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Table of contents :
Cover
A Brief Guide to Getting the Most from This Book
1 Read the Book
2 Work the Problems
3 Review for Quizzes and Tests
Title Page
Copyright
Contents
Preface
To the Student
Pearson’s Commitment to Diversity, Equity, and Inclusion
Applications Index
Chapter P. Fundamental Concepts of Algebra
P.1 Algebraic Expressions, Mathematical Models, and Real Numbers
P.2 Exponents and Scientific Notation
P.3 Radicals and Rational Exponents
P.4 Polynomials
MidChapter Check Point
P.5 Factoring Polynomials
P.6 Rational Expressions
Summary, Review, and Test
Review Exercises
Chapter P Test
Chapter 1. Equations and Inequalities
1.1 Graphs and Graphing Utilities
1.2 Linear Equations and Rational Equations
1.3 Models and Applications
1.4 Complex Numbers
1.5 Quadratic Equations
MidChapter Check Point
1.6 Other Types of Equations
1.7 Linear Inequalities and Absolute Value Inequalities
Summary, Review, and Test
Review Exercises
Chapter 1 Test
Chapter 2. Functions and Graphs
2.1 Basics of Functions and Their Graphs
2.2 More on Functions and Their Graphs
2.3 Linear Functions and Slope
2.4 More on Slope
MidChapter Check Point
2.5 Transformations of Functions
2.6 Combinations of Functions; Composite Functions
2.7 Inverse Functions
2.8 Distance and Midpoint Formulas; Circles
Summary, Review, and Test
Review Exercises
Chapter 2 Test
Cumulative Review Exercises (Chapters 1–2)
Chapter 3. Polynomial and Rational Functions
3.1 Quadratic Functions
3.2 Polynomial Functions and Their Graphs
3.3 Dividing Polynomials; Remainder and Factor Theorems
3.4 Zeros of Polynomial Functions
MidChapter Check Point
3.5 Rational Functions and Their Graphs
3.6 Polynomial and Rational Inequalities
3.7 Modeling Using Variation
Summary, Review, and Test
Review Exercises
Chapter 3 Test
Cumulative Review Exercises (Chapters 1–3)
Chapter 4. Exponential and Logarithmic Functions
4.1 Exponential Functions
4.2 Logarithmic Functions
4.3 Properties of Logarithms
MidChapter Check Point
4.4 Exponential and Logarithmic Equations
4.5 Exponential Growth and Decay; Modeling Data
Summary, Review, and Test
Review Exercises
Chapter 4 Test
Cumulative Review Exercises (Chapters 1–4)
Chapter 5. Trigonometric Functions
5.1 Angles and Radian Measure
5.2 Right Triangle Trigonometry
5.3 Trigonometric Functions of Any Angle
5.4 Trigonometric Functions of Real Numbers; Periodic Functions
MidChapter Check Point
5.5 Graphs of Sine and Cosine Functions
5.6 Graphs of Other Trigonometric Functions
5.7 Inverse Trigonometric Functions
5.8 Applications of Trigonometric Functions
Summary, Review, and Test
Review Exercises
Chapter 5 Test
Cumulative Review Exercises (Chapters 1–5)
Chapter 6. Analytic Trigonometry
6.1 Verifying Trigonometric Identities
6.2 Sum and Difference Formulas
6.3 DoubleAngle, PowerReducing, and HalfAngle Formulas
MidChapter Check Point
6.4 ProducttoSum and SumtoProduct Formulas
6.5 Trigonometric Equations
Summary, Review, and Test
Review Exercises
Chapter 6 Test
Cumulative Review Exercises (Chapters 1–6)
Chapter 7. Additional Topics in Trigonometry
7.1 The Law of Sines
7.2 The Law of Cosines
7.3 Polar Coordinates
7.4 Graphs of Polar Equations
MidChapter Check Point
7.5 Complex Numbers in Polar Form; DeMoivre’s Theorem
7.6 Vectors
7.7 The Dot Product
Summary, Review, and Test
Review Exercises
Chapter 7 Test
Cumulative Review Exercises (Chapters 1–7)
Chapter 8. Systems of Equations and Inequalities
8.1 Systems of Linear Equations in Two Variables
8.2 Systems of Linear Equations in Three Variables
8.3 Partial Fractions
8.4 Systems of Nonlinear Equations in Two Variables
MidChapter Check Point
8.5 Systems of Inequalities
8.6 Linear Programming
Summary, Review, and Test
Review Exercises
Chapter 8 Test
Cumulative Review Exercises (Chapters 1–8)
Chapter 9. Matrices and Determinants
9.1 Matrix Solutions to Linear Systems
9.2 Inconsistent and Dependent Systems and Their Applications
9.3 Matrix Operations and Their Applications
MidChapter Check Point
9.4 Multiplicative Inverses of Matrices and Matrix Equations
9.5 Determinants and Cramer’s Rule
Summary, Review, and Test
Review Exercises
Chapter 9 Test
Cumulative Review Exercises (Chapters 1–9)
Chapter 10. Conic Sections and Analytic Geometry
10.1 The Ellipse
10.2 The Hyperbola
10.3 The Parabola
MidChapter Check Point
10.4 Rotation of Axes
10.5 Parametric Equations
10.6 Conic Sections in Polar Coordinates
Summary, Review, and Test
Review Exercises
Chapter 10 Test
Cumulative Review Exercises (Chapters 1–10)
Chapter 11. Sequences, Induction, and Probability
11.1 Sequences and Summation Notation
11.2 Arithmetic Sequences
11.3 Geometric Sequences and Series
MidChapter Check Point
11.4 Mathematical Induction
11.5 The Binomial Theorem
11.6 Counting Principles, Permutations, and Combinations
11.7 Probability
Summary, Review, and Test
Review Exercises
Chapter 11 Test
Cumulative Review Exercises (Chapters 1–11)
Appendix: Where Did That Come From? Selected Proofs
Answers to Selected Exercises
Subject Index
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T
U
V
W
X
Y
Z
Credits
Definitions, Rules, and Formulas
GLOBAL EDITION
Algebra and Trigonometry SEVENTH EDITION
ROBERT F. BLITZER
A Brief Guide to Getting the Most from This Book 1 Read the Book Feature SectionOpening Scenarios
Description
Benefit
Every section opens with a scenario presenting a unique application of algebra or trigonometry in your life outside the classroom.
Realizing that algebra and trigonometry are everywhere will help motivate your learning. (See page 136.)
Examples are clearly written and provide stepbystep solutions. No steps are omitted, and each step is thoroughly explained to the right of the mathematics.
The blue annotations will help you understand the solutions by providing the reason why every algebraic or trigonometric step is true. (See page 138.)
Interesting applications from nearly every discipline, supported by uptodate realworld data, are included in every section.
Ever wondered how you’ll use algebra and trigonometry? This feature will show you how they can solve real problems. (See pages 294–296.)
Answers to students’ questions offer suggestions for problem solving, point out common errors to avoid, and provide informal hints and suggestions.
By seeing common mistakes, you’ll be able to avoid them. This feature should help you not to feel anxious or threatened when asking questions in class. (See page 139.)
BRIEF REVIEW
Brief Reviews cover skills you already learned but may have forgotten.
Having these refresher boxes easily accessible will help ease anxiety about skills you may have forgotten. (See page 508.)
BLITZER BONUS
Yet even more proof that math is an These enrichment essays provide interesting and dynamic discipline! historical, interdisciplinary, and otherwise interesting connections to the (See page 195.) algebra or trigonometry under study.
EXAMPLE
Applications Using RealWorld Data
GREAT QUESTION
Voice balloons help to demystify algebra Does math ever look foreign to you? This feature often translates math into everyday and trigonometry. They translate math English. (See page 193.) into plain English, clarify problemsolving procedures, and present alternative ways of understanding.
WHAT YOU'LL LEARN 1 Learning
Objective
TECHNOLOGY
Every section begins with a list of objectives. Each objective is restated in the margin where the objective is covered.
The objectives focus your reading by emphasizing what is most important and where to find it. (See page 178.)
The screens displayed in the technology boxes show how graphing utilities verify and visualize algebraic or trigonometric results.
Even if you are not using a graphing utility in the course, this feature will help you understand different approaches to problem solving. (See page 140.)
2 Work the Problems Feature
Description
Benefit
Each example is followed by a matched problem, called a Check Point, that offers you the opportunity to work a similar exercise. The answers to the Check Points are provided in the answer section.
You learn best by doing. You’ll solidify your understanding of worked examples if you try a similar problem right away to be sure you understand what you’ve just read. (See page 138.)
Achieving Success boxes offer strategies for persistence and success in college mathematics courses.
Follow these suggestions to help achieve your full academic potential in college mathematics. (See page 149.)
These shortanswer questions, mainly fillintheblank and true/false items, assess your understanding of the definitions and concepts presented in each section.
It is difficult to learn algebra and trigonometry without knowing their special language. These exercises test your understanding of the vocabulary and concepts. (See page 149.)
An abundant collection of exercises is included in an Exercise Set at the end of each section. Exercises are organized within several categories. Your instructor will usually provide guidance on which exercises to work. The exercises in the first category, Practice Exercises, follow the same order as the section’s worked examples.
The parallel order of the Practice Exercises lets you refer to the worked examples and use them as models for solving these problems. (See page 150.)
Practice PLUS
This category of exercises contains more challenging problems that often require you to combine several skills or concepts.
It is important to dig in and develop your problemsolving skills. Practice PLUS Exercises provide you with ample opportunity to do so. (See page 437.)
Retaining the Concepts
Beginning with Chapter 2, each Exercise Set contains review exercises under the header “Retaining the Concepts.”
These exercises improve your understanding of the topics and help maintain mastery of the material. (See page 284.)
Preview Exercises
Each Exercise Set concludes with three These exercises let you review previously covered material that you’ll need to be problems to help you prepare for the successful for the forthcoming section. next section. Some of these problems will get you thinking about concepts you’ll soon encounter. (See page 153.)
CHECK POINT
ACHIEVING SUCCESS
CONCEPT AND VOCABULARY CHECK
EXERCISE SET
3 Review for Quizzes and Tests Feature
Description
Benefit
At approximately the midway point in the chapter, an integrated set of review exercises allows you to review the skills and concepts you learned separately over several sections.
By combining exercises from the first half of the chapter, the MidChapter Check Points give a comprehensive review before you move on to the material in the remainder of the chapter. (See page 311.)
Each chapter contains a review chart that summarizes the definitions and concepts in every section of the chapter. Examples that illustrate these key concepts are also referenced in the chart.
Review this chart and you’ll know the most important material in the chapter! (See page 236.)
A comprehensive collection of review exercises for each of the chapter’s sections follows the review chart.
Practice makes perfect. These exercises contain the most significant problems for each of the chapter’s sections. (See page 239.)
Each chapter contains a practice test with approximately 25 problems that cover the important concepts in the chapter. Take the practice test, check your answers, and then watch the Chapter Test Prep Videos to see workedout solutions for any exercises you miss.
You can use the chapter test to determine whether you have mastered the material covered in the chapter. (See page 243.)
Chapter Test Prep Videos
These videos contain workedout solutions to every exercise in each chapter test and can be found in MyLab Math and on YouTube at youtube.com/user/pearsonmathstats (playlist “Blitzer Algebra and Trigonometry 7e”).
The videos let you review any exercises you miss on the chapter test.
Objective Videos
These fresh, interactive videos walk you through the concepts from every objective of the text.
The videos provide you with active learning at your own pace.
Beginning with Chapter 2, each chapter concludes with a comprehensive collection of mixed cumulative review exercises. These exercises combine problems from previous chapters and the present chapter, providing an ongoing cumulative review.
Ever forget what you’ve learned? These exercises ensure that you are not forgetting anything as you move forward. (See page 373.)
MidChapter Check Point
Chapter Review Chart
Chapter Review
Exercise Set
Chapter Test
Cumulative Review
7th EDITION GLOBAL EDITION
Algebra and Trigonometry Robert Blitzer Miami Dade College
Product Management: Gargi Banerjee and K. K. Neelakantan Content Strategy: Shabnam Dohutia, Amrita Naskar, and Shahana Bhattacharya Product Marketing: Wendy Gordon, Ashish Jain, and Ellen Harris Supplements: Bedasree Das Production and Digital Studio: Vikram Medepalli, Naina Singh, and Niharika Thapa Rights and Permissions: Rimpy Sharma and Nilofar Jahan Please contact https://support.pearson.com/getsupport/s/ with any queries on this content. Cover Image: Shutterstock/iconstudio
Pearson Education Limited KAO Two KAO Park Hockham Way Harlow CM17 9SR United Kingdom and Associated Companies throughout the world
Visit us on the World Wide Web at: www.pearsonglobaleditions.com © Pearson Education Limited 2023 The rights of Robert Blitzer to be identified as the author of this work have been asserted by him in accordance with the Copyright, Designs and Patents Act 1988.
Authorized adaptation from the United States edition, entitled Algebra and Trigonometry, 7th Edition, ISBN 9780136922179 by Robert Blitzer published by Pearson Education © 2022. 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 or otherwise, without either the prior written permission of the publisher or a license permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS. For information regarding permissions, request forms and the appropriate contacts within the Pearson Education Global Rights & Permissions department, please visit www.pearsoned.com/permissions/. Acknowledgments of thirdparty content appear on page 1317, which constitute an extension of this copyright page. PEARSON, ALWAYS LEARNING, and MYLAB are exclusive trademarks owned by Pearson Education, Inc. or its affiliates in the U.S. and/or other countries. Unless otherwise indicated herein, any thirdparty trademarks that may appear in this work are the property of their respective owners and any references to thirdparty trademarks, logos or other trade dress are for demonstrative or descriptive purposes only. Such references are not intended to imply any sponsorship, endorsement, authorization, or promotion of Pearson’s products by the owners of such marks, or any relationship between the owner and Pearson Education, Inc. or its affiliates, authors, licensees, or distributors. This eBook may be available as a standalone product or integrated with other Pearson digital products like MyLab and Mastering. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material in this eBook at any time. ISBN 10 (print): 1292443448 ISBN 13 (print): 9781292443447 ISBN 13 (uPDF eBook): 9781292443393 British Library CataloguinginPublication Data A catalogue record for this book is available from the British Library
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Contents Preface 9 About the Author 18 To the Student 19 Applications Index 21
P
Fundamental Concepts of Algebra 29
P.1
Algebraic Expressions, Mathematical Models, and Real Numbers 30
P.2
Exponents and Scientific Notation 48
P.3
Radicals and Rational Exponents 63
P.4
Polynomials 79
MidChapter Check Point 90 P.5
Factoring Polynomials 91
P.6
Rational Expressions 103
Summary, Review, and Test 118 Review Exercises 119 Chapter P Test 121
1
Equations and Inequalities 123
1.1
Graphs and Graphing Utilities 124
1.2
Linear Equations and Rational Equations 136
1.3
Models and Applications 154
1.4
Complex Numbers 168
1.5
Quadratic Equations 178
MidChapter Check Point 201 1.6
Other Types of Equations 203
1.7
Linear Inequalities and Absolute Value Inequalities 219
Summary, Review, and Test 236 Review Exercises 239 Chapter 1 Test 243
5
6
Contents MidChapter Check Point 440 3.5
Rational Functions and Their Graphs 441
3.6
Polynomial and Rational Inequalities 461
3.7
Modeling Using Variation 474
Summary, Review, and Test 484 Review Exercises 487 Chapter 3 Test 490 Cumulative Review Exercises (Chapters 1–3) 492
4
Exponential and Logarithmic Functions 493
4.1
Exponential Functions 494
4.2
Logarithmic Functions 508
4.3
Properties of Logarithms 523
MidChapter Check Point 533 4.4
Exponential and Logarithmic Equations 534
4.5
Exponential Growth and Decay; Modeling Data 549
Summary, Review, and Test 563
2
Functions and Graphs 245
2.1
Basics of Functions and Their Graphs 246
2.2
More on Functions and Their Graphs 265
2.3
Linear Functions and Slope 285
2.4
More on Slope 301
MidChapter Check Point 311 2.5
Transformations of Functions 312
2.6
Combinations of Functions; Composite Functions 328
2.7
Inverse Functions 343
2.8
Distance and Midpoint Formulas; Circles 354
Summary, Review, and Test 364 Review Exercises 367 Chapter 2 Test 371 Cumulative Review Exercises (Chapters 1–2) 373
3
Polynomial and Rational Functions 375
3.1
Quadratic Functions 376
3.2
Polynomial Functions and Their Graphs 394
3.3
Dividing Polynomials; Remainder and Factor Theorems 412
3.4
Zeros of Polynomial Functions 425
Review Exercises 565 Chapter 4 Test 569 Cumulative Review Exercises (Chapters 1–4) 570
Contents
5
Trigonometric Functions 571
5.1
Angles and Radian Measure 572
5.2
Right Triangle Trigonometry 589
5.3
Trigonometric Functions of Any Angle 606
5.4
Trigonometric Functions of Real Numbers; Periodic Functions 619
Graphs of Sine and Cosine Functions 629
5.6
Graphs of Other Trigonometric Functions 650
5.7
Inverse Trigonometric Functions 664
5.8
Applications of Trigonometric Functions 682
Complex Numbers in Polar Form; DeMoivre’s Theorem 813
7.6
Vectors 826
7.7
The Dot Product 841
Summary, Review, and Test 851 Review Exercises 854 Chapter 7 Test 856 Cumulative Review Exercises (Chapters 1–7) 857
MidChapter Check Point 628 5.5
7.5
Summary, Review, and Test 693 Review Exercises 697 Chapter 5 Test 699 Cumulative Review Exercises (Chapters 1–5) 700
6
Analytic Trigonometry 701
6.1
Verifying Trigonometric Identities 702
6.2
Sum and Difference Formulas 714
6.3
DoubleAngle, PowerReducing, and HalfAngle Formulas 725
MidChapter Check Point 736 6.4
ProducttoSum and SumtoProduct Formulas 737
6.5
Trigonometric Equations 746
Summary, Review, and Test 761 Review Exercises 762 Chapter 6 Test 764 Cumulative Review Exercises (Chapters 1–6) 765
7
Additional Topics in Trigonometry 767
8
Systems of Equations and Inequalities 859
8.1
Systems of Linear Equations in Two Variables 860
8.2
Systems of Linear Equations in Three Variables 879
8.3
Partial Fractions 887
8.4
Systems of Nonlinear Equations in Two Variables 898
7.1
The Law of Sines 768
MidChapter Check Point 908
7.2
The Law of Cosines 780
8.5
Systems of Inequalities 909
7.3
Polar Coordinates 789
8.6
Linear Programming 922
7.4
Graphs of Polar Equations 801
Summary, Review, and Test 929
MidChapter Check Point 812
7
Review Exercises 931 Chapter 8 Test 934 Cumulative Review Exercises (Chapters 1–8) 934
8
9
Contents
Matrices and Determinants 937
9.1
Matrix Solutions to Linear Systems 938
9.2
Inconsistent and Dependent Systems and Their Applications 952
9.3
Matrix Operations and Their Applications 961
MidChapter Check Point 976 9.4 9.5
Multiplicative Inverses of Matrices and Matrix Equations 977 Determinants and Cramer’s Rule 991
Summary, Review, and Test 1004 Review Exercises 1006 Chapter 9 Test 1008 Cumulative Review Exercises (Chapters 1–9) 1009
10
Conic Sections and Analytic Geometry 1011
10.1
The Ellipse 1012
10.2
The Hyperbola 1027
10.3
The Parabola 1042
MidChapter Check Point 1056 10.4
Rotation of Axes 1058
10.5
Parametric Equations 1069
10.6
Conic Sections in Polar Coordinates 1079
Summary, Review, and Test 1089 Review Exercises 1092 Chapter 10 Test 1094 Cumulative Review Exercises (Chapters 1–10) 1095
11
Sequences, Induction, and Probability 1097
11.1
Sequences and Summation Notation 1098
11.2
Arithmetic Sequences 1109
11.3
Geometric Sequences and Series 1120
MidChapter Check Point 1135
11.4
Mathematical Induction 1136
11.5
The Binomial Theorem 1145
11.6
Counting Principles, Permutations, and Combinations 1153
11.7
Probability 1164
Summary, Review, and Test 1179 Review Exercises 1181 Chapter 11 Test 1184 Cumulative Review Exercises (Chapters 1–11) 1185 Appendix: Where Did That Come From? Selected Proofs 1187 Answers to Selected Exercises 1193 Subject Index 1299 Credits 1317
Preface I’ve written Algebra and Trigonometry, Seventh Edition, to help diverse students, with different backgrounds and future goals, to succeed. The book has three fundamental goals: 1. To help students acquire a solid foundation in algebra
and trigonometry, preparing them for other courses such as calculus, business calculus, and finite mathematics. 2. To show students how algebra and trigonometry can model and solve authentic realworld problems. 3. To enable students to develop problemsolving skills, while fostering critical thinking, within an interesting setting.
textbook. This has been a regular source of frustration for me and for my colleagues in the classroom. Anecdotal evidence gathered over years highlights two basic reasons that students do not take advantage of their textbook: • “I’ll never use this information.” • “I can’t follow the explanations.” I’ve written every page of the Seventh Edition with the intent of eliminating these two objections. The ideas and tools I’ve used to do so are described for the student in “A Brief Guide to Getting the Most from This Book,” which appears in the front matter of the book.
One major obstacle in the way of achieving these goals is the fact that very few students actually read their
What’s New in the Seventh Edition? The Seventh Edition contains 97 workedout examples and exercises based on new data sets and 168 updated examples and exercises. Many of the new and updated applications involve topics relevant to college students.
New Applications • Cost and Enrollment for Federal Social Programs (Section P.2, Exercises 115–117) • Educational Attainment and Probability of Divorce (Section 1.1, Example 6) • Number of Smartphone Users in the U.S. (Section 2.1, Figure 2.2) • Spending on PrePrimary Education and Child Care (Section 2.1, Exercises 99–100) • Internet Plans (Section 2.2, Example 6 and Exercises 95–96) • Trust in Government and the Media (Section 2.3, Exercises 87–88) • Accelerating Climate Change (Blitzer Bonus in Section 2.3, p. 296) • Living Arrangements of Young Adults (Section 2.4 opener and Example 3) • U.S. Population Projections by Age (Section 2.6, Exercises 97–98) • Time Spent Online (Cumulative Review for Chapters 1–2, Exercise 21)
• Rumbling Back: Steven Spielberg’s New West Side Story (Blitzer Bonus in Section 2.7, p. 343) • Addressing Leisure Time Parabolically (Blitzer Bonus in Section 3.1, p. 389) • COVID19 Pandemic (Section 3.2 opener; Cumulative Review for Chapters 1–3, Exercise 21; Section 4.5, Example 3; Cumulative Review for Chapters 1–8, Exercise 36) • AIDS: A Global Perspective (Blitzer Bonus in Section 3.2, p. 398) • Area Burned by Wildfires in the U.S. (Section 3.2, Exercise 76) • Costco Paid Membership (Chapter 3 Review, Exercise 68) • Mumps (Chapter 3 MidChapter Check Point, Exercise 28) • Putting Off Medical Treatment Because of Expenses (Section 4.2, Exercises 115–116) • Ecommerce Sales (Cumulative Review for Chapters 1–4, Exercise 81) • The Electromagnetic Spectrum (Blitzer Bonus in Section 5.5, p. 646) • Modeling Body Temperature, Heart Rate, and Respiratory Rate (Chapter 8 opener; Section 8.5 opener, Example 5, and Exercises 77–82) • Number of Men and Women in the U.S. House of Representatives (Section 8.1, Exercise 70) 9
10
Preface
• Share of U.S. Income by Top 10% and Bottom 90% of Americans (Section 8.1, Exercise 72) • The Late Elvis Presley’s Business Machine (Section 8.2, Exercise 45) • Use of Social Media by Age (Chapter 8 Review, Exercise 8) • Online Classes vs. FacetoFace Classroom Experiences (Chapter 9 opener) • Political Party Affiliation by Generation (Section 9.1 opener) • Interracial Married Couples in the U.S. (Cumulative Review for Chapters 1–10, Exercise 21) • Probable Majors of College Freshmen (Section 11.2, Exercises 61–62) • Nonbinary Gender Options (Blitzer Bonus in Section 11.7, p. 1174) • Electrical Charging Stations (Chapter 11 Review, Exercise 28)
Updated Applications • Cost of Tuition and Fees at Public and Private Colleges (Section P.1, Example 2 and Exercises 131–132) • The National Debt (Section P.2 opener, Example 12, and Exercises 118–120) • Student Loan Debt (Chapter P MidChapter Check Point, Exercise 31) • Different Race or Ethnicity for Two Randomly Selected Americans (Chapter P Review, Exercise 23) • Inflation (Section 1.2, Exercises 109–112) • Median Earnings by Educational Attainment (Section 1.3, Example 1) • Attitudes of College Freshmen (Section 1.3, Example 2) • Interest Rates (Section 1.3, Example 5; Section 4.1, Example 7; Section 4.4, Example 10; Section 11.3, Example 7 and Exercises 79–82) • Car Prices and Age of Cars on U.S. Roads (Section 1.3, Exercises 5–6) • Average Price of a Movie Ticket (Chapter 1 Review, Exercise 37) • Toll Options (Section 1.3 opener, Example 3, and Exercises 11–12) • HighestPaid TV Actors and Actresses (Section 2.1, Figure 2.1) • The Wage Gap between Men and Women (Section 2.1, Exercises 103–104) • Fuel Efficiency of New U.S. Cars (Section 2.2 opener) • Number of Births and Deaths in the U.S. (Section 2.6 opener and Example 4) • Political Orientation of U.S. College Freshmen (Chapter 2 Review, Exercise 53)
• OnePerson Households as a Percentage of the U.S. Total (Chapter 2 Test, Exercise 28) • AIDS Cases in the U.S. (Section 3.2, Example 3) • World Tiger Population (Section 3.2, Exercises 73–74) • Federal Budget Expenditures on Human Resources (Section 3.5, Exercise 107) • Amazon Deforestation (Chapter 3 Review, Exercise 14) • Gray Wolf Population (Section 4.1, Example 6) • Percentage of High School Seniors Applying to More Than Three Colleges (Section 4.1, Exercises 71–72) • Number of Pages in the Federal Tax Code (Section 4.1, Exercise 85) • Percentage of GDP Going Toward Health Care (Section 4.4, Exercises 115–116) • U.S. Population (Section 4.5, Example 1; Section 11.3, Example 3) • World Population (Section 4.5, Examples 5 and 6) • Populations of Various Countries (Section 4.5, Exercises 1–14) • Marital Status of U.S. Adults (Section 8.1, Exercise 67; Section 11.7, Example 9 and Exercises 1–10) • Rate of Violent Crime and Imprisonment in the U.S. (Section 8.4, Exercise 63) • Percentage of Men and Women Completing the Transition to Adulthood (Section 9.3, Exercise 61) • Hours per Day Spent on Digital Media (Section 11.1, Exercise 69) • Giving Up U.S. Citizenship (Section 11.1, Exercise 70) • Dormitory Charges (Section 11.2, Exercises 65–66)
Other Textbook Changes • Prior to the exercises in each section, the Annotated Instructor’s Edition provides a list of resources available for that section in MyLab Math. • The list of each section’s objectives, previously headed “What am I supposed to learn?” (which annoyed some reviewers) has been renamed “What You’ll Learn.” • Section P.6 includes new examples and exercises involving adding and subtracting rational expressions with different monomial denominators. (Section P.6, Example 8 and Exercises 43–50) • Section 1.4 contains a new objective involving simplifying powers of i. (Section 1.4, Example 7 and Exercises 55–60) • In Chapter 3, the standard form of a quadratic function, f (x) = a(x  h)2 + k, has been renamed the vertex form. • Section 5.7 includes a new objective covering the definitions, properties, and graphs of the inverse cotangent, inverse cosecant, and inverse secant functions. (Section 5.7, Example 5 and Exercises 19−26, 57−62)
Preface
• Section 6.1 has a new objective on rewriting expressions that contain both trigonometric and logarithmic expresssions as equivalent expressions using trigonometric identities and properties of logarithms. (Section 6.1, Example 9 and Exercises 61−66) • Section 6.5 builds on the new material in Section 6.1 with a new objective on solving equations that contain both trigonometric and logarithmic expressions. (Section 6.5, Example 13 and Exercises 117−128) • Section 7 .5 introduces the notation r cis u as an abbreviation for r(cos u + i sin u). (Section 7.5, Great Question! on p. 816) • Section 7.5 provides a second explanation of finding complex roots using DeMoivre’s Theorem. The new approach relies more on the relationships among the roots and less on the use of a formula. (Section 7.5, Great Question! on p. 820 and Example 9) • Section 9.5 presents an alternative to expansion by minors for evaluating a thirdorder determinant. (Section 9.5, pp. 997−998)
New in MyLab Math • Integrated Review Activities for selected topics provide handson work with important prerequisites.
GeoGebra Graphing Exercises are gradable graphing exercises that help students demonstrate their
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understanding. They enable students to interact directly with the graph in a manner that reflects how students would graph on paper. • Setup & Solve Exercises – We added more of these popular exercises, which require students to first describe how they will set up and approach the problem. This mirrors what students will be expected to do on a test. • Interactive Figures – For this revision, we added many more interactive figures (in editable GeoGebra format) to the Video & Resource Library. • Enhanced Assignments – These sectionlevel assignments have three unique properties (and are fully editable): 1. They help keep skills fresh with spaced practice of
previously learned concepts. 2. They have learning aids strategically turned off for some exercises to ensure that students understand how to work the exercises independently. 3. They contain personalized prerequisite skills
exercises for gaps identified in the chapterlevel Skills Check Quiz. • Video Assignments – These sectionlevel assignments are especially helpful for online classes or “flipped” classes, where some or all learning takes place independently. • PowerPoint slides are now animated. They also utilize Microsoft’s Equation Editor, making them more easily editable. • Personal Inventory Assessments are a collection of online exercises designed to promote selfreflection and engagement in students. These 33 assessments include topics such as a Stress Management Assessment, Diagnosing Poor Performance and Enhancing Motivation, and Time Management Assessment.
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Preface
What Familiar Features Have Been Retained in the Seventh Edition? • Graphing and Functions. Graphing is introduced in Chapter 1 and functions are introduced in Chapter 2, with an integrated graphing functional approach emphasized throughout the book. Graphs and functions that model data appear in nearly every section and Exercise Set. Examples and exercises use graphs of functions to explore relationships between data and to provide ways of visualizing a problem’s solution. Because functions are the core of this course, students are repeatedly shown how functions relate to equations and graphs. • Learning Objectives. Learning objectives are clearly stated at the beginning of each section under the heading “What You’ll Learn.” These objectives help students recognize and focus on the section’s most important ideas. The objectives are restated in the margin at their point of use. • ChapterOpening and SectionOpening Scenarios. Every chapter and every section open with a scenario presenting a unique application of mathematics in students’ lives outside the classroom. These scenarios are revisited in the course of the chapter or section in an example, discussion, or exercise. • Innovative Applications. A wide variety of interesting applications, supported by uptodate, realworld data, are included in every section.
Explanatory Voice Balloons • Explanatory Voice Balloons. Voice balloons are used in a variety of ways to demystify mathematics. They translate algebraic and trigonometric ideas into everyday English, help clarify problemsolving procedures, present alternative ways of understanding concepts, and connect problem solving to concepts students have already learned. • Detailed WorkedOut Examples. Each example is titled, making the purpose of the example clear. Examples are clearly written and provide students with detailed stepbystep solutions. No steps are omitted and each step is thoroughly explained to the right of the mathematics. CHECK POINT
• Check Point Examples. Each example is followed by a similar matched problem, called a Check Point, offering students the opportunity to test their understanding of the example by working a similar exercise. The answers to the Check Points are provided in the answer section.
• Great Question! This feature presents a variety of study tips in the context of students’ questions. Answers to questions offer suggestions for problem solving, point out common errors to avoid, and provide informal hints and suggestions. As a secondary benefit, this feature should help students not to feel anxious or threatened when asking questions in class.
BLITZER BONUS • Blitzer Bonuses. These enrichment essays provide historical, interdisciplinary, and otherwise interesting connections to the algebra and trigonometry under study, showing students that math is an interesting and dynamic discipline. • Concept and Vocabulary Checks. This feature offers shortanswer exercises, mainly fillintheblank and true/false items, that assess students’ understanding of the definitions and concepts presented in each section. The Concept and Vocabulary Checks precede the section Exercise Sets and have the prefix “C.” • Extensive and Varied Exercise Sets. An abundant collection of exercises is included in an Exercise Set at the end of each section. Exercises are organized within nine category types: Practice Exercises, Practice PLUS Exercises, Application Exercises, Explaining the Concepts, Technology Exercises, Critical Thinking Exercises, Group Exercises, Retaining the Concepts, and Preview Exercises. This format makes it easy to create wellrounded homework assignments. The order of the Practice Exercises is exactly the same as the order of the section’s worked examples. This parallel order enables students to refer to the titled examples and their detailed explanations to achieve success working the Practice Exercises. • Practice PLUS Problems. This category of exercises contains more challenging practice problems that often require students to combine several skills or concepts. With an average of ten Practice PLUS problems per Exercise Set, instructors are provided with the option of creating assignments that take Practice Exercises to a more challenging level. • Retaining the Concepts. Beginning with Chapter 2 each Exercise Set contains three or four review exercises under the header “Retaining the Concepts.” These exercises are intended for students to review previously covered objectives in order to improve their understanding of the topics and to help maintain their mastery of the material. If students are not certain how to solve a review exercise, they can turn to the section and worked example given in parentheses at the end of each exercise.
Preface
• MidChapter Check Points. At approximately the midway point in each chapter, an integrated set of Review Exercises allows students to review and assimilate the skills and concepts they learned separately over several sections. • Integration of Technology Using Graphic and Numerical Approaches to Problems. Sidebyside features in the technology boxes connect algebraic and trigonometric solutions to graphic and numerical approaches to problems. Although the use of graphing utilities is optional, students can use the explanatory voice balloons to understand different approaches to problems even if they are not using a graphing utility in the course. • Brief Reviews. Beginning with Chapter 1, the Brief Review boxes that appear throughout the book summarize mathematical skills, many of which are course prerequisites that students have learned but which many students need to review. This feature appears whenever a particular skill is first needed and eliminates the need for you to reteach that skill. For more detail, students are referred to the appropriate section and objective in a previous chapter where the topic is fully developed.
ACHIEVING SUCCESS • Achieving Success. The Achieving Success boxes, appearing at the end of many sections in Chapters 1 through 8, offer strategies for persistence and success in college mathematics courses. • Discovery. Discovery boxes, found throughout the text, encourage students to further explore algebraic and trigonometric concepts. These explorations are optional and their omission does not interfere with the continuity of the topic under consideration. • Chapter Summaries. Each chapter contains a review chart that summarizes the definitions and concepts in every section of the chapter. Examples that illustrate these key concepts are also referenced in the chart. • EndofChapter Materials. A comprehensive collection of Review Exercises for each of the chapter’s sections follows the Summary. This is followed by a Chapter Test that enables students to test their understanding of the material covered in the chapter. Beginning with Chapter 2, each chapter concludes with a comprehensive collection of mixed Cumulative Review Exercises.
MyLab™ Math Resources for Success MyLab Math (https://mlm.pearson.com/global) is available to accompany Pearson’s marketleading text options, including this text (access code required). MyLab Math is the teaching and learning platform that empowers you to reach every student. It combines trusted author content— including full eText and online homework with immediate feedback—with digital tools and a flexible platform to personalize the learning experience and improve results for each student.
MyLab Math Student Resources Each student learns at a different pace. Personalized learning pinpoints the precise areas where each student needs practice, giving all students the support they need— when and where they need it—to be successful. Exercises with Immediate Feedback – The exercises in MyLab Math reflect the approach and learning style of this text and regenerate algorithmically to give students
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unlimited opportunity for practice and mastery. Most exercises include learning aids, such as guided solutions and sample problems, and they offer helpful feedback when students enter incorrect answers.
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Preface
GeoGebra Exercises are gradable graphing exercises that help students demonstrate their understanding. They enable students to interact directly with the graph in a manner that reflects how students would graph on paper. NEW!
• Setup & Solve exercises require students to first describe how they will set up and approach a problem. This reinforces conceptual understanding of the process applied in approaching the problem, promotes longterm retention of the skill, and mirrors what students will be expected to do on a test. • Concept & Vocabulary exercises require students to demonstrate understanding of key ideas.
Interactive Figures bring mathematical concepts to life, helping students see the concepts through directed explorations and purposeful manipulation. These figures are assignable in MyLab Math and encourage active learning, critical thinking, and conceptual understanding. NEW! For this revision, we added many more interactive figures (in editable GeoGebra format) to the Video & Resource Library.
Instructional Videos – Highquality instructional videos are included for every objective in the text. Many of these feature builtin interactive quizzes. Chapter Test Prep Videos correspond to each exercise
in the Chapter Test in the textbook, enabling students to effectively prepare for highstakes testing. These are available in MyLab Math and www.youtube. com/user/pearsonmathstats (playlist “Blitzer Algebra & Trigonometry 7e”). Learning Guide consists of four parts: 1. Learning Guide worksheets for each section of the
text. These worksheets start with a catchy headline and motivating realworld connection followed by numerous “Solved Problems” and accompanying “Pencil Problems.” 2. Classroom Activities for selected sections contain recommended group size, material needed, and time to complete. 3. Integrated Review worksheets for every
prerequisite objective. These feature both instruction and practice. 4. NEW! Integrated Review Activities for selected topics provide handson work with important prerequisites.
Mindset videos and assignable, openended exercises foster a growth mindset in students. This material encourages them to maintain a positive attitude about learning, value their own ability to grow, and view mistakes as learning opportunities—so often a hurdle for math students. These videos are one of many Study Skills and CareerReadiness Resources that address the nonmathrelated issues that can affect student success. NEW! Personal Inventory Assessments are a collection of online exercises designed to promote selfreflection and engagement in students. These 33 assessments include topics such as a Stress Management Assessment, Diagnosing Poor Performance and Enhancing Motivation, and Time Management Assessment.
Preface eText – Available in a “reflowable” format for use on
tablets and smartphones. It is also fully accessible using screen readers. Student Solutions Manual – Fully worked solutions to oddnumbered exercises. Available for download from within MyLab Math.
MyLab Math Instructor Resources Your course is unique. So whether you’d like to build your own assignments, teach multiple sections, or set prerequisites, MyLab gives you the flexibility to easily create your course to fit your needs. PreBuilt Assignments are designed to make the
homework experience as effective as possible for students. All of these assignments are fully editable. •
NEW!
Enhanced Assignments – These sectionlevel
assignments have three unique properties: 1. They help keep skills fresh with spaced practice of
previously learned concepts. 2. They have learning aids strategically turned off for some exercises to ensure that students understand how to work the exercises independently. 3. They contain personalized prerequisite skills exercises for gaps identified in the chapterlevel Skills Check Quiz. •
NEW! Video Assignments – These sectionlevel assignments are especially helpful for online classes or “flipped” classes, where some or all learning takes place independently.
Learning Catalytics – With Learning Catalytics™, you’ll hear from every student when it matters most. You pose a variety of questions in class (choosing from preloaded
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questions or questions of your own making) that help students recall ideas, apply concepts, and develop criticalthinking skills. Your students respond using their own smartphones, tablets, or laptops. Accessibility – Pearson works continuously to ensure
our products are as accessible as possible to all students. Currently we work toward achieving WCAG 2.0 AA for our existing products (2.1 AA for future products) and Section 508 standards, as expressed in the Pearson Guidelines for Accessible Educational Web Media (https://www.pearson.com/uk/accessibility.html). Other instructor resources include: • Mini Lecture Notes contain additional examples and helpful teaching tips for each section of the text. • Instructor Solution Manual contains workedout solutions for every exercise in the text. • PowerPoint Lecture Slides are fully editable and included for each section of the text. UPDATED! Slides are now animated. They also utilize Microsoft’s Equation Editor, making them more easily editable. • TestGen® enables instructors to build, edit, print, and administer tests using a computerized bank of questions developed to cover all the objectives of the text. TestGen is algorithmically based, allowing instructors to create multiple but equivalent versions of the same question or test with the click of a button. Instructors can also modify test bank questions or add new questions. The software and test bank are available for download from Pearson’s online catalog. • Test Bank features printable PDFs containing all of the test exercises available in TestGen.
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Preface
Acknowledgments An enormous benefit of authoring a successful series is the broadbased feedback I receive from the students, dedicated users, and reviewers. Every change to this edition is the result of their thoughtful comments and suggestions. I would like to express my appreciation to all the reviewers, whose collective insights form the backbone of this revision. In particular, I would like to thank the following people for reviewing College Algebra, Algebra and Trigonometry, Precalculus, and Trigonometry. (An asterisk * indicates reviewers for the current edition.) *Margaret Adams, South Georgia South College Karol Albus, South Plains College *Alvina Atkinson, Georgia Gwinnett College Kayoko Yates Barnhill, Clark College *Melissa Battis, Greenhill School Timothy Beaver, Isothermal Community College Jaromir Becan, University of Texas, San Antonio Imad Benjelloun, Delaware Valley College Lloyd Best, Pacific Union College *Joseph Bittner, Fort Wayne North Side High School *Elisa Bolotin, Saint Stephen’s Episcopal School *Marie Borrazzo, College Academy David Bramlett, Jackson State University Natasha BrewleyCorbin, Georgia Gwinnett College Denise Brown, Collin College, Spring Creek Campus *Julie Brown, University of Texas of the Permian Basin David Britz, Raritan Valley Community College Bill Burgin, Gaston College Jennifer Cabaniss, Central Texas College *Ryan Caimano, Oklahoma Bible Academy *Kate Calendrillo, Northwest Catholic High School *Jaime Castrataro, TriWest High School Jimmy Chang, St. Petersburg College Teresa Chasing Hawk, University of South Dakota *Janie Coker Pritchard, Tyler Junior College Diana Colt, University of Minnesota, Duluth Shannon Cornell, Amarillo College Wendy Davidson, Georgia Perimeter College, Newton Donna Densmore, Bossier Parish Community College
*Deanna Ettore, Washington Township High School *Rafat Ewais, Clifton High School *Rebecca Faber, Notre Dame High School *Ed Fischer, Northern Nash High School Nancy Fisher, University of Alabama *Vickie Flanders, Baton Rouge Community College *Sheila Fleming, Marble Falls High School Donna Gerken, Miami Dade College *David Ghazvini, Lee College Cynthia Glickman, Community College of Southern Nevada Sudhir Kumar Goel, Valdosta State University Donald Gordon, Manatee Community College David L. Gross, University of Connecticut Jason W. Groves, South Plains College Joel K. Haack, University of Northern Iowa Jeremy Haefner, University of Colorado Joyce Hague, University of Wisconsin at River Falls *Tawfik Haj, San Jacinto College Mike Hall, University of Mississippi *Leo Hartsock, Northland Preparatory Academy Mahshid Hassani, Hillsborough Community College Tom Hayes, Montana State University Christopher N. HayJahans, University of South Dakota Angela Heiden, St. Clair Community College *Baron Heinemann, Episcopal High School Celeste Hernandez, Richland College *Ann Ho, Taipei American School Alysmarie Hodges, Eastfield College Amanda Hood, CopiahLincoln Community College Jo Beth Horney, South Plains College Heidi Howard, Florida State College at Jacksonville, South Campus *S. Larue Huckaby, Shorter University Winfield A. Ihlow, SUNY College at Oswego *Dale Johanson, Northeast Community College
*Kevin Dibert, NSU University School *Barbara Dobbs, Academy of the Holy Cross *Juan Du, Broward College
Nancy Raye Johnson, Manatee Community College *Kimberly Jones, Dakota State University *Kevin Jones, Jr., McCluer High School *Cheryl Kerns, Blue Valley West High School *Clayton Kitchings, University of North Georgia
*Marcial Echenique, College of the Florida Keys Disa Enegren, Rose State College Keith A. Erickson, Georgia Gwinnett College
*Theo Koupelis, Broward College *Anthony Lamanna, St. John’s Preparatory Dennine Larue, Fairmont State University
Preface
*Tina Lee, Haywood Community College Mary Leesburg, Manatee Community College Christine Heinecke Lehman, Purdue University North Central Alexander Levichev, Boston University
*Brian Southworth, Independence Community College Caroline Spillman, Georgia Perimeter College, Clarkston *Laura Spunt, Posnack School Jonathan Stadler, Capital University
*Qingxia Li, Fisk University *Ethan Lightfoot, Gateway STEM High School Zongzhu Lin, Kansas State University Benjamin Marlin, Northwestern Oklahoma State University
Franotis R. Stallworth, Gwinnett Technical College John David Stark, Central Alabama Community College Chris Stump, Bethel College *Cynthia Lee Suplizio, Colorado Mountain College
Marilyn Massey, Collin County Community College Yvelyne McCarthyGermaine, University of New Orleans *Kayri McCartin, Grafton High School David McMann, Eastfield College Owen Mertens, Missouri State University, Springfield James Miller, West Virginia University Martha Nega, Georgia Perimeter College, Decatur *Youssef Oumanar, Greenhill School Shahla Peterman, University of Missouri, St. Louis Debra A. Pharo, Northwestern Michigan College *Janice Phillipp, Texas Southmost College Gloria Phoenix, North Carolina Agricultural and Technical State University Katherine Pinzon, Georgia Gwinnett College David Platt, Front Range Community College Juha Pohjanpelto, Oregon State University *David Quesnell, Immaculate High School Brooke Quinlan, Hillsborough Community College *Corrie Ramage, Kosciusko High School Janice Rech, University of Nebraska at Omaha Joseph W. Rody, Arizona State University *Lee Ann Roberts, Georgia Gwinnett College Behnaz Rouhani, Georgia Perimeter College, Dunwoody Judith Salmon, Fitchburg State University *Ryan Sasaki, Iolani School Michael Schramm, Indian River State College Cynthia Schultz, Illinois Valley Community College *Brittney Seale, Colorado Mountain College *Juanita Self, Central Texas College *Patricia Senn, Lurleen B. Wallace Community College *Olimpia Simeón Monet, Miami Dade College *Muhammad Naeem Sharif, Kuwait Technical College Pat Shelton, North Carolina Agricultural and Technical State University Jed Soifer, Atlantic Cape Community College
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Scott Sykes, University of West Georgia Richard Townsend, North Carolina Central University Pamela Trim, Southwest Tennessee Community College *Ruth Trubnik, Delaware Valley University Chris Turner, Arkansas State University Richard E. Van Lommel, California State University, Sacramento Dan Van Peursem, University of South Dakota Philip Van Veldhuizen, University of Nevada at Reno Jeffrey Weaver, Baton Rouge Community College *Felice Weiner, South Mountain Community College *Aaron Wernet, McLennan Community College Amanda Wheeler, Amarillo College David White, The Victoria College *Molly Whittington, Holy Cross School Tracy Wienckowski, University of Buffalo *Stacy Yarnell, Colorado Mountain College Additional acknowledgments are extended to: • Brad Davis, for contributing new and updated data, providing the book’s annos and answer section, and serving as (an amazing!) answer checker. • Dan Miller for preparing the solution manuals and the Learning Guide. • Brian Morris at Scientific Illustrators for superbly illustrating the book. • Jerilyn DiCarlo and Annie Brace for their brilliant design work. • Francesca Monaco, project manager, and Tamela Ambush, production editor, whose collective talents kept every aspect of this complex project moving through its many stages. (Special thanks to Sharon Cahill for seamlessly stepping in during Francesca’s absence.) • Dawn Murrin, Jeff Weidenaar, and Jon Krebs, my editors at Pearson, who guided and coordinated the book from manuscript through production. • Chelsea Kharakozova, product manager, for the steady financial guidance for the project.
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• Stacey Sveum and Peggy Lucas, marketing managers, for their innovative marketing efforts. • Shana Siegmund, MyLab producer, for coordinating the many digital aspects of the project. • Eric Gregg and Dominick Frank for splendid work on the MyLab exercises that support the texts. • Finally, thanks to the Pearson sales force, for their confidence and enthusiasm about the book.
I hope that my passion for teaching, as well as my respect for the diversity of students I have taught and learned from over the years, is apparent throughout this new edition. By connecting algebra and trigonometry to the whole spectrum of learning, it is my intent to show students that their world is profoundly mathematical, and indeed, P is in the sky.
Bob Blitzer Acknowledgments for the Global Edition Pearson would like to acknowledge and thank the following for the Global Edition:
Contributors José Luis Zuleta Estrugo, École polytechnique fédérale de Lausanne Haitham S. Solh, American University in Dubai
Reviewers Olcay Coşkun, Boğaziçi University Yanghong Huang, The University of Manchester Saadia Khouyibaba, American University of Sharjah
ABOUT THE AUTHOR Bob Blitzer is a native of Manhattan and received a Bachelor of Arts degree with dual majors in mathematics and psychology (minor: English literature) from the City College of New York. His unusual combination of academic interests led him toward a Master of Arts in mathematics from the University of Miami and a doctorate in behavioral sciences from Nova University. Bob’s love for teaching mathematics was nourished for nearly 30 years at Miami Dade College, where he received numerous teaching awards, including Innovator of the Year from the League for Innovations in the Community College and an endowed chair based on excellence in the classroom. In addition to Algebra and Trigonometry, Bob has written textbooks covering developmental mathematics, introductory algebra, intermediate algebra, trigonometry, precalculus, and liberal arts mathematics, all published by Pearson. When not secluded in his Northern California writer’s cabin, Bob can be found hiking the beaches and trails of Point Reyes National Seashore and tending to the chores required by his beloved entourage of horses, chickens, and irritable roosters.
The bar graph shows some of the qualities that students say make a great teacher. It was my goal to incorporate each of these qualities throughout the pages of this book.
Explains Things Clearly I understand that your primary purpose in reading Algebra and Trigonometry is to acquire a solid understanding of the required topics in this course. In order to achieve this goal, I’ve carefully explained each topic. Important definitions and procedures are set off in boxes, and workedout examples that present solutions in a stepbystep manner appear in every section. Each example is followed by a similar matched problem, called a Check Point, for you to try so that you can actively participate in the learning process as you read the book. (Answers to all Check Points appear in the back of the book.)
Funny & Entertaining Who says that an algebra and trigonometry textbook can’t be entertaining? From our unusual cover to the photos in the chapter and section openers, prepare to expect the unexpected. I hope some of the book’s enrichment essays, called Blitzer Bonuses, will put a smile on your face from time to time.
To the Student Qualities That Make a Great Teacher Explains Things Clearly
70%
Funny/ Entertaining
47% 40%
Helpful Passionate About Their Subject
22% 10% 30% 50% 70% Percentage of Teens Stating That the Quality Makes a Great Teacher
90%
Source: Avanta Learning System
Helpful I designed the book’s features to help you acquire knowledge of algebra and trigonometry, as well as to show you how algebra and trigonometry can solve authentic problems that apply to your life. These helpful features include: • Explanatory Voice Balloons: Voice balloons are used in a variety of ways to make math less intimidating. They translate algebraic and trigonometric language into everyday English, help clarify problemsolving procedures, present alternative ways of understanding concepts, and connect new concepts to concepts you have already learned. • Great Question!: The book’s Great Question! boxes are based on questions students ask in class. The answers to these questions give suggestions for problem solving, point out common errors to avoid, and provide informal hints and suggestions. • Achieving Success: The book’s Achieving Success boxes give you helpful strategies for success in learning algebra and trigonometry, as well as suggestions that can be applied for achieving your full academic potential in future college coursework. • Chapter Summaries: Each chapter contains a review chart that summarizes the definitions and concepts in every section of the chapter. Examples from the chapter that illustrate these key concepts are also referenced in the chart. Review these summaries and you’ll know the most important material in the chapter!
Passionate about the Subject I passionately believe that no other discipline comes close to math in offering a more extensive set of tools for application and development of your mind. I wrote the book in Point Reyes National Seashore, 40 miles north of San Francisco. The park consists of 75,000 acres with miles of pristine surfwashed beaches, forested ridges, and bays bordered by white cliffs. It was my hope to convey the beauty and excitement of mathematics using nature’s unspoiled beauty as a source of inspiration and creativity. Enjoy the pages that follow as you empower yourself with the algebra and trigonometry needed to succeed in college, your career, and your life. Regards,
Bob Blitzer 19
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We embrace the many dimensions of diversity, including but not limited to race, ethnicity, gender, socioeconomic status, ability, age, sexual orientation, and religious or political beliefs. Education is a powerful force for equity and change in our world. It has the potential to deliver opportunities that improve lives and enable economic mobility. As we work with authors to create content for every product and service, we acknowledge our responsibility to demonstrate inclusivity and incorporate diverse scholarship so that everyone can achieve their potential through learning. As the world’s leading learning company, we have a duty to help drive change and live up to our purpose to help more people create a better life for themselves and to create a better world. Our ambition is to purposefully contribute to a world where: • Everyone has an equitable and lifelong opportunity to succeed through learning.
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APPLICATIONS INDEX A Accidents, automobile accidents per day, age of driver and, 1008 age of driver and, 199 alcohol use and, 541–542, 547 probability of accident while intoxicated, 1179 Acid rain, 547 Actors, highestpaid, on television, 246–249 Actors, selection of, 1162, 1183 Adulthood, transition to, 975 Adult residential community costs, 1109, 1116 Advertising, 479–480, 484 African Americans, with high school diploma, 568 Age(s) arrests and drunk driving as function of, 460 average number of awakenings during night by, 134 bodymass index and, 920 body temperature variation and, 921 calories needed to maintain energy by, 117 chances of surviving to various, 263 of driver, accidents per day and, 1008 fatal crashes and, 199 height as function of, 305, 308, 326 perceived length of time period and, 483 percent body fat in adults by, 283 racial prejudice and, 89 social media use by, 931 systolic blood pressure and, 193–194 tax refund likelihood and, 77 U.S. population projection by, 341 weight of human fetus and, 242 Aging rate, space travel and, 63, 75, 78 AIDS/HIV cases diagnosed (U.S.), 398 global perspective on, 398 Aircraft/airplanes approaching runway, vector describing, 838 direction angle of, given speed, 840 distance and angle of elevation of, 659 distance flown by, 602 ground speed of, 840 height of, 626 leaving airport at same time, distance between, 779, 783–784, 854 linear speed of propeller, 692 Mach speed of, 735 runway departure lineup, 1183 speed with/against the wind, 877 true bearing of, 839–840 vector describing flight of, 838 velocity vector of, 836 weight/volume constraints, 923–925 wind speed and direction angle exerted on, 839–840 Airports, distance between, 786 Alcohol use and accident risk, 541–542, 547 drunk driving arrests, 460 moderate wine consumption and heart disease, 299–300 number of moderate users in U.S., 568 by U.S. high school seniors, 134 Alligator(s) population of, 202 tail length given body length, 482 Altitude gained by hiker climbing incline, 692 increase on inclined road of, 602 Altitude and atmospheric pressure, 567
Amazon deforestation, 487–488 American Idol, ratings of, 392–393 Amusia (tone deafness), sound quality and, 714, 716 Angle(s) in architecture, 570 clock hands forming, 570, 571 of elevation, 597–599, 601 603, 626, 659, 686, 687, 692, 695, 777–778 Angular speed of audio records, 583 of carousel, 582 of hard drive in computer, 582 of propeller on wind generator, 692 Annuities compound interest on, 1126–1127, 1133 value of, 1133, 1182 Antenna, height on top of building, 694 Apartment, rent, 241 Apogee/perigee of satellite’s orbit, 1026 Applause, decibel level of, 287 Arch bridge, 1092 Archer’s arrow, path of, 386 Architecture angles in, 570 conic sections in, 1027, 1037 Archway. See Semielliptical archway and truck clearance Area maximum, 388, 391 440, 487 of oblique triangle, 769 of plane figure, 90 of region under curve, 676 of shaded region, 89, 102 of triangle, 784, 1003 Area code possibilities, 1162 Arrests, drunk driving, 460 Artists, in documentary, 1157–1158 Aspirin, halflife of, 560, 1041 Asteroid detection, 898 Atmospheric pressure and altitude, 567 Audio records, angular speed and linear speed of, 583 Autism cases diagnosed, 1107 Automobiles accidents per day, age of driver and, 1008 alcohol use and accident risk, 541–542, 547 annual price increases of, 165 average age, on U.S. roads, 165 computing work of pushing, 847, 849 depreciation, 165, 263 drunk driving arrests as function of age, 460 electric, 1182 fatal accidents and driver’s age, 199 leaving city at same time, distance between, 854 possible race finishes, 1162 probability of accident while intoxicated, 1179 purchase options, 1162 rentals, 219–220, 243 repair estimates for, 235 required stopping distance, 471–472 travel by bicycle vs., 352–353 value over time, 1108 Average cost function, 454–455, 458, 489, 491 Average rate of change, 305–306, 326
B Backpack, price before reduction, 563 Ball angle of elevation and throwing distance of, 764
location of thrown, 1069 maximum height and throwing distance of, 887 thrown upward and outward, 391 Ball (attached to spring) finding amplitude and period of motion of, 723 simple harmonic motion of, 681–682, 760, 764 Ball (height above ground), 885, 887, 950, 1186 baseball, 570 bounce height, 482 football, 45, 384–386, 487, 950 maximum height, 487, 857, 887 when thrown from rooftop, 471 when thrown from top of Leaning Tower of Pisa, 469 Ballots, 1162 Banking angle and turning radius of bicycle, 482 Baseball angle of elevation and throwing distance of, 760 diamond diagonal length, 199 height of ball above ground, 570 path of, 1069, 1077–1078 pitcher’s angle of turn to throw ball, 788 position as function of time, 1077–1078 Baseball contract, 1097, 1132 Baseball diamond, distance from pitcher’s mound to bases on, 787 Basketball, hang time in air when shooting, 217 Basketball court, dimensions of, 162 Bass in lake over time, 489 Bearings, 680–681 695 of boat, 681 687, 787 distance at certain, 687, 694 to fire from two fire stations, 775, 777, 1089 of jet from control tower, 687 true, of plane, 839–840 between two cities, 694 Beauty, symmetry and, 268 Benefit concert lineup possibilities, 1162 Berlin Airlift, 922, 928 Bias, Implicit Association Test for, 79, 89 Bicycle(s) banking angle, 482 manufacturing, 263, 458, 874, 928 travel by car vs., 352–353 Biorhythms, 569, 604, 624, 645–647 Bird species population decline, 560 Birth(s), in U.S. from 2000 through 2009, 328, 333–334 Birthday, probability of same, 353, 1179 Birthday cake, 78 Birthday date sharing, 737 Bloodalcohol concentration, 43–44, 47, 541–542, 547 Blood pressure, systolic, age and, 193–194 Blood volume and body weight, 121 475–476 Boats/ships bearing of, 681 687 changing, 787 distance traveled at certain, 687 to sail into harbor, 687 direction angle of, 856 distance from lighthouse, 694, 779 distance from pier, 787 ground speed, 856 leaving harbor at same time, distance between after three hours, 786 location between two radio towers, 1092 on tilted ramp, vector components of force on, 845, 849 velocity of, 856 velocity vector of, 836
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22
Applications Index
Body fat in adults by age and gender, percent, 283 Bodymass index, 482, 920 Body temperature, variation in, 693, 921 Books arranging on shelf, 1157 selections, 1162, 1185 Bouquet, mixture of flowers in, 908 Box dimensions, 423 Brain growth of the human, 555 modeling brain activity, 659 Breakeven analysis, 695, 868–869, 874, 908. See also Cost and revenue functions/breakeven points Breathing cycle, 627 modeling, 641–642 velocity of air flow in, 759 Bridge arch, 1092 George Washington Bridge, 1093 suspension, parabolas formed by, 1093 Bridge coupon book/toll passes, 154, 157–159, 165, 202, 230–231 235, 342, 660 Budgeting, groceries vs. health care, 309 Building height of, 597–598, 686, 687, 694, 695, 779 shadow cast by, 760 Building’s shadow, 242 Business ventures, 874 Butterflies, symmetry of, 789
C Cable car, distance covered by, 778–779 Cable lengths between vertical poles, 218 Cable service, 1095 Cable television deals, 1095 Calculator manufacturing costs, 489 Call of Duty video game, retail sales of, 548 Calorienutrient information, 932 Calories needed by age groups and activity levels, 975 needed to maintain energy balance, 117 Camera price reductions, 1179 viewing angle for, 675 Camera, price before reduction, 159 Canoe manufacturing, 874 Car(s). See Automobiles Carbon14 dating, 552, 559–560 Carbon dioxide, atmospheric global warming and, 245, 294–296, 648, 660 Carbon dioxide, atmospheric levels of, 296 Cardboard, length/width for box, 907 Cards. See Deck of 52 cards, probability and Carousel, linear speed and angular speed of animals on, 582, 586 Cave paintings, prehistoric, 560 CD selection for vacation trip, 1183 Celebrity earnings, 246–249 Cell phones, 264, 875 Celsius/Fahrenheit temperature interconversions, 45, 234, 369 Centrifugal force, 480–481 Chaos, patterns of, 767, 813 Charging stations, for electric vehicles, 1182 Checking accounts, 235 Chernobyl nuclear power plant accident, 505 Chess moves, 1154 Chess tournament, roundrobin, 198 Child care, spending on, 262 Children modeling height of, 515, 522, 543 respiratory rates of, 919 Cholesterol and dietary restrictions, 919 intake, 919
Cigarette consumption. See Smoking Cigarette tax, 1107 Circle, finding length of arc on, 694, 744 Class structure of the United States, 1006–1007 Cliff, distance of ship from base of, 686 Climate change, 296 Clock(s) angles formed by hands of, 570, 571 degrees moved by minute hand on, 585 distance between tip of hour hand and ceiling, 646 distance between tips of hands at 10:00, 788 minute hand movement in terms of p, 585 Club membership fees, 961 Club officers, choosing, 1162, 1183 Coding, 977, 986–987, 989, 990 Coffee consumption, sleep and, 570 Coin tosses, 234, 1167, 1174–1175, 1177, 1178 College(s) attendance, 1183 average dormitory changes at, 1118–1119 majors of freshmen in, 1118 percentage of U.S. high school seniors applying to more than three, 506 projected enrollment, 165, 241 242 salary after, 240 College education availability of, to qualified students, 165 average yearly earnings and, 164–165 cost of, 30, 32–33, 47–48, 393 excuses for not meeting assignment deadlines, 240 government aid decreases, 242 women vs. men, 121 College graduates among people ages 25 and older, in U.S., 1118–1119 median starting salaries for, 155–156 College majors campus mergers and, 167 of freshmen, 1118 College students excuses for not meeting assignment deadlines, 240 freshmen attitudes about life goals, 156–157 with A average in high school, 153 claiming no religious affiliation, 249–250 grade inflation, 152 majors of, 1118 political orientation, 369 gender ratios and campus mergers, 877 hours of study per week, by major, 885–886 interactive online games played by, 877 loan debt, 91 majors and campus mergers, 877 procrastination and symptoms of physical illness among, 860, 876 sleep and, 877 study abroad destinations, 1041 College tuition government aid decreases, 242 student loan debt, 91 Collinear points, 1003 Comedians, net worth of, 522 Comedy act schedule, 1162 Comets Halley’s Comet, 1022, 1037, 1088 intersection of planet paths and, 906, 1037 Committee formation, 1158, 1160, 1162 Commuters, toll discount passes, 157–159, 165, 235, 660 Compound interest, 1108, 1133, 1182 on annuity, 1126–1127, 1133 choosing between investments, 503 compounding periods, 507
continuously compounded, 542, 546, 568, 569, 765, 935 formula for, 542 investments, 566, 688 savings accounts, 545–546 value of Manhattan Island and, 506 Computer(s) angular speed of hard drive in, 582 assembly, time required for, 490 computergenerated animation, 312 discounts, 334–335, 342 prices, 344 ratio of students to computers in U.S. public schools, 398 sale price, 102 Computer graphics, 937, 961 970, 971 Concentration of mixture, 153. See also Mixture problems Cone volume, 481 Conference attendees, choosing, 1160, 1162 Constraints, 923–927, 933 Continuously compounded interest, 542, 546, 568, 569, 765, 935 Cookies, supply and demand for, 875 Coronary heart disease, 561 Coronavirus cases modeling, in U.S., 553–554 in New York, 935 in U.S., 394–396 worldwide, 492 Corporate income tax, 202 Corporation officers, choosing, 1157, 1162 Cost(s). See also Manufacturing costs of college education, 30, 32–33, 47–48, 393 of federal social programs, 62 of industrial cleanup, 103–104 of medical treatment, 521 minimizing, 928 of public transportation infrastructure maintenance, 300 of raising child born in U.S., 1102–1103 truck rental, 1095 Cost and revenue functions/breakeven points, 868–869, 874, 908, 934 average, 454–455, 458, 489, 491 bike manufacturing, 458 breakeven points, 874, 934 graphing calculator manufacturing, 489 for PDA manufacturing, 908 roast beef sandwiches, 392 running shoe manufacturing, 458, 869 tablet case manufacturing, 341 virtual reality headset manufacturing, 454–455 wheelchair manufacturing, 455 Costco memberships, 489 Course schedule, options in planning, 1154 Cove, distance across, 787 Crane lifting boulder, computing work of, 849 Crate, computing work of dragging, 856 Credit card balances, 241 Crew (rowing), 877 Crime, 907 Cryptograms, 986–987, 990. See also Coding Cycloid, 1078
D Daylight, number of hours of, 604, 624, 644, 646, 659, 759 Deadlines, excuses for not meeting, 240 Dead Sea Scrolls, carbon14 dating of, 552 Death penalty, sentences rendered by U.S. juries, 411 Death rate, hours of sleep and, 879, 883
Applications Index Deaths in the 20th century, 932 from 2000 through 2009, 328, 333–334 by snakes, mosquitoes, and snails, 264 Debt national, 48, 59–60, 62, 63, 120 student loan, 91 Decay model for carbon14, 559–560 Decibels. See Sound intensity Deck of 52 cards, probability and, 1168–1169, 1171–1172, 1177, 1183, 1185 Decoding a word or message, 987, 989, 990 Deforestation, Amazon, 487–488 Degreedays, 1120 Delicate Arch, angle of elevation to determine height of, 603 Depreciation, 165, 263 Depression exercise and, 312 sense of humor and, 136–137, 148–149 Desk manufacturing, 950 Die rolling outcomes, 1167–1168, 1177, 1178, 1183 Digital camera, price reduction for, 1179 Digital photography, 961 970–971 974, 975, 1007 Dinosaur bones, potassium12 dating of, 560 Dinosaur footprints, pace angle and stride indicated by, 780, 786 Direction, 826–827 Disposable cups, reusable vs., 663 Distance across cove, 787 across lake, 598, 601 692, 786, 787 from base to top of Leaning Tower of Pisa, 777 braking, 885 between cars leaving city at same time, 854 of forest ranger from fire, 687 between houses at closest point, 1040 of island from coast, 686 of marching band from person filming it, 659 of oil platform from ends of beach, 777 between pairs of cities, 362–363 of rotating beam of light from point, 658, 659 safe, expressway speed and, 119 of ship from base of cliff, 686 of ship from base of Statue of Liberty, 686 of ship from lighthouse, 694 of ship from radio towers on coast, 1040 of stolen car from point directly below helicopter, 686 that skydiver falls in given time, 1135 traveled by plane, 602 traveled by walking and bus travel, 47 between two points on Earth, 585 between two points on opposite banks of river, 777 between two trains leaving station at same time, 812 Diver’s height above water, 471 Diversity index, 119 Diving board motion, modeling, 659 Divorce, educational attainment and probability of, 130–132 DNA, structure of, 587 Domed ceiling, light reflectance and parabolic surface of, 1057 “Don’t ask, don’t tell” policy, 309–310 Drinks, order possibilities for, 1162 Drivers, age of. See Age(s) Driving accidents. See Accidents, automobile Driving rate and time for trip, 477 Drug concentration, 307, 458 Drug dosage, child vs. adult, 765 Drug experiment volunteer selection, 1162 Drug tests, mandatory, probability of accurate results, 1178 Drug use among teenagers, 561 Drunk driving, probability of accidents, 1179
Drunk driving arrests, age as function of, 460 Dual investments, 47, 160, 166, 202, 284, 373, 484, 935
E Eagle, height and time in flight, 369 Earnings. See Salary(ies) Earth angular velocity of point on, 586 distance between two points on, 585 finding radius of, 688 motion of Moon relative to, 587 Earth, age of, 56 Earthquake epicenter, 362–363 intensity, 508, 516, 566 relief from, 922–925 simple harmonic motion from, 684 Ecommerce sales, 567 Economic impact of factory on town, 1134, 1183 Education. See also College education level of, U.S. population, 1177 online, 937 percentage of U.S. adults completing high school, 568 spending on preprimary, 262 unemployment and years of, 489 Educational attainment, divorce and, 130–132 Election ballots, 1162 Electrical resistance, 177, 483, 1186 Electric vehicles, 1182 Electromagnetic spectrum, 646 Elephant’s weight, 547 Elevation, angle of, 597–599, 601 603, 626, 659, 686–687, 692, 695, 777–778 Elevator capacity, 235, 919 Elk population, 569–570 Elliptical ceiling, 1024 Elliptipool, 1024 Encoding a message, 977, 986–987, 989, 990 Endangered species, 560 Equator, linear velocity of point on, 585 Equilibrium, forces in, 839 Ethnic diversity, 119 Exam grades, 235, 243, 975 Excuses, for not meeting college assignment deadlines, 240 Exercise depression and, 312 heart rate and, 31 target heart rate ranges for, 46 Explosions, location of, 1037–1038, 1040, 1057 Exponential decay model, 560, 568, 569, 1009 Exponential growth, 695 Expressway speeds and safe distances, 119 Eye color and gender, 1185
F Factory, economic impact on town, 1134 Fahrenheit/Celsius temperature interconversions, 45, 234, 369 Family, independent events in, 1175, 1177, 1184 Federal budget deficit, 62 (See also National debt) expenditures on human resources, 459 Federal Express, aircraft purchase decisions by, 928 Federal income tax, 264 Federal social programs, cost and enrollment, 62 Fencing for enclosure, 903–904 maximum area inside, 388, 391 393 for plot of land, 934
23
Ferris wheel, 363 height above ground when riding, 625 linear speed of, 586 Fetal weight, age and, 242 Field, dimensions of, 933, 1186 Field’s dimensions, 241 Films. See Movies Films, Oscarwinning, 343 Financial aid, college student, 242 Fire distance of fire station from, 1089 distance of forest ranger from, 687 locating, from tower, 768, 775, 777, 812, 856, 1089 Flagpole height of, finding, 765 leaning, angle made with ground, 779 on top of building, height of, 687 Flood, probability of, 1184 Floor dimensions, 906 Floor space, length and width of, 242 Flu epidemic, 506 inoculation costs, 117 outbreak on campus, 1134 timetemperature scenario, 265–266 vaccine mixture, 263 Food cost per item, 309, 886 Food stamps program, 62 Food truck vendor, profit and cost for, 392 Football height above ground, 950 maximum height of, 1093 position as function of time, 1093 vector describing thrown, 838 Football field dimensions, 161–162 Football’s height above ground, 45, 384–386 Force(s) on body leaning against wall, 826, 829 in equilibrium, 839 pulling cart up incline, 826 resultant, 839, 855, 856 Foreignborn population in U.S., 202 Frame dimensions, 166 Freedom 7 spacecraft flight, 354 Freefalling object’s position, 468–469, 471 490 Frequency, length of violin string and, 479 Freshmen. See under College students Friendship 7, distance from Earth’s center, 1088 Fuel efficiency, 265
G Galaxies, elliptical, 1145 Games, online, college students and, 877 Garbage, daily perpound production of, 91 Garden, width of path around, 200 Gasoline gallons of premium sold, 849, 874–875 gallons of regular sold, 849 Gas pressure in can, 478 Gay marriage, U.S. public opinion on, 561 Gay service members discharged from military, 309–310 Gender average number of awakenings during night by, 134 bachelor’s degrees awarded and, 121 calories needed to maintain energy by, 117 college gender ratios and campus mergers, 877 eye color and, 1185 firstyear U.S. college students claiming no religious affiliation by, 249–250 of House of Representatives members, 876 labor force participation by, 217 life expectancy by year of birth and, 298 percent body fat in adults by, 283 wage gap by, 263
24
Applications Index
George Washington Bridge, height of cable between towers of, 1093 Global warming, 245, 294–296 Golden rectangles, 77 Government, trust in, 298 Government financial aid, college tuition, 242 Grade inflation, 152 Gravitational force, 480 Gravity model, 483 Groceries, budgeting for, 309 Ground speed, 840 Groups fitting into van, 1162 Gutter crosssectional area, 200, 391 Guy wire attached to pole, angle made with ground and, 679 Gym membership fees, 244
H Halflife aspirin, 560, 1041 radioactive elements, 560, 568, 1009 Xanax, 560, 1079 Halley’s Comet, 1022, 1037, 1088 Happiness average level of, at different times of day, 353 per capita income and national, 299 Harmonic motion, simple. See Simple harmonic motion HDTV screen dimensions, 194–195, 473 Headlight parabolic surface of, 1093, 1094 unit design, 1093, 1094 Health care budgeting for, 309 gross domestic product (GDP) spent on, 546 savings needed for expenses during retirement, 561 Health club membership fees, 165 Heart beats over lifetime, 63 Heart disease coronary, 561 moderate wine consumption and, 299–300 smoking and, 460 Heart rate exercise and, 31 46 life span and, 490 before and during panic attack, 410 Heat generated by stove, 483 Heating and cooling systems, 800 Heat loss of a glass window, 483 Height above ground of building, shadow cast and, 242 of antenna on top of building, 694 of ball above ground (See Ball [height above ground]) of building, 597–598, 686, 687, 694, 695, 779 child’s height modeled, 515, 522, 543 diver’s height above water, 471 of eagle, in terms of time in flight, 369 on Ferris wheel while riding, 625 of flagpole, 687, 765 as function of age, 305, 308, 326 healthy weight region for, 920 of leaning wall, finding, 778 maximum, 857, 1094, 1186 of Mt. Rushmore sculpture, 680 percentage of adult height attained by girl of given age, 521 543 of plane, 626 of tower, finding, 678, 686, 825 of tree, finding, 812 weight and height recommendations/ calculations, 166, 482 Higher education costs, 1119 High school education, percentage of U.S. adults completing, 568
Hiking trails, finding bearings on, 681 Hill, magnitude of force required to keep car from sliding down, 839 Hispanic Americans, population growth, 568 HIV/AIDS cases diagnosed (U.S.), 398 global perspective on, 398 HIV infection number of Americans living with, 398 T cell count and, 246, 255–256 Home ownership, 301 304–305 Hotair balloon, distance traveled by ascending, 679, 687, 724 Hotel room types, 877 Households, mixed religious beliefs in, 234 House of Representatives, gender of members, 876 House sales prices, 264, 1135 House value, inflation rate and, 506 Housework, 1182 Hubble Space Telescope, 484 Human resources, federal budget expenditures on, 459 Humor, sense of, depression and, 136–137, 148–149 Hurricanes barometric air pressure and, 547 probability, 1178 Hydrogen ion concentration, 546–547
I Ice cream flavor combinations, 1158, 1162 Identical twins, distinguishing between, 878 Illumination intensity, 482, 483 Imaginary number joke, 177 Implicit Association Test, 79, 89 Imprisonment, crime rate and, 907 Income of Americans, 876 highestpaid TV celebrities, 246–249 median yearly income of fulltime workers, 60 Income tax, federal, 264 Individual Retirement Account (IRA), 1126–1127, 1133, 1134, 1182 Industrial accidents, cost of cleaning up, 103–104 Inflation, cost of, 152 Inflation rate, 506 Influenza. See Flu Inn charges, before tax, 166 Inoculation costs for flu, 117 Insulation, rate of heat lost through, 695 Insurance, pet, 283 Intelligence quotient (IQ) and mental/ chronological age, 482 Interactive online games, college students and, 877 Internet service providers, 241 Internet streaming services, 203 Interracial marriage percentage of Americans in favor of laws prohibiting, 242–243 rates of, 1095–1096 Investment(s) accumulated value of, 502–503, 506, 542 amounts invested per rate, 886 choosing between, 503 compound interest, 502–503, 506, 507, 542, 546, 547, 566, 568, 569, 688, 765, 935, 1133 for desired return, 243 dual, 47, 160, 166, 202, 284, 373, 484, 935 in greeting cards, 874 and interest rates, 47, 241 maximizing expected returns, 929 money divided between high and lowrisk, 919 in play, 874 possibility of stock changes, 1183 IQ (intelligence quotient) and mental/ chronological age, 482
IRA. See Individual Retirement Account Island, distance from coast of, 686
J Jeans, price of, 342 Jet ski manufacturing, 934 Job applicants, filling positions with, 1184 Job offers, 1119, 1120, 1132 Jokes about books, 1163
K Kidney stone disintegration, 1022, 1057 Kinetic energy, 483 Kite, angle made with ground of flying, 679, 1068
L Labor force, participation by gender, 217 Labrador retrievers, color of, 87 Ladder’s reach, 199 Lake, distance across, 598, 601 692, 786, 787 Land fencing for (See Fencing) rectangular plot, 934 triangular plot, 788, 854 Lead, halflife of, 560 Leaning Tower of Pisa, distance from base to top of, 777 Leaning wall, finding height of, 778 Learning curve, 152 Learning rate and amount learned, measuring, 857 Learning theory project, 554 Lemon tree, maximum yield, 393 Length of violin string and frequency, 479 Letter arrangements, 1162 LGBTQ marriage, U.S. public opinion on, 876 License plates, 1155 Life, most timeconsuming activities during, 164 Life events, sense of humor and response to, 136–137, 148–149 Life expectancy, 164, 298 Life span, heart rate and, 490 Light intensity, 491 545 Light reflectance and parabolic surface, 1057, 1093, 1094 Light waves, modeling, 693 Linear speed, 586 of airplane propeller, 692 of animals on carousel, 582, 586 of wind machine propeller, 583 Line formation, 1163 Literacy and child mortality, 285, 299 Little League baseball team batting order, 1155–1156 Living alone, number of Americans, 372 Living arrangements, of young adults, 301 304–305 Longdistance telephone charges, 166 Lottery numbers selection, 1162, 1169–1170 probability of winning, 1153, 1169–1170, 1177, 1178, 1184, 1185 LOTTO numbers selection, 1162 probability of winning, 1178 Loudness, 287, 483, 490, 521 532, 561 569 Love, course of over time, 234 Luggage, volume of carryon, 437–438 Lunch menus, 928, 1162
M Mach speed of aircraft, 735 Mailing costs, 283
Applications Index Mall browsing time and average amount spent, 494, 495 Mammography screening data, 1165 Mandatory drug testing, probability of accurate results, 1178 Mandelbrot set, 813, 823, 825 Manufacturing and testing, hours needed for, 960 Manufacturing constraints, 922, 924, 925 Manufacturing costs. See also Cost and revenue functions/breakeven points bicycles, 263 calculator, 489 PDAs, 908 portable satellite radio players, 491 tents, 933 virtual reality headsets, 441 454–455 wheelchair, 455 Maps, making, 599 Marching band, 878 Marijuana use by U.S. high school seniors, 134 Marital status U.S. adults (1970–2013), 875 U.S. population, ages 15 or older (2010), 1075, 1173–1174 Markup, 166 Marriage, interracial percentage of Americans in favor of laws prohibiting, 242–243 rates of, 1095–1096 Marriage equality, U.S. public opinion on, 561 876 Mass attached to spring, simple harmonic motion of, 683–684 Mathematics department personnel, random selection from, 1178 Mathematics exam problems, 1163 Mauna Loa Observatory, 296 Maximum area, 388, 391 440, 487 Maximum height, 857, 1094, 1186 Maximum product, 391 440, 490–491 Maximum profit, 440, 490–491 925, 934 Maximum scores, 928 Maximum yield, 393 Media, trust in, 298 Median age. See Age(s) Median yearly income, 60 Medicaid, 62 Medical treatment, delaying, 521 Medicare, 62 Medication dosage, adult vs. child/infant, 765 Memory retention, 506, 522, 523, 546, 567 Men, in House of Representatives, 876 Merrygoround linear speed of horse on, 626 polar coordinates of horses on, 799 Miles per gallon, 265 Military, gay service members discharged from, 309–310 Minimum product, 387, 487 Miscarriages, by age, 561 Mixture problems, 153, 263, 870–872, 877, 934, 961 Modernistic painting consisting of geometric figures, 887 Moiré patterns, 1041 Moon weight of person given Earth weight, 482 Moth eggs and abdominal width, 423 Mountain, measuring height of, 587, 599, 778–779 Movies ranking, 1162 ticket price of, 241 top ten Oscarwinning, 343 Movie theater, finding best viewing angle in, 661 675, 676 Mt. Rushmore sculpture, height of, 680 Multiplechoice test, 1154–1155, 1162, 1185 Multiplier effect, 1130
Mumps, U.S. cases of, 440 Music amplitude and frequency of note’s sine wave, 741 amusia and, 714, 716 modeling musical sounds, 683, 688
N National debt, 48, 59–60, 62, 63, 120 National diversity index, 119 Natural disaster relief, 928 Nature, Fibonacci numbers found in, 1098 Navigation, 587. See also Bearings Negative life events, sense of humor and response to, 136–137, 148–149 Negative numbers, square roots of, 168 Negative square roots, 177 Net worth, of wealthiest Americans, 48 Neurons in human vs. gorilla brain, 91 Newton’s Law of Cooling, 563 New York, coronavirus cases in, 935 Nutritional content, 950, 960
O Oculus Rift headset manufacturing costs, 441 454–455 Officers for Internet marketing consulting firm, choosing, 1156 Ohm’s law, 177 Oneperson households. See Living alone, number of Americans Online, time spent, 373 Online education, 937 Online games, college students and, 877 Open box lengths and widths, 200 Orbit(s) of comets, 906, 1022, 1037, 1041 1088 modeling, 1079 perigee/apogee of satellite’s orbit, 1026 of planets, 906, 1021 1026 Oscarwinning films, top ten, 343
P Package, forces exerted on held, 835 Pads, cost of, 1186 Panic attack, heart rate before and during, 410 Paragraph formation, 1162 Park, pedestrian route around, 199 Parking lot, dimensions of, 199 Parthenon at Athens, as golden rectangle, 77 Party affiliation, of voters, 938 Passwords, 1162, 1163 Path around swimming pool, dimensions of, 166 Pay. See Salary(ies) Paycheck size, 166 Payroll spent in town, 1183 PDA manufacturing costs and revenues, 908 Pedestrian route around park, 199 Pendulum swings, 1133 Pens color choices, 1162 cost of, 1186 Per capita income and national happiness, 299 Perceived length of time period and age, 483 Perigee/apogee of satellite’s orbit, 1026 Periodic rhythms, 746 Pesteradication program, 1134 Pets insurance for, 283 spending on, 1110 pH of human mouth after eating sugar, 458 scale, 546–547 Phone calls between cities, 474, 483
Phonograph records, angular speed and linear speed of, 583 Photography. See Digital photography Physician visits, 284 Piano keyboard, Fibonacci numbers on, 1098 Pier, finding length of, 778 Pitch of a musical tone, 490 Plane(s). See Aircraft/airplanes Planets elliptical orbits, 1021 1026 length of years for, 217 modeling motion of, 1086, 1088 Playground, dimensions of, 391 Playing cards. See Deck of 52 cards, probability and Play production, breakeven analysis of, 695 Pliocene Epoch, 296 Poker hands, 1160 Pole, angle made by rope anchoring circus tent and, 695 Political affiliation, academic major and, 1178 Political identification, of college freshmen, 369 Pollution, air, 1006 Pollution removal costs, 104 Pool. See Swimming pool Pool dimensions, 166, 199 Pool table, elliptical, 1092 Population Africa, 551 alligator, 202 Asia, 569 bird species in danger of extinction, 560 Bulgaria, 559 California, 545, 1132 Canada, 563 Colombia, 559 elk, 569–570 Europe, 908 exponential growth modeling, 559, 560 Florida, 921 1183 foreignborn (U.S.), 202, 887 geometric growth in, 1122 Germany, 559, 569 gray wolf, 500–501 Hispanic, 568 Hungary, 548 India, 505, 559 Iraq, 559 Israel, 559 Japan, 559 Madagascar, 559 Mexico, 560 New Zealand, 560 Nigeria, 562 over age 65 (U.S.), 562 Pakistan, 559 Palestinian, 559 Philippines, 559 Russia, 559 in scientific notation, 58 single, 304–305 Texas, 545, 1133 tigers, worldwide, 409 Uganda, 563 United States by age, 341 age 65 and older, 562, 625 modeling growth of, 551–552 and national debt, 60 by race/ethnicity, 1113 and walking speed, 555 world, 342, 549, 556–558, 561 569 racial and ethnic breakdown of, 950–951 Population projections, 77, 165, 559 Potassium40, 560 Powerball, probability of winning, 1169–1170 Preprimary education spending, 262
25
26
Applications Index
Presley, Elvis, 886 Price(s) advertising and, 479–480, 484 backpack, 563 computer, 344 of a house, 264, 1135 jeans, 342 of movie ticket, 241 Price reductions, 159, 166, 167, 202, 244, 344, 1179 Pricing options, 236 Prisons, crime rate and, 907 Problem solving, payments for, 167 Problemsolving time, 480 Profit function, 392, 869–870, 874, 908, 922 Profits department store branches, 341 maximizing, 392, 440, 490–491 928, 933, 934 maximum, 490–491 maximum daily, 925, 951 maximum monthly, 928 on newsprint/writing paper, 933 production and sales for gains in, 235 total monthly, 928 Projectiles, paths of, 376, 490, 1077–1078, 1093. See also Ball (height above ground); Freefalling object’s position Propeller of airplane, linear speed of, 692 on wind generator, angular speed of, 692 Public transportation infrastructure, cost of maintaining, 300 Pyramid volume, 490
R Racial diversity, 119 Racial prejudice, Implicit Association Test for, 79, 89 Radiation intensity and distance of radiation machine, 482 Radio(s), production/sales of, 874 Radio show programming, 1162 Radio stations call letters of, 1162 locating illegal, 777 Radio towers on coast, distance of ship from, 1040 Radio waves, simple harmonic motion of, 687 Raffle prizes, 1162, 1163 Railway crossing sign, length of arcs formed by cross on, 585 Rain gutter crosssectional area, 200, 391 Ramp computing work of pulling box along, 849 force and weight of box being pulled along, 839 magnitude of force required to keep object from sliding down, 839 vector components of force on boat on tilted, 845, 849 wheelchair, angle of elevation of, 687 Rate of travel airplane rate, 877 average rate and time traveled, 263 average rate on a roundtrip commute, 117 rowing rate, 877 and time for trip, 477 Razor blades sold, 886 Real estate agents, commission for, 241 Realestate sales and prices (U.S.), 1135 Records, angular speed and linear speed of, 583 Rectangle area of, 77 dimensions of, 199, 202, 242, 327, 472, 878, 903–904, 906, 907, 932, 934, 935, 1004, 1135 dimensions of, maximizing enclosed area, 388 golden, 77 perimeter of, 77, 117, 153
Rectangular box dimensions, 423 Rectangular carpet dimensions, 244 Rectangular garden dimensions of, 373 width of path around, 200 Rectangular sign dimensions, 200 Rectangular solid, volume of, 89 Redwood trees, finding height of, 778 Reflections, 317 Refund, likelihood of tax, 77 Relativity theory, space exploration and, 63, 75, 78 Religious affiliation firstyear U.S. college students claiming no, 249–250 spouses with different, 234 Rental car, 219–220, 243 truck, 235, 1095 Repair bill cost of parts and labor on, 166 estimate, 235 Residential community costs, adult, 1109, 1116 Resistance, electrical, 177, 483, 1186 Respiratory rates, of children, 919 Restaurant tables and maximum occupancy, 877 Resultant forces, 839, 840, 855, 856 Reusable cups, disposable vs., 663 Revenue functions. See Cost and revenue functions/breakeven points Reversibility of thought, 91 Right triangle, isosceles, 200 Roads to expressway, length of, 218 Rolling motion, 1075 Roof of Aframe cabin, finding length of, 854 Rotating beam of light, distance from point, 658, 659 Roulette wheel, independent events on, 1175 Rowing, speed with/against current, 877 Royal flush (poker hand), probability of, 1162 Rug’s length and width, 906 Runner’s pulse, 547
S Sailing angle to 10knot wind, sailing speed and, 799, 810 Salary(ies) after college, 240 anticipated earnings, 1133 choosing between pay arrangements, 373 college education and, 164–165 college graduates with undergraduate degrees, 155–156 comparing, 1118 gross amount per paycheck, 166 lifetime computation, 1126, 1133 and paycheck size, 166 salesperson’s earnings/commissions, 1186 in sixth year, 1182 summer sales job, 373 total, 1119, 1133, 1182, 1184 total weekly earnings, 928 wage gap in, by gender, 263 weekly, 153 Sale prices, 102. See also Price reductions Sales figures, 479–480, 484 Salesperson’s earnings, 1186 Sales volume/figures real estate, 1135 theater ticket, 886 Satellite, perigee/apogee of orbit, 1026 Satellite dish, placement of receiver for, 1093 Satellite radio players, manufacturing costs of, 491 Savings and compound interest, 545–546 geometric sequencing, 1132, 1133
needed for healthcare expenses during retirement, 561 total, 1133 Scattering experiments, 1040 Scheduling appearances, ways of, 1162, 1163 Semielliptical archway and truck clearance, 1022, 1025, 1057, 1092 Sense of humor, depression and, 136–137 Service contract, HVAC company, 460 Shaded region areas, 89, 102 Shading process, 1134 Shadow, length of, 838 Ship(s). See Boats/ships Shipping cost, 283 Ship tracking system, 906 “Shortest time” problems, 1075 Shot put angle and height of, 390–391 path of, given angle, 199 throwing distance, 735, 778 Shower, water used when taking, 1041 Simple harmonic motion, 857, 1186 ball attached to spring, 681–682, 760, 764 earthquake, 684 modeling, 681–684, 687, 694, 695 radio waves, 687 tuning fork, 687 Skeletons, carbon14 dating of, 560 Skydiver’s fall, 476–477, 490, 1135 Sled, pulling computing work of, 848 forces exerted, 838 Sleep average number of awakenings during night, by age and gender, 134 coffee consumption and, 570 college students and, 877 death rate and hours of, 879, 883 hours of, on typical night, 1164 Smoking deaths and disease incidence ratios, 459, 1150–1151 and heart disease, 460 Soccer field dimension, 166 Social media use by age, 931 Social Security benefits/costs, 62, 244 Sonic boom, hyperbolic shape of, 1037 Sound amplitude and frequency of, 741 from touching buttons on touchtone phone, 737, 743 Sound intensity, 287, 483, 490, 521 532, 561 569, 741 Sound quality, amusia and, 714, 716 Space exploration and relativity theory, 63, 75, 78 Space flight/travel aging rate and, 63, 75, 78 Freedom 7 spacecraft, 354 Hubble Space Telescope, 484 relativity theory and, 63, 75, 78 Spaceguard Survey, 1041 Speaker loudness, 490 Speed. See also Rate of travel angular, 582, 692 linear, 586 of airplane propeller, 692 of animals on carousel, 582, 586 of wind machine propeller, 583 Mach speed of aircraft, 735 Spinner, probability of pointer landing in specific way, 1173, 1177, 1184, 1185 Spouses with different faiths, 234 Spring(s) simple harmonic motion of object attached to, 681–684 ball, 681–682, 760, 764 distance from rest position, 685, 694
Applications Index frequency of, 685 maximum displacement of, 685 phase shift of motion, 685 time required for one cycle, 685 Spring, force required to stretch, 482 Square, length of side of, 200 Stadium seats, 1119 Standbys for airline seats, 1162 Statue of Liberty, distance of ship from base of, 686 Stereo speaker loudness, 490 Stolen plants, 167 Stomach acid, pH of, 547 Stonehenge, raising stones of, 602 Stopping distances for car, 471–472 for motorcycles at selected speeds, 489–490 for trucks, 472 Stories, matching graphs with, 135 Stress levels, 389 String length and frequency, 479 Strontium90, 552 Student government elections, 1158 Student loan debt, 91 Students, probability of selecting specific, 1185 Study, hours per week, 885–886 Sun, finding angle of elevation of, 598–599, 601 626, 687, 692 Sunscreen, exposure time without burning and, 30 Supply and demand, 874–875 Supplyside economics, 424 Surface sunlight, intensity beneath ocean’s surface, 545 Surveying bearings in, 680–681 to find distance between two points on opposite banks of river, 777 Sushi, population who won’t try, 561 Suspension bridges, parabolas formed by, 1093 Swimming pool path around, 166, 200 tile border, 201 Synthesizers, musical sounds modeled by, 677, 683 Systolic blood pressure, age and, 193–194
T Tablet case manufacturing, 341 Table tennis table, dimensions of, 932 Talent contest, picking winner and runnerup in, 1163 Target, probability of hitting, 1178 Target heart rate for exercise, 46 Task mastery, 532, 567 Taxes bills, 235 cigarette, 1107 federal tax rate schedule for tax owed, 283 income corporate, 202 federal, 264 inn charges before, 166 rebate and multiplier effect, 1130, 1134 refund likelihood and age, 77 tax rate percentage and revenue, 424 Taxi rates, 165 Teacher’s aide, hourly pay for, 927 Teenage drug use, 561 Telephone(s) land lines vs. cell phones, 875 sound from touching buttons on, 737, 743 Telephone numbers sevendigit, 1185 total possible, in United States, 1155 Telephone plans cellular plans, 264 texting plans, 164
Telephone pole angle between guy wire and, 602 tilted, finding length of, 778 Television highestpaid actors and actresses, 246–249 internet streaming services, 203 manufacturing profits and constraints, 927 programming of movies, 1162 sale prices, 102 screen area, 195 screen dimensions, 194–195, 473, 906 viewing, by annual income, 214 Temperature average monthly, 646, 647 body, variation in, 693, 921 of cooling cup of coffee, 566 degreedays, 1120 and depth of water, 482 in enclosed vehicle, increase in, 517–518 FahrenheitCelsius interconversions, 45, 234, 369 global warming, 245, 294–296 home temperature as function of time, 326–327 increase in an enclosed vehicle, 561 as magnitude, 826 Newton’s Law of Cooling, 563 timetemperature flu scenario, 265–266 Tennis club payment options, 167 Tennis court dimensions, 166 Texting plans, 164 Theater attendance, maximizing revenue from, 928 Theater seats, 1119, 1182 Theater ticket sales, 886 Thefts in U.S., 488 Thorium229 560 Throwing distance, 725, 735 angle of elevation of, 760, 764 maximum height of thrown ball, 857 shot put, 735, 778 Ticket prices/sales, 886 movie ticket prices, 241 Tides, behavior of, 604, 621 624 modeling cycle of, 643 modeling water depth and, 646 Tigers, worldwide population, 409 Time, perceived length of, 483 Time spent online, in U.S., 373 Time traveled, average rate and, 263 Tobacco use. See Smoking Tolls, 154, 157–159, 165, 202, 230–231 235, 342, 660 Touchtone phone, sounds from touching buttons on, 737, 743 Tower angle of elevation between point on ground and top of, 695, 825 height of, finding, 678, 686, 687, 825 length of two guy wires anchoring, 787 Traffic control, 952, 956–961 1007 Trains leaving station at same time, distance between, 812 Transformations of an image, 971–972, 974, 1007 Tree, finding height of, 812 Triangle area of, 769, 784, 1003 dimensions of, 976, 990, 1026, 1079, 1179 isosceles, 200, 877 oblique, 769 right, 897 Triangular piece of land cost of, 788, 854 length of sides of, 854 Trucks clearance under semielliptical archway, 1022, 1024, 1057, 1092 rental costs, 1095 stopping distances required for, 472 Trust, in government and media, 298
27
Tugboats towing ship, resultant force of two, 839, 840 Tuition, government aid for, 242 Tuning fork eardrum vibrations from, 724 simple harmonic motion on, 687 Tutoring, hourly pay for, 927 TV. See Television
U Unemployment and years of education, 489
V Vacation lodgings, 919 Vacation plan packages, cost of, 932 Vaccine, mixture for flu, 263 Value of an annuity, 1133, 1182 of house, inflation rate and, 506 of investments, 502–503, 506, 542 Van, groups fitting into, 1162 Vehicle fatalities, driver’s age and, 199 Velocity vector of boat, 836 of plane, 836 of wind, 835, 836, 839–840 Vertical pole supported by wire, 202, 244 Video games, retail sales of, 548 Videos rented, number of oneday and threeday, 849 Violent crime, imprisonment and, 907 Violin string length and frequency, 479 Virtual reality headset manufacturing costs, 441 454–455 Vitamin content, 950, 960 Volume (sound). See Sound intensity Volume (space) of carryon luggage, 437–438 of cone, 481 for given regions, 102 of open box, 89 of solid, 439 Voters age and gender of, 975 age and party affiliation of, 938
W Wage gap, 263, 393 Wages. See Salary(ies) Wagon, computing work of pulling, 847, 849, 856 Walking speed and city population, 555 Wardrobe selection, 1153 Washington Monument, angle of elevation to top of, 602 Water pressure and depth, 474–475 temperature and depth, 482 used in a shower, 476 used when taking a shower, 1041 Water pipe diameter, number of houses served and size of, 482 Water supply produced by snowpack, 490 Water wheel, linear speed of, 586 Wealthiest Americans, net worth of, 48 Weight blood volume and body, 121 475–476 elephant’s, age and, 547 of great white shark, cube of its length and, 477 healthy, for height and age, 920 and height recommendations/calculations, 482 of human fetus, age and, 242 moon weight of person given Earth weight, 482
28
Applications Index
Weightlifting, 562, 841 850 West Side Story (movie), 343 Wheelchair business manufacturing costs, 455 profit function for, 870 revenue and cost functions for, 868–870 Wheelchair ramp angle of elevation of, 687 vertical distance of, 199 Wheel rotation, centimeters moved with, 585 Whispering gallery, 1021 1026, 1094 Whitecollar jobs, in U.S., 568 White House, rooms, bathrooms, fireplaces, and elevators in, 951 Wildfires, area burned by, 410 Will distribution, 167
Wind, velocity vector of, 835, 836, 839–840 Wind force, 483 Wind generator angular speed of propeller on, 692 linear speed of propeller of, 583 Wind pressure, 483 Wine consumption, heart disease and, 299–300 Wing span of jet fighter, finding, 779 Wire length, 200 Witch of Agnesi, 1078 Women. See also Gender average level of happiness at different times of day, 353 college gender ratios and campus mergers, 877
in House of Representatives, 876 in the labor force, 217 Work, 847–849 crane lifting boulder, 849 dragging crate, 856 pulling box up ramp, 849 pulling wagon, 847, 849, 856 pushing car, 847, 849 of weightlifter, 841 850 Writing pads, cost of, 1186
X Xanax, halflife of, 560, 1079
PREREQUISITES
P
Fundamental Concepts of Algebra
What can algebra possibly have to tell me about • the skyrocketing cost of a college education? • studentloan debt? • my workouts? • the effects of alcohol? • the meaning of the national debt that is more than $25 trillion? • time dilation on a futuristic highspeed journey to a nearby star? • racial bias? • ethnic diversity in the United States? • the widening imbalance between numbers of women and men on college campuses? This chapter reviews fundamental concepts of algebra that are prerequisites for the study of college algebra. Throughout the chapter, you will see how the special language of algebra describes your world.
Here’s where you’ll find these applications: • College costs: Section P.1, Example 2; Exercise Set P.1, Exercises 131–132 • Studentloan debt: MidChapter Check Point, Exercise 31 • Workouts: Exercise Set P.1, Exercises 129–130 • The effects of alcohol: Blitzer Bonus beginning on page 43 • The national debt: Section P.2, Example 12 • Time dilation: Blitzer Bonus on page 75 • Racial bias: Exercise Set P.4, Exercises 91–92 • U.S. ethnic diversity: Chapter P Review, Exercise 23 • College gender imbalance: Chapter P Test, Exercise 32.
30
Chapter P Prerequisites: Fundamental Concepts of Algebra
SECTION P.1
Algebraic Expressions, Mathematical Models, and Real Numbers
4 Find the union of two sets.
How would your lifestyle change if a gallon of gas cost $9.15? Or if the price of a staple such as milk was $15? That’s how much those products would cost if their prices had increased at the same rate college tuition has increased since 1980. (Source: Center for College Affordability and Productivity) In this section, you will learn how the special language of algebra describes your world, including the skyrocketing cost of a college education.
5 Recognize subsets of the
Algebraic Expressions
WHAT YOU’LL LEARN 1 Evaluate algebraic expressions.
2 Use mathematical models. 3 Find the intersection of two sets.
real numbers.
6 Use inequality symbols. 7 Evaluate absolute value.
8 Use absolute value to express distance.
9 Identify properties of the real numbers.
10 Simplify algebraic expressions.
Algebra uses letters, such as x and y, to represent numbers. If a letter is used to represent various numbers, it is called a variable. For example, imagine that you are basking in the sun on the beach. We can let x represent the number of minutes that you can stay in the sun without burning with no sunscreen. With a number 6 sunscreen, exposure time without burning is six times as long, or 6 times x. This can be written 6 # x, but it is usually expressed as 6x. Placing a number and a letter next to one another indicates multiplication. Notice that 6x combines the number 6 and the variable x using the operation of multiplication. A combination of variables and numbers using the operations of addition, subtraction, multiplication, or division, as well as powers or roots, is called an algebraic expression. Here are some examples of algebraic expressions: x , 3x + 5, x 2  3, 1x + 7. 6 Many algebraic expressions involve exponents. For example, the algebraic expression x + 6, x  6, 6x,
x 2 + 361x + 3193 approximates the average cost of tuition and fees at public U.S. colleges for the school year ending x years after 2000. The expression x 2 means x # x and is read “x to the second power” or “x squared.” The exponent, 2, indicates that the base, x, appears as a factor two times. The negative sign in front of x 2 indicates that x 2 is multiplied by 1. Exponential Notation If n is a counting number (1, 2, 3, and so on), 'ZRQPGPVQT2QYGT
b n = b ∙ b ∙ b ∙ … ∙ b. $CUG
b appears as a factor n times.
bn is read “the nth power of b” or “b to the nth power.” Thus, the nth power of b is defined as the product of n factors of b. The expression bn is called an exponential expression. Furthermore, b1 = b. For example,
8 2 = 8 # 8 = 64, 53 = 5 # 5 # 5 = 125, and 24 = 2 # 2 # 2 # 2 = 16.
Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers
1 Evaluate algebraic expressions.
31
Evaluating Algebraic Expressions Evaluating an algebraic expression means to find the value of the expression for a given value of the variable. Many algebraic expressions involve more than one operation. Evaluating an algebraic expression without a calculator involves carefully applying the following order of operations agreement: The Order of Operations Agreement 1. Perform operations within the innermost parentheses and work outward. If the algebraic expression involves a fraction, treat the numerator and the denominator as if they were each enclosed in parentheses. 2. Evaluate all exponential expressions. 3. Perform multiplications and divisions as they occur, working from left to right. 4. Perform additions and subtractions as they occur, working from left to right. Evaluating an Algebraic Expression
EXAMPLE 1
Evaluate 7 + 5(x  4)3 for x = 6. Solution 7 + 5(x  4)3 = = = = =
7 + 7 + 7 + 7 + 47
CHECK POINT 1
2 Use mathematical models.
5(6  4)3 5(2)3 5(8) 40
Replace x with 6. First work inside parentheses: 6 − 4 = 2. Evaluate the exponential expression: 23 = 2 # 2 # 2 = 8. Multiply: 5(8) = 40. Add.
Evaluate 8 + 6(x  3)2 for x = 13.
Formulas and Mathematical Models An equation is formed when an equal sign is placed between two algebraic expressions. One aim of algebra is to provide a compact, symbolic description of the world. These descriptions involve the use of formulas. A formula is an equation that uses variables to express a relationship between two or more quantities. Here are two examples of formulas related to heart rate and exercise.
Working It
CouchPotato Exercise H= *GCTVTCVGKP DGCVURGTOKPWVG
KU
1 5
(220 − a) QH
VJGFKHHGTGPEGDGVYGGP CPF[QWTCIG
*GCTVTCVGKP DGCVURGTOKPWVG
H=
9 (220 10
KU
QH
− a) VJGFKHHGTGPEGDGVYGGP CPF[QWTCIG
Chapter P Prerequisites: Fundamental Concepts of Algebra
The process of finding formulas to describe realworld phenomena is called mathematical modeling. Such formulas, together with the meaning assigned to the variables, are called mathematical models. We often say that these formulas model, or describe, the relationships among the variables.
EXAMPLE 2
Modeling the Cost of Attending a Public College
The bar graph in Figure P.1 shows the average cost of tuition and fees for public fouryear colleges, adjusted for inflation. The formula T = x 2 + 361x + 3193 models the average cost of tuition and fees, T, for public U.S. colleges for the school year ending x years after 2000. a. Use the formula to find the average cost of tuition and fees at public U.S. colleges for the school year ending in 2020. b. By how much does the formula underestimate or overestimate the actual cost shown in Figure P.1?
Average Cost of Tuition and Fees at Public FourYear U.S. Colleges
Tuition and Fees
32
$11,000 $10,500 $10,000 $9500 $9000 $8500 $8000 $7500 $7000 $6500 $6000 $5500 $5000 $4500 $4000 $3500 $3000
10,440
8778
9037
8312 7713 6717 5943 5351 4587 3349
3735
2000 2002 2004 2006 2008 2010 2012 2014 2016 2018 2020 Ending Year in the School Year
Figure P.1 Source: The College Board
Solution a. Because 2020 is 20 years after 2000, we substitute 20 for x in the given formula. Then we use the order of operations to find T, the average cost of tuition and fees for the school year ending in 2020.
1PN[KUCHHGEVGF D[VJGGZRQPGPV 5SWCTGCPF EQR[VJGPGICVKXG
T = −x2 + 361x + 3193 T = −(20)2 + 361(20) + 3193 T = −(400) + 361(20) + 3193
Evaluate the exponential expression:
T = −400 + 7220 + 3193
Multiply from left to right: − (400) = − 114002
This is the given mathematical model. Replace each occurrence of x with 20. 202 = 20 # 20 = 400.
= − 400 and 361(20) = 7220.
T = 10,013
Add.
The formula indicates that for the school year ending in 2020, the average cost of tuition and fees at public U.S. colleges was $10,013. b. Figure P.1 shows that the average cost of tuition and fees for the school year ending in 2020 was $10,440. The cost obtained from the formula, $10,013, underestimates the actual data value by +10,440  +10,013, or by $427.
Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers
BLITZER BONUS
33
Is College Worthwhile?
“Questions have intensified about whether going to college is worthwhile,” says Education Pays, released by the College Board Advocacy & Policy Center. “For the typical student, the investment pays off very well over the course of a lifetime, even considering the expense.” Among the findings in Education Pays: • Median (middlemost) fulltime earnings with a bachelor’s degree in 2018 were $65,400, which is $24,900 more than high school graduates. • Compared with a high school graduate, a fouryear college graduate who enrolled in a public university at age 18 will break even by age 33. The college graduate will have earned enough by then to compensate for being out of the labor force for four years and for borrowing enough to pay tuition and fees, shown in Figure P.1. CHECK POINT 2
DISCOVERY Using the formula from Example 2 and Check Point 2, find T for x = 100, x = 200, x = 300, and x = 400. What happens to the values of T over time? Do you see how model breakdown has occurred?
a. Use the formula T = x 2 + 361x + 3193, described in Example 2, to find the average cost of tuition and fees at public U.S. colleges for the school year ending in 2016. b. By how much does the formula underestimate or overestimate the actual cost shown in Figure P.1? Sometimes a mathematical model gives an estimate that is not a good approximation or is extended to include values of the variable that do not make sense. In these cases, we say that model breakdown has occurred. For example, it is not likely that the formula in Example 2 would give a good estimate of tuition and fees in 2050 because it is too far in the future. Thus, model breakdown would occur.
Sets
GREAT QUESTION Can I use symbols other than braces when writing sets using the roster method? No. Grouping symbols such as parentheses, ( ), and square brackets, [ ], are not used to represent sets in the roster method. Furthermore, only commas are used to separate the elements of a set. Separators such as colons or semicolons are not used.
Before we describe the set of real numbers, let’s be sure you are familiar with some basic ideas about sets. A set is a collection of objects whose contents can be clearly determined. The objects in a set are called the elements of the set. For example, the set of numbers used for counting can be represented by {1, 2, 3, 4, 5, c}. The braces, { }, indicate that we are representing a set. This form of representation, called the roster method, uses commas to separate the elements of the set. The symbol consisting of three dots after the 5, called an ellipsis, indicates that there is no final element and that the listing goes on forever. A set can also be written in setbuilder notation. In this notation, the elements of the set are described but not listed. Here is an example: {x x is a counting number less than 6}. 6JGUGVQHCNNx
UWEJVJCV
xKUCEQWPVKPIPWODGTNGUUVJCP
The same set written using the roster method is {1, 2, 3, 4, 5}.
3 Find the intersection of two sets.
If A and B are sets, we can form a new set consisting of all elements that are in both A and B. This set is called the intersection of the two sets. Definition of the Intersection of Sets The intersection of sets A and B, written A ¨ B, is the set of elements common to both set A and set B. This definition can be expressed in setbuilder notation as follows: A ¨ B = {x x is an element of A AND x is an element of B}.
34
Chapter P Prerequisites: Fundamental Concepts of Algebra
A
B
Figure P.2 shows a useful way of picturing the intersection of sets A and B. The figure indicates that A ¨ B contains those elements that belong to both A and B at the same time.
EXAMPLE 3
A∩B Figure P.2 Picturing the intersection of two sets
Finding the Intersection of Two Sets
Find the intersection: {7, 8, 9, 10, 11} ¨ {6, 8, 10, 12}. Solution 10. Thus,
The elements common to {7, 8, 9, 10, 11} and {6, 8, 10, 12} are 8 and {7, 8, 9, 10, 11} ¨ {6, 8, 10, 12} = {8, 10}.
CHECK POINT 3
Find the intersection: {3, 4, 5, 6, 7} ¨ {3, 7, 8, 9}.
If a set has no elements, it is called the empty set, or the null set, and is represented by the symbol ∅. Here is an example that shows how the empty set can result when finding the intersection of two sets: {2, 4, 6} ∩ {3, 5, 7} = ∅. 6JGUGUGVUJCXGPQ EQOOQPGNGOGPVU
4 Find the union of two sets.
6JGKTKPVGTUGEVKQP JCUPQGNGOGPVU CPFKUVJGGORV[UGV
Another set that we can form from sets A and B consists of elements that are in A or B or in both sets. This set is called the union of the two sets.
Definition of the Union of Sets The union of sets A and B, written A ∪ B, is the set of elements that are members of set A or of set B or of both sets. This definition can be expressed in setbuilder notation as follows: A ∪ B = {x x is an element of A OR x is an element of B}.
A
B
A∪B
Figure P.3 shows a useful way of picturing the union of sets A and B. The figure indicates that A ∪ B is formed by joining the sets together. We can find the union of set A and set B by listing the elements of set A. Then we include any elements of set B that have not already been listed. Enclose all elements that are listed with braces. This shows that the union of two sets is also a set.
Figure P.3 Picturing the union of two sets
EXAMPLE 4 GREAT QUESTION How can I use the words union and intersection to help me distinguish between these two operations? Union, as in a marriage union, suggests joining things or uniting them. Intersection, as in the intersection of two crossing streets, brings to mind the area common to both, suggesting things that overlap.
Finding the Union of Two Sets
Find the union: {7, 8, 9, 10, 11} ∪ {6, 8, 10, 12}. Solution To find {7, 8, 9, 10, 11} ∪ {6, 8, 10, 12}, start by listing all the elements from the first set, namely, 7, 8, 9, 10, and 11. Now list all the elements from the second set that are not in the first set, namely, 6 and 12. The union is the set consisting of all these elements. Thus, {7, 8, 9, 10, 11} ∪ {6, 8, 10, 12} = {6, 7, 8, 9, 10, 11, 12}. #NVJQWIJCPFCRRGCTKPDQVJUGVU
CHECK POINT 4
FQPQVNKUVCPFVYKEG
Find the union: {3, 4, 5, 6, 7} ∪ {3, 7, 8, 9}.
Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers
5 Recognize subsets of the real numbers.
35
The Set of Real Numbers The sets that make up the real numbers are summarized in Table P.1. We refer to these sets as subsets of the real numbers, meaning that all elements in each subset are also elements in the set of real numbers.
Table P.1 Important Subsets of the Real Numbers Name/Symbol
Description
Examples
Natural numbers
{1, 2, 3, 4, 5, c}
ℕ
These are the numbers that we use for counting.
Whole numbers
{0, 1, 2, 3, 4, 5, c }
𝕎
The set of whole numbers includes 0 and the natural numbers.
2, 3, 5, 17 0, 2, 3, 5, 17 17, 5, 3, 2, 0, 2, 3, 5, 17
Integers
{c, 5, 4, 3, 2,  1, 0, 1, 2, 3, 4, 5, c}
ℤ
The set of integers includes the negatives of the natural numbers and the whole numbers.
Rational numbers
{ ba a and b are integers and b ≠ 0 }
ℚ
 17 =
The set of rational numbers is the set of all numbers that can be expressed as a quotient of two integers, with the denominator not 0. Rational numbers can be expressed as terminating or repeating decimals. The set of irrational numbers is the set of all numbers whose decimal representations are neither terminating nor repeating. Irrational numbers cannot be expressed as a quotient of integers.
𝕀
5 =
5 1 ,
 3,  2,
0, 2, 3, 5, 17,
6JKUOGCPUVJCVbKUPQVGSWCNVQ\GTQ
Irrational numbers
 17 1 ,
2 5 = 0.4, 2 3 = 0.6666
c= 0.6
12 ≈ 1.414214  13 ≈ 1.73205 p ≈ 3.142  p2 ≈ 1.571
Notice the use of the symbol ≈ in the examples of irrational numbers. The symbol means “is approximately equal to.” Thus, 22 ≈ 1.414214.
TECHNOLOGY A calculator with a square root key gives a decimal approximation for 22, not the exact value.
Real numbers Rational numbers Integers
Irrational numbers
Whole numbers Natural numbers
We can verify that this is only an approximation by multiplying 1.414214 by itself. The product is very close to, but not exactly, 2: 1.414214 * 1.414214 = 2.000001237796. Not all square roots are irrational. For example, 225 = 5 because 52 = 5 # 5 = 25. Thus, 225 is a natural number, a whole number, an integer, and a rational number 1 225 = 51 2 . The set of real numbers is formed by taking the union of the sets of rational numbers and irrational numbers. Thus, every real number is either rational or irrational, as shown in Figure P.4.
Real Numbers The set of real numbers is the set of numbers that are either rational or irrational: {x x is rational or x is irrational}.
Figure P.4 Every real number is either rational or irrational.
The symbol ℝ is used to represent the set of real numbers. Thus, ℝ = {x x is rational} ∪ {x x is irrational}.
36
Chapter P Prerequisites: Fundamental Concepts of Algebra
Recognizing Subsets of the Real Numbers
EXAMPLE 5
Consider the following set of numbers: 3 e 7,  , 0, 0.6,25, p, 7.3, 281 f. 4 List the numbers in the set that are a. natural numbers. b. whole numbers. d. rational numbers. e. irrational numbers.
c. integers. f. real numbers.
Solution a. Natural numbers: The natural numbers are the numbers used for counting. The only natural number in the set 5 7,  34 , 0, 0.6, 15, p, 7.3, 181 6 is 181 because 181 = 9. (9 multiplied by itself, or 9 2, is 81.) b. Whole numbers: The whole numbers consist of the natural numbers and 0. The elements of the set 5 7,  34, 0, 0.6, 15, p, 7.3, 181 6 that are whole numbers are 0 and 181. c. Integers: The integers consist of the natural numbers, 0, and the negatives of the natural numbers. The elements of the set 5 7,  34, 0, 0.6, 15, p, 7.3, 181 6 that are integers are 181, 0, and 7. d. Rational numbers: All numbers in the set 5 7,  34, 0, 0.6, 15, p, 7.3, 181 6 that can be expressed as the quotient of integers are rational numbers. These include 7 1 7 = 17 2 ,  34, 0 1 0 = 01 2 , and 181 1 181 = 91 2 . Furthermore, all numbers in the set that are terminating or repeating decimals are also rational numbers. These include 0.6 and 7.3. e. Irrational numbers: The irrational numbers in the set 5 7,  34, 0, 0.6, 15, p, 7.3, 181 6 are 15 1 15 ≈ 2.236 2 and p(p ≈ 3.14). Both 15 and p are only approximately equal to 2.236 and 3.14, respectively. In decimal form, 15 and p neither terminate nor have blocks of repeating digits. f. Real numbers: All the numbers in the given set 5 7,  34, 0, 0.6, 15, p, 7.3, 181 6 are real numbers. CHECK POINT 5
Consider the following set of numbers: e 9, 1.3, 0, 0.3,
p , 29 , 210 f. 2
List the numbers in the set that are a. natural numbers. b. whole numbers. d. rational numbers. e. irrational numbers.
c. integers. f. real numbers.
The Real Number Line The real number line is a graph used to represent the set of real numbers. An arbitrary point, called the origin, is labeled 0. Select a point to the right of 0 and label it 1. The distance from 0 to 1 is called the unit distance. Numbers to the right of the origin are positive and numbers to the left of the origin are negative. The real number line is shown in Figure P.5.
Negative direction
–7
–6
–5
–4
–3
–2
Negative numbers Figure P.5 The real number line
–1
0
1
2
3
4
5
Positive numbers
6
7
Positive direction
Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers
GREAT QUESTION How did you locate !2 as a precise point on the number line in Figure P.6? We used a right triangle with two legs of length 1. The remaining side has a length measuring 12. 1
Real numbers are graphed on a number line by placing a dot at the correct location for each number. The integers are easiest to locate. In Figure P.6, we’ve graphed six rational numbers and three irrational numbers on a real number line. 4CVKQPCN PWODGTU
–2
1
2 !
=
0
–1
1
2
=
3
!≈
–!≈–
√2 0
–
–
+TTCVKQPCN PWODGTU
1
37
!= 4
5
p≈
Figure P.6 Graphing numbers on a real number line
We’ll have lots more to say about right triangles later in the book.
Every real number corresponds to a point on the number line and every point on the number line corresponds to a real number. We say that there is a onetoone correspondence between all the real numbers and all points on a real number line.
6 Use inequality symbols.
Ordering the Real Numbers On the real number line, the real numbers increase from left to right. The lesser of two real numbers is the one farther to the left on a number line. The greater of two real numbers is the one farther to the right on a number line. Look at the number line in Figure P.7. The integers 4 and 1 are graphed.
–5
–4
–3
–2
–1
0
1
2
3
4
5
Figure P.7
Observe that 4 is to the left of 1 on the number line. This means that 4 is less than 1. –4 < –1
–KUNGUUVJCP–DGECWUG–KUVQ VJGNGHVQH–QPVJGPWODGTNKPG
In Figure P.7, we can also observe that 1 is to the right of 4 on the number line. This means that 1 is greater than 4. –1 > –4
–KUITGCVGTVJCP–DGECWUG–KUVQ VJGTKIJVQH–QPVJGPWODGTNKPG
The symbols 6 and 7 are called inequality symbols. These symbols always point to the lesser of the two real numbers when the inequality statement is true. –KUNGUUVJCP–
–4 < –1
The symbol points to − 4, the lesser number.
–KUITGCVGTVJCP–
–1 > –4
The symbol still points to − 4, the lesser number.
The symbols 6 and 7 may be combined with an equal sign, as shown in the following table:
6JKUKPGSWCNKV[KUVTWG KHGKVJGTVJGRCTVQT VJG=RCTVKUVTWG
Symbols
Meaning
Examples
Explanation
a≤b
a is less than or equal to b.
2≤9 9≤9
Because 2 < 9 Because 9 = 9
b≥a
b is greater than or equal to a.
9≥2 2≥2
Because 9 > 2 Because 2 = 2
38
Chapter P Prerequisites: Fundamental Concepts of Algebra
7 Evaluate absolute value.
–3 = 3
5 = 5
Absolute Value The absolute value of a real number a, denoted by a , is the distance from 0 to a on the number line. This distance is always taken to be nonnegative. For example, the real number line in Figure P.8 shows that 3 = 3 and
–5 –4 –3 –2 –1 0 1 2 3 4 5 Figure P.8 Absolute value as the distance from 0
5 = 5.
The absolute value of 3 is 3 because 3 is 3 units from 0 on the number line. The absolute value of 5 is 5 because 5 is 5 units from 0 on the number line. The absolute value of a positive real number or 0 is the number itself. The absolute value of a negative real number, such as 3, is the number without the negative sign. We can define the absolute value of the real number x without referring to a number line. The algebraic definition of the absolute value of x is given as follows: Definition of Absolute Value x = b
x x
if x Ú 0 if x 6 0
If x is nonnegative (that is, x Ú 0), the absolute value of x is the number itself. For example, 5 = 5
p = p
= 1 3
1 3
0 = 0.
CHECK POINT 3
a.
5 Add and subtract square roots.
48x3 = "8x2 = "4x2"2 = "4"x2"2 = 2x"2 Ä 6x
25 A 16
Simplify: b.
2150x 3 22x
.
Adding and Subtracting Square Roots Two or more square roots can be combined using the distributive property provided that they have the same radicand. Such radicals are called like radicals. For example,
GREAT QUESTION
7"11 + 6"11 = (7 + 6)"11 = 13"11.
Should like radicals remind me of like terms?
USWCTGTQQVUQHRNWUUSWCTGTQQVUQHTGUWNVKPUSWCTGTQQVUQH
Yes. Adding or subtracting like radicals is similar to adding or subtracting like terms: 7x + 6x = 13x and 7211 + 6211 = 13211.
EXAMPLE 4
Adding and Subtracting Like Radicals
Add or subtract as indicated: a. 722 + 522 b. 25x  725x. Solution a. 722 + 522 = (7 + 5) 22
Apply the distributive property.
= 1222
Simplify.
b. 25x  725x = 125x  725x = (1  7) 25x = 625x CHECK POINT 4
a. 8213 + 9213
Write 15x as 115x . Apply the distributive property. Simplify.
Add or subtract as indicated: b. 217x  20217x.
In some cases, radicals can be combined once they have been simplified. For example, to add 22 and 28, we can write 28 as 24 # 2 because 4 is a perfect square factor of 8. 22 + 28 = 22 + 24 # 2 = 122 + 222 = (1 + 2) 22 = 322
Chapter P Prerequisites: Fundamental Concepts of Algebra
68
EXAMPLE 5
Combining Radicals That First Require Simplification
Add or subtract as indicated: a. 723 + 212 b. 4250x  6232x. Solution a. 723 + 212
= 723 + 24 # 3
Split 12 into two factors such that one is a perfect square. 24 # 3 = 24 23 = 223
= 723 + 223 = (7 + 2) 23
Apply the distributive property. You will find that this step is usually done mentally.
= 923
Simplify.
b. 4250x  6232x = 4225 # 2x  6216 # 2x = 4 # 522x  6 # 422x
225 # 2x = 22522x = 522x and 216 # 2x = 21622x = 422x.
Multiply: 4 # 5 = 20 and 6 # 4 = 24.
= 2022x  2422x = (20  24) 22x = 422x CHECK POINT 5
Apply the distributive property. Simplify.
Add or subtract as indicated:
a. 5227 + 212
6 Rationalize denominators.
b. 6218x  428x.
Rationalizing Denominators
1 13 The calculator screen in Figure P.11 shows approximate values for and . The 3 13 two approximations are the same. This is not a coincidence:
1
=
√3
1
and
23 . 3
GREAT QUESTION What exactly does rationalizing a denominator do to an irrational number in the denominator? Rationalizing a numerical denominator makes that denominator a rational number.
1 √3
∙
√3 √3
=
√3 √9
=
√3 . 3
#P[PWODGTFKXKFGFD[KVUGNHKU /WNVKRNKECVKQPD[FQGUPQVEJCPIG VJGXCNWGQH √
Figure P.11 The calculator screen shows approximate values for 23
25 is the greatest perfect square factor of 50x and 16 is the greatest perfect square factor of 32x.
This process involves rewriting a radical expression as an equivalent expression in which the denominator no longer contains any radicals. The process is called rationalizing the denominator. If the denominator consists of the square root of a natural number that is not a perfect square, multiply the numerator and the denominator by the smallest number that produces the square root of a perfect square in the denominator.
EXAMPLE 6
Rationalizing Denominators
Rationalize the denominator: 15 12 a. b. . 26 28
Section P.3 Radicals and Rational Exponents
GREAT QUESTION
Solution
by 26, the 26 denominator becomes 26 # 26 = 236 = 6. Therefore, we multiply by 1, 26 choosing for 1. 26
Rationalizing the denominator, or even the numerator, is used to simplify fractional expressions containing radicals in this and other math classes. However, the simplest reason to rationalize a denominator has to do with number sense. Let’s say you know that 1 1 . You may not have Then 12 ≈ 1.4 a good sense of how big a number 1 like 1.4 is. Now let’s rationalize the 1 denominator of 12 : 1 12
# 12 12
=
12 2
≈
1.4 2
15
=
"6
15 "6
∙
"6 "6
=
15"6 "36
=
15"6 5"6 = 6 2 5KORNKH[ =
/WNVKRN[D[
÷ ÷
=
b. The smallest number that will produce the square root of a perfect square in 12 the denominator of is 22 , because 28 # 22 = 216 = 4. We multiply 28 22 by 1, choosing for 1. 22 12
= 0.7.
Notice how it’s easier to divide by 2 in your head than to divide by 1.4, making it easier to make a 1 mental approximation of 12 .
15
a. If we multiply the numerator and the denominator of
What is the point of rationalizing the denominator of a fraction?
12 ≈ 1.4.
69
12
28 CHECK POINT 6
a.
5 23
=
# 22
28 22
1222 =
216
=
1222 = 322 4
Rationalize the denominator: 6 b. . 212
Radical expressions that involve the sum and difference of the same two terms are called conjugates. Thus, 1a + 2b and
1a  2b
are conjugates. Conjugates are used to rationalize denominators because the product of such a pair contains no radicals: A!a + √bBA!a − √bB
/WNVKRN[GCEJVGTOQH√a–√b D[GCEJVGTOQH√a+√b
= !a A!a − √bB + √b A!a − √bB &KUVTKDWVG√a QXGT√a–√b
&KUVTKDWVG√b QXGT√a–√b
= !a ∙ !a − !a ∙ √b + √b ∙ !a − √b ∙ √b = A!aB − √ab + √ab − A√bB 2
2
–√ab+√ab=
= A!aB − A√bB 2
2
= a − b.
Multiplying Conjugates
1 1a
+ 2b 2 1 1a  2b 2 =
1 1a 2 2

