ƒ 5 in.

81. The opposite of angle y is 3 (not 4).

53. 120°

55. 144°

83. Secant is the reciprocal of cosine (not sine).

57. 60 min or 1 h 61. 241 ft

59. !7300 < 85 ft

65. 17 ft

67. 48 ft 3 28 ft

69. 38 ft

71. 225 ft

73. 20 points

75. 50.004 ft

77. 7.8 cm

79. 4 cm

85. true

63. 9.8 ft

81. The length opposite the 60° angle is !3 times (not twice) the length opposite 30° angle.

87. sin 30° 5 89. tan 30° 5

1 !3 and cos 30° 5 2 2 !3 and tan 60° 5 !3 3

91. sec 45° 5 !2 and csc 45° 5 !2 93. 0

95. 0

99.

83. false

85. true

3!5 97. 5

87. true

89. 110°

101.

91. DC 5 3

93. 25

105. a. 2.92398  b. 2.92380; (b) is more accurate.

95. The triangles are isosceles right triangles 45°245°290°. 97. 8 101. x 5 2

99. 2!3 103. 28.89 ft, 33.36 ft

Section 6.2 1. 45 5. 43 9. !5

2!10 13. 7

17.

3!10 20

13 9

103.

2!13 13 97 65

107. a. 0.70274  b. 0.70281; (b) is more accurate.

Section 6.3 1. three

3. two

3.

5 4

5. 47°

7. 55°

7.

!5 5

9. 83° 11. a < 14 in.

11. 2 15.

7 3

19. 30°

13. a < 18 ft 15. a < 5.50 mi 17. c < 12 km 19. c < 20.60 cm 21. a < 50° 23. a < 62° 25. a < 82.12 yd 27. c < 10.6 km 29. c < 19,293 km 31. b < 58°, a < 6.4 ft, b < 10 ft

21. 90° 2 x

23. 60°

33. a < 18°, a < 3.0 mm, b < 9.2 mm

25. cos 1 90° 2 x 2 y 2

27. sin 1 70° 2 A 2

35. b < 35.8°, b < 80.1 mi, c < 137 mi

29. tan 1 45° 1 x 2

31. sec 1 30° 1 u 2

33. a 35. b

37. b < 61.62°, a < 936.9 ft, c < 1971 ft 39. a < 56.0°, b < 34.0°, c < 51.3 ft

37. c

39.

!3 3

41. a < 55.480°, b < 34.520°, b < 24,235 km

41. !3

43.

2!3 3

45. c 5 4.16 cm, b 5 1.15 cm, a 5 73.90°

Young_AT_6160_ANS_SE-pg1211-1260.indd 1211

1211

43. c 5 27.0 in., a 5 24.4 in., a 5 64.6°

09/12/16 2:26 PM

1212 

Answers to Odd-Numbered Exercises

47. 286 ft

27.

49. a 5 88 ft

51. 260 ft (262 rounded to two significant digits) 53. 11° (she is too low) 55. 80 ft

57. 170 m

59. 0.000016° 61. 26 ft 63. 4400 ft

65. 8°

67. 136.7°

69. 91.99 ft 29.

71. 27 ft 73. 0 point ( just misses the entire target) 75. 121 ft 77. 24 ft 79. d < 3.5 ft 83. Tan

21

81. 4.7 in.

should have been used (not tan).

85. true

87. false

89. false

91. 2 mi

93. 0.5

95.

97. 40°

99. 0.8

101. u

31.

2 7

103. 35°; u

105. 0.785

Section 6.4 1. QI

3. QII

5. QIV

7. negative y-axis

9. x-axis

11. QIII

13. QI

15. QIII

17. QII

19. QII

33.

21. QIV 23. 35.

25.

37. c 39. e

41. f 43. 52°

45. 268°

49. 150°

sin u 53. 55. 57. 59.

Young_AT_6160_ANS_SE-pg1211-1260.indd 1212

2 !5 5 2 !5 5

4 !41 41 2 !5 5

47. 330° cos u !5 5 !5 5

5 !41 41 2

!5 5

51. 320.0001°

tan u

cot u

2

1 2

2

1 2

4 5

5 4

22

2

1 2

sec u !5 !5

!41 5 2!5

csc u !5 2 !5 2

!41 4 !5 2

09/12/16 2:26 PM

Answers to Odd-Numbered Exercises  

61. 2 63. 65.

7 !65 65

!15 5

!6 2 4

67. 2 69. 2 71. 73. 75. 77. 79.

2 !29 29

2

2

!5 5

!5 5

5 !29 29

2 !5 5 2

2 !5 5

2 !5 5

!5 5

!5 5

2 !5 5

!10 2 10 2 !13 13

!10 5

!10 2 4 2

3 !10 10 2

81. 1

7 4

4 !65 65

3 !13 13

2

!6 2

2

!15 5 2 5

4 7

2

!6 3

2

!15 3 5 2

2

1 2

22

2

1 2

22

2

1 2

1 2

2

1 2 3

23

2

2 3

2

!65 4

!10 2

!65 7

!15 3

2 !10 2 5

2 !6 2 3

2

2

!29 5

!5 2 2

!5 2

!5

!5 2

!10 3

3 2

2

2

!13 3

!29 2

2 !5 !5 !5 2 !5

2 !10 !13 2

25. 1

!3 3

29. 0

23.

2!3 3

27. 21 31. 1

33. 1 35. not possible 37. possible

39. not possible

41. possible

43. possible

45. possible

47. 212

49. 2 53. 1

!3 2

!3 3 55. 22

51.

57. 1

59. 30° and 330°

61. 210° and 330°

63. 90° and 270°

65. 270°

67. 20.8387

69. 211.4301

71. 2.0627

73. 21.0711

75. 22.6051

77. 110°

79. 143°

0

Undefined

0

Undefined

83. 21

0

Undefined

0

Undefined

21

85. 1

0

Undefined

0

Undefined

1

81. 322°

83. 140°

87. 21

0

Undefined

0

Undefined

21

85. 340°

87. 1°

89. 21

0

Undefined

0

Undefined

21

89. 335° 91. 1.3

91. 1

0

Undefined

0

Undefined

1

!3 3

22

93. 2

!3 2

2

1 2

95. 21200°

!3

97. 23240°

99. Don and Ron end up in the same place, 180° from where they started. 7 24

1

21. 2

2

2 !3 3

93. 12° 95. Lower leg is bent at knee at an angle of 15°. 97. 75.5° 99. The reference angle is made with the terminal side and the negative x-axis (not the y-axis). 101. true

103. false

105. false

107. 2

"a2 2 b2 109. 2 b

111. QI

101. 1440°

103. tan u 5

105. 120 ft

107. profit

109. 1.4 cm 113. false

111. r 5 !5 (not 5)

113. QI

117. true

119. false

Section 6.6

123. m 5 tan u

1. 15 or 0.2

127. y 5 2 1 tan u 21 x 2 a 2

1 5. 50 or 0.02 1 9. 5 or 0.2

121.

2 35

125. y 5 2!2 129. cos 270° 5 0

115. true

131. 0

13.

133. 0 135. Does not exist because we are dividing by 0 137. 21

17. 21.

Section 6.5 1. QIV

3. QII

25.

5. QI

7. QI

29.

a 2

"a 1 b2

115. QI

3.

2 11

7.

1 8

11. p 15. 6 7p 19. 4 17p 23. 18 7p 2 27. 6 30° 31.

or < 0.18

or 0.125

1 24

or 0.042 p 4 5p 12 13p 3

220p 135°

9. QIII

11. sin u 5 2 45

33. 67.5° 35. 75°

13. tan u 5 2 60 11

15. sin u 5 2 84 85

37. 1620°

17. tan u 5 !3

19. tan u 5 2

Young_AT_6160_ANS_SE-pg1211-1260.indd 1213

!11 5

1213

39. 171°

41. 284° 43. 229.18° 45. 48.70° 47. 2160.37°

09/12/16 2:26 PM

1214 

Answers to Odd-Numbered Exercises

49. 198.48° 51. 0.820 53. 1.95 55. 0.986 57. 12 mm 61.

5 2

65.

200p 3

59.

in. 63.

2p 3

ft

11p 5

mm

km 67. 12.8 sq ft

69. 2.85 sq km

71. 5 sq yd

73. 8.62 sq cm

75. 0.0236 sq ft

77.

2 5

m/ sec 79. 272 km / hr

81. 9.8 m 5p rad 85. 2 sec

83. 1.5 mi 2p rad 87. 9 sec

89. 6p in. / sec

91.

93. 26.2 cm

p mm / sec 4 95. 653 in. (or 54.5 ft)

97. 5262 km

99. 37 ft

41. p, 3p

43.

3p 5p , 4 4 p 3p 49. , 2 2 p 11p 53. , 6 6 57. 99.05°F 45.

3p 7p , 4 4

47. 0, p, 2p 51.

7p 11p , 6 6

55. 22.9°F 59. 2.6 ft

61. 135 lb

63. 13,000 guests

65. 10.7 mcg / mL

67. 35°C

69. Used the x-coordinate for sine and the y-coordinate for cosine; should have done the opposite 71. true

73. false

75. true p 5p 79. , 4 4

77. odd

85. Yes, u 5

83. sec u p 3p 5p 7p , , , 4 4 4 4

101. 200p < 628 ft

103. 50°

105. 60 in. (5 ft)

107. 911 mi

109. 157 sq ft

111. 78 sq in.

87. sin 1 423° 2 < 0.891 and sin 1 2423° 2 < 20.891

113. 74 sq mi

115. 1.4 nm

93. 0.8660

117. 70 mph

121. 16 ft / sec

123. 66 23 p rad/ min

125. 12.05 mph

127. 640 rev / min

129. 10.11 rad / sec or 1.6 rotations per second 133. 40p rad/min

135. 32p ft 137. Angular speed needs to be in radians (not degrees) per second. 139. true

141. true

143. 2p < 6.3 cm/sec

145. tripled

147.

15 2 p

square units

91. 0.5

Section 6.8

119. 1037.51 mph; 1040.35 mph

131. 17.59 m /s

89. 6.314, 26.314

1. c 3. a

5. h 7. b 3 2p 9. e 11. A 5 , p 5 2 3 2p 2 4p 13. A 5 1, p 5 15. A 5 , p 5 5 3 3 17. A 5 3, p 5 2 19. A 5 5, p 5 6 21.

149. 68°45r44s

Section 6.7 1. 2

!3 2

3. 2

!3 2

23.

!2 7. 21 2 9. 2 !2 11. !3 !3 13. 2 15. 2 2 !3 !2 17. 2 19. 2 2 2 5.

21. 2

!3 2

25. 1 29. 0 !3 3 4p 5p 37. , 3 3 33.

23.

!2 2

25.

!2 2 31. 22 27.

35.

p 11p , 6 6

39. 0, p, 2p, 3p, 4p

Young_AT_6160_ANS_SE-pg1211-1260.indd 1214

09/12/16 2:27 PM

Answers to Odd-Numbered Exercises  

27.

39.

29.

41. y 5 2sin 2x p 45. y 5 22 sin a xb 2

1215

43. y 5 cos px 47. y 5 sin 8px

49. Amplitude: 2; period: 2; phase shift:

1 (right) p

2p ; phase shift: 2 23 (left) 3 53. Amplitude: 6; period: 2; phase shift: 22 (left) p 55. Amplitude: 3; period: p; phase shift: 2 (left) 2 51. Amplitude: 5; period:

31.

57. Amplitude: 14 ; period: 8p; phase shift: 2p (right) 59. Amplitude: 2; period: 4; phase shift: 4 (right) 61.

33. 63.

35. 65.

37. 67.

Young_AT_6160_ANS_SE-pg1211-1260.indd 1215

09/12/16 2:27 PM

1216 

Answers to Odd-Numbered Exercises

69.

81.

71.

83.

73.

85.

75.

87.

77.

89.

91. 79.

Young_AT_6160_ANS_SE-pg1211-1260.indd 1216

09/12/16 2:27 PM

Answers to Odd-Numbered Exercises  

93.

1217

135. As t increases, the amplitude goes to zero for Y3.

95. 3500 widgets

137. a. Y2 is Y1 shifted upward by 1 unit.

97. Amplitude: 1 mg / L; period 8 weeks

b. Y2 is Y2 shifted downward by 1 unit.

99. Amplitude: 4 cm; Mass: 4 g 1 101. cycles per second 4p 103. Amplitude: 0.005 cm; Frequency: 256 hertz

139. 5

105. Amplitude: 0.008 cm; Frequency: 375 hertz 107. 660 m / sec

109. 660 m / sec

Section 6.9 1. b 3. h

5. c 7. d

9.

2p pt 1 t 2 1 2 d or y 5 25 2 25 cos a b 111. y 5 25 1 25 sin c 4 2

113. Forgot to reflect about the x-axis 115. true

117. false

119. 10, A2 np 121. x 5 , where n is an integer B A 123. a0, 2 b 2

11.

p 1 3 1 4n 2 125. a , 0b, where n an integer. 2B

127. c 2 129. no

5A 3A , d 2 2

13.

131. coincide

15.

p . 3 p b. Y2 is Y1 shifted to the right . 3 133. a. Y2 is Y1 shifted to the left

Young_AT_6160_ANS_SE-pg1211-1260.indd 1217

09/12/16 2:27 PM

1218 

Answers to Odd-Numbered Exercises

17.

29.

19.

31.

21.

33.

23.

35.

25.

37.

39. 27.

Young_AT_6160_ANS_SE-pg1211-1260.indd 1218

09/12/16 2:27 PM

Answers to Odd-Numbered Exercises  

1219

41.

53.

43.

55.

45.

57. Domain: all real numbers x, such that x 2 n, where n is an integer Range: all real numbers 59. Domain: all real numbers x, such that 2n 1 1 x2 p, where n is an integer 10 Range: 1 2q, 22 4 ∪ 3 2, q 2

61. Domain: all real numbers x, such that x 2 2np, where n is an integer

47.

Range: 1 2q, 1 4 ∪ 3 3, q 2

63. Domain: 5 x : x 2 4n 1 6, n an integer6 Range: all real numbers

65. Domain: 5 x : x 2 n, n an integer6

Range: A2q, 2 52 D ∪ C 2 32 , qB 67. 3 mi

69. 55.4 sq in.

71. Forgot the amplitude of 3

49.

73. true 75. n 5 integer p p 77. x 5 2p, 2 , 0, , p 2 2 79. a

np 2 C , 0b, where n an integer B

81. Infinitely many. The tangent function is an increasing, periodic function and y 5 x is increasing. 83. A 5 !2

85. p

Review Exercises 1. a. 62°  b. 152° 51. 5. a. 0.99°  b. 90.99°

3. a. 55°  b. 145° 7. g 5 25°

9. a 5 140°, b 5 20°, g 5 20° 11. b 5 !128 5 8!2 15. 12!2 yd

13. c 5 !65

17. Leg: 3!3 ft; Hypotenuse: 6 ft

Young_AT_6160_ANS_SE-pg1211-1260.indd 1219

19. F 5 4

21. C 5 147.6 km

23. 10!3 in.

25. 150°

09/12/16 2:27 PM

1220 

Answers to Odd-Numbered Exercises

27. 32 m

29.

2!13 13

33.

3 2

!13 2 35. cos 60°

37. cot 45°

39. csc 60°

41. b

43. b

45. c

47. 0.6691

49. 0.9548

51. 1.5399

53. 1.5477

31.

123.

Amplitude

55. 75 ft sin u 57. 2 59. 61.

4 5

!10 10

1 2

63. 2 65. 2

cos u

!5 5

!7.2 3

3 5 2

3 !10 10 !3 2

2 !5 5 2

!7.2 6

tan u 2

4 3

2

2

1 3

23

!3 3 2

2!3 3

75. 1.0355 79. 20.6494 83.

1 2

11p 6

sec u

3 4

!3

5 3 2

1 2

69. 2

!10 3

2 !3 3 !5 2

22

2

67. 2 12 71. 2

cot u

2

!7.2 1.2

csc u 2

5 4

!10

Period

Phase Shift

Vertical Shift

125. 3

2p

p (right) 2

12

127. 4

2p 3

2

p (left) 4

22

2

1 (right) 2p

129.

2

2!5 2

1 3

1 (down) 2

131.

!7.2 2.4

!3 3

73. 20.2419 77. 22.7904 3p 81. 4 6p 85. 5

87. 9p

89. 60°

91. 225°

93. 100°

133.

95. 1800° 97. 240p in./min < 754 in./min !3 99. 2 3 103. 1

101.

2 12

105. 21 1 2

107. 21

109.

111. 2p

113. y 5 4 cos x

115. 5 117. Amplitude: 2; Period: 1 2p 119. Amplitude: 15 ; Period: 3

135. Domain: all real numbers such that x 2 np, where n is an integer; Range: all real numbers 137. Domain: all real numbers such that 2n 1 1 x2 p, where n is an integer; 4 Range: 1 2q, 23 4 ∪ 3 3, q 2 139. Domain: e x : x 2

6n 1 7 , n an integer f 6

Range: A2q, 2 34 D ∪ C 2 14 , qB.

121. 141.

Young_AT_6160_ANS_SE-pg1211-1260.indd 1220

09/12/16 2:27 PM

Answers to Odd-Numbered Exercises  

1221

21. true

143.

23. y 5 4 sin 3 2 1 x 1 32 24 212 25.

145.

27. a. 1.34409 b. 1.34352, (b) is more accurate.

Cumulative Test 1. 147. 71.57 ft, 82.64 ft

153. 21

155. QIV

157. 100°46r45s

7. y 5 0.5x 1 3.4

9. 2 58

11. 1 13. ƒ 1 x 2 5 22x 2 1 7 15. Factors of a0 5 10; 61, 62, 65, 610;

159. 20.9659

Factors of an 5 2; 61, 62

p unit. 6 p b. Y2 is Y1 shifted to the right unit. 6 163. 5 161. a. Y2 is Y1 shifted to the left by

Possible rational zeros: 61, 62, 65, 610, 612 , 652 Testing the zeros: P 1 25 2 5 0, P 1 21 2 5 0, P A 12 B 5 0, P 1 2 2 5 0 17. y-intercept: 10, 02; domain: 1 2q, q 2 ;

Practice Test 1. 6000 ft 3. The first is an exact value of the cosine function; the second is an approximation. 5. QIV 7. 585° 11. Amplitude: 5; Period:

3. x 2 21, 0; no solution

5. 1 2q, 0 4 ∪ 1 1, 3 4

149. a. 1.02041  b. 1.02085, (b) is more accurate. 151. 2.612

x21 5

9.

range: 1 21, q 2 ; horizontal asymptote: y 5 21

15p sq in. 4

2p 3

13. 19.

15. x 5

np np or a , 0b, where n is an integer. 2 2

17. 1 2q, 24 4 ∪ 3 2, q 2

21. 0.435

23. sin u 5 19.

3!13 2!13 3 , cos u 5 2 , tan u 5 2 , 13 13 2

cot u 5 2 23 , sec u 5 2

!13 !13 , csc u 5 2 3

25. Amplitude: 4; Period: p; p Phase shift: 2 (left) 2

Young_AT_6160_ANS_SE-pg1211-1260.indd 1221

09/12/16 2:27 PM

1222 

Answers to Odd-Numbered Exercises

CHAPTER 7

Section 7.2

Section 7.1

1. 1

3. csc  x

5. 21

7. sec2 x

8 1. 7

3. 2

1 5. 2 5

2!5 7. 5

9. 2

5!7 7

11. 2

!3 13. 2 3

4 15. 3

17.

19.

!11 5

5 3

b a

25. 25

27. 2

37. 2 41.

15!11 44

!6 2

51. cos u 5 2 53. sin u 5 2

57.

1 sin u

61.

cos2 u sin u

1a 2 02

!3 2

35. 2

8!161 161

39. 2

2!13 13

57. conditional

59. conditional

61. identity

63. conditional

65.  a  cos u

71. false

73. QI, QIV

75. QIII and QIV

77. a2 1 b2

79. No, let A 5 308 and B 5 608. p 81. No, take A 5 . 4 83. cos1A 1 B2 5 cos A cos B 2 sin A sin B

87. cos2 A 5

2!29 5!29 , cos u 5 29 29

89. Y1 5 Y3

4!17 !17 , cos u 5 17 17

1 1 cos 1 2A 2 2

Section 7.3 1.

59. 21 1 sin u

1 1 2 sin u cos u 67. cos2 u

69. 1

55. conditional

85. sin1A 1 B2 5 sin A cos B 1 cos A sin B

5!34 3!34 , sin u 5 2 34 34

65. 1 1 2 sin u cos u

53. identity

69. This is a conditional equation. Just because the equation is true for p does not mean it is true for all x. 4

!5 2!5 , sin u 5 2 5 5

63.

51. conditional

67. The cos x and sin x terms do not cancel. The numerators become cos2 x and sin2 x, respectively.

3 4 47. cos u 5 2 , sin u 5 5 5

49. cos u 5 2

55. sin u 5

21. 2cos x 23. 1

43. !6

!62 45. 2 8

19. csc2 x

17. sin  x

31. 2 !17

33. 2 !5

15. 1

2

24.–  50.  See Instructor’s Solution Manual.

23. 28

!21 5

13. sec x

1 4

5 21. 64

29. 2

11. sin2 x 2 cos2 x

9. 1

!6 2 !2 4

5. 22 1 !3 9. 2 1 !3

13. !2 2 !6

3. 7.

!6 2 !2 4 !2 1 !6 4

11. 2 1 !3

15.

4 5 !6 2 !2 !2 1 1 1 !3 2

17. cos x 19. 2sin x

71.      r 5 8 sin u11 2 sin2 u2 5 8 sin u 2 8 sin3 u u 5 30°, r 5 3

21. 0 23. 22 cos1A 2 B2

u 5 60°, r 5 !3

25. 22 sin1A 1 B2 27. tan126°2

u 5 90°, r 5 0

73. Each dollar spent on costs produces $1.33 in revenue. 75. Cosine is negative in quadrant III. 77. false

79. 90°

83. 8 k cos u k

"1 2 sin u sin u

85. 5 k sec u k

89. yes  a. 0.3746  b. 0.9272  c. 0.4040 d. 0.4040 (c) and (d) are the same.

1 1 2!30 26!6 1 4 31. 12 25

33.

3 2 4!15 192 2 25!15 5 2119 4 1 3!15

35. identity

2

81. cot u 5 6

29.

37. conditional

39. identity 41. identity 87. true

43. identity 45. conditional 47. identity 49. identity 51. conditional

91. (a) and (c)

Young_AT_6160_ANS_SE-pg1211-1260.indd 1222

09/12/16 2:27 PM

Answers to Odd-Numbered Exercises  

53. y 5 sinax 1

p b 3

73. false

1223

75. false

79. A 5 np and B 5 mp, where n and m are integers, or A and B, where B 5 A 6 2np. 81. a.

55. y 5 cos ax 2

p b 4

c.

57. y 5 2sin14x2

59. y 5 tanax 1

61. y 5 tana

b.

p b 4

p 1xb 6

The difference quotients of y 5 sin x better approximate y 5 cos x as h goes to zero. 83. a.

b.

c.

z 67. cos1kz 2 ct2 5 cos1kz2cos1ct2 1 sin1kz2sin1ct2; when 5 integer, l then kz 5 2pn and the sin kz term goes to zero. p 69. T1t2 5 38 2 2.5 sin a tb 6 71. Tangent of a sum is not the sum of the tangents. Needed to use the tangent of a sum identity.

Young_AT_6160_ANS_SE-pg1211-1260.indd 1223

The graphs better approximate y 5 2 cos 2x as h goes to zero.

09/12/16 2:27 PM

1224 

Answers to Odd-Numbered Exercises

Section 7.4

49. y 5 2 sin x 2 3 cos12x2

4 1. 2 5 120 5. 169 9.

!19 10

13. tan130°2 5

120 3. 119 4 7. 2 3 11. !3 3

15.

119 120 1 p !2 sina b 5 2 4 4 !3 3

17. cos14x2

19. 2

!3 21. 2 2

!3 23. 2 2

41. y 5 cot x

51. C1t2 5 2 1 10 cos 2t 53. 22,565,385 lb 55. !2 ft

57. Sine is negative (not positive). 59. false

61. false

63. true

65. true

67. false 69. tan 1 24x 2 5 71. no

4 tan x 1 tan2 x 2 1 2 1 1 2 tan2 x 2 2 2 4 tan2 x

72.–76.  See Instructor’s Solution Manual. 43. y 5 sec12x2

77. Y1 and Y3

Section 7.5 1.

"2 2 !3 2

3. 2

"2 2 !3 2 2 9. 2 "2 1 !2 5.

2

13. 2 1 45. y 5 sin 1 4x 2 2

17. 21.

"2 2 !2

2!13 13

!5 2 1 3 2 !5 or 2 Å 2

25. 2 29. 2 1 47. y 5 12 sin 1 2x 2 2



"2 1 !3 2

7. "3 1 2!2

11. 1 2 !2 or 15. 1 19. 23.

21 "3 1 2!2

3!13 13 Å

3 1 2!2 1 1 !2  or  6 !6

!15 1 2 24/ !601 27. 5 Å 2

1 2 !0.91 7 31. Å 1 1 !0.91 Å3

33. cos a 37. tana

5p b 35. tan175°2 12

5p b 8

51. y 5 2 1 2 cos x

Young_AT_6160_ANS_SE-pg1211-1260.indd 1224

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Answers to Odd-Numbered Exercises  

53. y 5 cos x

27. 2 cos a

p 5p xbcos a xb 24 24

29. 2tan x 33. cot a

3x b 2

1225

31. tan12x2 p 4p 43. R 1 t 2 5 !3 cos a t 1 b 6 3

45. Average frequency: 443 Hz; Beat frequency: 102 Hz

55. y 5 1 2 2 cos x

47. 2 sin c

2pct 1 1 1 a b106 d 2 1.55 0.63

49. 2 sin c

2p 1 1979 2 t 2p 1 439 2 t d cos c d 5 2 sin11979pt2cos1439pt2 2 2

cos c

51.