1 2b 2 2
= a  b
70
Chapter P Prerequisites: Fundamental Concepts of Algebra
How can we rationalize a denominator if the denominator contains two terms with one or more square roots? Multiply the numerator and the denominator by the conjugate of the denominator. Here are three examples of such expressions:
•
7
•
5 + "3
6JGEQPLWICVGQHVJG FGPQOKPCVQTKU–√
1 1a
8
•
3"2 − 4
6JGEQPLWICVGQHVJG FGPQOKPCVQTKU√+
+ 2b 2 1 1a  2b 2 =
h "x + h − "x
6JGEQPLWICVGQHVJG FGPQOKPCVQTKU√x+h+√x
1 1a 2 2
1 2b 2 2
The product of the denominator and its conjugate is found using the formula
The simplified product will not contain a radical.
EXAMPLE 7

= a  b.
Rationalizing a Denominator Containing Two Terms
Rationalize the denominator:
7 5 + 23
.
Solution The conjugate of the denominator is 5  23. If we multiply the numerator and denominator by 5  23, the simplified denominator will not 5  23 contain a radical. Therefore, we multiply by 1, choosing for 1. 5  23 7
=
5 + "3
7
∙
5 + "3
5 − "3
5 − "3
/WNVKRN[D[
=
=
7A5 − "3B
52 − A"3B2
=
7A5 − "3B 25 − 3
A√a+√bBA√a–√bB =A√aB–A√bB
7A5 − "3B 35 − 7"3 or 22 22 +PGKVJGTHQTOQHVJGCPUYGTVJGTG KUPQTCFKECNKPVJGFGPQOKPCVQT
CHECK POINT 7
7 Evaluate and perform
operations with higher roots.
Rationalize the denominator:
8 4 + 25
.
Other Kinds of Roots n
We define the principal nth root of a real number a, symbolized by 1a, as follows: Definition of the Principal nth Root of a Real Number n
1a = b means that bn = a. If n, the index, is even, then a is nonnegative (a Ú 0) and b is also nonnegative (b Ú 0). If n is odd, a and b can be any real numbers. For example, 3 2 64 = 4 because 43 = 64 and
5 2 32 = 2 because ( 2)5 = 32.
The same vocabulary that we learned for square roots applies to nth roots. The n symbol 1 is called a radical and the expression under the radical is called the radicand.
Section P.3 Radicals and Rational Exponents
GREAT QUESTION Should I know the higher roots of certain numbers by heart? Some higher roots occur so frequently that you might want to memorize them. Cube Roots 3
11 = 1
3 1 125 = 5
3 1 8 = 2
3 1 216 = 6
3 1 27 = 3
3 1 1000 = 10
A number that is the nth power of a rational number is called a perfect nth power. For example, 8 is a perfect third power, or perfect cube, because 8 = 23. Thus, 3 3 3 2 8 = 2 2 = 2. In general, one of the following rules can be used to find the nth root of a perfect nth power: Finding nth Roots of Perfect nth Powers n
If n is odd, 2an = a. n If n is even, 2an = a . For example, 3
"(–2)3 = –2
3 1 64 = 4
Fourth Roots
11 = 1
4
116 = 2
5 1 32 = 2
4 1 81 = 3
5 1 243 = 3
11 = 1
4 1 256 = 4 4 1 625 = 5
4
"(–2)4 = –2 = 2.
and
#DUQNWVGXCNWGKUPQVPGGFGFYKVJQFF TQQVUDWVKUPGEGUUCT[YKVJGXGPTQQVU
Fifth Roots
4
71
5
The Product and Quotient Rules for Other Roots The product and quotient rules apply to cube roots, fourth roots, and all higher roots. The Product and Quotient Rules for nth Roots For all real numbers a and b, where the indicated roots represent real numbers, n
n
n
√ab = √a ∙ √b
n
n
n
√a ∙ √b = √ab
and
6JGnVJTQQVQHCRTQFWEV KUVJGRTQFWEVQHVJGnVJTQQVU
6JGRTQFWEVQHVYQnVJTQQVU KUVJGnVJTQQVQHVJGRTQFWEV QHVJGTCFKECPFU
n
n
a √a = n ,b≠0 Äb √b n
√a n a = , b ≠ 0. n Ä b √b
and
6JGnVJTQQVQHCSWQVKGPV KUVJGSWQVKGPVQHVJGnVJTQQVU
6JGSWQVKGPVQHVYQnVJTQQVU KUVJGnVJTQQVQHVJGSWQVKGPV QHVJGTCFKECPFU
Simplifying, Multiplying, and Dividing Higher Roots
EXAMPLE 8 Simplify: 3 a. 2 24
4 b. 2 8
# 24 4
c.
81 . A 16 4
Solution
a. 224 = 28 # 3 3
3
3 28 # 2 3 3
= 3 = 22 3
Find the greatest perfect cube that is a factor of 24. 23 = 8, so 8 is a perfect cube and is the greatest perfect cube factor of 24. 2 ab = 1 a # 2 b n
4
n
3 2 8 = 2
b. 28 # 24 = 28 # 4 4 = 2 32 4 = 216 # 2 4
n
4
4 4 = 2 16 # 2 2 4 = 22 2
1a # 2 b = 2 ab n
n
n
Find the greatest perfect fourth power that is a factor of 32. 24 = 16, so 16 is a perfect fourth power and is the greatest perfect fourth power that is a factor of 32. 2 ab = 2 a # 2 b n
n
4 2 16 = 2
n
72
Chapter P Prerequisites: Fundamental Concepts of Algebra 4
c.
81 281 = 4 A 16 216 3 = 2
n
a 1a = n Ab 2b
4
n
4 4 2 81 = 3 because 34 = 81 and 2 16 = 2 because 24 = 16.
Simplify:
CHECK POINT 8
b. 28 # 28
3
5
a. 240
5
c.
125 . A 27 3
We have seen that adding and subtracting square roots often involves simplifying terms. The same idea applies to adding and subtracting higher roots. Combining Cube Roots
EXAMPLE 9
3
3
Subtract: 5216  1122. Solution 3
3 52 16  1122
= 528 # 2  1122 3
3
Factor 16. 8 is the greatest perfect cube factor: 3 23 = 8 and 2 8 = 2.
= 5 # 222  1122 3
3 3 3 3 2 8#2 = 2 82 2 = 22 2
3
3
Multiply: 5 # 2 = 10.
3
= 1022  1122 3
= (10  11) 22 3
Apply the distributive property.
3
= 122 or  22 CHECK POINT 9
8 Understand and use rational exponents.
Simplify.
3 3 Subtract: 3281  423.
Rational Exponents We define rational exponents so that their properties are the same as the properties for integer exponents. For example, we know that exponents are multiplied when an exponential expression is raised to a power. For this to be true,
1 72 2 2
1#
1
1 27 2 2
We also know that
= 72
2
1
= 7 = 7.
= 27 # 27 = 249 = 7. 1
Can you see that the square of both 72 and 27 is 7? It is reasonable to conclude that 1
72
27.
means
1
We can generalize the fact that 72 means 27 with the following definition: 1
The Definition of an n
If 1a represents a real number, where n Ú 2 is an integer, then 1
n
a n = !a.
6JGFGPQOKPCVQTQHVJGTCVKQPCN GZRQPGPVKUVJGTCFKECNoUKPFGZ
Furthermore, a
1 n
1 = a
1 n
=
1 a ≠ 0. n , 1a
Section P.3 Radicals and Rational Exponents
73
1
TECHNOLOGY This graphing utility screen shows that 1 2
264 = 8 and 64 = 8.
EXAMPLE 10
Using the Definition of an
Simplify: 1
1
a. 642
1
1
c. 164
b. 1253
d. ( 27)3

1
e. 64 3.
Solution 1
a. 642 = 264 = 8 1
3
b. 125 3 = √125 = 5 6JGFGPQOKPCVQTKUVJGKPFGZ 1
4
c. –16 4 = –A√16B = –2 6JGDCUGKUCPFVJGPGICVKXGUKIPKUPQVCHHGEVGFD[VJGGZRQPGPV 1
3
d. (–27) 3 = √–27 = –3 2CTGPVJGUGUUJQYVJCVVJGDCUGKU–CPFVJCVVJGPGICVKXGUKIPKUCHHGEVGFD[VJGGZRQPGPV
e. 64

1 3
1 =
1 1 3
=
64
3
264
=
1 4
CHECK POINT 10 Simplify: 1
a. 252
1
1
1
c. 814
b. 8 3

d. ( 8)3
1
e. 27 3.
In Example 10 and Check Point 10, each rational exponent had a numerator of 1. If the numerator is some other integer, we still want to multiply exponents when raising a power to a power. For this reason, 2
1
a 3 = Qa 3 R
2
2
6JKUOGCPU√a
6JKUOGCPU !a
1 13 a 2 2
Thus, 2
a3 =
1
a 3 = Aa2B 3 .
and
3 2 = 2 a.
Do you see that the denominator, 3, of the rational exponent is the same as the index of the radical? The numerator, 2, of the rational exponent serves as an exponent in each of the two radical forms. We generalize these ideas with the following definition: m
The Definition of a n m is a positive rational number, n Ú 2, then n
n
If 1 a represents a real number and a Also, Furthermore, if a
m n
= m
m n
1 1n a 2 m. n
a n = 2am. is a nonzero real number, then a
m n
1 =
m.
an
74
Chapter P Prerequisites: Fundamental Concepts of Algebra m n
The first form of the definition of a , shown again below, involves taking the root first. This form is often preferable because smaller numbers are involved. Notice that the rational exponent consists of two parts, indicated by the following voice balloons: 6JGPWOGTCVQTKUVJGGZRQPGPV m
n
m
a n = A!aB .
6JGFGPQOKPCVQTKUVJGTCFKECNoUKPFGZ
m
Using the Definition of a n
EXAMPLE 11
3
2
Simplify: a. 273 TECHNOLOGY
Solution 2
Here are the calculator keystroke 3 sequences for 81 4: Many Scientific Calculators 81 yx ( 3 + /  , 4 ) =
a. 273 = 3
b. 9 2 = 3 4
c. 81
3 4
81 ^ ( (  ) 3 , 4 ) ENTER.
= 32 = 9
1 281 2 1
=
81
3
c. 81 4.
= 33 = 27
1
Many Graphing Calculators
The parentheses around the exponent may not be necessary on your graphing calculator. However, their use here illustrates the correct order of operations for this computation. Try the keystrokes with the parentheses and then again without them. If you get different results, then you should always enclose rational exponents in parentheses.
3 12 27 2 2
1 29 2 3
=

b. 9 2
4
3
=
1 1 = 3 27 3
CHECK POINT 11 Simplify: 3
4
a. 273

b. 42
2
c. 32 5.
Properties of exponents can be applied to expressions containing rational exponents.
EXAMPLE 12
Simplifying Expressions with Rational Exponents
Simplify using properties of exponents: a.
5
3 1 a5x 2 b a7x 4 b
b.
32x 3 3
.
16x 4
Solution
a. a5x 2 b a7x 4 b = 5 # 7x 2 # x 4 1
3
1
1 3 +4
= 35x 2 5
= 35x 4 32 x3 b£ 3≥ b. 3 = a 16 x4 16x 4
3
Group numerical factors and group variable factors with the same base. When multiplying expressions with the same base, add the exponents. 1 2
+
3 4
=
2 4
+
3 4
=
5 4
5
5
32x 3
5 3 4
= 2x 3
11
= 2x 12
Group numerical factors and group variable factors with the same base. When dividing expressions with the same base, subtract the exponents. 5 3
−
3 4
=
20 12
−
9 12
=
11 12
CHECK POINT 12 Simplify using properties of exponents:
a. a 2x 3 b a 5x 3 b 4
8
b.
20x 4 5x
3 2
.
Section P.3 Radicals and Rational Exponents
75
Rational exponents are sometimes useful for simplifying radicals by reducing the index.
EXAMPLE 13 Simplify:
Reducing the Index of a Radical
9
2x 3.
Solution
3
9
1
3
2x 3 = x 9 = x 3 = 2x CHECK POINT 13 Simplify:
BLITZER BONUS
6 3 2 x.
A Radical Idea: Time Is Relative The Persistence of Memory (1931), Salvador Dali:. © 2011 MOMA/ARS.
What does travel in space have to do with radicals? Imagine that in the future we will be able to travel at velocities approaching the speed of light (approximately 186,000 miles per second). According to Einstein’s theory of special relativity, time would pass more quickly on Earth than it would in the moving spaceship. The specialrelativity equation Ra = Rf
v 2 1  a b B c
gives the aging rate of an astronaut, Ra, relative to the aging rate of a friend, Rf , on Earth. In this formula, v is the astronaut’s speed and c is the speed of light. As the astronaut’s speed approaches the speed of light, we can substitute c for v. 1  a
v 2 b c
Ra = Rf
B
Ra = Rf
c 2 1  a b B c
Einstein’s equation gives the aging rate of an astronaut, Ra , relative to the aging rate of a friend, Rf , on Earth. The velocity, v , is approaching the speed of light, c , so let v = c.
= Rf 21  1
c 2 a b = 12 = 1 # 1 = 1 c
= Rf 20
Simplify the radicand: 1 − 1 = 0.
= Rf # 0 = 0
20 = 0 Multiply: Rf # 0 = 0.
Close to the speed of light, the astronaut’s aging rate, Ra, relative to a friend, Rf , on Earth is nearly 0. What does this mean? As we age here on Earth, the space traveler would barely get older. The space traveler would return to an unknown futuristic world in which friends and loved ones would be long gone.
Chapter P Prerequisites: Fundamental Concepts of Algebra
76
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. The symbol 1 is used to denote the nonnegative, or , square root of a number. C2. 164 = 8 because
= 64.
C3. 2a2 =
C8. The conjugate of 7 + 13 is
.
C9. We rationalize the denominator of 110 5 12 by multiplying the numerator and denominator by 3
C4. The product rule for square roots states that if a and b are . nonnegative, then 1ab =
C10. In the expression 164, the number 3 is called the and the number 64 is called the . 5 C11. 2  32 = 2 because n
C5. The quotient rule for square roots states that if a and b are a = . nonnegative and b ≠ 0, then Ab
C12. If n is odd, 1an =
C6. 823 + 1023 =
C13.
1 an
C14.
3 164
C7. 13 + 175 = 13 + 125 # 3 = 13 + =
13
=  32. .
n
If n is even, 1an = = =
4 12 16 2 3
.
= (__)3 =
P.3 EXERCISE SET Practice Exercises
In Exercises 33–44, add or subtract terms whenever possible.
Evaluate each expression in Exercises 1–12, or indicate that the root is not a real number.
33. 723 + 623
34. 825 + 1125
35. 6217x  8217x
36. 4213x  6213x 38. 220 + 625
1. 236
2. 225
37. 28 + 322
3.  236
4. 2 25
39. 250x  28x
40. 263x  228x
5. 236
6. 2 25
41. 3218 + 5250
42. 4212  2275
7. 225  16
8. 225  16
43. 328  232 + 3272  275
9. 225  216 2
11. 2( 13)
10. 2144 + 225 2
12. 2(  17)
Use the product rule to simplify the expressions in Exercises 13–22. In Exercises 17–22, assume that variables represent nonnegative real numbers. 13. 250 15. 245x 17.
22x # 26x
21. 22x 2 # 26x
16. 2125x 18.
2
210x # 2 8x
1 A 81 49 25. A 16
22. 26x # 23x 2
1 A 49 121 26. A 9
23.
31.
47.
24.
248x 3
28.
23x 2150x
4
30.
23x 2200x 210x
3
1
32.
53.
22
50.
3 + 211 7
52.
25  2 6
27 23 3 3 + 27 5 23  1 11
54. 25 + 23 27  23 Evaluate each expression in Exercises 55–66, or indicate that the root is not a real number. 3
28x
48.
25 13
3 55. 2125
272x 3
3 56. 2 8
3 57. 2 8
4
58. 2 125
59. 2 16
4 60. 2 81
4 61. 2 (  3) 4
4 62. 2 (  2) 4
5 63. 2 ( 3)5
5 64. 2 (  2)5
5 1 65. 2  32
6 1 66. 2 64
Simplify the radical expressions in Exercises 67–74, if possible.
224x 4
3 67. 232
23x
3
70. 2x 5
2500x 210x
49. 51.
20. 2y3
Use the quotient rule to simplify the expressions in Exercises 23–32. Assume that x 7 0.
29.
In Exercises 45–54, rationalize the denominator. 1 2 45. 46. 27 210
14. 227 2
19. 2x 3
27.
44. 3254  2224  296 + 4263
3
1
73.
5 2 64x 6 5 2 2x
3 68. 2 150
71. 29 # 26 3
74.
3
4 2 162x 5 4 2 2x
3 4 69. 2 x
3 3 72. 2 12 # 2 4
.
Section P.3 Radicals and Rational Exponents
a. What percentage of 25yearolds must pay more taxes?
In Exercises 75–82, add or subtract terms whenever possible. 5
5
5
5
75. 422 + 322
76. 623 + 223
3 3 77. 52 16 + 2 54
3 3 78. 32 24 + 2 81
3
3
3
79. 254xy3  y2128x
b. Rewrite the formula by rationalizing the denominator.
3
80. 224xy3  y281x
3
c. Use the rationalized form of the formula from part (b) to find the percentage of 25yearolds who must pay more taxes. Do you get the same answer as you did in part (a)? If so, does this prove that you correctly rationalized the denominator? Explain.
3 82. 23 + 2 15
81. 22 + 28
In Exercises 83–90, evaluate each expression without using a calculator. 1
1
1
83. 362
84. 1212
85. 8 3
1 273
2 1253
2 83
86.
89. 32
87.
4 5
90. 16
88.
116. America is getting older. The graph shows the elderly U.S. population for ages 65–84 and for ages 85 and older in 2010, with projections for 2020 and beyond.
5 2
Projected Elderly United States Population Ages 65–84
In Exercises 91–100, simplify using properties of exponents.
93.
1
1 20x 2
94.
1 5x 4
1 x3 2 3 1
96. 1
1 x5 2 5
1
98. (125x 9 y6)3
2
1 3 3y4 1 y12
1 3
4
97. (25x 4 y6)2 99.
3
3 72x 4
9x
2
95.
1 3x3 2 1 4x4 2 2
92.
Projected Population (millions)
1 7x3 2 1 2x4 2 1
91.
100.
1
2
1 4 2y 5 3 y10
4 101. 252
4 2 102. 2 7
3 6 103. 2 x
4 12 104. 2 x
6 4 105. 2 x
9 6 106. 2 x
108. 3x 4y8
In Exercises 109–110, evaluate each expression. 3 4 109. 3 216 + 2625 3
3 3 110. 43 2169 + 29 + 3 2 1000 + 2 216
In Exercises 111–114, simplify each expression. Assume that all variables represent positive numbers.
113. °
5 1 x 4 y3 3 x 4
¢
6
1 xy2 2
112. (8x 6 y3)3 1 x 6y
1
5
1
114. °
1 7 x2y 4 5 y 4
¢
65.8
64.7
61.9
60 47.3
50 40
34.1
30 20 10
6.1
7.3 2020
15.4
9.6 2030 Year
2040
20.9
2050
Source: U.S. Census Bureau
a. Use the formula to find the projected increase in the number of Americans ages 65–84, in millions, from 2020 to 2050. Express this difference in simplified radical form. b. Use a calculator and write your answer in part (a) to the nearest tenth. Does this rounded decimal overestimate or underestimate the difference in the projected data shown by the bar graph? By how much?
Practice PLUS
1 2
70
The formula E = 52x + 34.1 models the projected number of Americans ages 65–84, E, in millions, x years after 2010.
12
111. (49x 2 y4)
Ages 85+
80
2010
In Exercises 101–108, simplify by reducing the index of the radical.
9 6 3 107. 2 x y
77
4
 13
26
117. The early Greeks believed that the most pleasing of all rectangles were golden rectangles, whose ratio of width to height is w 2 . = h 25  1 The Parthenon at Athens fits into a golden rectangle once the triangular pediment is reconstructed.
Application Exercises 115. Do you expect to pay more taxes than were withheld? Would you be surprised to know that the percentage of taxpayers who receive a refund and the percentage of taxpayers who pay more taxes vary according to age? The formula P =
x113 + 2x2 52x
models the percentage, P, of taxpayers who are x years old who must pay more taxes.
Rationalize the denominator of the given ratio, called the golden ratio. Then use a calculator and find the ratio of width to height, correct to the nearest hundredth, in golden rectangles.
78
Chapter P Prerequisites: Fundamental Concepts of Algebra
118. Use Einstein’s specialrelativity equation v 1  a b , B c 2
Ra = Rf
described in the Blitzer Bonus on page 75, to solve this exercise. You are moving at 90% of the speed of light. Substitute 0.9c for v, your velocity, in the equation. What is your aging rate, correct to two decimal places, relative to a friend on Earth? If you are gone for 44 weeks, approximately how many weeks have passed for your friend? The perimeter, P, of a rectangle with length l and width w is given by the formula P = 2l + 2w. The area, A, is given by the formula A = lw. In Exercises 119–120, use these formulas to find the perimeter and area of each rectangle. Express answers in simplified radical form. Remember that perimeter is measured in linear units, such as feet or meters, and area is measured in square units, such as square feet, ft 2, or square meters, m2.
Critical Thinking Exercises Make Sense? In Exercises 129–132, determine whether each statement makes sense or does not make sense, and explain your reasoning. 129. The unlike radicals 322 and 523 remind me of the unlike terms 3x and 5y that cannot be combined by addition or subtraction. 130. Using my calculator, I determined that 67 = 279,936, so 6 must be a seventh root of 279,936. 131. I simplified the terms of 2220 + 4275, and then I was able to add the like radicals. m 132. When I use the definition for a n , I usually prefer to first raise a to the m power because smaller numbers are involved. In Exercises 133–136, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. 133. 72 # 72 = 49 134. 8 3 = 2 135. The cube root of 8 is not a real number. 220 210 136. = 8 4 1
119. √125 feet
2√20 feet
15
1