2pct 1 1 2 a b106 d 2 1.55 0.63

25 1 !6 2 !3 2 < 5.98 ft2 3

53. cos A cos B 2 cos1AB2 and sin A sin B 2 sin1AB2. Should have used the product-to-sum identity. 55. false

57. y 5 tan x

57. true

59. Answers will vary. Here is one approach.

f

sin A sin B sin C

1 2  3cos1A 2 B22cos 1A 1 B24

59.

1 2

cos 1 A 2 B 2 sin C

 3sin1A2B1C21sin1C2A1B24

sin  A sin B sin C 5

1 3

60.–  62.  See Instructor’s Solution Manual. x 63. sina b is positive, not negative, in this case. 2 65. false 67. false 68.–72.  See Instructor’s Solution Manual. 73. 0 , x , p

75. yes

77. identity

79. Y1 and Y3

2

cos 1 A 1 B 2 sin C ¥ f

1 ≥ 2

f

5

1 2

 3sin1A2B1C21sin1C2A1B24

1 3 sin 1 A 2 B 1 C 2 1 sin 1 C 2 A 1 B 24 4

2sin1A1 B 1 C2 2 sin1C2 A 2 B24

60.–  62.  See Instructor’s Solution Manual. 3 7p 3 5p 63. y 5 1 2 cos a xb 1 cos a xb 2 6 2 6

Section 7.6 1. 12 3sin13x2 1 sin x4

3. 52 3cos12x2 2 cos110x24

5. 23cos x 1 cos13x24

1 3p 1 p 65. y 5 2 cos a xb 2 cos a xb 2 2 2 6

7. 12 3cos x 2 cos14x24

1 2x 9. c cos a b 1 cos 1 2x 2 d 2 3

11. 232 3cos11.9x2 1 cos11.1x24

13. 2 3 sin 1 2!3x 2 2 sin 1 4!3x 24

15. 2 cos14x2cos x 19. 22 sin x

cosA 32 xB

23. 2 sin10. 5x2cos10.1x2

Young_AT_6160_ANS_SE-pg1211-1260.indd 1225

17. 2 sin x cos12x2 21. 2 cosA 32 xBcosA 56 xB 25. 22 sinA !5xBcosA2!5xB

09/12/16 2:27 PM

1226 

Answers to Odd-Numbered Exercises

101. April or October

67. y 5 sin14x2

103. in month 3 105. t < 0.026476 sec or 26 ms 107. 173.4; June 22–23 109. 11.3 < 11 years 7 1 1 x x 8x 111. tan u 5 5 2 7 1 x 27 12 ⋅ x x 113. 0.70 m; 0.24 m

69. Y1 and Y3

Section 7.7 p 1. 4 3p 5. 4 p 9. 6 13. 0

3. 2 7.

p 3

p 6

11. 2

p 3

19. 45°

21. 120°

23. 30°

25. 230°

27. 135°

29. 290°

31. 90°

33. 57.10°

35. 62.18°

37. 48.10°

39. 215.30°

41. 166.70°

43. 20.63

45. 1.43

47. 0.92

49. 2.09 5p 53. 12 p 57. 6

51. 0.31

65. not possible p 69. 2 4

150 300 b 2 tan21 a b x 200 2 x

117. The wrong interval for the identity was used. The correct p p domain is c 2 , d . 2 2 119. cot21 x 2

1 tan21 x

121. false

15. p

17. 60°

61. !3

115. u 5 p 2 tan21 a

55. not possible 2p 3 p 63. 3

59.

67. 0 71. not possible

73.

2p 3

75. 2

77.

!7 4

79.

12 13

p 4

125.

1 2

127.

"x 2 2 1 x

123. false

is not in the domain of the inverse secant function.

129. 0

p 5p 131. a.  a , b 4 4

b. ƒ21 1 x 2 5 cos21 1 x 2 3 2 1

133. a.  a

p 7p , b 12 12

p Domain: 32, 44 4

p 1 1 arccot 14x 2 82 12 2 Domain: All real numbers b. ƒ 21 1 x 2 5

135. The identity sin 1 sin21 x 2 5 x only holds for 21 # x # 1. p p 137. c 2 , 0b ∪ a0, d 2 2 139. a. 

720   b.  0.42832  c.  yes 1681

Section 7.8

81.

3 4

83.

5!23 23

3p 5p , 4 4 7p 11p 19p 23p 3. , , , 6 6 6 6 5. np, where n is an integer

85.

4!15 15

87.

11 60

7.

7p 11p 19p 23p , , , 12 12 12 12

89.

24 25

91.

56 65

9.

7p 11p 1 4np or 1 4np, where n is an integer 3 3

93.

24 25

95.

120 119

11. 2

4p 5p 11p p p 2p 7p 5p ,2 ,2 ,2 , , , , 3 6 6 3 6 3 6 3

99.

"1 2 u2 u

13. 2

4p 2p ,2 3 3

97. "1 2 u2

Young_AT_6160_ANS_SE-pg1211-1260.indd 1226

1.

09/12/16 2:27 PM

Answers to Odd-Numbered Exercises  

15.

p np 1 , where n is an integer 6 4

17. 2p, 2

5p p ,2 3 3

p p 7p 4p 21. , , , 6 3 6 3

p 7p 11p 19. 2 , 2 , 2 2 6 6 p 7p 13p 19p 23. , , , 12 12 12 12

25.

p 2p 4p 5p , , , 3 3 3 3

27.

2p 3

29.

p 3p 5p 7p , , , 4 4 4 4

31.

p 3p p 5p , , , 2 2 3 3

3p 7p 11p 33. , , 2 6 6

p 35. 2

37. 0, p 39.

p 5p 7p 11p 13p 17p 19p 23p , , , , , , , 12 12 12 12 12 12 12 12

41. 115.83°, 154.17°, 295.83°, 334.17° 43. 333.63°

45. 29.05°, 209.05°

47. 200.70°, 339.30°

49. 41.41°, 318.59°

107. 2016 3 111. sec 4

1227

109. 24°

p 3 5p 3 113. 1 0, 1 2 , a , b, 1 p, 23 2 , a , b 3 2 3 2 115. March and September

117. Around 1 am and 11 am 119. Extraneous solution. Forgot to check. 121. Can’t divide by cos x. Must factor. 123. false p 5p 7p 11p 127. , , , 6 6 6 6 p 129. or 30° 6

125. true

131. no solution

133. 5 1 2n, where n is an integer 135. x 5

p 5p < 0.524 and x 5 < 2.618 6 6

51. 56.31°, 126.87°, 236.31°, 306.87° 53. 9.74°, 80.26°,101.79°,168.21°,189.74°, 260.26°, 281.79°, 348.21° 55. 80.12°, 279.88° 57. 64.93°, 121.41°, 244.93°, 301.41° 59. 15°, 45°, 75°,105°,135°,165°,195°, 225°, 255°, 285°, 315°, 345° p 5p 61. , 4 4 p 65. 6 69.

63. p 67.

p 3

p 3p 5p 7p , , , 4 4 4 4

71.

p 3p , 2 2

p 7p , p, 4 4

75.

p 5p 7p 11p p 3p , , , , , 6 6 6 6 2 2

79.

p 5p 7p 11p , , , 6 6 6 6

83.

2p 4p , 3 3

73. 0,

p p 7p 4p 77. , , , 6 3 6 3 81.

3p 2

85.

p 5p , ,p 3 3

87.

p 5p 13p 17p 25p 29p 37p 41p , , , , , , , 24 24 24 24 24 24 24 24

137. no solution

139. no solution

89. 57.47°, 122.53°, 323.62°, 216.38° 91. 30°, 150°, 199.47°, 340.53° 93. 14.48°, 165.52°, 270° 95. 111.47°, 248.53°

141.

p 5p 97. , 3 3 99. Fourth quarter of 2017, second quarter of 2018, and fourth quarter of 2019 101. around 9 pm 103. March 105. A 5 12 h 1 b1 1 b2 2   5 12 1 x sin u 21 x 1 1 x cos u 1 x 1 x cos u 22   5 x 2 sin u 1 1 1 cos u 2

Young_AT_6160_ANS_SE-pg1211-1260.indd 1227

09/12/16 2:28 PM

1228 

Answers to Odd-Numbered Exercises

143. x < 1.3

59.

7 671 336 61. 63. 25 1800 625

65.

!3 3 67. 2 2

  68.–74.  See Instructor’s Solution Manual. 75. 2

"2 2 !2 2 2

145. 2.21911

79. 2

Review Exercises

5 83. 2 4

1. 211 7

8 3. 215

4 5. 27

7. 3!3

7 9. 224

5 11. 212

13. sin u 5 2

!3 1 , cos u 5 2 2 2

15.

cos u 17. sin  u sin2 u

19.

1 1 cos u sin u

"2 1 !3



77. !2 2 1 81.

7!2 10

85. sina

p b 12

  86.–90.  See Instructor’s Solution Manual. 91. y 5 ` sina

p xb ` 24

21. sec2 x

23. sec2 u

25. cos2 x 2

27. 214 1 2 csc x 1 csc  x2   28.–34.  See Instructor’s Solution Manual.

x 93. y 5 2tana b 2

35. identity 37. conditional 39. identity 41.

!2 2 !6 4

!3 2 3 5 !3 2 2 45. sin  x 3 1 !3 117 47. tan x 49. 44 897 51. 2 53. identity 1025 43.

55. y 5 cos ax 1

p b 2

95. 33sin17x2 1 sin13x24 97. 22 sin14x2sin x x 99. 2 sin a b cos x 3 101. cot13x2

102.–106.  See Instructor’s Solution Manual. p p 107. 109. 4 2 p 111. 113. 290° 6 115. 60° 117. 260° 57. y 5 tana

2x b 3

119. 237.50°

121. 22.50°

123. 1.75 p 127. 2 4 p 131. 3

125. 20.10

135.

7 6

139. July 143. 2

Young_AT_6160_ANS_SE-pg1211-1260.indd 1228

3p p ,2 2 2

129. 2 !3 133.

60 61

137.

6!35 35

141.

2p 5p 5p 11p , , , 3 6 3 6

145.

9p 21p , 4 4

09/12/16 2:28 PM

Answers to Odd-Numbered Exercises  

147.

Cumulative Test

p 2p 4p 5p , , , 3 3 3 3

1. x 5 3 6 2!5

3p 7p 11p 15p 149. , , , 8 8 8 8 151. 0, p,

3. 1x 1 3221 y 1 122 5 25 5. even

3p 7p , 4 4

7. ƒ + g 5

153. 80.46°, 170.46°, 260.46°, 350.46°

157. 17.62°, 162.38°

163. 0,

11. Q1x2 5 5x 2 4, r 1x2 5 25x 1 7

p 161. , p 3

p 11p , p, 6 6

167. p

165. 169.

1 2 1; domain is x 2 0 x3

6 3 9. a2 , 2 b 5 5

155. 90°, 138.59°, 221.41°, 270° p 5p 159. , 4 4

1229

13. HA: y 5 0.7; VA: x 5 22, x 5 3 15. 1 23, q 2

3p 2 p 3p , 2 2

171. 90°, 135°, 270°, 315°

17. 4

19. 0.4695

21. 2

23. conditional 25.

7p 12

5 12

27. 1.3994

173. 0° 175. 60°, 90°, 270°, 300°

CHAPTER 8

177. (a) and (c)

179. Y1 5 Y2

181. 3 cos13x2

183. Y1 and Y3

Section 8.1

185. Y1 and Y3

187. Y1 and Y3

1. SSA

3. SSS

189. a. 235   b. 20.6  c. yes

7. g 5 75°,  b 5 12.2 m,  c 5 13.7 m

191. 0.5787

9. b 5 62°,  a 5 163 cm,  c 5 215 cm

5. ASA

11. b 5 116.1°,  a 5 80.2 yd,  b 5 256.6 yd

Practice Test 1 2n 1 1 2 1. x 5 p, where n is an integer 2 "2 2 !2 3 !30 5. 5 3. 2 2 Å 10 10

7. cos110x2 9. cos a 11. 20 cos x cos 3

a1b b 2

4p 1 2np 3 , where n is an integer 13. µ 5p 1 2np 3 15. 14.48°, 90°, 165.52°, 270° 17. conditional 19. 2

!26 p p 21. cot a x 1 b 26 6 8

23. One-to-one on the interval 1 1 1 x2a c c c 2 , cb ∪ ac, c 1 d . f 21 1 x 2 5 csc21 a b 2  . p p 2 2 b

25. 83 27.

1 8n,

16 3

1 8n, where n is an integer

p 7p 1 6np, 1 6np, where n is an integer 2 2

3!10   b.  0.94868  c.  yes 29. a.  10

13. g 5 120°,  a 5 7 m,  b 5 7 m 15. a 5 97°,  a 5 118 yd,  b 5 52 yd 17. b1 5 20°,  g1 5 144°,  c1 5 9; b2 5 160°,  g2 5 4°,  c2 5 1 19. a 5 40°,  b 5 100°,  b 5 18 21. no triangle 23. b 5 90°,  g 5 60°,  c 5 16 25. b 5 23°,  g 5 123°,  c 5 15 27. b1 5 21.9°,  g1 5 136.8°,  c1 5 11.36 b2 5 158.1°,  g1 5 0.6°,  c1 5 0.17 29. b 5 62°,  g 5 2°,  c 5 0.3 31. b1 5 77°,  a1 5 63°,  a 5 457 b2 5 103°,  a2 5 37°,  a 5 309 33. a 5 31°,  g 5 43°,  c 5 2 35. 1246 ft 37. 1.7 mi 39. 1.3 mi 41. 26 ft 43. 270 ft 45. 63.83° 47. 76.5 ft

49. 60.14 ft

51. 1.2 cm

53. No solution because b . a and a , 90° 55. false 57. true 59. true 61.–70.  See Instructor’s Solution Manual.

Section 8.2 1. C 3. S

5. S 7. C

9. b 5 5,  a 5 47°,  g 5 33° 11. a 5 5,  b 5 158°,  g 5 6°

Young_AT_6160_ANS_SE-pg1211-1260.indd 1229

09/12/16 2:28 PM

1230 

Answers to Odd-Numbered Exercises

13. a 5 2,  b 5 80°,  g 5 80°

Section 8.4

15. b 5 5,  a 5 43°,  g 5 114°

1. !13 3. 5!2

17. b 5 7,  a 5 30°,  g 5 90°

7. !73; u 5 69.4° 9. !26; u 5 348.7°

19. a 5 93°,  b 5 39°,  g 5 48°

11. !17; u 5 166.0° 13. 8; u 5 180°

21. a 5 51.3°,  b 5 51.3°,  g 5 77.4°

15. 2!3; u 5 60° 17. 822, 229

23. a 5 75°,  b 5 57°,  g 5 48°

19. 8212, 99 21. 80, 2149

25. no triangle

23. 8236, 489 25. 86.3, 3.09

27. a 5 67°,  b 5 23°,  g 5 90°

27. 822.8, 15.89 29. 82.6, 23.19

29. g 5 105°,  b 5 5,  c 5 9

31. 88.2, 23.89 33. 821, 1.79

31. b 5 12°,  g 5 137°,  c 5 16

5 35. X213 , 212 13 Y

33. a 5 66°,  b 5 77°,  g 5 37°

39.

35. g 5 2°,  a 5 168°,  a 5 13 37. b 5 11.16,  a 5 42.40°,  g 5 85.91°

24

X25 ,

43. h

39. a 5 46.76°,  b 5 58.45°,  g 5 74.79°

37.

7 225 Y

!10 3!10 i , 10 10

41. X235 , 245 Y 45. 7i 1 3j

51. 2i 1 0j 53. 25i 1 5j

43. 2710 mi 45. 1280 mi 47. 63.7 ft

55. 7i 1 0j

51. 0.8 mi 53. 26°

57. /QRP 5 151.03°, /RQP 5 3.97°

57. Vertical: 1100 ft /sec Horizontal: 1905 ft /sec

59. 0.5°

59. 2801 lb

55. 21.67° 61. about 83°

63. Should have used the smaller angle b in Step 2

61. 11.7 mph; 31° west of due north

65. false 67. true 69. true

63. 52.41° east of north; 303 mph

70.–72.  See Instructor’s Solution Manual. 73.

Ç

65. 250 lb

1 2 cosA2 cos21 A 14 B B

67. Vertical: 51.4 ft /sec Horizontal: 61.3 ft /sec

2

75.–80.  See Instructor’s Solution Manual.

69. 29.93 yd

Section 8.3

71. 10.9°

1. 55.4 3. 0.5

5. 23.6 7. 6.4

9. 4408.4 11. 9.6

13. 97.4 15. 25.0

73. 1156 lb 75. Magnitude: 351.16; Angle: 23.75° from 180° N force

17. 26.7 19. 111.64 21. 111,632,076 23. no triangle 29. 11.98

77. Tension: 5!2; k S uk 5 5

25. 13.15 27. 174.76 31. 19.21 33. 312,297 nm

35. 10,591 ft2 37. 16° or 164° 39. 312.4 mi2 41. a.  41,842 sq ft  b. $89,123 43. 47,128 sq ft

60 11

X11 , 61 Y

47. 5i 2 3j 49. 2i 1 0j

41. a 5 1.09,  b 5 61.327°,  g 5 47.460° 49. 16 ft

5. 25

45. 23.38 ft2

2

S

79. a.  0 81. 8.97 Nm

b. 10 1 3!2 1 !10 units

83. 31.95 Nm

47. Area of PQ1Q2 5 12

85. Magnitude: 8.67; Direction: 18.05° counterclockwise of S

Area of PQ2Q3 5 22.55

87. 19°

Area of Q1Q2Q3 5 15.21 Area of PQ1Q3 5 49.76 The sum of the areas of the smaller three triangles does not equal the area of the outer triangle. 49. 10.86 51. Semiperimeter is half the perimeter.

89. Magnitude is never negative. Should not have factored out the negative but instead squared it in finding the magnitude. 91. false 93. true

53. true 55. true 57. false

95. vector

58.–60.  See Instructor’s Solution Manual.

97. "a2 1 b2

61. 0.69 63.–70.  See Instructor’s Solution Manual.

Young_AT_6160_ANS_SE-pg1211-1260.indd 1230

  98.–100.  See Instructor’s Solution Manual.

09/12/16 2:28 PM

Answers to Odd-Numbered Exercises  

101. 826, 49 5 12 828, 49 2281, 219

102.–104.  See Instructor’s Solution Manual. 105. Directly

1231

7p 7p b 1 i sina b d 5 4 4 !2 1 cos 315° 1 i sin 315° 2 9. !2 c cos a

p p 11. 2 c cos a b 1 i sina b d 5 3 3 1 2 2 cos 60° 1 i sin 60°

3p 3p b 1 i sina b d 5 4 4 4!2 1 cos 135° 1 i sin 135° 2

13. 4!2 c cos a

107.

5p 5p b 1 i sina b d 5 3 3 2!3 1 cos 300° 1 i sin 300° 2

15. 2!3 c cos a

17. 3 1 cos 0 1 i sin 0 2 5 3 1 cos 0° 1 i sin 0° 2 19.

109. 183, 100.4°

Section 8.5 1. 2 3. 23

5. 42 7. 11

!2 5p 5p c cos a b 1 i sina b d 5 2 4 4

!2 1 cos 225° 1 i sin 225° 2 2 21. 5.32 1 cos 0 1 i sin 0 2

23. !58 1 cos 293.2° 1 i sin 293.2° 2

9. 213a 11. 21.4

13. 98° 15. 109°

17. 3° 19. 30°

21. 105° 23. 180°

25. no 27. yes

29. no 31. yes

33. yes 35. yes

37. 400 ft-lb 39. 80,000 ft-lb

27. 13 1 cos 112.6° 1 i sin 112.6° 2

41. 1299 ft-lb

43. 148 ft-lb

31.

45. 1607 lb

47. 694,593 ft-lb

S S

49. u ⋅ v 5 49,300, and it represents the total cost of buying the prescribed number of 10-packs of both types of battery.

25. !61 1 cos 140.2° 1 i sin 140.2° 2 29. 10 1 cos 323.1° 1 i sin 323.1° 2

!13 1 cos 123.7° 1 i sin 123.7° 2 4

33. 5.59 1 cos 24.27° 1 i sin 24.27° 2

       50.–54.  See Instructor’s Solution Manual.

35. !17 1 cos 212.84° 1 i sin 212.84° 2

55. a.  v1 5 214.72  b. 64.57°

39. 25

S

S S

S

57. n 5 8r, 09; u ⋅ n 5 r k u k

43. 22 2 2!3i

59. 22

61. The dot product of two vectors is a scalar (not a vector). Should have summed the products of components. 63. false 65. true 67. 17   68.–74.  See Instructor’s Solution Manual. S

37. 4.54 1 cos 332.31° 1 i sin 332.31° 2

47. 1 1 i

41. !2 2 !2i

3 !3 45. 2 1 i 2 2

49. 2.1131 2 4.5315i 51. 20.5209 1 2.9544i 53. 5.3623 2 4.4995i 55. 22.8978 1 0.7765i

S

75. a.  22u   b.  2cu

77. Any vector that is perpendicular to u

57. 0.6180 2 1.9021i

79. 26 is minimum and 6 is maximum.

59. 100 1 cos 0° 1 i sin0° 2 1

S

81. 21083 83. 31.43° 85. 47°

Section 8.6 The answers to Exercises 1 to 8 are all plotted on the same graph, below.

120 1 cos 30° 1 i sin30° 2 ; 212.6 lb 61. 80 1 cos 0° 1 i sin0° 2 1

150 1 cos 30° 1 i sin30° 2 ; 19.7° 63. S u 5 8 1 cos 150° 1 i sin150° 2 ; S v 5 6 1 cos 45° 1 i sin45° 2 ;

S S w 5 kw k 1 cos u 1 i sinu 2 , where

S w 5 " 1 4!3 2 3!2 2 2 1 1 24 2 3!2 2 2;

u 5 tan21 a

Young_AT_6160_ANS_SE-pg1211-1260.indd 1231

24 2 3!2 b 4!3 2 3!2

09/12/16 2:28 PM

1232  S

Answers to Odd-Numbered Exercises S

65. R 5 k R k 1 cos u 1 i sinu 2 , where S

 R  5 " 1 100!2 1 180 2 2 1 1 100!2 2 2; u 5 tan21 a S

35. 21cos 20° 1 i sin 20°2, 21cos 140° 1 i sin 140°2, 21cos 260° 1 i sin 260°2

100!2 b 100!2 1 180

 R  5 351.16 1 cos 1 23.75° 2 1 i sin 1 23.75° 22 67. The point is in QIII (not QI). 69. true

71. true

73. 0° 75. k b k 77. a!5 1 cos 296.6° 1 i sin 296.6° 2

3 37. ! 2 1cos 110° 1 i sin 110°2, 3 ! 2 1cos 230° 1 i sin 230°2, 3 ! 2 1cos350° 1 i sin 350°2

1 !3 79. pa 2 ib 2 2

81. z 5 8.79 1 cos 28° 1 i sin 28° 2

83. 1.414, 45°, 1.414 1 cos 45° 1 i sin 45° 2 85. 2.236 1 cos 26.6° 1 i sin 26.6° 2 87. 35,323°

Section 8.7 1. 26 1 6!3i

3. 24!2 2 4!2i

5. 0 1 8i

7.

9. 0 1 12i 13. 2 !2 1 !2i 17.

3 3!3 1 i 2 2

9!2 9!2 1 i 2 2

3 3!3 11. 1 i 2 2 15. 0 2 2i

39. 2 1cos 78.75° 1 i sin 78.75°2, 2 1cos 168.75° 1 i sin 168.75°2, 2 1cos 258.75° 1 i sin 258.75°2, 2 1cos 348.75° 1 i sin 348.75°2

5 5!3 19. 2 2 i 2 2

21. 4 2 4i

23. 264 1 0i

25. 28 1 8!3i

27. 1,048,576 1 0i

29. 21,048,576!3 2 1,048,576i 31. 2 1cos 150° 1 i sin 150°2 and 2 1cos 330° 1 i sin 330°2

41. x 5 62, x 5 62i 43. 22, 1 2 !3i, 1 1 !3i

45. !2 2 !2i, !2 1 !2i, 2 !2 2 !2i, 2 !2 1 !2i 47. 1, 21, 49. 2 33. !6 1cos 157.5° 1 i sin 157.5°2, !6 1cos 337.5° 1 i sin 337.5°2

1 !3 1 !3 1 !3 1 !3 1 i, 2 i, 2 1 i, 2 2 i 2 2 2 2 2 2 2 2

!2 !2 !2 !2 1 i, 2 i 2 2 2 2

4 51. ! 2 c cos a

p pk p pk 1 b 1 i sina 1 bd, 8 2 8 2

k 5 0, 1, 2, 3 53. 2 c cos a

p 2pk p 2pk 1 b 1 i sina 1 bd, 5 5 5 5

k 5 0, 1, 2, 3, 4 55. p2 c cos a

p 2pk p 2pk 1 b 1 i sina 1 bd, 14 7 14 7

k 5 0, 1, 2, 3, 4, 5, 6

Young_AT_6160_ANS_SE-pg1211-1260.indd 1232

09/12/16 2:28 PM

Answers to Odd-Numbered Exercises  

57. 1cos 45° 1 i sin 45°2, 1cos 117° 1 i sin 117°2, 1cos 189° 1 i sin 189°2, 1cos 261° 1 i sin 261°2, 1cos 333° 1 i sin 333°2

1233

Section 8.8 The answers to Exercises 1 to 10 are all plotted on the same graph, below.

5 6 11 12

3 4

7 12

2 3

1

3 4

7

8

6

5 12

2

6

4

12 0

13 12 7 6

59. c cos a5

p pk p pk 1 1 b 1 i sin a5 bd, 18 3 18 3

k 5 0, 1, 2, 3, 4, 5

11. a4,

52 5 4

10

9 4 3

17 12

29. a 35.