2 15
2
1
In Exercises 137–138, fill in each box to make the statement true. 137.
+ 2
138. 2 x
120. 4√20 feet √80 feet
Explaining the Concepts
121. Explain how to simplify 210 # 25. 122. Explain how to add 23 + 212.
123. Describe what it means to rationalize a denominator. Use 1 1 both and in your explanation. 25 5 + 25 3 124. What difference is there in simplifying 2(  5)3 and 4 4 2( 5) ? m
125. What does a n mean? 126. Describe the kinds of numbers that have rational fifth roots. 127. Why must a and b represent nonnegative numbers when we write 1a # 2b = 2ab? Is it necessary to use this restriction 3 3 3 in the case of 2a # 2b = 2ab? Explain.
128. Read the Blitzer Bonus on page 75. The future is now: You have the opportunity to explore the cosmos in a starship traveling near the speed of light. The experience will enable you to understand the mysteries of the universe in deeply personal ways, transporting you to unimagined levels of knowing and being. The downside: You return from your twoyear journey to a futuristic world in which friends and loved ones are long gone. Do you explore space or stay here on Earth? What are the reasons for your choice?
= 5x
 2
7
= 22
7 139. Find the exact value of 13 + 22 + without the A 3 + 22 use of a calculator. 140. Place the correct symbol, 7 or 6, in the shaded area between the given numbers. Do not use a calculator. Then check your result with a calculator. 1
1
a. 32 33 b. 27 + 218 27 + 18 141. a. A mathematics professor recently purchased a birthday cake for her son with the inscription Happy 1 22 # 24 , 24 2 th Birthday. 5
3
1
How old is the son? b. The birthday boy, excited by the inscription on the cake, tried to wolf down the whole thing. Professor Mom, concerned about the possible metamorphosis of her son into a blimp, exclaimed, “Hold on! It is your birthday, so why not take 8
4 3 
3
+ 22 of the cake? I’ll eat half of what’s
16 4 + 21 left over.” How much of the cake did the professor eat?
Preview Exercises Exercises 142–144 will help you prepare for the material covered in the next section. 142. Multiply: (2x 3 y2)(5x 4 y7). 143. Use the distributive property to multiply: 2x 4(8x 4 + 3x). 144. Simplify and express the answer in descending powers of x: 2x(x 2 + 4x + 5) + 3(x 2 + 4x + 5).
Section P.4 Polynomials
SECTION P.4
79
Polynomials
2 Add and subtract
Can that be Axl, your author’s yellow lab, sharing a special moment with a baby chick? And if it is (it is), what possible relevance can this have to polynomials? An answer is promised before you reach the Exercise Set. For now, we open the section by defining and describing polynomials.
3 Multiply polynomials.
How We Define Polynomials
4 Use FOIL in polynomial
More than 2 million people have tested their racial prejudice using an online version of the Implicit Association Test. Most groups’ average scores fall between “slight” Old Dog ... New Chicks and “moderate” bias, but the differences among age groups are intriguing. In this section’s Exercise Set (Exercises 91 and 92), you will be working with a model that measures bias:
WHAT YOU’LL LEARN 1 Understand the vocabulary of polynomials. polynomials.
multiplication.
5 Use special products in
polynomial multiplication.
6 Perform operations with polynomials in several variables.
S = 0.3x 3  2.8x 2 + 6.7x + 30. In this model, S represents the score on the Implicit Association Test. (Higher scores indicate stronger bias.) The variable x represents age group. The algebraic expression that appears on the right side of the model is an example of a polynomial. A polynomial is a single term or the sum of two or more terms containing variables with wholenumber exponents. This particular polynomial contains four terms. Equations containing polynomials are used in such diverse areas as science, business, medicine, psychology, and sociology. In this section, we review basic ideas about polynomials and their operations.
1 Understand the vocabulary of polynomials.
How We Describe Polynomials Consider the polynomial 7x 3  9x 2 + 13x  6. We can express 7x 3  9x 2 + 13x  6 as 7x 3 + ( 9x 2) + 13x + ( 6). The polynomial contains four terms. It is customary to write the terms in the order of descending powers of the variable. This is the standard form of a polynomial. Some polynomials contain only one variable. Each term of such a polynomial in x is of the form ax n. If a ≠ 0, the degree of ax n is n. For example, the degree of the term 7x 3 is 3. The Degree of ax n If a ≠ 0, the degree of ax n is n. The degree of a nonzero constant is 0. The constant 0 has no defined degree.
GREAT QUESTION Why doesn’t the constant 0 have a degree? We can express 0 in many ways, including 0x, 0x 2, and 0x 3. It is impossible to assign a single exponent on the variable. This is why 0 has no defined degree.
Here is an example of a polynomial and the degree of each of its four terms: 6x 4 − 3x 3 − 2x − 5. FGITGG
FGITGG
FGITGG
FGITGGQHPQP\GTQEQPUVCPV
Notice that the exponent on x for the term 2x, meaning 2x 1, is understood to be 1. For this reason, the degree of 2x is 1. You can think of 5 as 5x 0; thus, its degree is 0.
80
Chapter P Prerequisites: Fundamental Concepts of Algebra
A polynomial is simplified when it contains no grouping symbols and no like terms. A simplified polynomial that has exactly one term is called a monomial. A binomial is a simplified polynomial that has two terms. A trinomial is a simplified polynomial with three terms. Simplified polynomials with four or more terms have no special names. The degree of a polynomial is the greatest degree of all the terms of the polynomial. For example, 4x 2 + 3x is a binomial of degree 2 because the degree of the first term is 2, and the degree of the other term is less than 2. Also, 7x 5  2x 2 + 4 is a trinomial of degree 5 because the degree of the first term is 5, and the degrees of the other terms are less than 5. Up to now, we have used x to represent the variable in a polynomial. However, any letter can be used. For example, • 7x 5  3x 3 + 8 • 6y3 + 4y2  y + 3 • z7 + 22
is a polynomial (in x) of degree 5. Because there are three terms, the polynomial is a trinomial. is a polynomial (in y) of degree 3. Because there are four terms, the polynomial has no special name. is a polynomial (in z) of degree 7. Because there are two terms, the polynomial is a binomial.
We can tie together the threads of our discussion with the formal definition of a polynomial in one variable. In this definition, the coefficients of the terms are represented by an (read “a sub n”), an  1 (read “a sub n minus 1”), an  2 , and so on. The small letters to the lower right of each a are called subscripts and are not exponents. Subscripts are used to distinguish one constant from another when a large and undetermined number of such constants are needed.
Definition of a Polynomial in x A polynomial in x is an algebraic expression of the form anx n + an  1x n  1 + an  2x n  2 + g + a1x + a0 , where an , an  1 , an  2 , c , a1 , and a0 are real numbers, an ≠ 0, and n is a nonnegative integer. The polynomial is of degree n, an is the leading coefficient, and a0 is the constant term.
2 Add and subtract polynomials.
Adding and Subtracting Polynomials Polynomials are added and subtracted by combining like terms. For example, we can combine the monomials 9x 3 and 13x 3 using addition as follows: –9x3 + 13x3 = (–9 + 13)x3 = 4x3. 6JGUGNKMGVGTOUDQVJ EQPVCKPxVQVJG VJKTFRQYGT
EXAMPLE 1
#FFEQGHƂEKGPVU CPFMGGRVJGUCOG XCTKCDNGHCEVQTx
Adding and Subtracting Polynomials
Perform the indicated operations and simplify: a. ( 9x 3 + 7x 2  5x + 3) + (13x 3 + 2x 2  8x  6) b. (7x 3  8x 2 + 9x  6)  (2x 3  6x 2  3x + 9).
Section P.4 Polynomials
GREAT QUESTION Can I use a vertical format to add and subtract polynomials? Yes. Arrange like terms in columns and combine vertically: 7x 3  8x 2 + 9x  6  2x 3 + 6x 2 + 3x  9 5x 3  2x 2 + 12x  15 The like terms can be combined by adding their coefficients and keeping the same variable factor. To subtract, change the sign of each term in the polynomial being subtracted and then add.
Solution a. ( 9x 3 + 7x 2  5x + 3) + (13x 3 + 2x 2  8x  6) = ( 9x 3 + 13x 3) + (7x 2 + 2x 2) + ( 5x  8x) + (3  6) = 4x 3 + 9x 2 + ( 13x) + ( 3) 3
2
= 4x + 9x  13x  3 3
2
81
Group like terms. Combine like terms. Simplify.
3
2
b. (7x − 8x + 9x − 6) − (2x − 6x − 3x + 9) %JCPIGVJGUKIPQHGCEJEQGHƂEKGPV
Rewrite subtraction as addition of the additive inverse.
= (7x3 − 8x2 + 9x − 6) + (–2x3 + 6x2 + 3x − 9) = (7x3 − 2x3) + (–8x2 + 6x2) + (9x + 3x) + (–6 − 9) 3
2
= 5x + (–2x ) + 12x + (–15) = 5x3 − 2x2 + 12x − 15
Group like terms. Combine like terms. Simplify.
Perform the indicated operations and simplify: a. ( 17x 3 + 4x 2  11x  5) + (16x 3  3x 2 + 3x  15) CHECK POINT 1
b. (13x 3  9x 2  7x + 1)  ( 7x 3 + 2x 2  5x + 9).
3 Multiply polynomials.
Multiplying Polynomials The product of two monomials is obtained by using properties of exponents. For example, (–8x6)(5x3) = –8 ∙ 5x6+3 = –40x9. /WNVKRN[EQGHƂEKGPVUCPFCFFGZRQPGPVU
Furthermore, we can use the distributive property to multiply a monomial and a polynomial that is not a monomial. For example, GREAT QUESTION Because monomials with the same base and different exponents can be multiplied, can they also be added? No. Don’t confuse adding and multiplying monomials.
3x4(2x3 − 7x + 3) = 3x4 ∙ 2x3 − 3x4 ∙ 7x + 3x4 ∙ 3 = 6x7 − 21x5 + 9x4. /QPQOKCN
6TKPQOKCN
How do we multiply two polynomials if neither is a monomial? For example, consider
Addition: 5x 4 + 6x 4 = 11x 4
(2x + 3)(x2 + 4x + 5).
Multiplication:
(5x 4)(6x 4) = (5 # 6)(x 4 # x 4) = 30x 4 + 4 = 30x
$KPQOKCN
One way to perform (2x + 3)(x 2 + 4x + 5) is to distribute 2x throughout the trinomial
8
Only like terms can be added or subtracted, but unlike terms may be multiplied. Addition: 5x 4 + 3x 2 cannot be simplified.
6TKPQOKCN
2x(x 2 + 4x + 5) and 3 throughout the trinomial 3(x 2 + 4x + 5). Then combine the like terms that result.
Multiplication:
(5x 4)(3x 2) = (5 # 3)(x 4 # x 2) = 15x 4 + 2 = 15x 6
Multiplying Polynomials When Neither Is a Monomial Multiply each term of one polynomial by each term of the other polynomial. Then combine like terms.
82
Chapter P Prerequisites: Fundamental Concepts of Algebra
Multiplying a Binomial and a Trinomial
EXAMPLE 2
Multiply: (2x + 3)(x 2 + 4x + 5). Solution (2x + 3)(x 2 + 4x + 5) = 2x(x 2 + 4x + 5) + 3(x 2 + 4x + 5)
Multiply the trinomial by each
= 2x # x 2 + 2x # 4x + 2x # 5 + 3x 2 + 3 # 4x + 3 # 5 = 2x 3 + 8x 2 + 10x + 3x 2 + 12x + 15
term of the binomial. Use the distributive property. Multiply monomials: Multiply coefficients and add exponents.
= 2x 3 + 11x 2 + 22x + 15
Combine like terms: 8x 2 + 3x 2 = 11x 2 and 10x + 12x = 22x.
Another method for performing the multiplication is to use a vertical format similar to that used for multiplying whole numbers. x2 + 4x + 5 2x + 3 3x2 + 12x + 15 3 2x + 8x2 + 10x
Write like terms in the same column.
3
CHECK POINT 2
multiplication.
x x+x+
2
2x + 11x + 22x + 15
Combine like terms.
4 Use FOIL in polynomial
x+x+
Multiply: (5x  2)(3x 2  5x + 4).
The Product of Two Binomials: FOIL Frequently, we need to find the product of two binomials. One way to perform this multiplication is to distribute each term in the first binomial through the second binomial. For example, we can find the product of the binomials 3x + 2 and 4x + 5 as follows: (3x + 2)(4x + 5) = 3x(4x + 5) + 2(4x + 5) &KUVTKDWVGx QXGTx+
&KUVTKDWVG QXGTx+
= 3x(4x) + 3x(5) + 2(4x) + 2(5) = 12x2 + 15x + 8x + 10. 9G NNEQODKPGVJGUGNKMGVGTOUNCVGT (QTPQYQWTKPVGTGUVKUKPJQYVQQDVCKP GCEJQHVJGUGHQWTVGTOU
We can also find the product of 3x + 2 and 4x + 5 using a method called FOIL, which is based on our preceding work. Any two binomials can be quickly multiplied by using the FOIL method, in which F represents the product of the first terms in each binomial, O represents the product of the outside terms, I represents the product of the inside terms, and L represents the product of the last, or second, terms in each binomial. For example, we can use the FOIL method to find the product of the binomials 3x + 2 and 4x + 5 as follows: ﬁrst
last (
1
+
.
(3x + 2)(4x + 5) = 12x 2 + 15x + 8x + 10 inside outside
2TQFWEV QH (KTUV VGTOU
2TQFWEV QH 1WVUKFG VGTOU
2TQFWEV QH +PUKFG VGTOU
= 12x 2 + 23x + 10.
2TQFWEV QH .CUV VGTOU Combine like terms.
Section P.4 Polynomials
83
In general, here’s how to use the FOIL method to find the product of ax + b and cx + d:
Using the FOIL Method to Multiply Binomials ﬁrst
last (
1
+
.
(ax + b)(cx + d) = ax ∙ cx + ax ∙ d + b ∙ cx + b ∙ d inside outside
EXAMPLE 3
2TQFWEV QH (KTUV VGTOU
2TQFWEV QH 1WVUKFG VGTOU
2TQFWEV QH +PUKFG VGTOU
2TQFWEV QH .CUV VGTOU
Using the FOIL Method
Multiply: (3x + 4)(5x  3). Solution ﬁrst
last (
1
+
.
(3x + 4)(5x − 3) = 3x ∙ 5x + 3x(–3) + 4 ∙ 5x + 4(–3) = 15x 2 − 9x + 20x − 12 inside Combine like terms. = 15x 2 + 11x − 12 outside CHECK POINT 3
5 Use special products in
polynomial multiplication.
Multiply: (7x  5)(4x  3).
Multiplying the Sum and Difference of Two Terms We can use the FOIL method to multiply A + B and A  B as follows: (
1
+
.
(A + B)(A − B) = A2 − AB + AB − B2 = A2 − B2. Notice that the outside and inside products have a sum of 0 and the terms cancel. The FOIL multiplication provides us with a quick rule for multiplying the sum and difference of two terms, referred to as a specialproduct formula.
The Product of the Sum and Difference of Two Terms (A + B)(A − B) = A2 − B2 6JGRTQFWEVQHVJGUWO CPFVJGFKHHGTGPEGQH VJGUCOGVYQVGTOU
KU
VJGUSWCTGQHVJGƂTUV VGTOOKPWUVJGUSWCTG QHVJGUGEQPFVGTO
84
Chapter P Prerequisites: Fundamental Concepts of Algebra
EXAMPLE 4
Finding the Product of the Sum and Difference of Two Terms
Multiply: a. (4y + 3)(4y  3) Solution
b. (5a4 + 6)(5a4  6).
Use the specialproduct formula shown.
(A + B)(A − B)
=
A2
−
B2
(KTUV VGTO USWCTGF
−
5GEQPF VGTO USWCTGF
= (4y)2
a. (4y + 3)(4y − 3)
− 32
b. (5a4 + 6)(5a4 − 6) = (5a4)2 − 62
2TQFWEV
=
= 16y2 − 9 = 25a8 − 36
Multiply:
CHECK POINT 4
b. (2y3  5)(2y3 + 5).
a. (7x + 8)(7x  8)
The Square of a Binomial Let us find (A + B)2, the square of a binomial sum. To do so, we begin with the FOIL method and look for a general rule. (
1
+
.
(A + B)2 = (A + B)(A + B) = A ∙ A + A ∙ B + A ∙ B + B ∙ B = A2 + 2AB + B2. This result implies the following rule, which is another example of a specialproduct formula: The Square of a Binomial Sum (A + B)2
A2
= KU
6JGUSWCTG QHC DKPQOKCNUWO
+
ƂTUV VGTO USWCTGF
2AB
RNWU
B2
+ RNWU
VKOGU VJGRTQFWEV QHVJGVGTOU
NCUV VGTO USWCTGF
GREAT QUESTION When finding (x + 3)2, why can’t I just write x2 + 32, or x2 + 9? Caution! The square of a sum is not the sum of the squares. (A + B)2 ≠ A2 + B2
EXAMPLE 5 Multiply: a. (x + 3)2 Solution
6JGOKFFNGVGTO ABKUOKUUKPI
(x + 3)2 ≠ x2 + 9 +PEQTTGEV
Show that (x + 3)2 and x 2 + 9 are not equal by substituting 5 for x in each expression and simplifying.
Finding the Square of a Binomial Sum b. (3x + 7)2.
Use the specialproduct formula shown.
(A + B)2 =
A2
+
2AB
+
B2
(First Term)2
+
2 # Product of the Terms
+
a. (x + 3)2 =
x2
+
2#x#3
(Last Term)2
+
32
= x 2 + 6x + 9
b. (3x + 7)2 =
(3x)2
+
2(3x)(7)
+
72
= 9x 2 + 42x + 49
CHECK POINT 5
a. (x + 10)2
Multiply: b. (5x + 4)2.
= Product
Section P.4 Polynomials
85
A similar pattern occurs for (A  B)2, the square of a binomial difference. Using the FOIL method on (A  B)2, we obtain the following rule: The Square of a Binomial Difference (A − B)2 KU
6JGUSWCTGQH CDKPQOKCN FKHHGTGPEG
2AB
− OKPWU
ƂTUV VGTO USWCTGF
B2
+ RNWU
VKOGU VJGRTQFWEV QHVJGVGTOU
NCUV VGTO USWCTGF
Finding the Square of a Binomial Difference
EXAMPLE 6 Multiply: a. (x  4)2 Solution
A2
=
b. (5y  6)2.
Use the specialproduct formula shown.
(A  B)2 =
A2

2AB
+
B2
(First Term)2
−
2 # Product of the Terms
+
a. (x  4)2 =
x2

2#x#4
(Last Term)2
+
42
= x 2  8x + 16
b. (5y  6)2 =
(5y)2

2(5y)(6)
+
62
= 25y2  60y + 36
CHECK POINT 6 2
a. (x  9)
= Product
Multiply: b. (7x  3)2.
`
Special Products There are several products that occur so frequently that it’s convenient to memorize the form, or pattern, of these formulas. GREAT QUESTION Do I have to memorize the special products shown in the table on the right? Not necessarily. Although it’s convenient to memorize these forms, the FOIL method can be used on all five examples in the box. To cube x + 4, you can first square x + 4 using FOIL and then multiply this result by x + 4. We suggest memorizing these special forms because they let you multiply far more rapidly than using the FOIL method.
Special Products Let A and B represent real numbers, variables, or algebraic expressions. Special Product
Example
Sum and Difference of Two Terms 2
(2x + 3)(2x  3) = (2x)2  32 = 4x 2  9
2
(A + B)(A  B) = A  B
Squaring a Binomial 2
2
(y + 5)2 = y2 + 2 # y # 5 + 52 = y2 + 10y + 25
2
(A + B) = A + 2AB + B
(A  B)2 = A2  2AB + B2
(3x  4)2 = (3x)2  2 # 3x # 4 + 42 = 9x 2  24x + 16
Cubing a Binomial 3
3
2
2
3
(A + B) = A + 3A B + 3AB + B
(x + 4)3 = x 3 + 3x 2(4) + 3x(4)2 + 43 = x 3 + 12x 2 + 48x + 64
(A  B)3 = A3  3A2 B + 3AB2  B3
(x  2)3 = x 3  3x 2(2) + 3x(2)2  23 = x 3  6x 2 + 12x  8
86
Chapter P Prerequisites: Fundamental Concepts of Algebra
6 Perform operations with polynomials in several variables.
Polynomials in Several Variables A polynomial in two variables, x and y, contains the sum of one or more monomials in the form ax n ym. The constant, a, is the coefficient. The exponents, n and m, represent whole numbers. The degree of the monomial ax n ym is n + m. Here is an example of a polynomial in two variables: 6JGEQGHƂEKGPVUCTG––CPF
7x2y3 &GITGGQH OQPQOKCN +=
17x4y2
−
&GITGGQH OQPQOKCN +=
+
xy
6y2
−
&GITGGQH OQPQOKCNAxy B +=
+
&GITGGQH OQPQOKCNA–xy B +=
9. &GITGGQH OQPQOKCNAx y B +=
The degree of a polynomial in two variables is the highest degree of all its terms. For the preceding polynomial, the degree is 6. Polynomials containing two or more variables can be added, subtracted, and multiplied just like polynomials that contain only one variable. For example, we can add the monomials 7xy2 and 13xy2 as follows: –7xy2 + 13xy2 = (–7 + 13)xy2 = 6xy2. 6JGUGNKMGVGTOUDQVJEQPVCKP VJGXCTKCDNGHCEVQTUxCPFy
EXAMPLE 7
#FFEQGHƂEKGPVUCPFMGGRVJG UCOGXCTKCDNGHCEVQTUxy
Subtracting Polynomials in Two Variables
Subtract: (5x 3  9x 2 y + 3xy2  4)  (3x 3  6x 2 y  2xy2 + 3). Solution (5x3 − 9x2y + 3xy2 − 4) − (3x3 − 6x2y − 2xy2 + 3) %JCPIGVJGUKIPQHGCEJEQGHƂEKGPV
= (5x3 − 9x2y + 3xy2 − 4) + (–3x3 + 6x2y + 2xy2 − 3)
Add the opposite of the polynomial being subtracted.
3
3
2
2
2
2
= (5x  3x ) + ( 9x y + 6x y) + (3xy + 2xy ) + ( 4  3)
Group like terms.
= 2x 3  3x 2 y + 5xy2  7
Combine like terms by adding coefficients and keeping the same variable factors.
CHECK POINT 7
EXAMPLE 8
Subtract: (x 3  4x 2 y + 5xy2  y3)  (x 3  6x 2 y + y3).
Multiplying Polynomials in Two Variables
Multiply: a. (x + 4y)(3x  5y)
b. (5x + 3y)2.
Solution We will perform the multiplication in part (a) using the FOIL method. We will multiply in part (b) using the formula for the square of a binomial sum, (A + B)2.
Section P.4 Polynomials
a. (x + 4y)(3x − 5y) (
87
Multiply these binomials using the FOIL method.
1
+
.
= (x)(3x) + (x)(–5y) + (4y)(3x) + (4y)(–5y) = 3x2 − 5xy + 12xy − 20y2 = 3x2 + 7xy − 20y2
Combine like terms.
A+B=A+∙A∙B+B
b. (5x + 3y)2 = (5x)2 + 2(5x)(3y) + (3y)2 = 25x2 + 30xy + 9y2 Multiply: a. (7x  6y)(3x  y) CHECK POINT 8
BLITZER BONUS
b. (2x + 4y)2.
Labrador Retrievers and Polynomial Multiplication The color of a Labrador retriever is determined by its pair of genes. A single gene is inherited at random from each parent. The blackfur gene, B, is dominant. The yellowfur gene, Y, is recessive. This means that labs with at least one blackfur gene (BB or BY) have black coats. Only labs with two yellowfur genes (YY) have yellow coats. Axl, your author’s yellow lab, inherited his genetic makeup from two black BY parents. 5GEQPFBYRCTGPVCDNCEMNCDYKVJCTGEGUUKXG[GNNQYHWTIGPG
(KTUVBYRCTGPVCDNCEMNCD YKVJCTGEGUUKXG[GNNQYHWTIGPG
B
Y
B
BB
BY
Y
BY
YY
6JGVCDNGUJQYUVJGHQWTRQUUKDNG EQODKPCVKQPUQHEQNQTIGPGUVJCVBY RCTGPVUECPRCUUVQVJGKTQHHURTKPI
Because YY is one of four possible outcomes, the probability that a yellow lab like Axl will be the offspring of these black parents is 14 . The probabilities suggested by the table can be modeled by the expression 1 12 B + 12 Y 2 2. 2
2
Q 2 B + 2 YR = Q 2 BR + 2 Q 2 BRQ 2 YR + Q 2 YR 1
1
1
1
= 4 BB 6JGRTQDCDKNKV[QHC DNCEMNCDYKVJVYQ FQOKPCPVDNCEMIGPGUKU
1
+
1
1 BY 2
1
+
6JGRTQDCDKNKV[QHC DNCEMNCDYKVJC TGEGUUKXG[GNNQYIGPGKU
2
1 YY 4
6JGRTQDCDKNKV[QHC [GNNQYNCDYKVJVYQ TGEGUUKXG[GNNQYIGPGUKU
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. A polynomial is a single term or the sum of two or more terms containing variables with exponents that are numbers.
C2. It is customary to write the terms of a polynomial in the order of descending powers of the variable. This is called the form of a polynomial.
88
Chapter P Prerequisites: Fundamental Concepts of Algebra
C3. A simplified polynomial that has exactly one term is called a/an . C4. A simplified polynomial that has two terms is called a/an . C5. A simplified polynomial that has three terms is called a/an . C6. If a ≠ 0, the degree of ax n is
.
C7. Polynomials are added by combining
terms.
C8. To multiply 7x 3(4x 5  8x 2 + 6), use the property to multiply each term of the trinomial _______________ by the monomial ______. C9. To multiply (5x + 3)(x 2 + 8x + 7), begin by multiplying each term of x 2 + 8x + 7 by . Then multiply each term of x 2 + 8x + 7 by . Then combine terms.
C10. When using the FOIL method to find (x + 7)(3x + 5), the product of the first terms is , the product of the outside terms is , the product of the inside terms is , and the product of the last terms is . C11. (A + B)(A  B) = . The product of the sum and difference of the same two terms is the square of the first term the square of the second term. C12. (A + B)2 = . The square of a binomial sum is the first term plus 2 times the plus the last term . C13. (A  B)2 = . The square of a binomial difference is the first term squared 2 times the the last term squared. plus or minus?
C14. If a ≠ 0, the degree of ax nym is
.
P.4 EXERCISE SET Practice Exercises In Exercises 1–4, is the algebraic expression a polynomial? If it is, write the polynomial in standard form. 1. 2x + 3x 2  5 2x + 3 3. x
2. 2x + 3x 1  5 4. x 2  x 3 + x 4  5
In Exercises 5–8, find the degree of the polynomial. 5. 3x 2  5x + 4 6. 4x 3 + 7x 2  11 7. x 2  4x 3 + 9x  12x 4 + 63 8. x 2  8x 3 + 15x 4 + 91 In Exercises 9–14, perform the indicated operations. Write the resulting polynomial in standard form and indicate its degree. 9. 10. 11. 12. 13. 14.
( 6x 3 + 5x 2  8x + 9) + (17x 3 + 2x 2  4x  13) ( 7x 3 + 6x 2  11x + 13) + (19x 3  11x 2 + 7x  17) (17x 3  5x 2 + 4x  3)  (5x 3  9x 2  8x + 11) (18x 4  2x 3  7x + 8)  (9x 4  6x 3  5x + 7) (5x 2  7x  8) + (2x 2  3x + 7)  (x 2  4x  3) (8x 2 + 7x  5)  (3x 2  4x)  ( 6x 3  5x 2 + 3)
In Exercises 15–58, find each product. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33.
(x + 1)(x 2  x + 1) (2x  3)(x 2  3x + 5) (x + 7)(x + 3) (x  5)(x + 3) (3x + 5)(2x + 1) (2x  3)(5x + 3) (5x 2  4)(3x 2  7) (8x 3 + 3)(x 2  5) (x + 3)(x  3) (3x + 2)(3x  2)
16. 18. 20. 22. 24. 26. 28. 30. 32. 34.
(x + 5)(x 2  5x + 25) (2x  1)(x 2  4x + 3) (x + 8)(x + 5) (x  1)(x + 2) (7x + 4)(3x + 1) (2x  5)(7x + 2) (7x 2  2)(3x 2  5) (7x 3 + 5)(x 2  2) (x + 5)(x  5) (2x + 5)(2x  5)
35. 37. 39. 41. 43. 45. 47. 49. 51. 53. 55. 57.
(5  7x)(5 + 7x) (4x 2 + 5x)(4x 2  5x) (1  y5)(1 + y5) (x + 2)2 (2x + 3)2 (x  3)2 2 (4x 2  1) 2 (7  2x) (x + 1)3 (2x + 3)3 (x  3)3 (3x  4)3
36. 38. 40. 42. 44. 46. 48. 50. 52. 54. 56. 58.
(3  4z)(3 + 4z) (3x 2 + 4x)(3x 2  4x) (1  y6)(1 + y6) (x + 5)2 (3x + 2)2 (x  4)2 2 (5x 2  3) 2 (9  5x) (x + 2)3 (3x + 4)3 (x  1)3 (2x  3)3
In Exercises 59–66, perform the indicated operations. Indicate the degree of the resulting polynomial. 59. 60. 61. 62. 63. 64. 65. 66.
(5x 2 y  3xy) + (2x 2 y  xy) (  2x 2 y + xy) + (4x 2 y + 7xy) (4x 2 y + 8xy + 11) + ( 2x 2 y + 5xy + 2) (7x 4 y2  5x 2 y2 + 3xy) + ( 18x 4 y2  6x 2 y2  xy) (x 3 + 7xy  5y2)  (6x 3  xy + 4y2) (x 4  7xy  5y3)  (6x 4  3xy + 4y3) (3x 4 y2 + 5x 3 y  3y)  (2x 4 y2  3x 3 y  4y + 6x) (5x 4 y2 + 6x 3 y  7y)  (3x 4 y2  5x 3 y  6y + 8x)
In Exercises 67–82, find each product. 67. 69. 71. 73. 74. 75. 76.
(x + 5y)(7x + 3y) (x  3y)(2x + 7y) (3xy  1)(5xy + 2) (7x + 5y)2 (9x + 7y)2 2 (x 2 y2  3) 2 (x 2 y2  5)
68. (x + 9y)(6x + 7y) 70. (3x  y)(2x + 5y) 72. (7x 2 y + 1)(2x 2 y  3)
Section P.4 Polynomials 77. 79. 81. 82.
(x  y)(x 2 + xy + y2) (3x + 5y)(3x  5y) (7xy2  10y)(7xy2 + 10y) (3xy2  4y)(3xy2 + 4y)
78. (x + y)(x 2  xy + y2) 80. (7x + 3y)(7x  3y)
+ (0.2x 3  1.6x 2 + 4.3x + 13).
In Exercises 83–90, perform the indicated operation or operations.
89.
(3x (5x (5x (3x (2x (3x
+ + + + +
92. a. The data can be described by the following polynomial model of degree 3: S = 0.1x 3  1.2x 2 + 2.4x + 17
Practice PLUS 83. 84. 85. 86. 87. 88.
89
4y)2  (3x  4y)2 2y)2  (5x  2y)2 7)(3x  2)  (4x  5)(6x  1) 5)(2x  9)  (7x  2)(x  1) 5)(2x  5)(4x 2 + 25) 4)(3x  4)(9x 2 + 16)
(2x  7)5
90.
3
(2x  7)
In this polynomial model, S represents the score on the Implicit Association Test for age group x. Simplify the model. b. Use the simplified form of the model from part (a) to find the score on the Implicit Association Test for the group in the 55–64 age range. How well does the model describe the score displayed by the bar graph? The volume, V, of a rectangular solid with length l, width w, and height h is given by the formula V = lwh. In Exercises 93–94, use this formula to write a polynomial in standard form that models, or represents, the volume of the open box.
(5x  3)6 (5x  3)4
93. x
Application Exercises
8 − 2x
The bar graph shows the differences among age groups on the Implicit Association Test that measures levels of racial prejudice. Higher scores indicate stronger bias. Exercises 91–92 are based on the information in the graph. In these exercises, the possible values of the variable x are the group numbers shown in the voice balloons below the bars in the graph.
10 − 2x
94. x
8 − 2x 5 − 2x
Score on the Implicit Association Test
Measuring Racial Prejudice, by Age 44 42 40 38 36 34 32 30 28
Key: 7.
Because 5  2x 7 7 means 5  2x 6 7 or 5  2x 7 7, we solve 5  2x 6 7 and 5  2x 7 7 separately. Then we take the union of their solution sets. 5  2x 6 7
5  2x 7 7
or
These are the inequalities without absolute value bars.
5  5  2x 6 7  5
5  5  2x 7 7  5
2x 6 12
2x 7 2
2x 12 7 2 2
2x 2 6 2 2
x 7 6 GREAT QUESTION The graph of the solution set in Example 10 consists of two intervals. When is the graph of the solution set of an absolute value inequality a single interval and when is it the union of two intervals? If u is a linear expression and c 7 0, the graph of the solution set for u 7 c will be two intervals whose union cannot be represented as a single interval. The graph of the solution set for u 6 c will be a single interval. Avoid the common error of rewriting u 7 c as  c 6 u 7 c.
Subtract 5 from both sides. Simplify. Divide both sides by − 2 and change the sense of each inequality.
x 6 1
Simplify.
The solution set consists of all numbers that are less than 1 or greater than 6. The solution set is {x x 6 1 or x 7 6}, or, in interval notation (  ∞, 1) ∪ (6, ∞). The graph of the solution set is shown as follows: x –3
–2
–1
0
1
2
3
4
5
6
7
8
CHECK POINT 10 Solve and graph the solution set on a number line:
18 6 6  3x .
Applications In Section 1.3, we solved equations to determine when two different pricing options would result in the same cost. With inequalities, we can look at the same situations and ask when one of the pricing options results in a lower cost than the other. Our next example shows how to use an inequality to select the better deal between our two options for paying the bridge toll from Example 3 in Section 1.3. We use our strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality.
EXAMPLE 11
Selecting the Better Deal
You still have two options for paying the toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is tollbyplate; with this option, you pay the full toll of $4.25 and a $3 administrative fee each time you cross the bridge, for a total of $7.25 for each crossing. Find the number of times you would need to cross the bridge for the transponder option to be the better deal.
Section 1.7 Linear Inequalities and Absolute Value Inequalities
231
Solution Step 1 Let x represent one of the unknown quantities. We are looking for the number of times you must cross the bridge to make the transponder option the better deal. Thus, let x = the number of times you cross the bridge. Step 2 Represent other unknown quantities in terms of x. We are not asked to find another quantity, so we can skip this step. Step 3 Write an inequality in x that models the conditions. The transponder is a better deal than tollbyplate if the total cost with the transponder is less than the total cost of tollbyplate. The total cost with the transponder
is less than
the total cost of tollbyplate.
TECHNOLOGY Graphic Connections The graphs of the cost models for the toll options
20
RNWU
+
VJGPWODGT QHETQUUKPIU
VKOGU
3.25
∙
x
VJGPWODGT QHETQUUKPIU
VKOGU
7.25
f (x2); f is decreasing on I.
x
y
Increasing, Decreasing, and Constant Functions
Increasing
x
x
x1 I (3) For x1 and x2 in I, f (x1) = f (x2); f is constant on I.
x2
x
x
Section 2.2 More on Functions and Their Graphs
GREAT QUESTION Do you use x@coordinates or y@coordinates when describing where functions increase, decrease, or are constant? The open intervals describing where functions increase, decrease, or are constant use x@coordinates and not y@coordinates. Furthermore, points are not used in these descriptions.
EXAMPLE 1
267
Intervals on Which a Function Increases, Decreases, or Is Constant
State the intervals on which each given function is increasing, decreasing, or constant. a.
b.
y 5 4 3 2 1 –5 –4 –3 –2 –1–1
f x=x
–x
DGNQPIUVQ VJGITCRJQHf 5
FQGUPQV 4 3
y
y=f x 1 1 2 3 4 5
x –5 –4 –3 –2 –1–1
1 2 3 4 5
x
–2 –3 –4 –5
–2 –3 –4 –5
Solution a. The function is decreasing on the interval (  ∞, 0), increasing on the interval (0, 2), and decreasing on the interval (2, ∞). b. Although the function’s equations are not given, the graph indicates that the function is defined in two pieces. The part of the graph to the left of the y@axis shows that the function is constant on the interval (  ∞, 0). The part to the right of the y@axis shows that the function is increasing on the interval (0, ∞). CHECK POINT 1 State the intervals on which the given function is increasing, decreasing, or constant.
y 25 20 15 f x=x–x 10 5 –5 –4 –3 –2 –1–5
1 2 3 4 5
x
–10 –15 –20 –25
2 Use graphs to locate relative maxima or minima.
Relative Maxima and Relative Minima The points at which a function changes its increasing or decreasing behavior can be used to find any relative maximum or relative minimum values of the function. Definitions of Relative Maximum and Relative Minimum
GREAT QUESTION Is there an informal way to think about relative maximum and relative minimum? You may find it helpful to relate the definitions to a graph’s peaks and valleys. A relative maximum is the value at a peak. A relative minimum is the value in a valley.
1. A function value f (a) is a relative maximum of f if there exists an open interval containing a such that f (a) 7 f (x) for all x ≠ a in the open interval. 2. A function value f (b) is a relative minimum of f if there exists an open interval containing b such that f (b) 6 f (x) for all x ≠ b in the open interval. The word local is sometimes used instead of relative when describing maxima or minima.
y 4GNCVKXG OKPKOWO
4GNCVKXG OCZKOWO
(a, f (a))
(b, f(b))
b
a
x
268
Chapter 2 Functions and Graphs
GREAT QUESTION What’s the difference between where a function has a relative maximum or minimum and the relative maximum or minimum? • If f (a) is greater than all other values of f near a, then f has a relative maximum at the input value, a. The relative maximum is the output value, f (a). Equivalently, f (a) is a relative maximum value of the function. • If f (b) is less than all other values of f near b, then f has a relative minimum at the input value, b. The relative minimum is the output value, f (b). Equivalently, f (b) is a relative minimum value of the function. y 4GNCVKXG OCZKOWO )TCRJQHf xKPVGTEGRV QT\GTQQHf
If the graph of a function is given, we can often visually locate the number(s) at which the function has a relative maximum or a relative minimum. For example, the graph of f in Figure 2.17 shows that
p a , 1b 2
1
x –p
– a–
p p 2 xKPVGTEGRV QT\GTQQHf
p 2
p , –1b 2
–1
p . 2
p The relative maximum is f a b = 1. 2
4GNCVKXG OKPKOWO
• f has a relative minimum at 
Figure 2.17 Using a graph to locate where a function has a relative maximum or minimum
3 Test for symmetry.
The relative minimum is f a
p . 2
p b = 1. 2
Tests for Symmetry
y
(–x, y)
• f has a relative maximum at
(x, y) x
Is beauty in the eye of the beholder? Or are there certain objects (or people) that are so well balanced and proportioned that they are universally pleasing to the eye? What constitutes an attractive human face? In Figure 2.18, we’ve drawn lines between paired features. Notice how the features line up almost perfectly. Each half of the face is a mirror image of the other half through the vertical line. If we superimpose a rectangular coordinate system on the attractive face, notice that a point (x, y) on the right has a mirror image at ( x, y) on the left. The attractive face is said to be symmetric with respect to the yaxis. Did you know that graphs of some equations exhibit exactly the kind of symmetry shown by the attractive face in Figure 2.18, as well as other kinds of symmetry? The word symmetry comes from the Greek symmetria, meaning “the same measure.” Figure 2.19 shows three graphs, each with a common form of symmetry. Notice that the graph in Figure 2.19(a) shows the yaxis symmetry found in the attractive face. y
Figure 2.18 To most people, an attractive face is one in which each half is an almost perfect mirror image of the other half.
(–x, y)
y
y
(x, y)
(x, y) x
x (x, –y)
(a) Symmetric with respect to the yaxis
(x, y)
(b) Symmetric with respect to the xaxis
Figure 2.19 Graphs illustrating the three forms of symmetry
x (–x, –y)
(c) Symmetric with respect to the origin
Section 2.2 More on Functions and Their Graphs
269
GREAT QUESTION The graph in Figure 2.19(b) is not a function. How can it have symmetry? Good observation! The graph of the relation with xaxis symmetry is not the graph of a function; it fails the vertical line test. Other than f (x) = 0 or any graph that lies entirely on the xaxis, graphs with xaxis symmetry do not represent functions: Each point on the graph above the xaxis has a corresponding point below the xaxis with the same xcoordinate and opposite ycoordinate, violating the condition for being a function. However, even the graphs of relations that are not functions may have symmetry with respect to the yaxis, the xaxis, or the origin. Later in this chapter we will study circles that exhibit all three types of symmetry but are not graphs of functions.
Table 2.2 defines three common forms of symmetry and gives rules to determine if the graph of an equation is symmetric with respect to the yaxis, the xaxis, or the origin. Table 2.2 Definitions and Tests for Symmetry Consider an equation in the variables x and y. Definition of Symmetry
0QVWUWCNN[ VJGITCRJ QHC HWPEVKQP
Test for Symmetry
The graph of the equation is symmetric with respect to the yaxis if for every point 1x, y2 on the graph, the point 1 x, y2 is also on the graph.
Substituting  x for x in the equation results in an equivalent equation.
The graph of the equation is symmetric with respect to the xaxis if for every point 1x, y2 on the graph, the point 1x,  y2 is also on the graph.
Substituting  y for y in the equation results in an equivalent equation.
The graph of the equation is symmetric with respect to the origin if for every point 1x, y2 on the graph, the point 1 x, y2 is also on the graph.
Substituting  x for x and  y for y in the equation results in an equivalent equation.
EXAMPLE 2
Testing for Symmetry
Determine whether the graph of x = y2  1 is symmetric with respect to the yaxis, the xaxis, or the origin. Solution Test for symmetry with respect to the yaxis. Replace x with x and see if this results in an equivalent equation. 6JGUG GSWCVKQPU CTGPQV GSWKXCNGPV
x = y2 − 1 –x = y2 − 1 x = –y2 + 1
This is the given equation. Replace x with − x. Multiply both sides by − 1 and solve for x.
Replacing x with x does not give the original equation. Thus, the graph of x = y2  1 is not symmetric with respect to the yaxis. Test for symmetry with respect to the xaxis. Replace y with y and see if this results in an equivalent equation. 6JGUG GSWCVKQPU CTG GSWKXCNGPV
x = y2 − 1 x = (–y)2 − 1 x = y2 − 1
This is the given equation. Replace y with − y. ( − y)2 = ( − y)( − y) = y2
Replacing y with y gives the original equation. Thus, the graph of x = y2  1 is symmetric with respect to the xaxis.
270
Chapter 2 Functions and Graphs
Test for symmetry with respect to the origin. Replace x with x and y with y and see if this results in an equivalent equation. x = y2 − 1 –x = (–y)2 − 1
6JGUG GSWCVKQPU CTGPQV GSWKXCNGPV
2
–x = y − 1 x = –y2 + 1
This is the given equation. Replace x with − x and y with − y. ( − y)2 = ( − y)( − y) = y2 Multiply both sides by − 1 and solve for x.
Replacing x with x and y with y does not give the original equation. Thus, the graph of x = y2  1 is not symmetric with respect to the origin. Our symmetry tests reveal that the graph of x = y2  1 is symmetric with respect to the xaxis only. The graph of x = y2  1 in Figure 2.20(a) shows the symmetry with respect to the xaxis that we determined in Example 2. Figure 2.20(b) illustrates that the graph fails the vertical line test. Consequently, y is not a function of x. We have seen that if a graph has xaxis symmetry, y is usually not a function of x.
y 5 4 3 2 1 –5 –4 –3 –2 –1–1
y 5 4 3 2 1
x=y–
1 2 3 4 5
x –5 –4 –3 –2 –1–1
–2 –3 –4 –5
1 2 3 4 5
x
–2 –3 –4 –5
(a) The graph is symmetric with respect to the xaxis.
(b) y is not a function of x. Two values of y correspond to an xvalue.
Figure 2.20 The graph of x = y2  1
Determine whether the graph of y = x 2  1 is symmetric with respect to the yaxis, the xaxis, or the origin. CHECK POINT 2
EXAMPLE 3
Testing for Symmetry
Determine whether the graph of y = x 3 is symmetric with respect to the yaxis, the xaxis, or the origin. Solution Test for symmetry with respect to the yaxis. Replace x with x and see if this results in an equivalent equation. 6JGUG GSWCVKQPU CTGPQV VJGUCOG
y = x3
This is the given equation.
y = (–x) y = –x
3
3
Replace x with − x. ( − x)3 = ( − x)( − x)( − x) = − x3
Replacing x with x does not give the original equation. Thus, the graph of y = x 3 is not symmetric with respect to the yaxis.
Section 2.2 More on Functions and Their Graphs
271
Test for symmetry with respect to the xaxis. Replace y with y and see if this results in an equivalent equation. 6JGUG GSWCVKQPU CTGPQV VJGUCOG
y = x3 –y = x3 y = –x3
This is the given equation. Replace y with − y. Multiply both sides by − 1 and solve for y.
Replacing y with y does not give the original equation. Thus, the graph of y = x 3 is not symmetric with respect to the xaxis. Test for symmetry with respect to the origin. Replace x with x and y with y and see if this results in an equivalent equation.
y 2
y=x
6JGUG GSWCVKQPU CTG VJGUCOG
1 –2
1
–1
x
2
y = x3 –y = (–x)3 –y = –x3 y = x3
This is the given equation. Replace x with − x and y with − y. ( − x)3 = ( − x)( − x)( − x) = − x3 Multiply both sides by − 1 and solve for y.
–1
Replacing x with x and y with y gives the original equation. Thus, the graph of y = x 3 is symmetric with respect to the origin.
–2
Figure 2.21 The graph is symmetric with respect to the origin.
Our symmetry tests reveal that the graph of y = x 3 is symmetric with respect to the origin only. The symmetry is shown in Figure 2.21. Determine whether the graph of y5 = x 3 is symmetric with respect to the yaxis, the xaxis, or the origin. CHECK POINT 3
GREAT QUESTION y
Can the graph of an equation have more than one form of symmetry? Yes. Consider the graph of x 2 + y2 = 25. Test for yaxis symmetry. Replace x with –x.
Test for xaxis symmetry. Replace y with –y.
x2 + y2 = 25 (–x)2 + y2 = 25 x2 + y2 = 25
x2 + y2 = 25 same
x2 + (–y)2 = 25 x2 + y2 = 25
Test for origin symmetry. Replace x with –x and y with –y. x2 + y2 = 25
same
(–x)2 + (–y)2 = 25 x2 + y2 = 25
same
The graph of x 2 + y2 = 25, shown in Figure 2.22, is symmetric with respect to the yaxis, the xaxis, and the origin. This “perfectly symmetric” graph is a circle, which we shall study in Section 2.8.
6 5 4 3 2 1 –6 –5 –4 –3 –2 –1–1
x+y=
1 2 3 4 5 6
x
–2 –3 –4 –5 –6 Figure 2.22 The graph, a circle, has three kinds of symmetry.
4 Identify even or odd
functions and recognize their symmetries.
Even and Odd Functions We have seen that if a graph is symmetric with respect to the xaxis, it usually fails the vertical line test and is not the graph of a function. However, many functions have graphs that are symmetric with respect to the yaxis or the origin. We give these functions special names. A function whose graph is symmetric with respect to the yaxis is called an even function. If the point (x, y) is on the graph of the function, then the point ( x, y) is also on the graph. Expressing this symmetry in function notation provides the definition of an even function.
272
Chapter 2 Functions and Graphs
Even Functions and Their Symmetries The function f is an even function if f ( x) = f (x)
for all x in the domain of f.
The graph of an even function is symmetric with respect to the yaxis.
y (–3, 9) (–x, f (x))
9 8 7 6 5 4 3 2 1
An example of an even function is f (x) = x 2. The graph, shown in Figure 2.23, is symmetric with respect to the yaxis. Let’s show that f ( x) = f (x).
(3, 9) (x, f (x))
f (x) = x 2
This is the function’s equation.
f ( x) = ( x)
Replace x with − x.
f ( x) = x 2
( − x)2 = ( − x)( − x) = x2
2
f x=x 1 2 3 4 5
–5 –4 –3 –2 –1–1
x
Figure 2.23 The graph of f (x) = x 2, an even function
Because f (x) = x 2 and f ( x) = x 2, we see that f ( x) = f (x). This algebraically identifies the function as an even function. A function whose graph is symmetric with respect to the origin is called an odd function. If the point (x, y) is on the graph of the function, then the point ( x, y) is also on the graph. Expressing this symmetry in function notation provides the definition of an odd function. Odd Functions and Their Symmetries The function f is an odd function if f ( x) = f (x)
for all x in the domain of f.
The graph of an odd function is symmetric with respect to the origin.
An example of an odd function is f (x) = x 3. The graph, shown in Figure 2.24, is symmetric with respect to the origin. Let’s show that f ( x) = f (x).
y 8 7 6 5 4 3 2 1 –5 –4 –3 –2 –1–1 (–x, –f (x))
(–2, –8)
(2, 8) f x=x
f (x) = x 3
(x, f (x))
1 2 3 4 5
This is the function’s equation.
f ( x) = ( x)
Replace x with − x.
f ( x) = x 3
( − x)3 = ( − x)( − x)( − x) = − x3
3
x
–2 –3 –4 –5 –6 –7 –8
Figure 2.24 The graph of f (x) = x 3, an odd function
Because f (x) = x 3 and f ( x) = x 3, we see that f ( x) = f (x). This algebraically identifies the function as an odd function.
EXAMPLE 4
Identifying Even or Odd Functions from Graphs
Determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd. a.
b.
y
x
c.
y
x
y
x
Section 2.2 More on Functions and Their Graphs
273
Solution Note that each graph passes the vertical line test and is therefore the graph of a function. In each case, use inspection to determine whether or not there is symmetry. If the graph is symmetric with respect to the yaxis, it is that of an even function. If the graph is symmetric with respect to the origin, it is that of an odd function. a.
b.
y (x, f (x))
c.
y
(–x, –f (x))
x
y
(x, f (x))
(x, f (x))
x
x
(–x, –f (x)) The graph is symmetric with respect to the yaxis. Therefore, the graph is that of an even function.
The graph is symmetric with respect to the origin. Therefore, the graph is that of an odd function.
The graph is neither symmetric with respect to the yaxis nor the origin. Therefore, the graph is that of a function which is neither even nor odd.
CHECK POINT 4 Determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.
a.
b.
y
x
c.
y
y
x
x
In addition to inspecting graphs, we can also use a function’s equation and the definitions of even and odd to determine whether the function is even, odd, or neither.
Identifying Even or Odd Functions from Equations • Even function: f ( x) = f (x) The right side of the equation of an even function does not change if x is replaced with x. • Odd function: f ( x) = f (x) Every term on the right side of the equation of an odd function changes sign if x is replaced with x. • Neither even nor odd: f ( x) ≠ f (x) and f ( x) ≠ f (x) The right side of the equation changes if x is replaced with x, but not every term on the right side changes sign.
EXAMPLE 5
Identifying Even or Odd Functions from Equations
Determine whether each of the following functions is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the yaxis, the origin, or neither. a. f (x) = x 3  6x
b. g(x) = x 4  2x 2
c. h(x) = x 2 + 2x + 1
274
Chapter 2 Functions and Graphs
Solution In each case, replace x with x and simplify. If the right side of the equation stays the same, the function is even. If every term on the right side changes sign, the function is odd. a. We use the given function’s equation, f (x) = x 3  6x, to find f ( x). Use f (x) = x3 − 6x. 4GRNCEGxYKVJ–x
f (–x) = (–x)3 − 6(–x) = (–x)(–x)(–x) − 6(–x) = –x3 + 6x There are two terms on the right side of the given equation, f (x) = x 3  6x, and each term changed sign when we replaced x with x. Because f ( x) = f (x), f is an odd function. The graph of f is symmetric with respect to the origin. b. We use the given function’s equation, g(x) = x 4  2x 2, to find g( x). Use g(x) = x4 − 2x2. 4GRNCEGxYKVJ–x
g(–x) = (–x)4 − 2(–x)2 = (–x)(–x)(–x)(–x) − 2(–x)(–x) = x4 − 2x2 The right side of the equation of the given function, g(x) = x 4  2x 2, did not change when we replaced x with x. Because g( x) = g(x), g is an even function. The graph of g is symmetric with respect to the yaxis. c. We use the given function’s equation, h(x) = x 2 + 2x + 1, to find h( x). GREAT QUESTION In the solution on the right, we keep getting ( −x) to various powers. Is there a way to avoid having to write ( −x) over and over again? Yes. If you have ( x) to an even power, the negative sign is eliminated. ( x)even power = x even power If you have ( x) to an odd power, the result has a negative sign. ( x)odd power =  x odd power
Use h(x) = x2 + 2x + 1. 4GRNCEGxYKVJ–x
h(–x) = (–x)2 + 2(–x) + 1 = x2 − 2x + 1 The right side of the equation of the given function, h(x) = x 2 + 2x + 1, changed when we replaced x with x. Thus, h( x) ≠ h(x), so h is not an even function. The sign of each of the three terms in the equation for h(x) did not change when we replaced x with x. Only the second term changed signs. Thus, h( x) ≠ h(x), so h is not an odd function. We conclude that h is neither an even nor an odd function. The graph of h is neither symmetric with respect to the yaxis nor with respect to the origin.
CHECK POINT 5 Determine whether each of the following functions is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the yaxis, the origin, or neither.
a. f (x) = x 2 + 6
b. g(x) = 7x 3  x
c. h(x) = x 5 + 1
Section 2.2 More on Functions and Their Graphs
275
GREAT QUESTION Is there a relationship between the exponents in the terms of a function’s equation and whether or not the function is even, odd, or neither? Yes. Consider the functions in Example 5, where the equations all contained polynomials in x. • f (x) = x3 − 6x : odd function 'XGT[VGTOEQPVCKPUx VQCPQFFRQYGT
• g(x) = x4 − 2x2 : even function 'XGT[VGTOEQPVCKPUx VQCPGXGPRQYGT
• h(x) = x2 + 2x + 1 : neither an even function nor an odd function 6JGTGKUCEQODKPCVKQPQHGXGPCPFQFFRQYGTU
A constant term has an understood even power of x, x 0. Thus, the function in Check Point 5(c), h(x) = x 5 + 1 is a combination of odd and even powers. This function is neither even nor odd.
Piecewise Functions
5 Understand and use
piecewise functions.
A cellphone company offers the following plan for a mobile hotspot: • $40 per month includes 15 GB of data. • Additional data costs $0.60 per GB. We can represent this plan mathematically by writing the total monthly cost, C, as a function of the amount of data used, g.
C(g) 80
C g=+ g– KHg>
40 if 0 ≤ g ≤ 15 C(g) = e 40 + 0.60(g − 15) if g > 15
60
KPENWFGU )$
40 20
C g= KH≤g≤ 10
Figure 2.25
20
30
40
50
g
RGT )$
VKOGUVJGCOQWPV QHFCVCKPGZEGUU QH)$
6JGEQUVKUHQTWRVQ CPFKPENWFKPI)$ 6JGEQUVKU RNWURGT)$ HQTFCVCGZEGGFKPI)$
A function that is defined by two (or more) equations over a specified domain is called a piecewise function. Many Internet plans can be represented with piecewise functions. The graph of the piecewise function described above is shown in Figure 2.25.
EXAMPLE 6
Evaluating a Piecewise Function
Use the function that describes the Internet plan C(g) = b
40 40 + 0.60(g  15)
if 0 … g … 15 if g 7 15
to find and interpret each of the following: a. C(10) b. C(45). Solution a. To find C(10), we let g = 10. Because 10 lies between 0 and 15, we use the first line of the piecewise function. C(g) = 40 C(10) = 40
This is the function’s equation for 0 " g " 15. Replace g with 10. Regardless of this function’s input, the constant output is 40.
This means that with 10 GB of data, the monthly cost is $40. This can be visually represented by the point (10, 40) on the first piece of the graph in Figure 2.25.
276
Chapter 2 Functions and Graphs
b. To find C(45), we let g = 45. Because 45 is greater than 15, we use the second line of the piecewise function.
C(g) 80
C g=+ g– KHg>
C(g) = 40 C(45) = 40 = 40 = 40
60 40 20
C g= KH≤g≤ 10
20
30
40
50
+ + + +
0.60(g  15) 0.60(45  15) 0.60(30) 18
= 58
g
This is the function’s equation for g + 15. Replace g with 45. Subtract within parentheses: 45 − 15 = 30. Multiply: 0.60(30) = 18. Add: 40 + 18 = 58.
Thus, C(45) = 58. This means that with 45 GB of data, the monthly cost is $58. This can be visually represented by the point (45, 58) on the second piece of the graph in Figure 2.25.
Figure 2.25 (repeated)
Use the function in Example 6 to find and interpret each of the following: b. C(50). a. C(5) Identify your solutions by points on the graph in Figure 2.25. CHECK POINT 6
EXAMPLE 7
GREAT QUESTION
Graphing a Piecewise Function
Graph the piecewise function defined by
When I graphed the function in Example 7, here’s what I got:
f (x) = b
y 5 4 3 2 1 –5 –4 –3 –2 –1–1
x + 2 4
if if
x … 1 x 7 1.
Solution We graph f in two parts, using a partial table of coordinates to create each piece. The tables of coordinates and the completed graph are shown in Figure 2.26.
1 2 3 4 5
–2 –3 –4 –5
Is my graph ok? No. You incorrectly ignored the domain for each piece of the function and graphed each equation as if its domain was (  ∞ , ∞ ).
x
)TCRJf x=HQTx> &QOCKP= ∞
)TCRJf x=x+HQTx≤ &QOCKP= –∞?
x (x ≤ 1)
f (x) = x + 2
(x, f (x))
x (x > 1)
f (x) = 4
(x, f (x))
1
f (1) = 1 + 2 = 3
(1, 3)
1.5
f(1.5) = 4
(1.5, 4)
0
f (0) = 0 + 2 = 2
(0, 2)
2
f(2) = 4
(2, 4)
–1
f (–1) = –1 + 2 = 1
(–1, 1)
3
f(3) = 4
(3, 4)
–2
f (–2) = –2 + 2 = 0
(–2, 0)
4
f(4) = 4
(4, 4)
–3
f (–3) = –3 + 2 = –1
(–3, –1)
5
f(5) = 4
(5, 4)
–4
f (–4) = –4 + 2 = –2
(–4, –2)
y 5 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
–2 –3 –4 –5
x
6JGQRGPFQVTGRTGUGPVUCRQKPV
PQVKPENWFGFKPVJGITCRJ 0QXGTVKECNNKPGKPVGTUGEVUVJG ITCRJOQTGVJCPQPEG 6JKUKUVJGITCRJQHCHWPEVKQP
Figure 2.26 Graphing a piecewise function
We can use the graph of the piecewise function in Figure 2.26 to find the range of f. The range of the blue piece on the left is {y y … 3}. The range of the red horizontal piece on the right is {y y = 4}. Thus, the range of f is {y y … 3} ∪ {y y = 4}, or (  ∞, 3] ∪ {4}. CHECK POINT 7
Graph the piecewise function defined by f (x) = b
3 x  2
if if
x … 1 x 7 1.
Section 2.2 More on Functions and Their Graphs
277
Some piecewise functions are called step functions because their graphs form discontinuous steps. One such function is called the greatest integer function, symbolized by int(x) or Œ xœ , where int(x) = the greatest integer that is less than or equal to x.
For example, int(1) = 1,
int(1.3) = 1,
int(1.5) = 1,
int(1.9) = 1.
KUVJGITGCVGUVKPVGIGTVJCVKUNGUUVJCPQTGSWCNVQCPF
Here are some additional examples:
y
int(2) = 2,
5 4 3 2 1 –5 –4 –3 –2 –1 –3 –4 –5
int(2.3) = 2,
int(2.5) = 2,
int(2.9) = 2.
KUVJGITGCVGUVKPVGIGTVJCVKUNGUUVJCPQTGSWCNVQCPF
1 2 3 4 5
x
f (x) = int(x)
Figure 2.27 The graph of the greatest integer function
Notice how we jumped from 1 to 2 in the function values for int(x). In particular, If 1 … x 6 2, then int(x) = 1. If 2 … x 6 3, then int(x) = 2. The graph of f (x) = int(x) is shown in Figure 2.27. The graph of the greatest integer function jumps vertically one unit at each integer. However, the graph is constant between each pair of consecutive integers. The rightmost horizontal step shown in the graph illustrates that If 5 … x 6 6, then int(x) = 5. In general, If n … x 6 n + 1, where n is an integer, then int(x) = n.
6 Find and simplify a function’s difference quotient.
Functions and Difference Quotients In Section 2.4, we will be studying the average rate of change of a function. A ratio, called the difference quotient, plays an important role in understanding the rate at which functions change. Definition of the Difference Quotient of a Function The expression f (x + h)  f (x) h for h ≠ 0 is called the difference quotient of the function f.
EXAMPLE 8
Evaluating and Simplifying a Difference Quotient
If f (x) = 2x  x + 3, find and simplify each expression: f (x + h)  f (x) a. f (x + h) b. , h ≠ 0. h 2
Solution a. We find f (x + h) by replacing x with x + h each time that x appears in the equation. f (x) = 2x2 4GRNCEGx YKVJx+h
4GRNCEGx YKVJx+h
−
x 4GRNCEGx YKVJx+h
f (x + h) = 2(x + h)2 − (x + h) 2
+3 %QR[VJG6JGTGKU PQxKPVJKUVGTO
+3
2
= 2(x + 2xh + h ) − x − h + 3 = 2x2 + 4xh + 2h2 − x − h + 3
278
Chapter 2 Functions and Graphs
b. Using our result from part (a), we obtain the following: 6JKUKUf x+h HTQORCTV C
6JKUKUf xHTQO VJGIKXGPGSWCVKQP
f (x + h) − f(x) 2x2 + 4xh + 2h2 − x − h + 3 − (2x2 − x + 3) = h h 2x2 + 4xh + 2h2 − x − h + 3 − 2x2 + x − 3 h 2 2 (2x − 2x ) + (–x + x) + (3 − 3) + 4xh + 2h2 − h = h 4xh + 2h2 − 1h = 9GYTQVG–hCU–hVQCXQKFRQUUKDNG h GTTQTUKPVJGPGZVHCEVQTKPIUVGR =
=
h(4x + 2h − 1) h
Remove parentheses and change the sign of each term in the parentheses. Group like terms.
Simplify.
Factor h from the numerator.
= 4x + 2h − 1
Divide out identical factors of h in the numerator and denominator.
CHECK POINT 8
a. f (x + h)
If f (x) = 2x 2 + x + 5, find and simplify each expression: b.
f (x + h)  f (x) , h ≠ 0. h
ACHIEVING SUCCESS Read ahead. You might find it helpful to use some of your homework time to read (or skim) the section in the textbook that will be covered in your professor’s next lecture. Having a clear idea of the new material that will be discussed will help you to understand the class a whole lot better.
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C7. If f is an odd function, then f (  x) = . The graph of an odd function is symmetric with respect to the .
C1. Assume that f is a function defined on an open interval I and x1 and x2 are any elements in the interval I. f is increasing on I if f (x1) when x1 6 x2. f is decreasing on I if f (x1) when x1 6 x2. f is constant on I if f (x1) .
C8. A function defined by two or more equations over a specified domain is called a/an function.
C2. If f (a) 7 f (x) in an open interval containing a, x ≠ a, then the function value f (a) is a relative of f. If f (b) 6 f (x) in an open interval containing b, x ≠ b, then the function value f (b) is a relative of f.
C9. The greatest integer function is defined by int(x) = the greatest integer that is . For example, int(2.5) = , int(  2.5) = , and int(0.5) = .
C3. The graph of an equation is symmetric with respect to the if substituting x for x in the equation results in an equivalent equation. C4. The graph of an equation is symmetric with respect to the if substituting y for y in the equation results in an equivalent equation. C5. The graph of an equation is symmetric with respect to the if substituting x for x and y for y in the equation results in an equivalent equation. . The C6. If f is an even function, then f ( x) = graph of an even function is symmetric with respect to the .
C10. The expression f (x + h)  f (x) h
, h ≠ 0,
is called the of the function f. We find this expression by replacing x with each time x appears in the function’s equation. Then we subtract . After simplifying, we factor from the numerator and divide out identical factors of in the numerator and denominator. C11. True or false: f (x + h) = f (x) + h C12. True or false: f (x + h) = f (x) + f (h)
Section 2.2 More on Functions and Their Graphs
279
2.2 EXERCISE SET Practice Exercises In Exercises 1–12, use the graph to determine
y
2.
4 3 2 1 1 2 3 4
x
4 3 2 1 1 2 3 4 5
1 2 3 4 5
–3 –2 –1–1
x
1 2 3 4 5
–3 –2 –1–1
y
1 2 3 4
1 2 3 4
–4 –3 –2 –1–1
1 2 3 4
x
–2 –3 –4
x
–2 –3 –4 y
y
10.
3 2 1
–2 –3 –4 –5
–4 –3 –2 –1–1
x
–2 –3 –4
–4 –3 –2 –1–1
4 3 2 1
4 3 2 1
–4 –3 –2 –1–1
y
14.
y
8.
4 3 2 1
9.
x
–2 –3 –4
–2 –3
7.
1 2 3 4
–4 –3 –2 –1–1
x
–2 –3 –4
4 3 2 1 x
1 2 3 4
–4 –3 –2 –1–1
y
6.
5 4 3 2 1
y 4 3 2 1
–2 –3 –4 y
1 2 3 4
x
1 2 3
–2 –3 –4
13.
x
–2 –3 –4
–5 –4 –3 –2 –1–1
a. The numbers, if any, at which f has a relative maximum. What are these relative maxima? b. The numbers, if any, at which f has a relative minimum. What are these relative minima?
4 3 2 1
–3 –2 –1–1
x
In Exercises 13–16, the graph of a function f is given. Use the graph to find each of the following:
y
4.
1 2 3 4 5
x
–2 –3 –4
y
5.
1 2 3 4
–4 –3 –2 –1–1
–2 –3 –4
4 3 2 1
–2 –3 –4
4 3 2 1
–4 –3 –2 –1–1
3.
–3 –2 –1–1
y
y
12.
4 3 2 1
a. intervals on which the function is increasing, if any. b. intervals on which the function is decreasing, if any. c. intervals on which the function is constant, if any. 1.
y
11.
15.
f x=x+x–x+
4 3 2 1 –4 –3 –2 –1–1 –2 –3 –4
1 2 3 4 5
x
[–4, 4, 1] by [–15, 25, 5]
x
280 16.
Chapter 2 Functions and Graphs y
36.
f x=x–x+x+
4 (–1, 3) 3 (0, 2) 2 (1, 1) 1 –4 –3 –2 –1–1
In Exercises 17–32, determine whether the graph of each equation is symmetric with respect to the yaxis, the xaxis, the origin, more than one of these, or none of these. y = x2 + 6 x = y2 + 6 y2 = x 2 + 6 y = 2x + 3 x 2  y3 = 2 x 2 + y2 = 100 x 2y2 + 3xy = 1 y4 = x 3 + 6
18. 20. 22. 24. 26. 28. 30. 32.
y = x2  2 x = y2  2 y2 = x 2 + 12 y = 6x + 7 x 3  y2 = 5 x 2 + y2 = 49 x 2y2 + 5xy = 2 y5 = x 4 + 2
In Exercises 33–36, use possible symmetry to determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd. y
33.
Q–2,
4 R 5
In Exercises 37–48, determine whether each function is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the yaxis, the origin, or neither. 37. f (x) = x 3 + x 38. f (x) = x 3  x 39. g(x) = x 2 + x 40. g(x) = x 2  x 41. h(x) = x 2  x 4 42. h(x) = 2x 2 + x 4 43. f (x) = x 2  x 4 + 1 44. f (x) = 2x 2 + x 4 + 1 45. f (x) = 51 x 6  3x 2 46. f (x) = 2x 3  6x 5 47. f (x) = x21  x 2
5 (0, 4) 4 3 2 Q2, 45 R 1
48. f (x) = x 2 21  x 2 49. Use the graph of f to determine each of the following. Where applicable, use interval notation. y
x
1 2 3 4
–4 –3 –2 –1–1
5 4 3 2 1
–2 –3
y
34.
–5 –4 –3 –2 –1–1
(1, 4)
4 3 2 1 –4 –3 –2 –1–1 (–4, –1) –2
x
–2 –3 –4
[–2, 6, 1] by [–15, 35, 5]
17. 19. 21. 23. 25. 27. 29. 31.
1 2 3 4
–2 –3 –4 –5
(4, 1) x
1 2 3 4
1 2 3 4 5 6 7 8 9
x
y=f x
a. the domain of f
–3 (–1, –4) –4
b. the range of f c. the x@intercepts d. the y@intercept
y
35. 4 3 2 (–1, 1) 1 –4 –3 –2 –1–1 –2 –3 –4
e. intervals on which f is increasing f. intervals on which f is decreasing g. intervals on which f is constant
(0, 0) 1 2 3 4 (1, –1)
h. the number at which f has a relative minimum x
i. the relative minimum of f j. f (  3) k. the values of x for which f (x) = 2 l. Is f even, odd, or neither?
Section 2.2 More on Functions and Their Graphs 50. Use the graph of f to determine each of the following. Where applicable, use interval notation.
52. Use the graph of f to determine each of the following. Where applicable, use interval notation.
y
y
5 4 3 2 1
5 4 3 2 1
y=f x
1 2 3 4 5
–5 –4 –3 –2 –1–1
281
x
–5 –4 –3 –2 –1–1
–2 –3 –4 –5
y=f x
1 2 3 4 5
x
–2 –3
a. the domain of f a. the domain of f
b. the range of f
b. the range of f
c. the zeros of f
c. the x@intercepts
d. f (0)
d. the y@intercept
e. intervals on which f is increasing
e. intervals on which f is increasing
f. intervals on which f is decreasing
f. intervals on which f is decreasing
g. intervals on which f is constant
g. values of x for which f (x) … 0
h. values of x for which f (x) 7 0
h. the numbers at which f has a relative maximum
i. values of x for which f (x) = 2
i. the relative maxima of f
j. Is f (4) positive or negative?
j. f (  2)
k. Is f even, odd, or neither?
k. the values of x for which f (x) = 0
l. Is f (2) a relative maximum?
l. Is f even, odd, or neither? 51. Use the graph of f to determine each of the following. Where applicable, use interval notation.
53. f (x) = b
y
3x + 5 4x + 7
a. f (  2)
5 4 y=f x 3 2 1 –5 –4 –3 –2 –1–1
In Exercises 53–58, evaluate each piecewise function at the given values of the independent variable.
54. f (x) = b
1 2 3 4 5
x
–2 –3 –4 –5
a. the domain of f b. the range of f c. the zeros of f d. f (0) e. intervals on which f is increasing f. intervals on which f is decreasing g. values of x for which f (x) … 0 h. any relative maxima and the numbers at which they occur i. the value of x for which f (x) = 4 j. Is f (  1) positive or negative?
b. f (0) 6x  1 7x + 3
a. f (  3) 55. g(x) = b
x + 3 (x + 3)
x  9 57. h(x) = c x  3 6
c. g( 3)
if x Ú  5 if x 6  5 b. g( 6)
a. g(0)
c. f (4)
if x Ú  3 if x 6  3 b. g( 6)
x + 5 (x + 5)
c. f (3)
if x 6 0 if x Ú 0 b. f (0)
a. g(0) 56. g(x) = b
if x 6 0 if x Ú 0
c. g( 5)
2
a. h(5) x 2  25 58. h(x) = c x  5 10 a. h(7)
if x ≠ 3 if x = 3 b. h(0)
c. h(3)
if x ≠ 5 if x = 5 b. h(0)
c. h(5)
282
Chapter 2 Functions and Graphs
In Exercises 59–70, the domain of each piecewise function is (  ∞ , ∞ ). a. Graph each function. b. Use your graph to determine the function’s range. x 59. f (x) = b x
if if
x 6 0 x Ú 0
x x
if if
x 6 0 x Ú 0
2x 2
if if
x … 0 x 7 0
1 x 62. f (x) = b 2 3
if if
x … 0 x 7 0
60. f (x) = b 61. f (x) = b
64. f (x) = b
x + 2 x  2
if if
x 6 3 x Ú 3
65. f (x) = b
3 3
if if
x … 1 x 7 1
4 66. f (x) = b 4
if if
x … 1 x 7 1
–2 –3 –4 –5
2f (  2.5)  f (1.9)  [f (  p)]2 + f ( 3) , f (1) # f (p).
 12 x 2 2x + 1
if if
x 6 1 x Ú 1
A cable company offers the following highspeed Internet plans. Also given are the piecewise functions that model these plans. Use this information to solve Exercises 95–96. Plan A • $40 per month includes 400 GB of data. • Additional data costs $0.20 per GB.
x 6 4 4 … x 6 0 x Ú 0
C(g) = b
x 6 3 3 … x 6 0 x Ú 0
f (x + h)  f (x)
,h ≠ 0
71. f (x) = 4x
72. f (x) = 7x
73. f (x) = 3x + 7
74. f (x) = 6x + 1
75. f (x) = x
if 0 … g … 400 if g 7 400
• $60 per month includes 1000 GB of data. • Additional data costs $0.20 per GB.
for the given function.
2
40 40 + 0.20(g  400)
Plan B
In Exercises 71–92, find and simplify the difference quotient
h
x
94. Find
2x  1
if if if
1 2 3 4 5
2f (  1.5) + f (  0.9)  [f (p)]2 + f ( 3) , f (1) # f ( p).
x 6 1 x Ú 1
0 70. f (x) = c  x x2  1
–5 –4 –3 –2 –1–1
93. Find
if if
if if if
92. f (x) = 2x  1
5 4 3 2 1
x 6 2 x Ú 2
0 69. f (x) = c x x2
91. f (x) = 1x
1 2x
y
if if
68. f (x) = c
90. f (x) =
In Exercises 93–94, let f be defined by the following graph:
x + 3 x  3
1 2 2x
88. f (x) = 7
1 89. f (x) = x
Practice PLUS
63. f (x) = b
67. f (x) = b
87. f (x) = 6
C(g) = b
60 60 + 0.20(g  1000)
if 0 … g … 1000 if g 7 1000
95. Simplify the algebraic expression in the second line of the piecewise function for plan A. Then use pointplotting to graph the function. 96. Simplify the algebraic expression in the second line of the piecewise function for plan B. Then use pointplotting to graph the function.
76. f (x) = 2x 2
77. f (x) = x 2  4x + 3
78. f (x) = x 2  5x + 8
79. f (x) = 2x + x  1
80. f (x) = 3x + x + 5
81. f (x) = x 2 + 2x + 4
82. f (x) = x 2  3x + 1
83. f (x) = 2x 2 + 5x + 7
84. f (x) =  3x 2 + 2x  1
85. f (x) = 2x 2  x + 3
86. f (x) = 3x 2 + x  1
2
2
In Exercises 97–98, write a piecewise function that models each highspeed Internet plan. Then graph the function. 97. $50 per month includes 600 GB. Additional data costs $0.30 per GB. 98. $80 per month includes 2000 GB. Additional data costs $0.35 per GB.
283
Section 2.2 More on Functions and Their Graphs
Application Exercises With aging, body fat increases and muscle mass declines. The line graphs show the percent body fat in adult women and men as they age from 25 to 75 years. Use the graphs to solve Exercises 99–106. Percent Body Fat in Adults
Percent Body Fat
40 36
The preceding tax table can be modeled by a piecewise function, where x represents the taxable income of a single U.S. taxpayer and T(x) is the tax owed: 0.10x 987.50 4617.50 T(x) = g 14,605.50 33,271.50
9QOGP
32
+ + + +
0.12(x 0.22(x 0.24(x 0.32(x ? ?