3 2

19 12

5 3

7 4

23 12 11 6

p 4p 3p b 13. a2, b 15. a4!2, b 3 3 4

17. 13, 02 19. a2, 23. a

3

7p b 21. 1 2, 22!3 2 6

!3 1 , 2 b 25. 1 0, 0 2 27. 1 21, 2 !3 2 2 2 !2 !2 ,2 b 2 2

31. d 33. a

61. Reversed the order of angles being subtracted 63. Use De Moivre’s theorem; in general, 1 a 1 b 2 6 2 a6 1 b6.

65. true

67. false 69. true 70.–78.  See Instructor’s Solution Manual. n1m p 2 4

79. 2

e

37.

1m2n2 i

81. cos 66° 1 i sin 66°, cos 138° 1 i sin 138°, cos 210° 1 i sin 210°, cos 282° 1 i sin 282°, cos 354° 1 i sin 354° 83. cos 40° 1 i sin 40°, cos 100° 1 i sin 100°, cos 160° 1 i sin 160°, cos 220° 1 i sin 220°, cos 280° 1 i sin 280°, cos 340° 1 i sin 340° 85. 3 1cos15° 1 sin15°2, 3 1cos135° 1 sin135°2, 3 1cos 255° 1 sin 255°2

Young_AT_6160_ANS_SE-pg1211-1260.indd 1233

39.

09/12/16 2:28 PM

1234 

Answers to Odd-Numbered Exercises

41.

55.

43.

57.

45.

59.

47.

61.

49.

63.

51. Line: y 5 22x 1 1 53. Circle: 1x 2 122 1 y2 5 9

Young_AT_6160_ANS_SE-pg1211-1260.indd 1234

09/12/16 2:28 PM

Answers to Odd-Numbered Exercises  

1235

85. x3 2 y3 2 2axy 5 0

65.

87. It is a circle of radius a and center 1a cos b, a sin b2. 89. The inner loop is generated beginning with u 5 with 3p 2.

p 2

and ending

91. The very tip of the inner loop begins with u 5 p2 ; then it crosses the origin (the first time) at u 5 cos21 A213 B, winds around, and eventually ends with u 5 3p 2. 93. 67.5°, 157.5°, 247.5°, 337.5° 67.

Review Exercises 1. g 5 150°, c 5 12, b 5 8 3. g 5 130°, a 5 1, b 5 9 5. b 5 158°, a 5 11, b 5 22

69. For (a)–(c), all three graphs generate the same set of points, as seen below:

7. b 5 90°, a 5 !2, c 5 !2

9. b 5 146°, b 5 266, c 5 178 11. b 5 26°, g 5 134°, c 5 15 or b 5 154°, g 5 6°, c 5 2 13. b 5 127°, g 5 29°, b 5 20 or b 5 5°, g 5 151°, b 5 2 15. no triangle 17. b 5 15°, g 5 155°, c 5 10 or b 5 165°, g 5 5°, c 5 2 19. 12.2 mi 21. a 5 42°, b 5 88°, c 5 46 23. a 5 51°, b 5 54°, g 5 75° 25. a 5 42°, b 5 48°, g 5 90°

Note that all three graphs are figure eights. Extending the domain in (b) results in twice as fast movement, while doing so in (c) results in movement that is four times as fast. 71. 6 times 73. a.  r 5 8 cos 3u b. 50 times because the period is 2p 75. The point is in QIII; the angle found was the reference angle (needed to add p).

29. b 5 68°, g 5 22°, a 5 11 31. a 5 51°, b 5 59°, g 5 70° 33. b 5 37°, g 5 43°, a 5 26 35. b 5 4°, g 5 166°, a 5 28 37. no triangle 39. b 5 10°, g 5 155°, c 5 10.3 41. 141.8 43. 51.5 45. 89.8

77. true 79. r 5

27. b 5 28°, g 5 138°, a 5 4

a cos u

81. 12a, u 6 180°2

83. The graphs intersect when p 2p 4p 5p u5 , , , . 3 3 3 3

47. 41.7 49. 5.2 in. 51. 13 53. 13 55. 26; 112.6° 57. 20; 323.1° 59. 82, 119 61. 838, 279 63. 82.6, 9.79 65. 823.1, 11.69 67. h

69. (5i 1 j)

71. 26 73. 29

75. 16 77. 59°

79. 49° 81. 166°

83. no 85. yes

87. no 89. no

91. 5i 4i 3i 2i i

–6 + 2i

–8

Young_AT_6160_ANS_SE-pg1211-1260.indd 1235

!2 !2 ,2 i 2 2

–6

–4

–2 –i –2i –3i –4i –5i

Imaginary axis

Real axis 1 2

09/12/16 2:28 PM

1236 

Answers to Odd-Numbered Exercises

139.

93. 2 1cos 315° 1 i sin 315°2 95. 8 1cos 270° 1 i sin 270°2 97. 61 1cos 169.6° 1 i sin 169.6°2 99. 17 1cos 28.1° 1 i sin 28.1°2 101. 3 2 3!3i

103. 21 1 i

105. 23.7588 2 1.3681i

107. 212i

21 21!3 109. 2 2 i 2 2 113. 26

1 !3 1 i 2 2 115. 2324

111. 2

117. 16 2 16!3i 141.

119. 2 1cos 30° 1 i sin 30°2, 2 1cos 210° 1 i sin 210°2

143. Z 5 30.4°, Y 5 107.6°, B 5 45.0 121. 4 1cos 45° 1 i sin 45°2,

145. C 5 4.0, X 5 74.1°, Y 5 64.3°

4 1cos 135° 1 i sin 135°2,

149. 65, 293° 151. 140° 153. 12,246°

4 1cos 225° 1 i sin 225°2,

155. 2 1cos 30° 1 sin 30°2, 2 1cos 120° 1 sin 120°2,

4 1cos 315° 1 i sin 315°2

2 1cos 210° 1 sin 210°2, 2 1cos 300° 1 sin 300°2

157. 10°, 50°, 130°, 170°, 250°, 290° 123. 26, 3 1 3!3i, 3 2 3!3i 125.

Practice Test

!2 !2 !2 !2 !2 !2 !2 !2 1 i, 2 i, 2 1 i, 2 2 i, 2 2 2 2 2 2 2 2

1. a 5 7.8, c 5 14.6, and g 5 110°

All the points from Exercises 127–132 are graphed on the single graph below. 3p 127. a2!2, b 4 129. a10,

7p b 6

3p b 2 3 3!3 133. a2 , b 2 2 131. a2,

135. 1 1, !3 2

1 !3 137. a2 , 2 b 2 2

Young_AT_6160_ANS_SE-pg1211-1260.indd 1236

3. a 5 35.4°, b 5 48.2°, and g 5 96.4° 5. no triangle 7. b 5 1.82, c 5 4.08, g 5 50°

5 6 11 12 13 12 7 6

3 4

2 3

7 12

2

5 12

9. 57 3

11. magnitude 5 13, u 5 112.6°

4 6

127

130 132

131 129 5 4

4 3

12

128 17 12

3 2

19 12

5 3

7 4

0 23 12 11 6

13. a.  8214, 59;  b.  216 15. 32,768 C 21 1 i !3 D 17. a

23!3 3 ,2 b 2 2

09/12/16 2:28 PM

Answers to Odd-Numbered Exercises  

23. 13, 12

19.

29. A1,

25. 12, 52 27. 123, 42

227 B 31.

5 2 2a b 5

39. infinitely many solutions: aa,

1.25 2 0.02a b 0.05

37. 122, 12

7 41. A 75 32 , 16 B 43. c

 20.–22.  See Instructor’s Solution Manual.

11 A 19 7 , 35 B

33. infinitely many solutions: aa, 35. 14, 02

1237

45. d

47. 10, 02

49. 121, 212

51. 10, 262

53. no solution

25. !2

23. 220

27. a 5 5.49, b 5 4.48, c 5 4.23, d 5 1.16 29. 4 c cos a

3p pk 3p pk 1 1 b 1 i sina bd, 8 2 8 2

k 5 0, 1, 2, 3 31. 112°

Cumulative Test 1. x 5 2 6 4i 3. 2 5. 22x 1 h 7. undefined 9. x 5 22, multiplicity 3; x 5 0, multiplicity 2

55. 6 AusPens per kit

11. VA: x 5 2; slant asymptote: y 5 x 1 2

57. 5 Montblanc pens, 64 Cross pens 59. 15.86 ml of 8% HCl, 21.14 ml of 15% HCl 61. $300,000 of sales 63. 169 highway miles, 180.5 city miles 65. plane speed: 450 mph, wind speed: 50 mph 67. 10% stock: $3500, 14% stock: $6500

13. $37,250

15. 0.001

69. 8 CD players

17. 15!2 ft < 21.21 ft

71. Every term in the first equation is not multiplied by 21 correctly. The equation should be 22x 2 y 5 3, and the resulting solution should be x 5 11, y 5 225.

19.

73. Did not distribute 21 correctly. In Step 3, the calculation should be 2123y 2 42 5 3y 1 4. 75. false

77. false

79. A 5 24, B 5 7

81. 2% drink: 8 cups, 4% drink: 96 cups 3 1 8!6 21. 2 25

83. 18.9, 6.42

23. c 5 6, a 5 132°, b 5 27° 25. 22 1 2i 27. 12°, 78°, 192°, 248°

CHAPTER 9 Section 9.1 1. 11, 02

3. 18, 212 5. 11, 212

7. 11, 22 9. u 5

32 17 ,

v5

11 17

11. no solution

85. infinitely many solutions: aa,



7 2 23a b 5

13. infinitely many solutions: 1a, 4a212 5a 2 15 15. infinitely many solutions: aa, b 3 17. 11, 32 19. 16, 82 21. 11.2, 0.042

Young_AT_6160_ANS_SE-pg1211-1260.indd 1237

09/12/16 2:28 PM

1238 

Answers to Odd-Numbered Exercises

87. 12.426, 20.0592

1 1 2 x x1 2 19. 1 x23 4 23. 2 x13 15.

27.

Section 9.2 1. x 5

232 ,

y 5 23, z 5

9 2

3. x 5 22, y 5

5. x 5 5, y 5 3, z 5 21

7. x 5

90 31 ,

y5

9 2,

z5

103 31 ,

1 2

z5

9 31

31. 35.

5 1 9. x 5 213 4 , y 5 2 , z 5 22 11. x 5 22, y 5 21, z 5 0

13. x 5 2, y 5 5, z 5 21

15. no solution

17. no solution 19. x 5 1 2 a, y 5 2 1 a 1 12 2 , z 5 a

21. x 5 41 1 4a, y 5 31 1 3a, z 5 a

23. x1 5 212 , x2 5 74 , x3 5 234 25. no solution 27. x1 5 1, x2 5 21 1 a, x3 5 a 29. x 5 23 a 1 83 , y 5 213 a 2

10 3,

z5a

31. 100 basic widgets, 100 midprice widgets, and 100 top-of-the-line widgets produced. 33. 8 touchdowns, 6 extra points, 4 field goals

1

7 x15 15 1x 1 322

22 7x 1 2 x14 x 13

29.

x 9x 2 2 1 x 19 x 1 922 2

33.

1 1 23x 2 1 1 1 x21 21x 1 12 2 1 x2 1 1 2

37.

3 2x 1 5 1 2 x21 x 1 2x 2 1

41.

1 1 1 1 5 d0 di f

39.

22 4x 2 3 1 2 x27 3x 2 7x 1 5 2x 2 3 5x 1 1 1 2 2 1 x 11 x 1 122 1 12x 1 2 x21 x 1x11

43. The form of the decomposition is incorrect. It should be A Bx 1 C 1 2 . Once this correction is made, the correct x x 11 1 2x 1 3 decomposition is 1 2 . x x 11 1 1 1 45. false 47. 2 1 x21 x12 x22 1 1 1 1 2 3 x x11 x 53. yes 49.

35. 6 Sriracha Chicken sandwiches 3 Tuna Salad sandwiches 5 Roast Beef sandwiches

1 x21 3 4 21. 1 1x 2 122 x21 3 1 2 25. 1 1 1x 2 522 x11 x25 17.



x 2x x12 2 2 1 2 1x 1 122 1x 1 123 x2 1 1 55. no 51.

37. h0 5 0, v0 5 52, a 5 232 39. y 5 20.0625x2 1 5.25x 2 50 41. money market: $10,000, mutual fund: $4000, stock: $6000 43. 33 regular model skis, 72 trick skis, 5 slalom skis 45. game 1: 885 points, game 2: 823 points, game 3: 883 points

57. yes

47. Equation 122 and Equation 132 must be added correctly—should be 2x 2 y 1 z 5 2. Also, should begin by eliminating one variable from Equation 112. 49. true 53. a 5

51. a 5 4, b 5 22, c 5 24

255 24 ,

b5

214 ,

c5

223 24 ,

d 5 14 , e 5 44

55. no solution 57. x1 5 22, x2 5 1, x3 5 24, x4 5 5

Section 9.4

59. x 5 41 1 4a, y 5 31 1 3a, z 5 a 61. same as answer in Exercise 59

80 48 63. A280 7,27, 7 B

Section 9.3

5.

9.

1. d

3. a

A B 7. 1 x25 x14 11. 2x 2 6 1

1. d 3. b

3x 1 33 x2 1 x 1 5

7.

11.

5. b A B C 9. 1 1 2 x24 x x

13.

Young_AT_6160_ANS_SE-pg1211-1260.indd 1238

Ax 1 B Cx 1 D 1 2 2 x 2 1 10 Ax 1 10B

09/12/16 2:29 PM

1239

Answers to Odd-Numbered Exercises  

13.

15.



17.

19.



45. no solution 47.

49. 4 units2

51. 7.5 units2

x $ 0, y $ 0 53. d � x 1 20y # 2400 25x 1 150y # 6000

21.

23. no solution



55. a. 275 # 10x 1 20y 125 # 15x 1 10y 200 # 20x 1 15y x $ 0, y $ 0 b. 

y 14 12 10

y

25.

27.

5 4 3 2 1 –5

–3

–1

8 6 4

x

2

x

1 2 3 4 5

2

–3 –4 –5

29.



31.

4

6

8

10

12

14

16

18

c.  Two possible diet combinations are 2 ounces of food A and 14 ounces of food B or 10 ounces of food A and 10 ounces of food B. 57. a. x $ 2y x 1 y $ 1000 x $ 0, y $ 0 b. 1000

y

900 800

33.



35.

700 600 500 400 300 200 100 x 100

37.

39.

200

300

400

500

600

700

800

900

c.  Two possible solutions would be for the manufacturer to produce 700 USB wireless mice and 300 Bluetooth mice or 800 USB wireless mice and 300 Bluetooth mice. P # 80 2 0.01x 59. • P $ 60 61. 20,000 x$0

41.

43.

63. The shading should be above the line. 65. true

67. false

69. shaded rectangle

Young_AT_6160_ANS_SE-pg1211-1260.indd 1239

09/12/16 2:29 PM

1240 

Answers to Odd-Numbered Exercises

71.

27. y 5 20.0050x2 1 0.4486x 2 3.8884

73.

29. 31.

75.

A B C D 1 1 1 1x 2 122 x21 x13 x25 A B C D 1 1 1 1 2x 1 1 2 2 x 4x 1 5 2x 1 1

33.

A B 1 x23 x14

35.

37.

4 5 1 x21 x17

39.

41.

22 2 1 x11 x

43.

Ax 1 B Cx 1 D 1 2 2 x 2 1 17 Ax 1 17B

1 15 3 1 2 2x 21x 2 52 21x 1 52 5 27 2 1x 1 222 x12

45.

47.

49.



Section 9.5 1. ƒ1x, y2 5 z 5 2x 1 3y  ƒ121, 42 5 10 ƒ12, 42 5 16 1MAX2 ƒ122, 212 5 27 1MIN2 ƒ11, 212 5 21

51.

3. ƒ1x, y2 5 z 5 1.5x 1 4.5y ƒ121, 42 5 16.5 ƒ12, 42 5 21 1MAX2 ƒ122, 212 5 27.5 1MIN2 ƒ11, 212 5 23

5. minimize at ƒ10, 02 5 0

7. no maximum

9. minimize at ƒ10, 02 5 0

53. no solution

11. maximize at ƒ11, 62 5 53 20 5 2.65

55.

13. Frances T-shirts: 130 Charley T-shirts: 50 1  profit $9502

15. laptops: 25, desktops: 0 1  profit $75002

17. first-class cars: 3, second-class cars: 27 19. 200 of each type of ski 21. Should compare the values of the objective function at the vertices rather than comparing the y-values of the vertices

57.

23. false 25. maximum at 10, a2 and is a 27. minimum at 10, 02 and is 0

29. maximum at 16.7, 4.52 and is 176.9

31. maximum at A 49 , 115 9 B and is approximately 25

59. minimum value of z is 0, occurs at 10, 02

Review Exercises 1. 13, 02 9. 122, 12

13. 13, 222

19 13 3. A 13 4 , 8B 5. 12, 12 7. A 8 , 8 B

11. A12, 35 6B

15. 121, 22 17. c

61. maximum value of z is 25.6, occurs at 10, 82 63. minimum value of z is 230, occurs at 10, 62 65. ocean watercolor: 10 geometric shape: 30 1  profit $3902

69.  13.6, 3, 0.82 71. yes

67. 12, 232

19. d

21. 6% NaCl: 10.5 ml, 18% NaCl: 31.5 ml 23. x 5 21, y 5 2a 1 2, z 5 a 25. no solution

Young_AT_6160_ANS_SE-pg1211-1260.indd 1240

09/12/16 2:29 PM

Answers to Odd-Numbered Exercises  

1241

15. possible rational zeros:

73.

61, 62, 65, 610, 612 , 652

rational zeros: 25, 21, 12 , 2 17. domain: 12q, q2 75. maximum is 12.06, occurs at 11.8, 0.62

range: 121, q2

y-intercept: 10, 02 HA: y 5 21

19. Shift the graph of y 5 ln x left one unit, and then down three units.

Practice Test 1. 17, 32

3. x 5 a, y 5 a 2 2

5. x 5 1, y 5 25, z 5 3

7. x 5 21, y 5 3, z 5 4

5 3 9. 2 x x11 13.

1 2 1 1 2 3x x13 31x 2 32

15.

11.

21. 0.435 23. x 5 2, y 5 0, z 5 5 25.

27. yes  

7 9 2 1x 1 222 x12

17.

CHAPTER 10 Section 10 . 1 1. 2 3 3 1 9.

7. c

23. money market: $14,000, aggressive stock: $8500, conservative stock: $7500 25. 111, 19, 12

15. e 19. e

x21 ,  x 2 56 5

5. 1 2q, 0 4 ∪ 1 1, 3 4

9. 2 85

3. no solution 7. y 5 0.5x 1 3.4 11. 1

2x 5 4 17. • 7x 1 9y 1 3z 5 23 4x 1 6y 2 5z 5 8



x5a y5b

21. not reduced form 25. not reduced form

27. reduced form

29. row-echelon form 1 22 21 3 33. £ 0 5 21 3 0 § 3 22 5 8

1 22 23 ` d 0 7 5

1 22 0 1 35. D 0 22 0 0

Cumulative Test

24 3 5 2 13. £ 2 23 22 3 23 § 22 4 3 1

23. reduced form

31. c

1.

1 0 3 0 21 3 2 § 1 1 5

23x 1 7y 5 2 x 1 5y 5 8

5. 1 3 1

2 23 4 23 9. £ 21 1 2 3 1§ 5 22 23 7

3 22 7 ` d 24 6 23

1 11. £ 1 0 2 1. minimum value of z is 7, occurs at 10, 12

3. 1 3 4

1 0 37. D 0 0

5 1 1 1

21 2 23 4 3 T 22 5 21 26

0 5 210 1 2 23 0 27 6 0 8 210

25 4 22 T 3 29

2

13. ƒ1x2 5 22x 1 7

Young_AT_6160_ANS_SE-pg1211-1260.indd 1241

09/12/16 2:29 PM

1242 

Answers to Odd-Numbered Exercises

1 0 39. D 0 0 41. c

0 1 0 0

4 2 1 0

119. a. 

0 27 0 4 211 T 0 21 1 23

     y 5 20.24x 2 1 0.93x 1 6.09

1 43. £ 0 0

1 0 28 ` d 0 1 6

1 45. £ 0 0

0 1 0

2 0 0 3 5§ 1 21

1 49. C 0 0

0 1 0

1 1 1 3 212 S 0 0

47. c

1 0

0 1 0

0 22 0 3 21 § 1 0

0 22 1 ` d 1 22 2

51. x 5 27, y 5 5

53. x 2 2y 5 23 or x 5 2a 2 3, y 5 a 55. no solution 59. x1 5

212 , x 2

57. x 5 4a 1 41, y 5 3a 1 31, z 5 a 5

7 4, x 3

5

234

61. no solution

63. x1 5 1, x2 5 a 2 1, x3 5 a

65. x 5 23 a 1 83 , y 5 213 a 2 67. no solution

73. no solution

77. 13, 22, 22 a 4

83. x 5

72 2 11a 14 ,

z5a

69. x1 5 22, x2 5 1, x3 5 24, x4 5 5

71. 11, 222 81. x 5

10 3,

79. no solution

1 3, y 5

7a 4

y5

75. 122, 1, 32

1 1, z 5 a

13a 1 4 14 ,

87. 8 touchdowns, 5 extra points, 1 two-point conversion, and 2 field goals 89. 2 Egg Salad, 2 Club, 8 Turkey, 2 Roast Beef 91. initial height: 0 ft, initial velocity: 50 ft/sec, acceleration: 232 ft/sec2 93. y 5 20.053x2 1 4.58x 2 34.76 95. about 88 ml of 1.5% solution and 12 ml of the 30% solution 97. 200 basic widgets, 100 midprice widgets, and 75 top-of-the-line widgets produced 99. money market: $5500, mutual fund: $2500, stock: $2000 101. 25 units product x, 40 units product y, 6 units product z 103. general admission: 25, reserved: 30, end zone: 45 280 44 105. a 5 222 17 , b 5 217 , c 5 2 17

107. Need to line up a single variable in a given column before forming the augmented matrix. The correct matrix is 2 21 1 1 1 0 0 2 £ 1 1 22 3 23 § , after reducing, £ 0 1 0 3 1 § 6 1 1 1 0 0 1 3 109. Row 3 is not inconsistent. It implies z 5 0.

115. ƒ 1 x 2 5

113. true 4 211 6 x

1

44 3 3 x

2

223 2 6 x

117.

     y 5 20.24x 2 1 0.93x 1 6.09

Section 10.2 1. 2 3 3

3. 2 3 2

7. 1 3 1

9. 4 3 4

1

94 3 x

1 44

5. 3 3 3

11. x 5 25, y 5 1

13. x 5 23, y 5 22, z 5 3

15. x 5 6, y 5 3

17. c

22 4 19. £ 2 22 § 21 3

3 6 35. £ 22 22 § 17 24 7 43. £ 0 23

22 13

12 2

23 33. £ 24 13

21 7 14

1 d 5 0 d 2

3 d 14 6 1§ 9

37. not defined

10 28 15 5§ 0 27

24 47. £ 24 § 216 51. A 5 c

6 8

29. c

3 d 5

39. 3 0 60 4

5 2

22 4

25. c

0 25 27. £ 210 5§ 215 25 8 11

21 5

21. not defined

23. not defined

31. c

z5a

85. x 5 1, y 5 2, z 5 23, w 5 1

111. false

b.  b.

41. 3 26 45. c

12 30

1 29 4

20 d 42

49. not defined

0.70 0.89 d ,   B 5 c d 0.30 0.84

a.  46A 5 c

32.2 d ,  out of 46 million people, 13.8

b.  46B 5 c

40.94 d ,   out of 46 million people, 38.64

  3 2.2 million said that they had tried to quit smoking, while 13.8 million said that they had not.



  4 0.94 million believed that smoking would increase the chance of getting lung cancer, and 38.64 million believed that smoking would shorten their lives.

53. A 5 c

0.693 0.597

AB 5 c

0.729 100M d, B 5 c d 0.637 110M

149.49M d 149.49 million registered voters, of those 129.77M

129.77 million actually vote.

Young_AT_6160_ANS_SE-pg1211-1260.indd 1242

09/12/16 2:29 PM

Answers to Odd-Numbered Exercises  

Section 10 . 3

55. A 5 3 0.575

0.5

1.00 4 ,

230 430 A5 D 290 330

3 19 5 5

44 46 45 47

9 20 T 19 24

460 860 2A 5 D 580 660

6 38 10 10

88 92 90 94

18 40 T, nutritional 38 48

7523 B 5 £ 2700 § AB 5 $20,875.73 15200

57.

22 5 x 10 1. c d c d 5 c d 7 22 y 24 3 5. £ 1 21

information corresponding to 2 sandwiches. 115 215 0.5A 5 D 145 165

1.5 22 4.5 9.5 23 10 T , nutritional 2.5 22.5 9.5 2.5 23.5 12

information corresponding to one-half of a sandwich. $0.228 59. AB 5 £ $0.081 § , total cost per mile to run each type of $0.015 automobile 61.

2 N 5 £1§ 0

10 XN 5 £ 16 § 20

The nutritional content of the meal is 10 g of carbohydrates, 16 g of protein, and 20 g of fat. 200 63. N 5 £ 25 § 0

9.25 XN 5 £ 13.25 § 15.75

65. Not multiplying correctly. It should be:

67. false 71. 5 c

2 # 21 d c 4 22

3 27 d 5 c 5 29

a211 1 a12a21 a21a11 1 a22a21

73. A 5 c

1 1

33 35 296 282 77. D T 31 19 146 138 5 81. £ 2 26

24 215 4

69. true

a11a12 1 a12a22 d a222 1 a21a12

1 2 d , A2 5 c 1 2

75. must have m 5 p

4 23 § 28

Young_AT_6160_ANS_SE-pg1211-1260.indd 1243

19 d 23

2 d , An 5 2n21 A, n $ 1 2

3 7. £ 0 1 9. yes

11. yes

17. no

13. no 19. c

20.1618 0.2284 23. c d 0.5043 20.1237 27. A

21

31.

1 22 x 8 dc d 5 c d 23 1 y 6

0 1 x 10 1 22 § £ y § 5 £ 4 § 2 0 z 6

1 2 £ 34 1 4

1 2 1 4 3 4

29. £

0 212 § 212

15. yes

1 213 0 21 d 21. c 20 1 2 39

25. £

does not exist.

1 2 1 2

1 2

0

212

1 2

212 1 2

0 21

5 2 232 §

1

33. x 5 2, y 5 21 37. x 5 4, y 5 23

39. x 5 0, y 5 0, z 5 1

41. A21 does not exist

43. x 5 21, y 5 1, z 5 27

45. x 5 3, y 5 5, z 5 4

47. a.  c

8 39 4 d 2117

1 0 2§ 0 212 212

35. x 5 12 , y 5 13

$50 d   b.  sweatshirt: $50, t-shirt: $20 $20 51. LEG

8 55. X 5 £ 6 10

4 10 4

53. EYE

6 21 18 1 5 § £ 21 § 5 £ 1 § 8 22 1

The combination of one serving each of foods A, B, and C will create a meal of 18 g of carbohydrates, 21 g of protein, and 22 g of fat. 0.03 57. X 5 £ 0.04 0.05

0.06 0.05 0.07

0.15 21 49.50 350 0.18 § £ 52.00 § 5 £ 400 § 0.13 58.50 100

The employee’s normal monthly usage is 350 minutes talking, 400 text messages, and 100 megabytes of data usage. 59. A is not invertible because the identity matrix was not reached. 61. false

63. x 5 9

65. A⋅A21 5 c

79. not defined

3. c

5 21 x 2 0 2 § £ y § 5 £ 17 § 1 21 z 4

49. JAW

Company 1 would charge $9.25, Company 2 would charge $13.25, and Company 3 would charge $15.75, respectively, for 200 minutes of talking and 25 text messages. The better cell phone provider for this employee would be Company 1. 3 c 1

1243



5

1 ad 2 bc

ad 2 bc 2 bc

c ad

0

a c

ac

a c 0

b 1 d d ⋅a c d ad 2 bc 2c

b d 2b 1 ad 2 bc 0 d ⋅c db 5 c d d 2c a ad 2 bc 0 ad 2 bc

ad 2 bc d ad 2 bc

67. ad 2 bc 5 0

2b db a

5 c

1 0

0 d 5I 1

09/12/16 2:29 PM

1244 

Answers to Odd-Numbered Exercises

115 26008 411 6008 57 751 429 26008

69. E

431 6008 2391 6008 28 751 145 6008

21067 6008 731 6008 85 2751 1035 6008

9. c

73. 13.7, 22.4, 9.32

71. x 5 1.8, y 5 21.6

Section 10 . 4 1. 22

5. no

103 751 22 2751 3 U 751 12 751

3. 31

5. 228

7. 20.6

13. c

7. no 1 24 3 21 11. £ 0 22 3 3 22 § 0 1 24 8

1 22 1 ` d 0 1 21 1 0

1 15. £ 0 0

3 0 ` 51 d 1 25

17. x 5 54 , y 5 78

11. x 5 5, y 5 26

21. x 5

13. x 5 22, y 5 1

15. x 5 23, y 5 24

23. x 5 1, y 5 3, z 5 25

17. x 5 22, y 5 5

19. x 5 2, y 5 2

25. x 5 237 a 2 2, y 5 27 a 1 2, z 5 a

21. D 5 0, inconsistent or dependent system

25. x 5

27. x 5 1.5, y 5 2.1

y 5 21

3 4

29. x 5 0, y 5 7

31. x 5 13 , y 5

33. 7

35. 225

37. 2180

39. 0

41. 238

43. 0

45. 95.7

47. x 5 2, y 5 3, z 5 5

49. x 5 22, y 5 3, z 5 5

51. x 5 2, y 5 23, z 5 1

55. D 5 0, inconsistent or dependent system

63. 6 units2

65. y 5 2x

67. I1 5 72 , I2 5 52 , I3 5 1

69. The second determinant should be subtracted; that is, it should be 21 `

2 `. 21

23 1

6 d should replace the column 23 corresponding to the variable that is being solved for in each case. 6 3 Precisely, Dx should be ` ` and Dy 23 21 2 6 should be ` `. 21 23 71. In Dx and Dy the column c

a2 a3

75. false

77. abc

c2 a1 ` 1 b2 ` c3 a3

c1 a1 ` 2 b3 ` c3 a2

79. 2419

c1 ` c2

5 2b1 C 1 a2 21 c3 2 2 1 a3 21 c2 2 D 1 b2 C 1 a1 21 c3 2

9 24 d 9 9

35. c

41. c

10 213 d 18 220

43. c

45. c

2 1 a3 21 c1 2 b3 2 C 1 a1 21 c2 2 2 1 a2 21 c1 2 D

3 7

5 8

4 18

13 d 11

2 d 1

0 219 27 211 28 d 39. c d 218 29 3 7 2

10 9 22 24

49. yes

61. 6 units

81. 2b1 `

33. c

20 d 2

17 28 33 0

2

215

51. c 53

10

0 212 53. c d 1 0 0 225 57. £ 1 225 3 212 10

216

55. £

1 5 1 5§ 1 10

1 2 1 6

1 d 10

7 12 214 1 212

1 212 214 § 5 212

59. x 5 5, y 5 4

61. x 5 8, y 5 12

63. x 5 1, y 5 2, z 5 3

65. 28

67. 5.4

71. x 5 6, y 5 0

73. x 5 90, y 5 155

69. x 5 3, y 5 1

77. 2abd 79. x 5 1, y 5 1, z 5 2

75. 11 81. x 5

18 d 42

47. yes

59. x 5 2, y 5 23, z 5 5

2

73. true

z5

237

29. not defined 31. c

37. c

53. D 5 0, inconsistent or dependent system 57. x 5 23, y 5 1, z 5 4

y5

273 21 ,

27. y 5 20.005x2 1 0.45x 2 3.89

23. D 5 0, inconsistent or dependent system 1 2,

0 24 0 3 8§ 1 24

19. x 5 2, y 5 1

9. 0

274 21 ,

0 1 0

215 7

,y5

225 7

,z5

19 14 83.

1 unit2

85. a. 

   y 5 20.16x 2 2 0.05x 2 4.10 b. 

5 2a2b1c3 1 a3b1c2 1 a1b2c3 2 a3b2c1 2 a1b3c2 1 a2b3c1 83. 2180

85. 21019

87. x 5 26.4, y 5 1.5, z 5 3.4

   y 5 0.16x 2 2 0.05x 2 4.10

Review Exercises 5 7 2 1. c ` d 3 24 22

Young_AT_6160_ANS_SE-pg1211-1260.indd 1244

2 3. £ 0 1

0 21 3 1 23 3 22 § 0 4 23

2238 87. £ 2113 40

206 159 230

50 135 § 0

89. x 5 2.25, y 5 24.35

91. x 5 29.5, y 5 3.4

09/12/16 2:29 PM

Answers to Odd-Numbered Exercises  

Practice Test 1. c

1 5. £ 0 0 9. c

6 9 3. £ 2 23 10 12

1 22 1 ` d 22 3 2

3 5 1 211 § 7 15 19 d 8

211 26

27. vertex: 10, 02 focus: A212 , 0B

1 5 1 3 3§ 2 9

directrix: x 5

11.

3 19 4 d 219

13. no inverse

15. x 5 23, y 5 1, z 5 7

17. 231

19. x 5 1, y 5 21, z 5 2

1 2

length of latus rectum: 2

7. x 5 213 a 1 76 , y 5 19 a 2 29 , z 5 a 1 c 195 19

1245

21. money market: $3500, conservative stock: $4500, aggressive stock: $7000 23. 1 12.5, 26.4 2

29. vertex: 10, 02 focus: 10, 42 directrix: y 5 24 length of latus rectum: 16

31. vertex: 10, 02 focus: 11, 02 directrix: x 5 21 length of latus rectum: 4

Cumulative Test 3. 1 x 1 3 2 2 1 1 y 1 1 2 2 5 25

1. x 5 3 6 2!5 5. even 7. ƒ 1 g 1 x 22 5 A 1x B 3 2 1 5

1 x3 2

9. A265 , 235 B

1 domain: 12q, 0 2 ∪ 1 0, q2

33. vertex: 123, 2 2

35. vertex: 1 3, 21 2

37. vertex: 125, 0 2

39. vertex: 10, 22

41. vertex: 1 23, 21 2

43. vertex: A 12 , 54 B



11. Q1x2 5 5x 2 4, r1x2 5 25x 1 7 13. VA: x 5 22, x 5 3  HA: y 5 0.7 15. 1 23, q 2

17. x 5 4 10 21. Maximum: z A 10 7, 7B 5

19. no solution 78 210 23. c 26 0 27. c

35 219

40 d 14

25. x 5 7

12 d , c 131 19 214 2262

3 11 ,

120 7

2 y 5 211



6 131 35 d 2262

CHAPTER 11



Section 11 . 1 1. hyperbola

3. circle

5. hyperbola

7. ellipse

9. parabola

11. circle

Section 11 . 2 1. c

3. d

2

9. x 5 12y

5. c 2

11. y 5 220x

13. 1 x 2 3 2 2 5 8 1 y 2 5 2

7. a

45. 10, 22, receiver placed 2 feet from vertex

47. opens up: y 5 18 x 2, for any x in [22.5, 2.5] opens right: x 5 18 y 2, for any y in [22.5, 2.5]

15. 1 y 2 4 2 2 5 28 1 x 2 2 2

49. x 2 5 4 1 40 2 y 5 160y

19. 1 y 1 1 2 5 4 1 1 21 x 2 2 2 5 4 1 x 2 2 2

53. 374.25 feet, x 2 5 1497y

17. 1 x 2 2 2 2 5 4 1 3 21 y 2 1 2 5 12 1 y 2 1 2 2

51. yes, opening height 18.75 feet, mast 17 feet

21. 1 y 2 2 2 2 5 8 1 x 1 1 2

55. The maximum profit of $400,000 is achieved when 3000 units are produced.

25. vertex: 10, 02 focus: 10, 22 directrix: y 5 22 length of latus rectum: 8

57. If the vertex is at the origin and the focus is at 13, 02, then the parabola must open to the right. So, the general equation is y 2 5 4px, for some p . 0.

23. 1 x 2 2 2 2 5 28 1 y 1 1 2

59. true

Young_AT_6160_ANS_SE-pg1211-1260.indd 1245

61. false

09/12/16 2:29 PM

1246 

Answers to Odd-Numbered Exercises

63. Equate d1 and d2 and simplify:  " 1 x 2 0 2 1 1 y 2 p 2 5 0 y 1 p 0 2

2

           x 2 5 4py

15. center: 10, 02 vertices: 1 62, 0 2 , 1 0, 6 !2 2

17.

x2 36

1

y2 20

5 1

19.

y2 x2 7 1 16 5 1

21.

x2 4

1

y2 16

5 1

23.

x2 9

25. c 65.

1

y2 49

51

27. b

29. center: 11, 22

vertices: 123, 22, 15, 22, 11, 02, 11, 42

6 7. vertex: 12.5, 23.52 69. vertex: 11.8, 1.52 opens right opens left

31. center: 123, 42 vertices: A22!2 2 3, 4B, A2!2 2 3, 4B, A23, 4 1 4!5B, A23, 4 2 4!5B

Section 11.3 1. d

3. a

5. center: 10, 02 vertices: 1 65, 0 2 , 1 0, 64 2

33. center: 10, 32

vertices: 122, 32, 12, 32, 10, 32, 10, 42

7. center: 10, 02 vertices: 1 64, 0 2 , 1 0, 68 2 35. center: 11, 12 9. center: 10, 02 vertices: 1 610, 0 2 , 1 0, 61 2

11. center: 10, 02 vertices:

A632 ,

0B, A0,

619 B

13. center: 10, 02 vertices: 1 62, 0 2 , 1 0, 64 2

vertices: A162!2, 1 2 , 11, 32, 11, 212

37. center: 122, 232

vertices: A226 !10, 23 2 , A22, 2365!2B

39. 43.

Young_AT_6160_ANS_SE-pg1211-1260.indd 1246

1x 2 222 25 1x 2 3 4

22

1 1

1y 2 522 9 1y 2 222 16

5 1 5 1

41. 45.

1x 2 422 7 1x 1 1 9

22

1 1

1y 1 422 16 1y 1 422 25

51 51

09/12/16 2:29 PM

Answers to Odd-Numbered Exercises  

47.

x2 225

1

y2 5625

2

13.

51

1247

15.

y2

x 1 400 5 1  b.  The width of the track at the end of the 49. a.  5625 field is 24 yards, so it will NOT encompass the football field, which is 30 yards wide.

51.

x2 5,914,000,0002

53.

x2 150,000,0002

1

1

y2 5,729,000,0002

y2 146,000,0002

51 2

y2

x 2 20 5 1 19. 1 7. 16

51

55. straight line y2 x2 5 1  b.  42 inches   c.  1509 steps 57. a.  1 64 25

y2 9 2

y 4

21. x 2 2 y 2 5 a2

23.

25. c

27. b

29.

31.

33.

35.

2

x2 7

51

2 x 2 5 b2

59. It should be a2 5 6, b2 5 4, so that a 5 6 !6, b 5 62. 61. false

63. true

65. Pluto: e > 0.25  Earth: e > 0.02 67. As c increases, ellipse becomes more elongated.

69. As c increases, circle gets smaller.

71. As c decreases, the major axis becomes longer.

37.

3 9.

1x 2 222 16

2

1y 2 522 9

41.

5 1

1y 1 422 9

2

1x 2 422 7

51

43. Ship will come ashore between the two stations: 28.5 miles from one and 121.5 miles from the other. 45. 0.000484 seconds 47. y 2 2 45 x 2 5 1 49. 275 feet

Section 11.4 1. b 5.

3. d 7.

51. The transverse axis should be vertical. The points are 13, 02, 123, 02, and the vertices are 10, 22, 10, 222. 53. false 2

55. true

2

2

2

2

57. x 2 y 5 a or y 2 x 5 a2 59. As c increases, the graphs become more squeezed down toward the x-axis.

9.

Young_AT_6160_ANS_SE-pg1211-1260.indd 1247

11.

09/12/16 2:29 PM

1248 

Answers to Odd-Numbered Exercises

61. As c decreases, the vertices are located at A61c , 0B and are moving away from the origin.

y 5 x2 1 1 . Any system in which the linear equation y51 is the tangent line to the parabola at its vertex will have only one solution. 57. Consider e

59. no solution

61. 121.57, 21.642



Section 11.5 1. 12, 62, 121, 32

3. 11, 02

7. 10, 12

5. no solution

9. 1 0.63, 21.61 2 , 120.63, 21.61 2

11. no solution 13. 11, 12

63. 11.067, 4.1192, 11.986, 0.6382, 121.017, 24.7572

15. A2!2, !2B, A22!2, 2 !2B, A !2, 2!2B, A2 !2, 22!2B

17. 126, 332, 12, 12

19. 13, 42, 122, 212

Section 11.6

23. 121, 21 2 , A 14 , 32 B

13.

15.

17.

19.

21.

23.

21. 1 0, 23 2 , A 25 , 211 5B

1. b

25. 121, 24 2 , 1 4, 1 2 27. 11, 32, 121, 23 2 29. 12, 42

33. no solution

3. j

5. h 7. c

9. d

11. k

31. A 12 , 13 B, A 21 ,213 B

35.

37.

(1,2)

( 1, 1)

2 5. 27.

39. 3 and 7 41. 8 and 9, 28 and 29 43. 8 cm 3 10 cm 45. 400 ft 3 500 ft or 1000 3 ft 3 600 ft 47. professor: 2

m

2 9. 31.

m

/sec, Kirani James: 10 /sec

2

49. In general, y 2 y 2 0. Must solve this system using substitution. 51. false

53. false

55. 2n

Young_AT_6160_ANS_SE-pg1211-1260.indd 1248

09/12/16 2:30 PM

Answers to Odd-Numbered Exercises  

33.

35.

3 7. 39.

63.

1249

65.

Section 11.7 3!3 3 1. A3!2, !2B 3. 1 1, 1 !3b a2 2 2 1 3!3 !3 3 3!3 3 5. a2 2 , 2 b 7. a , b 2 2 2 2 2 2

4 1. 43.

9. hyperbola;

X2 Y2 2 51 2 2

4 5. 47. 11. parabola; 2X2 2 2Y 2 1 5 0

4 9.

13. hyperbola;

X2 Y2 2 51 6 2

51. 92 p units2 53. There is no common region here—it is empty, as is seen in the graph below:

X2 Y2 15. ellipse; 1 51 2 1 5 9. 61.

55. false

Young_AT_6160_ANS_SE-pg1211-1260.indd 1249

57. 0 # a # b

09/12/16 2:30 PM

1250 

Answers to Odd-Numbered Exercises

17. parabola; 2X2 2 2Y 2 1 5 0

19. ellipse;

X2 Y2 2 5 1; 9 1 rotation of 45° 41.

X2 Y2 1 51 1 9 43. X2 2 Y 2 3 5 0; rotation of 30°

21. hyperbola;

X2 Y2 2 51 3 2 X2 Y2 1 5 1; 25 4 rotation of 30°

45.

23. parabola; Y2 2 X 2 4 5 0 47. X2 1 4X 2 Y 5 0; rotation of 45°

25. 45°

27. 60°

29. 30°

31. 45°

33. 15°

35. < 40.3°

37. < 50.7° X2 Y2 39. 1 5 1; 4 9 rotation of 60°

49. true

51. true

y2 x2 1 2 5 1. 2 b a b.  For 180°, the original equation results. 53. a. For 90°, the new equation is

55. a , 0 hyperbola; a 5 0 parabola; a . 0, a 2 1 ellipse; a 5 1 circle. 57. Amount of rotation is 30°. a. 

Young_AT_6160_ANS_SE-pg1211-1260.indd 1250

09/12/16 2:30 PM

Answers to Odd-Numbered Exercises  

b. 

1251

27. parabola; e 5 1

59. The amount of rotation is about 36°. a. 

29. hyperbola; e 5 2 b. 

61. a. The amount of rotation is about 63°. It is a parabola. 31. ellipse; e 5

1 2



b.  The amount of rotation is about 19°. It is a hyperbola.

c.  The amount of rotation is about 26.5°.

33. parabola; e 5 1

Section 11.8 1. r 5

5 8 3. r5 2 2 sin u 1 1 2 sin u

5. r 5

1 1 1 cos u

7. r 5

9. r 5

12 3 2 4 cos u

11. r 5

13. r 5

18 15. parabola 5 1 3 sin u

17. ellipse

6 4 1 3 cos u

35. ellipse; e 5

1 3

3 1 2 sin u

19. hyperbola

21. ellipse 23. parabola 25. hyperbola

Young_AT_6160_ANS_SE-pg1211-1260.indd 1251

09/12/16 2:30 PM

1252 

Answers to Odd-Numbered Exercises

37. hyperbola; e 5

5.

3 2

7. 39. parabola; e 5 1

9.

41. r 5 43. r 5

57,934,508 1 1 2 0.2062 2 1 2 0.206 cos u 75,000,000 1 1 2 0.2232 2 1 2 0.223 cos u

45. Conic becomes more elliptic as e S 1 and more circular as e S 0. 46.–  48.  See Instructor’s Solution Manual.

49.

2ep 1 2 e2



11.

ep ep 2 12e 11e 51. ± ,p ≤ 2

52.–  60.  See Instructor’s Solution Manual.

61. a. With u step 5 p/3, plot points 12, 02, 114.93, p/32,

114.93, 2p/32, 12, p2, 11.07, 4p/32, 11.07, 5p/32, and 12, 2p2.

b. With u step 5 0.8p, plot points 12, 02, 14.85, 0.8p2, 11.03, 1.6p2, 140.86, 2.4p2, 11.26, 3.2p2, and 12, 4p2.

13.

Section 11.9 1.

15.

3.

Young_AT_6160_ANS_SE-pg1211-1260.indd 1252

09/12/16 2:30 PM

Answers to Odd-Numbered Exercises  

17.

29.

19.

31. y 5

1253

1 33. y5x22 x2

x1y52 35. y 5 "x 2 1 1 37.

39. x 1 4y 5 8 41. 17.7 sec 43. yes 45. Height: 5742 ft; Horizontal distance: 13,261 ft 47. 125 sec 21.

49. 51.  t

x

y

0

A1B

0

p 2

0

A1B

p

2A 2B

0

3p 2

0

2A 2B

2p

A1B

0

23. Arrow in different directions, depending on t

53. The original domain must be t $ 0; therefore, only the part of the parabola where y $ 0 is part of the plane curve. 55. false

57. Quarter circle in QI

58.–  60.  See Instructor’s Solution Manual. 61. 65.

x1y y2x 1 5 1 2a 2b

a 63. y 5 b !x

25.

67.

27. 69. a 5 2, 0 # t , 2p a 5 3, 0 # t , 2p

Young_AT_6160_ANS_SE-pg1211-1260.indd 1253

09/12/16 2:30 PM

1254 

Answers to Odd-Numbered Exercises

Review Exercises

33. 35.

1. false

3. true

2

7. y 2 5 220x

5. y 5 12x 9. 1 x 2 2 2 2 5 8 1 y 2 3 2

11. 1 x 2 1 2 2 5 4 1 21 21 y 2 6 2 5 24 1 y 2 6 2 13. vertex: 10, 02 focus: 10, 232 directrix: y 5 3 latus rectum: 12

37.

1x 2 322 25

39.

x2 778,300,0002

1

1y 2 322 9

1

51

y2 777,400,0002

41.

15. vertex: 10, 02 focus: A 14 , 0B directrix: x 5 214 latus rectum: 1

17. vertex: 12, 222 focus: 13, 222 directrix: x 5 1 latus rectum: 4

2

51 43.

y2

47. 4 5. x9 2 16 5 1 49.

19. vertex: 123, 12 focus: 123, 212 directrix: y 5 3 latus rectum: 8

y2 9

2 x2 5 1

51.

1x 2 422

1y 2 322

2 9 51 5 3. 16 55. Ship will come ashore between the two stations 65.36 miles from one and 154.64 miles from the other. 57. 122, 272, 11, 242 61. no solution

21. vertex: A252 , 275 8B

67. A 12 ,

focus: A252 , 279 8B

25 8

x2 25

1 B, A 12 , !7

1 1 2 !7 B, A212 , 2 !7 B

65. 12, 32, 123, 222

5 3.125 feet from the center

25. 27.

29.

A212 ,

63. no solution

69. 71.

directrix: y 5 271 8 latus rectum: 2

23.

1 B, !7

59. 11, 22, 121, 22

1

y2 16

5 1

31.

Young_AT_6160_ANS_SE-pg1211-1260.indd 1254

x2 9

73.

1

y2 64

75.

51

09/12/16 2:30 PM

Answers to Odd-Numbered Exercises  

77. 79.

97.

3 23!3 1 1b 8 1. a2 1 !3, 2 2

99. x 5 4 2 y2

X2 Y2 83. 2 51 4 4

1255

101. y 5 2x 1 4

103. The vertex is located at 10.6, 21.22. The parabola opens down.

105. a. y 5 21.46 !8.81 2 3x

85. 60° 87. X 2 1 4 5 Y

b. Vertex at 12.94, 21.42, opens to the left. c. Yes, (a) and (b) agree with each other.

107. As c increases, the minor axis along the x-axis decreases.

89. r 5

21 7 2 3 sin u

91. hyperbola

93. e 5 12 ; vertices A 43 , 0B, 1 4, p 2 or in rectangular form A 43 , 0B, 1 24, 0 2

109. As c increases, the vertices of the hyperbolas located at a6

1 , 0b are moving toward the origin. 2c

95. 111. 10.635, 2.4802, 120.635, 2.4802, 121.245, 0.6452, 11.245, 0.6452

Young_AT_6160_ANS_SE-pg1211-1260.indd 1255

09/12/16 2:30 PM

1256 

Answers to Odd-Numbered Exercises

113.

19.

115. a. The amount of rotation is about 22.5°. It is an ellipse. 21.

b.  The amount of rotation is about 11.6°. It is a hyperbola.

23. x 2 5 6y; 2 2 # x # 2 25.

117. With u step 5 a1.06,

p , points 12, 02, 4

p p 3p b, a0.89, b, a1.06, b, 1 2, p 2 , 4 2 4

a17.22,

5p 3p 7p b, a28, b, a17.22, b, 4 2 4

27.

and 12, 2p2 are plotted.

119. a 5 2, b 5 3, t in 30, 2p4

29. Ellipse: e 5 a 5 3, b 5 2, t in 30, 2p4

31. 5.3 sec, 450 ft 33.

Practice Test

3 5. a. 

1. c

3. d

5. f

7. y 2 5 216x

9. 1 x 1 1 2 2 5 212 1 y 2 5 2 1

y2 16

15. x 2

y2 4

11.

x2 7

2

5 1 5 1

2 3

13.

1x 2 222 20

17.

y2 16

Young_AT_6160_ANS_SE-pg1211-1260.indd 1256

2

y2 36

51

1x 2 222 20

51

1

b. The vertex is located at 122.1, 1.22. The parabola opens upward. c. Yes

09/12/16 2:30 PM

Answers to Odd-Numbered Exercises   6 1 21 2 n 69. a n n50 2

Cumulative Test 1. 26, 3

3. 27

5. ƒ 1 x 2 5

1 3 1x

9. 21.9626 13.

2 7 2 1 7 2

p 2p 4p 5p , , , 3 3 3 3

71. a 1 21 2 n21n q

n51

5 1n 1 12! 73. a 5 a n1n 1 12 n51 1 n 2 1 2 ! n51 6

7. 24, 4 11.

1257

1 1 sin u cos u

q q x n21 xn 75. a 1 21 2 n21 5 a 1 21 2 n 1n 2 12! n! n51 n50

15. 825.1, 14.19

77. $28,640.89; total balance in account after 6 years (or 72 months)

17. Soda: $1.29; Soft pretzel: $1.45

79. sn 5 20 1 2n; a paralegal with 20 years experience would make $60 per hour

19.