9875) 40,125) 85,525) 163,300)
if 0 6 x … 9875 if 9875 6 x … 40,125 if 40,125 6 x … 85,525 if 85,525 6 x … 163,300 if 163,300 6 x … 207,350 if 207,350 6 x … 518,400 x 7 518,400 if
Use this information to solve Exercises 107–108.
28 24
107. Find and interpret T(35,000). 108. Find and interpret T(50,000).
/GP
20
In Exercises 109–110, refer to the preceding tax table. 35
45 Age
55
65
75
109. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (207,350, 518,400]. 110. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (518,400, ∞ ).
Source: Thompson et al., The Science of Nutrition, Benjamin Cummings, 2008 99. State the intervals on which the graph giving the percent body fat in women is increasing and decreasing. 100. State the intervals on which the graph giving the percent body fat in men is increasing and decreasing. 101. For what age does the percent body fat in women reach a maximum? What is the percent body fat for that age? 102. At what age does the percent body fat in men reach a maximum? What is the percent body fat for that age? 103. Use interval notation to give the domain and the range for the graph of the function for women. 104. Use interval notation to give the domain and the range for the graph of the function for men. 105. The function p(x) = 0.002x 2 + 0.15x + 22.86 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men. 106. The function p(x) = 0.004x 2 + 0.25x + 33.64 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men. Here is the Federal Tax Rate Schedule X that specifies the tax owed by a single taxpayer for 2020. If Your Taxable Income Is Over
The Tax You Owe Is
But Not Over
Of the Amount Over
The figure shows the cost of mailing a firstclass letter, f (x), as a function of its weight, x, in ounces, in June 2020. Use the graph to solve Exercises 111–114. y Cost (dollars)
25
1.00 0.85 0.70 0.55 1
2 Weight (ounces)
3
111. Find f (3). What does this mean in terms of the variables in this situation? 112. Find f (3.75). What does this mean in terms of the variables in this situation? 113. What is the cost of mailing a letter that weighs 1.5 ounces? 114. What is the cost of mailing a letter that weighs 1.8 ounces? 115. Furry Finances. A pet insurance policy has a monthly rate that is a function of the age of the insured dog or cat. For pets whose age does not exceed 4, the monthly cost is $20. The cost then increases by $2 for each successive year of the pet’s age. Age Not Exceeding
Monthly Cost $20
10%
$
0
987.50 + 12,
$
9875
5
$22
$ 4617.50 + 22,
$ 40,125
6
$24
$163,300
$14,605.50 + 24,
$ 85,525
$163,300
$ 207,350
$33,271.50 + 32,
$163,300
$ 207,350
$518,400
$47,367.50 + 35,
$207,350
$518,400
−
$ 156,235 + 37,
$518,400
0
$
9,875
$
9875
$ 40,125
$
$ 40,125
$ 85,525
$ 85,525
x
Source: Lynn E. Baring, Postmaster, Inverness, CA
4
$
4
The cost schedule continues in this manner for ages not exceeding 10. The cost for pets whose ages exceed 10 is $40. Use this information to create a graph that shows the monthly cost of the insurance, f (x), for a pet of age x, where the function’s domain is [0, 14].
284
Chapter 2 Functions and Graphs
Explaining the Concepts
Critical Thinking Exercises
116. What does it mean if a function f is increasing on an interval? 117. Suppose that a function f whose graph contains no breaks or gaps on (a, c) is increasing on (a, b), decreasing on (b, c), and defined at b. Describe what occurs at x = b. What does the function value f (b) represent? 118. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the yaxis? 119. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the xaxis? 120. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the origin? 121. If you are given a function’s graph, how do you determine if the function is even, odd, or neither? 122. If you are given a function’s equation, how do you determine if the function is even, odd, or neither? 123. What is a piecewise function? 124. Explain how to find the difference quotient of a function f , f (x + h)  f (x) , if an equation for f is given. h
Make Sense? In Exercises 133–136, determine whether each statement makes sense or does not make sense, and explain your reasoning.
Technology Exercises 125. The function f (x) =  0.00002x 3 + 0.008x 2  0.3x + 6.95 models the number of annual physician visits, f (x), by a person of age x. Graph the function in a [0, 100, 5] by [0, 40, 2] viewing rectangle. What does the shape of the graph indicate about the relationship between one’s age and the number of annual physician visits? Use the TABLE or minimum function capability to find the coordinates of the minimum point on the graph of the function. What does this mean? In Exercises 126–131, use a graphing utility to graph each function. Use a [  5, 5, 1] by [  5, 5, 1] viewing rectangle. Then find the intervals on which the function is increasing, decreasing, or constant. 126. f (x) = x 3  6x 2 + 9x + 1
133. My graph is decreasing on (  ∞ , a) and increasing on (a, ∞ ), so f (a) must be a relative maximum. 134. This work by artist Scott Kim has the same kind of symmetry as an even function.
“DYSLEXIA,” 1981 135. I graphed f (x) = b
2 3
if if
x ≠ 4 x = 4
and one piece of my graph is a single point. 136. I noticed that the difference quotient is always zero if f (x) = c, where c is any constant. 137. Sketch the graph of f using the following properties. (More than one correct graph is possible.) f is a piecewise function that is decreasing on (  ∞ , 2), f (2) = 0, f is increasing on (2, ∞ ), and the range of f is [0, ∞ ). 138. Define a piecewise function on the intervals (  ∞ , 2], (2, 5), and [5, ∞ ) that does not “jump” at 2 or 5 such that one piece is a constant function, another piece is an increasing function, and the third piece is a decreasing function. f (x) 139. Suppose that h(x) = . The function f can be even, odd, g(x) or neither. The same is true for the function g. a. Under what conditions is h definitely an even function? b. Under what conditions is h definitely an odd function?
Retaining the Concepts
127. g(x) = 4  x 2
140. You invested $80,000 in two accounts paying 1.5% and 1.7% annual interest. If the total interest earned for the year was $1320, how much was invested at each rate? (Section 1.3, Example 5) 141. Solve for A: C = A + Ar.
128. h(x) = x  2 + x + 2 1
129. f (x) = x 3 (x  4) 2
130. g(x) = x 3 2
(Section 1.3, Example 8)
131. h(x) = 2  x 5 n
132. a. Graph the functions f (x) = x for n = 2, 4, and 6 in a [ 2, 2, 1] by [ 1, 3, 1] viewing rectangle. b. Graph the functions f (x) = x n for n = 1, 3, and 5 in a [ 2, 2, 1] by [ 2, 2, 1] viewing rectangle. c. If n is positive and even, where is the graph of f (x) = x n increasing and where is it decreasing? d. If n is positive and odd, what can you conclude about the graph of f (x) = x n in terms of increasing or decreasing behavior? e. Graph all six functions in a [ 1, 3, 1] by [ 1, 3, 1] viewing rectangle. What do you observe about the graphs in terms of how flat or how steep they are?
142. Solve by the quadratic formula: 5x 2  7x + 3 = 0. (Section 1.5, Example 8)
Preview Exercises Exercises 143–145 will help you prepare for the material covered in the next section. y2  y1 143. If (x1 , y1) = ( 3, 1) and (x2 , y2) = ( 2, 4), find . x2  x1 144. Find the ordered pairs ( , 0) and (0, ) satisfying 4x  3y  6 = 0. 145. Solve for y: 3x + 2y  4 = 0.
285
Section 2.3 Linear Functions and Slope
SECTION 2.3
Linear Functions and Slope Is there a relationship between literacy and child mortality? As the percentage of adult females who are literate increases, does the mortality of children under five decrease? Figure 2.28 indicates that this is, indeed, the case. Each point in the figure represents one country.
WHAT YOU’LL LEARN 1 Calculate a line’s slope. 2 Write the pointslope form of the equation of a line.
3 Write and graph the
slopeintercept form of the equation of a line.
4 Graph horizontal or vertical lines.
5 Recognize and use the
y
general form of a line’s equation.
7
Model data with linear functions and make predictions.
350 UnderFive Mortality (per thousand)
6
Use intercepts to graph the general form of a line’s equation.
Literacy and Child Mortality
300 250 200 150 100 50 0
10 20 30 40 50 60 70 80 90 100 Percentage of Adult Females Who Are Literate
x Figure 2.28 Source: United Nations
Data presented in a visual form as a set of points is called a scatter plot. Also shown in Figure 2.28 is a line that passes through or near the points. A line that best fits the data points in a scatter plot is called a regression line. By writing the equation of this line, we can obtain a model for the data and make predictions about child mortality based on the percentage of literate adult females in a country. Data often fall on or near a line. In this section, we will use functions to model such data and make predictions. We begin with a discussion of a line’s steepness.
1 Calculate a line’s slope.
The Slope of a Line Mathematicians have developed a useful measure of the steepness of a line, called the slope of the line. Slope compares the vertical change (the rise) to the horizontal change (the run) when moving from one fixed point to another along the line. To calculate the slope of a line, we use a ratio that compares the change in y (the rise) to the corresponding change in x (the run). Definition of Slope The slope of the line through the distinct points (x1 , y1) and (x2 , y2) is 8GTVKECNEJCPIG Rise Change in y = *QTK\QPVCNEJCPIG Run Change in x y2 − y1 = , x2 − x1
where x2  x1 ≠ 0.
4WP x−x
y
y2 4KUG y−y y1
(x2, y2) (x1, y1) x1
x2
x
286
Chapter 2 Functions and Graphs
It is common notation to let the letter m represent the slope of a line. The letter m is used because it is the first letter of the French verb monter, meaning “to rise” or “to ascend.”
4WPWPKV
4KUG WPKVU –5 –4
––
EXAMPLE 1
y
Using the Definition of Slope
Find the slope of the line passing through each pair of points:
5 4 3 2 1
a. ( 3, 1) and ( 2, 4)
–
x
1 2 3 4 5
–2 –1–1 –2 –3 –4 –5
b. ( 3, 4) and (2, 2).
Solution a. Let (x1 , y1) = ( 3, 1) and (x2 , y2) = ( 2, 4). We obtain the slope as follows: m =
4  ( 1) Change in y y2  y1 4 + 1 5 = = = = = 5. x2  x1 Change in x 2  ( 3) 2 + 3 1
The situation is illustrated in Figure 2.29. The slope of the line is 5. For every vertical change, or rise, of 5 units, there is a corresponding horizontal change, or run, of 1 unit. The slope is positive and the line rises from left to right.
Figure 2.29 Visualizing a slope of 5
GREAT QUESTION When using the definition of slope, how do I know which point to call (x1, y1) and which point to call (x2, y2)? When computing slope, it makes no difference which point you call (x1 , y1) and which point you call (x2 , y2). If we let (x1 , y1) = ( 2, 4) and (x2 , y2) = (  3, 1), the slope is still 5: m =
Change in y Change in x
=
y2  y1 1  4 5 = = = 5. x2  x1 3  (  2) 1
However, you should not subtract in one order in the numerator (y2  y1) and then in a different order in the denominator (x1  x2). y
–
4KUG –WPKVU –5 –4
1  4 5 = =  5. 2  (  3) 1
5 4 3 2 1 –2 –1–1
1 2 3 4 5
–
x
–3 4WPWPKVU –4 –5 Figure 2.30 Visualizing a slope of  65
GREAT QUESTION Is it OK to say that a vertical line has no slope? Always be clear in the way you use language, especially in mathematics. For example, it’s not a good idea to say that a line has “no slope.” This could mean that the slope is zero or that the slope is undefined.
Incorrect! The slope is not − 5.
b. To find the slope of the line passing through ( 3, 4) and (2, 2), we can let (x1 , y1) = ( 3, 4) and (x2 , y2) = (2, 2). The slope of the line is computed as follows: m =
Change in y y2  y1 2  4 6 6 = = = =  . x2  x1 Change in x 2  ( 3) 5 5
The situation is illustrated in Figure 2.30. The slope of the line is  65 . For every vertical change of 6 units (6 units down), there is a corresponding horizontal change of 5 units. The slope is negative and the line falls from left to right. Find the slope of the line passing through each pair of points: a. ( 3, 4) and ( 4, 2) b. (4, 2) and ( 1, 5). CHECK POINT 1
Example 1 illustrates that a line with a positive slope is increasing and a line with a negative slope is decreasing. By contrast, a horizontal line is a constant function and has a slope of zero. A vertical line has no horizontal change, so x2  x1 = 0 in the formula for slope. Because we cannot divide by zero, the slope of a vertical line is undefined. This discussion is summarized in Table 2.3.
Section 2.3 Linear Functions and Slope
287
Table 2.3 Possibilities for a Line’s Slope Positive Slope
Negative Slope
Zero Slope
Undeﬁned Slope
y
y
y
y m= mKU WPFGƂPGF
m x
x
Line rises from left to right.
Line falls from left to right.
BLITZER BONUS
x
x
Line is horizontal.
Line is vertical.
Slope and Applauding Together
Using a decibel meter, sociologist Max Atkinson found a piecewise function that models the intensity of a group’s applause, d(t), in decibels, over time, t, in seconds. • Applause starts very fast, reaching a full crescendo of 30 decibels in one second. (m = 30) • Applause remains level at 30 decibels for 5.5 seconds. (m = 0) • Applause trails off by 15 decibels per second for two seconds. (m =  15) These verbal conditions can be modeled by a piecewise function. 30t d(t) = c 30 –15t + 127.5
0≤t≤1 1 < t ≤ 6.5 6.5 < t ≤ 8.5
+PVGPUKV[QHCRRNCWUGd t KPFGEKDGNUCUCHWPEVKQP QHVKOGtKPUGEQPFU
The graph of this piecewise function is shown below. The function suggests how strongly people work to coordinate their applause with one another, careful to start clapping at the “right” time and careful to stop when it seems others are stopping as well. Applauding Together
Intensity of Applause (decibels)
d(t) 40 30
m=– #RRNCWUGVTCKNUQHH HCKTN[HCUV
20 10 1
Source: The Sociology Project 2.0, Pearson, 2016
2 Write the pointslope form of the equation of a line.
m= #RRNCWUGTGOCKPUNGXGN HQTUGEQPFU
2
3 4 5 6 7 Time (seconds)
8
9
t
m= #RRNCWUGUVCTVUXGT[HCUV TGCEJKPIHWNNETGUEGPFQ KPQPGUGEQPF
The PointSlope Form of the Equation of a Line We can use the slope of a line to obtain various forms of the line’s equation. For example, consider a nonvertical line that has slope m and that contains the point (x1 , y1).
288
Chapter 2 Functions and Graphs y
6JKURQKPVKUCTDKVTCT[ 5NQRGKUm
(x, y) y − y1
The line in Figure 2.31 has slope m and contains the point (x1, y1). Let (x, y) represent any other point on the line. Regardless of where the point (x, y) is located, the steepness of the line in Figure 2.31 remains the same. Thus, the ratio for the slope stays a constant m. This means that for all points (x, y) along the line
(x1, y1) x − x1
m=
6JKURQKPVKUƂZGF x Figure 2.31 A line passing through (x1 , y1) with slope m
Change in y y − y1 = . x − x1 Change in x
9GWUGF xyKPUVGCFQH
xyDGECWUG xy TGRTGUGPVUCƂZGFRQKPV CPF xyKUPQVCƂZGF RQKPVQPVJGNKPG
We can clear the fraction by multiplying both sides by x  x1 , the least common denominator. y x y m(x  x1) = x m =