81. an 5 1.03an21; a0 5 30,000 83. an11 5 1000 2 75n; approximately 10.7 years 85. A1 5 100, A2 5 200.10, A3 5 300.30, A4 5 400.60, A36 5 3663.72 87. 7; 7.38906

89. 0.095310; 0.095310

91. The mistake is that 6! 2 3!2!, but rather 6! 5 6⋅5⋅4⋅3⋅2⋅1. 7 21. c 18 23.

216 23

1, n 5 1, 3, 5, . . . 21, n 5 2, 4, 6, . . . So, the terms should all be the opposite sign.

33 d 217

93. 1 21 2 n11 5 e

1y 1 222 1x 2 622 1 51 9 25

95.  true

97.  false

99.  C, C 1 D, C 1 2D, C 1 3D

25. 12, 42 14, 22

101. 1 and 1

27.

103. < 2.705; < 2.717; < 2.718; 105. 109 15

Section 12.2 1. arithmetic, d 5 3

3. not arithmetic

5. arithmetic, d 5 20.03 7. arithmetic, d 5

CHAPTER 12

9. not arithmetic

Section 12 . 1

11. 3, 1, 21, 23; arithmetic; d 5 22 5. 12 , 23 , 34 , 45

1. 1, 2, 3, 4 3. 1, 3, 5, 7 7. 2, 2, 43 , 23 9. 2x 2, x 3, 2x 4, x 5 1 1 1 1 11. 216 , 12 , 220 , 30 13. 512

17.

10201 10000

23. an 5

29. 72

31. 812

35. 83, 156, 160

37.

39. 1 2n 1 3 21 2n 1 2 2 41. 7, 10, 13, 16 43. 1, 2, 6, 24

45. 100, 50,

17. 0, 10, 20, 30; arithmetic; d 5 10 19. 21, 2, 23, 4; not arithmetic 21. an 5 11 1 1 n 2 1 2 5 5 5n 1 6

1 21 2 n2n 1 25. an 5 1n 1 12n 3n

1 5852

13. 1, 4, 9, 16; not arithmetic 15. 2, 7, 12, 17; arithmetic; d 5 5

1 15. 2420

19. 23 21. an 5 2n

5 1.0201

27. an 5 1 21 2 n11 33.

2 3

1 n1n 1 12

25 25 3 , 72

23. an 5 24 1 1 n 2 1 21 2 2 5 26 1 2n 25. an 5 0 1 1 n 2 1 2 23 5 23 n 2

2 3

27. an 5 0 1 1 n 2 1 2 e 5 en 2 e 29. 124

31. 2684

33.

16 3

35. a5 5 44, a17 5 152; an 5 8 1 1 n 2 1 2 9 5 9n 2 1

47. 1, 2, 2, 4 49. 1, 21, 22, 5

37. a7 5 21, a17 5 241; an 5 23 1 1 n 2 1 21 24 2 5 24n 1 27

51. 10

41. 552

53. 30 2

3

2

3

55. 36

59. 1 2 x 1 x 2 x 63. 1 1 x 1 65.

61.

109 15

4

x x x 1 1 2 6 24

20 9

Young_AT_6160_ANS_SE-pg1211-1260.indd 1257

57. 5

39. a4 5 3, a22 5 15; an 5 1 1 1 n 2 1 2 23 5 23 n 1 49. 3875 53. 630

67. not possible

43. 2780

51. 55. S43 5

45. 51

21 1 2 C6

43 4 15

2

13 2D

1 3

47. 416

5

21 1 2 39 2A 6 B

5 2133 2

2 43 2 5 2817 2 57. 1368

59. Colin: $347,500; Camden: $340,000

61. 850 seats

09/12/16 2:30 PM

1258 

Answers to Odd-Numbered Exercises

63. 1101 glasses on the bottom row, each row had 20 fewer glasses than the one before 65. 1600 feet

89.

67. 210 oranges

1 1  for 0 x 0 , 1 2 2x 2

69. a. 23 seats in the first row  b.  1125 seats 71. an 5 a1 1 1 n 2 1 2 d, not a1 1 nd



73. There are 11 terms, not 10. So, n 5 11, and thus, 11 S11 5 1 1 1 21 2 5 121. 2 75. false

Section 12.4

77. true

                1.–24.   See Instructor’s Solution Manual.

1 n 1 1 21 2a 1 nb 2 79. 81. 27,420 2 83. 5050 85. 2500 87. 18,850

25. 7 steps

    29.–30.   See Instructor’s Solution Manual. 31. false

Section 12.3 3. no 5. yes, r 5 12

1. yes, r 5 3

7. yes, r 5 1.7 9. 6, 18, 54, 162, 486

    33.–37.  See Instructor’s Solution Manual. 39.

255 256 ,

yes

11. 1, 24, 16, 264, 256

Section 12.5

13. 10,000, 10,600, 11,236, 11,910.16, 12,624.77

1. 35

1 1 15. 23 , 13 , 16 , 12 , 24 17. an 5 5 1 2 2 n21 16 1 n21 3 A24 B

25. a7 5 2128

29. a15 5 6.10 3 10216 33. 59,048

31.

39. 16,383 41. 2

27. a13 5

4096 3

13. y 5 2 15y 4 1 90y 3 2 270y 2 1 405y 2 243 15. x 5 1 5x 4y 1 10x 3y 2 1 10x 2y 3 1 5xy 4 1 y 5

37. 6560

17. x 3 1 9x 2y 1 27xy 2 1 27y 3

43. 214

19. 125x 3 2 150x 2 1 60x 2 8

45. not possible, diverges

21.

47. 227 10,526 51. 2 49. 53. 100

7. 1

11. x 4 1 8x 3 1 24x 2 1 32x 1 16

8191 3

35. 2.2

3. 45 5. 1

9. 17,296

19. an 5 1 1 23 2 n21 21. an 5 1000 1 1.07 2 n21 23. an 5

27. 31 steps

2 3

55. $44,610.95

20y 150y 2 500y 3 1 1 3 1 1 1 625y 4 4 2 x x x x

23. x 8 1 4x 6y 2 1 6x 4y 4 1 4x 2y 6 1 y 8 25. a5x 5 1 5a4bx 4y 1 10a3b2x 3y 2 1 10a2b2x 2y 3 1 5ab4xy 4 1 b5y 5

57. an 5 2000 1 0.5 2 ; a4 5 125, a7 5 16

27. x 3 1 12x 5@2 1 60x 2 1 160x 3@2 1 240x 1 192x 1@2 1 64

59. 17 feet

61. 58,640 students

29. a3 1 4a9@4b1@4 1 6a3@2b1@2 1 4a3@4b3@4 1 b

63. 67 days; $9618

65. $3877.64

31. x 1 8x 3@4y 1@2 1 24x 1@2y 1 32x 1@4y 3@2 1 16y 2

n

33. r 4 2 4r 3s 1 6r 2s2 2 4rs3 1 s4

67. 26 weeks: $13,196.88 52 weeks: $26,811.75

1 2

35. a6x 6 1 6a5bx 5y 1 15a4b2x 4y 2 1 20a3b3x 3y 3

69. $367,987 71. 51 1 2 12 1 73. should be r 5 23

1 15a2b4x 2y 4 1 6ab5xy 5 1 b6y 6 37. 3360

39. 5670

41. 22,680

75. should use r 5 23 all the way through the calculation; also, a1 2 12 (not 4)

43. 70

45. 3,838,380

47. 2,598,960

77. false 83.

47 99

1 87. a x 5 12x n50 q

n

a 79. true 81. 0 b 0 , 1, 12b 85. 237, 529, 996, 894, 754

for 0 x 0 # 1



Young_AT_6160_ANS_SE-pg1211-1260.indd 1258

7 7! 7! 7⋅6⋅5! 5 5 21 49. a b 2 , but rather 5 5! 5! 2! 5! 1 2⋅1 2 51. false

53. true

 55.–56.   See Instructor’s Solution Manual. 57. 1 2 3x 1 3x 2 2 x 3



09/12/16 2:30 PM

Answers to Odd-Numbered Exercises  

59. Graphs of the respective functions get closer to the graph of y4 5 1 1 2 x 2 3 when 1 , x , 2; when x . 1, no longer true.

1259

49. 2,598,960 51.

1287 2,598,960

53. The events aren’t mutually exculsive. So, probability 5

1

13 52

1 2   52     5 f

4 52

16 52

5

4 13 .

2 of spades

55. true

57. false



59. 0.0027

61.

61. graph of the curve better approximation to the graph of y 5 A1 1 1x B 3, for 1 , x , 2; no, does not get closer to this graph if 0 , x , 1

    63.–64.  Use random number generator and compare.

Section 12.6 1. 360

3. 15,120

7. 1716

9. 252

13. 1 19. 24

11. 15,890,700 17. 215,553,195

21. 12

23. 10,000

29. 100,000; 81,000

31. 2.65 3 1032

33. 59,280

35. 997,002,000

37. 22,957,480

39. 2,598,960

41. 1326 43. 4.9 3 1014 47. 15,625

61. answers the same 63. a. 5,040

1 n11

b. 5,040

21. 15

23. 69

q xn 25. a n21 27. a n51 2 n50 n! 7

1 21 2 n

c. yes

1 2

39. an 5 1 1 1 n 2 1 2 A223 B 5 223 n 1

5 3

41. a1 5 5, d 5 2, an 5 5 1 1 n 2 1 21 2 2 5 2n 1 3

43. a1 5 10, d 5 6, an 5 10 1 1 n 2 1 2 6 5 6n 1 4 45.  630

d. nPr 5 r!nCr

47. 420

49. Bob: $885,000 Tania: $990,000 51. geometric, r 5 22 53. geometric, r 5

Section 12.7 3. 5 BBBB, BBBG, BBGB, BBGG, BGBB, BGBG, BGGB, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG 6 5. 5 RR, RB, RW, BB, BR, BW, WR, WB 6 9. 78

7. 18

19.

1 2

55. 3, 6, 12, 24, 48

1. 5 2, 3, . . . , 12 6

4 13

27.    a.  270,725   b. 

1 32

15.

37. an 5 24 1 1 n 2 1 21 5 2 5 5n 2 9

n! 59. C 1 n, r 2 ⋅r! 5 1n 2 r2!

33.

9. an 5 1 21 2 n113n 11. an 5 1 21 2 n

35. arithmetic, d 5 1

49. 400

55. false

2 13

1 7. a15 5 23600

< 0.13

33. arithmetic, d 5

45. 256

r11 nCr 5 57. C n 2r n r11

5

32 243

31. arithmetic, d 5 22

53. true

8 52

5. a5 5

29. $36,639.90; amount in the account after 5 years

51. The combination formula nCr should be used instead.

29.

Review Exercises

17. 5, 3, 1, 21 19. 1, 2, 4, 32

27. Each of 20 questions has four answer choices. So, there are 420 < 1.1 3 1012 possible ways to answer the questions on the exam.

17.

65. 0.2907

13. 56

25. 32,760

10 13

< 0.333

1. 1, 8, 27, 64 3. 5, 8, 11, 14

5. 40,320

15. 27,405

1 3

< 15.4%

< 3.1%

4 < 5.03% 37. A 18 38 B

41. 25%

11.

1 18

21.

3 4

715 270,725

13. 12 23.

3 4

> 0.26%   c. 

31. 35.

4 663 31 32

15.

5 12

25. 0 13 270,725

< 0.005%

< 0.6%

< 96.9%

48 1326

< 3.6%

45. 0.001526% 47. a.  {(Brown, Blue), (Brown, Brown), (Blue, Brown), (Blue, Blue)} b. 1@4 c. 3@4

Young_AT_6160_ANS_SE-pg1211-1260.indd 1259

59. an 5 a1r n21 5 7⋅2n21

61. an 5 1 22 2 n21

63. a25 5 33,554,432

65. a12 5 22.048 3 1026

67. 4920.50

69. 16,400

71. 3 73. $60,875.61     75.–78.  See Instructor’s Solution Manual. 79. 165 4

81. 1 3

2

83. x 2 20x 1 150x 2 500x 1 625 85. 8x 3 2 60x 2 1 150x 2 125

39. 20% 43.

57. 100, 2400, 1600, 26400, 25,600

87. x 5@2 1 5x 2 1 10x 3@2 1 10x 1 5x 1@2 1 1 89. r 5 2 5r 4s 1 10r 3s2 2 10r 2s3 1 5rs4 2 s5 91. 112 97. 840

93. 37,500

95. 22,957,480

99. 95,040

09/12/16 2:30 PM

1260 

Answers to Odd-Numbered Exercises

101. 792 105. 30

103. 1

11. 2520

119.

2 3

113. 20,358,520

123.

2 13

127.

5369 3600

131.

1 1 1 2x

7 12

< 58.3%

125.

31 32

< 96.88%

129.

34,875 14

< 66.7% 121. < 15.4%

4

5 x2

15. 2184 1

1 x5

19. more permutations than combinations since order is taken into account when determining the number of permutations

30 < 6.25% 117. 36 < 83.3%

115.

1 16

7

17. x 1 5x 1 10x 1 10x 1

107. 5040

109. 120 seating arrangements, 15 years 111. 94,109,400

13. 455

10

21.

18 38

18 < 0.47 23. 38 > 0.47

25.

4 13

< 0.308

27. 184,756

Cumulative Test

5 6 3!7i 4 5. x 5 28 7. 2x 1 h 2 3 1. 2

3.

9. 115, 62 11. VA: x 5 3 HA: y 5 25 13. 2.585

15. no solution

17. z 1 1, 4 2 5 24 133. 99,900, yes 135. Graphs become better approximations of the graph of y 5 1 1 1 2x 2 4 for 20.1 , x , 0.1; does not get closer for 0.1 , x , 1. nPr 137.  a. 11,440 b. 11,440 c. yes d. 5 nCr r! 139. 0.0722

Young_AT_6160_ANS_SE-pg1211-1260.indd 1260

19. c

99 29

18 4

23. 3

Practice Test

1 2 xn 1. x n21 3. Sn 5 12x 7. 1 2 A 14 B 10 < 1 9. 24,950



5. 0 x 0 , 1

29 216 1 x 2 3 2 5 1 y 2 5 2 2 d 21. 7 25. 210 5 1024

27. 23432

09/12/16 2:30 PM

[Applications Index[ Agriculture pasture areas, 345, 924 Animals bird wings, 853 cat food amounts, 412 deer population, 751 dog runs, 295, 1048 dog training, 136 gray wolf population, 472 horse paddock, 1048 Archaeology dating of fossils, 453 Art/Music/Theater/Entertainment American Idol, 1149 candy costs, 104 chord frequencies, 105 Dancing with the Stars, 1149 frequencies of sound, 448–450 murals, 795 musical tones and frequencies, 711–712, 714, 747–748 New York Philharmonic sound levels, 472 phonographs, 582 The Price is Right, 583 rock concert sound levels, 447, 473 Survivor, 1149 television sizes, 118 theater seating, 1115, 1116 viewing paintings in a museum, 734 Automotive antifreeze concentration, 104 braking, 819 car depreciation, 1125 car options, 1164 car repair costs, 233 car value, 148, 158, 200, 482, 489 driving under a bridge, 1013, 1089 gas and electricity costs, 959 gas mileage, 346, 879, 922 gas/oil mixtures, 879 gasoline mixtures, 99–100 gasoline prices, 99, 104, 171–172, 248, 962 hybrid cars, 962 license plates, 1149 NASCAR race, 530 NASCAR revenue, 171 new car models, 472–473, 482 production levels, 890 rental car costs, 91, 135, 199 service costs, 89 speed of a car, 576 tire size and speedometer readings, 578, 581, 582 towing power, 819

traffic, 534 trip speeds, 370 van depreciation, 472 van value, 472 windshield wipers, 581 Aviation airplane slides, 788 bearing, 534, 787 distance between planes, 787, 1095 elevation of a rocket, 537 glide path, 533–534 heading and airspeed, 810 length of an aircraft wing, 787 locating airplanes, 532 midair refueling, 533 NASA astronaut weights, 136 NASA centrifuge, 582 NASA “vomit comet,” 346 sonic booms, 620 wind speed, 101, 105, 879–880 Biology/Life Sciences Archimedes spiral, 853 average height, 200 bacterial growth, 481, 489, 490, 1107 dating of fossils, 424, 453 DNA structure, 582 eye color, 1156, 1158 human anatomy, 318, 1166 human body temperature, 155, 310, 752 human body weight, 136, 200, 345, 361 sex of offspring, 1152, 1158, 1164 volume of air in the lungs, 751 yo-yo dieting, 591 Budgeting business, 103 car repair costs, 233 cost of flower seed, 795 costs, 270 event planning, 103, 252, 270, 879, 909 long-distance call costs, 252, 310 monthly driving costs, 199 personal, 104 race entry fee, 270, 310 salaries, 1116, 1123 service charges, 196–197 Business break-even point, 45, 118, 386, 751–752, 880 budgeting, 103 cash flow of a stock fund, 695 cell phone plan, 318 cell phone provision, 962, 974 computer assembly, 919 computer sales, 695

costs, 89, 91, 104, 127, 135, 137, 200–201, 270, 295–296, 341, 345, 714, 879 deduction for business expenses, 961 defective products, 1158 demand, 619 job applicants, 961 job offers, 876 leadership positions, 1164 losses, 550 markups, 103, 137 NASCAR revenue, 171 prices, 103, 104, 118, 148, 171, 270, 295, 318, 328, 889 production levels, 890, 912, 919, 923, 948 profits, 34, 45, 118, 136, 148, 160, 184, 270, 285, 345, 346, 386, 394–395, 413, 418, 422, 482, 550, 1013 real estate, 148, 437, 1107, 1148 revenues, 136, 155, 171, 347, 361, 714, 795–796, 819 salaries, 285, 310, 325, 326, 437, 879, 1106–1107, 1115, 1116, 1123, 1125, 1126, 1162, 1163 sales, 270, 318, 481–482, 522, 591, 663, 734, 751, 948, 1104, 1107 stock values, 115–116, 118, 127, 146, 148, 361, 488, 887 supply and demand, 127 time to pay off debt, 483 tipping, 961–962 typing speed, 412 van depreciation, 472 wages, 318, 327 working together, 105 Chemistry acidity/basicity of a solution, 452–453 enzyme reactions, 455 mixtures, 99–100, 104, 158, 879, 881, 922, 948 pH of a solution, 452–453, 488 structure of molecules, 535 temperature and pressure, 319 water molecules, 565 Communications cell phone costs, 91, 137, 318, 962, 974 cell phone coverage, 207 cell phone plans, 481, 879 cell phone provision by business, 962, 974 cell phone purchases, 482 cell phone towers, 171, 207 cell phone triangulation, 788 cell phone use, 25, 347 communication with astronauts, 531 laser communications, 25, 104, 137, 735

1261

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1262 

Applications Index

Communications (continued ) long-distance call cost, 270 monthly phone costs, 200 parabolic antennas, 1010–1011 postage rates, 270–271 satellite dish receivers, 1010, 1013, 1089 telephone infrastructure, 1132 touch-tone dialing, 714 10-digit dialing, 1142 Computers building a system, 1148 computer business, 919 depreciation, 437, 1125 e-mail passwords, 1164 human brain vs., 890 Internet costs, 91, 137, 328 passwords, 1148 production levels, 912 Scrabble game, 890 website hits, 1126 Construction and Home Improvement amount of fencing to buy, 1045–1046, 1048 area of a floor, 525 bathroom remodeling, 199 budgeting, 104 building a treehouse, 536 dog runs, 207 gardening, 119, 413 garden maze, 853 great room dimensions, 796 housecleaning, 102, 119 house painting, 105 hurricane preparations, 918–919 ladder size, 104 landscaping, 104, 105 pitch of a roof, 500 rain gutters, 751 sprinkler system, 270, 328 sprinkling grass, 184 window height bathroom remodeling, 788 Consumer Situations baseball card values, 252 cell phone costs, 91, 137, 318, 962, 974 cell phone plans, 481, 879 cell phone purchases, 482 cell phone use, 25, 347 comparative shopping, 135 depreciation of goods, 437 discount prices, 103, 236 food costs, 103, 104 revolving restaurant, 507 sales tax, 318, 327 search and rescue, 534 television sizes, 118 wedding invitations, 1148 Demographics (US/World) adult smokers in the United States, 961 average marriage age, 196, 951 cell phones in United States, 25, 347 HIV/AIDS infection rate, 481 population growth, 433, 437, 475, 481, 488, 489, 490

Young_AT_6160_SE_Index_pp1261-1276.indd 1262

registered voters, 961 time to register for a class, 819 underage smoking, 347 U.S. population, 15 women in science, 961 Design champagne fountain, 1115 field of tulips, 1115 home décor, 789 new car models, 472–473, 482 nuclear cooling towers, 1037 suspension bridge, 1013 university campus, 207 Earth Sciences earthquakes, 448, 452, 463, 472, 473, 490, 735 Economics cash flow of a stock fund, 695 exchange rates, 252 gasoline prices, 171–172 house prices, 137, 296, 481 market price, 295 oil consumption, 490 price increases, 148 prices, 118 profit function, 346–347 stock prices, 115–116, 118, 127, 146, 148, 361, 488, 887 supply and demand, 127, 184, 271, 318, 438, 913 U.S. national debt, 15 Education/Learning band uniforms, 270 bookstore markups, 103 class seating, 1149 fraternity elections, 1149 grades, 89, 105, 127, 137, 158, 160, 233, 483 leadership, 1148 memorization by professor, 412 multiple-choice tests, 1149 number of students on college campus, 478 paper margins, 118 research grants, 104 salaries, 879 sleep hours, 103 sorority financials, 119 sorority T-shirt costs, 270 sound level in college stadium, 452 sound level of teacher speaking, 488 travel time, 105 tutoring costs, 325 university population growth, 1126 Electricity/Electronics/Optics alternating current, 734 band-pass filters, 155 cardioid microphones, 853 cell phone towers, 171, 207 circuit theory, 988 dish TV, 534 electrical circuits, 68 electrical fields, 35, 45

electromagnetic spectrum, 453 electromagnetic wave propagation, 687 eyeglass lens, 1013, 1093 focal length of a lens, 58, 91, 105–106 gas and electricity costs, 959 laser beams, 25, 184 laser communication, 25, 104, 137 lens law, 891 light cast by a lamp, 1037 optical signals, 714 radio waves, 184 refraction of light, 522, 747, 751 resistance in circuits, 58 speed of a lighthouse beacon, 576–577 speed of an electric saw, 582 square waves, 271 television sizes, 118 Engineering Archimedes spiral, 853 bridge construction, 778 cell phone towers, 207 Christmas lights, 508 electrical fields, 35, 45 gears, 581 height of a projectile, 45 nuclear cooling towers, 1037 solar cooker, 1013 solar furnace, 1013 Environment carrying capacity, 482–483 dry erase markers, 879 envelope waste, 253 fires, 207 forestry, 550 gas mileage, 346 gasoline and electricity consumption, 962 global climate change, 253 height of a mountain, 639 hours of daylight, 734 insect infestation, 795 oil spill, 296 oxygen level fluctuations, 620 paper use, 118 plastic bag use, 200 pollen levels, 734, 750 sprinkler coverage, 581 tides, 591 toxic fume emissions, 581 Finance average credit card debt, 200 depreciation, 437, 472, 1125 federal funds rate, 253, 361–362 federal income tax, 137, 285, 961 insurance costs, 127 investments. See Investments loan payments, 58 mortgage rates, 422 paying off credit cards, 479–480 PIN number combinations, 1148 raffles, 1149 sorority financials, 119 time to pay off debt, 483

10/12/16 4:24 PM

Applications Index 

Food/Nutrition amount of meat to buy, 877 for cats, 412 coffee costs, 104 costs, 103, 104 diets, 103, 158, 889–890, 912, 942, 947 dinner combinations, 1148 donut store products, 919 energy drinks, 881 exercise and nutrition, 948 nutrient levels in diet, 962, 974 oranges display, 1116 pasta making, 583 pizza slices, 574 sandwich consumption, 962 solar cooker, 1013 Forensic Science blood splatter, 521–522 time of death, 482 Geography average temperatures, 134, 155, 160, 200, 252 Bermuda Triangle, 764, 785, 795 Earth’s circumference, 25 surveying glaciers, 789 Zip codes, 1149 Geometry angle between two lines, 687 angle traced by hands of a clock, 507, 550 areas, 34, 96, 119, 148, 326, 341–342, 345, 370, 413, 418, 419, 575, 639, 714, 795, 796, 912, 987 circles, 96, 103, 295, 326, 328, 891, 948, 1048 circumference, 96 cylinders, 34, 252 decagons, 796 depth of treasure, 537 distance traveled by hands of a clock, 580–581 graphs of quadratic function, 948 height of a flagpole, 508 height of a lighthouse, 508 height of a man, 508 height of a tree, 505, 508, 651 hexagons, 796, 839 lines, 987 octagons, 839 parallelograms, 796 pentagons, 839 perimeters, 34, 45, 91, 96, 103, 118, 158, 413, 924 polygon angles, 1132 quadrilaterals, 687 rectangles, 45, 96–97, 103, 119, 148, 158, 249, 326, 345, 370, 413, 418, 419, 924, 1048 rotation of combination lock dial, 550 semicircles, 104 squares, 119, 295, 839 surface areas, 34 tangent and slope related, 34

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Tower of Pisa, 775 triangles, 96, 103, 119, 158, 418, 550, 714, 795, 796, 987 vectors, 819 volumes, 34, 45, 119, 252, 296, 326, 419 Government DUI levels, 160 federal funds rate, 361–362 FEMA hurricane response, 864, 917–918 gun laws, 951 income taxes, 137, 285, 961 law enforcement, 788 postage rates, 270–271 registered voters, 961 sales tax, 318, 327 U.S. national debt, 15 Health/Medicine anesthesia, 472, 481 average height, 200 blood alcohol level, 160 body surface area, 127, 286 body temperature, 155, 310, 591, 592, 752 body weight, 136 bunions, 550–551 diets, 103, 158, 889–890, 912, 942, 947 drug concentrations, 592, 687 drug conversions/dosages, 91, 285, 347, 386–387, 395, 412, 413, 438 drug mixtures, 879 elbow joint torque, 788 elliptical trainers, 1026 exercise heart rate, 137, 472 flu outbreak model, 118 health care costs, 253 herd immunity, 228 HIV/AIDS, 481, 1107 insurance costs, 127 IV solutions, 104–105 malaria outbreak, 252–253 muscle force, 811 orthotic knee braces, 565 pollen levels, 734, 750 radiation therapy, 509 sleep hours, 103 smoking, 347, 961 traction, 536 velocity of air in trachea, 361 viral spread, 482, 490 weight loss/gain, 345, 361 Investments allocating principal, 104, 158, 880, 890, 922, 924, 948, 993 annuities, 1107, 1126 compound interest, 434–435, 437–438, 470–471, 472, 487, 488, 490, 1106, 1107, 1124, 1126, 1162 debt accrual, 734 future value, 1107 interest rates, 104, 158, 319 investments, 98–99 multiple, 98–99 real estate, 148, 437

1263

saving, 734 simple interest, 97–98 stock values, 115–116, 118, 127, 146, 148, 361, 488, 887 Math and Numbers approximating ex, 1107 approximating functions, 1107 biofilium, 663 cryptography, 926, 972 data curve-fitting, 890, 922, 943, 947–948, 993 difference quotients, 687 Fibonacci sequence, 1108 finding numbers, 1048 linear regression, 158 lottery numbers, 1139, 1149, 1164 optimization, 819 partial-fraction decomposition, 891 pursuit theory, 820 raffles, 1164 random number generator, 1158 safe combinations, 1164 sum and product, 103, 118 temperature conversions, 91, 285, 347, 386–387, 395, 412, 413, 438 Mental Exercises memorization by professor, 412 memorization of cards, 413 Personal finance average credit card debt, 200 budgeting, 104 car repair costs, 233 federal income tax, 137, 285, 961 investments. See Investments paying off credit cards, 479–480 safes, 1149 Physics and Astronomy asteroids, 1026, 1076 braking, 819 bullet speed, 148, 809 bungee jumping, 1126 calibrating recording devices, 1037 closing doors, 819 communication with astronauts, 531 distance traveled by a bullet, 1084 Earth’s orbit, 234 Earth’s rotation, 582 electromagnetic spectrum, 453 falling objects and gravity, 34, 68, 119, 252, 271, 345, 418, 890, 947, 1115–1116, 1126, 1163 fireworks, 296 flight of a projectile, 1084 frequencies of sound, 448–450 frequency of oscillations, 620 Halley’s Comet orbit, 853, 1026 height of a hot-air balloon, 777–778 interplanetary distances, 25 ISS orbit, 573 Jupiter’s rotation, 582 laser beams, 483 lifting weights, 819 low earth orbit satellite orbits, 580

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1264 

Applications Index

Physics and Astronomy (continued ) missile fired from a ship, 809 Moon’s orbit, 25 musical tones and frequencies, 711 NASA astronaut weights, 136 NASA centrifuge, 582 NASA “vomit comet,” 346 Newton’s Law of Heating and Cooling, 439, 482 parabolic telescope, 1013 path of a projectile, 809 path of a punted football, 342–343, 345 pendulums, 68, 127, 853–854 planetary orbits, 68, 1025–1026, 1075, 1089, 1093 Pluto’s orbit, 853 pre-crusher wheels, 695–696 radioactive decay, 434, 437, 476–477, 482, 489, 490 raising wrecks, 819 resultant force, 810, 811, 827 rocket tracking, 778 satellite orbits, 1021–1022 sliding box, 810 solar cooker, 1013 solar furnace, 1013 solar radiation and distance, 319 sound levels, 447, 452, 453, 463, 472, 473, 488, 490 sound wave amplitude and frequency, 620 Space Shuttle escape basket, 777 speed of light, 91 speed of sound, 104, 127, 318 spring height, 620 spring stretch and force, 318 temperature conversions, 91, 200, 292–293, 295, 310, 328 theory of relativity, 127–128 torque, 811 towing power, 819 work, 819 x-ray crystallography, 523 Puzzles cryptography, 974 distance puzzles, 103 sleep hours, 103 Sports and Recreation AKC field trials, 783 amusement park rides, 507, 581, 582, 583, 1025, 1084 archery, 508–509, 523, 535, 778

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aviation, 105 balancing a seesaw, 105 baseball, 119, 252, 787, 810, 1084 basketball, 270, 889, 947, 974, 1149 bicycle gears, 581 bicycle speed, 582 blackjack, 1096, 1149, 1158, 1164 boating, 105, 1163 bowling, 778, 788–789 bungee jumping, 1126 canasta, 1140, 1164 cards, 1149, 1152, 1155, 1158, 1159, 1164 child’s swing, 611–613 club sign-up, 811 dog field trials, 507–508 dog training, 136 drawing cards, 1164, 1166 exercise and nutrition, 948 exercise equipment, 810 firecrackers, 252 fireworks, 296 football, 34, 105, 171, 232, 330, 342–343, 345, 370, 422, 452, 810, 889, 947, 948, 1022, 1025, 1106, 1149, 1158, 1162, 1164 footraces, 1048 golf, 82, 137, 155, 533, 1082–1083 Ironman Triathlon, 1149 jogging, 105 lifeguard posts, 778 Little League football, 477–478 marbles, 1166 marching band formation, 1113–1114 memorization of cards, 413 motorcycle manufacture, 341 NASCAR, 171 NASCAR race, 530 Olympic Decathlon, 209–210, 228 party tent, 508 peg game, 1147 poker, 1140, 1159 rock climbing, 778, 788 rolling dice, 1153 roulette, 1158, 1166 rowing, 270, 310 running, 200 Samurai sword artistry, 853 Scrabble game, 890 seasonal sales at Magic Kingdom, 591 season tickets, 1149 skiing, 534 soccer, 346

swimming pool dimensions, 249 swimming pool volume, 296 tennis, 534–535, 778, 819 tossing a coin, 1126 Tower of Hanoi game, 1131–1132 track, 550, 1025 video games, 853 volleyball, 1145 walking, 105 weightlifting, 809 Zim-Zam, 550 Transportation aviation. See Aviation boating, 105, 805–806, 807–808, 809–810, 1013 bypass around town, 232 ship navigation, 1034, 1037, 1090 shortcuts, 522 train, 919 trip speeds, 370 Travel blood alcohol level and DUI, 160 distance traveled, 158, 171 travel time, 105 trip lengths, 94, 105 trip speeds, 370 Weather atmospheric temperature, 591 average rainfall, 200 average temperatures, 134, 155, 160, 200, 252, 760 FEMA hurricane response, 864, 917–918 global climate change, 253 global warming, 271 humidity, 137 hurricanes, 912, 917–919 raindrops, 326 staking down trees before a storm, 502, 508 temperature conversions, 200 Zoology alligator length, 104 bacterial growth, 481, 489, 490, 1107 carrying capacity, 482–483, 489 deer population, 751 firefly bioluminescence, 620–621 fish population, 489 gray wolf population, 472 hamster on a wheel, 583 insect infestation, 795 phytoplankton growth, 481 snake length, 104

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[Subject Index[ Abscissa, 166 Absolute value, 11 definition of, 149 properties of, 150 Absolute value equations, 150–152 definition of, 150 quadratic, 152 Absolute value function, 257 Absolute value inequalities, 152–154 properties of, 152 Accuracy, 525 Actual velocity, 804 Acute angles, 495 calculating trigonometric function values for, 545–547 Acute triangles, 766 Addition. See also Sum(s) associative property of, 9 commutative property of, 9 of complex numbers, 71 of functions, 287–288 of matrices, 952–954 order of operations and, 6–9 of ordinates, 614–616 of polynomials, 28 of rational expressions, 51–53 of real numbers, 9–10 of vectors, 799, 801–802 Addition method. See Elimination method Additive identity, for a matrix, 954 Additive identity property definition of, 9 for matrices, 954 of real numbers, 9 Additive inverse definition of, 9 for a matrix, 954 Additive inverse property for matrices, 954 of real numbers, 9 Algebraic expressions definition of, 8, 84 evaluating, 8–9 Algebraic properties of vectors, 802 Algebraic signs, of trigonometric functions, 552–554 Algebraic techniques, solving trigonometric equations using, 741–742 Alternating sequences, 1099 Amplitude, of sinusoidal functions, 600 Angle(s), 494–496 acute, 495 acute, calculating trigonometric function values for, 545–547

central, 567 complementary, 496 coterminal, 541–542 degree measure of, 494–496 of depression, 529, 530 of elevation, 529 of inclination, 530 negative, 494 nonacute. See Nonacute angles obtuse, 495 positive, 494 quadrantal, 539 quadrantal, calculating trigonometric function values for, 547–548 radian measure of, 567–569 rays of, 494 reference, 556–557, 558 right, 495 of rotation, transforming a general second-degree equation into an equation of a conic and, 1060–1064 in standard position, 538–541 straight, 495 supplementary, 496 of a triangle, finding, 497 between vectors, 814–818 vertex of, 494 Angle of rotation formula, 1061 Angular speed, 576–577, 578 Apparent velocity, 804 Applications. See also Applications index involving composite functions, 289, 292–293 involving ellipses, 1021–1022 involving exponential functions, 432–435 involving finite arithmetic sequences, 1113–1114 involving functions, 248–249 involving geometric series, 1123–1124 involving hyperbolas, 1034 involving inverses of square matrices, 972 involving linear equations, 93–103, 195–197 involving logarithmic equations, 469–471 involving logarithms, 446–450 involving matrices, 959, 972 involving matrix multiplication, 959 involving parabolas, 1010–1011 involving quadratic equations, 115–116 involving quadratic functions, 340–343 involving series, 1104 involving systems of linear equations in two variables, 876–877 involving systems of linear inequalities in two variables, 909–911

involving systems of nonlinear equations, 1045–1046 involving trigonometric equations, 747–748 solving application problems using modeling, 93–95 Approximations of decimals, 6 of exponential functions, 426 of real zeros, bisection method for, 382–383 Arc length, 572–573 Area of a circular sector, 573–575 of a triangle, 791–794 Arguments, variable, evaluating functions with, 244 Arithmetic operations. See also Addition; Division; Multiplication; Subtraction order of, 6–9 Arithmetic sequences, 1109–1114 common difference in, 1109–1110 definition of, 1109 finite. See Finite arithmetic sequences general term of, 1110–1111 sum of, 1111–1116 Association, direction of, in scatterplots, 213–214 Associative property of addition for matrices, 954 of real numbers, 9 Associative property of multiplication, of real numbers, 9 Asymptotes, of rational functions, 398–403, 406–408 Augmented matrices, 930–932 row operations on, 931–932 Average rate of change, 260–263 Axis(es) of an ellipse, 1015 of a graph, 166 imaginary, 821 polar, 841 real, 821 rotation of, 1057–1064 transverse, of a hyperbola, 1028 Axis of symmetry of a parabola, 1003 of a quadratic function, 333 Back substitution, Gaussian elimination with, 933–935 Base(s) b, exponential functions with, 931–932 of an exponential function, 426 exponents and, 17

1265

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1266 

Subject Index

Base(s) (continued ) of a logarithmic function, 440 natural (e), 431–432 Basic properties of real numbers, 9–13 Basic rules of algebra, 9 Bearing, 531, 532 Bell-shaped curve, 477 Best fit line. See Linear regression Binomial(s), 27, 1133 cubing, 32 FOIL method for multiplying, 29–33 Binomial coefficients, 1133–1134, 1135 definition of, 1134 Pascal’s triangle and, 1136–1137 Binomial expansions finding a particular term of, 1138 Pascal’s triangle in, 1137 patterns of, 1133–1134 Binomial theorem, 1133–1138 binomial coefficients in, 1133–1134, 1135 binomial expansion and. See Binomial expansions Bisection method, for approximating real zeros, 382–383 Bounded graphs, 911 Bounded regions, 908 Brackets ([]) for grouping, 8 in inequalities, 129 Branches, of hyperbolas, 1028 Calculator use. See TI-83+/TI-84 calculators Cardioids, 847–848 Carrying capacity, 478 Cartesian coordinate system, 166, 538 Cartesian plane, 166, 543–548 quadrants of, 538–539 Center of a circle, 202, 203–204, 205 of an ellipse, 1015 of a hyperbola, 1028 Central angle, 567 Change-of-base formula, 460–461 Circles, 202–205 area of, formula for, 96 center of, 202, 203–204, 205 circumference of, formula for, 96 defined, 1000 equations of, transforming equations to standard form, 205 general form of equation of, 204 graphing, 202–204 radius of, 202, 203, 205 standard equation of, 202–204, 205 unit, 202, 584–589 Circular functions, 585–589 domains of, 588 even, 588, 589 odd, 588, 589 properties of, 587–589 ranges of, 588 translations of, 633–636

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Circular sectors, area of, 573–575 Closed intervals, inequalities and, 129–130 Coefficients binomial, 1134 correlation (r), 216–219 definition of, 8 leading, of a polynomial, 27 leading, of a polynomial function, 332 of a monomial, 27 Cofactor of a square matrix, 977–978 definition of, 977 Cofunction(s), trigonometric, 515–516 Cofunction identities, 516, 680 Cofunction theorem, 516 Column index, of a matrix, 929 Column matrices, 930 Combinations, 1145–1146 definition of, 1146–1147 permutations distinguished from, 1143 Combined inequalities, 129. See also Double inequalities Combined variation, 315, 316 Combining like radicals, 63 Common difference, in arithmetic sequences, 1109–1110 Common logarithm(s) (base 10), properties of, 457 Common logarithmic function, 442 Common ratio, of a geometric sequence, 1117–1118 Commutative property of addition for matrices, 954 of real numbers, 9 Commutative property of multiplication, of real numbers, 9 Complement(s), of events, 1153 Complementary angles, 496 Completing the square quadratic equations and, 110–112 quadratic functions and, 336–337 transforming equations of circles to standard form by, 205 Complex conjugate(s), 72–73 Complex conjugate pairs, 389–392 Complex conjugate zeros theorem, 389–392 Complex numbers, 70–74 adding, 71 definition of, 71 dividing, 72–73 equality of, 71 imaginary unit i and, 70–71 modulus of, 821–822, 823 multiplying, 72 polar (trigonometric) form of, 822–825 powers of, 832–833 products of, 829–830 quotients of, 830–831 raising to integer powers, 73–74 real part of, 71 in rectangular form, 821–822 roots of, 833–837 subtracting, 71 Complex plane, 821

Complex rational expressions, 54–56 definition of, 54 simplifying, 54–56 Complex zeros in conjugate pairs, 389–392 factoring polynomials with, 390–391 Composite functions, applications involving, 289, 292–293 Composition of functions, 236, 289–293 Compounding, 434 Compound interest, 434–435 definition of, 97 Cones, double, 1000 Conic sections, 1000–1035. See also Circle; Circles; Ellipses; Hyperbolas; Parabolas alternative definition of, 1067–1069 angle of rotation necessary to transform a general second-degree equation into an equation of, 1060–1064 definitions of, 1000–1001 determining type of, 1001–1002 graphing by point-plotting, 1073 graphing from its equation, 1071–1072 names of, 1000 polar equations of, 1067–1074 rotated, graphing, 1063–1064 Conjugates complex, 72–73 of the denominator, 63–64 Constant(s), 8 of variation (proportionality), 312, 313, 314 Constant functions, 255–256, 259 Constant term, of a polynomials, 27 Constraints, in linear programming, 914 Consumer surplus, 909–911 Continuous compounding, 435 Continuous graphs, 351 Converging series, 1104 Corner points, in graphs, 911 Correlation coefficient (r), 216–219 computing using Excel 2010, 217–218 computing using TI-83+/TI-84 calculators, 216–217 Cosecant function graphing, 627–628, 631, 632–633 inverse, 725–727 Cosine(s), Law of, 780–786 Cosine function, 513 cofunction identity for, 680 graphing, 597–598 graphing sums of, 616, 628 inverse, 720–722 sum and difference identities for, 676–680 summary of, 598 Cotangent function graphing, 625, 628, 629, 630–631 inverse, 725–727, 728 Coterminal angles, 541–542 Cramer’s rule, 980–985 for solving systems of linear equations in three variables, 982–985 for solving systems of linear equations in two variables, 981–982

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Subject Index 

Cryptography, matrices in, 926 Cube(s) of binomials, 32 difference of, factoring, 39 sum of, factoring, 38 Cube function, 256 Cube root(s), 62 Cube root function, 257 Current velocity, 804 Curves, plane, 1078 Cycloids, 1080–1081 Damped harmonic motion, 609, 612–613 Decay, exponential, 474, 476–477 Decibels (dBs), 446–447 definition of, 447 Decimals approximation of, 6 converting between scientific notation and, 27–33 rational and irrational, 4, 5 Decreasing functions, 259–260 Degree of a monomial, 27 of a polynomial, 27 Degree measure, 494–496 Degrees (of angles), converting between radians and, 569–571 De Moivre’s theorem, 832–833 Denominators, 4 conjugates of, 63–64 least common, 12 rationalizing, 63–65 Dependent systems of equations, 866, 884–885 matrices and, 939, 940, 941 Dependent variables, 209, 240 Descartes’ rule of signs, 375, 378–379 Determinant of a matrix, 968 Cramer’s rule and, 980–985 definition of, 978 expanding by the first row, 978 finding, 978–980 of an n 3 n matrix, 977–980 of a 2 3 2 matrix, 976–977 Difference(s). See also Subtraction common, in arithmetic sequences, 1109–1110 of two cubes, factoring, 39 of two squares, factoring, 37–38 Difference function, 287–288 Difference identities, trigonometric, 675–685 Difference quotient, 246 calculating, 263 definition of, 263 Directed line segments, 798 Direction along a curve, 1078 of vectors, 798–801 Direction angle of a vector, 800–801 of vectors, 800–801

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Directrix, of a parabola, 1003 Direct variation, 312–314 with powers, 313–314 Discriminants, 115 Distance formula, definition of, 168 Distance-rate-time problems, 100–102 graphing, 167–168 Distinguishable permutations, 1146–1147 Distributive property, of real numbers, 9, 10, 11 Diverging series, 1104 Dividend, 363 Division. See also Quotient(s) of complex numbers, 72–73 of functions, 287–289 long, of polynomials, 363–367 order of operations and, 6–9 of rational expressions, 49–51 synthetic, of polynomials, 367–369 Division algorithm, 366 Division sign (/), 6, 7 Divisor, 363 Domains of circular functions, 588 explicit, 247 of functions, 239, 247–249, 288–289, 291, 301–302 implicit, 247 of logarithmic functions, finding, 443–444 of rational expressions, 46–47 of rational functions, 396–397 of a relation, 238 Dot product, 802, 813–818 Double-angle identities, 689–694 applying, 691–693 derivation of, 690 Double cones, 1000 Double inequalities, 129 linear, solving, 133–134 Doubling time growth model, 432–434 Eccentricity alternative definition of conics and, 1067–1069 of ellipses, 1015, 1017 Elements of matrices, 929 of a set, 4 Elimination method Gaussian, with back substitution, 933–935 Gauss-Jordan, 935–938 solving systems of linear equations in three variables using, 934–937 solving systems of linear equations in two variables using, 869–872 solving systems of nonlinear equations using, 1040–1043 Ellipses, 1015–1023 applications involving, 1021–1022 centered at point (h,k), 1019–1021 centered at the origin, 1015–1019 center of, 1015 defined, 1000, 1015

1267

eccentricity of, 1015, 1017 equations of, 1016–1017, 1019, 1020 foci of, 1015 major axis of, 1015 minor axis of, 1015 rotating, 1059–1060 vertices of, 1015 Ellipsis (punctuation), 4 Empty set, 4 e (natural base), 431–432 End behavior, of a polynomial function, 357 Endpoints in interval notation, 129 of a ray, 494 Equality of complex numbers, 71 of matrices, 950–952 Equal sign (5), 7 Equal vectors, 798–799, 801 Equations absolute value, 150–152 of a circle, 202–204, 205 of a conic, angle of rotation necessary to transform a general second-degree equation into, 1060–1064 definition of, 84 dependent systems of. See Dependent systems of equations of an ellipse, 1016–1017, 1019, 1020 equivalent, generating, 84 exponential. See Exponential equations first-degree, 85 functions defined by, 240–242 general form. See General form graphs of, 174–175 of a hyperbola, 1029–1030, 1033, 1034 inconsistent systems of. See Inconsistent systems of equations independent systems of, 866 linear. See Linear equations; Systems of linear equations in three variables; Systems of linear equations in two variables logarithmic. See Logarithmic equations matrix, 964–965 nonlinear, systems of. See Systems of nonlinear equations with no solution, 84 of a parabola, 1004, 1007 parametric. See Parametric equations polar, of conic sections, 1067–1074 polynomial. See Polynomial equations quadratic. See Quadratic equations quadratic in form, 123–125 radical, 121–123 rational. See Rational equations solution set of, 84 solutions (roots) of, 84 solving, 84 standard form. See Standard form trigonometric. See Trigonometric equations using complex roots, solving, 836–837

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1268 

Subject Index

Equilateral triangles, 497 Equilibrants, 807 Equilibrium point, 909 Equivalent equations, generating, 84 Evaluation of algebraic expressions, 8–9 Evaluation of functions difference quotient and, 246 exponential, 426–427 logarithmic, 440–442 with negatives, 245 with quotients, 245–246 by substitution, 243 with sums, 245 with variable arguments, 244 Even functions, 257–259, 588, 589 Even trigonometric identities, 666 Events complements of, 1153 definition of, 1151 independent, 1156–1157 mutually exclusive, 1154–1156 not occurring, probability of, 1153–1154 probability of, 1152–1153 union of, probability of, 1154–1156 Excel 2010 computing correlation coefficients using, 217–218 creating scatterplots using, 212–213 finding line of best fit using, 222–224 Experiments, 1151 Explicit domains, 247 Exponent(s) fractional, equations quadratic in form with, solving, 125 integer, 17–19 natural-number, 17 negative, equations quadratic in form with, solving, 124 properties of, 456 rational. See Rational exponents Exponential decay models, 474, 476–477 Exponential equations, 465–471 inverse properties and, 465 one-to-one properties and, 465 strategies for solving, 465 Exponential expressions, simplification of, 20–21 Exponential functions, 426–436 applications of, 432–435 definition of, 426 evaluating, 426–427 graphs of, 427–430 natural base e and, 431–432 Exponential growth models, 474, 475 Expressions algebraic, 8–9, 84 exponential, simplification of, 20–21 radical, simplified form of, 65 rational. See Rational expressions Extraneous solutions, 121, 122 definition of, 87

Young_AT_6160_SE_Index_pp1261-1276.indd 1268

Factor(s). See also Partial-fraction decomposition linear, in partial fraction decomposition, 894–897 of polynomials, 36 Factorial(s), definition of, 1100 Factorial notation, 1100–1101 Factoring, 36 of expressions with rational exponents, 67 of polynomial functions, 377–378 of polynomials. See Factoring polynomials of quadratic equations, 107–109 solving equations with, 125–126 of trinomials, as product of two binomials, 39–42 Factoring polynomials, 36–44, 390–393 with complex zeros, 390–391 factor theorem for, 373–375 greatest common factor and, 36–37 by grouping, 42–43 over the integers, 36 real zeros and, 380–381 special forms of, 37–39 strategy for, 43 trinomials as product of two binomials and, 39–42 upper and lower bound rules and, 381 Factor theorem, 373–375 Feasible solutions, linear programming and, 914 Fibonacci sequence, 1101–1102 Finite arithmetic sequences, 1111–1114 application involving, 1113–1114 definition of, 1112 Finite geometric series, 1119–1120 Finite sequences, 1098. See also Finite arithmetic sequences Finite series, 1102, 1103–1104. See also Finite geometric series First-degree equations, 85. See also Linear equations Focus/foci of ellipses, 1015 of a parabola, 1003 FOIL method, for multiplying binomials, 29–30 Force vectors, 804 Formulas angle of rotation, 1061 for area of a circle, 96 for area of a rectangle, 96 for area of a triangle, 96 change-of-base, 460–461 distance, definition of, 168 geometric, 96 Heron’s, 793–794 midpoint, definition of, 169 or circumference of a circle, 96 partial-sum, proving with mathematical induction, 1130 for perimeter of a rectangle, 96 for perimeter of a triangle, 96 quadratic, 112–115

recursion, 1101–1102 rotation of axes, 1057–1060 458 angles evaluating trigonometric functions exactly for, 518–519 in standard position, 540 458-458-908 triangles, 499–500 Four-leaved rose, 847 Fraction(s) linear inequalities with, solving, 132–133 partial, 893. See also Partial fraction decomposition properties of, 12–13 Fractional exponents, equations quadratic in form with, solving, 125 Function(s), 238–249 absolute value, 257 adding, 287–288 applications involving, 248–249 argument of, 243 average rate of change and, 260–263 common logarithmic, 442 composite, 289, 292–293 composition of, 236, 289–293 constant, 255–256, 259 cube, 256 cube root, 257 decreasing, 259–260 defined by equations, 240–242 definition of, 238–239 difference, 287–288 dividing, 287–289 domain of, 239, 247–249, 288–289, 291, 301–302 evaluating. See Evaluation of functions even, 257–259 exponential. See Exponential functions expressing, 242 function notation and, 242–246 graphing, 243–244 greatest integer, 266 identity, 256 increasing, 259–260 inverse. See Inverse functions linear, 255 logarithmic. See Logarithmic functions multiplying, 287–288 objective, in linear programming, 914–916 odd, 257–259 one-to-one, 298–300 periodic, 594 piecewise-defined. See Piecewise-defined functions polynomial. See Polynomial functions power. See Power functions product, 287–288 quadratic. See Quadratic functions quotient, 287–289 range of, 239 rational. See Rational functions reciprocal, 257 relations and, 238–240 sinusoidal. See Sinusoidal functions