y1 x1 y1 # (x  x1) x1
m(x  x1) = y  y1
This is the slope of the line in Figure 2.31.
Multiply both sides by x − x1. Simplify:
y − y1 x − x1
~ (x − x1 ) = y − y1.
Now, if we reverse the two sides, we obtain the pointslope form of the equation of a line. GREAT QUESTION When using y − y1 = m(x − x1), for which variables do I substitute numbers? When writing the pointslope form of a line’s equation, you will never substitute numbers for x and y. You will substitute values for x1 , y1 , and m.
PointSlope Form of the Equation of a Line The pointslope form of the equation of a nonvertical line with slope m that passes through the point (x1 , y1) is y  y1 = m(x  x1). For example, the pointslope form of the equation of the line passing through (1, 5) with slope 2 (m = 2) is y  5 = 2(x  1). We will soon be expressing the equation of a nonvertical line in function notation. To do so, we need to solve the pointslope form of a line’s equation for y. Example 2 illustrates how to isolate y on one side of the equal sign.
EXAMPLE 2
Writing an Equation for a Line in PointSlope Form
Write an equation in pointslope form for the line with slope 4 that passes through the point ( 1, 3). Then solve the equation for y. Solution We use the pointslope form of the equation of a line with m = 4, x1 = 1, and y1 = 3. y  y1 = m(x  x1) y  3 = 4[x  ( 1)]
This is the pointslope form of the equation. Substitute the given values: m = 4 and (x1, y1) = ( − 1, 3).
y  3 = 4(x + 1)
We now have an equation in pointslope form for the given line.
Now we solve this equation for y. 9GPGGFVQ KUQNCVGy
y − 3 = 4(x + 1) y − 3 = 4x + 4 y = 4x + 7
This is the pointslope form of the equation. Use the distributive property. Add 3 to both sides.
Section 2.3 Linear Functions and Slope
289
CHECK POINT 2 Write an equation in pointslope form for the line with slope 6 that passes through the point (2, 5). Then solve the equation for y.
y
–
EXAMPLE 3
6 5 4 3 2 1
–5 –4 –3 –2 –1–1 –2 –3 –4
Writing an Equation for a Line in PointSlope Form
Write an equation in pointslope form for the line passing through the points (4, 3) and ( 2, 6). (See Figure 2.32.) Then solve the equation for y.
1 2 3 4 5
x
Solution To use the pointslope form, we need to find the slope. The slope is the change in the y@coordinates divided by the corresponding change in the x@coordinates. m =
–
6  ( 3) 9 3 = = 2  4 6 2
We can take either point on the line to be (x1 , y1). Let’s use (x1 , y1) = (4, 3). Now, we are ready to write the pointslope form of the equation.
Figure 2.32 Write an equation in pointslope form for this line.
y  y1 = m(x  x1) y  ( 3) = 
3 2 (x
 4)
y + 3 =  32 (x  4)
You can use either point for (x1 , y1) when you write a pointslope equation for a line. Rework Example 3 using (  2, 6) for (x1 , y1). Once you solve for y, you should still obtain y = 
This is the pointslope form of the equation. Substitute: (x1, y1) = (4, − 3) and m = − 32 . Simplify.
We now have an equation in pointslope form for the line shown in Figure 2.32. Now, we solve this equation for y.
DISCOVERY
3 2x
This is the definition of slope using (4, − 3) and ( − 2, 6).
9GPGGFVQ KUQNCVGy
3
y + 3 = – 2 (x − 4) 3
y + 3 = – 2x + 6
This is the pointslope form of the equation. Use the distributive property.
3
y = – 2x + 3
Subtract 3 from both sides.
+ 3.
Write an equation in pointslope form for the line passing through the points ( 2, 1) and ( 1, 6). Then solve the equation for y. CHECK POINT 3
The SlopeIntercept Form of the Equation of a Line
3 Write and graph the
slopeintercept form of the equation of a line.
Let’s write the pointslope form of the equation of a nonvertical line with slope m and y@intercept b. The line is shown in Figure 2.33. Because the y@intercept is b, the line passes through (0, b). We use the pointslope form with x1 = 0 and y1 = b. y − y1 = m(x − x1)
y yKPVGTEGRVKUb(KZGFRQKPV
xyKU b (0, b)
.GVy=b
.GVx=
We obtain
5NQRGKUm
y  b = m(x  0).
6JKURQKPV KUCTDKVTCT[ (x, y)
Figure 2.33 A line with slope m and y@intercept b
Simplifying on the right side gives us x
y  b = mx. Finally, we solve for y by adding b to both sides. y = mx + b
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Chapter 2 Functions and Graphs
Thus, if a line’s equation is written as y = mx + b with y isolated on one side, the coefficient of x is the line’s slope and the constant term is the y@intercept. This form of a line’s equation is called the slopeintercept form of the line.
SlopeIntercept Form of the Equation of a Line The slopeintercept form of the equation of a nonvertical line with slope m and y@intercept b is y = mx + b.
The slopeintercept form of a line’s equation, y = mx + b, can be expressed in function notation by replacing y with f (x): f (x) = mx + b. We have seen that functions in this form are called linear functions. Thus, in the equation of a linear function, the coefficient of x is the line’s slope and the constant term is the y@intercept. Here are two examples: 1
y = 2x − 4 6JGUNQRGKU
f (x) = 2 x + 2. 6JGUNQRGKU
6JGyKPVGTEGRVKU–
6JGyKPVGTEGRVKU
If a linear function’s equation is in slopeintercept form, we can use the y@intercept and the slope to obtain its graph.
GREAT QUESTION If the slope is an integer, such as 2, why should I express it as 21 for graphing purposes? Writing the slope, m, as a fraction allows you to identify the rise (the fraction’s numerator) and the run (the fraction’s denominator).
Graphing y = mx + b Using the Slope and y@Intercept 1. Plot the point containing the y@intercept on the y@axis. This is the point (0, b). 2. Obtain a second point using the slope, m. Write m as a fraction, and use rise over run, starting at the point containing the y@intercept, to plot this point. 3. Use a straightedge to draw a line through the two points. Draw arrowheads at the ends of the line to show that the line continues indefinitely in both directions.
EXAMPLE 4
Graphing Using the Slope and yIntercept
Graph the linear function:
3 f (x) =  x + 2. 2
Solution The equation of the line is in the form ƒ(x) = mx + b. We can find the slope, m, by identifying the coefficient of x. We can find the y@intercept, b, by identifying the constant term. 3 f (x) = – x + 2 2 6JGUNQRGKU–
6JGyKPVGTEGRVKU
Now that we have identified the slope,  32 , and the y@intercept, 2, we use the threestep procedure to graph the equation.
291
Section 2.3 Linear Functions and Slope
Step 1 Plot the point containing the y@intercept on the y@axis. The y@intercept is 2. We plot (0, 2), shown in Figure 2.34.
y 5 4 3 4KUG=– 2 1
Step 2 Obtain a second point using the slope, m. Write m as a fraction, and use rise over run, starting at the point containing the y@intercept, to plot this point. The slope,  32 , is already written as a fraction.
yKPVGTEGRV 1
–5 –4 –3 –2 –1–1 –2 –3 –4 –5
3 4 5
x
m = 
4WP=
We plot the second point on the line by starting at (0, 2), the first point. Based on the slope, we move 3 units down (the rise) and 2 units to the right (the run). This puts us at a second point on the line, (2, 1), shown in Figure 2.34.
Figure 2.34 The graph of f (x) =  32 x + 2
Step 3 Use a straightedge to draw a line through the two points. The graph of the linear function f (x) =  32 x + 2 is shown as a blue line in Figure 2.34. CHECK POINT 4
4 Graph horizontal or vertical lines.
Graph the linear function:
f (x) = 35 x + 1.
Equations of Horizontal and Vertical Lines If a line is horizontal, its slope is zero: m = 0. Thus, the equation y = mx + b becomes y = b, where b is the y@intercept. All horizontal lines have equations of the form y = b.
EXAMPLE 5
y
–5 –4 –3 –2 –1–1 –2 –3 –5 –6
Graphing a Horizontal Line
Graph y = 4 in the rectangular coordinate system.
4 3 2 1
(–2, –4)
3 3 Rise = = 2 2 Run
1 2 3 4 5
x
yKPVGTEGRVKU– (3, –4) (0, –4)
Figure 2.35 The graph of y =  4 or f (x) =  4
Solution All ordered pairs that are solutions of y = 4 have a value of y that is always 4. Any value can be used for x. In the table on the right, we have selected three of the possible values for x: 2, 0, and 3. The table shows that three ordered pairs that are solutions of y = 4 are ( 2, 4), (0, 4), and (3, 4). Drawing a line that passes through the three points gives the horizontal line shown in Figure 2.35. CHECK POINT 5
x
y = −4
(x, y)
–2
–4
(–2, –4)
0
–4
(0, –4)
3
–4
(3, –4)
(QTCNN EJQKEGUQHx
yKUC EQPUVCPV–
Graph y = 3 in the rectangular coordinate system.
Equation of a Horizontal Line A horizontal line is given by an equation of the form y = b, where b is the y@intercept of the line. The slope of a horizontal line is zero.
y
b
yKPVGTEGRVb x
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Chapter 2 Functions and Graphs
Because any vertical line can intersect the graph of a horizontal line y = b only once, a horizontal line is the graph of a function. Thus, we can express the equation y = b as f (x) = b. This linear function is often called a constant function. Next, let’s see what we can discover about the graph of an equation of the form x = a by looking at an example.
y
EXAMPLE 6
5 4 (2, 3) 3 xKPVGTEGRV 2 KU 1 (2, 0) –3 –2 –1–1 –2 –3 –4 –5
1 2 3 4 5 6 7 (2, –2)
Figure 2.36 The graph of x = 2
Graphing a Vertical Line
Graph the linear equation: x = 2.
x
Solution All ordered pairs that are solutions of x = 2 have a value of x that is always 2. Any value can be used for y. In the table on the right, we have selected three of the possible values for y: 2, 0, and 3. The table shows that three ordered pairs that are solutions of x = 2 are (2, 2), (2, 0), and (2, 3). Drawing a line that passes through the three points gives the vertical line shown in Figure 2.36.
(QTCNNEJQKEGUQHy
xKU CNYC[U
x=2
y
(x, y)
2
–2
(2, –2)
2
0
(2, 0)
2
3
(2, 3)
Equation of a Vertical Line y
A vertical line is given by an equation of the form x = a,
a
where a is the x@intercept of the line. The slope of a vertical line is undefined.
x xKPVGTEGRVa
Does a vertical line represent the graph of a linear function? No. Look at the graph of x = 2 in Figure 2.36. A vertical line drawn through (2, 0) intersects the graph infinitely many times. This shows that infinitely many outputs are associated with the input 2. No vertical line represents a linear function. CHECK POINT 6
5 Recognize and use the general form of a line’s equation.
Graph the linear equation: x = 3.
The General Form of the Equation of a Line The vertical line whose equation is x = 5 cannot be written in slopeintercept form, y = mx + b, because its slope is undefined. However, every line has an equation that can be expressed in the form Ax + By + C = 0. For example, x = 5 can be expressed as 1x + 0y  5 = 0, or x  5 = 0. The equation Ax + By + C = 0 is called the general form of the equation of a line.
General Form of the Equation of a Line Every line has an equation that can be written in the general form Ax + By + C = 0, where A, B, and C are real numbers, and A and B are not both zero.
Section 2.3 Linear Functions and Slope
GREAT QUESTION In the general form Ax + By + C = 0, can I immediately determine that the slope is A and the y@intercept is B? No. Avoid this common error. You need to solve Ax + By + C = 0 for y before finding the slope and the y@intercept.
293
If the equation of a nonvertical line is given in general form, it is possible to find the slope, m, and the y@intercept, b, for the line. We solve the equation for y, transforming it into the slopeintercept form y = mx + b. In this form, the coefficient of x is the slope of the line and the constant term is its y@intercept.
EXAMPLE 7
Finding the Slope and the yIntercept
Find the slope and the y@intercept of the line whose equation is 3x + 2y  4 = 0. Solution The equation is given in general form. We begin by rewriting it in the form y = mx + b. We need to solve for y. 3x + 2y − 4 = 0 1WTIQCNKU VQKUQNCVGy
This is the given equation.
2y = –3x + 4
Isolate the term containing y by adding − 3x + 4 to both sides.
2y –3x + 4 = 2 2 3 y=– x+2 2 UNQRG
yKPVGTEGRV
Divide both sides by 2.
On the right, divide each term in the numerator by 2 to obtain slopeintercept form.
The coefficient of x,  32 , is the slope and the constant term, 2, is the y@intercept. This is the form of the equation that we graphed in Figure 2.34. CHECK POINT 7 Find the slope and the y@intercept of the line whose equation is 3x + 6y  12 = 0. Then use the y@intercept and the slope to graph the equation.
6 Use intercepts to graph
the general form of a line’s equation.
Using Intercepts to Graph Ax + By + C = 0 Example 7 and Check Point 7 illustrate that one way to graph the general form of a line’s equation is to convert to slopeintercept form, y = mx + b. Then use the slope and the y@intercept to obtain the graph. A second method for graphing Ax + By + C = 0 uses intercepts. This method does not require rewriting the general form in a different form.
Using Intercepts to Graph Ax + By + C = 0 1. Find the x@intercept. Let y = 0 and solve for x. Plot the point containing the x@intercept on the x@axis. 2. Find the y@intercept. Let x = 0 and solve for y. Plot the point containing the y@intercept on the y@axis. 3. Use a straightedge to draw a line through the two points containing the intercepts. Draw arrowheads at the ends of the line to show that the line continues indefinitely in both directions. As long as none of A, B, and C is zero, the graph of Ax + By + C = 0 will have distinct x and yintercepts, and this threestep method can be used to graph the equation.
294
Chapter 2 Functions and Graphs
EXAMPLE 8
Using Intercepts to Graph a Linear Equation
Graph using intercepts: 4x  3y  6 = 0. Solution Step 1 Find the x@intercept. Let y = 0 and solve for x.
y 5 4 3 2 1 –5 –4 –3 –2 –1–1
–
–2 –3 –4 –5
xKPVGTEGRV x 1 2 3 4 5 yKPVGTEGRV–
Figure 2.37 The graph of 4x  3y  6 = 0
4x  3 # 0  6 = 0 4x  6 = 0 4x = 6 x =
Replace y with 0 in 4x − 3y − 6 = 0. Simplify. Add 6 to both sides.
6 3 = 4 2
Divide both sides by 4.
The x@intercept is 32 , so the line passes through 1 32 , 0 2 or (1.5, 0), as shown in Figure 2.37. Step 2 Find the y@intercept. Let x = 0 and solve for y. 4 # 0  3y  6 3y  6 3y y
= = = =
0 0 6 2
Replace x with 0 in 4x − 3y − 6 = 0. Simplify. Add 6 to both sides. Divide both sides by − 3.
The y@intercept is 2, so the line passes through (0, 2), as shown in Figure 2.37. Step 3 Graph the equation by drawing a line through the two points containing the intercepts. The graph of 4x  3y  6 = 0 is shown in Figure 2.37. CHECK POINT 8
Graph using intercepts: 3x  2y  6 = 0.
We’ve covered a lot of territory. Let’s take a moment to summarize the various forms for equations of lines. Equations of Lines
7 Model data with linear functions and make predictions.
1. Pointslope form
y  y1 = m(x  x1)
2. Slopeintercept form
y = mx + b or f (x) = mx + b
3. Horizontal line
y = b or f (x) = b
4. Vertical line
x = a
5. General form
Ax + By + C = 0
Applications Linear functions are useful for modeling data that fall on or near a line.
EXAMPLE 9
Modeling Global Warming
The amount of carbon dioxide in the atmosphere, measured in parts per million, has been increasing as a result of the burning of oil and coal. The buildup of gases and particles traps heat and raises the planet’s temperature. The bar graph in Figure 2.38(a) gives the average atmospheric concentration of carbon dioxide and the average global temperature for six selected years. The data are displayed as a set of six points in a rectangular coordinate system in Figure 2.38(b). a. Shown on the scatter plot in Figure 2.38(b) is a line that passes through or near the six points. Write the slopeintercept form of this equation using function notation.
Section 2.3 Linear Functions and Slope
295
Carbon Dioxide Concentration and Global Temperature
58.2° 58.0°
57.8°
57.6° 57.4° 57.2°
Average Global Temperature
Average Global Temperature (degrees Fahrenheit)
y
57.99
57.64 57.67
57.35
57.04 57.06
57.0° 317 326 339 354 369 385 Average Carbon Dioxide Concentration (parts per million)
58.2°
58.0° 57.8° 57.6° 57.4° 57.2°
57.0° 310
Figure 2.38(a) Source: National Oceanic and Atmospheric Administration
320 330 340 350 360 370 380 390 Average Carbon Dioxide Concentration
Figure 2.38(b)
b. The preindustrial concentration of atmospheric carbon dioxide was 280 parts per million. The United Nations’ Intergovernmental Panel on Climate Change predicts global temperatures will rise between 2°F and 5°F if carbon dioxide concentration doubles from the preindustrial level. Compared to the average global temperature of 57.99°F for 2009, how well does the function from part (a) model this prediction? TECHNOLOGY You can use a graphing utility to obtain a model for a scatter plot in which the data points fall on or near a straight line. After entering the data in Figure 2.38(a), a graphing utility displays a scatter plot of the data and the regression line, that is, the line that best fits the data.
Solution a. The line in Figure 2.38(b) passes through (326, 57.06) and (385, 57.99). We start by finding its slope. m =
Change in y 57.99  57.06 0.93 = = ≈ 0.02 Change in x 385  326 59
The slope indicates that for each increase of one part per million in carbon dioxide concentration, the average global temperature is increasing by approximately 0.02°F. Now we write the line’s equation in slopeintercept form. y  y1 = m(x  x1) y  57.06 = 0.02(x  326)
Begin with the pointslope form. Either ordered pair can be (x1, y1). Let (x1, y1) = (326, 57.06). From above, m ? 0.02.
y  57.06 = 0.02x  6.52 [310, 390, 10] by [56.8, 58.4, 0.2]
Also displayed is the regression line’s equation.
Apply the distributive property: 0.02(326) = 6.52.
y = 0.02x + 50.54
Add 57.06 to both sides and solve for y.
A linear function that models average global temperature, f (x), for an atmospheric carbon dioxide concentration of x parts per million is f (x) = 0.02x + 50.54. b. If carbon dioxide concentration doubles from its preindustrial level of 280 parts per million, which many experts deem very likely, the concentration will reach 280 * 2, or 560 parts per million. We use the linear function to predict average global temperature at this concentration.
x
296
Chapter 2 Functions and Graphs
f (x) = 0.02x + 50.54 f (560) = 0.02(560) + 50.54
Use the function from part (a). Substitute 560 for x.
= 11.2 + 50.54 = 61.74 Our model projects an average global temperature of 61.74°F for a carbon dioxide concentration of 560 parts per million. Compared to the average global temperature of 57.99°F for 2009 shown in Figure 2.38(a) on the previous page, this is an increase of 61.74°F  57.99°F = 3.75°F. This is consistent with a rise between 2°F and 5°F as predicted by the Intergovernmental Panel on Climate Change. CHECK POINT 9 Use the data points (317, 57.04) and (354, 57.64), shown, but not labeled, in Figure 2.38(b) to obtain a linear function that models average global temperature, f (x), for an atmospheric carbon dioxide concentration of x parts per million. Round m to three decimal places and b to one decimal place. Then use the function to project average global temperature at a concentration of 600 parts per million.
ACHIEVING SUCCESS Read your lecture notes before starting your homework. Often homework problems, and later the test problems, are variations of the ones done by your professor in class.
BLITZER BONUS
Accelerating Climate Change
In a grim milestone for our warming planet, atmospheric carbon dioxide levels have reached their highest level in more than three million years. In 2020, sensors at the Mauna Loa Observatory in Hawaii recorded the level at 415 parts per million. The last time there was this much carbon dioxide in the atmosphere was during the Pliocene Epoch, between 5.3 million and 2.6 million years ago, when sea levels were about 50 feet higher than they are now. Scientists warn that if greenhouse gas emissions are not curbed, carbon dioxide concentrations could hit 500 parts per million within 30 years. “None of these specific numbers are really thresholds in the sense that anything particular happens when we cross them. But as we go through them, we are putting our foot on the accelerator of climate change, and impacts and damage will continue to rise.” –Gavin Schmidt, director of NASA’s Goddard Institute for Space Studies
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. Data presented in a visual form as a set of points is called a/an . A line that best fits this set of points is called a/an line.
C8. The slopeintercept form of the equation of a line is , where m represents the b represents the .
C2. The slope, m, of a line through the distinct points (x1, y1)
C9. In order to graph the line whose equation is 2 y = x + 3, begin by plotting the point . 5 From this point, we move units up (the rise) and units to the right (the run).
and (x2, y2) is given by the formula m =
.
C3. If a line rises from left to right, the line has C4. If a line falls from left to right, the line has C5. The slope of a horizontal line is
.
C6. The slope of a vertical line is
.
slope. slope.
C7. The pointslope form of the equation of a nonvertical line with slope m that passes through the point (x1, y1) is .
C10. The graph of the equation y = 3 is a/an C11. The graph of the equation x =  2 is a/an
and
line. line.
C12. The equation Ax + By + C = 0, where A and B are not both zero, is called the form of the equation of a line.
Section 2.3 Linear Functions and Slope
297
2.3 EXERCISE SET Practice Exercises In Exercises 1–10, find the slope of the line passing through each pair of points or state that the slope is undefined. Then indicate whether the line through the points rises, falls, is horizontal, or is vertical. 1. 3. 5. 7. 9.
(4, 7) and (8, 10) ( 2, 1) and (2, 2) (4,  2) and (3,  2) ( 2, 4) and (  1, 1) (5, 3) and (5, 2)
2. 4. 6. 8. 10.
(2, 1) and (3, 4) ( 1, 3) and (2, 4) ( 2, 4) and (  6, 4) (6,  4) and (4,  2) (3,  4) and (3, 5)
In Exercises 11–38, use the given conditions to write an equation for each line in pointslope form and slopeintercept form. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38.
Slope = 2, passing through (3, 5) Slope = 4, passing through (1, 3) Slope = 6, passing through ( 2, 5) Slope = 8, passing through (4,  1) Slope =  3, passing through (  2,  3) Slope =  5, passing through (  4,  2) Slope =  4, passing through (  4, 0) Slope =  2, passing through (0, 3) Slope =  1, passing through 1  12 ,  2 2 Slope =  1, passing through 1  4,  14 2 Slope = 12 , passing through the origin Slope = 13 , passing through the origin Slope =  23 , passing through (6,  2) Slope =  35 , passing through (10, 4) Passing through (1, 2) and (5, 10) Passing through (3, 5) and (8, 15) Passing through (  3, 0) and (0, 3) Passing through (  2, 0) and (0, 2) Passing through (  3,  1) and (2, 4) Passing through (  2,  4) and (1, 1) Passing through (  3,  2) and (3, 6) Passing through (  3, 6) and (3,  2) Passing through (  3,  1) and (4, 1) Passing through (  2,  5) and (6, 5) Passing through (2, 4) with x@intercept = 2 Passing through (1, 3) with x@intercept =  1 x@intercept =  12 and y@intercept = 4 x@intercept = 4 and y@intercept =  2
In Exercises 39–48, give the slope and yintercept of each line whose equation is given. Then graph the linear function. 39. y = 2x + 1 41. f (x) = 2x + 1 3 43. f (x) = x  2 4 3 45. y =  x + 7 5 1 47. g(x) =  x 2
40. y = 3x + 2 42. f (x) = 3x + 2 3 44. f (x) = x  3 4 2 46. y =  x + 6 5 1 48. g(x) =  x 3
In Exercises 49–58, graph each equation in a rectangular coordinate system. 50. y = 4 49. y =  2 53. y = 0 54. x = 0 57. 3x  18 = 0
51. x = 3 52. x = 5 55. f (x) = 1 56. f (x) = 3 58. 3x + 12 = 0
In Exercises 59–66, a. Rewrite the given equation in slopeintercept form. b. Give the slope and yintercept. c. Use the slope and yintercept to graph the linear function. 59. 61. 63. 65.
3x 2x 8x 3y
+ + 
y  5 = 0 3y  18 = 0 4y  12 = 0 9 = 0
60. 62. 64. 66.
4x 4x 6x 4y
+ + +
y  6 = 0 6y + 12 = 0 5y  20 = 0 28 = 0
In Exercises 67–72, use intercepts to graph each equation. 67. 6x  2y  12 = 0 69. 2x + 3y + 6 = 0 71. 8x  2y + 12 = 0
68. 6x  9y  18 = 0 70. 3x + 5y + 15 = 0 72. 6x  3y + 15 = 0
Practice PLUS In Exercises 73–76, find the slope of the line passing through each pair of points or state that the slope is undefined. Assume that all variables represent positive real numbers. Then indicate whether the line through the points rises, falls, is horizontal, or is vertical. 74. ( a, 0) and (0, b) 76. (a  b, c) and (a, a + c)
73. (0, a) and (b, 0) 75. (a, b) and (a, b + c)
In Exercises 77–78, give the slope and y@intercept of each line whose equation is given. Assume that B ≠ 0. 77. Ax + By = C
78. Ax = By  C
In Exercises 79–80, find the value of y if the line through the two given points is to have the indicated slope. 79. (3, y) and (1, 4), m =  3 80. (  2, y) and (4,  4), m =
1 3
In Exercises 81–82, graph each linear function. 82. 6x  5f (x)  20 = 0 81. 3x  4f (x)  6 = 0 83. If one point on a line is (3,  1) and the line’s slope is 2, find the y@intercept. 84. If one point on a line is (2, 6) and the line’s slope is  32 , find the y@intercept. Use the figure to make the lists in Exercises 85–86. y y = m1x + b1 y = m2x + b2 x y = m3x + b3 y = m4x + b4
85. List the slopes m1 , m2 , m3 , and m4 in order of decreasing size. 86. List the y@intercepts b1 , b2 , b3 , and b4 in order of decreasing size.
298
Chapter 2 Functions and Graphs
Application Exercises Americans’ trust in government and the media has generally been on a downward trend since pollsters first asked questions on these topics in the second half of the twentieth century. Trust in government hit an alltime low of 14% in 2014, while trust in the media bottomed out at 32% in 2016. The bar graph shows the percentage of Americans trusting in the government and the media for five selected years. The data are displayed as two sets of five points each, one scatter plot for the percentage of Americans trusting in the government and one for the percentage of Americans trusting in the media. Also shown for each scatter plot is a line that passes through or near the five points. Use these lines to solve Exercises 87–88. Americans’ Trust in Government and the Media y
Percentage of Americans
60%
Trust in Media
54
50%
46
40%
36
44
41
40
31
30%
20
20%
19
17
10% 2003
2007
2011 Year
2015
2019
70% Percentage of Americans
Trust in Government
70%
60%
50%
Media
40% 30%
20%
Government
10% 0
4
8
12
16
20
x
Years after 2003
Sources: Gallup, Pew Research
87. In this exercise, you will use the red line for trust in government shown on the scatter plot to develop a model for the percentage of Americans trusting in government. a. Use the two points whose coordinates are shown by the voice balloons to find the pointslope form of the equation of the line that models the percentage of Americans trusting in government, y, x years after 2003. Round the slope to two decimal places. b. Write the equation from part (a) in slopeintercept form. Use function notation. c. Use the linear function to predict the percentage of Americans trusting in government in 2025. Round to the nearest percent.
88. In this exercise, you will use the blue line for trust in media shown on the scatter plot to develop a model for the percentage of Americans trusting in media. a. Use the two points whose coordinates are shown by the voice balloons to find the pointslope form of the equation of the line that models the percentage of Americans trusting in media, y, x years after 2003. Round the slope to two decimal places. b. Write the equation from part (a) in slopeintercept form. Use function notation. c. Use the linear function to predict the percentage of Americans trusting in media in 2025. Round to the nearest percent.
The bar graph gives the life expectancy for American men and women born in six selected years. In Exercises 89–90, you will use the data to obtain models for life expectancy and make predictions about how long American men and women will live in the future.
89. Use the data for males shown in the bar graph at the bottom of the previous column to solve this exercise. a. Let x represent the number of birth years after 1960 and let y represent male life expectancy. Create a scatter plot that displays the data as a set of six points in a rectangular coordinate system. b. Draw a line through the two points that show male life expectancies for 1980 and 2000. Use the coordinates of these points to write a linear function that models life expectancy, E(x), for American men born x years after 1960. c. Use the function from part (b) to project the life expectancy of American men born in 2020. 90. Use the data for females shown in the bar graph at the bottom of the previous column to solve this exercise. a. Let x represent the number of birth years after 1960 and let y represent female life expectancy. Create a scatter plot that displays the data as a set of six points in a rectangular coordinate system.
Life Expectancy in the United States, by Year of Birth Males Females 90 74.3 79.7
76.2 81.0
1960
1970
70.0 77.4
2010
60
67.1 74.7
2000
70
66.6 73.1
1980 1990 Birth Year Source: National Center for Health Statistics
Life Expectancy
71.8 78.8
80
50 40 30 20 10
299
Section 2.3 Linear Functions and Slope b. Draw a line through the two points that show female life expectancies for 1970 and 2000. Use the coordinates of these points to write a linear function that models life expectancy, E(x), for American women born x years after 1960. Round the slope to two decimal places. c. Use the function from part (b) to project the life expectancy of American women born in 2020. 91. Shown, again, is the scatter plot that indicates a relationship between the percentage of adult females in a country who are literate and the mortality of children under five. Also shown is a line that passes through or near the points. Find a linear function that models the data by finding the slopeintercept form of the line’s equation. Use the function to make a prediction about child mortality based on the percentage of adult females in a country who are literate. y
Literacy and Child Mortality
UnderFive Mortality (per thousand)
350 300
Explaining the Concepts 93. What is the slope of a line and how is it found? 94. Describe how to write the equation of a line if the coordinates of two points along the line are known. 95. Explain how to derive the slopeintercept form of a line’s equation, y = mx + b, from the pointslope form y  y1 = m(x  x1). 96. Explain how to graph the equation x = 2. Can this equation be expressed in slopeintercept form? Explain. 97. Explain how to use the general form of a line’s equation to find the line’s slope and y@intercept. 98. Explain how to use intercepts to graph the general form of a line’s equation. 99. Take another look at the scatter plot in Exercise 91. Although there is a relationship between literacy and child mortality, we cannot conclude that increased literacy causes child mortality to decrease. Offer two or more possible explanations for the data in the scatter plot.
250 200 150
Technology Exercises
100
Use a graphing utility to graph each equation in Exercises 100–103. Then use the TRACE feature to trace along the line and find the coordinates of two points. Use these points to compute the line’s slope. Check your result by using the coefficient of x in the line’s equation.
50 0
10 20 30 40 50 60 70 80 90 100 Percentage of Adult Females Who Are Literate Source: United Nations
x
92. Just as money doesn’t buy happiness for individuals, the two don’t necessarily go together for countries either. However, the scatter plot does show a relationship between a country’s annual per capita income and the percentage of people in that country who call themselves “happy.”
101. y = 3x + 6 100. y = 2x + 4 1 102. y =  2 x  5 103. y = 34 x  2 104. Is there a relationship between wine consumption and deaths from heart disease? The table gives data from 19 developed countries. (TCPEG
y
Percentage of People Calling Themselves “Happy”
100 90 80
Per Capita Income and National Happiness
Chile
Uruguay Argentina Croatia Hungary Algeria South Africa
Brazil
Egypt India
60 50
Iran Jordan Albania
Pakistan
40 30
Turkey Latvia Belarus Romania
Ukraine Russia Zimbabwe
Switzerland
New Zealand Czech RepublicPortugal
Philippines China
70
Ireland Netherlands
Nigeria Colombia Venezuela Vietnam Mexico Indonesia
Slovakia Poland Estonia
Bulgaria
Spain Israel Slovenia
Norway Italy
Japan
France
U.S. Canada
Country
A
B
C
D
E
F
G
Liters of alcohol from drinking wine, per person per year (x)
2.5
3.9
2.9
2.4
2.9
0.8
9.1
Deaths from heart disease, per 100,000 people per year (y)
211 167 131 191 220 297
71
Denmark Austria
South Korea Greece
Belgium Australia Germany Finland Sweden Singapore Britain
Moldova Tanzania
$5000 $15,000 $25,000 $35,000 Annual Per Capita Income (dollars) Source: Richard Layard, Happiness: Lessons from a New Science, Penguin, 2005
75 x
Draw a line that fits the data so that the spread of the data points around the line is as small as possible. Use the coordinates of two points along your line to write the slopeintercept form of its equation. Express the equation in function notation and use the linear function to make a prediction about national happiness based on per capita income.
Country
H
I
J
K
L
M
N
O
P
Q
R
S
(x)
0.8 0.7 7.9 1.8 1.9 0.8 6.5 1.6 5.8 1.3 1.2 2.7
(y)
211 300 107 167 266 227 86 207 115 285 199 172
Source: New York Times
a. Use the statistical menu of your graphing utility to enter the 19 ordered pairs of data items shown in the table. b. Use the scatter plot capability to draw a scatter plot of the data.
300
Chapter 2 Functions and Graphs c. Select the linear regression option. Use your utility to obtain values for a and b for the equation of the regression line, y = ax + b. You may also be given a correlation coefficient, r. Values of r close to 1 indicate that the points can be described by a linear relationship and the regression line has a positive slope. Values of r close to  1 indicate that the points can be described by a linear relationship and the regression line has a negative slope. Values of r close to 0 indicate no linear relationship between the variables. In this case, a linear model does not accurately describe the data. d. Use the appropriate sequence (consult your manual) to graph the regression equation on top of the points in the scatter plot.
Critical Thinking Exercises Make Sense? In Exercises 105–108, determine whether each statement makes sense or does not make sense, and explain your reasoning.
Group Exercise 117. In Exercises 87–88, we used the data in a bar graph to develop linear functions that modeled trust in government and trust in media. For this group exercise, you might find it helpful to pattern your work after Exercises 87 and 88. Group members should begin by consulting an almanac, newspaper, magazine, or the Internet to find data that appear to lie approximately on or near a line. Working by hand or using a graphing utility, group members should construct scatter plots for the data that were assembled. If working by hand, draw a line that approximately fits the data in each scatter plot and then write its equation as a function in slopeintercept form. If using a graphing utility, obtain the equation of each regression line. Then use each linear function’s equation to make predictions about what might occur in the future. Are there circumstances that might affect the accuracy of the prediction? List some of these circumstances.
105. The graph of my linear function at first increased, reached a maximum point, and then decreased. 106. A linear function that models tuition and fees at public fouryear colleges from 2000 through 2020 has negative slope. 107. Because the variable m does not appear in Ax + By + C = 0, equations in this form make it impossible to determine the line’s slope. 108. The federal minimum wage was $7.25 per hour from 2009 through 2020, so f (x) = 7.25 models the minimum wage, f (x), in dollars, for the domain {2009, 2010, 2011, c, 2020}.
Retaining the Concepts
In Exercises 109–112, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.
x + 3 x  2 Ú + 1 4 3 (Section 1.7, Example 5) 120. 3 2x + 6  9 6 15
109. The equation y = mx + b shows that no line can have a y@intercept that is numerically equal to its slope. 110. Every line in the rectangular coordinate system has an equation that can be expressed in slopeintercept form. 111. The graph of the linear function 5x + 6y  30 = 0 is a line passing through the point (6, 0) with slope  56 . 112. The graph of x = 7 in the rectangular coordinate system is the single point (7, 0). In Exercises 113–114, find the coefficients that must be placed in each shaded area so that the function’s graph will be a line satisfying the specified conditions. y  12 = 0; x@intercept = 2; y@intercept = 4 1 114. x + y  12 = 0; y@intercept =  6; slope = 2 115. Prove that the equation of a line passing through (a, 0) and y x (0, b)(a ≠ 0, b ≠ 0) can be written in the form + = 1. a b Why is this called the intercept form of a line? 116. Excited about the success of celebrity stamps, post office officials were rumored to have put forth a plan to institute two new types of thermometers. On these new scales, °U represents degrees Usher and °R represents degrees Rihanna. If it is known that 40°U = 25°R, 280°U = 125°R, and degrees Usher is linearly related to degrees Rihanna, write an equation expressing U in terms of R. 113.
x +
118. According to the U.S. Office of Management and Budget, the cost of maintaining existing public transportation infrastructure in 2020 was $103.4 billion and is projected to increase by $0.9 billion each year. By which year is the cost of maintaining existing public transportation infrastructure expected to reach $116 billion? (Section 1.3, Example 2) In Exercises 119–120, solve and graph the solution set on a number line. 119.
(Section 1.7, Example 9)
Preview Exercises Exercises 121–123 will help you prepare for the material covered in the next section. 121. Write the slopeintercept form of the equation of the line passing through (  3, 1) whose slope is the same as the line whose equation is y = 2x + 1. 122. Write an equation in general form of the line passing through (3,  5) whose slope is the negative reciprocal (the reciprocal with the opposite sign) of  14 . 123. If f (x) = x 2, find f(x2)  f(x1) x2  x1 where x1 = 1 and x2 = 4.
,
Section 2.4 More on Slope
SECTION 2.4
301
More on Slope Living Arrangements of U.S. Young Adults, Ages 25–34
1 Find slopes and
equations of parallel and perpendicular lines.
2 Interpret slope as rate of change.
3 Find a function’s average rate of change.
Percentage with Living Arrangement
WHAT YOU'LL LEARN 50 45 40 35 30 25 20 15 10 5
Own Home
2000
Live with Parents 2005
2010 Year
2015
2020
Figure 2.39 Source: urban.org
As housing prices skyrocket, fewer U.S. young adults are buying homes and more are living with their parents. Figure 2.39 shows that in 2017, 38.4% of 25to34yearold young adults owned homes, a decrease from the percentage displayed for 2000, and 22.0% lived with parents, nearly doubling the 2000 percentage. Take a second look at Figure 2.39. The red graph is going down from left to right, indicating a negative rate of change in home ownership among young adults. The green graph is going up from left to right, indicating a positive rate of change in young adults living with parents. In this section, you will learn how to interpret slope as a rate of change. You will also explore the relationships between parallel and perpendicular lines.
Parallel and Perpendicular Lines Two nonintersecting lines that lie in the same plane are parallel. If two lines do not intersect, the ratio of the vertical change to the horizontal change is the same for both lines. Because two parallel lines have the same “steepness,” they must have the same slope.
Slope and Parallel Lines 1. If two nonvertical lines are parallel, then they have the same slope. 2. If two distinct nonvertical lines have the same slope, then they are parallel. 3. Two distinct vertical lines, both with undefined slopes, are parallel.
1 Find slopes and equations
of parallel and perpendicular lines.
EXAMPLE 1
Writing Equations of a Line Parallel to a Given Line
Write an equation of the line passing through ( 3, 1) and parallel to the line whose equation is y = 2x + 1. Express the equation in pointslope form and slopeintercept form.
302
Chapter 2 Functions and Graphs
Solution The situation is illustrated in Figure 2.40. We are looking for the equation of the red line passing through ( 3, 1) and parallel to the blue line whose equation is y = 2x + 1. How do we obtain the equation of this red line? Notice that the line passes through the point ( 3, 1). Using the pointslope form of the line’s equation, we have x1 = 3 and y1 = 1.
6JGGSWCVKQPQHVJKUNKPGKUIKXGPy=x+ y
(–3, 1)
–7 –6 –5 –4 –3 –2 –1–1
y − y1 = m(x − x1) y=
5 4 3 2 1 1 2 3
x
–2 –3 –4 –5
x=–
With (x1, y1) = ( 3, 1), the only thing missing 9GOWUVYTKVGVJGGSWCVKQPQHVJKUNKPG from the equation of the red line is m, the slope. Do we know anything about the slope of either Figure 2.40 line in Figure 2.40? The answer is yes; we know the slope of the blue line on the right, whose equation is given. y = 2x + 1 6JGUNQRGQHVJGDNWGNKPGQP VJGTKIJVKP(KIWTGKU
Parallel lines have the same slope. Because the slope of the blue line is 2, the slope of the red line, the line whose equation we must write, is also 2: m = 2. We now have values for x1, y1, and m for the red line. y − y1 = m(x − x1) y=
m=
x=–
The pointslope form of the red line’s equation is y  1 = 2[x  ( 3)] or y  1 = 2(x + 3). Solving for y, we obtain the slopeintercept form of the equation. y  1 = 2x + 6 y = 2x + 7
Apply the distributive property. Add 1 to both sides. This is the slopeintercept form, y = mx + b, of the equation. Using function notation, the equation is f (x) = 2x + 7.
Write an equation of the line passing through ( 2, 5) and parallel to the line whose equation is y = 3x + 1. Express the equation in pointslope form and slopeintercept form. CHECK POINT 1
Two lines that intersect at a right angle (90°) are said to be perpendicular, shown in Figure 2.41. The relationship between the slopes of perpendicular lines is not as obvious as the relationship between parallel lines. Figure 2.41 shows line AB, with slope dc . Rotate line AB counterclockwise 90° to the left to obtain line A′B′, perpendicular to line AB. The figure indicates that the rise and the run of the new line are reversed from the original line, but the former rise, the new run, is now negative. This means that the slope of the new line is  dc. Notice that the product of the slopes of the two perpendicular lines is 1:
y B′ 5NQRG=– d
d c d
A′ –c
B
c A
5NQRG=
c d x
Figure 2.41 Slopes of perpendicular lines
c d a b a  b = 1. c d
This relationship holds for all perpendicular lines and is summarized in the box at the top of the next page.
Section 2.4 More on Slope
303
Slope and Perpendicular Lines 1. If two nonvertical lines are perpendicular, then the product of their slopes is 1. 2. If the product of the slopes of two lines is 1, then the lines are perpendicular. 3. A horizontal line having zero slope is perpendicular to a vertical line having undefined slope.
An equivalent way of stating this relationship is to say that one line is perpendicular to another line if its slope is the negative reciprocal of the slope of the other line. For example, if a line has slope 5, any line having slope  15 is perpendicular to it. Similarly, if a line has slope  34, any line having slope 43 is perpendicular to it.
EXAMPLE 2
Writing Equations of a Line Perpendicular to a Given Line
a. Find the slope of any line that is perpendicular to the line whose equation is x + 4y  8 = 0. b. Write the equation of the line passing through (3, 5) and perpendicular to the line whose equation is x + 4y  8 = 0. Express the equation in general form. Solution a. We begin by writing the equation of the given line, x + 4y  8 = 0, in slopeintercept form. Solve for y. x + 4y − 8 = 0 4y = –x + 8 y=
– 14 x
+2
This is the given equation. To isolate the yterm, subtract x and add 8 on both sides. Divide both sides by 4.
5NQRGKU–
The given line has slope  14. Any line perpendicular to this line has a slope that is the negative reciprocal of  14. Thus, the slope of any perpendicular line is 4. b. Let’s begin by writing the pointslope form of the perpendicular line’s equation. Because the line passes through the point (3, 5), we have x1 = 3 and y1 = 5. In part (a), we determined that the slope of any line perpendicular to x + 4y  8 = 0 is 4, so the slope of this particular perpendicular line must also be 4: m = 4. y − y1 = m(x − x1) y=–
m=
x=
The pointslope form of the perpendicular line’s equation is y  ( 5) = 4(x  3) or y + 5 = 4(x  3).
304
Chapter 2 Functions and Graphs
How can we express this equation, y + 5 = 4(x  3), in general form (Ax + By + C = 0)? We need to obtain zero on one side of the equation. Let’s do this and keep A, the coefficient of x, positive. y + 5 = 4(x  3)
This is the pointslope form of the line’s equation.
y + 5 = 4x  12 y  y + 5  5 = 4x  y  12  5
Apply the distributive property. To obtain 0 on the left, subtract y and subtract 5 on both sides.
0 = 4x  y  17
Simplify.
In general form, the equation of the perpendicular line is 4x  y  17 = 0. CHECK POINT 2
a. Find the slope of any line that is perpendicular to the line whose equation is x + 3y  12 = 0. b. Write the equation of the line passing through ( 2, 6) and perpendicular to the line whose equation is x + 3y  12 = 0. Express the equation in general form.
Slope as Rate of Change
2 Interpret slope as rate of change.
Slope is defined as the ratio of a change in y to a corresponding change in x. It describes how fast y is changing with respect to x. For a linear function, slope may be interpreted as the rate of change of the dependent variable per unit change in the independent variable. Our next example shows how slope can be interpreted as a rate of change in an applied situation. When calculating slope in applied problems, keep track of the units in the numerator and the denominator.
EXAMPLE 3
Slope as a Rate of Change
The line graphs for the living arrangements of young adults are shown again in Figure 2.42. Find the slope of the line segment for the percentage of young adults, age 25 to 34, owning a home. Describe what this slope represents. Solution We let x represent a year and y the percentage of young adults owning a home in that year. The two points shown on the line segment for home ownership have the following coordinates:
Percentage with Living Arrangement
Living Arrangements of U.S. Young Adults, Ages 25–34 50 45 40 35 30 25 20 15 10 5
+PQYPGFJQOGU
Own Home
2000
(2000, 45.4)
(2017, 38.4). +PQYPGFJQOGU
Now we compute the slope:
Live with Parents 2005
Figure 2.42 Source: urban.org
Change in y 38.4 − 45.4 m= = 2017 − 2000 Change in x
2010 Year
2015
2020
=
–0.41 percent –7 ≈ . year 17
6JGWPKVKPVJG PWOGTCVQTKURGTEGPV 6JGWPKVKPVJG FGPQOKPCVQTKU[GCT
The slope indicates that the percentage of U.S. young adults, age 25 to 34, owning a home decreased at a rate of approximately 0.41 each year for the period from 2000 to 2017. The rate of change is  0.41% per year.
Section 2.4 More on Slope
305
CHECK POINT 3 Use the ordered pairs in Figure 2.42 to find the slope of the green line segment for young adults, age 25 to 34, living with parents. Express the slope correct to two decimal places and describe what it represents.
rate of change.
The Average Rate of Change of a Function If the graph of a function is not a straight line, the average rate of change between any two points is the slope of the line containing the two points. This line is called a secant line. For example, Figure 2.43 shows the graph of a particular man’s height, in inches, as a function of his age, in years. Two points on the graph are labeled: (13, 57) and (18, 76). At age 13, this man was 57 inches tall and at age 18, he was 76 inches tall.
y
80
5GECPVNKPG
70 Height (inches)
3 Find a function’s average
60 50 40 30 20 10 0
0
5
10 Age (years)
15
20
x
Figure 2.43 Height as a function of age
The man’s average growth rate between ages 13 and 18 is the slope of the secant line containing (13, 57) and (18, 76): m =
Change in y 76  57 19 4 = = = 3 . Change in x 18  13 5 5
This man’s average rate of change, or average growth rate, from age 13 to age 18 was 3 45, or 3.8, inches per year.
The Average Rate of Change of a Function Let (x1, f (x1)) and (x2, f (x2)) be distinct points on the graph of a function f. (See Figure 2.44.) The average rate of change of f from x1 to x2 is f (x2)  f (x1) . x2  x1 y
xf x
xf x y = f (x)
5GECPVNKPG x1
x2
x Figure 2.44
306
Chapter 2 Functions and Graphs
EXAMPLE 4
Finding the Average Rate of Change
Find the average rate of change of f (x) = x 2 from a. x1 = 0 to x2 = 1 b. x1 = 1 to x2 = 2
c. x1 = 2 to x2 = 0.
Solution a. The average rate of change of f (x) = x 2 from x1 = 0 to x2 = 1 is f (x2)  f (x1) f (1)  f (0) 12  02 = = = 1. x2  x1 1  0 1 Figure 2.45(a) shows the secant line of f (x) = x 2 from x1 = 0 to x2 = 1. The average rate of change is positive and the function is increasing on the interval (0, 1). b. The average rate of change of f (x) = x 2 from x1 = 1 to x2 = 2 is f (x2)  f (x1) f (2)  f (1) 22  12 = = = 3. x2  x1 2  1 1 Figure 2.45(b) shows the secant line of f (x) = x 2 from x1 = 1 to x2 = 2. The average rate of change is positive and the function is increasing on the interval (1, 2). Can you see that the graph rises more steeply on the interval (1, 2) than on (0, 1)? This is because the average rate of change from x1 = 1 to x2 = 2 is greater than the average rate of change from x1 = 0 to x2 = 1. c. The average rate of change of f (x) = x 2 from x1 = 2 to x2 = 0 is 2
f (x2)  f (x1) f (0)  f ( 2) 02  ( 2) 4 = = = = 2. x2  x1 0  ( 2) 2 2 Figure 2.45(c) shows the secant line of f (x) = x 2 from x1 = 2 to x2 = 0. The average rate of change is negative and the function is decreasing on the interval ( 2, 0).
y 4
4
3
3
2 1 1
4
–
3 5GECPVNKPG
x
2
–2
Figure 2.45(a) The secant line of f (x) = x 2 from x1 = 0 to x2 = 1
2 1
1
–1
5GECPV NKPG
2
5GECPVNKPG
–2
y
y
–1
1
2
Figure 2.45(b) The secant line of f (x) = x 2 from x1 = 1 to x2 = 2
x
–2
–1
1
Figure 2.45(c) The secant line of f (x) = x 2 from x1 =  2 to x2 = 0
Find the average rate of change of f (x) = x 3 from a. x1 = 0 to x2 = 1 b. x1 = 1 to x2 = 2 c. x1 = 2 to x2 = 0. CHECK POINT 4
2
x
Section 2.4 More on Slope
307
Suppose we are interested in the average rate of change of f from x1 = x to x2 = x + h. In this case, the average rate of change is f (x2)  f (x1) f (x + h)  f (x) f (x + h)  f (x) = = . x2  x1 x + h  x h Do you recognize the last expression? It is the difference quotient that you used in Section 2.2. Thus, the difference quotient gives the average rate of change of a function from x to x + h. In the difference quotient, h is thought of as a number very close to 0. In this way, the average rate of change can be found for a very short interval.
EXAMPLE 5 Drug Concentration in the Blood (milligrams per 100 milliliters)
y 0.06
When a person receives a drug injected into a muscle, the concentration of the drug in the body, measured in milligrams per 100 milliliters, is a function of the time elapsed after the injection, measured in hours. Figure 2.46 shows the graph of such a function, where x represents hours after the injection and f (x) is the drug’s concentration at time x. Find the average rate of change in the drug’s concentration between 3 and 7 hours.
0.05 0.04
y=f x
0.03
Solution At 3 hours, the drug’s concentration is 0.05 and at 7 hours, the concentration is 0.02. The average rate of change in its concentration between 3 and 7 hours is
0.02 0.01 0
2
4 6 8 10 12 Time (hours)
Figure 2.46 Concentration of a drug as a function of time
Finding the Average Rate of Change
x
f (x2)  f (x1) f (7)  f (3) 0.02  0.05 0.03 = = = = 0.0075. x2  x1 7  3 7  3 4 The average rate of change is 0.0075. This means that the drug’s concentration is decreasing at an average rate of 0.0075 milligram per 100 milliliters per hour.
GREAT QUESTION Can you clarify how you determine the units that you use when describing slope as a rate of change? Units used to describe x and y tend to “pile up” when expressing the rate of change of y with respect to x. The unit used to express the rate of change of y with respect to x is the unit used to describe y +P(KIWTGyQT FTWIEQPEGPVTCVKQPKU FGUETKDGFKPOKNNKITCOU RGTOKNNKNKVGTU
per
the unit used to describe x. +P(KIWTGxQT VKOGKUFGUETKDGFKPJQWTU
In Example 5, the rate of change is described in terms of milligrams per 100 milliliters per hour.
Use Figure 2.46 to find the average rate of change in the drug’s concentration between 1 hour and 3 hours. CHECK POINT 5
308
Chapter 2 Functions and Graphs
BLITZER BONUS
How Calculus Studies Change
Take a rapid sequence of still photographs of a moving scene and project them onto a screen at thirty shots a second or faster. Our eyes see the results as continuous motion. The small difference between one frame and the next cannot be detected by the human visual system. The idea of calculus likewise regards continuous motion as made up of a sequence of still configurations. Calculus masters the mystery of movement by “freezing the frame” of a continuously changing process, instant by instant. For example, Figure 2.47 shows a male’s changing height over intervals of time. Over the period of time from P to D, his average rate of growth is his change in height—that is, his height at time D minus his height at time P—divided by the change in time from P to D. This is the slope of secant line PD. The secant lines PD, PC, PB, and PA shown in Figure 2.47 have slopes that show average growth rates for successively shorter periods of time. Calculus makes these time frames so small that they approach a single point—that is, a single instant in time. This point is shown as point P in Figure 2.47. The slope of the line that touches the graph at P gives the male’s growth rate at one instant in time, P.
A
B
C
D
P
Figure 2.47 Analyzing continuous growth over intervals of time and at an instant in time
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. If two nonvertical lines are parallel, then they have slope.
C5. The slope of the line through the distinct points (x1, y1) and (x2, y2) can be interpreted as the rate of change in with respect to .
C2. If two nonvertical lines are perpendicular, then the product of their slopes is .
C6. If (x1, f (x1)) and (x2, f (x2)) are distinct points on the graph of a function f, the average rate of change of f from x1 to
C3. Consider the line whose equation is y =  31x + 5. The slope of any line that is parallel to this line is The slope of any line that is perpendicular to this line is .
.
x2 is
.
C4. Consider the line whose equation is 2x + y  6 = 0. The slope of any line that is parallel to this line is . The slope of any line that is perpendicular to this line is .
2.4 EXERCISE SET Practice Exercises In Exercises 1–4, write an equation for line L in pointslope form and slopeintercept form. 1.
y y = 2x
L
2.
y = –2x y
3.
L
4.
y y = 2x
y = –2x y L
4
4
2 –4 –2
4
–2 –4
L is parallel to y = 2x.
x –4 –2
(2, 4)
4
2
(4, 2) 2
(3, 4)
2 2
4
–2 –4
L is parallel to y = –2x.
x –4 –2
L 2
4
x
(–1, 2) x
–2 –4
L is perpendicular to y = 2x.
L is perpendicular to y = –2x.
Section 2.4 More on Slope
The bar graph shows that as costs changed over the decades, Americans devoted less of their budget to groceries and more to health care.
6. Passing through ( 8, 7) and parallel to the line whose equation is y =  4x + 4
12. Passing through (5, 9) and perpendicular to the line whose equation is x + 7y  12 = 0 In Exercises 13–18, find the average rate of change of the function from x1 to x2. 13. f (x) = 3x from x1 = 0 to x2 = 5 14. f (x) = 8x from x1 = 0 to x2 = 3 15. f (x) = x 2 + 2x from x1 = 3 to x2 = 5 16. f (x) = x 2  2x from x1 = 3 to x2 = 6 17. f (x) = 1x from x1 = 4 to x2 = 9 18. f (x) = 1x from x1 = 9 to x2 = 16
Practice PLUS In Exercises 19–24, write an equation in slopeintercept form of a linear function f whose graph satisfies the given conditions.
24. The graph of f is perpendicular to the line whose equation is 4x  y  6 = 0 and has the same y@intercept as this line.
7%
2010
1950
1990
3%
3%
Health Care
In Exercises 25–26, find a linear function in slopeintercept form that models the given description. Each function should model the percentage of total spending, p(x), by Americans x years after 1950. 25. In 1950, Americans spent 22% of their budget on food. This has decreased at an average rate of approximately 0.25% per year since then. 26. In 1950, Americans spent 3% of their budget on health care. This has increased at an average rate of approximately 0.22% per year since then. The stated intent of the 1994 “don’t ask, don’t tell” policy was to reduce the number of LGBTQ discharges from the U.S. military. Nearly 14,000 activeduty LGBTQ servicemembers were dismissed under the policy, which officially ended in 2011, after 18 years. The line graph shows the number of discharges under “don’t ask, don’t tell” from 1994 through 2010. Use the data displayed by the graph to solve Exercises 27–28. Number of ActiveDuty LGBTQ Servicemembers Discharged from the U.S. Military 1300
1200 Number of Discharged ActiveDuty Servicemembers
23. The graph of f is perpendicular to the line whose equation is 3x  2y  4 = 0 and has the same y@intercept as this line.
7%
6%
Source: Time, October 10, 2011
20. The graph of f passes through (  2, 6) and is perpendicular to the line whose equation is x = 4.
22. The graph of f passes through (  5, 6) and is perpendicular to the line that has an x@intercept of 3 and a y@intercept of 9.
10%
9%
Food
19. The graph of f passes through (  1, 5) and is perpendicular to the line whose equation is x = 6.
21. The graph of f passes through (  6, 4) and is perpendicular to the line that has an x@intercept of 2 and a y@intercept of 4.
16% 13%
12%
10. Passing through ( 1, 3) and parallel to the line whose equation is 3x  2y  5 = 0 11. Passing through (4, 7) and perpendicular to the line whose equation is x  2y  3 = 0
16%
15%
1970
9. Passing through ( 2, 2) and parallel to the line whose equation is 2x  3y  7 = 0
18%
1950
In Exercises 9–12, use the given conditions to write an equation for each line in pointslope form and general form.
22%
21%
2010
8. Passing through ( 4, 2) and perpendicular to the line whose equation is y = 13 x + 7
24% Percentage of Total Spending
7. Passing through (2, 3) and perpendicular to the line whose equation is y = 15 x + 6
Percentage of Total Spending in the United States on Food and Health Care
1990
5. Passing through (  8,  10) and parallel to the line whose equation is y = 4x + 3
Application Exercises
1970
In Exercises 5–8, use the given conditions to write an equation for each line in pointslope form and slopeintercept form.
309
1100 1000 900 800 700 600
500 400 300 200 100 ’94 ’95 ’96 ’97 ’98 ’99 ’00 ’01 ’02 ’03 ’04 ’05 ’06 ’07 ’08 ’09 ’10 Year
Source: General Accountability Office
310
Chapter 2 Functions and Graphs
Critical Thinking Exercises
(In Exercises 27–28, be sure to refer to the graph at the bottom of the previous page.) 27. Find the average rate of change, rounded to the nearest whole number, from 1994 through 1998. Describe what this means. 28. Find the average rate of change, rounded to the nearest whole number, from 2001 through 2006. Describe what this means. The function f (x) = 1.1x 3  35x 2 + 264x + 557 models the number of discharges, f (x), under “don’t ask, don’t tell” x years after 1994. Use this model and its graph, shown below, to solve Exercises 29–30.
1200
f
f
Number of Discharges
1100 1000 900 800
f x=x–x+x+
700 600
f
f 0
1
2
3
4
5 6 7 8 9 Years after 1994
10 11 12 13
38. I computed the slope of one line to be  35 and the slope of a second line to be  35, so the lines must be perpendicular. 39. I have linear functions that model changes for men and women over the same time period. The functions have the same slope, so their graphs are parallel lines, indicating that the rate of change for men is the same as the rate of change for women. 40. The graph of my function is not a straight line, so I cannot use slope to analyze its rates of change. 41. According to the Blitzer Bonus on page 308, calculus studies change by analyzing slopes of secant lines over successively shorter intervals. 42. What is the slope of a line that is perpendicular to the line whose equation is Ax + By + C = 0, A ≠ 0 and B ≠ 0? 43. Determine the value of A so that the line whose equation is Ax + y  2 = 0 is perpendicular to the line containing the points (1,  3) and (  2, 4).
Graph of a Model for Discharges under “Don’t Ask, Don’t Tell”
y
Make Sense? In Exercises 38–41, determine whether each statement makes sense or does not make sense, and explain your reasoning.
x
29. a. Find the slope of the secant line, rounded to the nearest whole number, from x1 = 0 to x2 = 4. b. Does the slope from part (a) underestimate or overestimate the average yearly increase that you determined in Exercise 27? By how much? 30. a. Find the slope of the secant line, rounded to the nearest whole number, from x1 = 7 to x2 = 12. b. Does the slope from part (b) underestimate or overestimate the average yearly decrease that you determined in Exercise 28? By how much?
Explaining the Concepts 31. If two lines are parallel, describe the relationship between their slopes. 32. If two lines are perpendicular, describe the relationship between their slopes. 33. If you know a point on a line and you know the equation of a line perpendicular to this line, explain how to write the line’s equation. 34. A formula in the form y = mx + b models the average retail price, y, of a new car x years after 2000. Would you expect m to be positive, negative, or zero? Explain your answer. 35. What is a secant line? 36. What is the average rate of change of a function?
Technology Exercises 37. a. Why are the lines whose equations are y = 13 x + 1 and y =  3x  2 perpendicular? b. Use a graphing utility to graph the equations in a [ 10, 10, 1] by [ 10, 10, 1] viewing rectangle. Do the lines appear to be perpendicular? c. Now use the zoom square feature of your utility. Describe what happens to the graphs. Explain why this is so.
Retaining the Concepts 44. Solve and check: 24 + 3 (x + 2) = 5(x  12). (Section 1.2, Example 2) 45. After a 30% price reduction, you purchase a 50 ˝ 4K UHD TV for $245. What was the television’s price before the reduction? (Section 1.3, Example 4) 2 1 46. Solve: 2x 3  5x 3  3 = 0. (Section 1.6, Example 7)
Preview Exercises Exercises 47–49 will help you prepare for the material covered in the next section. In each exercise, graph the functions in parts (a) and (b) in the same rectangular coordinate system. 47. a. Graph f (x) = x using the ordered pairs ( 3, f ( 3)), (  2, f (  2)), ( 1, f ( 1)), (0, f (0)), (1, f (1)), (2, f (2)), and (3, f (3)). b. Subtract 4 from each y@coordinate of the ordered pairs in part (a). Then graph the ordered pairs and connect them with two linear pieces. c. Describe the relationship between the graph in part (b) and the graph in part (a). 48. a. Graph f (x) = x 2 using the ordered pairs ( 3, f ( 3)), (  2, f (  2)), ( 1, f (  1)), (0, f (0)), (1, f (1)), (2, f (2)), and (3, f (3)). b. Add 2 to each x@coordinate of the ordered pairs in part (a). Then graph the ordered pairs and connect them with a smooth curve. c. Describe the relationship between the graph in part (b) and the graph in part (a). 49. a. Graph f (x) = x 3 using the ordered pairs ( 2, f ( 2)), (  1, f (  1)), (0, f (0)), (1, f (1)), and (2, f (2)). b. Replace each x@coordinate of the ordered pairs in part (a) with its opposite, or additive inverse. Then graph the ordered pairs and connect them with a smooth curve. c. Describe the relationship between the graph in part (b) and the graph in part (a).
Section 2.4 More on Slope
311
MidChapter Check Point
CHAPTER 2
WHAT YOU KNOW: We learned that a function is a relation in which no two ordered pairs have the same first component and different second components. We represented functions as equations and used function notation. We graphed functions and applied the vertical line test to identify graphs of functions. We determined the domain and range of a function from its graph, using inputs on the x@axis for the domain and outputs on the y@axis for the range. We used graphs to identify intervals on which functions increase, decrease, or are constant, as well as to locate relative maxima or minima. We determined when graphs of equations are symmetric with respect to the yaxis (no change when x is substituted for x), the xaxis (no change when y is substituted for y), and the origin (no change when x is substituted for x and y is substituted for y). We identified even functions [f ( x) = f (x): y@axis symmetry] and odd functions [f ( x) = f (x): origin symmetry]. Finally, we studied linear functions and slope, using slope (change in y divided by change in x) to develop various forms for equations of lines: 2QKPVUNQRGHQTO
5NQRGKPVGTEGRVHQTO
y − y1 = m(x − x1)
*QTK\QPVCNNKPG
y = f (x) = b
y = f(x) = mx + b
8GTVKECNNKPG
)GPGTCNHQTO
Ax + By + C = 0.
x=a
We saw that parallel lines have the same slope and that perpendicular lines have slopes that are negative reciprocals. For linear functions, slope was interpreted as the rate of change of the dependent variable per unit change in the independent variable. For nonlinear functions, the slope of the secant line between (x1, f (x1)) and (x2, f (x2)) described the average rate f (x2)  f (x1) of change of f from x1 to x2 : . x2  x1 In Exercises 1–6, determine whether each relation is a function. Give the domain and range for each relation.
Use the graph of f to solve Exercises 9–24. Where applicable, use interval notation.
1. {(2, 6), (1, 4), (2, 6)} y
2. {(0, 1), (2, 1), (3, 4)} y=f x
3.
4.
y
y
2
4
4
2 –4 –2
2 2
4
–4 –2
–4
2
4
x
2
x
–2 –4
–2 –4
6.
y
–4 –2
–6 –4 –2
x
–2
5.
4
y
4
4
2
2 2
–2 –4
4
x –4 –2
2
4
x
–2 –4
In Exercises 7–8, determine whether each equation defines y as a function of x. 7. x 2 + y = 5 8. x + y2 = 5
9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24.
Explain why f represents the graph of a function. Find the domain of f. Find the range of f. Find the x@intercept(s). Find the y@intercept. Find the interval(s) on which f is increasing. Find the interval(s) on which f is decreasing. At what number does f have a relative maximum? What is the relative maximum of f ? Find f (  4). For what value or values of x is f (x) = 2? For what value or values of x is f (x) = 0? For what values of x is f (x) 7 0? Is f (100) positive or negative? Is f even, odd, or neither? Find the average rate of change of f from x1 = 4 to x2 = 4.
Chapter 2 Functions and Graphs
In Exercises 25–26, determine whether the graph of each equation is symmetric with respect to the yaxis, the xaxis, the origin, more than one of these, or none of these. 25. x = y2 + 1
26. y = x 3  1
46. Exercise is useful not only in preventing depression, but also as a treatment. The following graphs show the percentage of patients with depression in remission when exercise (brisk walking) was used as a treatment. (The control group that engaged in no exercise had 11% of the patients in remission.)
In Exercises 27–38, graph each equation in a rectangular coordinate system. 29. x + y =  2 32. 4x  2y = 8 35. f (x) = x  4
a. Find f ( x). Is f even, odd, or neither? f (x + h)  f (x) , h ≠ 0. b. Find h 30 if 40. Let C(x) = b 30 + 0.40(x  200) if a. Find C(150).
0 … x … 200 . x 7 200
b. Find C(250).
In Exercises 41–44, write a function in slopeintercept form whose graph satisfies the given conditions. 41. Slope = 2, passing through ( 4, 3) 42. Passing through (  1, 5) and (2, 1) 43. Passing through (3,  4) and parallel to the line whose equation is 3x  y  5 = 0 44. Passing through (  4, 3) and perpendicular to the line whose equation is 2x  5y  10 = 0 45. Determine whether the line through (2, 4) and (7, 0) is parallel to a second line through (  4, 2) and (1, 6).
SECTION 2.5
Exercise and Percentage of Patients with Depression in Remission 60% Percentage with Depression in Remission
y =  2x 28. y =  2 y = 13 x  2 31. x = 3.5 f (x) = x 2  4 34. f (x) = x  4 5y =  3x 37. 5y = 20 1 if x … 0 38. f (x) = b 2x + 1 if x 7 0 39. Let f (x) =  2x 2 + x  5. 27. 30. 33. 36.
60%
50%
42
40% 30%
26
20% 10% 80 180 Amount of Brisk Walking (minutes)
1 Recognize graphs of common functions.
2 Use vertical shifts to graph functions.
3 Use horizontal shifts to graph functions.
4 Use reflections to graph functions.
5 Use vertical stretching and
shrinking to graph functions.
6 Use horizontal stretching and shrinking to graph functions.
7 Graph functions involving a sequence of transformations.
50% 40% 30% 20%
10% 60 100 140 180 Amount of Brisk Walking (minutes)
Source: Newsweek, March 26, 2007
a. Find the slope of the line passing through the two points shown by the voice balloons. Express the slope as a decimal. b. Use your answer from part (a) to complete this statement: For each minute of brisk walking, the percentage of patients with depression in remission increased by %. The rate of change is % per . 47. Find the average rate of change of f (x) = 3x 2  x from x1 =  1 to x2 = 2.
Transformations of Functions
WHAT YOU’LL LEARN
Percentage with Depression in Remission
312
The films Terminator 2, The Mask, and The Matrix were among the first films to use spectacular effects in which a character or object having one shape was transformed in a fluid fashion into a quite different shape. The name for such a transformation is morphing. The effect allows a real actor to be seamlessly transformed into a computergenerated animation. The animation can be made to perform impossible feats before it is morphed back to the conventionally filmed image. Like transformed movie images, the graph of one function can be turned into the graph of a different function. To do this, we need to rely on a function’s equation. Knowing that a graph is a transformation of a familiar graph makes graphing easier.
313
Section 2.5 Transformations of Functions
Graphs of Common Functions
1 Recognize graphs of common functions.
Table 2.4 gives names to seven frequently encountered functions in algebra. The table shows each function’s graph and lists characteristics of the function. Study the shape of each graph and take a few minutes to verify the function’s characteristics from its graph. Knowing these graphs is essential for analyzing their transformations into more complicated graphs.
Table 2.4 Algebra’s Common Graphs Constant Function
Identity Function
Absolute Value Function
y
y
y
f x=x
f x=c
2 1 –2
–1
1
2
2
2
1
1
x –2
1
–1
2
x –2
1
–1
–1
–1
–1
–2
–2
–2
• Domain: (–∞, ∞) • Range: the single number c • Constant on (–∞, ∞)
• Domain: (–∞, ∞) • Range: (–∞, ∞) • Increasing on (–∞, ∞)
• Even function
• Odd function
f x=x
x
2
• Domain: (–∞, ∞) • Range: [0, ∞) • Decreasing on (–∞, 0) and increasing on (0, ∞) • Even function
Standard Quadratic Function
Square Root Function
Standard Cubic Function
Cube Root Function
y
y
y
y
2
2 f x=x
1 –2
1
–1
2
f x=√x
2
1
x –2
1 1
–1
2
f x=x
2
x –2
1 1
–1
f x=√x
2
x –2
1
–1
2
–1
–1
–1
–1
–2
–2
–2
–2
• Domain: [0, ∞) • Range: [0, ∞) • Increasing on (0, ∞)
• Domain: (–∞, ∞) • Range: (–∞, ∞) • Increasing on (–∞, ∞)
• Domain: (–∞, ∞) • Range: (–∞, ∞) • Increasing on (–∞, ∞)
• Neither even nor odd
• Odd function
• Odd function
• Domain: (–∞, ∞) • Range: [0, ∞) • Decreasing on (–∞, 0) and increasing on (0, ∞) • Even function
DISCOVERY The study of how changing a function’s equation can affect its graph can be explored with a graphing utility. Use your graphing utility to verify the handdrawn graphs as you read this section.
x
314
Chapter 2 Functions and Graphs
2 Use vertical shifts to graph functions.
Let’s begin by looking at three graphs whose shapes are the same. Figure 2.48 shows the graphs. The black graph in the middle is the standard quadratic function, f (x) = x 2. Now, look at the blue graph on the top. The equation of this graph, g(x) = x 2 + 2, adds 2 to the right side of f (x) = x 2. The y@coordinate of each point of g is 2 more than the corresponding y@coordinate of each point of f. What effect does this have on the graph of f ? It shifts the graph vertically up by 2 units.
y 5 4 3 2 1 –5 –4 –3 –2 –1–1
Vertical Shifts
6JGITCRJQHg g x=x+ f x=x h x=x– x 1 2 3 4 5
–2 –3 –4 –5
UJKHVUVJGITCRJQHfWRWPKVU
g(x) = x2 + 2 = f (x) + 2
Finally, look at the red graph on the bottom in Figure 2.48. The equation of this graph, h(x) = x 2  3, subtracts 3 from the right side of f (x) = x 2. The y@coordinate of each point of h is 3 less than the corresponding y@coordinate of each point of f. What effect does this have on the graph of f ? It shifts the graph vertically down by 3 units. 6JGITCRJQHh
UJKHVUVJGITCRJQHfFQYPWPKVU
h(x) = x2 − 3 = f (x) − 3
In general, if c is positive, y = f (x) + c shifts the graph of f upward c units and y = f (x)  c shifts the graph of f downward c units. These are called vertical shifts of the graph of f.
Figure 2.48 Vertical shifts
Vertical Shifts Let f be a function and c a positive real number. • The graph of y = f (x) + c is the graph of y = f (x) shifted c units vertically upward. GREAT QUESTION If I’m using the graph of a familiar function, how do I actually obtain the graph of a transformation?
• The graph of y = f (x)  c is the graph of y = f (x) shifted c units vertically downward. y
To keep track of transformations and obtain their graphs, identify a number of points on the given function’s graph. Then analyze what happens to the coordinates of these points with each transformation.
y
y=f x+c
y=f x
c c y=f x
x
x
y=f x–c
EXAMPLE 1
Vertical Shift Downward
Use the graph of f (x) = x to obtain the graph of g(x) = x  4. Solution The graph of g(x) = x  4 has the same shape as the graph of f (x) = x . However, it is shifted down vertically 4 units. $GIKPYKVJVJGITCRJQH f x=x9GoXGKFGPVKƂGF VJTGGRQKPVUQPVJGITCRJ y (–4, 4) f x=x
5 4 3 2 1
–5 –4 –3 –2 –1–1
(4, 4)
1 2 3 4 5
–2 (0, 0) –3 –4 –5
6JGITCRJQH g x=x– )TCRJg x=x– 5JKHVfWPKVUFQYP 5WDVTCEVHTQOGCEJ yEQQTFKPCVG
x
y
(–4, 0)
5 4 3 2 1
–5 –4 –3 –2 –1–1
(4, 0) 1 2 3 4 5
x
–2 g x=x– –3 –4 (0, –4) –5
Section 2.5 Transformations of Functions
315
Use the graph of f (x) = x to obtain the graph of
CHECK POINT 1
g(x) = x + 3.
Horizontal Shifts
3 Use horizontal shifts to graph functions.
h x= x+
g x= x– y
8 7 6 5 4 3 2 1 –5 –4 –3 –2 –1–1 –2
We return to the graph of f (x) = x 2, the standard quadratic function. In Figure 2.49, the graph of function f is in the middle of the three graphs. By contrast to the vertical shift situation, this time there are graphs to the left and to the right of the graph of f. Look at the blue graph on the right. The equation of this graph, g(x) = (x  3)2, subtracts 3 from each value of x before squaring it. What effect does this have on the graph of f (x) = x 2? It shifts the graph horizontally to the right by 3 units. g(x) = (x − 3)2 = f (x − 3) 6JGITCRJQHg
2 3 4 5
x
f x=x
Figure 2.49 Horizontal shifts
UJKHVUVJGITCRJQHfWPKVUVQVJGTKIJV
Does it seem strange that subtracting 3 in the domain causes a shift of 3 units to the right? Perhaps a partial table of coordinates for each function will numerically convince you of this shift. x
f (x) = x2
x
g(x) = (x − 3)2
2
(  2)2 = 4
1
(1  3)2 = (  2)2 = 4
1
(  1)2 = 1
2
(2  3)2 = (  1)2 = 1
0
02 = 0
3
(3  3)2 =
02 = 0
1
12 = 1
4
(4  3)2 =
12 = 1
2
22 = 4
5
(5  3)2 =
22 = 4
Notice that for the values of f (x) and g(x) to be the same, the values of x used in graphing g must each be 3 units greater than those used to graph f. For this reason, the graph of g is the graph of f shifted 3 units to the right. Now, look at the red graph on the left in Figure 2.49. The equation of this graph, 2 h(x) = (x + 2) , adds 2 to each value of x before squaring it. What effect does this have on the graph of f (x) = x 2? It shifts the graph horizontally to the left by 2 units. h(x) = (x + 2)2 = f (x + 2) 6JGITCRJQHh
GREAT QUESTION When I use my intuition, it seems that f (x + c) should cause a shift to the right and f (x − c) should cause a shift to the left. Is my intuition on target when it comes to these horizontal shifts? No. On a number line, if x represents a number and c is positive, then x + c lies c units to the right of x and x  c lies c units to the left of x. This orientation does not apply to horizontal shifts: f (x + c) causes a shift of c units to the left and f (x  c) causes a shift of c units to the right.
UJKHVUVJGITCRJQHfWPKVUVQVJGNGHV
In general, if c is positive, y = f (x + c) shifts the graph of f to the left c units and y = f (x  c) shifts the graph of f to the right c units. These are called horizontal shifts of the graph of f. Horizontal Shifts Let f be a function and c a positive real number. • The graph of y = f (x + c) is the graph of y = f (x) shifted to the left c units. • The graph of y = f (x  c) is the graph of y = f (x) shifted to the right c units. y
y
y=f x+c
y=f x
c
c x
y=f x
x y=f x–c
316
Chapter 2 Functions and Graphs
EXAMPLE 2
Horizontal Shift to the Left
Use the graph of f (x) = 1x to obtain the graph of g(x) = 2x + 5. Solution Compare the equations for f (x) = 1x and g(x) = 2x + 5. The equation for g adds 5 to each value of x before taking the square root. y = g(x) = "x + 5 = f (x + 5) 6JGITCRJQHg
UJKHVUVJGITCRJQHfWPKVUVQVJGNGHV
The graph of g(x) = 2x + 5 has the same shape as the graph of f (x) = 1x. However, it is shifted horizontally to the left 5 units. 6JGITCRJQH g x=√x+
6JGITCRJQHf x=√x YKVJVJTGGRQKPVUKFGPVKƂGF y 5 f x=√x 4 3 2 (1, 1) (4, 2) 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
)TCRJg x=√x+ 5JKHVfWPKVU NGHV5WDVTCEVHTQO GCEJxEQQTFKPCVG
x
y 5 4 3 (–1, 2) 2 (–4, 1) 1 –5 –4 –3 –2 –1–1
–2 (0, 0) –3 –4 –5
(–5, 0)
g x=√x+
1 2 3 4 5
x
–2 –3 –4 –5
GREAT QUESTION What’s the difference between f (x) + c and f (x + c)? • y = f (x) + c shifts the graph of y = f (x) c units vertically upward. • y = f (x + c) shifts the graph of y = f (x) c units horizontally to the left. There are analogous differences between f (x)  c and f (x  c).
CHECK POINT 2
Use the graph of f (x) = 1x to obtain the graph of
g(x) = 2x  4. Some functions can be graphed by combining horizontal and vertical shifts. These functions will be variations of a function whose equation you know how to graph, such as the standard quadratic function, the standard cubic function, the square root function, the cube root function, or the absolute value function. In our next example, we will use the graph of the standard quadratic function, 2 f (x) = x 2, to obtain the graph of h(x) = (x + 1)  3. We will graph three functions: f (x) = x2 5VCTVD[ITCRJKPI VJGUVCPFCTF SWCFTCVKEHWPEVKQP
g(x) = (x + 1)2 5JKHVVJGITCRJ QHfJQTK\QPVCNN[ QPGWPKVVQVJGNGHV
h(x) = (x + 1)2 − 3. 5JKHVVJGITCRJ QHgXGTVKECNN[ FQYPWPKVU
Section 2.5 Transformations of Functions
EXAMPLE 3
317
Combining Horizontal and Vertical Shifts 2
Use the graph of f (x) = x 2 to obtain the graph of h(x) = (x + 1)  3. Solution 6JGITCRJQHf x=x YKVJVJTGGRQKPVUKFGPVKƂGF
6JGITCRJQHg x= x+
y
(–2, 4) f x=x
5 4 3 2 1
(2, 4)
1 2 3 4 5 (0, 0)
–5 –4 –3 –2 –1–1 –2 –3 –4 –5
)TCRJg x= x+ 5JKHVfJQTK\QPVCNN[ WPKVNGHV5WDVTCEV (–3, 4) HTQOGCEJxEQQTFKPCVG
6JGITCRJQHh x= x+– )TCRJh x= x+– 5JKHVgXGTVKECNN[FQYP WPKVU5WDVTCEVHTQO GCEJyEQQTFKPCVG
y 5 4 3 2 1
x –5 –4 –3 –2 –1–1 (–1, 0) –2 –3 –4 –5
(1, 4)
1 2 3 4 5
x
g x= x+
y
(–3, 1)
5 4 3 2 1
–5 –4 –3 –2 –1–1
(1, 1) 1 2 3 4 5
x
–2 –3 h x= x+– (–1, –3) –4 –5
DISCOVERY Work Example 3 by first shifting the graph of f (x) = x 2 three units down, graphing 2 g(x) = x 2  3. Now, shift this graph one unit left to graph h(x) = (x + 1)  3. Did you obtain the last graph shown in the solution of Example 3? What can you conclude?
CHECK POINT 3
Use the graph of f (x) = 1x to obtain the graph of
h(x) = 2x  1  2.
Reflections of Graphs
4 Use reflections to graph functions.
y 5 4 (–2, 4) 3 2 (–1, 1) 1 –5 –4 –3 –2 (–1, –1) (–2, –4)
–2 –3 –4 –5
f x=x (2, 4) (1, 1) 2 3 4 5 (1, –1)
x
(2, –4) g x=–x
Figure 2.50 Reflection about the x@axis
This photograph shows a reflection of an old bridge in a river. This perfect reflection occurs because the surface of the water is absolutely still. A mild breeze rippling the water’s surface would distort the reflection. Is it possible for graphs to have mirrorlike qualities? Yes. Figure 2.50 shows the graphs of f (x) = x 2 and g(x) = x 2. The graph of g is a reflection about the x@axis of the graph of f. For corresponding values of x, the y@coordinates of g are the opposites of the y@coordinates of f. In general, the graph of y = f (x) reflects the graph of f about the x@axis. Thus, the graph of g is a reflection of the graph of f about the x@axis because g(x) = x 2 = f (x). Reflection about the x@Axis The graph of y = f (x) is the graph of y = f (x) reflected about the x@axis.
318
Chapter 2 Functions and Graphs
EXAMPLE 4
Reflection about the x@Axis
3 3 Use the graph of f (x) = 1 x to obtain the graph of g(x) =  1 x. 3 3 Solution Compare the equations for f (x) = 1 x and g(x) =  1 x. The graph of g is a reflection about the x@axis of the graph of f because 3 g(x) =  1 x = f (x).
6JGITCRJQHf x=√x YKVJVJTGGRQKPVUKFGPVKƂGF
)TCRJg x=–√x 4GƃGEVfCDQWVVJG xCZKU4GRNCEGGCEJ yEQQTFKPCVGYKVJKVU QRRQUKVG
y 5 4 3 2 1 –8 –7 –6 –5 –4 –3 –2 –1–1 (–8, –2)
6JGITCRJQHg x=–√x
f x=√x (8, 2)
y 5 4 3 2 1
(–8, 2) x
1 2 3 4 5 6 7 8
–8 –7 –6 –5 –4 –3 –2 –1–1
–2 (0, 0) –3 –4 –5
–2 –3 –4 –5
CHECK POINT 4
(0, 0) x
1 2 3 4 5 6 7 8
(8, –2)
g x=–√x
Use the graph of f (x) = x to obtain the graph of g(x) =  x.
It is also possible to reflect graphs about the y@axis. Reflection about the y@Axis The graph of y = f ( x) is the graph of y = f (x) reflected about the y@axis. For each point (x, y) on the graph of y = f (x), the point (x, y) is on the graph of y = f (x).
EXAMPLE 5
Reflection about the y@Axis
Use the graph of f (x) = 1x to obtain the graph of h(x) = 1x. Solution Compare the equations for f (x) = 1x and h(x) = 1x. The graph of h is a reflection about the y@axis of the graph of f because h(x) = 1x = f ( x). 6JGITCRJQH h x=√–x
6JGITCRJQHf x=√x YKVJVJTGGRQKPVUKFGPVKƂGF y 5 f x=√x 4 3 2 (1, 1) (4, 2) 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
)TCRJh x=√–x 4GƃGEVfCDQWVVJG yCZKU4GRNCEGGCEJ xEQQTFKPCVGYKVJ KVUQRRQUKVG
x
5 4 (–1, 1) 3 2 (–4, 2) 1 –5 –4 –3 –2 –1–1
–2 (0, 0) –3 –4 –5
CHECK POINT 5
y
–2 –3 –4 –5
3
h x=√–x x
1 2 3 4 5 (0, 0)
3
Use the graph of f (x) = 1x to obtain the graph of h(x) = 1 x.
319
Section 2.5 Transformations of Functions
5 Use vertical stretching and
shrinking to graph functions. g x=x y 7 6 5 4 3 2 1
–5 –4 –3 –2 –1–1
f x=x h x= x 1 2 3 4 5
x
–2 –3 Figure 2.51 Vertically stretching and shrinking f (x) = x 2
Vertical Stretching and Shrinking Morphing does much more than move an image horizontally, vertically, or about an axis. An object having one shape is transformed into a different shape. Horizontal shifts, vertical shifts, and reflections do not change the basic shape of a graph. Graphs remain rigid and proportionally the same when they undergo these transformations. How can we shrink and stretch graphs, thereby altering their basic shapes? Look at the three graphs in Figure 2.51. The black graph in the middle is the graph of the standard quadratic function, f (x) = x 2. Now, look at the blue graph on the top. The equation of this graph is g(x) = 2x 2, or g(x) = 2f (x). Thus, for each x, the y@coordinate of g is two times as large as the corresponding y@coordinate on the graph of f. The result is a narrower graph because the values of y are rising faster. We say that the graph of g is obtained by vertically stretching the graph of f. Now, look at the red graph on the bottom. The equation of this graph is h(x) = 12 x 2, or h(x) = 12 f (x). Thus, for each x, the y@coordinate of h is onehalf as large as the corresponding y@coordinate on the graph of f. The result is a wider graph because the values of y are rising more slowly. We say that the graph of h is obtained by vertically shrinking the graph of f. These observations can be summarized as follows: Vertically Stretching and Shrinking Graphs Let f be a function and c a positive real number.
GREAT QUESTION Does vertical stretching or shrinking change a graph’s shape?
• If c 7 1, the graph of y = cf (x) is the graph of y = f (x) vertically stretched by multiplying each of its y@coordinates by c. • If 0 6 c 6 1, the graph of y = cf (x) is the graph of y = f (x) vertically shrunk by multiplying each of its y@coordinates by c.
Yes. A vertical stretch moves a function’s graph away from the x@axis. A vertical shrink compresses a function’s graph toward the x@axis. The other transformations we have discussed (vertical shifts, horizontal shifts, and reflections) change only the position of a function’s graph without changing the shape of the basic graph.
Stretching : c > 1
Shrinking : 0 < c < 1
y
y
y=cf x
y=f x
x
0 y=f x
EXAMPLE 6
x
0 y=cf x
Vertically Shrinking a Graph
Use the graph of f (x) = x 3 to obtain the graph of h(x) = 12 x 3. Solution The graph of h(x) = 12 x 3 is obtained by vertically shrinking the graph of f (x) = x 3. 6JGITCRJQHf x=x YKVJVJTGGRQKPVUKFGPVKƂGF y 10 8 6 4 2 –5 –4 –3 –2 –1–2
f x=x
(–2, –8)
(2, 8)
1 2 3 4 5
–4 (0, 0) –6 –8 –10
6JGITCRJQH h x=x )TCRJh x=x 8GTVKECNN[UJTKPMVJG ITCRJQHf/WNVKRN[ GCEJyEQQTFKPCVGD[
y 10 8 6 4 2
x –5 –4 –3 –2 –1–2 (–2, –4)
h x=x
(2, 4) 1 2 3 4 5
–4 (0, 0) –6 –8 –10
x
320
Chapter 2 Functions and Graphs CHECK POINT 6
Use the graph of f (x) = x to obtain the graph of
g(x) = 2x .
Horizontal Stretching and Shrinking
6 Use horizontal stretching and shrinking to graph functions.
It is also possible to stretch and shrink graphs horizontally.
Horizontally Stretching and Shrinking Graphs
GREAT QUESTION How does horizontal shrinking or stretching change a graph’s shape? A horizontal shrink compresses a function’s graph toward the y@axis. A horizontal stretch moves a function’s graph away from the y@axis.
Let f be a function and c a positive real number. • If c 7 1, the graph of y = f (cx) is the graph of y = f (x) horizontally shrunk by dividing each of its x@coordinates by c. • If 0 6 c 6 1, the graph of y = f (cx) is the graph of y = f (x) horizontally stretched by dividing each of its x@coordinates by c. Shrinking : c > 1
Stretching : 0 < c < 1
y
y y=f x
x
0 y=f cx
y (–2, 4)
(–4, 0)
EXAMPLE 7
5 4 y=f x 3 2 1 (0, 0) (4, 0)
–5 –4 –3 –2 –1–1 –2 –3 –4 –5
1 2 3 4 5 (2, –2)
y=f cx
x
0 y=f x
Horizontally Stretching and Shrinking a Graph
Use the graph of y = f (x) in Figure 2.52 to obtain each of the following graphs: a. g(x) = f (2x) b. h(x) = f 1 12 x 2 . Solution x
a. The graph of g(x) = f (2x) is obtained by horizontally shrinking the graph of y = f (x). 6JGITCRJQH g x=f x
6JGITCRJQHy=f x YKVJƂXGRQKPVUKFGPVKƂGF
Figure 2.52 y
(–4, 0)
(–2, 4) 5 4 y=f x 3 2 1 (0, 0) (4, 0)
–5 –4 –3 –2 –1–1 –2 –3 –4 –5
1 2 3 4 5 (2, –2)
)TCRJg x=f x *QTK\QPVCNN[UJTKPM VJGITCRJQHy=f x &KXKFGGCEJxEQQTFKPCVG D[
x
y (–1, 4)
(–2, 0)
5 4 g x=f x 3 2 (0, 0) 1 (2, 0)
–5 –4 –3 –2 –1–1 –2 –3 –4 –5
1 2 3 4 5 (1, –2)
x
321
Section 2.5 Transformations of Functions
b. The graph of h(x) = f 1 12 x 2 is obtained by horizontally stretching the graph of y = f (x). 6JGITCRJQHy=f x YKVJƂXGRQKPVUKFGPVKƂGF y y
(–2, 0)
5 4 (0, 3) 3 y=f x 2 1 (2, 0) (6, 0)
–3 –2 –1–1
1 2 3 4 5 6 7
–2 –3 –4 –5
(–4, 0)
(–2, 4) 5 4 y=f x 3 2 1 (0, 0) (4, 0)
–5 –4 –3 –2 –1–1 –2 –3 –4 –5
x
1 2 3 4 5
)TCRJh x=f x *QTK\QPVCNN[UVTGVEJ VJGITCRJQHy=f x &KXKFGGCEJxEQQTFKPCVG D[YJKEJKUVJGUCOG CUOWNVKRN[KPID[
6JGITCRJQHh x=f x
y (–4, 4)
(–8, 0)
5 4 3 2 (0, 0) 1
x –8 –7 –6 –5 –4 –3 –2 –1–1 –2 –3 –4 –5
(2, –2)
(8, 0)
1 2 3 4 5 6 7 8 (4, –2)
x
h x=f x
(4, –3)
CHECK POINT 7 Figure 2.53
Use the graph of y = f (x) in Figure 2.53 to obtain each of
the following graphs: a. g(x) = f (2x)
7 Graph functions involving a
sequence of transformations.
b. h(x) = f 1 12 x 2 .
Sequences of Transformations Table 2.5 summarizes the procedures for transforming the graph of y = f (x).
Table 2.5 Summary of Transformations In each case, c represents a positive real number. Draw the Graph of f and:
Changes in the Equation of y = f (x)
y = f (x) + c
Raise the graph of f by c units.
c is added to f (x).
y = f (x)  c
Lower the graph of f by c units.
c is subtracted from f (x).
y = f (x + c)
Shift the graph of f to the left c units.
x is replaced with x + c.
y = f (x  c)
Shift the graph of f to the right c units.
x is replaced with x  c.
Reflection about the x@axis
Reflect the graph of f about the x@axis.
f (x) is multiplied by  1.
Reflect the graph of f about the y@axis.
x is replaced with x.
y = cf (x), c 7 1
Multiply each y@coordinate of y = f (x) by c, vertically stretching the graph of f.
f (x) is multiplied by c, c 7 1.
y = cf (x), 0 6 c 6 1
Multiply each y@coordinate of y = f (x) by c, vertically shrinking the graph of f.
f (x) is multiplied by c, 0 6 c 6 1.
y = f (cx), c 7 1
Divide each x@coordinate of y = f (x) by c, horizontally shrinking the graph of f.
x is replaced with cx, c 7 1.
y = f (cx), 0 6 c 6 1
Divide each x@coordinate of y = f (x) by c, horizontally stretching the graph of f.
x is replaced with cx, 0 6 c 6 1.
To Graph: Vertical shifts
Horizontal shifts
y =  f (x) Reflection about the y@axis y = f (  x) Vertical stretching or shrinking
Horizontal stretching or shrinking
322
Chapter 2 Functions and Graphs
Order of Transformations A function involving more than one transformation can be graphed by performing transformations in the following order: 1. Horizontal shifting 2. Stretching or shrinking 3. Reflecting 4. Vertical shifting
EXAMPLE 8
Graphing Using a Sequence of Transformations
Use the graph of y = f (x) given in Figure 2.52 of Example 7, and repeated below, to graph y =  12 f (x  1) + 3. Solution Our graphs will evolve in the following order: 1. Horizontal shifting: Graph y = f (x  1) by shifting the graph of y = f (x) 1 unit to the right. 2. Shrinking: Graph y = 12 f (x  1) by shrinking the previous graph by a factor of 12. 3. Reflecting: Graph y =  12 f (x  1) by reflecting the previous graph about the x@axis. 4. Vertical shifting: Graph y =  12 f (x  1) + 3 by shifting the previous graph up 3 units. 6JGITCRJQHy=f x YKVJƂXGRQKPVUKFGPVKƂGF )TCRJy=f x– 5JKHVWPKVVQVJG TKIJV#FFVQGCEJ xEQQTFKPCVG
y
(–4, 0)
(–2, 4) 5 4 y=f x 3 2 1 (0, 0) (4, 0)
–5 –4 –3 –2 –1–1 –2 –3 –4 –5
6JGITCRJQHy=f x–
6JGITCRJQHy=f x–
1 2 3 4 5
(–1, 4)
(–3, 0)
x
)TCRJy=f x– 5JTKPMXGTVKECNN[D[C HCEVQTQH/WNVKRN[ GCEJyEQQTFKPCVGD[
y 5 4 3 2 1
–5 –4 –3 –2 –1–1 –2 –3 –4 –5
(2, –2)
y 5 4 (–1, 2) 3 2 1 (–3, 0)
y=f x– (1, 0)
(5, 0)
1 2 3 4 5
x
–5 –4 –3 –2 –1–1 –2 –3 –4 –5
(3, –2)
)TCRJy=–f x–4GƃGEV CDQWVVJGxCZKU4GRNCEGGCEJyEQQTFKPCVG YKVJKVUQRRQUKVG
)TCRJ y=–f x–+ 5JKHVWRWPKVU#FF VQGCEJyEQQTFKPCVG
y
(–3, 0)
5 4 3 2 1
–5 –4 –3 –2 –1–1
y=–f x–
(–1, –2)
–2 –3 –4 –5
y
(–3, 3)
(3, 1) (5, 0) 1 2 3 4 5 (1, 0)
6JGITCRJQHy=–f x–
x
(–1, 1)
5 4 3 2 1
–5 –4 –3 –2 –1–1
(3, 4) (5, 3) (1, 3) y=–f x–+ x 1 2 3 4 5
–2 –3 –4 –5 6JGITCRJQHy=–f x–+
y=f x–
(1, 0)
(5, 0)
1 2 3 4 5 (3, –1)
x
Section 2.5 Transformations of Functions
323
CHECK POINT 8 Use the graph of y = f (x) given in Figure 2.53 of Check Point 7 to graph y =  13 f (x + 1)  2.
EXAMPLE 9
Graphing Using a Sequence of Transformations 2
Use the graph of f (x) = x 2 to graph g(x) = 2(x + 3)  1. Solution Our graphs will evolve in the following order: 2 1. Horizontal shifting: Graph y = (x + 3) by shifting the graph of f (x) = x 2 three units to the left. 2 2. Stretching: Graph y = 2(x + 3) by stretching the previous graph by a factor of 2. 2 3. Vertical shifting: Graph g(x) = 2(x + 3)  1 by shifting the previous graph down 1 unit. 6JGITCRJQHf x=x YKVJVJTGGRQKPVUKFGPVKƂGF
6JGITCRJQHy= x+
y (–2, 4) f x=x
5 4 3 2 1
(2, 4)
1 2 3 4 5
–5 –4 –3 –2 –1–1 –2 –3 –4 –5
)TCRJy= x+ 5JKHVWPKVUVQVJGNGHV 5WDVTCEVHTQOGCEJ xEQQTFKPCVG
y (–5, 4) (–1, 4)
5 4 3 2 1
x –5 –4 –3 –2 –1–1 (–3, 0)
(0, 0)
y= x+ 1 2 3 4 5
x
–2 –3 –4 –5
)TCRJy= x+5VTGVEJXGTVKECNN[ D[CHCEVQTQH/WNVKRN[GCEJyEQQTFKPCVG D[
)TCRJg x= x+− 5JKHVFQYPWPKV 5WDVTCEVHTQOGCEJ yEQQTFKPCVG
y
(–5, 8)
(–1, 8)
y= x+
9 8 7 6 5 4 3 2 1
–7 –6 –5 –4 –3 –2 –1–1 (–3, 0)
y
(–5, 7)
y= x+−
1 2 3 4 5
–2 –3 –4 –5
x –7 –6 –5 –4 –3 –2 –1–1 (–3, –1) –2
1 2 3 4 5
x
–3 –4 –5
6JGITCRJQHy= x+
CHECK POINT 9
9 8 7 (–1, 7) 6 5 4 3 2 1
6JGITCRJQHg x= x+− 2
Use the graph of f (x) = x 2 to graph g(x) = 2(x  1) + 3.
324
Chapter 2 Functions and Graphs
ACHIEVING SUCCESS When using your professor’s office hours, show up prepared. If you are having difficulty with a concept or problem, bring your work so that your instructor can determine where you are having trouble. If you miss a lecture, read the appropriate section in the textbook, borrow class notes, and attempt the assigned homework before your office visit. Because this text has an accompanying video lesson for every objective, you might find it helpful to view the video covering the material you missed. It is not realistic to expect your professor to rehash all or part of a class lecture during office hours.
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. The graph of y = f (x)  5 is obtained by a/an shift of the graph of y = f (x) a distance of 5 units. C2. The graph of y = f (x  5) is obtained by a/an shift of the graph of y = f (x) a distance of 5 units. C3. The graph of y = f (x) is the graph of y = f (x) reflected about the . C4. The graph of y = f (  x) is the graph of y = f (x) reflected about the .
C5. The graph of y = 5f (x) is obtained by a/an stretch of the graph of y = f (x) by multiplying each of its coordinates by 5.
C6. The graph of y = f 1 15x 2 is obtained by a/an stretch of the graph of y = f (x) by multiplying each of its coordinates by 5.
C7. True or false: The graph of g(x) = 1x + 4 is the graph of f (x) = 1x shifted horizontally to the right by 4 units.
2.5 EXERCISE SET Practice Exercises In Exercises 1–16, use the graph of y = f (x) to graph each function g.
In Exercises 17–32, use the graph of y = f (x) to graph each function g. y
y y=f x 4 3 2 (–2, 2) 1 –5 –4 –3 –2 –1–1
(0, 2) (–4, 0) (2, 2) 1 2 3 4 5
x
–5 –4 –3 –2 –1–1 –2 (–2, –2) –3 –4
–2 –3 –4
23. g(x) =  f (x)
24. g(x) = f ( x)
10. g(x) = f (  x) + 3
25. g(x) = f (  x) + 1
26. g(x) = f (x) + 1
12. g(x) = 2f (x)
27. g(x) = 2f (x)
14. g(x) = f (2x)
29. g(x) = f (2x)
16. g(x) = f (2x)  1
31. g(x) = 2f (x + 2) + 1
30. g(x) = f 1 12 x 2
7. g(x) = f (  x)
8. g(x) = f (x)
+ 1
(0, 0)
22. g(x) = f (x + 1)  2
6. g(x) = f (x + 1) + 2
15. g(x) =
x
21. g(x) = f (x  1) + 2
5. g(x) = f (x  1)  2
1 2 1 2
1 2 3 4 5
20. g(x) = f (x + 1)
4. g(x) = f (x  1)
13. g(x) =
(4, 0)
18. g(x) = f (x) + 1
3. g(x) = f (x + 1)
11. g(x) =
y=f x
17. g(x) = f (x)  1
2. g(x) = f (x)  1
1 2 f (x) f 12 x  f 12 x
(2, 2)
19. g(x) = f (x  1)
1. g(x) = f (x) + 1
9. g(x) =  f (x) + 3
4 3 2 1
28. g(x) = 21 f (x)
32. g(x) = 2f (x + 2)  1
325
Section 2.5 Transformations of Functions In Exercises 33–44, use the graph of y = f (x) to graph each function g. y
81. g(x) = x + 4
4 3 (0, 0) 2 1 (–4, 0) –5 –4 –3 –2 –1–1 y=f x
33. 35. 37. 39. 41. 43.
g(x) g(x) g(x) g(x) g(x) g(x)
= = = = = =
x
1 2 3 4 5
–2 –3 (0, –2) –4
f (x) + 2 f (x + 2) f (x + 2)  12 f (x + 2)  12 f (x + 2)  2 1 2 f (2x)
34. 36. 38. 40. 42. 44.
(4, –2)
g(x) g(x) g(x) g(x) g(x) g(x)
= = = = = =
f (x)  2 f (x  2)  f (x  2)  12 f (x  2)  12 f (x  2) + 2 2f 1 12 x 2
In Exercises 45–52, use the graph of y = f (x) to graph each function g.
–5 –4 –3 –2 –1–1 (–4, –2)
82. g(x) = x + 3
83. g(x) = x + 4
84. g(x) = x + 3
85. h(x) = x + 4 2
86. h(x) = x + 3 2
87. h(x) =  x + 4
88. h(x) =  x + 3
89. g(x) =  x + 4 + 1
90. g(x) =  x + 4 + 2
91. h(x) = 2 x + 4
92. h(x) = 2 x + 3
93. g(x) =  2 x + 4 + 1
94. g(x) = 2 x + 3 + 2
In Exercises 95–106, begin by graphing the standard cubic function, f (x) = x 3. Then use transformations of this graph to graph the given function. 95. g(x) = x 3  3
96. g(x) = x 3  2 3
3
97. g(x) = (x  3)
98. g(x) = (x  2)
3
99. h(x) =  x 3
100. h(x) =  (x  2)
101. h(x) = 21 x 3
102. h(x) = 41 x 3 3
103. r(x) = (x  3) + 2 3
105. h(x) = 21 (x  3)  2
y 4 y=f x 3 2 1 (–2, 0)
In Exercises 81–94, begin by graphing the absolute value function, f (x) = x . Then use transformations of this graph to graph the given function.
1 2 3 4 5
x
–2 –3 –4
3 107. g(x) = 1x + 2
3 108. g(x) = 1 x  2
3 109. g(x) = 2 x + 2
3 110. g(x) = 2 x  2
111. h(x) = 113. r(x) =
1 3 2 2x 1 3 2 2x
+ 2
3 112. h(x) = 21 2 x  2
+ 2  2
3 114. r(x) = 21 2 x  2 + 2
3
45. 47. 49. 51.
g(x) g(x) g(x) g(x)
= = = =
f (x  1)  1 f (x  1) + 1 2f 1 12 x 2 1 2 f (x + 1)
46. 48. 50. 52.
g(x) g(x) g(x) g(x)
= = = =
f (x + 1) + 1 f (x + 1)  1 1 2 f (2x) 2f (x  1)
In Exercises 53–66, begin by graphing the standard quadratic function, f (x) = x 2. Then use transformations of this graph to graph the given function. 53. 55. 57. 59. 61. 63. 65.
g(x) g(x) h(x) h(x) g(x) h(x) h(x)
= = = = = = =
x2  2 2 (x  2) 2 (x  2) 2 (x  2) + 1 2 2(x  2) 2 2(x  2)  1 2 2(x + 1) + 1
54. 56. 58. 60. 62. 64. 66.
g(x) g(x) h(x) h(x) g(x) h(x) h(x)
= = = = = = =
3
106. h(x) = 51 (x + 5)  2
In Exercises 107–118, begin by graphing the cube root function, 3 f (x) = 1 x. Then use transformations of this graph to graph the given function.
(0, 2) (2, 2) (4, 0)
3
104. r(x) = (x  7) + 5
x2  1 2 (x  1) 2 (x  1) 2 (x  1) + 2 2 1 2 (x  1) 2 1 2 (x  1)  1 2 2(x + 2) + 1
115. h(x) =  2x + 2
3 116. h(x) =  2 x  2
3 117. g(x) = 2 x  2
3 118. g(x) = 2 x + 2
Practice PLUS In Exercises 119–122, use transformations of the graph of the greatest integer function, f (x) = int(x), to graph each function. (The graph of f (x) = int(x) is shown in Figure 2.27 on page 277.) 119. g(x) = 2 int (x + 1)
120. g(x) = 3 int (x  1)
121. h(x) = int(  x) + 1
122. h(x) = int( x)  1
In Exercises 123–126, write a possible equation for the function whose graph is shown. Each graph shows a transformation of a common function. 123.
124.
In Exercises 67–80, begin by graphing the square root function, f (x) = 1x. Then use transformations of this graph to graph the given function. 67. 69. 71. 73. 75. 77.
g(x) g(x) h(x) h(x) g(x) h(x)
= = = = = =
1x + 2 2x + 2  2x + 2 2 x + 2 1 2 2x
+ 2 2x + 2  2
79. g(x) = 22x + 2  2
68. 70. 72. 74. 76. 78.
g(x) g(x) h(x) h(x) g(x) h(x)
= = = = = =
1x + 1 2x + 1  2x + 1 2x + 1 22x + 1 2x + 1  1
80. g(x) = 22x + 1  1
[–3, 3, 1] by [–6, 6, 1]
[–2, 8, 1] by [–1, 4, 1]
125.
126.
[–5, 3, 1] by [–5, 10, 1]
[–1, 9, 1] by [–1, 5, 1]
326
Chapter 2 Functions and Graphs
Application Exercises 127. The function f (x) = 2.91x + 20.1 models the median height, f (x), in inches, of boys who are x months of age. The graph of f is shown. y
Boys’ Heights
Median Height (inches)
50 f x=√x+ 40 30
Explaining the Concepts
20 10 0
10
20 30 40 Age (months)
50
60
x
Source: Laura Walther Nathanson, The Portable Pediatrician for Parents
a. Describe how the graph can be obtained using transformations of the square root function f (x) = 1x. b. According to the model, what is the median height of boys who are 48 months, or four years, old? Use a calculator and round to the nearest tenth of an inch. The actual median height for boys at 48 months is 40.8 inches. How well does the model describe the actual height? c. Use the model to find the average rate of change, in inches per month, between birth and 10 months. Round to the nearest tenth. d. Use the model to find the average rate of change, in inches per month, between 50 and 60 months. Round to the nearest tenth. How does this compare with your answer in part (c)? How is this difference shown by the graph? 128. The function f (x) = 3.11x + 19 models the median height, f (x), in inches, of girls who are x months of age. The graph of f is shown. y
Girls’ Heights
50 Median Height (inches)
round to the nearest tenth of an inch. The actual median height for girls at 48 months is 40.2 inches. How well does the model describe the actual height? c. Use the model to find the average rate of change, in inches per month, between birth and 10 months. Round to the nearest tenth. d. Use the model to find the average rate of change, in inches per month, between 50 and 60 months. Round to the nearest tenth. How does this compare with your answer in part (c)? How is this difference shown by the graph?
f x=√x+ 40 30 20 10 0
10
20 30 40 Age (months)
50
60
x
Source: Laura Walther Nathanson, The Portable Pediatrician for Parents
a. Describe how the graph can be obtained using transformations of the square root function f (x) = 1x. b. According to the model, what is the median height of girls who are 48 months, or 4 years, old? Use a calculator and
129. What must be done to a function’s equation so that its graph is shifted vertically upward? 130. What must be done to a function’s equation so that its graph is shifted horizontally to the right? 131. What must be done to a function’s equation so that its graph is reflected about the x@axis? 132. What must be done to a function’s equation so that its graph is reflected about the y@axis? 133. What must be done to a function’s equation so that its graph is stretched vertically? 134. What must be done to a function’s equation so that its graph is shrunk horizontally?
Technology Exercises 135. a. Use a graphing utility to graph f (x) = x 2 + 1. b. Graph f (x) = x 2 + 1, g(x) = f (2x), h(x) = f (3x), and k(x) = f (4x) in the same viewing rectangle. c. Describe the relationship among the graphs of f, g, h, and k, with emphasis on different values of x for points on all four graphs that give the same y@coordinate. d. Generalize by describing the relationship between the graph of f and the graph of g, where g(x) = f (cx) for c 7 1. e. Try out your generalization by sketching the graphs of f (cx) for c = 1, c = 2, c = 3, and c = 4 for a function of your choice. 136. a. Use a graphing utility to graph f (x) = x 2 + 1. b. Graph f (x) = x 2 + 1, g(x) = f 1 12 x 2 , and h(x) = f 1 14 x 2 in the same viewing rectangle. c. Describe the relationship among the graphs of f, g, and h, with emphasis on different values of x for points on all three graphs that give the same y@coordinate. d. Generalize by describing the relationship between the graph of f and the graph of g, where g(x) = f (cx) for 0 6 c 6 1. e. Try out your generalization by sketching the graphs of f (cx) for c = 1, and c = 12, and c = 14 for a function of your choice.
Critical Thinking Exercises Make Sense? During the winter, you program your home thermostat so that at midnight, the temperature is 55°. This temperature is maintained until 6 A.M. Then the house begins to warm up so that by 9 A.M. the temperature is 65°. At 6 P.M. the house begins to cool. By 9 P.M., the temperature is again 55°.
Section 2.5 Transformations of Functions The graph illustrates home temperature, f (t), as a function of hours after midnight, t.
y
146. 5 4 3 2 1
Home Temperature as a Function of Time
y
Temperature (°F)
70° 65°
–3 –2 –1–1
y=f t
60°
50° 3
6
9 12 15 18 Hours after Midnight
21
24
t
137. I decided to keep the house 5° warmer than before, so I reprogrammed the thermostat to y = f (t) + 5. 138. I decided to keep the house 5° cooler than before, so I reprogrammed the thermostat to y = f (t)  5. 139. I decided to change the heating schedule to start one hour earlier than before, so I reprogrammed the thermostat to y = f (t  1). 140. I decided to change the heating schedule to start one hour later than before, so I reprogrammed the thermostat to y = f (t + 1). In Exercises 141–144, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. 141. If f (x) = x and g(x) = x + 3 + 3, then the graph of g is a translation of the graph of f 3 units to the right and 3 units upward. 142. If f (x) =  1x and g(x) = 1 x, then f and g have identical graphs. 143. If f (x) = x 2 and g(x) = 5(x 2  2), then the graph of g can be obtained from the graph of f by stretching f 5 units followed by a downward shift of 2 units. 3 144. If f (x) = x 3 and g(x) = (x  3)  4, then the graph of g can be obtained from the graph of f by moving f 3 units to the right, reflecting about the x@axis, and then moving the resulting graph down 4 units. In Exercises 145–148, functions f and g are graphed in the same rectangular coordinate system. If g is obtained from f through a sequence of transformations, find an equation for g. y 5 4 3 2 1 –7 –6 –5 –4 –3 –2 –1–1
g x=!
–2 –3 –4 –5
1 2 3 4 5 6 7
x
g x=!
y
147. 5 4 3 2 1
In Exercises 137–140, determine whether each statement makes sense or does not make sense, and explain your reasoning. If the statement makes sense, graph the new function on the domain [0, 24]. If the statement does not make sense, correct the function in the statement and graph the corrected function on the domain [0, 24].
145.
f x=x
–2 –3 –4 –5
55°
0
327
–2 –1–1
f x=√x g x=! 1 2 3 4 5 6 7 8
x
–2 –3 –4 –5 y
148.
f x=√–x
5 3 2 1 –5
–3 –2 –1–1
1 2 3
–2 –3 –4 –5
5
x
g x=!
For Exercises 149–152, assume that (a, b) is a point on the graph of f. What is the corresponding point on the graph of each of the following functions? 149. y = f (  x) 150. y = 2f (x) 151. y = f (x  3) 152. y = f (x)  3
Retaining the Concepts 153. The length of a rectangle exceeds the width by 13 yards. If the perimeter of the rectangle is 82 yards, what are its dimensions? (Section 1.3, Example 6) 154. Solve: 2x + 10  4 = x. (Section 1.6, Example 3) 155. Multiply and write the product in standard form: (3  7i)(5 + 2i). (Section 1.4, Example 2)
Preview Exercises f x=x 1 2 3
x
Exercises 156–158 will help you prepare for the material covered in the next section. In Exercises 156–157, perform the indicated operation or operations. 156. (2x  1)(x 2 + x  2) 2 157. (f (x))  2f (x) + 6, where f (x) = 3x  4 2 158. Simplify: . 3  1 x
328
Chapter 2 Functions and Graphs
SECTION 2.6
Combinations of Functions; Composite Functions
WHAT YOU'LL LEARN 1 Find the domain of a function.
2 Combine functions using the algebra of functions, specifying domains.
3 Form composite functions. 4 Determine domains for composite functions.
5 Write functions as compositions.
We’re born. We die. Figure 2.54 quantifies these statements by showing the number of births and deaths in the United States for nine selected years. In this section, we look at these data from the perspective of functions. By considering the yearly change in the U.S. population, you will see that functions can be subtracted using procedures that will remind you of combining algebraic expressions.
Number of Births and Deaths in the United States Births 4400
Number (thousands)
4100
4059
4266
4112
4022
Deaths
4248 3999
3988
3953
3947
3800 3500 3200 2900 2600
2403
2443
2398
2472
2426
2468
2543
2628
2744
2300 2000 2000
2002
2004
2006
2008 Year
2010
2012
2014
2016
Figure 2.54 Source: U.S. Department of Health and Human Services
1 Find the domain of a function.
The Domain of a Function We begin with two functions that model the data in Figure 2.54. B(x) = −2.6x2 + 33x + 4043 0WODGTQHDKTVJUB xKP VJQWUCPFUx[GCTUCHVGT
D(x) = 2x2 − 13x + 2428 0WODGTQHFGCVJUD xKP VJQWUCPFUx[GCTUCHVGT
The data in Figure 2.54 show evennumbered years from 2000 through 2016. Because x represents the number of years after 2000, Domain of B = {0, 2, 4, 6, c, 16} and Domain of D = {0, 2, 4, 6, c, 16}.
Section 2.6 Combinations of Functions; Composite Functions
329
Functions that model data often have their domains explicitly given with the function’s equation. However, for most functions, only an equation is given and the domain is not specified. In cases like this, the domain of a function f is the largest set of real numbers for which the value of f (x) is a real number. For example, consider the function f (x) =
1 . x  3
Because division by 0 is undefined, the denominator, x  3, cannot be 0. Thus, x cannot equal 3. The domain of the function consists of all real numbers other than 3, represented by Domain of f = {x x is a real number and x ≠ 3}. GREAT QUESTION Despite the voice balloons, the notation for the domain of f on the right is a mouthful! Will you be using set operations with interval notation in this section? What should I already know? Yes, you’ll be seeing the intersection and the union of sets that are expressed in interval notation. Recall that the intersection of sets A and B, written A ¨ B, is the set of elements common to both set A and set B. When sets A and B are in interval notation, to find the intersection, graph each interval and take the portion of the number line that the two graphs have in common. We will also be using notation involving the union of sets A and B, A ∪ B, meaning the set of elements in A or in B or in both. For more detail, see Section P.1, Objectives 3 and 4, and Section 1.7, Objective 2.
Using interval notation, Domain of f = (–∞, 3) ∪ (3, ∞). #NNTGCNPWODGTU NGUUVJCP
QT
#NNTGCNPWODGTU ITGCVGTVJCP
Now consider a function involving a square root: g(x) = 2x  3. Because only nonnegative numbers have square roots that are real numbers, the expression under the square root sign, x  3, must be nonnegative. We can use inspection to see that x  3 Ú 0 if x Ú 3. The domain of g consists of all real numbers that are greater than or equal to 3: Domain of g = {x x Ú 3} or [3,∞). Finding a Function’s Domain If a function f does not model data or verbal conditions, its domain is the largest set of real numbers for which the value of f(x) is a real number. Exclude from a function’s domain real numbers that cause division by zero and real numbers that result in an even root, such as a square root, of a negative number.
EXAMPLE 1
Finding the Domain of a Function
Find the domain of each function: a. f (x) = x 2  7x c. h(x) = 23x + 12
3x + 2 x  2x  3 3x + 2 d. j(x) = . 214  2x b. g(x) =
2
Solution The domain is the set of all real numbers, (  ∞, ∞), unless x appears in a denominator or in an even root, such as a square root. a. The function f (x) = x 2  7x contains neither division nor a square root. For every real number, x, the algebraic expression x 2  7x represents a real number. Thus, the domain of f is the set of all real numbers. Domain of f = (  ∞, ∞) 3x + 2 contains division. Because division by 0 is x 2  2x  3 undefined, we must exclude from the domain the values of x that cause the denominator, x 2  2x  3, to be 0. We can identify these values by setting x 2  2x  3 equal to 0.
b. The function g(x) =
330
Chapter 2 Functions and Graphs
x 2  2x  3 = 0 (x + 1)(x  3) = 0 x + 1 = 0 or x  3 = 0 x = 1 x = 3
Set the function’s denominator equal to 0. Factor. Set each factor equal to 0. Solve the resulting equations.
We must exclude 1 and 3 from the domain of g(x) =
3x + 2 . x  2x  3 2
Domain of g = (  ∞, 1) ∪ ( 1, 3) ∪ (3, ∞)
GREAT QUESTION When finding the domain of a function, when do I have to factor? In parts (a) and (b), observe when to factor and when not to factor a polynomial. • f (x) = x2 − 7x
• g(x) =
&QPQVHCEVQTx–xCPFUGVKVGSWCNVQ\GTQ 0QXCNWGUQHxPGGFDGGZENWFGFHTQOVJGFQOCKP
3x + 2 x2 − 2x − 3
&QHCEVQTx–x–CPFUGVKVGSWCNVQ\GTQ 9GOWUVGZENWFGXCNWGUQHxVJCVECWUGVJKUFGPQOKPCVQTVQDG\GTQ
c. The function h(x) = 23x + 12 contains an even root. Because only nonnegative numbers have real square roots, the quantity under the radical sign, 3x + 12, must be greater than or equal to 0. h x=√x+
3x + 12 Ú 0 3x Ú 12 x Ú 4
Set the function’s radicand greater than or equal to 0. Subtract 12 from both sides. Divide both sides by 3. Division by a positive number preserves the sense of the inequality.
&QOCKP=–∞
The domain of h consists of all real numbers greater than or equal to 4. Domain of h = [4,∞)
[–10, 10, 1] by [–10, 10, 1] Figure 2.55
The domain is highlighted on the x@axis in Figure 2.55. 3x + 2 d. The function j(x) = contains both an even root and division. 214  2x Because only nonnegative numbers have real square roots, the quantity under the radical sign, 14  2x, must be greater than or equal to 0. But wait, there’s more! Because division by 0 is undefined, 14  2x cannot equal 0. Thus, 14  2x must be strictly greater than 0. 14  2x 7 0 2x 7 14 x 6 7
Set the function’s radicand greater than 0. Subtract 14 from both sides. Divide both sides by − 2. Division by a negative number reverses the direction of the inequality.
The domain of j consists of all real numbers less than 7. Domain of j = (  ∞, 7) Find the domain of each function: 5x a. f (x) = x 2 + 3x  17 b. g(x) = 2 x  49 5x c. h(x) = 29x  27 d. j(x) = . 224  3x CHECK POINT 1
Section 2.6 Combinations of Functions; Composite Functions
2 Combine functions using the algebra of functions, specifying domains.
331
The Algebra of Functions We can combine functions using addition, subtraction, multiplication, and division by performing operations with the algebraic expressions that appear on the right side of the equations. For example, the functions f(x) = 2x and g(x) = x  1 can be combined to form the sum, difference, product, and quotient of f and g. Here’s how it’s done: (f + g)(x) = f (x) + g(x)
5WOf+ g
= 2x + (x − 1) = 3x − 1 &KHHGTGPEGf− g
(QT GCEJ HWPEVKQP f x=x CPF g x=x−
(f − g)(x) = f (x) − g(x) = 2x − (x − 1) = 2x − x + 1 = x + 1
2TQFWEVf g
(fg)(x) = f (x) ∙ g(x) = 2x(x − 1) = 2x2 − 2x
3WQVKGPV
f g
f f (x) 2x a b (x) = = , x ≠ 1. g g(x) x−1
The domain for each of these functions consists of all real numbers that are common to the domains of f and g. Using Df to represent the domain of f and Dg to represent the domain of g, the domain for each function is Df ¨ Dg. In the case of the f (x) quotient function , we must remember not to divide by 0, so we add the further g(x) restriction that g(x) ≠ 0. The Algebra of Functions: Sum, Difference, Product, and Quotient of Functions Let f and g be two functions. The sum f + g, the difference f  g, the product fg, f and the quotient are functions whose domains are the set of all real numbers g common to the domains of f and g (Df ¨ Dg), defined as follows: 1. Sum: 2. Difference: 3. Product: 4. Quotient:
EXAMPLE 2
(f + g)(x) = f (x) + g(x) (f  g)(x) = f (x)  g(x) (fg)(x) = f (x) # g(x) f (x) f a b (x) = , provided g(x) ≠ 0. g g(x) Combining Functions
Let f (x) = 2x  1 and g(x) = x 2 + x  2. Find each of the following functions: f a. (f + g)(x) b. (f  g)(x) c. (fg)(x) d. a b (x). g Determine the domain for each function. Solution a. (f + g)(x) = = = b. (f  g)(x) =
f (x) + g(x) (2x  1) + (x 2 + x  2) x 2 + 3x  3 f (x)  g(x)
This is the definition of the sum f + g. Substitute the given functions. Remove parentheses and combine like terms. This is the definition of the difference f − g.
= (2x  1)  (x + x  2) = 2x  1  x 2  x + 2
Substitute the given functions.
= x 2 + x + 1
Combine like terms and arrange terms in descending powers of x.
2
Remove parentheses and change the sign of each term in the second set of parentheses.
332
Chapter 2 Functions and Graphs
c. (f g)(x) = f (x) # g(x)
This is the definition of the product fg.
= (2x  1)(x 2 + x  2)
Substitute the given functions.
= 2x(x + x  2)  1(x + x  2) 2
2
Multiply each term in the second factor by 2x and − 1, respectively.
= 2x 3 + 2x 2  4x  x 2  x + 2 = 2x 3 + (2x 2  x 2) + ( 4x  x) + 2 GREAT QUESTION Should I simplify a quotient function before finding its domain? f No. If the function can be g simplified, determine the domain before simplifying. All values of x for which g(x) = 0 must be excluded from the domain. Example: f (x) = x2 − 4 and g(x) = x − 2 a
1 (x + 2)(x − 2) = =x+2 (x − 2) 1
Rearrange terms so that like terms are adjacent.
= 2x 3 + x 2  5x + 2 f (x) f d. a b (x) = g g(x) =
Combine like terms.
f This is the definition of the quotient . g
2x  1 x + x  2 2
Substitute the given functions. This rational expression cannot be simplified.
Because the equations for f and g do not involve division or contain even roots, the domain of both f and g is the set of all real numbers. Thus, the domain of f + g, f  g, and fg is the set of all real numbers, (  ∞, ∞). f The function contains division. We must exclude from its domain values of x g that cause the denominator, x 2 + x  2, to be 0. Let’s identify these values.
f x2 − 4 b (x) = g x−2 x≠6JGFQOCKPQH f KU –∞∪ ∞ g
Use the distributive property.
x2 + x  2 (x + 2)(x  1) x + 2 = 0 or x  1 x = 2 x
= = = =
0 0 0 1
Set the denominator of
f g
equal to 0.
Factor. Set each factor equal to 0. Solve the resulting equations.
f We must exclude 2 and 1 from the domain of . g Domain of CHECK POINT 2
f = (  ∞, 2) ∪ ( 2, 1) ∪ (1, ∞ ) g
Let f (x) = x  5 and g(x) = x 2  1. Find each of the
following functions: a. (f + g)(x)
b. (f  g)(x)
c. (fg)(x)
Determine the domain for each function.
EXAMPLE 3
f d. a b (x). g
Adding Functions and Determining the Domain
Let f (x) = 2x + 3 and g(x) = 2x  2. Find each of the following: a. (f + g)(x) b. the domain of f + g. Solution a. (f + g)(x) = f (x) + g(x) = 2x + 3 + 2x  2 b. The domain of f + g is the set of all real numbers that are common to the domain of f and the domain of g. Thus, we must find the domains of f and g before finding their intersection. • f(x) = √x + 3 x+OWUVDGPQPPGICVKXG x+≥Df==–∞
• g(x) = √x − 2 x–OWUVDGPQPPGICVKXG x–≥Dg==∞
Section 2.6 Combinations of Functions; Composite Functions Domain of f Domain of g –3
2
Domain of f+g
333
Now, we can use a number line to determine Df ¨ Dg, the domain of f + g. Figure 2.56 shows the domain of f , [3, ∞), in blue and the domain of g, [2, ∞), in red. Can you see that all real numbers greater than or equal to 2 are common to both domains? This is shown in purple on the number line. Thus, the domain of f + g is [2,∞).
Figure 2.56 Finding the domain of the sum f + g
TECHNOLOGY Graphic Connections The graph on the left is the graph of
y = 2x + 3 + 2x  2 in a [ 3, 10, 1] by [0, 8, 1] viewing rectangle. The graph reveals what we discovered algebraically in Example 3(b). The domain of this function is [2, ∞ ). &QOCKP=∞
CHECK POINT 3
Let f (x) = 2x  3 and g(x) = 2x + 1. Find each of the
following: a. (f + g)(x)
b. the domain of f + g.
EXAMPLE 4
Applying the Algebra of Functions
We opened the section with functions that model the number of births and deaths in the United States for selected years from 2000 through 2016: B(x) = −2.6x2 + 33x + 4043 0WODGTQHDKTVJUB xKP VJQWUCPFUx[GCTUCHVGT
D(x) = 2x2 − 13x + 2428. 0WODGTQHFGCVJUD xKP VJQWUCPFUx[GCTUCHVGT
a. Write a function that models the change in U.S. population, B(x)  D(x), for each year from 2000 through 2016. b. Use the function from part (a) to find the change in U.S. population in 2014. c. Does the result in part (b) overestimate or underestimate the actual population change in 2014 obtained from the data in Figure 2.54 on page 328? By how much? Solution a. The change in population is the number of births minus the number of deaths. Thus, we will find the difference function, B  D. (B  D)(x) = B(x)  D(x) = ( 2.6x 2 + 33x + 4043)  (2x 2  13x + 2428) = 2.6x 2 + 33x + 4043 + (  2x 2) + 13x  2428
Substitute the given functions. Remove parentheses and change the sign of each term in the second set of parentheses.
= ( 2.6x  2x ) + (33x + 13x) + (4043  2428) = 4.6x 2 + 46x + 1615 2
2
Group like terms. Combine like terms.
The function (B  D)(x) = 4.6x 2 + 46x + 1615 models the change in U.S. population, in thousands, x years after 2000.
334
Chapter 2 Functions and Graphs
b. Because 2014 is 14 years after 2000, we substitute 14 for x in the difference function (B  D)(x). (B  D)(x) = 4.6x 2 + 46x + 1615
Use the difference function B − D.
(B  D)(14) = 4.6(14) + 46(14) + 1615 = 4.6(196) + 46(14) + 1615 2
Substitute 14 for x. Evaluate the exponential expression: 142 = 196.
= 901.6 + 644 + 1615 = 1357.4
Perform the multiplications. Add from left to right.
We see that (B  D)(14) = 1357.4. The model indicates that there was a population increase of 1357.4 thousand, or approximately 1,357,400 people, in 2014. c. The data for 2014 in Figure 2.54 on page 328 show 3988 thousand births and 2628 thousand deaths. population change = births  deaths = 3988  2628 = 1360 The actual population increase was 1360 thousand, or 1,360,000. Our model gave us an increase of 1357.4 thousand. Thus, the model underestimates the actual increase by 1360  1357.4, or 2.6 thousand people. Use the birth and death models from Example 4. a. Write a function that models the total number of births and deaths in the United States for the years from 2000 through 2016. b. Use the function from part (a) to find the total number of births and deaths in the United States in 2000. c. Does the result in part (b) overestimate or underestimate the actual number of total births and deaths in 2000 obtained from the data in Figure 2.54 on page 328? By how much? CHECK POINT 4
3 Form composite functions.
Composite Functions There is another way of combining two functions. To help understand this new combination, suppose that your local computer store is having a sale. The models that are on sale cost either $300 less than the regular price or 85% of the regular price. If x represents the computer’s regular price, the discounts can be modeled with the following functions: f(x) = x − 300 6JGEQORWVGTKUQP UCNGHQTNGUU VJCPKVUTGIWNCTRTKEG
g(x) = 0.85x. 6JGEQORWVGTKUQP UCNGHQTQHKVU TGIWNCTRTKEG
At the store, you bargain with the salesperson. Eventually, she makes an offer you can’t refuse. The sale price will be 85% of the regular price followed by a $300 reduction: 0.85x − 300. QH VJGTGIWNCT RTKEG
HQNNQYGFD[ C TGFWEVKQP
Section 2.6 Combinations of Functions; Composite Functions
335
In terms of the functions f and g, this offer can be obtained by taking the output of g(x) = 0.85x, namely, 0.85x, and using it as the input of f: f(x) = x − 300 4GRNCEGxYKVJxVJGQWVRWVQHg x=x
f (0.85x) = 0.85x − 300. Because 0.85x is g(x), we can write this last equation as f (g(x)) = 0.85x  300. We read this equation as “f of g of x is equal to 0.85x  300.” We call f (g(x)) the composition of the function f with g, or a composite function. This composite function is written f ∘ g. Thus, (f ∘ g)(x) = f (g(x)) = 0.85x − 300. 6JKUECPDGTGCFpfQHgQHxq QTpfEQORQUGFYKVJgQHxq
Like all functions, we can evaluate f ∘ g for a specified value of x in the function’s domain. For example, here’s how to find the value of the composite function describing the offer you cannot refuse at 1400: (f ∘ g)(x) = 0.85x − 300 4GRNCEGxYKVJ
(f ∘ g)(1400) = 0.85(1400) − 300 = 1190 − 300 = 890. This means that a computer that regularly sells for $1400 is on sale for $890 subject to both discounts. We can use a partial table of coordinates for each of the discount functions, g and f, to verify this result numerically. %QORWVGToU TGIWNCTRTKEG
QHVJG TGIWNCTRTKEG
QHVJG TGIWNCTRTKEG
TGFWEVKQP
x
g(x) = 0.85x
x
1200
1020
1020
720
1300
1105
1105
805
1400
1190
1190
890
f(x) = x − 300
Using these tables, we can find (f ∘ g)(1400): (f ∘ g)(1400) = f(g(1400)) = f (1190) = 890.
6JGVCDNGHQTgUJQYU VJCVg =
6JGVCDNGHQTfUJQYU VJCVf =
This verifies that a computer that regularly sells for $1400 is on sale for $890 subject to both discounts. Before you run out to buy a computer, let’s generalize our discussion of the computer’s double discount and define the composition of any two functions.
336
Chapter 2 Functions and Graphs
The Composition of Functions The composition of the function f with g is denoted by f ∘ g and is defined by the equation (f ∘ g)(x) = f (g(x)). The domain of the composite function f ∘ g is the set of all x such that 1. x is in the domain of g and 2. g(x) is in the domain of f. The composition of f with g, f ∘ g, is illustrated in Figure 2.57. Step 1 Input x into g.
Step 2 Input g(x) into f. t tpu Ou x) g(
Function g
Function f
t tpu Ou (x)) g ( f
ut Inp ) g(x
ut Inp x
xOWUVDGKP VJGFQOCKPQHg
g xOWUVDGKP VJGFQOCKPQHf
Figure 2.57
The figure reinforces the fact that the inside function g in f (g(x)) is done first.
EXAMPLE 5
Forming Composite Functions
Given f (x) = 3x  4 and g(x) = x 2  2x + 6, find each of the following: a. ( f ∘ g)(x) b. (g ∘ f )(x) c. (g ∘ f )(1). Solution a. We begin with ( f ∘ g)(x), the composition of f with g. Because ( f ∘ g)(x) means f (g(x)), we must replace each occurrence of x in the equation for f with g(x). f (x) = 3x − 4
This is the given equation for f.
4GRNCEGxYKVJg x
(f ∘ g)(x) = f (g(x)) = 3g(x) − 4 = 3(x2 − 2x + 6) − 4 = 3x 2  6x + 18  4 = 3x 2  6x + 14
Because g(x) = x2 − 2x + 6, replace g(x) with x2 − 2x + 6. Use the distributive property. Simplify.
Thus, (f ∘ g)(x) = 3x 2  6x + 14. b. Next, we find (g ∘ f )(x), the composition of g with f. Because (g ∘ f )(x) means g( f (x)), we must replace each occurrence of x in the equation for g with f (x). g(x) = x2 − 2x + 6
This is the equation for g.
4GRNCEGxYKVJf x
(g ∘ f)(x) = g(f (x)) = (f (x))2 − 2f(x) + 6 = (3x − 4)2 − 2(3x − 4) + 6
Because f (x) = 3x − 4, replace f (x) with 3x − 4.
Section 2.6 Combinations of Functions; Composite Functions
= 9x 2  24x + 16  6x + 8 + 6 = 9x 2  30x + 30
337
Use 1 A − B 2 2 = A2 − 2AB + B2 to square 3x − 4. Simplify: − 24x − 6x = − 30x and 16 + 8 + 6 = 30.
Thus, (g ∘ f )(x) = 9x 2  30x + 30. Notice that ( f ∘ g)(x) is not the same function as (g ∘ f )(x). c. We can use (g ∘ f)(x) to find (g ∘ f )(1). (g ∘ f )(x) = 9x2 − 30x + 30 4GRNCEGxYKVJ
(g ∘ f )(1) = 9 ∙ 12 − 30 ∙ 1 + 30 = 9 − 30 + 30 = 9 It is also possible to find (g ∘ f )(1) without determining (g ∘ f )(x). (g ∘ f )(1) = g(f (1)) = g(–1) = 9 (KTUVƂPFf f x=x–UQ f =∙–=–
CHECK POINT 5
the following: a. (f ∘ g)(x)
4 Determine domains for
0GZVƂPFg – g x=x–x+UQ g –= –– –+ =++=
Given f (x) = 5x + 6 and g(x) = 2x 2  x  1, find each of b. (g ∘ f)(x)
c. (f ∘ g)( 1).
We need to be careful in determining the domain for a composite function.
composite functions.
Excluding Values from the Domain of (f ° g)(x) = f (g(x)) The following values must be excluded from the input x: • If x is not in the domain of g, it must not be in the domain of f ∘ g. • Any x for which g(x) is not in the domain of f must not be in the domain of f ∘ g.
EXAMPLE 6 A BRIEF REVIEW
Simplifying Complex Fractions One method for simplifying the complex fraction on the right is to find the least common denominator of all the rational expressions in the numerator and the denominator. Then multiply each term in the numerator and denominator by this least common denominator. The procedures for simplifying complex fractions can be found in Section P.6, Objective 6.
Forming a Composite Function and Finding Its Domain
2 3 and g(x) = , find each of the following: x x  1 a. (f ∘ g)(x) b. the domain of f ∘ g.
Given f (x) =
Solution a. Because (f ∘ g)(x) means f (g(x)), we must replace x in f (x) = (f ∘ g)(x) = f (g(x)) =
2 x 2x 2 2 = = = ∙ g(x) − 1 x 3−x 3 3 −1 −1 x x g x= x
Thus, (f ∘ g)(x) =
2x . 3  x
2 with g(x). x1
5KORNKH[VJGEQORNGZ HTCEVKQPD[OWNVKRN[KPI x D[QT x
338
Chapter 2 Functions and Graphs
b. We determine values to exclude from the domain of (f ∘ g)(x) in two steps. Applying the Rules to 2 3 f (x) = and g(x) = x − 1 x
Rules for Excluding Numbers from the Domain of ( f ° g)(x) = f (g(x))
3 , 0 is not in the x domain of g. Thus, 0 must be excluded from the domain of f ∘ g.
If x is not in the domain of g, it must not be in the domain of f ∘ g.
Because g(x) =
2 , we must g(x)  1 exclude from the domain of f ∘ g any x for which g(x) = 1.
Any x for which g(x) is not in the domain of f must not be in the domain of f ∘ g.
Because f (g(x)) =
3 = 1 x 3 = x
Set g(x) equal to 1. Multiply both sides by x.
3 must be excluded from the domain of f ∘ g.
We see that 0 and 3 must be excluded from the domain of f ∘ g. The domain of f ∘ g is (  ∞, 0) ∪ (0, 3) ∪ (3,∞).
CHECK POINT 6
Given f (x) =
a. (f ∘ g)(x)
5 Write functions as compositions.
4 1 and g(x) = , find each of the following: x x + 2
b. the domain of f ∘ g.
Decomposing Functions When you form a composite function, you “compose” two functions to form a new function. It is also possible to reverse this process. That is, you can “decompose” a given function and express it as a composition of two functions. Although there is more than one way to do this, there is often a “natural” selection that comes to mind first. For example, consider the function h defined by 5
h(x) = (3x 2  4x + 1) . GREAT QUESTION Is there a procedure I can use to write a function as a composition? Yes. Suppose the form of function h is power h(x) = (algebraic expression) . Function h can be expressed as a composition, f ∘ g, using
f (x) = x power g(x) = algebraic expression.
The function h takes 3x 2  4x + 1 and raises it to the power 5. A natural way to write h as a composition of two functions is to raise the function g(x) = 3x 2  4x + 1 to the power 5. Thus, if we let f (x) = x 5 and g(x) = 3x 2  4x + 1, then 5
(f ∘ g)(x) = f (g(x)) = f (3x 2  4x + 1) = (3x 2  4x + 1) .
EXAMPLE 7
Writing a Function as a Composition
Express h(x) as a composition of two functions: 3
h(x) = 2x 2 + 1. Solution The function h takes x 2 + 1 and takes its cube root. A natural way to write h as a composition of two functions is to take the cube root of the function g(x) = x 2 + 1. Thus, we let 3 f (x) = 1 x and g(x) = x 2 + 1.
339
Section 2.6 Combinations of Functions; Composite Functions
We can check this composition by finding (f ∘ g)(x). This should give the original 3 2 function, namely, h(x) = 2 x + 1. 3
(f ∘ g)(x) = f (g(x)) = f (x 2 + 1) = 2x 2 + 1 = h(x) Express h(x) as a composition of two functions: h(x) = 2x 2 + 5.
CHECK POINT 7
ACHIEVING SUCCESS Learn from your mistakes. Being human means making mistakes. By finding and understanding your errors, you will become a better math student. Source of Error
Remedy
Not Understanding a Concept
Review the concept by finding a similar example in your textbook or class notes. Ask your professor questions to help clarify the concept.
Skipping Steps
Show clear stepbystep solutions. Detailed solution procedures help organize your thoughts and enhance understanding. Doing too many steps mentally often results in preventable mistakes.
Carelessness
Write neatly. Not being able to read your own math writing leads to errors. Avoid writing in pen so you won’t have to put huge marks through incorrect work.
“You can achieve your goal if you persistently pursue it.” –Cha SaSoon, a 68yearold South Korean woman who passed her country’s written driver’slicense exam on her 950th try (Source: Newsweek)
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. We exclude from a function’s domain real numbers that cause division by .
C10. The notation f ∘ g, called the of the function f with g, is defined by (f ∘ g)(x) =
C2. We exclude from a function’s domain real numbers that result in a square root of a/an number.
C11. I find (f ∘ g)(x) by replacing each occurrence of x in the equation for with .
C3. (f + g)(x) =
C12. The notation g ∘ f , called the g with f , is defined by (g ∘ f )(x) =
C4. (f  g)(x) = f (x) = g
, provided
≠ 0
C7. The domain of f (x) = 5x + 7 consists of all real numbers, represented in interval notation as . 3 C8. The domain of g(x) = consists of all real x  2 numbers except 2, represented in interval notation as (  ∞ , 2) ∪
of the function .
C13. I find (g ∘ f)(x) by replacing each occurrence of x in the equation for with .
C5. (fg)(x) = C6.
.
.
1 7 + consists of all real x x  3 numbers except 0 and 3, represented in interval notation ∪ . as (  ∞ , 0) ∪
C9. The domain of h(x) =
C14. True or false: f ∘ g is the same function as g ∘ f . C15. True or false: f (g(x)) = f (x) # g(x) 3 8 and g(x) = , then 0 and g(x)  4 x must be excluded from the domain of f ∘ g.
C16. If f (g(x)) =
340
Chapter 2 Functions and Graphs
2.6 EXERCISE SET Practice Exercises In Exercises 1–30, find the domain of each function. 1. f (x) = 3(x  4) 3 3. g(x) = x  4 5. f (x) = x 2  2x  15 7. g(x) = 9. f (x) = 11. g(x) = 12. g(x) = 13. h(x) =
15. f (x) =
3x + 1 2x  4 , g(x) = 2 2 x  25 x  25 8x 6 9x 7 45. f (x) = , g(x) = 46. f (x) = , g(x) = x2 x+3 x4 x+8 47. f (x) = 2x + 4, g(x) = 2x  1 44. f (x) =
3 x 2  2x  15 1 3 + x + 7 x  9 1 1  2 2 x + 1 x  1 1 1  2 2 x + 4 x  4 4 3  1 x 1 4  2 x  1
2. f (x) = 2(x + 5) 2 4. g(x) = x + 5 6. f (x) = x 2 + x  12 30 x 2 + 7x  98 1 3 10. f (x) = + x + 8 x  10 8. f(x) =
48. f (x) = 2x + 6, g(x) = 2x  3 49. f (x) = 2x  2, g(x) = 22  x 50. f (x) = 2x  5, g(x) = 25  x In Exercises 51–66, find a. (f ∘ g)(x)
5 4  1 x 1 16. f (x) = 4  3 x  2 14. h(x) =
17. f (x) = 2x  3 1 19. g(x) = 2x  3
18. f (x) = 2x + 2 1 20. g(x) = 2x + 2
21. g(x) = 25x + 35
22. g(x) = 27x  70
23. f (x) = 224  2x
24. f (x) = 284  6x
25. h(x) = 2x  2 + 2x + 3
51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62.
f (x) f (x) f (x) f (x) f (x) f (x) f (x) f (x) f (x) f (x) f (x) f (x)
= = = = = = = = = = = =
63. f (x) = 64. f (x) =
26. h(x) = 2x  3 + 2x + 4 2x  2 x  5 2x 29. f (x) = 3 x  5x 2 7x 30. f (x) = 3 x  2x 2 27. g(x) =
28. g(x) = + + 
2x  3 x  6
7 4x + 20 2 9x + 18
In Exercises 31–50, find f + g, f  g, fg, and domain for each function.
66. f (x) =
c. (f ∘ g)(2)
d. (g ∘ f)(2).
2x, g(x) = x + 7 3x, g(x) = x  5 x + 4, g(x) = 2x + 1 5x + 2, g(x) = 3x  4 4x  3, g(x) = 5x 2  2 7x + 1, g(x) = 2x 2  9 x 2 + 2, g(x) = x 2  2 x 2 + 1, g(x) = x 2  3 4  x, g(x) = 2x 2 + x + 5 5x  2, g(x) =  x 2 + 4x  1 1x, g(x) = x  1 1x, g(x) = x + 2 x + 3 2x  3, g(x) = 2 x + 3 6x  3, g(x) = 6 1 1 , g(x) = x x 2 2 , g(x) = x x
In Exercises 67–74, find a. ( f ∘ g)(x) f g.
Determine the
f (x) = 2x + 3, g(x) = x  1 32. f (x) = 3x  4, g(x) = x + 2 f (x) = x  5, g(x) = 3x 2 34. f (x) = x  6, g(x) = 5x2 2 f (x) = 2x  x  3, g(x) = x + 1 f (x) = 6x 2  x  1, g(x) = x  1 f (x) = 3  x 2, g(x) = x 2 + 2x  15 f (x) = 5  x 2, g(x) = x 2 + 4x  12 f (x) = 1x, g(x) = x  4 40. f (x) = 1x, g(x) = x  5 1 1 41. f (x) = 2 + , g(x) = x x 1 1 42. f (x) = 6  , g(x) = x x 5x + 1 4x  2 43. f (x) = 2 , g(x) = 2 x  9 x  9 31. 33. 35. 36. 37. 38. 39.
65. f (x) =
b. (g ∘ f)(x)
67. f (x) = 68. f (x) = 69. f (x) = 70. f (x) = 71. f (x) =
b. the domain of f ∘ g.
2 , g(x) x + 3 5 , g(x) x + 4 x , g(x) x + 1 x , g(x) x + 5 1x, g(x) =
1 x 1 = x 4 = x 6 = x x  2 =
72. f (x) = 1x, g(x) = x  3 73. f (x) = x 2 + 4, g(x) = 21  x 74. f (x) = x 2 + 1, g(x) = 22  x In Exercises 75–82, express the given function h as a composition of two functions f and g so that h(x) = (f ∘ g)(x). 4
75. h(x) = (3x  1)
3
76. h(x) = (2x  5)
Section 2.6 Combinations of Functions; Composite Functions 3 2 77. h(x) = 2 x  9
78. h(x) = 25x 2 + 3
79. h(x) = 2x  5
80. h(x) = 3x  4
81. h(x) =
1 2x  3
82. h(x) =
Here are two functions that model the data in the graph at the bottom of the previous column:
1 4x + 5
f (x) = −0.004x2 + 0.61x + 194
2TQLGEVGF75RQRWNCVKQP WPFGTf xKPOKNNKQPU x [GCTUCHVGT
g(x) = −0.008x2 + 1.62x + 140.
2TQLGEVGF75RQRWNCVKQP CPFQNFGTg xKPOKNNKQPU x [GCTUCHVGT
Practice PLUS Use the graphs of f and g to solve Exercises 83–90. y
83. Find (f + g)( 3). 84. Find (g  f)( 2). 85. Find (fg)(2).
5 4 3 2 1
y=g x
y=f x
1 2 3 4 5
–5 –4 –3 –2 –1–1
Use the functions to solve Exercises 97–98.
86. Find a b(3). f g
x
–2
87. Find the domain of f + g. f 88. Find the domain of . g 89. Graph f + g. 90. Graph f  g.
In Exercises 91–94, use the graphs of f and g to evaluate each composite function. y
91. (f ∘ g)(  1) 92. ( f ∘ g)(1)
5 4 3 2 1
1 2 3 4 5
–5 –4 –3 –2 –1–1 y = g x
93. (g ∘ f)(0) 94. (g ∘ f )(  1)
y = f x x
–2 –3 –4 –5
In Exercises 95–96, find all values of x satisfying the given conditions. 95. f (x) = 2x  5, g(x) = x 2  3x + 8, and ( f ∘ g)(x) = 7. 96. f (x) = 1  2x, g(x) = 3x 2 + x  1, and ( f ∘ g)(x) = 5.
Application Exercises
R(x) = 65x.
45 and Older
Population (millions)
250 200 150
154
140
213
208
204
201
170
181
192
100 50
2020
2030
Source: U.S. Census Bureau
2040 Year
2050
99. A company that sells protective tablet cases has yearly fixed costs of $600,000. It costs the company $45 to produce each case. Each case will sell for $65. The company’s costs and revenue are modeled by the following functions, where x represents the number of tablet cases produced and sold: This function models the company’s costs.
U.S. Population Projections by Age
193
97. a. Write a function d that models the difference between the projected population under 45 and the projected population 45 and older for the years shown in the bar graph. b. Use the function from part (a) to find how many more people under 45 than 45 and older there are projected to be in 2060. c. Does the result in part (b) overestimate, underestimate, or give the actual difference between the under45 and 45andolder populations in 2060 shown by the bar graph? 98. a. Write a function r that models the ratio of the projected population 45 and older to the projected population under 45 for the years shown in the bar graph. b. Use the function from part (a) to find the ratio of the projected population 45 and older to the projected population under 45, correct to two decimal places, for 2040. c. Find the ratio of the projected population 45 and older to the projected population under 45, correct to two decimal places, shown by the bar graph. How does the rounded ratio in part (b) compare with this ratio?
C(x) = 600,000 + 45x
The bar graph shows U.S. population projections, by age, in millions, for five selected years. Under 45
341
2060
This function models the company’s revenue.
Find and interpret (R  C)(20,000), (R  C)(30,000), and (R  C)(40,000). 100. A department store has two locations in a city. From 2016 through 2020, the profits for each of the store’s two branches are modeled by the functions f (x) = 0.44x + 13.62 and g(x) = 0.51x + 11.14. In each model, x represents the number of years after 2016, and f and g represent the profit, in millions of dollars. a. What is the slope of f ? Describe what this means. b. What is the slope of g? Describe what this means. c. Find f + g. What is the slope of this function? What does this mean?
342
Chapter 2 Functions and Graphs
101. The regular price of a laptop is x dollars. Let f (x) = x  400 and g(x) = 0.75x. a. Describe what the functions f and g model in terms of the price of the laptop. b. Find (f ∘ g)(x) and describe what this models in terms of the price of the laptop. c. Repeat part (b) for (g ∘ f)(x). d. Which composite function models the greater discount on the laptop, f ∘ g or g ∘ f? Explain. 102. The regular price of a pair of jeans is x dollars. Let f (x) = x  5 and g(x) = 0.6x. a. Describe what functions f and g model in terms of the price of the jeans. b. Find (f ∘ g)(x) and describe what this models in terms of the price of the jeans. c. Repeat part (b) for (g ∘ f)(x). d. Which composite function models the greater discount on the jeans, f ∘ g or g ∘ f? Explain.
Explaining the Concepts 103. If a function is defined by an equation, explain how to find its domain. 104. If equations for f and g are given, explain how to find f  g. 105. If equations for two functions are given, explain how to obtain the quotient function and its domain. 106. Describe a procedure for finding (f ∘ g)(x). What is the name of this function? 107. Describe the values of x that must be excluded from the domain of (f ∘ g)(x).
Technology Exercises 108. Graph y1 = x 2  2x, y2 = x, and y3 = y1 , y2 in the same [ 10, 10, 1] by [  10, 10, 1] viewing rectangle. Then use the TRACE feature to trace along y3. What happens at x = 0? Explain why this occurs. 109. Graph y1 = 22  x, y2 = 1x, and y3 = 22  y2 in the same [  4, 4, 1] by [0, 2, 1] viewing rectangle. If y1 represents f and y2 represents g, use the graph of y3 to find the domain of f ∘ g. Then verify your observation algebraically.
112. I must have made a mistake in finding the composite functions f ∘ g and g ∘ f, because I notice that f ∘ g is not the same function as g ∘ f. 113. This diagram illustrates that f (g(x)) = x 2 + 4. 1st Output x2
f(x) 1st Input x
x2
g(x) 2nd Input x2
2nd Output x2 + 4
x+4
In Exercises 114–117, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. 114. If f (x) = x 2  4 and g(x) = 2x 2  4, then ( f ∘ g)(x) = x 2 and (f ∘ g)(5) = 25. 115. There can never be two functions f and g, where f ≠ g, for which (f ∘ g)(x) = (g ∘ f)(x). 116. If f (7) = 5 and g(4) = 7, then (f ∘ g)(4) = 35. 117. If f (x) = 1x and g(x) = 2x  1, then (f ∘ g)(5) = g(2). 118. Prove that if f and g are even functions, then fg is also an even function. 119. Define two functions f and g so that f ∘ g = g ∘ f.
Retaining the Concepts x  1 x + 3 x = 1  . 5 2 4 (Section 1.2, Example 3) 121. In July 2020, the toll for the Golden Gate Bridge was $8.40 for drivers making onetime payments for each crossing. Drivers who pay a onetime fee of $20.00 to purchase a FasTrak toll tag pay $7.70 for each crossing. How many times must a driver cross the Golden Gate Bridge for the cost of these two options to be the same? Round to the nearest whole number. Find the total cost of each option for the rounded number of crossings. (Section 1.3, Example 3)
120. Solve and check:
122. Solve for y: Ax + By = Cy + D. (Section 1.3, Example 8)
Critical Thinking Exercises Make Sense? In Exercises 110–113, determine whether each statement makes sense or does not make sense, and explain your reasoning. 110. I used a function to model data from 1990 through 2020. The independent variable in my model represented the number of years after 1990, so the function’s domain was {x x = 0, 1, 2, 3, c, 30}. 111. I have two functions. Function f models total world population x years after 2000 and function g models population of the world’s moredeveloped regions x years after 2000. I can use f  g to determine the population of the world’s lessdeveloped regions for the years in both function’s domains.
Preview Exercises Exercises 123–125 will help you prepare for the material covered in the next section. 123. Consider the function defined by {( 2, 4), ( 1, 1), (1, 1), (2, 4)}. Reverse the components of each ordered pair and write the resulting relation. Is this relation a function? 124. Solve for y: x =
5 + 4. y
125. Solve for y: x = y2  1, y Ú 0.
Section 2.7 Inverse Functions
SECTION 2.7
343
Inverse Functions
WHAT YOU’LL LEARN 1 Verify inverse functions. 2 Find the inverse of a function.
3 Use the horizontal line test to determine if a function has an inverse function.
4 Use the graph of a
onetoone function to graph its inverse function.
5 Find the inverse of a
function and graph both functions on the same axes.
Based on Shakespeare’s Romeo and Juliet, the film West Side Story swept the 1961 Academy Awards with ten Oscars. The top four movies to win the most Oscars are shown in Table 2.6. Table 2.6 Films Winning the Most Oscars
BLITZER BONUS Rumbling Back
Movie
Year
Number of Academy Awards
BenHur
1960
11
Titanic
1998
11
The Lord of the Rings: The Return of the King
2003
11
West Side Story
1961
10
Source: Russell Ash, The Top 10 of Everything, 2011
We can use the information in Table 2.6 to define a function. Let the domain of the function be the set of four movies shown in the table. Let the range be the number of Academy Awards for each of the respective films. The function can be written as follows: f : {(Ben@Hur, 11), (Titanic, 11), (The Lord of the Rings, 11), (West Side Story, 10)}. Ansel Elgort as Tony and Rachel Zegler as Maria in the 2020 remake of West Side Story
Rumbling into movie theaters in December 2020, Steven Spielberg’s $100M musical project explored the forbidden love and rivalry between the Jets and the Sharks, two teenage street gangs of different ethnic backgrounds (Puerto Rican and Polish, a change in the cultural aspect of the story). Tony Kushner (Angels in America) penned the new adaptation of the 1957 Broadway musical, which left the entire Leonard BernsteinStephen Sondheim score intact.
Now let’s “undo” f by interchanging the first and second components in each of the ordered pairs. Switching the inputs and outputs of f, we obtain the following relation: 5COGƂTUVEQORQPGPV
Undoing f : {(11, BenHur), (11, Titanic), (11, The Lord of the Rings), (10, West Side Story)}. &KHHGTGPVUGEQPFEQORQPGPVU
Can you see that this relation is not a function? Three of its ordered pairs have the same first component and different second components. This violates the definition of a function. If a function f is a set of ordered pairs, (x, y), then the changes produced by f can be “undone” by reversing the components of all the ordered pairs. The resulting relation, (y, x), may or may not be a function. In this section, we will develop these ideas by studying functions whose compositions have a special “undoing” relationship.
344
Chapter 2 Functions and Graphs
Inverse Functions Here are two functions that describe situations related to the price of a computer, x: f (x) = x  300
g(x) = x + 300.
Function f subtracts $300 from the computer’s price and function g adds $300 to the computer’s price. Let’s see what f (g(x)) does. Put g(x) into f: f(x) = x − 300
This is the given equation for f.
4GRNCEGxYKVJg x
f (g(x)) = g(x) − 300 = x + 300 − 300 = x.
6JKUKUVJGEQORWVGToU QTKIKPCNRTKEG
Because g(x) = x + 300, replace g(x) with x + 300.
Using f (x) = x  300 and g(x) = x + 300, we see that f (g(x)) = x. By putting g(x) into f and finding f (g(x)), the computer’s price, x, went through two changes: the first, an increase; the second, a decrease: x + 300  300. The final price of the computer, x, is identical to its starting price, x. In general, if the changes made to x by a function g are undone by the changes made by a function f, then f (g(x)) = x. Assume, also, that this “undoing” takes place in the other direction: g( f (x)) = x. Under these conditions, we say that each function is the inverse function of the other. The fact that g is the inverse of f is expressed by renaming g as f 1, read ;f@inverse.< For example, the inverse functions f (x) = x  300
g(x) = x + 300
are usually named as follows: f (x) = x  300
f 1(x) = x + 300.
We can use partial tables of coordinates for f and f 1 to gain numerical insight into the relationship between a function and its inverse function. %QORWVGToUTGIWNCTRTKEG
TGFWEVKQP
2TKEGYKVJTGFWEVKQP
RTKEGKPETGCUG
x
f(x) = x − 300
x
f −1(x) = x + 300
1200
900
900
1200
1300
1000
1000
1300
1400
1100
1100
1400
1TFGTGFRCKTUHQTf
1TFGTGFRCKTUHQTf –
The tables illustrate that if a function f is the set of ordered pairs (x, y), then its inverse, f 1, is the set of ordered pairs (y, x). Using these tables, we can see how one function’s changes to x are undone by the other function: (f –1 ∘ f)(1300) = f –1(f (1300)) = f –1(1000) = 1300.
6JGVCDNGHQTfUJQYU VJCVf =
6JGVCDNGHQTf –UJQYU VJCVf – =
The final price of the computer, $1300, is identical to its starting price, $1300.
Section 2.7 Inverse Functions
345
With these ideas in mind, we present the formal definition of the inverse of a function:
GREAT QUESTION Is the −1 in f −1 an exponent? The notation f 1 represents the inverse function of f. The  1 is not an exponent. The notation f 1 1 does not mean : f f
1 ≠ . f
1
Definition of the Inverse of a Function Let f and g be two functions such that f (g(x)) = x
for every x in the domain of g
g( f (x)) = x
for every x in the domain of f.
and The function g is the inverse of the function f and is denoted by f 1 (read ;f@inversebKPVJG ITGGPTGIKQP
y=b x
x y$GECWUGQH>VJGITGCVGT VJCPU[ODQNUJCFGVJGJCNHRNCPGVQVJGTKIJV QHVJGXGTVKECNNKPG y
Graph each inequality in a rectangular coordinate system: b. x … 2.
Graphing a Nonlinear Inequality in Two Variables Example 4 illustrates that a nonlinear inequality in two variables is graphed in the same way that we graph a linear inequality. Graphing a Nonlinear Inequality in Two Variables
EXAMPLE 4 2
2
Graph: x + y … 9. Solution Step 1 Replace the inequality symbol with = and graph the nonlinear equation. We need to graph x 2 + y2 = 9. The graph is a circle of radius 3 with its center at the origin. The graph is shown in Figure 8.19 as a solid circle because equality is included in the … symbol.
Section 8.5 Systems of Inequalities A BRIEF REVIEW
Circles Centered at the Origin The graph of
913
Step 2 Choose a test point from one of the regions and not from the circle. Substitute its coordinates into the inequality. The circle divides the plane into three parts—the circle itself, the region inside the circle, and the region outside the circle. We need to determine whether the region inside or outside the circle is included in the solution. To do so, we will use the test point (0, 0) from inside the circle.
x 2 + y2 = r 2
x 2 + y2 … 9
is a circle with center at the origin and radius r. For more detail, see Section 2.8, Objective 3.
0 + 0 …9 ? 0 + 0 …9 0 … 9
This is the given inequality.
2 ?
2
Test (0, 0) by substituting 0 for x and 0 for y. Square 0: 02 = 0. Add. This statement is true.
Step 3 If a true statement results, shade the region containing the test point. The true statement tells us that all the points inside the circle satisfy x 2 + y2 … 9. The graph is shown using green shading and a solid blue circle in Figure 8.20. y
y
5 4 3 2 1
5 4 3 2 1 1 2 3 4 5
–5 –4 –3 –2 –1–1
x
–2 –3 –4 –5
3 Use mathematical models
involving linear inequalities.
Table 8.1 (partially repeated) Normal Heart Rates for Children Ages 3 to 15 Age (years)
Awake Rate (beats per minute)
Figure 8.20 The graph of x 2 + y2 … 9
Graph: x 2 + y2 Ú 16.
Modeling with Systems of Linear Inequalities Just as two or more linear equations make up a system of linear equations, two or more linear inequalities make up a system of linear inequalities. A solution of a system of linear inequalities in two variables is an ordered pair that satisfies each inequality in the system. In our next example, we use a system of linear inequalities to model the normal heart rates of children from the section opener. Modeling Heart Rate with Inequalities
EXAMPLE 5
Table 8.1 showing normal heart rates for children ages 3 to 15 is partially repeated in the margin. Figure 8.21 shows a normal heart rate region for children ages 3 to 15 that approximately models the heart rate ranges shown in the table. y
65–110
120
6–11
60–95
110
12–15
55–85
Awake Heart Rate (beats per minute)
3–5
Source: emedicinehealth.com
x
–2 –3 –4 –5
Figure 8.19 Preparing to graph x 2 + y2 … 9
CHECK POINT 4
1 2 3 4 5
–5 –4 –3 –2 –1–1
Normal Awake Heart Rate, Ages 3 to 15
x+y=
100 90
B
80
Normal Heart Rate Region
70
A
60 50
x+y= x 3
5
7 9 11 Age (years)
13
15 Figure 8.21
Chapter 8 Systems of Equations and Inequalities
914 y
Normal Awake Heart Rate, Ages 3 to 15
If x represents the age of a child and y represents the awake heart rate, in beats per minute, the normal heart rate region can be described by the following system of linear inequalities:
Awake Heart Rate (beats per minute)
120 110
x+y=
100 B
90 80
x + y Ú 70 2.6x + y … 115.
Show that point A in Figure 8.21 is a solution of the system of inequalities that describes normal awake heart rate.
Normal Heart Rate Region
70
A
60 50
b
x+y= x 3
5
7 9 11 13 15 Age (years)
Solution Point A has coordinates (12, 70). This means that if a child is 12 years old and has an awake heart rate of 70 beats per minute, then that child’s heart rate is within the normal heart rate region. We can show that (12, 70) satisfies the system of inequalities by substituting 12 for x and 70 for y in each inequality in the system.
Figure 8.21 (repeated)
x + y Ú 70 ? 12 + 70 Ú 70 82 Ú 70,
2.6x + y … 115 ? 2.61 122 + 70 … 115 true
?
31.2 + 70 … 115 101.2 … 115,
true
The coordinates (12, 70) make each inequality true. Thus, (12, 70) satisfies the system for the normal awake heart rate region and is a solution of the system. CHECK POINT 5 Show that point B in Figure 8.21 is a solution of the system of inequalities that describes normal awake heart rate.
4 Graph a system of inequalities.
Graphing Systems of Linear Inequalities The solution set of a system of linear inequalities in two variables is the set of all ordered pairs that satisfy each inequality in the system. Thus, to graph a system of inequalities in two variables, begin by graphing each individual inequality in the same rectangular coordinate system. Then find the region, if there is one, that is common to every graph in the system. This region of intersection gives a picture of the system’s solution set.
EXAMPLE 6
Graphing a System of Linear Inequalities
Graph the solution set of the system: b
x  y 6 1 2x + 3y Ú 12.
Solution Replacing each inequality symbol with an equal sign indicates that we need to graph x  y = 1 and 2x + 3y = 12. We can use intercepts to graph these lines. x−y=1 5GVy=KP x−0=1 GCEJGSWCVKQP x=1 The line passes through (1, 0).
xintercept:
5GVx=KP 0−y=1 GCEJGSWCVKQP –y = 1 y = –1 The line passes through (0, –1).
yintercept:
2x + 3y = 12 2x + 3 ∙ 0 = 12 2x = 12 x=6 The line passes through (6, 0). xintercept:
2 ∙ 0 + 3y = 12 3y = 12 y=4 The line passes through (0, 4). yintercept:
Now we are ready to graph the solution set of the system of linear inequalities.
Section 8.5 Systems of Inequalities #FFVJGITCRJQHx+y≥6JGTGFNKPG x+y=KUUQNKF'SWCNKV[KUKPENWFGFKP x+y≥$GECWUG OCMGUVJGKPGSWCNKV[ HCNUG ∙+∙≥QT≥KUHCNUG UJCFGVJGJCNHRNCPGPQVEQPVCKPKPI WUKPI ITGGPXGTVKECNUJCFKPI
)TCRJx−yb
Major axis
'PFRQKPVUQHOCLQT CZKUCTGaWPKVUTKIJVCPF aWPKVUNGHVQHEGPVGT
b2
+
(h, k) x
8GTVGZ h−ak
Foci are c units right and c units left of center, where c2 = a2 − b2. (x − h)2
8GTVGZ h+ak
(QEWU h−ck
(y − k)2 a2
=1
(h, k)
Parallel to the yaxis, vertical
(h, k − a) (h, k + a)
(QEWU h+ck
8GTVGZ hk+a
y
(QEWU hk+c
a>b (h, k)
'PFRQKPVUQHVJGOCLQT CZKUCTGaWPKVUCDQXGCPF aWPKVUDGNQYVJGEGPVGT
(QEWU hk−c x
Foci are c units above and c units below the center, where c2 = a2 − b2.
8GTVGZ hk−a Major axis
EXAMPLE 4 Graph:
Graphing an Ellipse Centered at (h, k)
(x  1)2 (y + 2)2 + = 1. Where are the foci located? 4 9
Solution To graph the ellipse, we need to know its center, (h, k). In the standard forms of equations centered at (h, k), h is the number subtracted from x and k is the number subtracted from y. 6JKUKU x−h YKVJh=
(x − 1)2 4
6JKUKU y−k YKVJk=–
+
(y − (–2))2 9
=1
We see that h = 1 and k = 2. Thus, the center of the ellipse, (h, k), is (1, 2). We can graph the ellipse by locating endpoints on the major and minor axes. To do this, we must identify a2 and b2. (x − 1)2 4
+
b=6JKUKUVJG UOCNNGTQHVJGVYQ FGPQOKPCVQTU
(y + 2)2 9
=1
a=6JKUKUVJG NCTIGTQHVJGVYQ FGPQOKPCVQTU
The larger number is under the expression involving y. This means that the major axis is vertical and parallel to the y@axis.
Section 10.1 The Ellipse y 5 4 3 2 1 –5 –4 –3 –2 –1–1 (–1, –2)
1019
We can sketch the ellipse by locating endpoints on the major and minor axes. (x − 1)2 22 (1, 1) 1 2 3 4 5
x− y+ + = Figure 10.10 The graph of an ellipse centered at (1,  2)
(y + 2)2 32
=1
'PFRQKPVUQHVJGOCLQT CZKU VJGXGTVKEGUCTG WPKVUWRCPFFQYP HTQOVJGEGPVGT
'PFRQKPVUQHVJGOKPQT CZKUCTGWPKVUVQVJG TKIJVCPFNGHVQHVJG EGPVGT
x
–2 (1, –2) (3, –2) –3 –4 –5 (1, –5)
+
We categorize the observations in the voice balloons as follows: For a Vertical Major Axis with Center (1, −2) WPKVU CDQXGCPF DGNQYEGPVGT
Vertices
Endpoints of Minor Axis
(1, –2 + 3) = (1, 1)
(1 + 2, –2) = (3, –2)
(1, –2 − 3) = (1, –5)
(1 − 2, –2) = (–1, –2)
WPKVU TKIJVCPF NGHVQHEGPVGT
Using the center and these four points, we can sketch the ellipse shown in Figure 10.10. With c 2 = a2  b2, we have c 2 = 9  4 = 5. So the foci are located 25 units above and below the center, at 1 1, 2 + 25 2 and 1 1, 2  25 2 . CHECK POINT 4
A BRIEF REVIEW
Completing the Square To complete the square on x 2 + bx, take half the coefficient of x. Then square this number. By adding the square of half the coefficient of x, a perfect square trinomial will result. Once you’ve completed the square, remember that changes made on the left side of the equation must also be made on the right side of the equation. For more detail, see Section 1.5, Objective 3, and Section 2.8, Objective 5.
Graph:
(x + 1)2 (y  2)2 + = 1. Where are the foci located? 9 4
In some cases, it is necessary to convert the equation of an ellipse to standard form by completing the square on x and y. For example, suppose that we wish to graph the ellipse whose equation is 9x 2 + 4y2  18x + 16y  11 = 0. Because we plan to complete the square on both x and y, we need to rearrange terms so that • x@terms are arranged in descending order. • y@terms are arranged in descending order. • the constant term appears on the right. 9x 2 + 4y2  18x + 16y  11 = 0 (9x 2  18x) + (4y2 + 16y) = 11
This is the given equation. Group terms and add 11 to both sides.
9(x 2  2x + □) + 4(y2 + 4y + □) = 11 9GCFFGF∙QT VQVJGNGHVUKFG
9GCNUQCFFGF∙QT VQVJGNGHVUKFG
9(x2 − 2x + 1) + 4(y2 + 4y + 4) = 11 + 9 + 16 CPFCFFGFQPVJGNGHVUKFG OWUVCNUQDGCFFGFQPVJGTKIJVUKFG
9(x  1)2 + 4(y + 2)2 = 36 4(y + 2) 9(x  1) 36 + = 36 36 36 2 2 (x  1) (y + 2) + = 1 4 9 2
To complete the square, coefficients of x2 and y2 must be 1. Factor out 9 and 4, respectively. Complete each square by adding the square of half the coefficient of x and y, respectively. Factor.
2
Divide both sides by 36.
Simplify.
The equation is now in standard form. This is precisely the form of the equation that we graphed in Example 4.
1020
Chapter 10 Conic Sections and Analytic Geometry
EXAMPLE 5
Graphing an Ellipse Centered at (h, k) by Completing the Square
Graph: 4x 2 + 36y2 + 40x  288y + 532 = 0. Where are the foci located? Solution
We begin by completing the square on x and y.
4x 2 + 36y2 + 40x  288y + 532 = 0
14x 2 + 40x2 + 1 36y2  288y2 = 532
4(x 2 + 10x + □) + 36(y2  8y + □) = 532
This is the given equation. Group terms and subtract 532 from both sides. To complete the square, coefficients of x2 and y2 must be 1. Factor out 4 and 36, respectively.
9GCFFGF∙QT VQVJGNGHVUKFG
9GCFFGF∙QT VQVJGNGHVUKFG
4(x2 + 10x + 25) + 36(y2 − 8y + 16) = –532 + 100 + 576 CPFCFFGFQPVJGNGHVUKFG OWUVCNUQDGCFFGFQPVJGTKIJVUKFG
Complete the square by adding the square of half the coefficient of x and y, respectively.
41x + 52 2 + 361y  42 2 = 144
Factor. Simplify the right side.
41x + 52 2 361y  42 2 144 + = 144 144 144
Divide both sides by 144.
1 x + 52 2 1y  42 2 + = 1 36 4
Simplify.
The larger number is under the expression involving x. This means that the major axis is horizontal and parallel to the xaxis. We can express this last equation in the form 1x  h 2 2 a2
+
1y  k22
(x − (–5))2 6
'PFRQKPVUQHVJGOCLQT CZKU VJGXGTVKEGUCTG WPKVUVQVJGTKIJV CPFNGHVQHVJGEGPVGT
= 1.
k=
h=–
2
b2
+
(y − 4)2 22
=1
The center of the ellipse, (h, k), is ( − 5, 4).
'PFRQKPVUQHVJGOKPQT CZKUCTGWPKVU WRCPFFQYP HTQOVJGEGPVGT
We categorize the observations in the voice balloons as follows: For a Horizontal Major Axis with Center (−5, 4) WPKVU TKIJVCPF NGHVQHEGPVGT
Vertices
Endpoints of Minor Axis
(–5 + 6, 4) = (1, 4)
(–5, 4 + 2) = (–5, 6)
(–5 − 6, 4) = (–11, 4)
(–5, 4 − 2) = (–5, 2)
WPKVU CDQXGCPF DGNQYEGPVGT
Section 10.1 The Ellipse
1021
Using the center and these four points, we can sketch the ellipse shown in Figure 10.11. y 7 6 5 4 3 2 1
(–5, 6) (–5, 4) (–11, 4) (–5, 2) –12
–10
–8
–6
–4
–2 –1
(1, 4)
2
4
x
Figure 10.11 The graph of an ellipse centered at (  5, 4)
Now let’s locate the foci. c 2 = a 2  b2
Use the formula for locating the foci.
c = 36  4
a2 = 36 and b2 = 4.
c 2 = 32
Simplify.
2
c = 232 = 422
Take the principal square root and simplify: 232 = 216 # 2 = 21622 = 422.
The foci are located 422 units right and left of the center, at ( 5 + 422, 4) and ( 5  422, 4), or approximately at (0.7, 4) and ( 10.7, 4). CHECK POINT 5
Graph: x 2 + 4y2 + 10x + 24y + 45 = 0. Where are the
foci located?
4 Solve applied problems involving ellipses.
Applications Ellipses have many applications. German scientist Johannes Kepler (1571–1630) showed that the planets in our solar system move in elliptical orbits, with the Sun at a focus. Earth satellites also travel in elliptical orbits, with Earth at a focus.
Venus
Earth
Mars
Mercury
Planets move in elliptical orbits.
Whispering in an elliptical dome
One intriguing aspect of the ellipse is that a ray of light or a sound wave emanating from one focus will be reflected from the ellipse to exactly the other focus. A whispering gallery is an elliptical room with an elliptical, domeshaped ceiling. People standing at the foci can whisper and hear each other quite clearly, while persons in other locations in the room cannot hear them. Statuary Hall in the U.S. Capitol Building is elliptical. President John Quincy Adams, while a member of the House of Representatives, was aware of this acoustical phenomenon. He situated his desk at a focal point of the elliptical ceiling, easily eavesdropping on the private conversations of other House members located near the other focus.
1022
Chapter 10 Conic Sections and Analytic Geometry
The elliptical reflection principle is used in a procedure for disintegrating kidney stones. The patient is placed within a device that is elliptical in shape. The patient is placed so the kidney is centered at one focus, while ultrasound waves from the other focus hit the walls and are reflected to the kidney stone. The convergence of the ultrasound waves at the kidney stone causes vibrations that shatter it into fragments. The small pieces can then be passed painlessly through the patient’s system. The patient recovers in days, as opposed to up to six weeks if surgery is used instead. Ellipses are often used for supporting arches of bridges and in tunnel construction. This application forms the basis of our next example.
BLITZER BONUS Halley’s Comet
Halley’s Comet has an elliptical orbit with the Sun at one focus. The comet returns every 76.3 years. The first recorded sighting was in 239 b.c. It was last seen in 1986. At that time, spacecraft went close to the comet, measuring its nucleus to be 7 miles long and 4 miles wide. By 2024, Halley’s Comet will have reached the farthest point in its elliptical orbit before returning to be next visible from Earth as it loops around the Sun in 2062.
EXAMPLE 6
F2
Disintegrating kidney stones
An Application Involving an Ellipse
A semielliptical archway over a oneway road has a height of 10 feet and a width of 40 feet (see Figure 10.12). Your truck has a width of 10 feet and a height of 9 feet. Will your truck clear the opening of the archway?
40 ft 10 ft
Solution Because your truck’s width is 10 feet, to determine the clearance, we must find the height of the Figure 10.12 archway 5 feet from the center. If that height is 9 feet or A semielliptical archway less, the truck will not clear the opening. In Figure 10.13, we’ve constructed a coordinate system with the x@axis on the ground and the origin at the center of the archway. Also shown is the truck, whose height is 9 feet.
y (0, 10)
Truck (–20, 0)
5 20
Neptune
F1
9 ft (20, 0)
5
x
20
Figure 10.13
76 years
Earth Sun
Jupiter Uranus The elliptical orbit of Halley’s Comet
y2 x2 + = 1, we can express the equation of the blue archway a2 b2 y2 y2 x2 x2 in Figure 10.13 as 2 + 2 = 1, or + = 1. 400 100 20 10 As shown in Figure 10.13, the edge of the 10footwide truck corresponds to x = 5. We find the height of the archway 5 feet from the center by substituting 5 for x and solving for y. Using the equation
Section 10.1 The Ellipse
400a
y2 52 + = 1 400 100
Substitute 5 for x in
y2 25 + = 1 400 100
Square 5.
y2 25 + b = 400(1) 400 100
y2 x2 + = 1. 400 100
Clear fractions by multiplying both sides by 400.
25 + 4y2 = 400
Instructor Resources for Section 10.1 in MyLab Math
1023
Use the distributive property and simplify.
2
4y = 375 375 y2 = 4 375 y = A 4
Subtract 25 from both sides. Divide both sides by 4. Take only the positive square root. The archway is above the x@axis, so y is nonnegative.
≈ 9.68
Use a calculator.
Thus, the height of the archway 5 feet from the center is approximately 9.68 feet. Because your truck’s height is 9 feet, there is enough room for the truck to clear the archway. CHECK POINT 6 Will a truck that is 12 feet wide and has a height of 9 feet clear the opening of the archway described in Example 6?
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. The set of all points in a plane the sum of whose distances from two fixed points is constant is a/an . The two fixed points are called the . The midpoint of the line segment connecting the two fixed points is the .
C4. Consider an ellipse centered at the origin whose major axis is vertical. The equation of this ellipse in standard form indicates that a2 = 9 and b2 = 4. Thus, c 2 = . The foci are located at and .
C2. Consider the following equation in standard form: C5. The graph of y2 x2 + = 1. 25 9 The value of a2 is , so the x@intercepts are and . The graph passes through and , which are the vertices. The value of b2 is , so the y@intercepts are and . The graph passes through and . C3. Consider the following equation in standard form: 2
2
y x + = 1. 9 25 The value of a2 is , so the y@intercepts are and . The graph passes through and , which are the vertices. The value of b2 is , so the x@intercepts are and . The graph passes through and .
at
.
(x + 1)2 25
+
(y  4)2 9
= 1 has its center
C6. If the center of an ellipse is (3, 2), the major axis is horizontal and parallel to the x@axis, and the distance from the center of the ellipse to its vertices is a = 5 units, then the coordinates of the vertices are and . C7. If the foci of an ellipse are located at (1, 6) and (1, 12), then the coordinates of the center of the ellipse are . C8. In the equation 3(x 2 + 4x) + 4(y2  2y) = 32, we complete the square on x by adding within the first parentheses. We complete the square on y by adding within the second parentheses. Thus, we must add to the right side of the equation.
1024
Chapter 10 Conic Sections and Analytic Geometry
10.1 EXERCISE SET Practice Exercises
In Exercises 25–36, find the standard form of the equation of each ellipse satisfying the given conditions.
In Exercises 1–18, graph each ellipse and locate the foci. 1.
y2 x2 + = 1 16 4
2.
y2 x2 + = 1 25 16
3.
y2 x2 + = 1 9 36
4.
y2 x2 + = 1 16 49
5.
y2 x2 + = 1 25 64
6.
y2 x2 + = 1 49 36
7.
y2 x2 + = 1 49 81
8.
y2 x2 + = 1 64 100
9.
x2 9 4
+
y2 25 4
= 1
10.
x2 81 4
+
y2 25 16
25. 26. 27. 28. 29. 30. 31. 32.
= 1
33.
11. x 2 = 1  4y2
12. y2 = 1  4x 2
13. 25x 2 + 4y2 = 100
14. 9x 2 + 4y2 = 36
15. 4x 2 + 16y2 = 64
16. 4x 2 + 25y2 = 100
17. 7x 2 = 35  5y2
18. 6x 2 = 30  5y2
34. 35. 36.
In Exercises 19–24, find the standard form of the equation of each ellipse and give the location of its foci. y
19. 4 3 2 1
4 3 2 1 1 2 3 4
–4 –3 –2 –1–1
1 2 3 4
41. 42.
x
1 2 3 4
–4 –3 –2 –1–1
–2 –3 –4
x
–2 –3 –4 y
45.
4 3 2 1 1 2 3 4
x –4 –3 –2 –1–1 (–1, –1)
–2 –3 –4
43. 44.
y
24.
4 (–1, 1) 3 2 1
9 (x  1)2 16
+ +
(y  1)2 4 (y + 2)2 9
= 1 = 1
40. (x  3)2 + 9(y + 2)2 = 18 y
22.
38.
(x  2)2
39. (x + 3)2 + 4(y  2)2 = 16
4 3 2 1
–4 –3 –2 –1–1
–2 –3 –4
x
–2 –3 –4
4 3 2 1
–4 –3 –2 –1–1
1 2 3 4
–4 –3 –2 –1–1
y
23.
37.
x
–2 –3 –4
21.
In Exercises 37–50, graph each ellipse and give the location of its foci.
y
20.
Foci: (  5, 0), (5, 0); vertices: (  8, 0), (8, 0) Foci: (  2, 0), (2, 0); vertices: (  6, 0), (6, 0) Foci: (0,  4), (0, 4); vertices: (0,  7), (0, 7) Foci: (0,  3), (0, 3); vertices: (0,  4), (0, 4) Foci: (  2, 0), (2, 0); y@intercepts: 3 and 3 Foci: (0,  2), (0, 2); x@intercepts: 2 and 2 Major axis horizontal with length 8; length of minor axis = 4; center: (0, 0) Major axis horizontal with length 12; length of minor axis = 6; center: (0, 0) Major axis vertical with length 10; length of minor axis = 4; center: (  2, 3) Major axis vertical with length 6; length of minor axis = 4; center: (8, 6) Endpoints of major axis: (7, 9) and (7, 3) Endpoints of minor axis: (5, 6) and (9, 6) Endpoints of major axis: (  8, 9) and ( 8, 3) Endpoints of minor axis: (  6, 6) and ( 10, 6)
46.
1 2 3 4
x
47. 48.
(x  4)2 9 (x  3)2 9
+ +
(y + 2)2 25 (y + 1)2 16
= 1 = 1
(y  2)2 x2 + = 1 25 36 (x  4)2 4 (x + 3)2 9 (x + 2)2 16 (x  1)2 2 (x + 1)2 2
+
y2 = 1 25
+ (y  2)2 = 1 + (y  3)2 = 1 + +
(y + 3)2 5 (y  3)2 5
= 1 = 1
49. 9(x  1)2 + 4(y + 3)2 = 36 50. 36(x + 4)2 + (y + 3)2 = 36
1025
Section 10.1 The Ellipse In Exercises 51–60, convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci.
71. The elliptical ceiling in Statuary Hall in the U.S. Capitol Building is 96 feet long and 23 feet tall. y
51. 9x 2 + 25y2  36x + 50y  164 = 0 52. 4x 2 + 9y2  32x + 36y + 64 = 0
(0, 23)
53. 9x 2 + 16y2  18x + 64y  71 = 0 54. x 2 + 4y2 + 10x  8y + 13 = 0 55. 4x 2 + y2 + 16x  6y  39 = 0 56. 4x + 25y  24x + 100y + 36 = 0 2
2
57. 25x 2 + 4y2  150x + 32y + 189 = 0 58. 49x + 16y + 98x  64y  671 = 0 2
2
(48, 0)
x
a. Using the rectangular coordinate system in the figure shown, write the standard form of the equation of the elliptical ceiling.
59. 36x 2 + 9y2  216x = 0 60. 16x 2 + 25y2  300y + 500 = 0
Practice PLUS In Exercises 61–66, find the solution set for each system by graphing both of the system’s equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations. 61. e
(–48, 0)
62. e
x 2 + y2 = 1 x 2 + 9y2 = 9
x 2 + y2 = 25 25x 2 + y2 = 25
y2 x2 + = 1 63. • 25 9 y = 3
y2 x2 + = 1 64. • 4 36 x = 2
65. e
66. e
4x 2 + y2 = 4 2x  y = 2
4x 2 + y2 = 4 x + y = 3
In Exercises 67–68, graph each semiellipse. 67. y =  216  4x 2
68. y =  24  4x 2
b. John Quincy Adams discovered that he could overhear the conversations of opposing party leaders near the left side of the chamber if he situated his desk at the focus at the right side of the chamber. How far from the center of the ellipse along the major axis did Adams situate his desk? (Round to the nearest foot.) 72. If an elliptical whispering room has a height of 30 feet and a width of 100 feet, where should two people stand if they would like to whisper back and forth and be heard?
Explaining the Concepts 73. What is an ellipse? 74. Describe how to graph
y2 x2 + = 1. 25 16
75. Describe how to locate the foci for
y2 x2 + = 1. 25 16
76. Describe one similarity and one difference between the graphs of
y2 y2 x2 x2 + = 1 and + = 1. 25 16 16 25
77. Describe one similarity and one difference between the
Application Exercises 69. Will a truck that is 8 feet wide carrying a load that reaches 7 feet above the ground clear the semielliptical arch on the oneway road that passes under the bridge shown in the figure?
graphs of
(x  1)2 (y  1)2 y2 x2 + = 1 and + = 1. 25 16 25 16
78. An elliptipool is an elliptical pool table with only one pocket. A pool shark places a ball on the table, hits it in what appears to be a random direction, and yet it bounces off the edge, falling directly into the pocket. Explain why this happens.
10 ft 30 ft
70. A semielliptic archway has a height of 20 feet and a width of 50 feet, as shown in the figure. Can a truck 14 feet high and 10 feet wide drive under the archway without going into the other lane?
50 ft 20 ft
Technology Exercises 79. Use a graphing utility to graph any five of the ellipses that you graphed by hand in Exercises 1–18. 80. Use a graphing utility to graph any three of the ellipses that you graphed by hand in Exercises 37–50. First solve the given equation for y by using the square root property. Enter each of the two resulting equations to produce each half of the ellipse.
1026
Chapter 10 Conic Sections and Analytic Geometry
81. Use a graphing utility to graph any one of the ellipses that you graphed by hand in Exercises 51–60. Write the equation as a quadratic equation in y and use the quadratic formula to solve for y. Enter each of the two resulting equations to produce each half of the ellipse. 82. Write an equation for the path of each of the following elliptical orbits. Then use a graphing utility to graph the two ellipses in the same viewing rectangle. Can you see why early astronomers had difficulty detecting that these orbits are ellipses rather than circles? • Earth’s orbit: Length of major axis: 186 million miles Length of minor axis: 185.8 million miles • Mars’s orbit: Length of major axis: 283.5 million miles Length of minor axis: 278.5 million miles
Critical Thinking Exercises Make Sense? In Exercises 83–86, determine whether each statement makes sense or does not make sense, and explain your reasoning. 83. I graphed an ellipse with a horizontal major axis and foci on the y@axis. 84. I graphed an ellipse that was symmetric about its major axis but not symmetric about its minor axis. 85. You told me that an ellipse centered at the origin has vertices at (  5, 0) and (5, 0), so I was able to graph the ellipse. 86. In a whispering gallery at our science museum, I stood at one focus, my friend stood at the other focus, and we had a clear conversation, very little of which was heard by the 25 museum visitors standing between us. 87. Find the standard form of the equation of an ellipse with vertices at (0,  6) and (0, 6), passing through (2,  4). 88. An Earth satellite has an elliptical orbit described by 2
2
y x + = 1. (5000)2 (4750)2 Satellite
Apogee
Perigee
Write the equation for each circle shown in the figure. y
x
y2 x2 90. What happens to the shape of the graph of 2 + 2 = 1 as a b c S 0, where c 2 = a2  b2? a
Retaining the Concepts 91. Solve by eliminating variables: x  6y = 22 • 2x + 4y  3z = 29 3x  2y + 5z = 17. (Section 8.2, Example 3) 92. Graph the solution set of the system: 2x + y … 4 • x 7 3 y Ú 1. (Section 8.5, Example 8) 93. Where possible, find each product. 1 0 1 0 a. c dc d 0 1 0 1 b. c
1 0
0 1 dc 1 0
0 1
1 d 1
(Section 9.3, Example 7) 94. Use the Law of Sines to solve triangle ABC if A = 35°, a = 11, and b = 15. Assume B is acute. Round lengths of sides to the nearest tenth and angle A measures to the nearest degree. (Section 7.1, Example 3)
C b = 15
a = 11
35° B
Preview Exercises (All units are in miles.) The coordinates of the center of Earth are (16, 0). a. The perigee of the satellite’s orbit is the point that is nearest Earth’s center. If the radius of Earth is approximately 4000 miles, find the distance of the perigee above Earth’s surface. b. The apogee of the satellite’s orbit is the point that is the greatest distance from Earth’s center. Find the distance of the apogee above Earth’s surface. 89. The equation of the red ellipse in the figure shown is y2 x2 + = 1. 25 9
Exercises 95–97 will help you prepare for the material covered in the next section. 95. Divide both sides of 4x 2  9y2 = 36 by 36 and simplify. How does the simplified equation differ from that of an ellipse? y2 x2 = 1. 16 9 a. Find the x@intercepts. b. Explain why there are no y@intercepts.
96. Consider the equation
y2 x2 = 1. 9 16 a. Find the y@intercepts. b. Explain why there are no x@intercepts.
97. Consider the equation
Section 10.2 The Hyperbola
SECTION 10.2
The Hyperbola Conic sections are often used to create unusual architectural designs. The top of St. Mary’s Cathedral in San Francisco is a 2135cubicfoot dome with walls rising 200 feet above the floor and supported by four massive concrete pylons that extend 94 feet into the ground. Cross sections of the roof are parabolas and hyperbolas. In this section, we study the curve with two parts known as the hyperbola.
WHAT YOU’LL LEARN 1 Locate a hyperbola’s vertices and foci.
2 Write equations of
hyperbolas in standard form.
3 Graph hyperbolas centered
Definition of a Hyperbola
at the origin.
Figure 10.14 shows a cylindrical lampshade casting two shadows on a wall. These shadows indicate the distinguishing feature of hyperbolas: Their St. Mary’s Cathedral graphs contain two disjoint parts, called branches. Although each branch might look like a parabola, its shape is actually quite different. The definition of a hyperbola is similar to that of an ellipse. For an ellipse, the sum of the distances to the foci is a constant. By contrast, for a hyperbola the difference of the distances to the foci is a constant.
4 Graph hyperbolas not
centered at the origin.
5 Solve applied problems involving hyperbolas.
y
6TCPUXGTUG CZKU
8GTVGZ
Figure 10.14 Casting hyperbolic shadows
Definition of a Hyperbola 8GTVGZ x
(QEWU
1027
(QEWU
%GPVGT
A hyperbola is the set of points in a plane the difference of whose distances from two fixed points, called foci, is constant. Figure 10.15 illustrates the two branches of a hyperbola. The line through the foci intersects the hyperbola at two points, called the vertices. The line segment that joins the vertices is the transverse axis. The midpoint of the transverse axis is the center of the hyperbola. Notice that the center lies midway between the vertices, as well as midway between the foci.
Figure 10.15 The two branches of a hyperbola
Standard Form of the Equation of a Hyperbola
y
P(x, y) d2 F1(–c, 0)
d1 (0, 0)
%GPVGT
Figure 10.16
F2(c, 0)
x
The rectangular coordinate system enables us to translate a hyperbola’s geometric definition into an algebraic equation. Figure 10.16 is our starting point for obtaining an equation. We place the foci, F 1 and F 2 , on the x@axis at the points ( c, 0) and (c, 0). Note that the center of this hyperbola is at the origin. We let (x, y) represent the coordinates of any point, P, on the hyperbola. What does the definition of a hyperbola tell us about the point (x, y) in Figure 10.16? For any point (x, y) on the hyperbola, the absolute value of the difference of the distances from the two foci, d 2  d 1 , must be constant. We denote this constant by 2a, just as we did for the ellipse. Thus, the point (x, y) is on the hyperbola if and only if d 2  d 1 = 2a. 0 2(x + c) + (y  0)  2(x  c) + (y  0)2 0 = 2a 2
2
2
Use the distance formula.
1028
Chapter 10 Conic Sections and Analytic Geometry
After eliminating radicals and simplifying, we obtain (c 2  a2)x 2  a2 y2 = a2(c 2  a2). For convenience, let b2 = c 2  a2. Substituting b2 for c 2  a2 in the preceding equation, we obtain b2 x 2  a 2 y 2 = a 2 b2 . a2 y 2 a 2 b2 b2 x 2 = a 2 b2 a 2 b2 a 2 b2 y2 x2  2 = 1 2 a b
Divide both sides by a 2 b 2.
Simplify.
This last equation is called the standard form of the equation of a hyperbola centered at the origin. There are two such equations. The first is for a hyperbola in which the transverse axis lies on the x@axis. The second is for a hyperbola in which the transverse axis lies on the y@axis. GREAT QUESTION Which equation do I need to use for locating the foci of a hyperbola? The form c 2 = a2 + b2 is the one you should remember. When finding the foci, this form is easy to manipulate.
Standard Forms of the Equations of a Hyperbola The standard form of the equation of a hyperbola with center at the origin is y2 x2 = 1 a2 b2
or
y2 a2

x2 = 1. b2
Figure 10.17(a) illustrates that for the equation on the left, the transverse axis lies on the x@axis. Figure 10.17(b) illustrates that for the equation on the right, the transverse axis lies on the y@axis. The vertices are a units from the center and the foci are c units from the center. For both equations, b2 = c 2  a2. Equivalently, c 2 = a 2 + b2 . y
y
(0, c)
(–a, 0)
6TCPUXGTUG CZKU
(0, a) (a, 0) x
DISCOVERY Apply the tests for symmetry, found in Section 2.2, to each of the standard forms in Figure 10.17(a) and Figure 10.17(b). What can you conclude about the graphs of all hyperbolas centered at the origin?
(–c, 0)
(c, 0)
y x − = a b
6TCPUXGTUG CZKU
x (0, –a)
x y − = a b Figure 10.17(a) Transverse axis lies on the x@axis.
(0, –c) Figure 10.17(b) Transverse axis lies on the y@axis.
GREAT QUESTION How can I tell from a hyperbola’s equation whether the transverse axis is horizontal or vertical? When the x 2@term is preceded by a plus sign, the transverse axis is horizontal. When the y2@term is preceded by a plus sign, the transverse axis is vertical.
1 Locate a hyperbola’s vertices and foci.
Using the Standard Form of the Equation of a Hyperbola We can use the standard form of the equation of a hyperbola to find its vertices and locate its foci. Because the vertices are a units from the center, begin by identifying a2 in the equation. In the standard form of a hyperbola’s equation, a2 is the number
Section 10.2 The Hyperbola
GREAT QUESTION What’s the difference between the equations for locating an ellipse’s foci and locating a hyperbola’s foci? Notice the sign difference between the following equations:
1029
under the variable whose term is preceded by a plus sign ( +). If the x 2@term is preceded by a plus sign, the transverse axis lies along the x@axis. Thus, the vertices are a units to the left and right of the origin. If the y2@term is preceded by a plus sign, the transverse axis lies along the y@axis. Thus, the vertices are a units above and below the origin. We know that the foci are c units from the center. The substitution that is used to derive the hyperbola’s equation, c 2 = a2 + b2, is needed to locate the foci when a2 and b2 are known.
Finding an ellipse’s foci: c 2 = a2  b2
EXAMPLE 1
Finding a hyperbola’s foci: c 2 = a2 + b2.
Find the vertices and locate the foci for each of the following hyperbolas with the given equation: y2 y2 x2 x2 = 1 b. = 1. a. 16 9 9 16 Solution Both equations are in standard form. We begin by identifying a2 and b2 in each equation. y2 x2 a. The first equation is in the form 2  2 = 1. a b
y 5 4 8GTVGZ 3
– 2 1
y2 x2 − =1 16 9
8GTVGZ
x
1 2 3 4 5
–5 –4 –3 –2 –1–1 (QEWU
–
Finding Vertices and Foci from a Hyperbola’s Equation
–2 –3 –4 –5
(QEWU
a=6JKUKUVJGFGPQOKPCVQT QHVJGVGTORTGEGFGFD[CRNWUUKIP
b=6JKUKUVJGFGPQOKPCVQT QHVJGVGTORTGEGFGFD[COKPWUUKIP
Because the x 2@term is preceded by a plus sign, the transverse axis lies along the x@axis. Thus, the vertices are a units to the left and right of the origin. Based on the standard form of the equation, we know the vertices are ( a, 0) and (a, 0). Because a2 = 16, a = 4. Thus, the vertices are ( 4, 0) and (4, 0), shown in Figure 10.18. We use c 2 = a2 + b2 to find the foci, which are located at ( c, 0) and (c, 0). We know that a2 = 16 and b2 = 9; we need to find c 2 in order to find c.
Figure 10.18 The graph of y2 x2 = 1 16 9
c 2 = a2 + b2 = 16 + 9 = 25 Because c 2 = 25, c = 5. The foci are located at ( 5, 0) and (5, 0). They are shown in Figure 10.18. y 5 4 3 2 1
(QEWU
b. The second given equation is in the form
–5 –4 –3 –2 –1–1
1 2 3 4 5
–2 –3 –4 –5
8GTVGZ –
x
a=6JKUKUVJGFGPQOKPCVQT QHVJGVGTORTGEGFGFD[CRNWUUKIP
(QEWU – 9

x2 = 1. b2
x2 = 1 16
b=6JKUKUVJGFGPQOKPCVQT QHVJGVGTORTGEGFGFD[COKPWUUKIP
y2 x2 = 1 is preceded by a plus sign, the transverse 9 16 axis lies along the y@axis. Thus, the vertices are a units above and below the origin. Based on the standard form of the equation, we know the vertices are (0, a) and (0, a). Because a2 = 9, a = 3. Thus, the vertices are (0, 3) and (0, 3), shown in Figure 10.19. Because the y2@term in
y2
a2

x2 y2 − =1 9 16
8GTVGZ
Figure 10.19 The graph of
y2
1030
Chapter 10 Conic Sections and Analytic Geometry
y 5 4 3 2 1
We use c 2 = a2 + b2 to find the foci, which are located at (0, c) and (0, c).
(QEWU
c 2 = a2 + b2 = 9 + 16 = 25 Because c 2 = 25, c = 5. The foci are located at (0, 5) and (0, 5). They are shown in Figure 10.19.
8GTVGZ
–5 –4 –3 –2 –1–1
1 2 3 4 5
–2 –3 –4 –5
8GTVGZ –
x
CHECK POINT 1 Find the vertices and locate the foci for each of the following hyperbolas with the given equation:
a.
y2 x2 = 1 25 16
b.
y2 x2 = 1. 25 16
(QEWU –
In Example 1, we used equations of hyperbolas to find their foci and vertices. In the next example, we reverse this procedure.
Figure 10.19 (repeated) The graph y2 x2 of = 1 9 16
2 Write equations of hyperbolas in standard form.
y 5 4 3 2 1 –5 –4 –3 –2 –1–1 –2 –3 –4 –5
Solution Because the foci are located at (0, 3) and (0, 3), on the y@axis, the transverse axis lies on the y@axis. The center of the hyperbola is midway between the foci, located at (0, 0). Thus, the form of the equation is
8GTVGZ
(QEWU
–
Finding the Equation of a Hyperbola from Its Foci and Vertices
Find the standard form of the equation of a hyperbola with foci at (0, 3) and (0, 3) and vertices (0, 2) and (0, 2), shown in Figure 10.20.
(QEWU
1 2 3 4 5 8GTVGZ –
EXAMPLE 2
y2
x
a2

x2 = 1. b2
We need to determine the values for a2 and b2. The distance from the center, (0, 0), to either vertex, (0, 2) or (0, 2), is 2, so a = 2. y2 22
Figure 10.20

x2 = 1 b2
or
y2 x2  2 = 1 4 b
We must still find b2. The distance from the center, (0, 0), to either focus, (0, 3) or (0, 3), is 3. Thus, c = 3. Using c 2 = a2 + b2, we have 32 = 22 + b2 and b2 = 32  22 = 9  4 = 5. y2 x2  2 = 1 gives us the standard form of the hyperbola’s 4 b equation. The equation is
Substituting 5 for b2 in
y2 x2 = 1. 4 5 CHECK POINT 2 Find the standard form of the equation of a hyperbola with foci at (0, 5) and (0, 5) and vertices (0, 3) and (0, 3).
The Asymptotes of a Hyperbola As x and y get larger, the two branches of the graph of a hyperbola approach a pair of intersecting straight lines, called asymptotes. The asymptotes pass through the center of the hyperbola and are helpful in graphing hyperbolas.
1031
Section 10.2 The Hyperbola
Figure 10.21 shows the asymptotes for the graphs of hyperbolas centered at the origin. The asymptotes pass through the corners of a rectangle. Note that the dimensions of this rectangle are 2a by 2b. The line segment of length 2b is the conjugate axis of the hyperbola and is perpendicular to the transverse axis through the center of the hyperbola.
x y − = y a b
y x − = a b
#U[ORVQVG b y= ax
#U[ORVQVG a y= x b
(0, a)
(0, b)
(–a, 0)
y
(a, 0)
(–b, 0)
x
(b, 0)
x
(0, –b) (0, –a) #U[ORVQVG a x b
#U[ORVQVG b y=– ax
y=–
Figure 10.21 Asymptotes of a hyperbola
The Asymptotes of a Hyperbola Centered at the Origin The hyperbola
y2 x2 = 1 has a horizontal transverse axis and two asymptotes a2 b2 y =
The hyperbola
y2 a2

b x a
and
y = 
b x. a
x2 = 1 has a vertical transverse axis and two asymptotes b2 y =
a x b
and
y = 
a x. b
b Why are y = { x the asymptotes for a hyperbola whose transverse axis is a horizontal? The proof can be found in the appendix.
3 Graph hyperbolas centered at the origin.
Graphing Hyperbolas Centered at the Origin Hyperbolas are graphed using vertices and asymptotes. Graphing Hyperbolas 1. Locate the vertices. 2. Use dashed lines to draw the rectangle centered at the origin with sides parallel to the axes, crossing one axis at {a and the other at {b. 3. Use dashed lines to draw the diagonals of this rectangle and extend them to obtain the asymptotes. 4. Draw the two branches of the hyperbola by starting at each vertex and approaching the asymptotes.
1032
Chapter 10 Conic Sections and Analytic Geometry
EXAMPLE 3
Graphing a Hyperbola
Graph and locate the foci:
y2 x2 = 1. What are the equations of the asymptotes? 25 16
Solution
y2 x2 Step 1 Locate the vertices. The given equation is in the form 2  2 = 1, with a b a2 = 25 and b2 = 16.
a=
x2 y2 − =1 25 16
b=
Based on the standard form of the equation with the transverse axis on the x@axis, we know that the vertices are ( a, 0) and (a, 0). Because a2 = 25, a = 5. Thus, the vertices are ( 5, 0) and (5, 0), shown in Figure 10.22. Step 2 Draw a rectangle. Because a2 = 25 and b2 = 16, a = 5 and b = 4. We construct a rectangle to find the asymptotes, using 5 and 5 on the x@axis (the vertices are located here) and 4 and 4 on the y@axis. The rectangle passes through these four points, shown using dashed lines in Figure 10.22. Step 3 Draw extended diagonals for the rectangle to obtain the asymptotes. We draw dashed lines through the opposite corners of the rectangle, shown in Figure 10.22, to obtain the graph of the asymptotes. Based on the standard form of the hyperbola’s equation, the equations for these asymptotes are y = {
y
#U[ORVQVG
8GTVGZ –
TECHNOLOGY 2
2
y x Graph = 1 by solving 25 16 for y: y1 =
216x 2  400 5
y2 = 
216x 2  400 = y1 . 5
y=
√x−
–7 –6
b x a
or
4 y = { x. 5 y
#U[ORVQVG
5
5
3 2 1
3 2 1
–4 –3 –2 –1–1
1 2 3 4
8GTVGZ x 6 7
–7 –6
–4 –3 –2 –1–1
–2 –3
–2 –3
–5
–5
Figure 10.22 Preparing to graph
y2 x2 = 1 25 16
Figure 10.23 The graph of
y=–y
[–10, 10, 1] by [–6, 6, 1]
6 7
x
y2 x2 = 1 25 16
Step 4 Draw the two branches of the hyperbola by starting at each vertex and approaching the asymptotes. The hyperbola is shown in Figure 10.23. We now consider the foci, located at ( c, 0) and (c, 0). We find c using c 2 = a 2 + b2 . c 2 = 25 + 16 = 41
y=–y
1 2 3 4
Because c 2 = 41, c = 241 . The foci are located at approximately ( 6.4, 0) and (6.4, 0).
1  241, 0 2
and
1 241, 0 2 ,
y2 x2 Graph and locate the foci: = 1. What are the 36 9 equations of the asymptotes? CHECK POINT 3
Section 10.2 The Hyperbola
EXAMPLE 4
1033
Graphing a Hyperbola
Graph and locate the foci: 9y2  4x 2 = 36. What are the equations of the asymptotes? Solution We begin by writing the equation in standard form. The right side should be 1, so we divide both sides by 36. 9y2 4x 2 36 = 36 36 36 2 2 y x = 1 Simplify. The right side is now 1. 4 9 Now we are ready to use our fourstep procedure for graphing hyperbolas. Step 1 Locate the vertices. The equation that we obtained is in the form y2 x2  2 = 1, with a2 = 4 and b2 = 9. 2 a b y2 x2 − =1 4 9
a=
b=
Based on the standard form of the equation with the transverse axis on the y@axis, we know that the vertices are (0, a) and (0, a). Because a2 = 4, a = 2. Thus, the vertices are (0, 2) and (0, 2), shown in Figure 10.24. Step 2 Draw a rectangle. Because a2 = 4 and b2 = 9, a = 2 and b = 3. We construct a rectangle to find the asymptotes, using 2 and 2 on the y@axis (the vertices are located here) and 3 and 3 on the x@axis. The rectangle passes through these four points, shown using dashed lines in Figure 10.24. Step 3 Draw extended diagonals of the rectangle to obtain the asymptotes. We draw dashed lines through the opposite corners of the rectangle, shown in Figure 10.24, to obtain the graph of the asymptotes. Based on the standard form of the hyperbola’s equation, the equations of these asymptotes are a 2 x or y = { x. b 3 Step 4 Draw the two branches of the hyperbola by starting at each vertex and approaching the asymptotes. The hyperbola is shown in Figure 10.25. y = {
y #U[ORVQVG
5 4 3
y 5 4 3
8GTVGZ
1 –5 –4
–2 –1–1
#U[ORVQVG
–3 –4 –5
1 1 2
4 5
x –5 –4
1 2
4 5
x
–3 –4 –5
8GTVGZ –
Figure 10.24 Preparing to graph y2 x2 = 1 4 9
–2 –1–1
Figure 10.25 The graph of y2 x2 = 1 4 9
We now consider the foci, located at (0, c) and (0, c). We find c using c 2 = a 2 + b2 . Because c 2 = 13, c = 213. The foci are located at 1 0,  213 2 and 1 0, 213 2 , approximately (0, 3.6) and (0, 3.6). c 2 = 4 + 9 = 13
1034
Chapter 10 Conic Sections and Analytic Geometry
Graph and locate the foci: y2  4x 2 = 4. What are the equations of the asymptotes? CHECK POINT 4
4 Graph hyperbolas not
centered at the origin.
Translations of Hyperbolas The graph of a hyperbola can be centered at (h, k), rather than at the origin. Horizontal and vertical translations are accomplished by replacing x with x  h and y with y  k in the standard form of the hyperbola’s equation. Table 10.2 gives the standard forms of equations of hyperbolas centered at (h, k) and shows their graphs.
Table 10.2 Standard Forms of Equations of Hyperbolas Centered at (h, k) Equation (x − h)2 a2
−
(y − k)2 b2
Center
Transverse Axis
Vertices
(h, k)
Parallel to the xaxis; horizontal
(h − a, k) (h + a, k)
y
=1
8GTVKEGUCTGaWPKVUTKIJV CPFaWPKVUNGHVQHEGPVGT
a
2
−
(x − h)2 b
2
=1
(QEWU h−ck
(QEWU h+ck
8GTVGZ h−ak
8GTVGZ h+ak
x
Foci are c units right and c units left of center, where c2 = a2 + b2.
(y − k)2
Graph
%GPVGT hk
(h, k)
Parallel to the yaxis; vertical
(h, k − a) (h, k + a)
8GTVKEGUCTGaWPKVUCDQXG CPFaWPKVUDGNQYVJGEGPVGT
y (QEWU hk+c
8GTVGZ hk+a
%GPVGT hk x
Foci are c units above and c units below the center, where c2 = a2 + b2.
(QEWU hk−c
EXAMPLE 5
8GTVGZ hk−a
Graphing a Hyperbola Centered at (h, k)
(x  2) (y  3)2 = 1. Where are the foci located? What are the 16 9 equations of the asymptotes? 2
Graph:
Solution In order to graph the hyperbola, we need to know its center, (h, k). In the standard forms of equations centered at (h, k), h is the number subtracted from x and k is the number subtracted from y. 6JKUKU x–h YKVJh=
2
2
(x − 2) (y − 3) − =1 16 9
6JKUKU y–k YKVJk=
We see that h = 2 and k = 3. Thus, the center of the hyperbola, (h, k), is (2, 3). We can graph the hyperbola by using vertices, asymptotes, and our fourstep graphing procedure.
Section 10.2 The Hyperbola
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Step 1 Locate the vertices. To do this, we must identify a2.
a= y 7 6 a=4 8GTVGZ 2
– 1 –3 –2 –1–1
a=4 %GPVGT
8GTVGZ
1 2 3 4 5 6 7
x
–2 –3 Figure 10.26 Locating a hyperbola’s center and vertices
GREAT QUESTION Do I have to use the equations for the asymptotes of a hyperbola centered at the origin, b a y = t a x or y = t x, to b determine the equations of the asymptotes when the hyperbola’s center is (h, k)?
(x − 2)2 (y − 3)2 − =1 16 9
Based on the standard form of the equation with a horizontal transverse axis, the vertices are a units to the left and right of the center. Because a2 = 16, a = 4. This means that the vertices are 4 units to the left and right of the center, (2, 3). Four units to the left of (2, 3) puts one vertex at (2  4, 3), or ( 2, 3). Four units to the right of (2, 3) puts the other vertex y at (2 + 4, 3), or (6, 3). The vertices 8 are shown in Figure 10.26. 7 Step 2 Draw a rectangle. Because %GPVGT a2 = 16 and b2 = 9, a = 4 and 5 4 b = 3. The rectangle passes through 8GTVGZ – 8GTVGZ 3 points that are 4 units to the right 2 and left of the center (the vertices 1 are located here) and 3 units above x –5 –4 –3 –2 –1–1 1 2 3 4 5 6 7 8 9 and below the center. The rectangle –2 is shown using dashed lines in Figure 10.27. (x  2)2 (y  3)2 = 1 Step 3 Draw extended diagonals Figure 10.27 The graph of 16 9 of the rectangle to obtain the asymptotes. We draw dashed lines through the opposite corners of the rectangle, shown in Figure 10.27, to obtain the graph of the asymptotes. The equations of the y2 x2 b 3 asymptotes of the unshifted hyperbola = 1 are y = { x, or y = { x. a 16 9 4 Thus, the asymptotes for the hyperbola that is shifted 2 units to the right and 3 units up, namely, (x  2)2 (y  3)2 = 1 16 9
No. You can also use the pointslope form of a line’s equation y  y1 = m(x  x1) to find the equations of the asymptotes. The center of the hyperbola, (h, k), is a point on each asymptote, so x1 = h and b y1 = k. The slopes, m, are { a for a horizontal transverse axis and a { for a vertical transverse axis. b
b =
The form of this equation is (x − h)2 (y − k)2 − = 1. 2 a b2
have equations that can be expressed as 3 y  3 = { (x  2). 4 Step 4 Draw the two branches of the hyperbola by starting at each vertex and approaching the asymptotes. The hyperbola is shown in Figure 10.27. We now consider the foci, located c units to the right and left of the center. We find c using c 2 = a2 + b2. c 2 = 16 + 9 = 25 Because c 2 = 25, c = 5. This means that the foci are 5 units to the left and right of the center, (2, 3). Five units to the left of (2, 3) puts one focus at (2  5, 3), or ( 3, 3). Five units to the right of (2, 3) puts the other focus at (2 + 5, 3), or (7, 3). (x  3)2 (y  1)2 = 1. Where are the foci 4 1 located? What are the equations of the asymptotes? CHECK POINT 5
Graph:
In our next example, it is necessary to convert the equation of a hyperbola to standard form by completing the square on x and y.
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Chapter 10 Conic Sections and Analytic Geometry
EXAMPLE 6
Graphing a Hyperbola Centered at (h, k)
Graph: 4x  24x  25y2 + 250y  489 = 0. Where are the foci located? What are the equations of the asymptotes? 2
Solution
We begin by completing the square on x and y.
4x 2  24x  25y2 + 250y  489 = 0
This is the given equation.
(4x 2  24x) + ( 25y2 + 250y) = 489
Group terms and add 489 to both sides.
4(x 2  6x + □)  25(y2  10y + □) = 489
Factor out 4 and − 25, respectively, so coefficients of x2 and y2 are 1.
4(x2 − 6x + 9) − 25(y2 − 10y + 25) = 489 + 36 + (–625) 9GCFFGF∙QT VQVJGNGHVUKFG
9GCFFGF–∙QT –VQVJGNGHVUKFG
#FF+ – VQVJGTKIJVUKFG
GREAT QUESTION Shouldn’t the center of (y − 5)2 (x − 3)2 − = 1 4 25 be (5, 3)? No. The hyperbola’s center is (3, 5) because the equation shows that 3 is subtracted from x and 5 is subtracted from y. Many students tend to read the equation from left to right and get the center backward. The hyperbola’s center is not (5, 3).
x and y, respectively.
4(x  3)2  25(y  5)2 = 100
Factor.
4(x  3)2 25(y  5)2 100 = 100 100 100
Divide both sides by − 100.
(x  3)2 (y  5)2 + = 1 25 4 6JKUKU y–k YKVJk=
Complete each square by adding the square of half the coefficient of
(y − 5)2 (x − 3)2 − =1 4 25
Simplify.
6JKUKU x–h YKVJh=
Write the equation in standard (y − k)2 (x − h)2 form, − = 1. 2 a b2
We see that h = 3 and k = 5. Thus, the center of the hyperbola, (h, k), is (3, 5). Because the x 2@term is being subtracted, the transverse axis is vertical and the hyperbola opens upward and downward. We use our fourstep procedure to obtain the graph of
a=
(y − 5)2 (x − 3)2 − = 1. 4 25
Step 1 Locate the vertices. Based on the standard form of the equation with a vertical transverse axis, the vertices are a units above and below the center. Because a2 = 4, a = 2. This means that the vertices are 2 units above and below the center, (3, 5). This puts the vertices at (3, 7) and (3, 3), shown in Figure 10.28. Step 2 Draw a rectangle. Because a2 = 4 and b2 = 25, a = 2 and b = 5. The rectangle passes through points that are 2 units above and below the center (the vertices are located here) and 5 units to the right and left of the center. The rectangle is shown using dashed lines in Figure 10.28.
b =
y 9 8 7 6 5 4 3 2 1 –4 –3 –2 –1–1
8GTVGZ
%GPVGT
1 2 3 4 5 6 7 8 9 10 8GTVGZ
Figure 10.28 The graph of (y  5)2 (x  3)2 = 1 4 25
x
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Section 10.2 The Hyperbola
Step 3 Draw extended diagonals of the rectangle to obtain the asymptotes. We draw dashed lines through the opposite corners of the rectangle, shown in Figure 10.28, to obtain the graph of the asymptotes. The equations of the asymptotes y2 x2 a 2 of the unshifted hyperbola = 1 are y = { x, or y = { x. Thus, the 4 25 b 5 asymptotes for the hyperbola that is shifted 3 units to the right and 5 units up, namely, (y  5)2 (x  3)2 = 1 4 25 have equations that can be expressed as 2 y  5 = { (x  3). 5 Step 4 Draw the two branches of the hyperbola by starting at each vertex and approaching the asymptotes. The hyperbola is shown in Figure 10.28. We now consider the foci, located c units above and below the center, (3, 5). We find c using c 2 = a2 + b2. c 2 = 4 + 25 = 29 Because c 2 = 29, c = 229. The foci are located at 1 3, 5  229 2 .
1 3, 5
+ 229 2 and
Graph: 4x 2  24x  9y2  90y  153 = 0. Where are the foci located? What are the equations of the asymptotes? CHECK POINT 6
5 Solve applied problems involving hyperbolas.
The hyperbolic shape of a sonic boom
Applications Hyperbolas have many applications. When a jet flies at a speed greater than the speed of sound, the shock wave that is created is heard as a sonic boom. The wave has the shape of a cone. The shape formed as the cone hits the ground is one branch of a hyperbola. Halley’s Comet, a permanent part of our solar system, travels around the Sun in an elliptical orbit. Other comets pass through the solar system only once, following a hyperbolic path with the Sun as a focus. Hyperbolas are of practical importance in fields ranging from architecture to navigation. Cooling towers used in the design for nuclear power plants have cross sections that are both ellipses and hyperbolas. Threedimensional solids whose cross sections are hyperbolas are used in some rather unique architectural creations, including the former TWA building at Kennedy Airport in New York City and the St. Louis Science Center Planetarium.
EXAMPLE 7
An Application Involving Hyperbolas
An explosion is recorded by two microphones that are 2 miles apart. Microphone M1 received the sound 4 seconds before microphone M2. Assuming sound travels at 1100 feet per second, determine the possible locations of the explosion y relative to the location of the microphones. 8000 'NNKRVKECNQTDKV Sun
*[RGTDQNKEQTDKV Orbits of comets
Solution We begin by putting the microphones in a coordinate system. Because 1 mile = 5280 feet, we place M1 5280 feet on the xaxis to the right of the origin and M2 5280 feet on the xaxis to the left of the origin. Figure 10.29 illustrates that the two microphones are 2 miles apart.
P xyGZRNQUKQP
6000 4000 M – –6000
M
2000 –2000
2000
6000
x
–2000 Figure 10.29 a hyperbola
Locating an explosion on the branch of
1038
Chapter 10 Conic Sections and Analytic Geometry y 8000 P xyGZRNQUKQP
6000 4000 M – –6000
y2 x2 = 1 a2 b2
M
2000 –2000
We know that M2 received the sound 4 seconds after M1. Because sound travels at 1100 feet per second, the difference between the distance from P to M1 and the distance from P to M2 is 4400 feet. The set of all points P (or locations of the explosion) satisfying these conditions fits the definition of a hyperbola, with microphones M1 and M2 at the foci.
2000
x
6000
Use the standard form of the hyperbola’s equation. P(x, y), the explosion point, lies on this hyperbola. We must find a2 and b 2.
The difference between the distances, represented by 2a in the derivation of the hyperbola’s equation, is 4400 feet. Thus, 2a = 4400 and a = 2200.
–2000
y2 x2 = 1 (2200)2 b2
Figure 10.29 (repeated) Locating an explosion on the branch of a hyperbola
y2 x2  2 = 1 4,840,000 b
Substitute 2200 for a.
Square 2200.
We must still find b2. We know that a = 2200. The distance from the center, (0, 0), to either focus, ( 5280, 0) or (5280, 0), is 5280. Thus, c = 5280. Using c 2 = a2 + b2, we have 52802 = 22002 + b2 and b2 = 52802  22002 = 23,038,400. The equation of the hyperbola with a microphone at each focus is y2 x2 = 1. 4,840,000 23,038,400 Instructor Resources for Section 10.2 in MyLab Math
Substitute 23,038,400 for b 2.
We can conclude that the explosion occurred somewhere on the right branch (the branch closer to M1) of the hyperbola given by this equation. In Example 7, we determined that the explosion occurred somewhere along one branch of a hyperbola, but not exactly where on the hyperbola. If, however, we had received the sound from another pair of microphones, we could locate the sound along a branch of another hyperbola. The exact location of the explosion would be the point where the two hyperbolas intersect. CHECK POINT 7 Rework Example 7 assuming microphone M1 receives the sound 3 seconds before microphone M2.
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. The set of all points in a plane the difference of whose distances from two fixed points is constant is a/an . The two fixed points are called the . The line through these points intersects the graph at two points, called the , which are joined by the axis. y2 x2 C2. The vertices of = 1 are and . 25 9 The foci are located at and . y2 x2 = 1 are C3. The vertices of 25 9 The foci are located at
and and
. .
C4. The two branches of the graph of a hyperbola approach a pair of intersecting lines, called . These intersecting lines pass through the of the hyperbola. C5. The equation 9x 2  4y2 = 36 can be written in standard form by both sides by . y2 x2 = 1 are C6. The equations for the asymptotes of 4 9 and
.
y2 C7. The equations for the asymptotes of  x 2 = 1 are 4 and .
Section 10.2 The Hyperbola C8. If the center of a hyperbola with a horizontal transverse axis is (2, 3) and a2 = 25, then the coordinates of the vertices are and 2 2 (y + 2) (x  7) = 1 is C9. The center of 4 36
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C10. In the equation 9(x 2  8x)  16(y2 + 2y) = 16, we complete the square on x by adding within the first parentheses. We complete the square on y by adding within the second parentheses. Thus, we must add to the right side of the equation.
. .
10.2 EXERCISE SET Practice Exercises
17.
In Exercises 1–4, find the vertices and locate the foci of each hyperbola with the given equation. Then match each equation to one of the graphs that are shown and labeled (a)–(d). 2
1.
2
2
y x = 1 4 1
2.
y2 x2 3. = 1 4 1 y a.
y x = 1 1 4
22. 4x 2  25y2 = 100
23. 9y2  25x 2 = 225
24. 16y2  9x 2 = 144
25. y = { 2x  2
26. y = { 2x 2  8
In Exercises 27–32, find the standard form of the equation of each hyperbola. y y 27. 28.
1 2 3 4
1 2 3 4
–4 –3 –2 –1–1
4 3 2 1
x
–2 –3 –4
d.
y
–4 –3 –2 –1–1
–4
1 2 3 4
–4 –3 –2 –1–1
–2 –3 –4
–4 1 2
1 2 3 4
x
1 2
4
x
y
30.
4 3
–2 –3 –4
7 6
1 –4
–2 –1–1
1 2
4 3 2 1
x
4
–3 –4
–4
7. Foci: (  4, 0), (4, 0); vertices: ( 3, 0), (3, 0) 8. Foci: (  7, 0), (7, 0); vertices: ( 5, 0), (5, 0)
y
31.
In Exercises 13–26, use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes.
y
32. 4 3 2 1
2 1
11. Center: (4, 2); Focus: (7, 2); vertex: (6,  2) 12. Center: ( 2, 1); Focus: (  2, 6); vertex: (  2, 4)
x
–6 –7
9. Endpoints of transverse axis: (0, 6), (0, 6); asymptote: y = 2x 10. Endpoints of transverse axis: (0, 20), (0, 20); asymptote: y = 4x
–2 –1–1 –2 –3 –4
6. Foci: (0, 6), (0, 6); vertices: (0,  2), (0, 2)
y2 x2 = 1 16 25 y2 x2 16. = 1 144 81
4
–3 –4
y
29.
5. Foci: (0, 3), (0, 3); vertices: (0,  1), (0, 1)
y2 x2 = 1 9 25 y2 x2 15. = 1 100 64
1 2
–6
In Exercises 5–12, find the standard form of the equation of each hyperbola satisfying the given conditions.
13.
–2 –1–1
x
4
–2 –3 –4
4 3 2 1 x
1
–2 –1–1
y
4 3 2 1
4 3
6
x
–2 –3 –4
c.
21. 9x  4y = 36 2
y2 x2 4. = 1 1 4 y b.
y2 x2 = 1 25 64
20. 9y2  x 2 = 1
2
4 3 2 1
–4 –3 –2 –1–1
18.
19. 4y2  x 2 = 1 2
2
4 3 2 1
y2 x2 = 1 16 36
–2 –1 –2 –3 –4
1 2 3 4 5 6
x
–5 –4
–1–1
14.
–8
–5 –6
1 2 3
x
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Chapter 10 Conic Sections and Analytic Geometry
In Exercises 33–42, use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. 33.
(x + 4)2 9 (x + 3)2