10/12/16 4:24 PM

Subject Index 

square, 256 square root. See Square root function step, 266 subtracting, 287–288 sum, 287–288 trigonometric. See Trigonometric functions vertical line test for, 241–242 Function notation, 242–246, 288 Fundamental counting theorem, 1141–1143 Fundamental period, 594 Fundamental theorem of algebra, 388–389 Gaussian distribution models, 474, 477–478 Gaussian elimination, with back substitution, 933–935 Gauss-Jordan elimination, 935–938 General form of the equation of a circle, 204 of the equation of a quadratic function, 336–339 of the equation of a straight line, 186 General terms of an arithmetic sequence, 1110–1111 of a geometric sequence, 1118–1119 of a sequence, 1098–1099 Geometric formula, 96 Geometric sequences, 1117–1119 common ratio of, 1117–1118 definition of, 1117 general term of, 1118–1119 Geometric series, 1119–1124 applications involving, 1123–1124 finite, 1119–1120 infinite, 1121–1123 Geometry problems, 95–97. See also Applications index Geostationary satellites, 1021 Geosynchronous (GEO) orbit, 1021 Graphs/graphing, 164–231 asymptotes of rational functions and, 398–403, 406–408 best fit line and. See Linear regression bounded, 911 Cartesian coordinate system and, 166 of a circle, 202–204 compression of, vertical and horizontal, 280–282 of a conic from its equation, 1071–1072 continuous, 351 of curves defined by parametric equations, 1079–1080 of a cycloid, 1081 distance between two points and, 167–168 of ellipses, 1017–1018, 1020–1021 of an equation, 174–175 of a function, 243–244 of a hyperbola, 1031–1032, 1033 intercepts and, 176–177, 180–181, 186–188 of inverse functions, 303–304 of a line, 186–197 of linear equations, 190–197 of linear inequalities, 130–131, 904–905 linear regression and. See Linear regression

Young_AT_6160_SE_Index_pp1261-1276.indd 1269

of logarithmic functions, 443–446 with logarithmic scales, 449–450 midpoint of a line segment joining two points and, 168–169 of parabolas with vertex at (h,k), 1007–1009 of parabolas with vertex at the origin, 1005–1007 of piecewise functions, 264–266 point-plotting and, 166, 173–175 by point-plotting in polar coordinates, 1073 point-slope form and, 192–193 of polar equations, 843–850 of polynomial functions, 351–352, 355–359, 383–384 of polynomial inequalities, 139 of projectile motion, 1081–1083 of quadratic functions, 332–343 of rational functions, 404–409 reflection about the axes in, 278–280 of a rotated conic, 1063–1064 of scatterplots. See Scatterplots of shifted sinusoidal functions, 606–608 shifts of, horizontal and vertical, 274–278 of sinusoidal functions, 594–606 slope and, 188–190 slope-intercept form and, 190–191 smooth, 351 solving systems of linear equations in two variables using, 873–875 stretching of, vertical and horizontal, 280–282 of sums of functions, 614–616 symmetry and, 177–181 of systems of linear inequalities in two variables, 906–908 of systems of nonlinear inequalities, 1051–1052 transformations of, 274–283 unbounded, 911 Greatest common factor (GCF), 36–37 Greatest integer function, 266 Grouping factoring a polynomial by, 42–43 factoring by, solving polynomial equations using, 125–126 notations for, 8 Growth, logistic, 474, 478 Growth models doubling time, 432–434 exponential, 474, 475 logistic, 474, 478 Half-angle identities, 697–704 applying, 700–704 derivation of, 698–699 Half-life, in exponential decay, 476 Half open intervals, inequalities and, 129–130 Half-planes, 903 Harmonic motion, 609–614 damped, 609, 612–613 examples of, 610

1269

resonance and, 609, 614 simple, 609, 610–612 Heading, 531 Heaviside step function, 266 Heron’s formula, 793–794 Horizontal asymptotes, of rational functions, 398, 400–403, 406–407 Horizontal components, of vectors, 803 Horizontal line(s), finding equation of, 193 Horizontal line test, 299–300 definition of, 299 Horizontal shifts graphing exponential functions using, 429–430 graphing logarithmic functions using, 445 Hyperbolas, 1028–1035 applications involving, 1034 branches of, 1028 centered at point (h,k), 1032–1034 centered at the origin, 1028–1032 center of, 1028 defined, 1000 equation of, 1029–1030, 1033, 1034 graphs of, 1031–1032, 1033 transverse axis of, 1028 vertices of, 1028 Hypotenuse, of a right triangle, 497 Identities cofunctions, 516 defined, 654 definition of, 84 reciprocal, 513–515 trigonometric. See Trigonometric identities Identity function, 256 Imaginary axis, 821 Imaginary numbers, 70 pure, 71 Imaginary unit (i), of complex numbers, 70–71 Immediate value theorem, definition of, 354 Implicit domains, 247 Improper rational expressions, 893 Improper rational functions, 401 Inconsistent systems of equations, 866, 885–886 matrices and, 939 Increasing functions, 259–260 Independent events, 1156–1157 Independent systems of equations, 866 Independent variables, 209, 240 Index(es) of a root, 62 row and column, of a matrix, 929 of summation, 1102–1103 Inequalities absolute value. See Absolute value inequalities double (combined). See Double inequalities equations compared with, 129 linear. See Linear inequalities; Systems of linear inequalities in two variables nonlinear. See Nonlinear inequalities; Systems of nonlinear inequalities

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1270 

Subject Index

Inequalities (continued ) polynomial, 139–143 properties of, 132 quadratic, 140–143 rational, 144–146 strict, 129 Infinite geometric series, 1121–1123 Infinite sequences, 1098 Infinite series, 1102, 1104 Infinity (∞), 130 Initial point, of a vector, 798 Initial ray (side) of an angle, 494 Inspection, solving trigonometric equations by, 737–740 Integer(s), 4, 5 Integer exponents, 17–19 complex numbers and, 73–74 properties of, 19–21 Intercepts as graphing aid, 180–181 in graphs, 176–177, 180–181, 186–188 slope-intercept form and, 190–191 Interest. See also Applications index compound, 97, 434–435 simple, 97–98 solving problems involving, 97–99, 434–435 Interest rates, 434 Intermediate value theorem, 382–383 Intersection of sets, 131 Interval(s) closed and open, 129–130 half open, 129–130 test, on real number line, 139 Interval notation, linear inequalities and, 129–131 Inverse(s) additive, 9 additive, for a matrix, 954 multiplicative, 9 of square matrices. See Inverses of square matrices Inverse functions, 301–307 definition of, 301 finding, 304–307 graphical interpretation of, 303–304 trigonometric equations requiring the use of, solving, 742–744 Inverse properties, exponential equations and, 465 Inverses of square matrices, 926, 965–970 applications involving, 972 definition of, 966 solving systems of linear equations using, 970–972 Inverse trigonometric functions, 716–732 finding exact values for expressions involving, 728–731 Inverse variation, 314–315 Irrational numbers, 4, 5 Irreducible quadratic factors, in partial fraction decomposition, 897–899 Isosceles triangles, 497

Young_AT_6160_SE_Index_pp1261-1276.indd 1270

Joint variation, 315, 316 Latus rectum, 1005–1007 Law of Cosines, 780–786 Law of Sines, 769–771 Leading coefficient of a polynomial, 27 of a polynomial function, 332 Least common denominator (LCD), 12 adding and subtracting rational expressions using, 53 Least common multiple (LCM), 12 Legs, of a right triangle, 497 Like radicals, combining, 63 Like terms, combining, 28 Limaçons, 847 Line(s) best fit. See Linear regression defined, 494 equations of. See Linear equations graphing, 186–197 horizontal, finding equation of, 193 joining two points, midpoint of, 168–169 parallel, 194 parametric representation of, 885, 886 perpendicular, 195 slope of, 188–190 Linear equations, 84–90, 190–197 applications involving, 93–103, 195–197 definition of, 85 general form of, 186 in one variable, solving, 84–86 parallel, 194–195 perpendicular, 194–195 point-slope form of, 192–193 rational equations reducible to, solving, 87–89 slope-intercept form of, 190–191 solving using mathematical models, 93–95 systems of. See Systems of linear equations in three variables; Systems of linear equations in two variables of a vertical line, finding, 193 Linear factors, in partial fraction decomposition, 894–897 Linear functions, 255 Linear inequalities, 129–135 graphing, 130–131 interval notation for, 129–131 solving, 131–135 systems of. See Systems of linear inequalities in two variables in two variables. See Linear inequalities in two variables; Systems of linear inequalities in two variables Linear inequalities in two variables, 903–905 graphing, 904–905 systems of. See Systems of linear inequalities in two variables Linearity, in scatterplots, 214–216 Linear programming, 914–918 constraints and, 914 feasible solutions and, 914

objective function and, 914–916 solving an optimization problem using, 914–918 Linear regression, 219–225 determining best fit line and, 219–224 prediction using best fit line and, 224–225 Linear relationships, 214–215 strength of, 216–219 Linear speed, 575–576, 577–578 Line segments, 494 directed, 798 Logarithm(s), 440 applications of, 446–450 change-of-base formula and, 460–461 common (base 10), properties of, 457 evaluating, 440–442 finding value of, 441–442 natural (base e), properties of, 457 properties of, 456–460 Logarithmic equations, 468–471 applications of, 469–471 strategy for solving, 468 Logarithmic functions, 440–450 common, 442 definition of, 440 exponential form of, 440–441 finding domains of, 443–444 graphs of, 443–446 logarithmic form of, 440–441 natural, 442 properties of, 456–460 Logarithmic models, 474, 479–480 Logarithmic scales, 448–450 graphing using, 449–450 Logistic growth models, 474, 478 Long division, of polynomials, 363–367 Loran, 1034 Lower/ upper bound rules, for real zeros, 381 Lowest terms rational functions in, 399 reducing to. See Simplification Magnitude, of vectors, 798–801 Main diagonal entries, of matrices, 929 Major axis, of an ellipse, 1015 Mathematical induction, 1128–1130 principle pf, 1128 Matrix(ces), 926–992 applications involving, 959, 972 augmented, 930–931 column, 930 column index of, 929 in cryptography, 926 dependent systems of equations and, 939, 940, 941 determinants of. See Determinant of a matrix elements of, 929 equality of, 950–952 finding the order of, 929 Gaussian elimination with back substitution and, 933–935 Gauss-Jordan elimination and, 935–938

10/12/16 4:24 PM

Subject Index 

inconsistent systems of equations and, 939 inverses of, 926, 965–970 main diagonal entries of, 929 nonsingular, 967 reduced row-echelon form of, 932 row-echelon form of, 932–933 row index of, 929, 930 row operations on, 931–932 singular, 967 square. See Inverses of square matrices; Square matrices zero, 953–954 Matrix addition, 952–954 definition of, 952 properties of, 954 Matrix algebra, 951 solving systems of linear equations using, 970–972 Matrix equations, 964–965 Matrix multiplication, 955–959 application involving, 959 definition of, 956 by inverse, 966 properties of, 958–959 Matrix subtraction, 952–953 definition of, 952 Members, of a set, 4 Midpoint, of a line segment joining two points, 168–169 Midpoint formula, definition of, 169 Minor axis, of an ellipse, 1015 Minor of a square matrix, 977–978 definition of, 977 Mixture(s), definition of, 99 Mixture problems, 97 Modeling of doubling time growth, 432–434 of exponential decay, 474, 476–477 exponential decay models and, 474, 476–477 of exponential growth, 474, 475 exponential growth models and, 474, 475 Gaussian (normal) distribution models and, 474, 477–478 logarithmic models and, 474, 479–480 logistic growth models and, 474, 478 normal distribution models and, 474, 477–478 solving application problems using, 93–95 with a system of three linear equations, 887–888 with systems of linear equations in three variables, 887–888 using variation, 312–317 Modulus, of a complex number, 821–822, 923 Monomials, 27 coefficient of, 27 degree of, 27 multiplying with a polynomial, 29 Motion, projectile, 1081–1083 Multiplication. See also Product(s) associative property of, 9 of binomials, FOIL method for, 29–33

Young_AT_6160_SE_Index_pp1261-1276.indd 1271

commutative property of, 9 of complex numbers, 72 of functions, 287–288 of matrices. See Matrix multiplication order of operations and, 6–9 of polynomials, 28–33 of rational expressions, 49–51 scalar, of vectors, 802 of vectors, 813–818 Multiplication sign (?), 6, 7 Multiplicative identity matrix, 965–966 Multiplicative identity property, of real numbers, 9 Multiplicative inverse, 9 Multiplicative inverse property, of real numbers, 9 Multiplicity of a zero, 354–355, 357 definition of, 354 Mutually exclusive events, 1154–1156 Natural base (e), 431–432 Natural exponential function, 431 Natural logarithm(s) (base e), properties of, 457 Natural logarithmic function, 442 Natural number(s), 4, 5 Natural-number exponents, 17 Negative(s) evaluating functions with, 245 properties of, 10–11 Negative angles, 494 Negative association, in a scatterplot, 213, 214 Negative exponents, equations quadratic in form with, solving, 124 Negative infinity (2∞), 130 Negative-integer exponent property, 18 Nonacute angles calculating trigonometric function values for, 545–547 trigonometric functions of, 552–563 Nondistinguishable permutations, 1146–1147 Nonlinear equations, systems of. See Systems of nonlinear equations Nonlinear inequalities systems of, 1051–1054 in two variables, 1049–1051 Nonlinear relationships, 214, 215 Nonsingular matrices, 967 Normal distribution models, 474, 477–478 Notation factorial, 1100–1101 function, 242–246, 288 interval. See Interval notation scientific, 21–23 sigma (S) (summation), 1102–1103 nth partial sum of an arithmetic sequence, 1111–1114 definition of, 1102 nth roots, 62–65 principal, 62 nth root theorem, 833–834

1271

Null set, 4 Number(s) complex. See Complex numbers imaginary. See Imaginary numbers irrational, 4, 5 natural, 4, 5 rational, 4, 5. See also Fraction(s) real. See Real number(s) sets and subsets of, 4 whole, 4, 5 Number line. See Real number line Numerators, 4 n zeros theorem, 388–389 Objective functions, in linear programming, 914–916 Oblique triangles acute, 766 Law of Cosines to solve, 780–786 Law of Sines to solve, 769–771 obtuse, 766 solving, 766–776 Obtuse angles, 495 Obtuse triangles, 766 Odd functions, 257–259, 588, 589 Odd trigonometric identities, 666 One-to-one functions, 298–300 One-to-one properties, exponential equations and, 465 Open intervals, inequalities and, 129–130 Operations. See also Addition; Division; Multiplication; Subtraction order of, 6–9 Optimization problem, solving using linear programming, 914–918 Order of matrices, finding, 929 of operations, 6–9 Ordered pairs, 166 Ordinates, 166 addition of, 614–616 Orientation, along a curve, 1078 Origin of a graph, 166 in polar coordinate system, 841 Orthogonal vectors, 815–816 Outcomes, 1151 Parabolas, 330, 332–343, 1003–1011 applications involving, 1010–1011 axis of symmetry of, 1003 defined, 1000, 1003 directrix of, 1003 equation of, 1004, 1007 finding the equation of, 340–343 focus of, 1003 with vertex at the origin, 1003–1007 with vertex at the point (h,k), 1007–1009 vertex of, 1003 Parallel lines, definition of, 194 Parallel vectors, 815 Parameter(s), defined, 1078

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1272 

Subject Index

Parametric equations, 1078–1083 applications of, 1080–1083 defined, 1078 graphing curves defined by, 1079–1080 Parametric representation, of a line, 885, 886 Parentheses for grouping, 8 in inequalities, 129 Partial fraction(s), 893 Partial fraction decomposition, 892–901 with combinations of factors, 900 with distinct irreducible quadratic factors, 897–898 with distinct linear factors, 894–895 with repeated irreducible quadratic factors, 899 with repeated linear factors, 895–897 Partial-sum formulas, proving with mathematical induction, 1130 Pascal’s triangle, 1136–1137 Perfect squares, 111 factoring, 38 square roots of, 60–61 Periodic functions, 594 of sinusoidal functions, 602 Permutations, 1143–1145 combinations distinguished from, 1143 definition of, 1143 distinguishable, 1146–1147 with repetition (nondistinguishable), 1146–1147 Perpendicular lines, definition of, 195 Perpendicular vectors, 815–816 Piecewise-defined functions, 263–266 finding inverse of, 307 Plane Cartesian, 166 division into half-planes, 903 graph of linear equation as, 882 Plane curves, 1078 Point-plotting, 166, 173–175 graphic conics by, 1073 of polar coordinates, 841–842 Point-slope form, of linear equations, 192–193 Polar axis, 841 Polar coordinates, 841–842 converting between rectangular coordinates and, 842–843 equations of conic in, 1067–1074 point-plotting, 841–842 Polar equations converting between rectangular form equations and, 849–850 graphs of, 843–850 Polar form, of a complex number, 822–825 Pole, in polar coordinate system, 841 Polynomial(s), 27–33. See also Complex rational expressions; Rational expressions adding, 28 complex zeros of. See Complex zeros degree of, 27 dividing, 363–367

Young_AT_6160_SE_Index_pp1261-1276.indd 1272

factored completely, 36 factoring. See Factoring polynomials factors of, 36 greatest common factor of, 36–37 multiplying, 28–33 prime, 36, 42 real zeros of. See Real zeros special products of, 29–33 in standard form, 27 subtracting, 28 synthetic division of, 367–369 zeros of, 139. See also Complex zeros; Real zeros Polynomial equations, factoring by grouping to solve, 125–126 Polynomial functions, 349–359 definition of, 332, 349 factoring, 377–378 graphing, 351–352, 355–359, 383–384 identifying, 349–351 leading coefficient of, 332 quadratic. See Quadratic functions ratios of. See Rational functions real zeros of, 353–355 Polynomial inequalities, 139–143 Position vectors, 799 Positive angles, 494 Positive association, in a scatterplot, 213 Power(s). See also Exponent(s) of complex numbers, 832–833 direct variation with, 313–314 Power functions characteristics of, 352 definition of, 351 graphing polynomial functions using transformations of, 351–352 Power property, of integer exponents, 19 Prediction, using line of best fit, 224–225 Predictor variables, 209, 240 Prime polynomials, 36, 42 Principal (investment), 434 Principal nth root, 62 Principal square root, 60, 70, 109 Probability, 1151–1157 of an event, 1152–1153 of an event not occurring, 1153–1154 independent events and, 1156–1157 mutually exclusive events and, 1154–1156 sample space and, 1151 Producer surplus, 909–911 Product(s). See also Multiplication of complex numbers, 829–830 dot, 802, 813–818 special, 29–33 Product function, 287–288 Product property, of integer exponents, 19 Product to a power property, of integer exponents, 19 Product-to-sum identities, 708–709 Projectile motion, 1081–1083 Proper rational expressions, 893 Proper rational functions, 401 Proportionality. See Variation

Pure imaginary numbers, 71 Pythagorean identities, trigonometric, 657–660, 666 Pythagorean theorem, 498–499 Quadrant(s), of Cartesian plane, 166, 538–539 Quadrantal angles, 539 calculating trigonometric function values for, 547–548 Quadratic equations, 107–109 absolute value, 152 applications involving, 115–116 completing the square to solve, 110–112 factoring, 107–109 quadratic formula and, 112–115 square root method to solve, 109–110 Quadratic factors, irreducible, in partial fraction decomposition, 897–899 Quadratic formula, 112–115 Quadratic functions, 330, 332–343 application problems involving, 340–343 axis of symmetry of, 333 definition of, 333 in general form, graphing, 336–339 graphs of, 333–343 in standard form, graphing, 333–336 vertex of, 333 Quadratic inequalities, 140–143 Quadratic in form equations, 123–125 Quotient(s), 363. See also Division of complex numbers, 830–831 difference. See Difference quotient evaluating functions with, 245–246 Quotient function, 287–289 Quotient identities, trigonometric, 656–657, 666 Quotient property, of integer exponents, 19 Quotient to a power property, of integer exponents, 19 Radian(s), 567 converting between degrees and, 569–571 Radian measure, of an angle, 567–569 Radical(s), 62 like, combining, 63 properties of, 62 Pythagorean theorem with, 499 simplifying, 62 Radical equations, 121–123 Radical expressions, simplified form of, 65 Radical sign A " B, 60 Radicand, 60 Radius, of a circle, 202, 203, 205 Ranges of circular functions, 588 of a function, 239 of a relation, 238 of trigonometric functions, 555–556 Rate of change average, 260–263 slope as, 195–196

10/12/16 4:24 PM

Subject Index 

Ratio(s) right triangle. See Right triangle ratios setting up trigonometric functions as, 515 trigonometric, 512. See also Trigonometric functions Rational equations definition of, 87 reducible to linear equations, solving, 87–89 Rational exponents, 66–67 equations with, solving by factoring, 125 Rational expressions, 46–53 adding, 51–53 complex. See Complex rational expressions dividing, 49–51 domains of, 46–47 improper, 893 multiplying, 49–51 proper, 893 simplifying, 48–49 subtracting, 51–53 Rational functions, 396–410 asymptotes of, 398–403 definition of, 396 domain of, 396–397 graphing, 404–409 improper, 401 proper, 401 Rational inequalities, 144–146 Rationalizing denominators, 63–65 Rational numbers, 4, 5. See also Fraction(s) Rational root test, 375–378 Rational zero theorem, 375–378 Rays, 494 Real axis, 821 Real number(s) properties of, 9–13 set of, 4–5, 84 trichotomy property of, 129 Real number line absolute value and, 11 definition of, 5 distance between two points on, 150 test intervals on, 139 Real part, of a complex number, 71 Real zeros, 372–385 approximating, 382–383 Descartes’ rule of signs and, 375, 378–379 factoring polynomials and, 380–381 factor theorem and, 373–375 finding, 375 graphing polynomial functions and, 383–384 intermediate value theorem and, 382–383 number of, 375 of a polynomial function, 353–355 rational zero theorem and, 375–378 remainder theorem and, 372–373 upper and lower bound rules for, 381 Reciprocal(s), 9, 654 Reciprocal function, 257 Reciprocal identities, 513–515 trigonometric, 654–656, 665

Young_AT_6160_SE_Index_pp1261-1276.indd 1273

Rectangles area of, formula for, 96 perimeter of, formula for, 96 Rectangular coordinates, converting between polar coordinates and, 842–843 Rectangular coordinate system, 166, 538, 821 Rectangular form, of complex numbers, 821–822 Rectangular form equations, converting between polar equations and, 849–850 Recursion formulas, 1101–1102 Reduced row-echelon form, of a matrix, 932 Reducing to lowest terms. See Simplification Reference angles, 556–557, 558 Reference right triangles, 556, 557 Relations definition of, 238 domain of, 238 functions and, 238–240 range of, 238 Remainder(s), 363 Remainder theorem, 372–373 Repetition, permutations with, 1146–1147 Resonance, harmonic motion and, 609, 614 Response variables, 209, 240 Resultant vectors, 804–808 Resultant velocity, 804 Richter scale, definition of, 448 Right angles, 495 Right triangle(s), 497 458-458-908, 499–500 hypotenuse of, 497 legs of, 497 Pythagorean theorem and, 498–499 reference, 556, 557 solving, 525–532 308-608-908, 501–502 Right triangle ratios, 511–520 cofunctions and, 515–516 evaluating trigonometric functions exactly for, 517–519 reciprocal identities and, 513–515 Roots of complex numbers, 833–837 cube, 62 of equations, 84 square. See Square root(s) Rose, 847 Rotation of axes formulas, 1057–1060 Rounding, of decimals, 6 Row-echelon form, of a matrix, 932–933 Row index, of a matrix, 929 Row matrices, 930 Row operations, on matrices, 931–932 Sample space, 1151 Scalar(s), 798 Scalar multiplication, 954–955 definition of, 954 of vectors, 802 Scatterplots, 209–225 creating using Excel 2010, 212–213

1273

creating using TI-83+ or TI-84 calculators, 211 linear regression and. See Linear regression patterns in, 213–219 Scientific notation, 21–23 Secant function graphing, 626–627, 628, 631, 632 inverse, 725–727 Sectors, circular, area of, 573–575 Sequences, 1098–1102 alternating, 1099 arithmetic. See Arithmetic sequences; Finite arithmetic sequences definition of, 1098 factorials and, 1101–1102 Fibonacci, 1101–1102 finite and infinite, 1098. See also Finite arithmetic sequences geometric. See Geometric sequences terms of, 1098–1099 Series, 1102–1104 application involving, 1104 converging and diverging, 1104 definition of, 1102 finite and infinite, 1102, 1103–1104. See also Finite geometric series geometric. See Geometric series sigma (summation) notation and, 1102–1103 Sets, 4 empty (null), 4 intersection of, 131 of real numbers, 4–5, 84 union of, 131 Shifts graphing exponential functions using, 429–430 graphing logarithmic functions using, 445 Sigma (S) notation, 1102–1103 Sign(s), algebraic, of trigonometric functions, 552–554 Significant digits, 525–526 Similar triangles, applications involving, 505 Simple harmonic motion, 609, 610–612 Simple interest, 97–98 definition of, 97 Simplification of a complex rational expression, 54–56 of exponential expressions, 20–21 of expressions with rational exponents, 66–67 order of operations and, 7–8 of radical expressions, 65 of radicals using imaginary numbers, 70 of rational expressions, 48–49, 54–56 of square roots, 61 Simplified form, of radical expressions, 65 Sine(s), Law of, 769–771 Sine function, 513 cofunction identity for, 680 graphing, 594–596, 615–616, 628 inverse, 716–719 sum and difference identities for, 680–683 summary of, 596