(y + 3)2 16
= 1
2
34.
(x + 2)2 9 (x + 2)2

(y  1)2 25
63. An architect designs two houses that are shaped and positioned like a part of the branches of the hyperbola whose equation is 625y2  400x 2 = 250,000, where x and y are in yards. How far apart are the houses at their closest point?
= 1 y
2
y 35. = 1 25 16 (y + 2)2 (x  1)2 37. = 1 4 16 39. (x  3)2  4(y + 3)2 = 4
y 36. = 1 9 25 (y  2)2 (x + 1)2 38. = 1 36 49 40. (x + 3)2  9(y  4)2 = 9
41. (x  1)2  (y  2)2 = 3
42. (y  2)2  (x + 3)2 = 5
x
In Exercises 43–50, convert each equation to standard form by completing the square on x and y. Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. 43. 44. 45. 46. 47. 48. 49. 50.
x 2  y2  2x  4y  4 = 0 4x 2  y2 + 32x + 6y + 39 = 0 16x 2  y2 + 64x  2y + 67 = 0 9y2  4x 2  18y + 24x  63 = 0 4x 2  9y2  16x + 54y  101 = 0 4x 2  9y2 + 8x  18y  6 = 0 4x 2  25y2  32x + 164 = 0 9x 2  16y2  36x  64y + 116 = 0
64. Scattering experiments, in which moving particles are deflected by various forces, led to the concept of the nucleus of an atom. In 1911, the physicist Ernest Rutherford (1871–1937) discovered that when alpha particles are directed toward the nuclei of gold atoms, they are eventually deflected along hyperbolic paths, illustrated in the figure. If a particle gets as close as 3 units to the nucleus along a hyperbolic path with an asymptote given by y = 12 x, what is the equation of its path?
Practice PLUS
y
In Exercises 51–56, graph each relation. Use the relation’s graph to determine its domain and range. y2 x2 = 1 9 16 y2 x2 54. + = 1 25 4 51.
y2 x2 = 1 25 4 y2 x2 55. = 1 16 9
52.
In Exercises 57–60, find the solution set for each system by graphing both of the system’s equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations. 57. b
x 2  y2 = 4 x 2 + y2 = 4
9x 2 + y2 = 9 59. b 2 y  9x 2 = 9
58. b
y= x
y2 x2 + = 1 9 16 y2 x2 56. = 1 4 25
53.
x 2  y2 = 9 x 2 + y2 = 9
x
Moiré patterns, such as those shown in Exercises 65–66, can appear when two repetitive patterns overlap to produce a third, sometimes unintended, pattern. a. In each exercise, use the name of a conic section to describe the moiré pattern. b. Select one of the following equations that can possibly describe a conic section within the moiré pattern:
4x 2 + y2 = 4 60. b 2 y  4x 2 = 4
Application Exercises 61. An explosion is recorded by two microphones that are 1 mile apart. Microphone M1 received the sound 2 seconds before microphone M2. Assuming sound travels at 1100 feet per second, determine the possible locations of the explosion relative to the location of the microphones. 62. Radio towers A and B, 200 kilometers apart, are situated along the coast, with A located due west of B. Simultaneous radio signals are sent from each tower to a ship, with the signal from B received 500 microseconds before the signal from A. a. Assuming that the radio signals travel 300 meters per microsecond, determine the equation of the hyperbola on which the ship is located. b. If the ship lies due north of tower B, how far out at sea is it?
#NRJCRCTVKENG
0WENGWU
x 2 + y2 = 1; x 2  y2 = 1; x 2 + 4y2 = 4. 65.
66.
Section 10.2 The Hyperbola
Explaining the Concepts 67. What is a hyperbola? 68. Describe how to graph
y2 x2 = 1. 9 1
y2 x2 = 1. 9 1 one difference between the x2 = 1. 1 one difference between the  3)2 (y + 3)2 = 1. 9 1 ellipse from a hyperbola by
69. Describe how to locate the foci of the graph of
70. Describe one similarity and y2 y2 x2 graphs of = 1 and 9 1 9 71. Describe one similarity and (x y2 x2 graphs of = 1 and 9 1 72. How can you distinguish an looking at their equations? 73. In 1992, a NASA team began a project called Spaceguard Survey, calling for an international watch for comets that might collide with Earth. Why is it more difficult to detect a possible “doomsday comet” with a hyperbolic orbit than one with an elliptical orbit?
Technology Exercises 74. Use a graphing utility to graph any five of the hyperbolas that you graphed by hand in Exercises 13–26. 75. Use a graphing utility to graph any three of the hyperbolas that you graphed by hand in Exercises 33–42. First solve the given equation for y by using the square root property. Enter each of the two resulting equations to produce each branch of the hyperbola. 76. Use a graphing utility to graph any one of the hyperbolas that you graphed by hand in Exercises 43–50. Write the equation as a quadratic equation in y and use the quadratic formula to solve for y. Enter each of the two resulting equations to produce each branch of the hyperbola. y2 x2 77. Use a graphing utility to graph = 0. Is the graph a 4 9 y2 x2 hyperbola? In general, what is the graph of 2  2 = 0? a b y2 y2 x2 x2 78. Graph 2  2 = 1 and 2  2 =  1 in the same viewing a b a b rectangle for values of a2 and b2 of your choice. Describe the relationship between the two graphs. 79. Write 4x 2  6xy + 2y2  3x + 10y  6 = 0 as a quadratic equation in y and then use the quadratic formula to express y in terms of x. Graph the resulting two equations using a graphing utility in a [ 50, 70, 10] by [ 30, 50, 10] viewing rectangle. What effect does the xy@term have on the graph of the resulting hyperbola? What problems would you encounter if you attempted to write the given equation in standard form by completing the square? x x y y y2 x2 80. Graph = 1 and = 1 in the same 16 9 16 9 viewing rectangle. Explain why the graphs are not the same.
Critical Thinking Exercises Make Sense? In Exercises 81–84, determine whether each statement makes sense or does not make sense, and explain your reasoning. 81. I changed the addition in an ellipse’s equation to subtraction and this changed its elongation from horizontal to vertical. 82. I noticed that the definition of a hyperbola closely resembles that of an ellipse in that it depends on the distances
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between a set of points in a plane to two fixed points, the foci. 83. I graphed a hyperbola centered at the origin that had y@intercepts but no x@intercepts. 84. I graphed a hyperbola centered at the origin that was symmetric with respect to the x@axis and also symmetric with respect to the y@axis. In Exercises 85–88, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. 85. If one branch of a hyperbola is removed from a graph, then the branch that remains must define y as a function of x. 86. All points on the asymptotes of a hyperbola also satisfy the hyperbola’s equation. x 2 y2 2 = 1 does not intersect the line y =  x. 9 4 3 88. Two different hyperbolas can never share the same asymptotes. 87. The graph of
y2 x2 89. What happens to the shape of the graph of 2  2 = 1 as a b c S ∞ , where c 2 = a2 + b2? a 90. Find the standard form of the equation of the hyperbola with vertices (5,  6) and (5, 6), passing through (0, 9). 91. Find the equation of a hyperbola whose asymptotes are perpendicular.
Retaining the Concepts 92. The top destinations for U.S. college students studying abroad are the United Kingdom, Italy, and Spain. The number of students studying in the U.K. exceeds the number studying in Spain by 10 thousand. The number of students studying in Italy exceeds the number studying in Spain by 2 thousand. Combined, 93 thousand U.S. students study in the U.K., Italy, and Spain. Determine the number of students studying abroad, in thousands, in the U.K., Italy, and Spain. (Source: Institute of International Education) (Section 1.3, Example 1) 93. The number of gallons of water, W, used when taking a shower varies directly as the time, t, in minutes, in the shower. A shower lasting 5 minutes uses 30 gallons of water. How much water is used in a shower lasting 13 minutes? (Section 3.7, Example 1) 94. Use the exponential decay model, A = A0 e kt, to solve this exercise. The halflife of aspirin in your bloodstream is 12 hours. How long, to the nearest tenth of an hour, will it take for the aspirin to decay to 60% of the original dosage? (Section 4.5, Example 2) 95. Determine the amplitude, period, and phase shift of y = 3 sin(x + p). Then graph one period of the function. (Section 5.5, Example 4)
Preview Exercises Exercises 96–98 will help you prepare for the material covered in the next section. In Exercises 96–97, graph each parabola with the given equation. 97. y = 3(x  1)2 + 2 96. y = x 2 + 4x  5 98. Isolate the terms involving y on the left side of the equation: y2 + 2y + 12x  23 = 0. Then write the equation in an equivalent form by completing the square on the left side.
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Chapter 10 Conic Sections and Analytic Geometry
SECTION 10.3
The Parabola At first glance, this image looks like columns of smoke rising from a fire into a starry sky. Those are, indeed, stars in the background, but you are not looking at ordinary smoke columns. These stand almost 6 trillion miles high and are 7000 lightyears from Earth—more than 400 million times as far away as the Sun.
WHAT YOU’LL LEARN 1 Graph parabolas with vertices at the origin.
2 Write equations of
parabolas in standard form.
3 Graph parabolas with
vertices not at the origin.
4 Solve applied problems involving parabolas.
5 Identify conics without
completing the square.
This NASA photograph is one of a series of stunning images captured from the ends of the universe by the Hubble Space Telescope. The image shows infant star systems the size of our solar system emerging from the gas and dust that shrouded their creation. Using a parabolic mirror that is 94.5 inches in diameter, the Hubble has provided answers to many of the profound mysteries of the cosmos: How big and how old is the universe? How did the galaxies come to exist? Do other Earthlike planets orbit other Sunlike stars? In this section, we study parabolas and their applications, including parabolic shapes that gather distant rays of light and focus them into spectacular images.
Definition of a Parabola In Chapter 3, we studied parabolas, viewing them as graphs of quadratic functions in the form y = a(x  h)2 + k or y = ax 2 + bx + c.
A BRIEF REVIEW
Graphing Parabolas
Graphing y = a(x − h)2 + k and y = ax2 + bx + c • If a 7 0, the graph opens upward. If a 6 0, the graph opens downward. • The vertex of y = a(x  h)2 + k is (h, k). y
y y=a x−h+k a
x
x=h
• The x@coordinate of the vertex of y = ax 2 + bx + c is x = For more detail, see Section 3.1, Objective 2.
b . 2a
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Section 10.3 The Parabola 2CTCDQNC
&KTGEVTKZ (QEWU 8GTVGZ
#ZKUQH U[OOGVT[
Parabolas can be given a geometric definition that enables us to include graphs that open to the left or to the right, as well as those that open obliquely. The definitions of ellipses and hyperbolas involved two fixed points, the foci. By contrast, the definition of a parabola is based on one point and a line. Definition of a Parabola A parabola is the set of all points in a plane that are equidistant from a fixed line, the directrix, and a fixed point, the focus, that is not on the line (see Figure 10.30).
Figure 10.30
In Figure 10.30, find the line passing through the focus and perpendicular to the directrix. This is the axis of symmetry of the parabola. The point of intersection of the parabola with its axis of symmetry is called the vertex. Notice that the vertex is midway between the focus and the directrix.
Standard Form of the Equation of a Parabola y
The rectangular coordinate system enables us to translate a parabola’s geometric definition into an algebraic equation. Figure 10.31 is our starting point for obtaining an equation. We place the focus on the x@axis at the point (p, 0). The directrix has an equation given by x = p. The vertex, located midway between the focus and the directrix, is at the origin. What does the definition of a parabola tell us about the point (x, y) in Figure 10.31? For any point (x, y) on the parabola, the distance d 1 to the directrix is equal to the distance d 2 to the focus. Thus, the point (x, y) is on the parabola if and only if
M(–p, y)
d1 P(x, y) d2 x
&KTGEVTKZx=–p
(QEWU p
Figure 10.31
d 1 = d 2. 2(x + p) + (y  y)2 = 2(x  p)2 + (y  0)2 (x + p)2 = (x  p)2 + y2 2
Use the distance formula. Square both sides of the equation.
x + 2px + p = x  2px + p + y 2px = 2px + y2 2
2
2
2
2
Square x + p and x − p. Subtract x2 + p2 from both sides of the equation.
2
y = 4px
Solve for y2.
This last equation is called the standard form of the equation of a parabola with its vertex at the origin. There are two such equations, one for a focus on the x@axis and one for a focus on the y@axis. Standard Forms of the Equations of a Parabola The standard form of the equation of a parabola with vertex at the origin is y2 = 4px
or
x 2 = 4py.
Figure 10.32(a) at the top of the next page illustrates that for the equation on the left, the focus is on the x@axis, which is the axis of symmetry. Figure 10.32(b) illustrates that for the equation on the right, the focus is on the y@axis, which is the axis of symmetry.
1044
Chapter 10 Conic Sections and Analytic Geometry y
y
GREAT QUESTION
x=py
What is the relationship between the sign of p and the position of the focus from the vertex?
(QEWU p
y=px
8GTVGZ
8GTVGZ
It is helpful to think of p as the directed distance from the vertex to the focus. If p 7 0, the focus lies p units to the right of the vertex or p units above the vertex. If p 6 0, the focus lies p units to the left of the vertex or p units below the vertex.
1 Graph parabolas with
x
x
(QEWU p
&KTGEVTKZx=–p
&KTGEVTKZy=–p
Figure 10.32(b) Parabola with the y@axis as the axis of symmetry. If p 7 0, the graph opens upward. If p 6 0, the graph opens downward.
Figure 10.32(a) Parabola with the x@axis as the axis of symmetry. If p 7 0, the graph opens to the right. If p 6 0, the graph opens to the left.
Using the Standard Form of the Equation of a Parabola
vertices at the origin.
We can use the standard form of the equation of a parabola to find its focus and directrix. Observing the graph’s symmetry from its equation is helpful in locating the focus. y2 = 4px
x2 = 4py
6JGGSWCVKQPFQGUPQVEJCPIGKH yKUTGRNCEGFYKVJ–y6JGTGKU xCZKUU[OOGVT[CPFVJGHQEWUKU QPVJGxCZKUCV p
6JGGSWCVKQPFQGUPQVEJCPIGKH xKUTGRNCEGFYKVJ–x6JGTGKU yCZKUU[OOGVT[CPFVJGHQEWUKU QPVJGyCZKUCV p
Although the definition of a parabola is given in terms of its focus and its directrix, the focus and directrix are not part of the graph. The vertex, located at the origin, is a point on the graph of y2 = 4px and x 2 = 4py. Example 1 illustrates how you can find two additional points on the parabola.
EXAMPLE 1
Finding the Focus and Directrix of a Parabola
Find the focus and directrix of the parabola given by y2 = 12x. Then graph the parabola. Solution The given equation, y2 = 12x, is in the standard form y2 = 4px, so 4p = 12. 0QEJCPIGKHyKU TGRNCEGFYKVJ–y 6JGRCTCDQNCJCU xCZKUU[OOGVT[
y &KTGEVTKZ x=–
–5 –4
7 6 5 4 3 2 1 –2 –1–1 –2 –3 –4 –5 –6 –7
(3, 6)
y2 = 12x 6JKUKUp
We can find both the focus and the directrix by finding p. (QEWU 1 2 3 4 5 8GTVGZ
(3, –6)
Figure 10.33 The graph of y2 = 12x
x
4p = 12 p = 3
Divide both sides by 4.
Because p is positive, the parabola, with its x@axis symmetry, opens to the right. The focus is 3 units to the right of the vertex, (0, 0). Focus: Directrix:
(p, 0) = (3, 0) x = p; x = 3
The focus, (3, 0), and directrix, x = 3, are shown in Figure 10.33.
Section 10.3 The Parabola
1045
To graph the parabola, we will use two points on the graph that lie directly above and below the focus. Because the focus is at (3, 0), substitute 3 for x in the parabola’s equation, y2 = 12x. y2 = 12 # 3
Replace x with 3 in y2 = 12x.
2
y = 36 y = { 236 = {6
Simplify. Apply the square root property.
The points on the parabola above and below the focus are (3, 6) and (3, 6). The graph is sketched in Figure 10.33. Find the focus and directrix of the parabola given by y2 = 8x. Then graph the parabola. CHECK POINT 1
In general, the points on a parabola y2 = 4px that lie above and below the focus, (p, 0), are each at a distance 2p from the focus. This is because if x = p, then y2 = 4px = 4p2, so y = {2p. The line segment joining these two points is called the latus rectum; its length is 4p . TECHNOLOGY We graph y2 = 12x with a graphing utility by first solving for y. The screen shows the graphs of y = 212x and y =  212x. The graph fails the vertical line test. Because y2 = 12x is not a function, you were not familiar with this form of the parabola’s equation in Chapter 3.
The Latus Rectum and Graphing Parabolas The latus rectum of a parabola is a line segment that passes through its focus, is parallel to its directrix, and has its endpoints on the parabola. Figure 10.34 shows that the length of the latus rectum for the graphs of y2 = 4px and x 2 = 4py is 4p . y
y (QEWU p
x=py
y=px y=√x
.CVWUTGEVWO NGPIVJp x
y=–√x &KTGEVTKZx=–p
x
(QEWU p .CVWUTGEVWO NGPIVJp
[–6, 6, 1] by [–8, 8, 1]
&KTGEVTKZy=–p
Figure 10.34 Endpoints of the latus rectum are helpful in determining a parabola’s “width,” or how it opens.
EXAMPLE 2
Finding the Focus and Directrix of a Parabola
Find the focus and directrix of the parabola given by x 2 = 8y. Then graph the parabola. Solution The given equation, x 2 = 8y, is in the standard form x 2 = 4py, so 4p = 8. 0QEJCPIGKHxKU TGRNCEGFYKVJ–x 6JGRCTCDQNCJCU yCZKUU[OOGVT[
x2 = –8y 6JKUKUp
1046
Chapter 10 Conic Sections and Analytic Geometry y
8GTVGZ
We can find both the focus and the directrix by finding p.
5 4 3
&KTGEVTKZy=
1
.CVWU4GEVWO
–5 –4 –3 –2 –1–1 (–4, –2) –2 –3 –4 –5
x 1 2 3 4 5 (4, –2)
4p = 8 p = 2
Because p is negative, the parabola, with its y@axis symmetry, opens downward. The focus is 2 units below the vertex, (0, 0). Focus: Directrix:
(QEWU –
Figure 10.35 The graph of x 2 =  8y
x2 . The graph 8 passes the vertical line test. Because x 2 =  8y is a function, you were familiar with the parabola’s alternate algebraic 1 form, y =  x 2, in Chapter 3. 8 The alternate algebraic form is y = ax 2 + bx + c, with 1 a =  , b = 0, and c = 0. 8 for y: y = 
(0, p) = (0, 2) y = p; y = 2
The focus and directrix are shown in Figure 10.35. To graph the parabola, we will use the vertex, (0, 0), and the two endpoints of the latus rectum. The length of the latus rectum is
TECHNOLOGY Graph x 2 =  8y by first solving
Divide both sides by 4.
4p = 4( 2) = 8 = 8. Because the graph has y@axis symmetry, the latus rectum extends 4 units to the left and 4 units to the right of the focus, (0, 2). The endpoints of the latus rectum are ( 4, 2) and (4, 2). Passing a smooth curve through the vertex and these two points, we sketch the parabola, shown in Figure 10.35. CHECK POINT 2 Find the focus and directrix of the parabola given by x 2 = 12y. Then graph the parabola.
In Examples 1 and 2, we used the equation of a parabola to find its focus and directrix. In the next example, we reverse this procedure.
EXAMPLE 3
Finding the Equation of a Parabola from Its Focus and Directrix y
Find the standard form of the equation of a parabola with focus (5, 0) and directrix x = 5, shown in Figure 10.36. [–6, 6, 1] by [–6, 6, 1]
2 Write equations of parabolas in standard form.
Solution The focus is (5, 0). Thus, the focus is on the x@axis. We use the standard form of the equation in which there is x@axis symmetry, namely, y2 = 4px. We need to determine the value of p. Figure 10.36 shows that the focus is 5 units to the right of the vertex, (0, 0). Thus, p is positive and p = 5. We substitute 5 for p in y2 = 4px to obtain the standard form of the equation of the parabola. The equation is y2 = 4 # 5x
or
y2 = 20x.
7 6 5 4 3 2 1
&KTGEVTKZ x=–
–4 –3 –2 –1–1
(QEWU 1 2 3 4 5
x
–2 –3 –4 –5 –6 –7 Figure 10.36
CHECK POINT 3 Find the standard form of the equation of a parabola with focus (8, 0) and directrix x = 8.
3 Graph parabolas with
vertices not at the origin.
Translations of Parabolas The graph of a parabola can have its vertex at (h, k), rather than at the origin. Horizontal and vertical translations are accomplished by replacing x with x  h and y with y  k in the standard form of the parabola’s equation.
Section 10.3 The Parabola
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Table 10.3 gives the standard forms of equations of parabolas with vertex at (h, k). Figure 10.37 shows their graphs. Table 10.3 Standard Forms of Equations of Parabolas with Vertex at (h, k) Equation
Vertex
Axis of Symmetry
Focus
Directrix
(y  k)2 = 4p(x  h)
(h, k)
Horizontal
(h + p, k)
x = h  p
(x  h)2 = 4p(y  k)
(h, k)
Vertical
(h, k + p)
y = k  p
GREAT QUESTION
y
If y is the squared term, there is horizontal symmetry and the parabola’s equation is not a function. If x is the squared term, there is vertical symmetry and the parabola’s equation is a function. Continue to think of p as the directed distance from the vertex, (h, k), to the focus.
If p 7 0, opens to the right. If p 6 0, opens to the left. If p 7 0, opens upward. If p 6 0, opens downward. y
y–k
What are the main differences between parabolas whose equations contain y2 and parabolas whose equations contain x2?
Description
=p x–h
x–h=p y–k (QEWU
hk+p
&KTGEVTKZx=h–p 8GTVGZ hk
8GTVGZ hk x
x
(QEWU h+pk &KTGEVTKZy=k–p Figure 10.37 Graphs of parabolas with vertex at (h, k) and p 7 0
The two parabolas shown in Figure 10.37 illustrate standard forms of equations for p 7 0. If p 6 0, a parabola with a horizontal axis of symmetry will open to the left and the focus will lie to the left of the directrix. If p 6 0, a parabola with a vertical axis of symmetry will open downward and the focus will lie below the directrix.
EXAMPLE 4
Graphing a Parabola with Vertex at (h, k)
Find the vertex, focus, and directrix of the parabola given by (x  3)2 = 8(y + 1). Then graph the parabola. Solution In order to find the focus and directrix, we need to know the vertex. In the standard forms of equations with vertex at (h, k), h is the number subtracted from x and k is the number subtracted from y. (x − 3)2 = 8(y − (–1)) 6JKUKU x–h YKVJh=
6JKUKUy–k YKVJk=–
We see that h = 3 and k = 1. Thus, the vertex of the parabola is (h, k) = (3, 1). Now that we have the vertex, we can find both the focus and directrix by finding p. (x − 3)2 = 8(y + 1) 6JKUKUp
The equation is in the standard form (x − h)2 = 4p(y − k). Because x is the squared term, there is vertical symmetry and the parabola’s equation is a function.
Because 4p = 8, p = 2. Based on the standard form of the equation, the axis of symmetry is vertical. With a positive value for p and a vertical axis of symmetry,
1048
Chapter 10 Conic Sections and Analytic Geometry y 5 4 3 2
the parabola opens upward. Because p = 2, the focus is located 2 units above the vertex, (3, 1). Likewise, the directrix is located 2 units below the vertex. .CVWU4GEVWO
(–1, 1)
(7, 1)
–2 –1–1
Focus: (h, k + p) = (3, –1 + 2) = (3, 1)
(QEWU
1 2
4 5 6 7 8
–2
8GTVGZ –
–4 –5
&KTGEVTKZy=–
6JGXGTVGZ hk KU –
x
Directrix: y = k − p y = –1 − 2 = –3 6JGFKTGEVTKZKUWPKVU DGNQYVJGXGTVGZ –
Figure 10.38 The graph of (x  3)2 = 8(y + 1)
TECHNOLOGY
Thus, the focus is (3, 1) and the directrix is y = 3. They are shown in Figure 10.38. To graph the parabola, we will use the vertex, (3, 1), and the two endpoints of the latus rectum. The length of the latus rectum is 4p = 4 # 2 = 8 = 8.
Graph (x  3)2 = 8(y + 1) by first solving for y: 1 8 (x
 3)2 = y + 1
y = 18(x  3)2  1. The graph passes the vertical line test. Because (x  3)2 = 8(y + 1) is a function, you were familiar with the parabola’s alternate algebraic form, y = 18(x  3)2  1, in Chapter 3. The alternate algebraic form is y = a(x  h)2 + k with a = 18 , h = 3, and k = 1.
6JGHQEWUKUWPKVU CDQXGVJGXGTVGZ –
Because the graph has vertical symmetry, the latus rectum extends 4 units to the left and 4 units to the right of the focus, (3, 1). The endpoints of the latus rectum are (3  4, 1), or ( 1, 1), and (3 + 4, 1), or (7, 1). Passing a smooth curve through the vertex and these two points, we sketch the parabola, shown in Figure 10.38. Find the vertex, focus, and directrix of the parabola given by (x  2)2 = 4(y + 1). Then graph the parabola. CHECK POINT 4
In some cases, we need to convert the equation of a parabola to standard form by completing the square on x or y, whichever variable is squared. Let’s see how this is done.
EXAMPLE 5
Graphing a Parabola with Vertex at (h, k)
Find the vertex, focus, and directrix of the parabola given by y2 + 2y + 12x  23 = 0. [–3, 9, 1] by [–6, 6, 1]
Then graph the parabola. Solution We convert the given equation to standard form by completing the square on the variable y. We isolate the terms involving y on the left side. y2 + 2y + 12x  23 = 0 y2 + 2y = 12x + 23 y2 + 2y + 1 = 12x + 23 + 1
This is the given equation. Isolate the terms involving y. Complete the square by adding the square of half the coefficient of y.
(y + 1)2 = 12x + 24
Factor.
To express the equation (y + 1)2 = 12x + 24 in the standard form (y  k)2 = 4p(x  h), we factor out 12 on the right. The standard form of the parabola’s equation is (y + 1)2 = 12(x  2).
Section 10.3 The Parabola
We use (y + 1)2 = 12(x  2) to identify the vertex, (h, k), and the value for p needed to locate the focus and the directrix.
&KTGEVTKZx=
y 5 (–1, 5) 4 .CVWU 3 4GEVWO 2 1
[y − (–1)] 2 = –12(x − 2)
3 4
1
–3 –2 –2 (QEWU –3
–– –4 –5 –6 (–1, –7) –7
6 7
1049
x
8GTVGZ
–
6JKUKU y–k YKVJk=–
6JKUKU p
6JKUKUx–h YKVJh=
The equation is in the standard form (y − k)2 = 4p(x − h). Because y is the squared term, there is horizontal symmetry and the parabola’s equation is not a function.
We see that h = 2 and k = 1. Thus, the vertex of the parabola is (h, k) = (2, 1). Because 4p = 12, p = 3. Based on the standard form of the equation, the axis of symmetry is horizontal. With a negative value for p and a horizontal axis of symmetry, the parabola opens to the left. Because p = 3, the focus is located 3 units to the left of the vertex, (2, 1). Likewise, the directrix is located 3 units to the right of the vertex.
Figure 10.39 The graph of y2 + 2y + 12x  23 = 0, or (y + 1)2 =  12(x  2)
Focus: (h + p, k) = (2 + (–3), –1) = (–1, –1) 6JGXGTVGZ hk KU –
TECHNOLOGY Graph y2 + 2y + 12x  23 = 0 by solving the equation for y.
6JGHQEWUKUWPKVUVQVJG NGHVQHVJGXGTVGZ –
Directrix: x = h − p x = 2 − (–3) = 5
y2 + 2y + (12x − 23) = 0 a=
b=
Use the quadratic formula to solve for y and enter the resulting equations. y1 = y2 =
6JGFKTGEVTKZKUWPKVUVQVJG TKIJVQHVJGXGTVGZ –
c=x–
 2 + 24  4(12x  23)
Thus, the focus is ( 1, 1) and the directrix is x = 5. They are shown in Figure 10.39. To graph the parabola, we will use the vertex, (2, 1), and the two endpoints of the latus rectum. The length of the latus rectum is 4p = 4( 3) = 12 = 12.
2  2  24  4(12x  23) 2 y
y [–4, 8, 1] by [–8, 6, 1]
4 Solve applied problems involving parabolas.
Because the graph has horizontal symmetry, the latus rectum extends 6 units above and 6 units below the focus, ( 1, 1). The endpoints of the latus rectum are ( 1, 1 + 6), or ( 1, 5), and ( 1, 1  6), or ( 1, 7). Passing a smooth curve through the vertex and these two points, we sketch the parabola shown in Figure 10.39. CHECK POINT 5 Find the vertex, focus, and directrix of the parabola given by y2 + 2y + 4x  7 = 0. Then graph the parabola.
Applications Parabolas have many applications. Cables hung between structures to form suspension bridges form parabolas. Arches constructed of steel and concrete, whose main purpose is strength, are usually parabolic in shape. 2CTCDQNC
2CTCDQNC
Suspension bridge
Arch bridge
We have seen that comets in our solar system travel in orbits that are ellipses and hyperbolas. Some comets follow parabolic paths. Only comets with elliptical orbits, such as Halley’s Comet, return to our part of the galaxy.
1050
Chapter 10 Conic Sections and Analytic Geometry
If a parabola is rotated about its axis of symmetry, a parabolic surface is formed. Figure 10.40(a) shows how a parabolic surface can be used to reflect light. Light originates at the focus. Note how the light is reflected by the parabolic surface, so that the outgoing light is parallel to the axis of symmetry. The reflective properties of parabolic surfaces are used in the design of searchlights [see Figure 10.40(b)] and automobile headlights.
BLITZER BONUS The Hubble Space Telescope
1WVIQKPINKIJV #ZKUQHU[OOGVT[ The Hubble Space Telescope
.KIJVCVHQEWU
For decades, astronomers hoped to create an observatory above the atmosphere that would provide an unobscured view of the universe. This vision was realized with the 1990 launching of the Hubble Space Telescope. Barreling along at 17,500 miles per hour 350 miles above Earth, Hubble has led scientists to significant discoveries. Its more than one million images from the ends of the universe have helped to explain the birth of new stars and black holes, confirming the existence of hundreds of galaxies in deep space.
(QEWU
Figure 10.40(a) Parabolic surface reflecting light
Figure 10.40(b) Light from the focus is reflected parallel to the axis of symmetry.
Figure 10.41(a) shows how a parabolic surface can be used to reflect incoming light. Note that light rays strike the surface and are reflected to the focus. This principle is used in the design of reflecting telescopes, radar, and television satellite dishes. Reflecting telescopes magnify the light from distant stars by reflecting the light from these bodies to the focus of a parabolic mirror [see Figure 10.41(b)].
'[GRKGEG +PEQOKPINKIJV #ZKUQHU[OOGVT[
(QEWU
Figure 10.41(a) Parabolic surface reflecting incoming light
EXAMPLE 6 y (2, 2) 2 2 inches
1 –2
1
–1 4 inches
Figure 10.43
2
x
2CTCDQNKEUWTHCEG
Figure 10.41(b) Incoming light rays are reflected to the focus.
Using the Reflection Property of Parabolas
An engineer is designing a flashlight using a parabolic reflecting mirror and a light source, shown in Figure 10.42. The casting has a diameter of 4 inches and a depth of 2 inches. What is the equation of the parabola used to shape the mirror? At what point should the light source be placed relative to the mirror’s vertex? Solution We position the parabola with its vertex at the origin and opening upward (Figure 10.43). Thus, the focus is on the y@axis, located at (0, p). We use the
2 inches
4 inches Figure 10.42 Designing a flashlight
Section 10.3 The Parabola
1051
standard form of the equation in which there is y@axis symmetry, namely, x 2 = 4py. We need to find p. Because (2, 2) lies on the parabola, we let x = 2 and y = 2 in x 2 = 4py. 22 = 4p # 2 4 = 8p p = 12
Substitute 2 for x and 2 for y in x2 = 4py. Simplify. Divide both sides of the equation by 8 and reduce the resulting fraction.
We substitute 12 for p in x 2 = 4py to obtain the standard form of the equation of the parabola. The equation of the parabola used to shape the mirror is x 2 = 4 # 12 y
or
x 2 = 2y.
The light source should be placed at the focus, (0, p). Because p = 12, the light should be placed at 1 0, 12 2 , or 12 inch above the vertex.
CHECK POINT 6 In Example 6, suppose that the casting has a diameter of 6 inches and a depth of 4 inches. What is the equation of the parabola used to shape the mirror? At what point should the light source be placed relative to the mirror’s vertex?
BLITZER BONUS
Five Things Scientists Learned from the Hubble Space Telescope
• The universe is approximately 13.7 billion years old. • There is a high probability that the universe is expanding at everaccelerating rates. • New stars are created from clouds of gas and dust when giant galaxies collide. • Dust rings around stars transform into planets. • Planets are far more common than previously thought. Source: Listomania, Harper Design, 2011
Degenerate Conic Sections We opened the chapter by noting that conic sections are curves that result from the intersection of a cone and a plane. However, these intersections might not result in a conic section. Three degenerate cases occur when the cutting plane passes through the vertex. These degenerate conic sections are a point, a line, and a pair of intersecting lines, illustrated in Figure 10.44.
5 Identify conics without
completing the square.
Identifying Conic Sections without Completing the Square
Point
Line
Two intersecting lines
Figure 10.44 Degenerate conics
Conic sections can be represented both geometrically (as intersecting planes and cones) and algebraically. The equations of the conic sections we have considered in this chapter can be expressed in the form Ax 2 + Cy2 + Dx + Ey + F = 0, in which A and C are not both zero. You can use A and C, the coefficients of x 2 and y2, respectively, to identify a conic section without completing the square.
1052
Chapter 10 Conic Sections and Analytic Geometry
GREAT QUESTION Why do you skip “B” in the coefficients of the equation Ax2 + Cy2 + Dx + Ey + F = 0? Wouldn’t it make more sense to use A, B, C, D, and E as the coefficients? Actually, the general equation is Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0. At this point, we are focusing on equations without an xyterm. (Without an xyterm, we can complete the squares on x and y, as needed, to write the equation in standard form.) Thus, the value of B is zero, and we don’t write a term with a coefficient of 0. The conditions given in the box for identifying a conic section are only valid for equations without an xyterm. In the next section, you will study conic sections with equations that contain an xyterm.
Identifying a Conic Section without Completing the Square A nondegenerate conic section of the form Ax 2 + Cy2 + Dx + Ey + F = 0, in which A and C are not both zero, is • a circle if A = C, • a parabola if AC = 0, • an ellipse if A ≠ C and AC 7 0, and • a hyperbola if AC 6 0.
EXAMPLE 7
Identifying a Conic Section without Completing the Square
Identify the graph of each of the following nondegenerate conic sections: a. b. c. d.
4x 2  25y2  24x + 250y  489 = 0 x 2 + y2 + 6x  2y + 6 = 0 y2 + 12x + 2y  23 = 0 9x 2 + 25y2  54x + 50y  119 = 0.
Solution We use A, the coefficient of x 2, and C, the coefficient of y2, to identify each conic section. a. 4x2 − 25y2 − 24x + 250y − 489 = 0 A=
C=–
Because AC = 4( 25) = 100 6 0, the graph of the equation is a hyperbola. b. x2 + y2 + 6x − 2y + 6 = 0 A=
C=
Because A = C, the graph of the equation is a circle. c. We can write y2 + 12x + 2y  23 = 0 as 0x2 + y2 + 12x + 2y − 23 = 0. A=
C=
Because AC = 0(1) = 0, the graph of the equation is a parabola. Instructor Resources for Section 10.3 in MyLab Math
d. 9x2 + 25y2 − 54x + 50y − 119 = 0 A=
C=
Because AC = 9(25) = 225 7 0 and A ≠ C, the graph of the equation is an ellipse. CHECK POINT 7 Identify the graph of each of the following nondegenerate conic sections: a. 3x 2 + 2y2 + 12x  4y + 2 = 0 b. x 2 + y2  6x + y + 3 = 0 c. y2  12x  4y + 52 = 0 d. 9x 2  16y2  90x + 64y + 17 = 0.
Section 10.3 The Parabola
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CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. The set of all points in a plane that are equidistant from a fixed line and a fixed point is a/an . The fixed line is called the and the fixed point is called the .
Use the graph shown to answer Exercises C6–C9. y
Use the graph shown to answer Exercises C2–C5. y
x (–2, –1)
C6. The equation of the parabola is of the form x
a. (y  1)2 = 4p(x + 2) b. (y + 1)2 = 4p(x + 2) c. (x  2)2 = 4p(y  1) d. (x + 2)2 = 4p(y + 1). C7. If 4p = 4, then the coordinates of the focus are
C2. The equation of the parabola is of the form 2
C8. If 4p = 4, then the equation of the directrix is
2
C9. If 4p = 4, then the length of the latus rectum is endpoints of the latus rectum are and
a. y = 4px b. x = 4py. C3. If 4p = 28, then the coordinates of the focus are .
. . . The .
C10. A nondegenerate conic section in the form
C4. If 4p = 28, then the equation of the directrix is .
Ax 2 + Cy2 + Dx + Ey + F = 0, in which A and C are not both zero is a/an if AC = 0, a/an if A = C, a/an if A ≠ C and AC 7 0, and a/an if AC 6 0.
C5. If 4p = 28, then the length of the latus rectum is . The endpoints of the latus rectum are and .
10.3 EXERCISE SET Practice Exercises
c.
In Exercises 1–4, find the focus and directrix of each parabola with the given equation. Then match each equation to one of the graphs that are shown and labeled (a)–(d).
–4 –3 –2 –1–1
2. x 2 = 4y 4. y2 =  4x b.
y 4 3 2 1 –4 –3 –2 –1–1 –2 –3 –4
1 2 3 4
x –4 –3 –2 –1–1
y
In Exercises 5–16, find the focus and directrix of the parabola with the given equation. Then graph the parabola.
4 3 2 1 1 2 3 4
1 2 3 4
–2 –3 –4
–2 –3 –4
3. x 2 = 4y
y 4 3 2 1
4 3 2 1
1. y2 = 4x
a.
d.
y
x –4 –3 –2 –1–1 –2 –3 –4
1 2 3 4
x
5. 7. 9. 11. 13. 15.
y2 = 16x y2 =  8x x 2 = 12y x 2 = 16y y2  6x = 0 8x 2 + 4y = 0
6. 8. 10. 12. 14. 16.
y2 = 4x y2 = 12x x 2 = 8y x 2 = 20y x 2  6y = 0 8y2 + 4x = 0
x
1054
Chapter 10 Conic Sections and Analytic Geometry
17. Focus: (7, 0); Directrix: x =  7
In Exercises 43–48, convert each equation to standard form by completing the square on x or y. Then find the vertex, focus, and directrix of the parabola. Finally, graph the parabola.
18. Focus: (9, 0); Directrix: x =  9
43. x 2  2x  4y + 9 = 0
44. x 2 + 6x + 8y + 1 = 0
45. y2  2y + 12x  35 = 0
46. y2  2y  8x + 1 = 0
47. x + 6x  4y + 1 = 0
48. x 2 + 8x  4y + 8 = 0
In Exercises 17–30, find the standard form of the equation of each parabola satisfying the given conditions.
19. Focus: ( 5, 0); Directrix: x = 5
2
20. Focus: ( 10, 0); Directrix: x = 10
In Exercises 49–56, identify each equation without completing the square.
21. Focus: (0, 15); Directrix: y = 15 22. Focus: (0, 20); Directrix: y =  20
49. y2  4x + 2y + 21 = 0
23. Focus: (0,  25); Directrix: y = 25
50. y2  4x  4y = 0
24. Focus: (0,  15); Directrix: y = 15
51. 4x 2  9y2  8x  36y  68 = 0
25. Vertex: (2,  3); Focus: (2, 5)
52. 9x 2 + 25y2  54x  200y + 256 = 0
26. Vertex: (1,  3); Focus: (1, 6)
53. 4x 2 + 4y2 + 12x + 4y + 1 = 0
27. Focus: (3, 2); Directrix: x =  1
54. 9x 2 + 4y2  36x + 8y + 31 = 0
28. Focus: (2, 4); Directrix: x =  4
55. 100x 2  7y2 + 90y  368 = 0
29. Focus: ( 3, 4); Directrix: y = 2
56. y2 + 8x + 6y + 25 = 0
30. Focus: (7,  1); Directrix: y =  9
Practice PLUS
In Exercises 31–34, find the vertex, focus, and directrix of each parabola with the given equation. Then match each equation to one of the graphs that are shown and labeled (a)–(d).
In Exercises 57–62, use the vertex and the direction in which the parabola opens to determine the relation’s domain and range. Is the relation a function?
31. (y  1)2 = 4(x  1)
57. y2 + 6y  x + 5 = 0
58. y2  2y  x  5 = 0
32. (x + 1)2 = 4(y + 1)
59. y = x + 4x  3
60. y = x 2  4x + 4
61. x = 4(y  1)2 + 3
62. x = 3(y  1)2  2
2
33. (x + 1)2 = 4(y + 1)
In Exercises 63–68, find the solution set for each system by graphing both of the system’s equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations.
34. (y  1)2 = 4(x  1) a.
b.
y 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3
5 4 3 2 1
x
1 2 3 4
d.
5 4 3 2 1 1 2 3 4
–2 –3
64. b
(y  3)2 = x  2 x + y = 5
65. b
x = y2  3 x = y2  3y
66. b
x = y2  5 x 2 + y2 = 25
y 4 3 2 1
x
(y  2)2 = x + 4 1 c y =  x 2
x
–2 –3
y
–4 –3 –2 –1–1
63.
–4 –3 –2 –1–1
–2 –3 –4
c.
y
–5 –4 –3 –2 –1–1
1 2 3
–2 –3 –4
x
x = (y + 2)2  1 67. b (x  2)2 + (y + 2)2 = 1 68. b
x = 2y2 + 4y + 5 (x + 1)2 + (y  2)2 = 1
Application Exercises In Exercises 35–42, find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. 35. (x  2)2 = 8(y  1)
36. (x + 2)2 = 4(y + 1)
37. (x + 1)2 = 8(y + 1)
38. (x + 2)2 =  8(y + 2)
39. (y + 3)2 = 12(x + 1)
40. (y + 4)2 = 12(x + 2)
41. (y + 1)2 = 8x
42. (y  1)2 =  8x
69. The reflector of a flashlight is in the shape of a parabolic surface. The casting has a diameter of 4 inches and a depth of 1 inch. How far from the vertex should the light bulb be placed? 70. The reflector of a flashlight is in the shape of a parabolic surface. The casting has a diameter of 8 inches and a depth of 1 inch. How far from the vertex should the light bulb be placed?
Section 10.3 The Parabola 71. A satellite dish, like the one shown below, is in the shape of a parabolic surface. Signals coming from a satellite strike the surface of the dish and are reflected to the focus, where the receiver is located. The satellite dish shown has a diameter of 12 feet and a depth of 2 feet. How far from the base of the dish should the receiver be placed?
1055
75. The parabolic arch shown in the figure is 50 feet above the water at the center and 200 feet wide at the base. Will a boat that is 30 feet tall clear the arch 30 feet from the center?
100 ft y
100 ft 50 ft
12 feet
4GEGKXGT
76. A satellite dish in the shape of a parabolic surface has a diameter of 20 feet. If the receiver is to be placed 6 feet from the base, how deep should the dish be?
(6, 2) 2 feet
x
Explaining the Concepts
72. In Exercise 71, if the diameter of the dish is halved and the depth stays the same, how far from the base of the smaller dish should the receiver be placed? 73. The towers of the Golden Gate Bridge connecting San Francisco to Marin County are 1280 meters apart and rise 160 meters above the road. The cable between the towers has the shape of a parabola and the cable just touches the sides of the road midway between the towers. What is the height of the cable 200 meters from a tower? Round to the nearest meter.
y
2CTCDQNKE %CDNG
77. What is a parabola? 78. Explain how to use y2 = 8x to find the parabola’s focus and directrix. 79. If you are given the standard form of the equation of a parabola with vertex at the origin, explain how to determine if the parabola opens to the right, left, upward, or downward. 80. Describe one similarity and one difference between the graphs of y2 = 4x and (y  1)2 = 4(x  1). 81. How can you distinguish parabolas from other conic sections by looking at their equations? 82. Look at the satellite dish shown in Exercise 71. Why must the receiver for a shallow dish be farther from the base of the dish than for a deeper dish of the same diameter? 83. Explain how to identify the graph of Ax 2 + Cy2 + Dx + Ey + F = 0.
(640, 160)
Technology Exercises x
160 meters 1280 mete
rs
74. The towers of a suspension bridge are 800 feet apart and rise 160 feet above the road. The cable between the towers has the shape of a parabola and the cable just touches the sides of the road midway between the towers. What is the height of the cable 100 feet from a tower?
2CTCDQNKE %CDNG
(400, 160)
Road 800 feet
84. Use a graphing utility to graph any five of the parabolas that you graphed by hand in Exercises 5–16. 85. Use a graphing utility to graph any three of the parabolas that you graphed by hand in Exercises 35–42. First solve the given equation for y, possibly using the square root property. Use a graphing utility to graph the parabolas in Exercises 86–87. Write the given equation as a quadratic equation in y and use the quadratic formula to solve for y. Enter each of the equations to produce the complete graph. 86. y2 + 2y  6x + 13 = 0 87. y2 + 10y  x + 25 = 0 In Exercises 88–89, write each equation as a quadratic equation in y and then use the quadratic formula to express y in terms of x. Graph the resulting two equations using a graphing utility. What effect does the xyterm have on the graph of the resulting parabola? 88. 16x 2  24xy + 9y2  60x  80y + 100 = 0 89. x 2 + 223xy + 3y2 + 823x  8y + 32 = 0
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Chapter 10 Conic Sections and Analytic Geometry
Critical Thinking Exercises
102. Consider the system
Make Sense? In Exercises 90–93, determine whether each statement makes sense or does not make sense, and explain your reasoning. 90. I graphed a parabola that opened to the right and contained a maximum point. 91. Knowing that a parabola opening to the right has a vertex at (  1, 1) gives me enough information to determine its graph. 92. I noticed that depending on the values for A and C, assuming that they are not both zero, the graph of Ax 2 + Cy2 + Dx + Ey + F = 0 can represent any of the conic sections. 93. I’m using a telescope in which light from distant stars is reflected to the focus of a parabolic mirror. In Exercises 94–97, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. 94. The parabola whose equation is x = 2y  y2 + 5 opens to the right. 95. If the parabola whose equation is x = ay2 + by + c has its vertex at (3, 2) and a 7 0, then it has no y@intercepts. 96. Some parabolas that open to the right have equations that define y as a function of x. 97. The graph of x = a(y  k) + h is a parabola with vertex at (h, k). 98. Find the focus and directrix of a parabola whose equation is of the form Ax 2 + Ey = 0, A ≠ 0, E ≠ 0. 99. Write the standard form of the equation of a parabola whose points are equidistant from y = 4 and ( 1, 0).
Group Exercise 100. Consult the research department of your library or the Internet to find an example of architecture that incorporates one or more conic sections in its design. Share this example with other group members. Explain precisely how conic sections are used. Do conic sections enhance the appeal of the architecture? In what ways?
Retaining the Concepts 101. Solve the system:
e
y = x  7 x 2 + y2 = 13. 2
(Section 8.4, Example 4)
CHAPTER 10
x  y + z = 3  2y + z = 6 2x  3y = 10. a. Write the system as a matrix equation in the form AX = B. b. Solve the system using the fact that the inverse of c
1 C 0 2
1 2 3
1 3 1 S is C 2 0 4
3 2 5
1 1 S . 2
(Section 9.4, Example 5) 103. Use Cramer’s Rule (determinants) to solve the system: e
x  y = 5 3x + 2y = 0.
(Section 9.5, Example 2) 104. Verify the identity: sin a
3p  xb = cos x 2
(Section 6.2, Example 6)
Preview Exercises Exercises 105–107 will help you prepare for the material covered in the next section. 105. Simplify and write the equation in standard form in terms of x′ and y′ c
22 22 (x′  y′) d c (x′ + y′) d = 1. 2 2
106. a. Make a sketch of an angle u in standard position for which 7 cot 2u =  and 90° 6 2u 6 180°. 24 b. Use your sketch from part (a) to determine the value of cos 2u. c. Use the value of cos 2u from part (b) and the identities 1  cos 2u 1 + cos 2u sin u = and cos u = A 2 A 2 to determine the values of sin u and cos u. d. In part (c), why did we not write ± before the radical in each formula? 107. The equation 3x 2  223xy + y2 + 2x + 223y = 0 is in the form Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0. Use the equation to determine the value of B2  4AC.
MidChapter Check Point
WHAT YOU KNOW: We learned that the four conic sections are the circle, the ellipse, the hyperbola, and the parabola. Prior to this chapter, we graphed circles with center (h, k) and radius r: (x  h)2 + (y  k)2 = r 2.
In this chapter, you learned to graph ellipses centered at the origin and ellipses centered at (h, k): (x  h)2 a2
+
(y  k)2 b2
= 1 or
(x  h)2 b2
+
(y  k)2 a2
= 1, a2 7 b2.
Section 10.3 The Parabola
We saw that the larger denominator (a2) determines whether the major axis is horizontal or vertical. We used vertices and asymptotes to graph hyperbolas centered at the origin and hyperbolas centered at (h, k): (x  h)2 a2

(y  k)2 b2
(y  k)2
= 1 or
a2

(x  h)2 b2
= 1.
We used c 2 = a2  b2 to locate the foci of an ellipse. We used c 2 = a2 + b2 to locate the foci of a hyperbola. We used the vertex and the latus rectum to graph parabolas with vertices at the origin and parabolas with vertices at (h, k): (y  k)2 = 4p(x  h) or (x  h)2 = 4p(y  k). Finally, we learned to identify conic sections without completing the square. A nondegenerate conic section of the form Ax 2 + Cy2 + Dx + Ey + F = 0 is a circle if A = C, a parabola if AC = 0, an ellipse if A ≠ C and AC 7 0, and a hyperbola if AC 6 0. In Exercises 1–5, graph each ellipse. Give the location of the foci. 1. 3. 4.
y2 x2 + = 1 25 4 (x  2)2 16 (x + 2)2 25
+ +
2. 9x 2 + 4y2 = 36
(y + 1)2 25 (y  1)2 16
= 1 = 1
5. x 2 + 9y2  4x + 54y + 49 = 0 In Exercises 6–11, graph each hyperbola. Give the location of the foci and the equations of the asymptotes. 6.
x2  y2 = 1 9
7.
8. y2  4x 2 = 16 10.
(x  2)2 9