10/12/16 4:24 PM

1274 

Subject Index

Singular matrices, 967 Sinusoidal functions amplitude and period of graphs of, 599–600 graphs of, 594–606 shifted, graphing, 606–608 608 angles evaluating trigonometric functions exactly for, 519 in standard position, 540 Slant asymptotes, of rational functions, 403, 408 Slope of a line, 188–190 as rate of change, 195–196 of the secant line, 260–262 Slope-intercept form, of linear equations, 190–191 Smooth graphs, 351 Software. See Excel 2010 Solutions of equations, 84 extraneous, 87, 121, 122 feasible, linear programming and, 914 Solution sets of equations, 84 of a system of inequalities, 905 Special products, 29–33 Speed angular, 576–577, 578 defined, 575 linear, 575–576, 577–578 Spirals, 848 Square(s) completing, quadratic equations and, 110–112 difference of, factoring, 37–38 perfect. See Perfect squares in special products, 31 Square function, 256 Square matrices, 929 Cramer’s rule and, 980–985 inverse of, finding, 967–970 inverses of. See Inverses of square matrices Square root(s), 60–61 principal, 60, 70, 109 properties of, 61 simplifying, 61 Square root function, 257 inverse of, 305–307 Square root method, quadratic equations and, 109–110 Square root property, 109–110 Standard form of a complex number, 71 of complex numbers, 821–822 of equation of a hyperbola, 1029–1030, 1034 of equation of a circle, 202–204, 205 of equation of a straight line, 186 of polynomials, 27 quadratic functions in, graphing, 333–336

Young_AT_6160_SE_Index_pp1261-1276.indd 1274

Standard position, angles in, 538–541 Step functions, 266 Straight angles, 495 Stretching, vertical and horizontal, of graphs, 280–282 Strict inequalities, 129 Subsets, 4 Substitution evaluating functions by, 243 solving systems of linear equations in two variables using, 867–869 solving systems of nonlinear equations using, 1044–1045 u-substitution, 123–125 Substitution principle, evaluating algebraic expressions using, 8 Subtraction. See also Difference(s) of complex numbers, 71 of functions, 287–288 of matrices, 952–953 order of operations and, 6–9 of polynomials, 28 of rational expressions, 51–53 of real numbers, 9–10 Sum(s). See also Addition of an arithmetic sequence, 1111–1114 evaluating functions with, 245 of an infinite geometric series, 1121 nth partial sum, 1102 of two cubes, factoring, 38 Sum function, 287–288 Sum identities, trigonometric, 675–685 Summation notation, 1102–1103 Sum-to-product identities, 710–712 Supplementary angles, 496 Surplus, consumer and producer, 909–911 Symmetry of graphs, 177–179 as graphing aid, 180–181 tests for, 178–179 types of, 178 Synthetic division, of polynomials, 367–369 Systems of linear equations Cramer’s rule for solving, 981–985 in four variables, Gauss-Jordan elimination to solve, 937–938 matrices to write, 930 solving using matrix algebra and inverses of square matrices, 970–972 in three variables. See Systems of linear equations in three variables in two variables. See Systems of linear equations in two variables writing as matrix equations, 964–965 Systems of linear equations in three variables, 881–888 combining elimination and substitution methods to solve, 882–883 Cramer’s rule for solving, 982–985 dependent, 884–885 Gauss-Jordan elimination to solve, 936–937 inconsistent, 885–886 modeling with, 887–888 solving, 881–884

Systems of linear equations in two variables, 866–878 applications involving, 876–877 Cramer’s rule for solving, 981–982 dependent, 866 elimination method for solving, 869–872 Gaussian elimination with back-substitution to solve, 934 graphing method for solving, 873–875 identifying method to use for solving, 875–876 inconsistent, 866 independent, 866 solving, 866–875 substitution method for solving, 867–869 Systems of linear inequalities in two variables, 905–911 applications involving, 909–911 graphing, 906–908 solution set of, 905 Systems of nonlinear equations, 1039–1047 applications involving, 1045–1046 solving using elimination, 1040–1043 solving using substitution, 1044–1045 Systems of nonlinear inequalities, 1051–1054 graphs of, 1051–1052 solving, 1052–1054 Tangent function, 513 graphing, 623–625, 628, 629–630 inverse, 723–725 sum and difference identities for, 683–685 Technology use. See Excel 2010; TI-83+/ TI-84 calculators Term(s) of an algebraic expression, 8 of a binomial expansion, finding, 1138 constant, of a polynomial, 27 general. See General terms like, combining, 28 lowest. See Lowest terms; Simplification; Substitution of a polynomial, 27 of a sequence, 1098–1099 Terminal point, of a vector, 798 Terminal ray (side) of an angle, 494 Test intervals, on real number line, 139 30° angles evaluating trigonometric functions exactly for, 517–518, 519 in standard position, 540 30°-60°-90° triangles, 501–502 TI-83+/TI-84 calculators computing correlation coefficients using, 216–217 creating scatterplots using, 211 finding line of best fit using, 221–222 Transformations graphing logarithmic functions using, 446 of power functions, graphing polynomial functions using, 351–352 Transverse axis, of a hyperbola, 1028

10/12/16 4:24 PM

Subject Index 

Triangles, 497–505 angle sum of, 497 area of, 791–794 area of, formula for, 96 congruent, 503 equilateral, 497 finding angles of, 497 isosceles, 497 oblique. See Oblique triangles perimeter of, formula for, 96 right. See Right triangle(s); Right triangle ratios similar, 502–505 Trichotomy property, of real numbers, 129 Trigonometric equations, 737–749 applications involving, 747–748 linear, solving, 741 quadratic, solving, 741–742 requiring the use of inverse functions, solving, 742–744 solving by inspection, 737–740 solving using algebraic techniques, 741–742 solving using trigonometric identities, 744–748 Trigonometric form, of a complex number, 822–825 Trigonometric functions, 492–651. See also Cosine function; Sine function; Tangent function algebraic signs of, 552–554 angles and degree measure and, 494–496 angles in standard position and, 538–541 angular speed and, 576–578 arc length and, 572–573 area of a circular sector and, 573–575 cartesian plane and, 543–548 cofunctions and, 515–516 coterminal angles and, 541–542 evaluating exactly for special angles measures, 517–519 evaluating using calculators, 519–520 graphing, 623–636 graphs of sinusoidal functions and, 594–608 harmonic motion and, 609–614 inverse, 716–732 linear speed and, 575–576, 577–578 of nonacute angles, 552–563 radian measure and, 567–579 ranges of, 555–556 reciprocal identities and, 513–515 reference angles and reference right triangles and, 556–559 right triangle ratios and, 511–513 setting up as ratios, 515 solving right triangles and, 525–532 sums, of graphing, 614–616

Young_AT_6160_SE_Index_pp1261-1276.indd 1275

translations of, 633–636 triangles and, 497–505 unit circle approach and, 584–589 Trigonometric identities, 654–661 basic, 665–666 cofunction, 680 double-angle, 689–694 even-odd, 666 half-angle, 697–704 product-to-sum, 708–709 Pythagorean, 657–660, 666 quotient, 656–657, 666 reciprocal, 654–656, 665 solving trigonometric equations using, 744–748 sum and difference, 675–685 sum-to-product, 710–712 verifying, 668–671 Trigonometric ratios, 512. See also Trigonometric functions Trinomials, 27 factoring as product of two binomials, 39–42 Truncation, of decimals, 6 Unbounded graphs, 911 Union of sets, 131 of events, probability of, 1154–1156 Unit circle, 202, 584–589 Unit step function, 266 Unit vectors, 803–804 Upper bound rule, for real zeros, 381 Upper/lower bound rules, for real zeros, 381 u-substitution, 123–125 Variables, 8 dependent (response), 209, 240 independent (predictor), 209, 240 Variation, 312–317 combined, 315, 316 constant of, 312, 313, 314 direct, 312–314 inverse, 314–315 joint, 315, 316 Vectors, 798–808 addition of, 799, 801–802 algebraic interpretation of, 799–800 algebraic properties of, 802 angle between, 814–818 components of, 799 direction angle of, 800–801 direction of, 798–801 equal, 798–799, 801 force, 804 geometric interpretation of, 798–799 horizontal and vertical components of, 803 magnitude of, 798–801 multiplication of, 813–818

1275

parallel, 815 perpendicular (orthogonal), 815–816 position, 799 resultant, 804–808 scalar multiplication of, 802 in standard position, 799 unit, 803–804 velocity, 804 work problems and, 816–818 Velocity actual, 804 apparent, 804 current, 804 defined, 575 resultant, 804 Velocity vectors, 804 Vertex(ices) of an ellipse, 1015 in graphs, 911 of a hyperbola, 1028 of a parabola, 1003 of a quadratic function, 333 Vertical asymptotes, of rational functions, 398, 399–400 Vertical components, of vectors, 803 Vertical line(s), finding equation of, 193 Vertical line test, 241–242 Vertical shifts graphing exponential functions using, 429–430 graphing logarithmic functions using, 445 Whole numbers, 4, 5 Word problems, procedure for solving, 93–95 Work problems, vectors and, 816–818 x-axis, 166 x-coordinate, 166 x-intercepts, 176–177, 186–188 y-axis, 166 y-coordinate, 166 y-intercepts, 176–177, 186–188 Zero, properties of, 12 Zero(s) complex, of a polynomial. See Complex zeros multiplicity of. See Multiplicity of a zero of a polynomial, 139. See also Complex zeros; Real zeros real. See Real zeros Zero-exponent property, 18–19 Zero matrices, 953–954 Zero product property, 12 factoring quadratic equations and, 107–109 Zero row, in Pascal’s triangle, 1137

10/12/16 4:24 PM

A rithmetic O perations

ab 1 ac 5 a 1 b 1 c 2 b ab aa b 5 c c

a c ad 1 bc 1 5 b d bd

a1b a b 5 1 c c c

a2b b2a 5 c2d d2c

ab 1 ac 5 b 1 c, a 2 0 a

a a b b ad 5 bc c a b d

a x ax a b 5 x b b

"am 5 am/n 5 a"ab

a ac 5 b b a b c

a a b b a 5 c bc

E xponents and R adicals a0 5 1, a 2 0 a2x 5

1 ax

ax 5 ax2y ay 1 ax 2 y 5 axy

1 ab 2 x 5 axbx

axay 5 ax1y

n

n

"a 5 a1/2

n

x if x $ 0 2x if x , 0 2. If 0 x 0 5 c, then x 5 c or x 5 2c. 1 c . 0 2

1. Difference of two squares:

4. If 0 x 0 . c, then x , 2c or x . c. 1 c . 0 2 P roperties of L ogarithms 1. logb 1 MN 2 5 logb M 1 logb N

M 2. logb a b 5 logb M 2 logb N N

n

n

a "a a b 5 n Å b "b

n

"a 5 a1/n

S pecial F actorizations

3. If 0 x 0 , c, then 2c , x , c. 1 c . 0 2

n

m

"ab 5 "a"b

A bsolute V alue 1. 0 x 0 5 e

n

A2 2 B2 5 1 A 1 B 2 1 A 2 B 2



2. Perfect square trinomials:

A2 1 2AB 1 B2 5 1 A 1 B 2 2 A2 2 2AB 1 B2 5 1 A 2 B 2 2

3. Sum of two cubes:

A3 1 B3 5 1 A 1 B 2 1 A2 2 AB 1 B2 2

4. Difference of two cubes:

A3 2 B3 5 1 A 2 B 2 1 A2 1 AB 1 B2 2

3. logb M p 5 p logb M 4. logb M 5

loga M log M ln M 5 5 loga b ln b log b

5. logb bx 5 x; ln ex 5 x 6. blogb x 5 x; eln x 5 x, x . 0

Young_AT_6160_FEP-BEP_SE_pp1286-1292.indd 1286

13/12/16 9:27 pm

F ormulas /E quations The distance from 1 x1, y1 2 to 1 x2, y2 2 is " 1 x2 2 x1 2 2 1 1 y2 2 y1 2 2 .

Distance Formula Midpoint Formula Standard Equation of a Circle

The midpoint of the line segment with endpoints 1 x1, y1 2 and 1 x2, y2 2 is a

x1 1 x2 y1 1 y2 , b. 2 2

The standard equation of a circle of radius r with center at 1 h, k 2 is 1 x 2 h 2 2 1 1 y 2 k 2 2 5 r 2. The slope m of the line containing the points 1 x1, y1 2 and 1 x2, y2 2 is change in y y2 2 y1 slope 1 m 2 5 5    1 x 2 x2 2 x2 2 x1 1 change in x m is undefined  if x1 5 x2.

Slope Formula

The equation of a line with slope m and y-intercept 10, b2 is y 5 mx 1 b.

Slope-Intercept Equation of a Line Point-Slope Equation of a Line Quadratic Formula

The equation of a line with slope m containing the point 1 x1, y1 2 is y 2 y1 5 m 1 x 2 x1 2 . The solutions of the equation ax 2 1 bx 1 c 5 0, a 2 0, are x 5

2b 6 "b2 2 4ac . 2a

If b2 2 4ac . 0, there are two unequal real solutions. If b2 2 4ac 5 0, there is a repeated real solution. If b2 2 4ac , 0, there are two complex solutions that are not real.

F unctions Constant Function Linear Function Quadratic Function Polynomial Function Rational Function Exponential Function Logarithmic Function

ƒ1x2 5 b

ƒ 1 x 2 5 mx 1 b, where m is the slope and b is the y-intercept

ƒ 1 x 2 5 ax 2 1 bx 1 c, a 2 0 or ƒ 1 x 2 5 a 1 x 2 h 2 2 1 k parabola vertex 1 h, k 2 ƒ 1 x 2 5 anx n 1 an21x n21 1 c1 a1x 1 a0 R1x2 5

anx n 1 an21x n21 1 c1 a1x 1 a0 n1x2 5 d1x2 bmx m 1 am21x m21 1 c1 b1x 1 b0

ƒ 1 x 2 5 bx, b . 0, b 2 1

ƒ 1 x 2 5 logb x, b . 0, b 2 1

T ransformations In each case, c represents a positive real number. Function Vertical translations Horizontal translations Reflections

Draw the graph of f and: y 5 ƒ1x2 1 c e y 5 ƒ1x2 2 c

Shift f upward c units. Shift f downward c units.

y 5 2ƒ 1 x 2 e y 5 ƒ 1 2x 2

Reflect f about the x-axis. Reflect f about the y-axis.

y 5 ƒ1x 2 c2 e y 5 ƒ1x 1 c2

Young_AT_6160_FEP-BEP_SE_pp1286-1292.indd 1287

Shift f to the right c units. Shift f to the left c units.

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C onic S ections

S equences Parabola

y

y2 =

1. Infinite Sequence: y

4px



x 2 = 4 py Vertex

Focus (p, 0)

2. Summation Notation:

Focus (0, p) Directrix y = p

i51

x

3. nth Term of an Arithmetic Sequence:

Vertex

an 5 a1 1 1 n 2 1 2 d

4. Sum of First n Terms of an Arithmetic Sequence:

Ellipse y

y

Vertex (0, a)

y2

x2 + =1 a2 b2

Foci Vertex (a, 0)

x

Sn 5

x2 + y2 = 1 b2 a2

Vertex (0, a)

y Vertex (0, a) x

Focus Vertex (a, 0)

x2 y2 = 1  a2 b2

Vertex (0, a)

T he B inomial T heorem 1. n! 5 n 1 n 2 1 2 1 n 2 2 2 c3 ⋅ 2 ⋅ 1; 1! 5 1; 0! 5 1 n n! 2. a b 5 r r! 1 n 2 r 2 !

3. Binomial theorem:

n n 1 a 1 b 2 n 5 a b an 1 a b an21b 0 1 n n 1 a b an22b2 1 c1 a b bn 2 n

Young_AT_6160_FEP-BEP_SE_pp1286-1292.indd 1288

S5 y2 x 2 =1  a2 b2

a1 12r

Focus Center

Focus

a1 1 1 2 r n 2 1r 2 12 12r

7. Sum of an Infinite Geometric Series with  r  , 1:

Hyperbola

Vertex (a, 0)

an 5 a1r n21 6. Sum of First n Terms of a Geometric Sequence:

Center

y

Center

n 1 a 1 an 2 2 1

5. nth Term of a Geometric Sequence:

Center Foci

Sn 5

Major axis

x

Major axis

c1 an a ai 5 a1 1 a2 1 a3 1 n

x

Directrix x = p

Vertex (a, 0)

5 an 6 5 a1, a2, a3, c, an, c

x

Focus

P ermutations , C ombinations , and P robability 1. nPr , the number of permutations of n elements taken r at a time, is given by n! nPr 5 . 1n 2 r2!

2. nCr , the number of combinations of n elements taken r at a time, is given by n! nCr 5 . 1 n 2 r 2 !r! n1E2 3. Probability of an Event: P 1 E 2 5 , where n1S2 n 1 E 2 5 the number of outcomes in event E and n 1 S 2 5 the number of outcomes in the sample space.

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1289

3.1  

T rigonometric F unctions in the C artesian P lane

R ight T riangle T rigonometry  sin u 5

opp hyp

cos u 5

hyp opp

csc u 5

adj hyp

hyp adj

sec u 5

opp  tan u 5 adj

Hypotenuse c θ

adj cot u 5               opp

y sin u 5 r x cos u 5 r y tan u 5 x

b Opposite

a Adjacent

r y r sec u 5 x x cot u 5 y

y

csc u 5

(x, y)

r

y

θ

x x

E xact V alues of T rigonometric F unctions x degrees

x radians

sin x

cos x

tan x

0°

0

0

1

0

30° 45°

p 6

1 2

p 4

!2 2

90°

!2 2

!3 2

p 3 p 2

60°

!3 2

1 2

1

0

A ngle M easurement p radians 5 180°

!3 3

s

r

s 5 ru  A 5 12r 2u  1 u in radians2

θ

To convert from radians to degrees, multiply by

180° . p

r

To convert from degrees to p radians, multiply by . 180°

1 !3 —

O blique T riangle Law of Sines In any triangle, sin b sin g sin a 5 5 . a c b

S pecial R ight T riangles

30º 45º 1

45º

c

β

a2 5 b2 1 c2 2 2bc cos a b2 5 a2 1 c2 2 2ac cos b c2 5 a2 1 b2 2 2ab cos g

60º 1

1

α

a

Law of Cosines

2

√3

√2

b

γ

C ircular F unctions ( cos  u , sin  u )

(

)

2 2 –√ , √ 2

2

(0, 1)

y

90º π 2

3π 4

45º

135º

(

2 2 –√ , – √ 2 2

)

5π 4

2

2

0 225º

3π

270º 2

(0, –1)

Young_AT_6160_FEP-BEP_SE_pp1286-1292.indd 1289

(

2

,–

2

(

)

) (

(

)

(

(

– 1 , √3 2 2

)

x 0º 0 360º 2π (1, 0) 315º 7π 4 √2 √2

(–1, 0) π 180º

π 4

(

√2 , √2

)

)

(0, 1)

y

(

1 √3 2, 2

90º π 2 π 2π √3 , 1 3 1 3 3 120º –√ , 2 60º π 2 2 5π 2 30º 6 6 150º x (–1, 0) 0º 0 π 180º 360º 2π (1, 0) 0 330º 11π 7π 210º √3 , – 1 3 6 240º 300º 6 –√ , – 1 5π 4π 2 2 2 2 3π 3 3 270º 2

(

3 – 1, – √ 2 2

)

(0, –1)

(

1 √3 2,– 2

)

)

)

13/12/16 9:27 pm

1290 

CHAPTER 3  

I dentities Reciprocal Identities 1 1 csc x 5 sec x 5 cos x sin x

cot x 5

1 tan x

Quotient Identities sin x cos x tan x 5 cot x 5 cos x sin x Sum Identities sin 1 x 1 y 2 5 sin x cos y 1 cos x sin y cos 1 x 1 y 2 5 cos x cos y 2 sin x sin y tan x 1 tan y tan 1 x 1 y 2 5 1 2 tan x tan y

Difference Identities sin 1 x 2 y 2 5 sin x cos y 2 cos x sin y

cos 1 x 2 y 2 5 cos x cos y 1 sin x sin y tan x 2 tan y tan 1 x 2 y 2 5 1 1 tan x tan y Double-Angle Identities sin 1 2x 2 5 2 sin x cos x 2

2

cos x 2 sin x cos 1 2x 2 5 • 1 2 2 sin2 x 2 cos2 x 2 1 2 tan x 2 cot x 2 tan 1 2x 2 5 5 5 2 2 cot x 2 tan x 1 2 tan x cot x 2 1 Half-Angle Identities

x 1 2 cos x     sin a b 5 6 2 Å 2 x 1 1 cos x cos a b 5 6 2 Å 2

Sign 1 1/2 2 is determined by quadrant in which x /2 lies

x 1 2 cos x sin x 1 2 cos x    tan a b 5 5 56 2 sin x 1 1 cos x Å 1 1 cos x Identities for Reducing Powers 1 2 cos 1 2x 2 1 1 cos 1 2x 2 sin2 x 5     cos2 x 5 2 2 1 2 cos 1 2x 2 tan2 x 5 1 1 cos 1 2x 2

Young_AT_6160_FEP-BEP_SE_pp1286-1292.indd 1290

Identities for Negatives sin 1 2x 2 5 2sinx tan 1 2x 2 5 2tanx

cos 1 2x 2 5 cos x

Pythagorean Identities sin2 x 1 cos2 x 5 1 1 1 cot2 x 5 csc2 x

tan2 x 1 1 5 sec2 x

Cofunction Identities p aReplace with 90° if x is in degree measure.b 2 p p sina 2 xb 5 cos x cos a 2 xb 5 sin x 2 2 tana

p 2 xb 5 cot x 2

cot a

p   seca 2 xb 5 csc x 2

csca

p 2 xb 5 tan x 2

p 2 xb 5 sec x 2

Product-to-Sum Identities sin x cos y 5 12 3 sin 1 x 1 y 2 1 sin 1 x 2 y 2 4

cos x sin y 5 12 3 sin 1 x 1 y 2 2 sin 1 x 2 y 2 4

sin x sin y 5 12 3 cos 1 x 2 y 2 2 cos 1 x 1 y 2 4

cos x cos y 5 12 3 cos 1 x 1 y 2 1 cos 1 x 2 y 2 4

Sum-to-Product Identities x1y x2y sin x 1 sin y 5 2 sin a b cos a b 2 2 sin x 2 sin y 5 2 cos a

x1y x2y b sin a b 2 2

cos x 1 cos y 5 2 cos a

x1y x2y b cos a b 2 2

cos x 2 cos y 5 22 sin a

x1y x2y b sin a b 2 2

13/12/16 9:27 pm

3.1  

1291

G raphs of the T rigonometric F unctions y

y = sinx

y

1

y = cosx

y

y = tanx

5

1

3 x –2π

π

–π

1

x

2π

–2π

π

–π

2π

–2π

x π

–π –1

2π

–3 –1

–1

y

y = cscx

y

y = secx

y

5

5

5

3

3

3

1 –2π

–5

–π –1

1

x π

2π

–2π

–π –1

1

x π

y = cotx

2π

–2π

–π –1

–3

–3

–3

–5

–5

–5

x π

2π

A mplitude , P eriod , and P hase S hift

P olar C oordinates

y 5 A sin 1Bx 1 C 2

y x 5 r cos u y 5 r sin u (r, θ) 2 (x, y) r  5 x 2 1 y2 r y y tan u 5 x θ x x

y 5 A cos 1Bx 1 C 2 2p Amplitude 5  A   Period 5 B

C .0 C B Phase shift 5 µ B C right    if , 0 B left      if

y 5 A tan 1Bx 1 C 2

y 5 A cot 1Bx 1 C 2 p Period 5 B

C Phase shift 5 µ B

C .0 B C right    if , 0 B

Complex Numbers x 1 iy 5 r  1cos u 1 i sin u2 Rectangular Trigonometric form (polar) form

left      if

P owers and R oots of C omplex N umbers

H eron ’ s F ormula A rea

for

  zn 5 1x 1 iy 2 n    5 3r  1cos u 1 i sin u24 n    5 r n  3cos 1nu2 1 i sin 1nu24    n 5 1, 2, . . . n ! z 5 1x 1 iy 2 1/n    5 3r  1cos u 1 i sin u24 1/n u 1 2kp u 1 2kp    5 r 1/n c cos a b 1 i sina bd n n   k 5 0, 1, 2, . . . , n 2 1

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If the semiperimeter s of a triangle is a1b1c s5 2 then the area of that triangle is A 5 "s 1 s 2 a 2 1 s 2 b 2 1 s 2 c 2

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1292 

CHAPTER 3  

I nverse T rigonometric F unctions p p #y# 2 2

y 5 sin21 x

x 5 sin y

2

y 5 cos21 x

   x 5 cos y

y 5 tan21 x

 x 5 tan y

0#y#p p p 2 ,y, 2 2

y 5 cot21 x

 x 5 cot y

y 5 sec21 x

  x 5 sec y

  0 # y # p, y 2

y 5 csc21 x

  x 5 csc y

2

21 # x # 1 21 # x # 1 x is any real number x is any real number

0,y,p p 2

x # 21 or x $ 1

p p #y# ,y20 2 2

x # 21 or x $ 1

G raphs of the I nverse T rigonometric F unctions y = sin–1x or y = arc sin x π 2 π 4 –1

–0.5

–π 4 π – 2

y π

1

or

π y = arctan x 2 π 4 –3

π 4 –1

–0.5

–2

–1

x 0.5

y

y = tan–1 x

3π 4

x 0.5

y y = cos–1 x or y = arccos x

1

–π 4 π – 2

x 1

2

3

V ectors B

A

→

Vector v = AB

Vector Addition

For vectors u 5 8a, b9 and v 5 8c, d9, and real number k,

u 5 ai 1 bj



 u  5 "a2 1 b2

u

+

v v

u

v

+

v



u

u

Scalar Multiplication



ku

(0, 1)

u 1 v 5 8a 1 c, b 1 d9

j

x

ku 5 8ka, kb9

u # v 5 ac 1 bd u#v cos u 5  u  v 

Compv u 5  u cos u 5 ku

y

u#v  u

i

(1, 0)

u θ v

u (k > 0)

(k < 0)

Young_AT_6160_FEP-BEP_SE_pp1286-1292.indd 1292

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