y2  x2 = 1 9
9. 4x 2  49y2 = 196
(y + 2)2 16
= 1
11. 4x 2  y2 + 8x + 6y + 11 = 0 In Exercises 12–13, graph each parabola. Give the location of the focus and the directrix. 12. (x  2)2 =  12(y + 1)
13. y2  2x  2y  5 = 0
In Exercises 14–16, identify each equation without completing the square. 14. x 2  4y2  6x + 24y  11 = 0 15. 3y2  x  2y + 1 = 0 16. 9x 2 + 4y2 + 54x  8y + 49 = 0
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In Exercises 17–24, graph each equation. 17. 18. 19. 20. 21. 22. 23. 24.
x 2 + y2 = 4 x + y = 4 x 2  y2 = 4 x 2 + 4y2 = 4 (x + 1)2 + (y  1)2 = 4 x 2 + 4(y  1)2 = 4 (x  1)2  (y  1)2 = 4 (y + 1)2 = 4(x  1)
In Exercises 25–30, find the standard form of the equation of the conic section satisfying the given conditions. 25. Ellipse; Foci: (  4, 0), (4, 0); Vertices: ( 5, 0), (5, 0) 26. Ellipse; Endpoints of major axis: ( 8, 2), (10, 2); Foci: (  4, 2), (6, 2) 27. Hyperbola; Foci: (0, 3), (0, 3); Vertices: (0, 2), (0, 2) 28. Hyperbola; Foci: (  4, 5), (2, 5); Vertices: ( 3, 5), (1, 5) 29. Parabola; Focus: (4, 5); Directrix: y = 1 30. Parabola; Focus: (  2, 6); Directrix: x = 8 31. A semielliptical archway over a oneway road has a height of 10 feet and a width of 30 feet. A truck has a width of 10 feet and a height of 9.5 feet. Will this truck clear the opening of the archway? 32. A lithotriper is used to disentegrate kidney stones. The patient is placed within an elliptical device with the kidney centered at one focus, while ultrasound waves from the other focus hit the walls and are reflected to the kidney stone, shattering the stone. Suppose that the length of the major axis of the ellipse is 40 centimeters and the length of the minor axis is 20 centimeters. How far from the kidney stone should the electrode that sends the ultrasound waves be placed in order to shatter the stone? 33. An explosion is recorded by two forest rangers, one at a primary station and the other at an outpost 6 kilometers away. The ranger at the primary station hears the explosion 6 seconds before the ranger at the outpost. a. Assuming sound travels at 0.35 kilometer per second, write an equation in standard form that gives all the possible locations of the explosion. Use a coordinate system with the two ranger stations on the x@axis and the midpoint between the stations at the origin. b. Graph the equation that gives the possible locations of the explosion. Show the locations of the ranger stations in your drawing. 34. A domed ceiling is a parabolic surface. Ten meters down from the top of the dome, the ceiling is 15 meters wide. For the best lighting on the floor, a light source should be placed at the focus of the parabolic surface. How far from the top of the dome, to the nearest tenth of a meter, should the light source be placed?
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Chapter 10 Conic Sections and Analytic Geometry
Rotation of Axes
SECTION 10.4
WHAT YOU’LL LEARN 1 Use rotation of axes formulas.
2 Write equations of rotated conics in standard form.
3 Identify conics without rotating axes.
How’s that free map from the visitor’s center working out for you? Now that your GPS is out of range, you’re thinking the $10 map may have been more useful for an avid hiker like you. Each map imposes a coordinate system on the terrain you’re about to explore, but the one you have just isn’t cutting it in this situation. We have the same problem with our usual xycoordinate system. This system works well for identifying and graphing equations in the form Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0, as long as B = 0. When B ≠ 0, our usual xycoordinate system is not as useful. Our solution? Change the coordinate system.
1 Use rotation of axes formulas.
Rotation of Axes
y y′
Figure 10.45 shows the graph of
30° 2
x′
7x 2  623xy + 13y2  16 = 0.
1 30° –2
1
–1
x
2
–1 –2
The graph looks like an ellipse, although its major axis neither lies along the x@axis or y@axis, nor is parallel to the x@axis or y@axis. Do you notice anything unusual about the equation? It contains an xy@term. However, look at what happens if we rotate the x@ and y@axes through an angle of 30°. In the rotated x′y′@system, the major axis of the ellipse lies along the x′@axis. We can write the equation of the ellipse in this rotated x′y′@system as y′2 x′2 + = 1. 4 1
Figure 10.45 The graph of 7x 2  6 23xy + 13y2  16 = 0, a rotated ellipse
Observe that there is no x′y′@term in the equation. Except for degenerate cases, the general seconddegree equation Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0
y
y′
u x′
u O
(a) Rotating the x and yaxes through a positive angle u Figure 10.46 Rotating axes
x
represents one of the conic sections. However, due to the xy@term in the equation, these conic sections are rotated in such a way that their axes are no longer parallel to the x@ and y@axes. To reduce y these equations to forms of the conic sections with y′ P= xy= x′y′ which you are already familiar, we use a procedure called rotation of axes. Suppose that the x@ and y@axes are rotated r x′ through a positive angle u, resulting in a new a x′y′coordinate system. This system is shown in u x Figure 10.46(a). The origin in the x′y′@system O is the same as the origin in the xy@system. Point P in Figure 10.46(b) has coordinates (x, y) relative to the xy@system and coordinates (x′, y′) relative to the x′y′@system. Our goal is (b) Describing point P relative to the to obtain formulas relating the old and new xysystem and the rotated x′y′system coordinates. Thus, we need to express x and y in terms of x′, y′, and u.
Section 10.4 Rotation of Axes
1059
Look at Figure 10.46(b). Notice that r = the distance from the origin O to point P. a = the angle from the positive x′@axis to the ray from O through P. Using the definitions of sine and cosine, we obtain x′ : x′ = r cos a r 6JKUKUHTQOVJGTKIJVVTKCPING YKVJCNGICNQPIVJGx′CZKU y′ sin a = : y′ = r sin a r x cos(u + a) = : x = r cos(u + a) 6JKUKUHTQOVJGVCNNGT r TKIJVVTKCPINGYKVJCNGI y CNQPIVJGxCZKU sin(u + a) = : y = r sin(u + a). r cos a =
Thus, x = r cos(u + a)
This is the third of the preceding equations.
= r(cos u cos a  sin u sin a)
Use the formula for the cosine of the sum of two angles.
= (r cos a)cos u  (r sin a)sin u
Apply the distributive property and rearrange factors.
= x′ cos u  y′ sin u.
Use the first and second of the preceding equations: x′ = r cos A and y′ = r sin A.
Similarly, y = r sin(u + a) = r(sin u cos a + cos u sin a) = x′ sin u + y′ cos u.
Rotation of Axes Formulas Suppose an xy@coordinate system and an x′y′@coordinate system have the same origin and u is the angle from the positive x@axis to the positive x′@axis. If the coordinates of point P are (x, y) in the xy@system and (x′, y′) in the rotated x′y′@system, then x = x′ cos u  y′ sin u y = x′ sin u + y′ cos u.
EXAMPLE 1
Rotating Axes
Write the equation xy = 1 in terms of a rotated x′y′@system if the angle of rotation from the x@axis to the x′@axis is 45°. Express the equation in standard form. Use the rotated system to graph xy = 1. Solution With u = 45°, the rotation formulas for x and y are x = x′ cos u  y′ sin u = x′ cos 45°  y′ sin 45° = x′ ¢
22 22 22 (x′  y′) ≤  y′ ¢ ≤ = 2 2 2
y = x′ sin u + y′ cos u = x′ sin 45° + y′ cos 45° = x′ ¢
22 22 22 ≤ + y′ ¢ ≤ = (x′ + y′). 2 2 2
1060
Chapter 10 Conic Sections and Analytic Geometry
Now substitute these expressions for x and y in the given equation, xy = 1. xy = 1 J
22 22 (x′  y′) R J (x′ + y′) R = 1 2 2 2 (x′  y′)(x′ + y′) = 1 4 1 (x′ 2  y′ 2) = 1 2 x′2 y′2 − =1 2 2 a=
x′
y′
y 2
8GTVGZ
√
1 45° –2 8GTVGZ
–√
–1
1
2
x
–1
Substitute the expressions for x and y from the rotation formulas. Multiply:
22 # 22 2 = . 4 2 2
Reduce 24 and multiply the binomials.
Write the equation in standard form: x2 a2
−
y2 b2
= 1.
This equation expresses xy = 1 in terms of the rotated x′y′@system. Can you see that this is the standard form of the equation of a hyperbola? The hyperbola’s center is at (0, 0), with the transverse axis on the x′@axis. The vertices are ( a, 0) and (a, 0). Because a2 = 2, the vertices are 1  22, 0 2 and 1 22, 0 2 , located on the x′@axis. Based on the standard form of the hyperbola’s equation, the equations for the asymptotes are 22 b y′ = { x′ or y′ = { x′. a 22
–2
Figure 10.47 The graph of y′2 x′2 xy = 1 or = 1 2 2
b=
This is the given equation.
The equations of the asymptotes can be simplified to y′ = x′ and y′ = x′, which correspond to the original x@ and y@axes. The graph of the hyperbola is shown in Figure 10.47. Write the equation xy = 2 in terms of a rotated x′y′@system if the angle of rotation from the x@axis to the x′@axis is 45°. Express the equation in CHECK POINT 1
standard form. Use the rotated system to graph xy = 2.
2 Write equations of rotated conics in standard form.
GREAT QUESTION I’m more familiar with the tangent function. Is it ok if I rewrite the amount of rotation B formula as tan 2U = ? A  C No, we use cotangent instead of tangent for a good reason. Notice that we are assuming that B ≠ 0, so the expression on the right A  C , is of the given formula, B never undefined. Therefore, we can always solve for the angle, u. B , the In tan 2u = A  C expression on the right is undefined when A = C, in which case we are not able to solve for u. You might falsely conclude that there is no angle of rotation that will eliminate the xyterm.
Using Rotations to Transform Equations with xy Terms to Standard Equations of Conic Sections We have noted that the appearance of the term Bxy (B ≠ 0) in the general seconddegree equation indicates that the graph of the conic section has been rotated. A rotation of axes through an appropriate angle can transform the equation to one of the standard forms of the conic sections in x′ and y′ in which no x′y′@term appears. Amount of Rotation Formula The general seconddegree equation Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0, B ≠ 0 can be rewritten as an equation in x′ and y′ without an x′y′@term by rotating the axes through angle u, where cot 2u =
A  C . B
Before we learn to apply this formula, let’s see how it can be derived. We begin with the general seconddegree equation Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0, B ≠ 0.
Section 10.4 Rotation of Axes
1061
Then we rotate the axes through an angle u. In terms of the rotated x′y′@system, the general seconddegree equation can be written as 2
A(x′ cos u  y′ sin u) + B(x′ cos u  y′ sin u)(x′ sin u + y′ cos u) 2
+ C(x′ sin u + y′ cos u) + D(x′ cos u  y′ sin u) + E(x′ sin u + y′ cos u) + F = 0. After a lot of simplifying that involves expanding and collecting like terms, you will obtain the following equation: 9GYCPVCTQVCVKQPVJCVTGUWNVUKPPQx′y′VGTO
(A cos2 u + B sin u cos u + C sin2 u)x′2 + [B(cos2 u − sin2 u) + 2(C − A)(sin u cos u)]x′y′ + (A sin2 u − B sin u cos u + C cos2 u)y′2 + (D cos u + E sin u)x′ + (–D sin u + E cos u)y′ + F = 0. If this looks somewhat ghastly, take a deep breath and focus only on the x′y′@term. We want to choose u so that the coefficient of this term is zero. This will give the required rotation that results in no x′y′@term. B(cos2 u  sin2 u) + 2(C  A) sin u cos u = 0 B cos 2u + (C  A) sin 2u = 0
Set the coefficient of the x′y′@term equal to 0. Use the doubleangle formulas: cos 2U = cos2 U − sin2 U and sin 2U = 2 sin U cos U.
B cos 2u = (C  A) sin 2u
Subtract (C − A) sin 2U from both sides.
B cos 2u = (A  C) sin 2u
Simplify.
(A  C) sin 2u B cos 2u = B sin 2u B sin 2u cos 2u A  C = sin 2u B A  C cot 2u = B
Divide both sides by B sin 2U.
Simplify.
Apply a quotient identity: cot 2U =
GREAT QUESTION What do I do after substituting the expressions for x and y from the rotation formulas into the given equation? You must simplify the resulting equation by expanding and collecting like terms. Work through this process slowly and carefully, allowing lots of room on your paper. If your rotation equations are correct but you obtain an equation that has an x′y′@term, you have made an error in the algebraic simplification.
cos 2U . sin 2U
If cot 2u is positive, we will select u so that 0° 6 u 6 45°. If cot 2u is negative, we will select u so that 45° 6 u 6 90°. Thus u, the angle of rotation, is always an acute angle. Here is a stepbystep procedure for writing the equation of a rotated conic section in standard form: Writing the Equation of a Rotated Conic in Standard Form 1. Use the given equation Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0, B ≠ 0 to find cot 2u. A  C cot 2u = B 2. Use the expression for cot 2u to determine u, the angle of rotation. 3. Substitute u in the rotation formulas x = x′ cos u  y′ sin u and y = x′ sin u + y′ cos u and simplify. 4. Substitute the expressions for x and y from the rotation formulas in the given equation and simplify. The resulting equation should have no x′y′@term. 5. Write the equation involving x′ and y′ in standard form. Using the equation in step 5, you can graph the conic section in the rotated x′y′@system.
1062
Chapter 10 Conic Sections and Analytic Geometry
EXAMPLE 2
Writing the Equation of a Rotated Conic Section in Standard Form
Rewrite the equation 7x 2  623xy + 13y2  16 = 0 in a rotated x′y′@system without an x′y′@term. Express the equation in the standard form of a conic section. Graph the conic section in the rotated system. Solution Step 1 Use the given equation to find cot 2U. We need to identify the constants A, B, and C in the given equation. 7x2 − 6"3xy + 13y2 − 16 = 0 AKUVJG EQGHƂEKGPVQH VJGxVGTO A=
BKUVJG EQGHƂEKGPVQH VJGxyVGTO B=–√
CKUVJG EQGHƂEKGPVQH VJGyVGTO C=
The appropriate angle u through which to rotate the axes satisfies the equation 23 A  C 7  13 6 1 . = = = or B 3 623 623 23 Step 2 Use the expression for cot 2U to determine the angle of rotation. We 23 have cot 2u = . Based on our knowledge of exact values for trigonometric 3 functions, we conclude that 2u = 60°. Thus, u = 30°. Step 3 Substitute U in the rotation formulas x = x′ cos U − y′ sin U and y = x′ sin U + y′ cos U and simplify. Substituting 30° for u, cot 2u =
x = x′ cos 30°  y′ sin 30° = x′ ¢
23x′  y′ 23 1 ≤  y′ a b = 2 2 2
x′ + 23y′ 1 23 y = x′ sin 30° + y′ cos 30° = x′ a b + y′ ¢ ≤ = . 2 2 2 Step 4 Substitute the expressions for x and y from the rotation formulas in the given equation and simplify. 7x 2  623xy + 13y2  16 = 0 7¢
23x′  y′ 2 23x′  y′ x′ + 23y′ ≤  623¢ ≤¢ ≤ 2 2 2 + 13¢
7¢
This is the given equation.
x′ + 23y′ 2 ≤  16 = 0 2
Substitute the expressions for x and y from the rotation formulas.
3x′2  223x′y′ + y′2 23x′2 + 3x′y′  x′y′  23y′2 ≤  623¢ ≤ 4 4 + 13¢
x′2 + 223x′y′ + 3y′2 ≤  16 = 0 4
7 1 3x′2  223x′y′ + y′2 2  623 1 23x′2 + 2x′y′  23y′2 2
+ 13(x′2 + 223x′y′ + 3y′2)  64 = 0
21x′  1423x′y′ + 7y′  18x′  1223x′y′ + 18y′ 2
2
2
Multiply both sides by 4.
2
+ 13x′2 + 2623x′y′ + 39y′2  64 = 0 21x′2  18x′2 + 13x′2  1423x′y′  1223x′y′ + 2623x′y′ + 7y′2 + 18y′2 + 39y′2  64 = 0 16x′ + 64y′  64 = 0 2
Square and multiply.
2
Distribute throughout parentheses.
Rearrange terms. Combine like terms.
Section 10.4 Rotation of Axes
1063
Do you see how we “lost” the x′y′@term in the last equation? 1423x′y′  1223x′y′ + 2623x′y′ = 2623x′y′ + 2623x′y′ = 0x′y′ = 0 Step 5 Write the equation involving x′ and y′ in standard form. We can express 16x′2 + 64y′2  64 = 0, an equation of an ellipse, in the standard form y2 x2 + 2 = 1. 2 a b 16x′2 + 64y′2  64 = 0
y y′
30° 2
16x′2 + 64y′2 = 64
x′
64y′ 64 16x′2 + = 64 64 64
30° 1
–1
Add 64 to both sides.
2
1 –2
This equation describes the ellipse relative to a system rotated through 30°.
2
x
y′2 x′2 + = 1 4 1
–1 –2
Divide both sides by 64.
Simplify.
y′2 x′2 + = 1 is the standard form of the equation of an ellipse. The 4 1 major axis is on the x′@axis and the vertices are ( 2, 0) and (2, 0). The minor axis is on the y′@axis with endpoints (0, 1) and (0, 1). The graph of the ellipse is shown in Figure 10.45. Does this graph look familiar? It should—you saw it earlier in this section on page 1058.
The equation Figure 10.45 (repeated) The graph of 7x 2  6 23xy + 13y2  16 = 0 or y′2 x′2 = 1, a rotated ellipse 4 1
CHECK POINT 2
Rewrite the equation 2x 2 + 23xy + y2  2 = 0
in a rotated x′y′@system without an x′y′@term. Express the equation in the standard form of a conic section. Graph the conic section in the rotated system.
TECHNOLOGY In order to graph a general seconddegree equation in the form Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0 using a graphing utility, it is necessary to solve for y. Rewrite the equation as a quadratic equation in y. Cy2 + (Bx + E)y + (Ax 2 + Dx + F) = 0 By applying the quadratic formula, the graph of this equation can be obtained by entering y1 =
(Bx + E) + 2(Bx + E)2  4C(Ax 2 + Dx + F) 2C
and y2 =
 (Bx + E)  2(Bx + E)2  4C(Ax 2 + Dx + F) 2C
The graph of 7x 2  623xy + 13y2  16 = 0 is shown on the right in a [  3.2, 3.2, 1] by [  2, 2, 1] viewing rectangle. The graph was obtained by entering the equations for y1 and y2 shown above with A = 7, B = 623, C = 13, D = 0, E = 0, and F =  16.
.
1064
Chapter 10 Conic Sections and Analytic Geometry
In Example 2 and Check Point 2, we found u, the angle of rotation, directly because 23 we recognized as the value of cot 60°. What do we do if cot 2u is not the cotangent 3 of one of the familiar angles? We use cot 2u to find sin u and cos u as follows: • Use a sketch of cot 2u to find cos 2u. • Find sin u and cos u using the identities sin u =
1 − cos 2u Å 2
and
cos u =
$GECWUGuKUCPCEWVGCPINGVJGRQUKVKXG USWCTGTQQVUCTGCRRTQRTKCVG
1 + cos 2u . Å 2
The resulting values for sin u and cos u are used to write the rotation formulas that give an equation with no x′y′@term.
EXAMPLE 3
Graphing the Equation of a Rotated Conic
Graph relative to a rotated x′y′@system in which the equation has no x′y′@term: 16x 2  24xy + 9y2 + 110x  20y + 100 = 0. y
Solution
√ –+= 24 2u
–7 Figure 10.48 Using cot 2u to find cos 2u
x
Step 1 Use the given equation to find cot 2U. With A = 16, B = 24, and C = 9, we have A  C 16  9 7 cot 2u = = =  . B 24 24 Step 2 Use the expression for cot 2U to determine sin U and cos U. A rough sketch showing cot 2u is given in Figure 10.48. Because u is always acute and cot 2u is negative, 2u is in quadrant II. The third side of the triangle is found using r = 2x 2 + y2. Thus, r = 2( 7)2 + 242 = 2625 = 25. By the definition of the cosine function, x 7 7 cos 2u = = =  . r 25 25 Now we use identities to find values for sin u and cos u. 1  cos 2u sin u = = B 2 S
1  a2
7 b 25
25 32 7 + 4 25 25 25 32 16 = = = = = S 2 S2 B 50 B 25 5
cos u =
B
1 + cos 2u = 2 S
1 + a2
7 b 25
7 25 18 25 25 25 18 9 3 = = = = = S 2 S2 B 50 B 25 5 Step 3 Substitute sin U and cos U in the rotation formulas x = x′cos U − y′ sin U and y = x′ sin U + y′ cos U and simplify. Substituting 45 for sin u and 35 for cos u, 3x′  4y′ 3 4 x = x′a b  y′a b = 5 5 5 4x′ + 3y′ 4 3 y = x′a b + y′a b = . 5 5 5
Section 10.4 Rotation of Axes
1065
Step 4 Substitute the expressions for x and y from the rotation formulas in the given equation and simplify. 16x 2  24xy + 9y2 + 110x  20y + 100 = 0
16¢
3x′  4y′ 2 3x′  4y′ 4x′ + 3y′ 4x′ + 3y′ 2 ≤  24¢ ≤¢ ≤ + 9¢ ≤ 5 5 5 5 + 110¢
3x′  4y′ 4x′ + 3y′ ≤  20¢ ≤ + 100 = 0 5 5
This is the given equation. Substitute the expressions for x and y from the rotation formulas.
Work with the preceding equation. Take a few minutes to expand, multiply both sides of the equation by 25, and combine like terms. You should obtain y′2 + 2x′  4y′ + 4 = 0, an equation that has no x′y′@term. Step 5 Write the equation involving x′ and y′ in standard form. With only one variable that is squared, we have the equation of a parabola. We need to write the equation in the standard form (y  k)2 = 4p(x  h). y′2 + 2x′  4y′ + 4 = 0
This is the equation without an x′y′@term.
y′2  4y′ = 2x′  4
Isolate the terms involving y′.
y′2  4y′ + 4 = 2x′  4 + 4 (y′  2)2 = 2x′
Complete the square by adding the square of half the coefficient of y′. Factor.
The standard form of the parabola’s equation in the rotated x′y′@system is (y′ − 2)2 = –2x′. 6JKUKU y′–k YKVJk=
y′
8GTVGZ 4
2
–4 –2 .CVWU –2 4GEVWO
x′
y
4
u = sin–1 5 ≈ 53° x 2 4
–4
Figure 10.49 The graph of (y′  2)2 =  2x′ in a rotated x′y′@system
6JKUKU p
6JKUKUx′–h YKVJh=
We see that h = 0 and k = 2. Thus, the vertex of the parabola in the x′y′@system is (h, k) = (0, 2). We can use the x′y′@system to graph the parabola. Using a calculator to 4 4 solve sin u = , we find that u = sin1 ≈ 53°. Rotate the axes through 5 5 1 1 approximately 53°. With 4p = 2 and p =  , the parabola’s focus is unit to 2 2 the left of the vertex, (0, 2). Thus, the focus in the x′y′@system is 1  12, 2 2 . To graph the parabola, we use the vertex, (0, 2), and the two endpoints of the latus rectum. length of latus rectum = 4p = 2 = 2
The latus rectum extends 1 unit above and 1 unit below the focus, 1  12, 2 2 . Thus, the endpoints of the latus rectum in the x′y′@system are 1  12, 3 2 and 1  12, 1 2 . Using the rotated system, pass a smooth curve through the vertex and the two endpoints of the latus rectum. The graph of the parabola is shown in Figure 10.49. CHECK POINT 3
Graph relative to a rotated x′y′@system in which the equation
has no x′y′@term: 4x 2  4xy + y2  825x  1625y = 0.
1066
Chapter 10 Conic Sections and Analytic Geometry
3 Identify conics without rotating axes.
Identifying Conic Sections without Rotating Axes We now know that the general seconddegree equation Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0, B ≠ 0 can be rewritten as A′x′2 + C′y′2 + D′x′ + E′y′ + F′ = 0 in a rotated x′y′@system. A relationship between the coefficients of the two equations is given by B2  4AC = 4A′C′. We also know that A′ and C′ can be used to identify the graph of the rotated equation. Because B2  4AC = 4A′C′, we can also use B2  4AC to identify the graph of the general seconddegree equation.
Instructor Resources for Section 10.4 in MyLab Math
Identifying a Conic Section without a Rotation of Axes A nondegenerate conic section of the form Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0 is • a parabola if B2  4AC = 0, • an ellipse or a circle if B2  4AC 6 0, and • a hyperbola if B2  4AC 7 0.
TECHNOLOGY Graphic Connections The graph of is shown in a 3 1.6, 1.6, 0.25 4 by 3 1, 1, 0.25 4 viewing rectangle. The graph verifies that the equation represents a rotated hyperbola. 11x 2 + 1023xy + y2  4 = 0
EXAMPLE 4
Identifying a Conic Section without Rotating Axes
Identify the graph of 11x 2 + 1023xy + y2  4 = 0. Solution We use A, B, and C to identify the conic section. 11x2 + 10"3xy + y2 − 4 = 0 A=
B=√
C=
B2  4AC = (1023)2  4(11)(1) = 100 # 3  44 = 256 7 0 Because B2  4AC 7 0, the graph of the equation is a hyperbola. CHECK POINT 4
Identify the graph of 3x 2  223xy + y2 + 2x + 223y = 0.
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. The general seconddegree equation
C2. A nondegenerate conic section of the form
Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0, B ≠ 0 can be rewritten as an equation in x′ and y′ without an x′y′@term by rotating the axes through an acute angle u that satisfies the equation
.
Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0 is a/an or a/an a/an
if B2  4AC = 0, a/an if B2  4AC 6 0, and if B2  4AC 7 0.
Section 10.4 Rotation of Axes
1067
10.4 EXERCISE SET Practice Exercises
26. 32x 2  48xy + 18y2  15x  20y = 0
In Exercises 1–6, write each equation in terms of a rotated x′y′@system using u, the angle of rotation. Write the equation involving x′ and y′ in standard form.
27. x 2 + 4xy  2y2  1 = 0
1. xy = 1; u = 45°
28. 3xy  4y2 + 18 = 0 29. 34x 2  24xy + 41y2  25 = 0 30. 6x 2  6xy + 14y2  45 = 0
2. xy = 4; u = 45° 3. x 2  4xy + y2  3 = 0; u = 45°
In Exercises 31–36, identify each equation without applying a rotation of axes.
4. x 2 + 6xy + y2  8 = 0; u = 45°
31. 5x 2  2xy + 5y2  12 = 0
5. 23x 2 + 2623xy  3y2  144 = 0; u = 30° 6. 13x 2 + 1823xy  5y2  154 = 0; u = 30°
32. 10x 2 + 24xy + 17y2  9 = 0 33. 24x 2 + 1623xy + 8y2  x + 23y  8 = 0 34. 3x 2  223xy + y2 + 2x + 223y = 0 35. 23x 2 + 2623xy  3y2  144 = 0
In Exercises 7–18, write the appropriate rotation formulas so that in a rotated system the equation has no x′y′@term.
36. 4xy + 3y2 + 4x + 6y  1 = 0
7. x 2 + xy + y2  10 = 0 8. x 2 + 4xy + y2  3 = 0
Practice PLUS
9. 3x 2  10xy + 3y2  32 = 0
In Exercises 37–40,
10. 5x 2  8xy + 5y2  9 = 0 11. 11x 2 + 1023xy + y2  4 = 0 12. 7x 2  623xy + 13y2  16 = 0 13. 10x 2 + 24xy + 17y2  9 = 0 14. 32x 2  48xy + 18y2  15x  20y = 0
• If the graph of the equation is an ellipse, find the coordinates of the endpoints of the minor axis. • If the graph of the equation is a hyperbola, find the equations of the asymptotes. • If the graph of the equation is a parabola, find the coordinates of the vertex.
16. 3xy  4y2 + 18 = 0
Express answers relative to an x′y′@system in which the given equation has no x′y′@term. Assume that the x′y′@system has the same origin as the xy@system.
17. 34x 2  24xy + 41y2  25 = 0
37. 5x 2  6xy + 5y2  8 = 0
18. 6x 2  6xy + 14y2  45 = 0
38. 2x 2  4xy + 5y2  36 = 0
15. x 2 + 4xy  2y2  1 = 0
In Exercises 19–30, a. Rewrite the equation in a rotated x′y′@system without an x′y′@term. Use the appropriate rotation formulas from Exercises 7–18. b. Express the equation involving x′ and y′ in the standard form of a conic section. c. Use the rotated system to graph the equation. 19. x 2 + xy + y2  10 = 0 20. x 2 + 4xy + y2  3 = 0 21. 3x 2  10xy + 3y2  32 = 0
39. x 2  4xy + 4y2 + 525y  10 = 0 40. x 2 + 4xy  2y2  6 = 0
Explaining the Concepts 41. If there is a 60° angle from the positive x@axis to the positive x′@axis, explain how to obtain the rotation formulas for x and y. 42. How do you obtain the angle of rotation so that a general seconddegree equation has no x′y′@term in a rotated x′y′@system?
23. 11x 2 + 1023xy + y2  4 = 0
43. What is the most timeconsuming part in using a graphing utility to graph a general seconddegree equation with an xy@term?
24. 7x 2  623xy + 13y2  16 = 0
44. Explain how to identify the graph of
22. 5x 2  8xy + 5y2  9 = 0
25. 10x 2 + 24xy + 17y2  9 = 0
Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0.
1068
Chapter 10 Conic Sections and Analytic Geometry
Technology Exercises
Retaining the Concepts
In Exercises 45–51, use a graphing utility to graph each equation.
60. Use the graph of f to determine each of the following. Where applicable, use interval notation.
45. x + 4xy + y  3 = 0 2
2
y
46. 7x 2 + 8xy + y2  1 = 0 5 4 3 2 1
47. 3x 2 + 4xy + 6y2  7 = 0 48. 3x 2  6xy + 3y2 + 10x  8y  2 = 0 49. 9x + 24xy + 16y + 90x  130y = 0 2
2
50. x 2 + 4xy + 4y2 + 1025x  9 = 0
–5 –4 –3 –2 –1–1
51. 7x 2 + 6xy + 2.5y2  14x + 4y + 9 = 0
52. I graphed 2x 2  3y2 + 6y + 4 = 0 by using the procedure for writing the equation of a rotated conic in standard form. 53. In order to graph an ellipse whose equation contained an xy@term, I used a rotated coordinate system that placed the ellipse’s center at the origin. 54. Although the algebra of rotations can get ugly, the main idea is that rotation through an appropriate angle will transform a general seconddegree equation into an equation in x′ and y′ without an x′y′@term. 55. I can verify that 2xy  9 = 0 is the equation of a hyperbola by rotating the axes through 45° or by showing that B2  4AC 7 0. 56. Explain the relationship between the graph of 3x 2  2xy + 3y2 + 2 = 0 and the sound made by one hand clapping. Begin by following the directions for Exercises 19–30. (You will first need to write rotation formulas that eliminate the x′y′@term.) 57. What happens to the equation x 2 + y2 = r 2 in a rotated x′y′@system?
In Exercises 58–59, let Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0 be an equation of a conic section in an xy@coordinate system. Let A′x′2 + B′x′y′ + C′y′2 + D′x′ + E′y′ + F′ = 0 be the equation of the conic section in the rotated x′y′@coordinate system. Use the coefficients A′, B′, and C′, shown in the equation with the voice balloon pointing to B′ on page 1061, to prove the following relationships. 58. A′ + C′ = A + C 59. B′2  4A′C′ = B2  4AC
1 2 3 4 5
x
–2 –3 –4 –5
Critical Thinking Exercises Make Sense? In Exercises 52–55, determine whether each statement makes sense or does not make sense, and explain your reasoning.
y=f x
a. the domain of f b. the range of f c. the xintercepts d. the yintercept e. interval(s) where f is increasing f. interval(s) where f is decreasing g. interval(s) where f is constant h. f(4) i. the number at which f has a minimum j. the minimum value of f (Section 2.1, Examples 7 and 8; Section 2.2, Example 1, Figure 2.17) 61. Find the zeros of f(x) = (x + 3)2(2x  5)3 and give the multiplicity of each zero. State whether the graph crosses the xaxis or touches the xaxis and turns around at each zero. (Section 3.2, Example 7) 62. A kite flies at a height of 35 feet when 60 feet of string is out. If the string is in a straight line, find the angle that it makes with the ground. Round to the nearest tenth of a degree. (Section 5.8, Example 3) 63. Verify the identity: sec x = sin x. cot x + tan x (Section 6.1, Example 2)
Preview Exercises Exercises 64–66 will help you prepare for the material covered in the next section. In each exercise, graph the equation in a rectangular coordinate system. 1 64. y2 = 4(x + 1) 65. y = x 2 + 1, x Ú 0 2 y2 x2 66. + = 1 25 4
Section 10.5 Parametric Equations
SECTION 10.5
1069
Parametric Equations What a baseball game! You got to see the great Albert Pujols of the Los Angeles Angels blast a powerful homer. In less than 8 seconds, the parabolic path of his home run took the ball a horizontal distance of over 1000 feet. Is there a way to model this path that gives both the ball’s location and the time that it is in each of its positions? In this section, we look at ways of describing curves that reveal the where and the when of motion.
WHAT YOU'LL LEARN 1 Use point plotting to graph plane curves described by parametric equations.
2 Eliminate the parameter. 3 Find parametric equations for functions.
Plane Curves and Parametric Equations
4 Understand the
advantages of parametric representations.
You throw a ball from a height of 6 feet, with an initial velocity of 90 feet per second and at an angle of 40° with the horizontal. After t seconds, the location of the ball can be described by
x = (90 cos 40°)t
and
6JKUKUVJGDCNNoU JQTK\QPVCNFKUVCPEGKPHGGV
y = 6 + (90 sin 40°)t − 16t 2. 6JKUKUVJGDCNNoU XGTVKECNJGKIJVKPHGGV
Because we can use these equations to calculate the location of the ball at any time t, we can describe the path of the ball. For example, to determine the location when t = 1 second, substitute 1 for t in each equation: x = (90 cos 40°)t = (90 cos 40°)(1) ≈ 68.9 feet y = 6 + (90 sin 40°)t  16t 2 = 6 + (90 sin 40°)(1)  16(1)2 ≈ 47.9 feet. This tells us that after one second, the ball has traveled a horizontal distance of approximately 68.9 feet, and the height of the ball is approximately 47.9 feet. Figure 10.50 displays this information and the results for calculations corresponding to t = 2 seconds and t = 3 seconds.
y (feet)
t=UGE x≈HV y≈HV
t=UGE x≈HV y≈HV
60
t=UGE x≈HV y≈HV
40 20 40
80
120
160
200
240
x (feet)
Figure 10.50 The location of a thrown ball after 1, 2, and 3 seconds
The voice balloons in Figure 10.50 tell where the ball is located and when the ball is at a given point (x, y) on its path. The variable t, called a parameter, gives the various times for the ball’s location. The equations that describe where the ball is located express both x and y as functions of t and are called parametric equations.
1070
y (feet)
Chapter 10 Conic Sections and Analytic Geometry
t=UGE x≈HV y≈HV
t=UGE x≈HV y≈HV
60
x = (90 cos 40°)t t=UGE x≈HV y≈HV
6JKUKUVJGRCTCOGVTKE GSWCVKQPHQTx
y = 6 + (90 sin 40°)t − 16t 2 6JKUKUVJGRCTCOGVTKE GSWCVKQPHQTy
40 20 40
80
120
160
200
240
x (feet)
The collection of points (x, y) in Figure 10.50 is called a plane curve.
Figure 10.50 (repeated)
Plane Curves and Parametric Equations Suppose that t is a number in an interval I. A plane curve is the set of ordered pairs (x, y), where x = f(t), y = g(t) for t in interval I. The variable t is called a parameter, and the equations x = f (t) and y = g(t) are called parametric equations for the curve.
1 Use point plotting to graph plane curves described by parametric equations.
Graphing Plane Curves Graphing a plane curve represented by parametric equations involves plotting points in the rectangular coordinate system and connecting them with a smooth curve. Graphing a Plane Curve Described by Parametric Equations 1. Select some values of t on the given interval. 2. For each value of t, use the given parametric equations to compute x and y. 3. Plot the points (x, y) in the order of increasing t and connect them with a smooth curve. Take a second look at Figure 10.50. Do you notice arrows along the curve? These arrows show the direction, or orientation, along the curve as t increases. After graphing a plane curve described by parametric equations, use arrows between the points to show the orientation of the curve corresponding to increasing values of t.
EXAMPLE 1
Graphing a Curve Defined by Parametric Equations
Graph the plane curve defined by the parametric equations: x = t 2  1,
y = 2t,
2 … t … 2.
Solution Step 1 Select some values of t on the given interval. We will select integral values of t on the interval 2 … t … 2. Let t = 2, 1, 0, 1, and 2. Step 2 For each value of t, use the given parametric equations to compute x and y. We organize our work in a table. The first column lists the choices for the parameter t. The next two columns show the corresponding values for x and y. The last column lists the ordered pair (x, y). t
x = t2 − 1
y = 2t
(x, y)
2
(  2)  1 = 4  1 = 3
2(  2) =  4
(3, 4)
1
(  1)  1 = 1  1 = 0
2(  1) =  2
(0, 2)
0
0  1 = 1
2(0) = 0
( 1, 0)
1
1  1 = 0
2(1) = 2
(0, 2)
2
2  1 = 4  1 = 3
2(2) = 4
(3, 4)
2 2
2
2
2
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Section 10.5 Parametric Equations y
Step 3 Plot the points (x, y) in the order of increasing t and connect them with a smooth curve. The plane curve defined by the parametric equations on the given interval is shown in Figure 10.51. The arrows show the direction, or orientation, along the curve as t varies from 2 to 2.
t=
5 4 t= 3 2 t= – 1 –5 –4 –3 –2 –1–1
CHECK POINT 1 1 2 3 4 5
x
–2 t=– – –3 –4 –5
Graph the plane curve defined by the parametric equations: x = t 2 + 1,
y = 3t,
2 … t … 2.
Eliminating the Parameter t=– –
Figure 10.51 The plane curve defined by x = t 2  1, y = 2t,  2 … t … 2
2 Eliminate the parameter.
TECHNOLOGY A graphing utility can be used to obtain a plane curve represented by parametric equations. Set the mode to parametric and enter the equations. You must enter the minimum and maximum values for t and an increment setting for t (tstep). The setting tstep determines the number of points the graphing utility will plot. Shown below is the plane curve for x = t2  1 y = 2t in a [ 8, 8, 1] by [ 5, 5, 1] viewing rectangle with tmin = 2, tmax = 2, and tstep = 0.01.
The graph in Figure 10.51 shows the plane curve for x = t 2  1, y = 2t, 2 … t … 2. Even if we examine the parametric equations carefully, we may not be able to tell that the corresponding plane curve is a portion of a parabola. By eliminating the parameter, we can write one equation in x and y that is equivalent to the two parametric equations. The voice balloons illustrate this process. $GIKPYKVJVJG RCTCOGVTKEGSWCVKQPU
x = t2 − 1 y = 2t
5QNXGHQTt KPQPGQHVJGGSWCVKQPU
Using y = 2t, y t= . 2
5WDUVKVWVGVJGGZRTGUUKQPHQTt KPVJGQVJGTRCTCOGVTKEGSWCVKQP
Using t =
y 2
and x = t 2 − 1,
y 2 x = a b − 1. 2
y2  1, can be written as 4 2 y = 4(x + 1). This is the standard form of the equation of a parabola with vertex
The rectangular equation (the equation in x and y), x =
at ( 1, 0) and axis of symmetry along the x@axis. Because the parameter t is restricted to the interval [ 2, 2], the plane curve in Figure 10.51 and the technology box on the left shows only a part of the parabola. Our discussion illustrates a second method for graphing a plane curve described by parametric equations. Eliminate the parameter t and graph the resulting rectangular equation in x and y. However, you may need to change the domain of the rectangular equation to be consistent with the domain for the parametric equation in x. This situation is illustrated in Example 2.
EXAMPLE 2
Finding and Graphing the Rectangular Equation of a Curve Defined Parametrically
Sketch the plane curve represented by the parametric equations x = 1t and y = 12 t + 1 by eliminating the parameter. Solution We eliminate the parameter t and then graph the resulting rectangular equation. $GIKPYKVJVJG RCTCOGVTKEGSWCVKQPU
x = !t y=
1 2
t+1
5QNXGHQTt KPQPGQHVJGGSWCVKQPU
Using x = !t and squaring both sides, t = x2.
5WDUVKVWVGVJGGZRTGUUKQPHQTt KPVJGQVJGTRCTCOGVTKEGSWCVKQP
Using t = x2 and y = y=
1 2 x 2
1 2
t + 1,
+ 1.
Because t is not limited to a closed interval, you might be tempted to graph the entire bowlshaped parabola whose equation is y = 12 x 2 + 1. However, take a second look at the parametric equation for x: x = 1t.
1072
Chapter 10 Conic Sections and Analytic Geometry y
The equation x = 2t is defined only when t Ú 0. Thus, x is nonnegative. The plane curve is the parabola given by y = 12 x 2 + 1 with the domain restricted to x Ú 0. The plane curve is shown in Figure 10.52.
9 8 7 6 5 4 3 2 1
CHECK POINT 2
x = 1t and y = 2t  1 by eliminating the parameter. 1 2 3 4 5
–5 –4 –3 –2 –1–1
x
Eliminating the parameter is not always a simple matter. In some cases, it may not be possible. When this occurs, you can use point plotting to obtain a plane curve. Trigonometric identities can be helpful in eliminating the parameter. For example, consider the plane curve defined by the parametric equations
Figure 10.52 The plane curve for x = 1t and y = 12 t + 1, or y = 12 x 2 + 1, x Ú 0
t=
Sketch the plane curve represented by the parametric equations
x = sin t, y = cos t, 0 … t 6 2p.
y
We use the trigonometric identity sin2 t + cos2 t = 1 to eliminate the parameter. Square each side of each parametric equation and then add. x2 = sin2 t y2 = cos2 t
x t= p
p t=
t=p Figure 10.53 The plane curve defined by x = sin t, y = cos t, 0 … t 6 2p
x2 + y2 = sin2 t + cos2 t
6JKUKUVJGUWOQHVJGVYQGSWCVKQPU CDQXGVJGJQTK\QPVCNNKPGU
Using a Pythagorean identity, we write this equation as x 2 + y2 = 1. The plane curve is a circle with center (0, 0) and radius 1. It is shown in Figure 10.53.
EXAMPLE 3
Finding and Graphing the Rectangular Equation of a Curve Defined Parametrically
Sketch the plane curve represented by the parametric equations x = 5 cos t, y = 2 sin t, 0 … t … p by eliminating the parameter. Solution We eliminate the parameter using the identity cos2 t + sin2 t = 1. To apply the identity, divide the parametric equation for x by 5 and the parametric equation for y by 2. y x = cos t and = sin t 5 2 Square and add these two equations. x2 = cos2 t 25 y2 = sin2 t 4 y2 x2 + = cos2 t + sin2 t 25 4
6JKUKUVJGUWOQHVJGVYQGSWCVKQPU CDQXGVJGJQTK\QPVCNNKPGU
Using a Pythagorean identity, we write this equation as y2 x2 + = 1. 25 4
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Section 10.5 Parametric Equations
This rectangular equation is the standard form of the equation for an ellipse centered at (0, 0). y2 x2 + =1 25 4 a='PFRQKPVUQH OCLQTCZKUCTGWPKVUNGHV CPFTKIJVQHEGPVGT
b='PFRQKPVUQH OKPQTCZKUCTGWPKVU CDQXGCPFDGNQYEGPVGT
The ellipse is shown in Figure 10.54(a). However, this is not the plane curve. We are given that 0 … t … p. Because t is restricted to the interval [0, p], the plane curve is only a portion of the ellipse. Use the starting and ending values for t, 0 and p, respectively, and a value of t in the interval (0, p) to find which portion to include. $GIKPCVt=
x = 5 cos t = 5 cos 0 = 5 ∙ 1 = 5 y = 2 sin t = 2 sin 0 = 2 ∙ 0 = 0
p
+PETGCUGVQt=
'PFCVt=p
p x = 5 cos t = 5 cos = 5 ∙ 0 = 0 2 p y = 2 sin t = 2 sin = 2 ∙ 1 = 2 2
x = 5 cos t = 5 cos p = 5(–1) = –5 y = 2 sin t = 2 sin p = 2(0) = 0
Points on the plane curve include (5, 0), which is the starting point, (0, 2), and ( 5, 0), which is the ending point. The plane curve is the top half of the ellipse, shown in Figure 10.54(b). y
y
5 4 3 2 1 –5 –4 –3 –2 –1–1
5 4 3 2 1 1 2 3 4 5
x
–2 –3 –4 –5 Figure 10.54(a) The graph of y2 x2 + = 1 25 4
CHECK POINT 3
p
t=
–5 –4 –3 –2 –1–1
1 2 3 4 5
t=p – –2 –3 –4 –5
t=
x
Figure 10.54(b) The plane curve for x = 5 cos t, y = 2 sin t, 0 … t … p
Sketch the plane curve represented by the parametric equations x = 6 cos t, y = 4 sin t, p … t … 2p
by eliminating the parameter.
3 Find parametric equations for functions.
Finding Parametric Equations Infinitely many pairs of parametric equations can represent the same plane curve. If the plane curve is defined by the function y = f(x), here is a procedure for finding a set of parametric equations: Parametric Equations for the Function y = f 1 x 2 One set of parametric equations for the plane curve defined by y = f(x) is x = t and y = f(t), in which t is in the domain of f.
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Chapter 10 Conic Sections and Analytic Geometry
EXAMPLE 4
Finding Parametric Equations
Find a set of parametric equations for the parabola whose equation is y = 9  x 2. Solution Let x = t. Parametric equations for y = f (x) are x = t and y = f (t). Thus, parametric equations for y = 9  x 2 are x = t and y = 9  t 2. CHECK POINT 4 Find a set of parametric equations for the parabola whose equation is y = x 2  25.
You can write other sets of parametric equations for y = 9  x 2 by starting with a different parametric equation for x. Here are three more sets of parametric equations for y = 9  x 2: 2
6
• If x = t 3, y = 9  (t 3) = 9  t . Parametric equations are x = t 3 and y = 9  t 6. • If x = t + 1, y = 9  (t + 1)2 = 9  (t 2 + 2t + 1) = 8  t 2  2t. Parametric equations are x = t + 1 and y = 8  t 2  2t. t t 2 t2 • If x = , y = 9  a b = 9  . 2 2 4
Parametric equations are x =
t t2 and y = 9  . 2 4
When finding parametric equations for y = 9  x 2, can we start with any choice for the parametric equation for x? The answer is no. The substitution for x must be a function that allows x to take on all the values in the domain of the given rectangular equation. For example, the domain of the function y = 9  x 2 is the set of all real numbers. If you incorrectly let x = t 2, these values of x exclude negative numbers that are included in y = 9  x 2. The parametric equations
4 Understand the advantages
of parametric representations.
TECHNOLOGY The ellipse shown was obtained using the parametric mode and the radian mode of a graphing utility. x(t) = 2 + 3 cos t y(t) = 3 + 2 sin t We used a [ 3.6, 7.6, 1] by [ 1, 6, 1] viewing rectangle with tmin = 0, tmax = 6.3, and tstep = 0.1.
2
x = t 2 and y = 9  (t 2) = 9  t 4 do not represent y = 9  x 2 because only points for which x Ú 0 are obtained.
Advantages of Parametric Equations over Rectangular Equations We opened this section with parametric equations that described the horizontal distance and the vertical height of your thrown baseball after t seconds. Parametric equations are frequently used to represent the path of a moving object. If t represents time, parametric equations give the location of a moving object and tell when the object is located at each of its positions. Rectangular equations tell where the moving object is located but do not reveal when the object is in a particular position. When using technology to obtain graphs, parametric equations that represent relations that are not functions are often easier to use than their corresponding rectangular equations. It is far easier to enter the equation of an ellipse given by the parametric equations x = 2 + 3 cos t and y = 3 + 2 sin t than to use the rectangular equivalent (x  2)2 (y  3)2 + = 1. 9 4 The rectangular equation must first be solved for y and then entered as two separate equations before a graphing utility reveals the ellipse.
Section 10.5 Parametric Equations
BLITZER BONUS
The Parametrization of DNA DNA, the molecule of biological inheritance, is hip. At least that’s what a new breed of marketers would like you to believe. For $2500, you can spit into a test tube and a Webbased company will tell you your risks for heart attack and other conditions. It’s been more than 60 years since James Watson and Francis Crick defined the structure, or shape, of DNA. A knowledge of how a molecule is structured does not always lead to an understanding of how it works, but it did in the case of DNA. The structure, which Watson and Crick announced in Nature in 1953, immediately suggested how the molecule could be reproduced and how it could contain biological information. The structure of the DNA molecule reveals the vital role that trigonometric functions play in the genetic information and instruction codes necessary for the maintenance and continuation of life.
z
y x 6JG&0#OQNGEWNGUVTWEVWTGFNKMG CURKTCNGFNCFFGTEQPUKUVUQHVYQ RCTCNNGNJGNKEGU UKPIWNCTJGNKZ VJCVCTGKPVGTVYKPGF
1075
'CEJJGNKZECPDGFGUETKDGFD[CEWTXG KPVJTGGFKOGPUKQPUTGRTGUGPVGFD[ RCTCOGVTKEGSWCVKQPUKPxyCPFz x=aEQUty=aUKPtz=bt YJGTGaCPFbCTGRQUKVKXGEQPUVCPVU
A curve that is used in physics for much of the theory of light is called a cycloid. The path of a fixed point on the circumference of a circle as it rolls along a line is a cycloid. A point on the rim of a bicycle wheel traces out a cycloid curve, shown in Figure 10.55. If the radius of the circle is a, the parametric equations of the cycloid are x = a(t  sin t) and y = a(1  cos t). Linear functions and cycloids are used to describe rolling motion. The light at the rolling circle’s center shows that it moves linearly. By contrast, the light at the circle’s edge has rotational motion and traces out a cycloid.
Instructor Resources for Section 10.5 in MyLab Math
It is an extremely complicated task to represent the cycloid in rectangular form. Cycloids are used to solve problems that involve the “shortest time.” For example, Figure 10.56 shows a bead sliding down a wire. For the bead to travel along the wire in the shortest possible time, the shape of the wire should be that of an inverted cycloid.
y 2a
(x, y)
a a pa
2pa
x
Figure 10.55 The curve traced by a fixed point on the circumference of a circle rolling along a straight line is a cycloid.
Figure 10.56
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Chapter 10 Conic Sections and Analytic Geometry
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. The pair of equations x = 1t and y = 2t  1 are called equations and the common variable t is called the . The graph for this pair of equations is called a/an .
C3. In order to eliminate the parameter from x = 3 sin t and y = 2 cos t, isolate and , square the two equations, and then use the identity .
C2. Eliminating the parameter from x = f(t) and y = g(t) means eliminating from the pair of equations to obtain one equation in and only.
C4. True or false: There is more than one way for pairs of parametric equations to represent the same plane curve.
10.5 EXERCISE SET Practice Exercises In Exercises 1–8, parametric equations and a value for the parameter t are given. Find the coordinates of the point on the plane curve described by the parametric equations corresponding to the given value of t. 1. x = 3  5t, y = 4 + 2t; t = 1
In Exercises 21–40, eliminate the parameter t. Then use the rectangular equation to sketch the plane curve represented by the given parametric equations. Use arrows to show the orientation of the curve corresponding to increasing values of t. (If an interval for t is not specified, assume that  ∞ 6 t 6 ∞ .) 21. x = t, y = 2t
2. x = 7  4t, y = 5 + 6t; t = 1
22. x = t, y = 2t
3. x = t 2 + 1, y = 5  t 3; t = 2
23. x = 2t  4, y = 4t 2
4. x = t 2 + 3, y = 6  t 3; t = 2
24. x = t  2, y = t 2
5. x = 4 + 2 cos t, y = 3 + 5 sin t; t =
p 2
6. x = 2 + 3 cos t, y = 4 + 2 sin t; t = p 7. x = (60 cos 30°)t, y = 5 + (60 sin 30°)t  16t 2; t = 2 8. x = (80 cos 45°)t, y = 6 + (80 sin 45°)t  16t 2; t = 2 In Exercises 9–20, use point plotting to graph the plane curve described by the given parametric equations. Use arrows to show the orientation of the curve corresponding to increasing values of t. 9. x = t + 2, y = t 2; 2 … t … 2 10. x = t  1, y = t 2; 2 … t … 2 11. x = t  2, y = 2t + 1; 2 … t … 3 12. x = t  3, y = 2t + 2; 2 … t … 3 13. x = t + 1, y = 1t; t Ú 0
25. x = 1t, y = t  1 26. x = 1t, y = t + 1 27. x = 2 sin t, y = 2 cos t; 0 … t 6 2p 28. x = 19 sin t, y = 19 cos t; 0 … t 6 2p 29. x = 1 + 3 cos t, y = 2 + 3 sin t; 0 … t 6 2p 30. x = 4 + 5 cos t, y = 4 + 5 sin t; 0 … t 6 2p 31. x = 2 cos t, y = 3 sin t; 0 … t 6 2p 32. x = 3 cos t, y = 5 sin t; 0 … t 6 2p 33. x = 1 + 3 cos t, y =  1 + 2 sin t; 0 … t … p 34. x = 2 + 4 cos t, y =  1 + 3 sin t; 0 … t … p 35. x = sec t, y = tan t 36. x = 5 sec t, y = 3 tan t 37. x = t 2 + 2, y = t 2  2
14. x = 1t, y = t  1; t Ú 0
38. x = 1t + 2, y = 1t  2
15. x = cos t, y = sin t; 0 … t 6 2p
39. x = 2t, y = 2t; t Ú 0
16. x = sin t, y = cos t; 0 … t 6 2p
40. x = e t, y = e t; t Ú 0
17. x = t , y = t ;  ∞ 6 t 6 ∞ 2
3
18. x = t 2 + 1, y = t 3  1;  ∞ 6 t 6 ∞
In Exercises 41–43, eliminate the parameter. Write the resulting equation in standard form.
19. x = 2t, y = t  1 ;  ∞ 6 t 6 ∞
41. A circle: x = h + r cos t, y = k + r sin t
20. x = t + 1 , y = t  2;  ∞ 6 t 6 ∞
42. An ellipse: x = h + a cos t, y = k + b sin t 43. A hyperbola: x = h + a sec t, y = k + b tan t
Section 10.5 Parametric Equations 44. The following are parametric equations of the line through (x1, y1) and (x2, y2): x = x1 + t(x2  x1) and y = y1 + t(y2  y1). Eliminate the parameter and write the resulting equation in pointslope form. In Exercises 45–52, use your answers from Exercises 41–44 and the parametric equations given in Exercises 41–44 to find a set of parametric equations for the conic section or the line. 45. Circle: Center: (3, 5); Radius: 6 46. Circle: Center: (4, 6); Radius: 9 47. Ellipse: Center: (  2, 3); Vertices: 5 units to the left and right of the center; Endpoints of Minor Axis: 2 units above and below the center 48. Ellipse: Center: (4,  1); Vertices: 5 units above and below the center; Endpoints of Minor Axis: 3 units to the left and right of the center
61. x = t 2 + t + 1, y = 2t 62. x = t 2  t + 6, y = 3t In Exercises 63–68, sketch the function represented by the given parametric equations. Then use the graph to determine each of the following: a. intervals, if any, on which the function is increasing and intervals, if any, on which the function is decreasing. b. the number, if any, at which the function has a maximum and this maximum value, or the number, if any, at which the function has a minimum and this minimum value. 63. x = 2t, y = t 64. x = e t, y = t t 65. x = , y = 2t 2  8t + 3 2 t 66. x = , y =  2t 2 + 8t  1 2 67. x = 2(t  sin t), y = 2(1  cos t); 0 … t … 2p
49. Hyperbola: Vertices: (4, 0) and (  4, 0); Foci: (6, 0) and ( 6, 0)
68. x = 3(t  sin t), y = 3(1  cos t); 0 … t … 2p
50. Hyperbola: Vertices: (0, 4) and (0, 4); Foci: (0, 5) and (0,  5)
Application Exercises
51. Line: Passes through (  2, 4) and (1, 7) 52. Line: Passes through (3, 1) and (9, 12) In Exercises 53–56, find two different sets of parametric equations for each rectangular equation. 53. y = 4x  3 54. y = 2x  5 55. y = x 2 + 4 56. y = x 2  3 In Exercises 57–58, the parametric equations of four plane curves are given. Graph each plane curve and determine how they differ from each other. 57. a. x = t and y = t 2  4
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The path of a projectile that is launched h feet above the ground with an initial velocity of 0 feet per second and at an angle u with the horizontal is given by the parametric equations x = (v0 cos u)t and y = h + (v0 sin u)t  16t 2, where t is the time, in seconds, after the projectile was launched. The parametric equation for x gives the projectile’s horizontal distance, in feet. The parametric equation for y gives the projectile’s height, in feet. Use these parametric equations to solve Exercises 69–70. 69. The figure shows the path for a baseball hit by Albert Pujols. The ball was hit with an initial velocity of 180 feet per second at an angle of 40° to the horizontal. The ball was hit at a height 3 feet off the ground. y (feet) 200
b. x = t 2 and y = t 4  4 c. x = cos t and y = cos2 t  4 d. x = e t and y = e 2t  4 58. a. x = t, y = 24  t 2;  2 … t … 2 b. x = 24  t 2, y = t;  2 … t … 2 c. x = 2 sin t, y = 2 cos t; 0 … t 6 2p d. x = 2 cos t, y = 2 sin t; 0 … t 6 2p
Practice PLUS In Exercises 59–62, sketch the plane curve represented by the given parametric equations. Then use interval notation to give each relation’s domain and range. 59. x = 4 cos t + 2, y = 4 cos t  1 60. x = 2 sin t  3, y = 2 sin t + 1
500
1000
x (feet)
a. Find the parametric equations that describe the position of the ball as a function of time. b. Describe the ball’s position after 1, 2, and 3 seconds. Round to the nearest tenth of a foot. Locate your solutions on the plane curve. c. How long, to the nearest tenth of a second, is the ball in flight? What is the total horizontal distance that it travels before it lands? Is your answer consistent with the figure shown? d. You meet Albert Pujols and he asks you to tell him something interesting about the path of the baseball that he hit. Use the graph to respond to his request. Then verify your observation algebraically.
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Chapter 10 Conic Sections and Analytic Geometry
70. The figure shows the path for a baseball that was hit with an initial velocity of 150 feet per second at an angle of 35° to the horizontal. The ball was hit at a height of 3 feet off the ground. y (feet) 150 100 50 100
700
x (feet)
a. Find the parametric equations that describe the position of the ball as a function of time. b. Describe the ball’s position after 1, 2, and 3 seconds. Round to the nearest tenth of a foot. Locate your solutions on the plane curve. c. How long is the ball in flight? (Round to the nearest tenth of a second.) What is the total horizontal distance that it travels, to the nearest tenth of a foot, before it lands? Is your answer consistent with the figure shown? d. Use the graph to describe something about the path of the baseball that might be of interest to the player who hit the ball. Then verify your observation algebraically.
Use the equations for the path of a projectile given prior to Exercises 69–70 to solve Exercises 83–85. In Exercises 83–84, use a graphing utility to obtain the path of a projectile launched from the ground (h = 0) at the specified values of u and 0. In each exercise, use the graph to determine the maximum height and the time at which the projectile reaches its maximum height. Also use the graph to determine the range of the projectile and the time it hits the ground. Round all answers to the nearest tenth. 83. u = 55°, v0 = 200 feet per second 84. u = 35°, v0 = 300 feet per second 85. A baseball player throws a ball with an initial velocity of 140 feet per second at an angle of 22° to the horizontal. The ball leaves the player’s hand at a height of 5 feet. a. Write the parametric equations that describe the ball’s position as a function of time. b. Use a graphing utility to obtain the path of the baseball. c. Find the ball’s maximum height and the time at which it reaches this height. Round all answers to the nearest tenth. d. How long is the ball in the air? e. How far does the ball travel?
Explaining the Concepts 71. What are plane curves and parametric equations?
Critical Thinking Exercises
72. How is point plotting used to graph a plane curve described by parametric equations? Give an example with your description.
Make Sense? In Exercises 86–89, determine whether each statement makes sense or does not make sense, and explain your reasoning.
73. What is the significance of arrows along a plane curve?
86. Parametric equations allow me to use functions to describe curves that are not graphs of functions.
74. What does it mean to eliminate the parameter? What useful information can be obtained by doing this? 75. Explain how the rectangular equation y = 5x can have infinitely many sets of parametric equations. 76. Discuss how the parametric equations for the path of a projectile (see Exercises 69–70) and the ability to obtain plane curves with a graphing utility can be used by a baseball coach to analyze performances of team players.
Technology Exercises 77. Use a graphing utility in a parametric mode to verify any five of your handdrawn graphs in Exercises 9–40. In Exercises 78–82, use a graphing utility to obtain the plane curve represented by the given parametric equations. 78. Cycloid: x = 3(t  sin t), y = 3(1  cos t); [0, 60, 5] * [0, 8, 1], 0 … t 6 6p
87. Parametric equations let me think of a curve as a path traced out by a moving point. 88. I represented y = x 2  9 with the parametric equations x = t 2 and y = t 4  9. 89. I found alternate pairs of parametric equations for the same rectangular equation. 90. Eliminate the parameter: x = cos3 t and y = sin3 t. 91. The plane curve described by the parametric equations x = 3 cos t and y = 3 sin t, 0 … t 6 2p, has a counterclockwise orientation. Alter one or both parametric equations so that you obtain the same plane curve with the opposite orientation. 92. The figure shows a circle of radius a rolling along a horizontal line. Point P traces out a cycloid. Angle t, in radians, is the angle through which the circle has rolled. C is the center of the circle.
79. Cycloid: x = 2(t  sin t), y = 2(1  cos t); [0, 60, 5] * [0, 8, 1], 0 … t 6 6p
y
80. Witch of Agnesi: x = 2 cot t, y = 2 sin2 t; [  6, 6, 1] * [ 4, 4, 1], 0 … t 6 2p 81. Hypocycloid: x = 4 cos3 t, y = 4 sin3 t; [  8, 8, 1] * [ 5, 5, 1], 0 … t 6 2p
y
82. Lissajous Curve: x = 2 cos t, y = sin 2t; [  3, 3, 1] * [ 2, 2, 1], 0 … t 6 2p
O
P
x
a
C t
2a
B
A
x
Section 10.6 Conic Sections in Polar Coordinates Refer to the figure at the bottom of the previous page. Use the suggestions in parts (a) and (b) to prove that the parametric equations of the cycloid are x = a(t  sin t) and y = a(1  cos t). a. Derive the parametric equation for x using the figure and x = OA  xA. b. Derive the parametric equation for y using the figure and y = AC  BC.
Retaining the Concepts 93. Find all zeros of f(x) = 2x 3 + x 2  13x + 6. (Section 3.4, Example 3) 94. Use the exponential decay model, A = A0 e kt to solve this exercise. The halflife of the tranquilizer Xanax in the bloodstream is 36 hours. How long, to the nearest tenth of an hour, will it take for Xanax to decay to 70% of the original dosage? (Section 4.5, Example 2) 95. Let v = 2i  5j and w = 4i + 3j. Find each of the following. a. v + w b. 2v  w c. v # w d. v # v (Section 7.6, Examples 4–6; Section 7.7, Example 1) 96. Solve triangle ABC with C A = 39°, b = 5, and c = 7. Round lengths of sides to b=5 a the nearest tenth and angle measures to the nearest 39° B A degree. c=7 (Section 7.2, Example 1)
SECTION 10.6
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Preview Exercises Exercises 97–99 will help you prepare for the material covered in the next section. 4 97. Rewrite r = by dividing the numerator and the 2 + cos u denominator by 2. 98. Complete the table of coordinates below. Where necessary, round to two decimal places. Then plot the resulting points, (r, u), using a polar coordinate system.
0
U r =
p 2
2p 3
3p 4
5p 6
p
4 2 + cos U
99. a. Showing all steps, 9r 2 = (1 + 3r cos u)2.
rewrite
r =
1 3  3 cos u
as
b. Express 9r 2 = (1 + 3r cos u)2 in rectangular coordinates. Which conic section is represented by the rectangular equation?
Conic Sections in Polar Coordinates
WHAT YOU'LL LEARN
John Glenn made the first U.S.manned flight around Earth on Friendship 7.
1 Define conics in terms of a focus and a directrix.
2 Graph the polar equations of conics.
On the morning of February 20, 1962, millions of Americans collectively held their breath as the world’s newest pioneer swept across the threshold of one of our last frontiers. Roughly one hundred miles above Earth, astronaut John Glenn sat comfortably in the weightless environment of a 9 12@by@6@foot space capsule that offered the leg room of a Volkswagen “Beetle” and the aesthetics of a garbage can. Glenn became the first American to orbit Earth in a threeorbit mission that lasted slightly under 5 hours. In this section’s Exercise Set (Exercises 31–32), you will see how John Glenn’s historic orbit can be described using conic sections in polar coordinates. To obtain this model, we begin with a definition that permits a unified approach to the conic sections.
1080
Chapter 10 Conic Sections and Analytic Geometry
1 Define conics in terms of a
The FocusDirectrix Definitions of the Conic Sections
focus and a directrix.
The definition of a parabola is given in terms of a fixed point, the focus, and a fixed line, the directrix. By contrast, the definitions of an ellipse and a hyperbola are given in terms of two fixed points, the foci. It is possible to define each of these conic sections in terms of a point and a line. Figure 10.57 shows a conic section in the polar coordinate system. The fixed point, the focus, is at the pole. The fixed line, the directrix, is perpendicular to the polar axis.
&KTGEVTKZKU RGTRGPFKEWNCT VQVJGRQNCTCZKU D
P= ru r u F
FocusDirectrix Definitions of the Conic Sections
Polar axis
(QEWUKUCV VJGRQNG
Let F be a fixed point, the focus, and let D be a fixed line, the directrix, in a plane (Figure 10.57). A conic section, or conic, is the set of all points P in the plane such that PF = e, PD
Figure 10.57 A conic in the polar coordinate system
where e is a fixed positive number, called the eccentricity. If e = 1, the conic is a parabola. If e 6 1, the conic is an ellipse. If e 7 1, the conic is a hyperbola.
Figure 10.58 illustrates the eccentricity for each type of conic. Notice that if e = 1, the definition of the parabola is the same as the focusdirectrix definition with which you are familiar.
p 2
p 2
p 2
&KTGEVTKZ
D
D
P F (QEWUCVRQNG
0
P
D
F (QEWUCVRQNG
P F
0 P′
0 (QEWUCVRQNG
D′
&KTGEVTKZ &KTGEVTKZ Parabola
Ellipse
PF =e PD e=1
PF =e PD e1
Figure 10.58 The eccentricity for each conic
2 Graph the polar equations of conics.
Polar Equations of Conics By locating a focus at the pole, all conics can be represented by similar equations in the polar coordinate system. In each of these equations, • (r, u) is a point on the graph of the conic. • e is the eccentricity. (Remember that e 7 0.) • p is the distance between the focus (located at the pole) and the directrix.
Section 10.6 Conic Sections in Polar Coordinates
1081
Standard Forms of the Polar Equations of Conics Let the pole be a focus of a conic section of eccentricity e with the directrix p units from the focus. The equation of the conic is given by one of the four equations listed. ep 1 + e cos u
r=
r=
p
u = p2 or yaxis
u= 2 or yaxis &KTGEVTKZ x=p
&KTGEVTKZ x=–p (QEWU CVRQNG
Polar axis or xaxis
(QEWUCVRQNG
r=
ep 1 − e cos u
ep 1 + e sin u u = p2 or yaxis
r=
Polar axis or xaxis
ep 1 − e sin u p
&KTGEVTKZy=p
u= 2 or yaxis (QEWU CVRQNG
Polar axis or xaxis
Polar axis or xaxis
(QEWU CVRQNG &KTGEVTKZy=–p
&KTGEVTKZ D
P= ru r u F
p
Q (QEWUKUCV VJGRQNG
Polar axis
The graphs in the box illustrate two kinds of symmetry—symmetry with respect to the polar axis and symmetry with respect to the y@axis. If the equation contains cos u, the polar axis is an axis of symmetry. If the equation contains sin u, the line u = p2 , or the y@axis, is an axis of symmetry. Take a moment to verify these observations. We will derive one of the equations displayed in the box. The other three equations are obtained in a similar manner. In Figure 10.59, let P = (r, u) be any point on a conic section. PF = e PD
Figure 10.59
r = e PD r = e p + FQ r = e p + r cos u
By definition, the ratio of the distance between P and the focus to the distance between P and the directrix equals the positive constant e. Figure 10.59 shows that the distance from P to the focus, located at the pole, is r : PF = r. Figure 10.59 shows that the distance from P to the directrix is p + FQ: PD = p + FQ. FQ Using the triangle in the figure, cos U = and r FQ = r cos U.
1082
Chapter 10 Conic Sections and Analytic Geometry
r = e for r, we will obtain the desired equation. Clear fractions p + r cos u by multiplying both sides by the least common denominator.
By solving
r = e( p + r cos u)
Multiply both sides by p + r cos U.
r = ep + er cos u
Apply the distributive property.
r  er cos u = ep
Subtract er cos U from both sides to collect terms involving r on the same side.
r(1  e cos u) = ep r =
Factor out r from the two terms on the left.
ep 1  e cos u
Divide both sides by 1 − e cos U and solve for r.
In summary, the standard forms of the polar equations of conics are r=
ep 1 ± e cos u
and r =
ep . 1 ± e sin u
+PCNNHQTOUVJGEQPUVCPVVGTOKPVJGFGPQOKPCVQTKU
Graphing the Polar Equation of a Conic 1. If necessary, write the equation in one of the standard forms. 2. Use the standard form to determine values for e and p. Use the value of e to identify the conic. 3. Use the appropriate figure for the standard form of the equation shown in the box on page 1081 to help guide the graphing process.
EXAMPLE 1
Graphing the Polar Equation of a Conic
Graph the polar equation: r =
4 . 2 + cos u
Solution Step 1 Write the equation in one of the standard forms. The equation is not in standard form because the constant term in the denominator is not 1. r=
4 2 + cos u
6QQDVCKPKPVJKURQUKVKQPFKXKFG VJGPWOGTCVQTCPFFGPQOKPCVQTD[
The equation in standard form is r=
2 1 2
1 + cos u e=
ep=
.
This equation is in the form r =
ep 1 + e cos U
.
1083
Section 10.6 Conic Sections in Polar Coordinates
Step 2 Use the standard form to find e and p, and identify the conic. The voice balloons at the bottom of the previous page show that e = Thus, e =
1 2
1 2
and ep = 12 p = 2.
and p = 4. Because e =
1 2
6 1, the conic is an ellipse.
Step 3 Use the figure for the equation’s standard form to guide the graphing process. The figure for the conic’s standard form is shown in Figure 10.60(a). We have symmetry with respect to the polar axis. One focus is at the pole and the corresponding directrix is x = 4, located four units to the right of the pole.
u = p2 or yaxis &KTGEVTKZ x=p
TECHNOLOGY
8GTVGZ
p p
The graph of r =
4 2 + cos u
is obtained using the polar mode with angle measure in radians. To verify the handdrawn graph in Figure 10.60(b), we used a [ 4.8, 4.8, 1] by [ 3, 3, 1] viewing rectangle with umin = 0, p . umax = 2p, and ustep = 48
Polar axis or xaxis
(QEWUCVRQNG
5p 6
3p 4
p 2
2p 3
p 3
8GTVGZ
&KTGEVTKZ p 4
2
(QEWU
11p 6 5p 4 4p 3
(a) Using r = r =
2 1 +
1 2
1 + e cos u
0
4
7p 6
ep
p 6
3p 2
5p 3
7p 4
(b) The graph of 4 2 r = or r = 2 + cos u 1 + 12 cos u
to graph
cos u
Figure 10.60
Figure 10.60(a) indicates that the major axis is on the polar axis. Thus, we find the vertices by selecting 0 and p for u. The corresponding values for r are 43 and 4, respectively. Figure 10.60(b) shows the vertices, 1 43, 0 2 and (4, p). You can sketch the upper half of the ellipse by plotting some points from u = 0 to u = p. r =
4 2 + cos u
U
p 2
2p 3
3p 4
5p 6
r
2
2.7
3.1
3.5
Using symmetry with respect to the polar axis, you can sketch the lower half. The graph of the given equation is shown in Figure 10.60(b). Use the three steps shown in the preceding box on page 1082 to graph the polar equation: CHECK POINT 1
r =
4 . 2  cos u
1084
Chapter 10 Conic Sections and Analytic Geometry
EXAMPLE 2
Graphing the Polar Equation of a Conic
Graph the polar equation: r =
12 . 3 + 3 sin u
Solution Step 1 Write the equation in one of the standard forms. The equation is not in standard form because the constant term in the denominator is not 1. Divide the numerator and denominator by 3 to write the standard form.
r=
4 1 + 1 sin u
ep= This equation is in the form r =
ep 1 + e sin U
.
e=
Step 2 Use the standard form to find e and p, and identify the conic. The voice balloons show that e = 1 and ep = 1p = 4.
&KTGEVTKZ u = p2 or yaxis
5p 6
&KTGEVTKZy=p p Polar axis or xaxis (QEWU CVRQNG
(a) Using r =
ep 1 + e sin u
to graph r =
Figure 10.61
TECHNOLOGY The graph of 12 3 + 3 sin u was obtained using a [ 8, 8, 1] by [ 5, 5, 1] viewing rectangle with r =
3p 4
8GTVGZ
p 3
p
(4, p)
p 4
p 6
(4, 0) 2
(QEWU
4
7p 6
0
the directrix is y = 4, located four units above the pole. Figure 10.61(a) indicates that the p vertex is on the line u = , or the y@axis. 2 p Thus, we find the vertex by selecting 2 for u. The corresponding value for r is 2.
11p 6 5p 4 4p 3
4 1 + sin u
Thus, e = 1 and p = 4. Because e = 1, the conic is a parabola. Step 3 Use the figure for the equation’s standard form to guide the graphing process. Figure 10.61(a) indicates that p we have symmetry with respect to u = . 2 The focus is at the pole and, with p = 4,
p 2
2p 3
3p 2
5p 3
7p 4
12 (b) The graph of r = or 3 + 3 sin u 4 r = 1 + sin u
Figure 10.61(b) shows the vertex, a2,
p b. 2 To find where the parabola crosses the polar axis, select u = 0 and u = p. The corresponding values for r are 4 and 4, respectively. Figure 10.61(b) shows the points (4, 0) and (4, p) on the polar axis. You can sketch the right half of the parabola by plotting some points from p u = 0 to u = . 2
umin = 0, umax = 2p, p ustep = . 48
r =
12 3 + 3 sin u
U
p 6
p 4
p 3
r
2.7
2.3
2.1
p , you can sketch the left half. The graph of 2 the given equation is shown in Figure 10.61(b).
Using symmetry with respect to u =
1085
Section 10.6 Conic Sections in Polar Coordinates
CHECK POINT 2 Use the three steps shown in the preceding box on page 1082 to graph the polar equation: 8 r = . 4 + 4 sin u
EXAMPLE 3
Graphing the Polar Equation of a Conic
Graph the polar equation: r =
9 . 3  6 cos u
Solution Step 1 Write the equation in one of the standard forms. We can obtain a constant term of 1 in the denominator by dividing each term by 3. ep=
r=
3 1 − 2 cos u
6JKUGSWCVKQPKUKPVJGHQTO ep r= –eEQUu
e=
Step 2 Use the standard form to find e and p, and identify the conic. The voice balloons show that e = 2 and ep = 2p = 3. 3 2.
Thus, e = 2 and p = Because e = 2 7 1, the conic is a hyperbola. Step 3 Use the figure for the equation’s standard form to guide the graphing process. Figure 10.62(a) indicates that we have symmetry with respect to the polar axis. One focus is at the pole and, with p = 32, the corresponding directrix is x =  32, located 1.5 units to the left of the pole. Figure 10.62(a) indicates that the transverse axis is horizontal and the vertices lie on the polar axis. Thus, we find the vertices by selecting 0 and p for u. Figure 10.62(b) shows the vertices, ( 3, 0) and (1, p). p p 3p To find where the hyperbola crosses the line u = , select and for u. 2 2 2 3p p Figure 10.62(b) shows the points a3, b and a3, b on the graph. 2 2 u = p2 or yaxis
&KTGEVTKZ
&KTGEVTKZ x=–p
5p 6
(QEWUCVRQNG
Instructor Resources for Section 10.6 in MyLab Math
3p 4
2p 3
p 2 p 3
Q3,
8GTVGZ – p Polar axis or xaxis
1  e cos u
Figure 10.62
to graph r =
p 6
3p
Q3, 2 R 3p 2
0
4
(QEWU
5p 4 4p 3
ep
p 4
8GTVGZ p
7p 6
(a) Using r =
p R 2
5p 3
11p 6 7p 4
9 3 3 (b) The graph of r = or r = 3  6 cos u 1  2 cos u 1  2 cos u
1086
Chapter 10 Conic Sections and Analytic Geometry &KTGEVTKZ
5p 6
3p 4
2p 3
We sketch the hyperbola by plotting some points from u = 0 to u = p.
p 2 p 3 p
Q3, 2 R
8GTVGZ – p
r =
p 4
p 6
0
4
(QEWU 3p
Q3, 2 R
7p 6 5p 4 4p 3
3p 2
5p 3
p 6  4.1
U
8GTVGZ p
11p 6 7p 4
Figure 10.62(b) (repeated) The graph of 9 3 r = or r = 3  6 cos u 1  2 cos u
r
2p 3 1.5
5p 6 1.1
2p 5p p Figure 10.62(b) shows the points a 4.1, b , a1.5, b , and a1.1, b on the 6 3 6 p graph. Observe that a 4.1, b is on the lower half of the hyperbola. Using 6 symmetry with respect to the polar axis, we sketch the entire hyperbola. The graph of the given equation is shown in Figure 10.62(b). CHECK POINT 3 Use the three steps shown in the preceding box on page 1082 to graph the polar equation:
r =
BLITZER BONUS
3 1  2 cos u
9 . 3  9 cos u
Modeling Planetary Motion
Polish astronomer Nicolaus Copernicus (1473–1543) was correct in stating that planets in our solar system revolve around the Sun and not Earth. However, he incorrectly believed that celestial orbits move in perfect circles, calling his system “the ballet of the planets.”
Earth
Venus Mars
Earth
Mercury
Venus Mars
Sun Moon
Mercury Sun
Saturn Saturn Jupiter
Jupiter Copernican model
Ptolemaic model
Table 10.4 indicates that the planets in our solar system have orbits with eccentricities that are much closer to 0 than to 1. Most of these orbits are almost circular, which made it difficult for early astronomers to detect that they are actually ellipses. German scientist and mathematician Johannes Kepler (1571–1630) discovered that planets move in elliptical orbits with the Sun at one focus. The polar equation for these orbits is r =
(1  e 2)a 1  e cos u
Table 10.4 Eccentricities of Planetary Orbits Mercury
0.2056
Jupiter
0.0484
Venus
0.0068
Saturn
0.0543
Earth
0.0167
Uranus
0.0460
Mars
0.0934
Neptune
0.0082
,
where the length of the orbit’s major axis is 2a. Describing planetary orbits, Kepler wrote, “The heavenly motions are nothing but a continuous song for several voices, to be perceived by the intellect, not by the ear.”
Section 10.6 Conic Sections in Polar Coordinates
1087
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. A conic section is the set of all points in the plane such that the ratio of the distance from a fixed point, called the , to the distance from a fixed line, called the , equals a constant e, called the . If e = 1, the conic is a/an . If e 6 1, the conic is a/an . If e 7 1, the conic is a/an .
Standard Forms of the Polar Equations of Conics a.
b.
c.
d.
ep 1 + e cos u
Directrix is perpendicular to the polar axis at a distance p units to the right of the pole.
ep r = 1  e cos u
Directrix is perpendicular to the polar axis at a distance p units to the left of the pole.
r =
r =
r =
ep 1 + e sin u
Directrix is parallel to the polar axis at a distance p units above the pole.
ep 1  e sin u
Directrix is parallel to the polar axis at a distance p units below the pole.
Use the four equations in the previous column to solve Exercises C2–C4. C2. For all four equations shown in the previous column, the focus is at the and e represents the conic’s . C3. Consider the equation r =
3 . 1 + 3 cos u
e = , so this is the equation of a/an p = , so the directrix is axis at a distance unit(s) to the
. to the polar of the pole.
C4. Consider the equation r =
2 . 1  sin u
e = , so this is the equation of a/an . p = , so the directrix is to the polar axis at a distance unit(s) the pole. 12 is not in standard form 4  4 cos u because the constant term in the denominator is not . The equation can be written in standard form by .
C5. The equation r =
10.6 EXERCISE SET Practice Exercises
15. r =
6 2  2 sin u
16. r =
6 2 + 2 sin u
17. r =
8 2  4 cos u
18. r =
8 2 + 4 cos u
19. r =
12 3  6 cos u
20. r =
12 3  3 cos u
In Exercises 1–8, a. Identify the conic section that each polar equation represents. b. Describe the location of a directrix from the focus located at the pole. 3 3 1. r = 2. r = 1 + sin u 1 + cos u 6 6 4. r = 3. r = 3  2 cos u 3 + 2 cos u 8 8 5. r = 6. r = 2 + 2 sin u 2  2 sin u 12 10 8. r = 7. r = 2  4 cos u 8 + 9 sin u In Exercises 9–20, use the three steps shown in the box on page 1082 to graph each polar equation. 1 1 10. r = 9. r = 1 + sin u 1 + cos u 2 2 11. r = 12. r = 1  cos u 1  sin u 12 12 14. r = 13. r = 5 + 3 cos u 5  3 cos u
Practice PLUS In Exercises 21–28, describe a viewing rectangle, or window, such as [ 30, 30, 3] by [  8, 4, 1], that shows a complete graph of each polar equation and minimizes unused portions of the screen. 21. r =
15 3  2 cos u
22. r =
16 5  3 cos u
23. r =
8 1  cos u
24. r =
8 1 + cos u
25. r =
4 1 + 3 cos u
26. r =
16 3 + 5 cos u
27. r =
4 5 + 5 sin u
28. r =
2 3 + 3 sin u
1088
Chapter 10 Conic Sections and Analytic Geometry
Application Exercises
Explaining the Concepts
Halley’s Comet has an elliptical orbit with the Sun at one focus. Its orbit, shown in the figure below, is given approximately by
33. How are the conics described in terms of a fixed point and a fixed line?
r =
34. If all conics are defined in terms of a fixed point and a fixed line, how can you tell one kind of conic from another?
1.069 . 1 + 0.967 sin u
In the formula, r is measured in astronomical units. (One astronomical unit is the average distance from Earth to the Sun, approximately 93 million miles.) Use the given formula and the figure to solve Exercises 29–30. Round to the nearest hundredth of an astronomical unit and the nearest million miles. p 2
p
p 2
0 5WP *CNNG[oU%QOGV
'CTVJ
p
0
35. If you are given the standard form of the polar equation of a conic, how do you determine its eccentricity? 36. If you are given the standard form of the polar equation of a conic, how do you determine the location of a directrix from the focus at the pole? 1 37. Describe a strategy for graphing r = . 1 + sin u 38. You meet an astronaut and she asks you to tell her something of interest about the elliptical orbit of John Glenn’s first space voyage in 1962. Describe how to use the polar equation for orbits in the Blitzer Bonus on page 1086, the equation for John Glenn’s 1962 journey in Exercises 31–32, and a graphing utility to provide an interesting visual analysis.
Technology Exercises *CNNG[oU%QOGV 3p 2
29. Find the distance from Halley’s Comet to the Sun at its shortest distance from the Sun. 30. Find the distance from Halley’s Comet to the Sun at its greatest distance from the Sun. On February 20, 1962, John Glenn made the first U.S.manned flight around the Earth for three orbits on Friendship 7. With Earth at one focus, the orbit of Friendship 7 is given approximately by r =
4090.76 , 1  0.0076 cos u
where r is measured in miles from Earth’s center. Use the formula and the figure shown to solve Exercises 31–32. p 2
39. Use a graphing utility to verify any five of your handdrawn graphs in Exercises 9–20. In Exercises 40–42, identify the conic that each polar equation represents. Then use a graphing utility to graph the equation. 16 12 41. r = 40. r = 4  3 cos u 4 + 5 sin u 42. r =
0 ,QJP)NGPPoUQTDKV KP(TKGPFUJKR
3p 2
31. How far from Earth’s center was John Glenn at his greatest distance from the planet? Round to the nearest mile. If the radius of Earth is 3960 miles, how far was he from Earth’s surface at this point on the flight? 32. How far from Earth’s center was John Glenn at his closest distance from the planet? Round to the nearest mile. If the radius of Earth is 3960 miles, how far was he from Earth’s surface at this point on the flight?
18 6  6 cos u
In Exercises 43–44, use a graphing utility to graph the equation. Then answer the given question. 4
; How does the graph differ from the p 1  sin au  b 4 4 graph of r = ? 1  sin u 3 44. r = ; How does the graph differ from the p 2 + 6 cos au + b 3 3 graph of r = ? 2 + 6 cos u 45. Use the polar equation for planetary orbits,
43. r =
'CTVJ
p
Use the polar mode of a graphing utility with angle measure in radians to solve Exercises 39–42. Unless otherwise indicated, p use umin = 0, umax = 2p, and u step = . If you are not 48 satisfied with the quality of the graph, experiment with smaller values for u step.
r =
(1  e 2)a 1  e cos u
,
to find the polar equation of the orbit for Mercury and Earth. Mercury: e = 0.2056 and a = 36.0 * 106 miles Earth: e = 0.0167 and a = 92.96 * 106 miles Use a graphing utility to graph both orbits in the same viewing rectangle. What do you see about the orbits from their graphs that is not obvious from their equations?
Chapter 10 Summary, Review, and Test
Critical Thinking Exercises
Retaining the Concepts
Make Sense? In Exercxises 46–49, determine whether each statement makes sense or does not make sense, and explain your reasoning. 46. Eccentricity and polar coordinates enable me to see that ellipses, hyperbolas, and parabolas are a unified group of interrelated curves. ep 47. I graphed a conic in the form r = that was 1  e cos u symmetric with respect to the y@axis. 48. Given the focus is at the pole, I can write the polar equation of a conic section if I know its eccentricity and the rectangular equation of the directrix. 49. As long as I know how to graph in polar coordinates, a knowledge of conic sections is not necessary to graph the equations in Exercises 9–20. 50. Identify the conic and graph the equation: 4 sec u r = . 2 sec u  1 In Exercises 51–52, write a polar equation of the conic that is named and described. 51. Ellipse: a focus at the pole; vertex: (4, 0); e = 12 52. Hyperbola: a focus at the pole; directrix: x =  1; e = 32 53. Identify the conic and write its equation in rectangular 1 . coordinates: r = 2  2 cos u 54. Prove that the polar equation of a planet’s elliptical orbit is (1  e 2)a , r = 1  e cos u where e is the eccentricity and 2a is the length of the major axis.
55. Expand:
logb(x 3 2y). (Section 4.3, Example 4)
56. Solve the system: b
x + y = 1 x 2 + y2 = 25.
(Section 8.4, Example 2) 57. Verify the identity: sin 2x = 2 cot x sin2 x. (Section 6.3, Example 6) 58. Two firelookout stations are 10 miles apart with station B directly east of station A. Both stations spot a fire on a mountain to the north. The bearing from station A to the fire is N39°E (39° east of north). The bearing from station B to the fire is N42°W (42° west of north). How far, to the nearest tenth of a mile, is the fire from station A? (Section 7.1, Example 7) N W
E
C
S
b
a
39°
42° 51°
p 2
2NCPGV
A
48° c = 10 miles
B
Preview Exercises
r u
p
1089
0
5WPHQEWUCVRQNG
2a 3p 2
CHAPTER 10
Exercises 59–61 will help you prepare for the material covered in the first section of the next chapter. (  1)n 59. Evaluate n for n = 1, 2, 3, and 4. 3  1 60. Find the product of all positive integers from n down through 1 for n = 5. 61. Evaluate j 2 + 1 for all consecutive integers from 1 to 6, inclusive. Then find the sum of the six evaluations.
Summary, Review, and Test
Summary Definitions and Concepts
Examples
10.1 The Ellipse a. An ellipse is the set of all points in a plane the sum of whose distances from two fixed points, the foci, is constant. y2 x2 b. Standard forms of the equations of an ellipse with center at the origin are 2 + 2 = 1 a b y2 x2 [foci: ( c, 0), (c, 0)] and 2 + 2 = 1 [foci: (0, c), (0, c)], where c 2 = a2  b2 and a2 7 b2. b a See the box on page 1014 containing Figure 10.6.
Ex. 1, p. 1015; Ex. 2, p. 1016; Ex. 3, p. 1017
1090
Chapter 10 Conic Sections and Analytic Geometry
Definitions and Concepts
Examples
c. Standard forms of the equations of an ellipse centered at (h, k) are
(x  h)2
+
(y  k)2
= 1
Ex. 4, p. 1018
d. In some cases, it is necessary to convert the equation of an ellipse to standard form by completing the square on x and y.
Ex. 5, p. 1020
(x  h)
2
and
b
2
(y  k)
a2
2
+
a2
b2
= 1, a2 7 b2. See Table 10.1 on page 1018.
10.2 The Hyperbola a. A hyperbola is the set of all points in a plane the difference of whose distances from two fixed points, the foci, is constant. b. Standard forms of the equations of a hyperbola with center at the origin are [foci: ( c, 0), (c, 0)] and
y
2 2

a on page 1028 and Figure 10.17.
c. Asymptotes for
2
y2 x2 = 1 a2 b2
x = 1 [foci: (0, c), (0, c)], where c 2 = a2 + b2. See the box b2
Ex. 1, p. 1029; Ex. 2, p. 1030
y2 y2 x2 b a x2 = 1 are y = { x. Asymptotes for = 1 are y = { x. 2 2 2 2 a b a b a b
d. A procedure for graphing hyperbolas is given in the lower box on page 1031.
e. Standard forms of the equations of a hyperbola centered at (h, k) are (y  k)2 (x  h)2 and = 1. See Table 10.2 on page 1034. a2 b2
(x  h)2
Ex. 3, p. 1032; Ex. 4, p. 1033 
(y  k)2
= 1
Ex. 5, p. 1034
f. In some cases, it is necessary to convert the equation of a hyperbola to standard form by completing the square on x and y.
Ex. 6, p. 1036
a2
b2
10.3 The Parabola a. A parabola is the set of all points in a plane that are equidistant from a fixed line, the directrix, and a fixed point, the focus. b. Standard forms of the equations of parabolas with vertex at the origin are y2 = 4px [focus: (p, 0)] and x 2 = 4py [focus: (0, p)]. See the box on page 1043 and Figure 10.32 on page 1044.
Ex. 1, p. 1044; Ex. 3, p. 1046
c. A parabola’s latus rectum is a line segment that passes through its focus, is parallel to its directrix, and has its endpoints on the parabola. The length of the latus rectum for y2 = 4px and x 2 = 4py is 4p . A parabola can be graphed using the vertex and endpoints of the latus rectum.
Ex. 2, p. 1045
d. Standard forms of the equations of a parabola with vertex at (h, k) are (y  k)2 = 4p(x  h) and (x  h)2 = 4p(y  k). See Table 10.3 on page 1047 and Figure 10.37.
Ex. 4, p. 1047
e. In some cases, it is necessary to convert the equation of a parabola to standard form by completing the square on x or y, whichever variable is squared.
Ex. 5, p. 1048
f. A nondegenerate conic section of the form Ax 2 + Cy2 + Dx + Ey + F = 0 in which A and C are not both zero is 1. a circle if A = C; 2. a parabola if AC = 0; 3. an ellipse if A ≠ C and AC 7 0; 4. a hyperbola if AC 6 0.
Ex. 7, p. 1052
Chapter 10 Summary, Review, and Test
Definitions and Concepts
1091
Examples
10.4 Rotation of Axes a. Rotation of Axes Formulas u is the angle from the positive x@axis to the positive x′@axis. x = x′ cos u  y′ sin u and y = x′ sin u + y′ cos u
Ex. 1, p. 1059
b. Amount of Rotation Formula The general seconddegree equation Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0 can be rewritten in x′ and y′ without an x′y′@term by rotating the axes through angle u, where A  C and u is an acute angle. cot 2u = B c. If 2u in cot 2u is one of the familiar angles such as 30°, 45°, or 60°, write the equation of a rotated conic in standard form using the fivestep procedure in the box on page 1061.
Ex. 2, p. 1062
d. If cot 2u is not the cotangent of one of the more familiar angles, use a sketch of cot 2u to find cos 2u. Then use 1  cos 2u 1 + cos 2u sin u = and cos u = A 2 A 2 to find values for sin u and cos u in the rotation formulas.
Ex. 3, p. 1064
e. A nondegenerate conic section of the form Ax 2 + Bxy + Cy2 + Dx + Ey + F = 0 2 is 1. a parabola if B  4AC = 0; 2. an ellipse or a circle if B2  4AC 6 0; 3. a hyperbola if B2  4AC 7 0.
Ex. 4, p. 1066
10.5 Parametric Equations a. The relationship between the parametric equations x = f(t) and y = g(t) and plane curves is described in the first box on page 1070. b. Point plotting can be used to graph a plane curve described by parametric equations. See the second box on page 1070.
Ex. 1, p. 1070
c. Plane curves can be sketched by eliminating the parameter t and graphing the resulting rectangular equation. It is sometimes necessary to change the domain of the rectangular equation to be consistent with the domain for the parametric equation in x.
Ex. 2, p. 1071; Ex. 3, p. 1072
d. Infinitely many pairs of parametric equations can represent the same plane curve. One pair for y = f(x) is x = t and y = f(t), in which t is in the domain of f.
Ex. 4, p. 1074
10.6 Conic Sections in Polar Coordinates a. The focusdirectrix definitions of the conic sections are given in the box on page 1080. For all points on a conic, the ratio of the distance from a fixed point (focus) and the distance from a fixed line (directrix) is constant and is called its eccentricity. If e = 1, the conic is a parabola. If e 6 1, the conic is an ellipse. If e 7 1, the conic is a hyperbola. b. Standard forms of the polar equations of conics are ep ep r = and r = , 1 { e cos u 1 { e sin u in which (r, u) is a point on the conic’s graph, e is the eccentricity, and p is the distance between the focus (located at the pole) and the directrix. Details are shown in the box on page 1081. c. A procedure for graphing the polar equation of a conic is given in the box on page 1082.
Ex. 1, p. 1082; Ex. 2, p. 1084; Ex. 3, p. 1085
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Chapter 10 Conic Sections and Analytic Geometry
Chapter 10 Review Exercises 10.1
10.2
In Exercises 1–8, graph each ellipse and locate the foci.
In Exercises 15–22, graph each hyperbola. Locate the foci and find the equations of the asymptotes. x2 15.  y2 = 1 16
y2 x + = 1 1. 36 25 2
2.
y2 x2 + = 1 25 16
16.
3. 4x 2 + y2 = 16
17. 9x 2  16y2 = 144
4. 4x + 9y = 36 2
5. 6.
2
(x  1)2 16 (x + 1)2 9
+ +
(y + 2)2 9 (y  2)2 16
18. 4y2  x 2 = 16 =1 19. = 1
20.
7. 4x 2 + 9y2 + 24x  36y + 36 = 0
(x  2)2 25 (y + 2)2 25

(y + 3)2 16
=1
(x  3)2 16
= 1
21. y2  4y  4x 2 + 8x  4 = 0
8. 9x + 4y  18x + 8y  23 = 0 2
y2  x2 = 1 16
2
22. x 2  y2  2x  2y  1 = 0
In Exercises 9–11, find the standard form of the equation of each ellipse satisfying the given conditions. 9. Foci: (  4, 0), (4, 0); Vertices: (  5, 0), (5, 0)
In Exercises 23–24, find the standard form of the equation of each hyperbola satisfying the given conditions. 23. Foci: (0,  4), (0, 4); Vertices: (0,  2), (0, 2)
10. Foci: (0,  3), (0, 3); Vertices: (0, 6), (0, 6)
24. Foci: (  8, 0), (8, 0); Vertices: (  3, 0), (3, 0)
11. Major axis horizontal with length 12; length of minor
25. Explain why it is not possible for a hyperbola to have foci at (0,  2) and (0, 2) and vertices at (0, 3) and (0, 3). 26. Radio tower M2 is located 200 miles due west of radio tower M1. The situation is illustrated in the figure shown, where a coordinate system has been superimposed. Simultaneous radio signals are sent from each tower to a ship, with the signal from M2 received 500 microseconds before the signal from M1. Assuming that radio signals travel at 0.186 mile per microsecond, determine the equation of the
axis = 4; center: (  3, 5). 12. A semielliptical arch supports a bridge that spans a river 20 yards wide. The center of the arch is 6 yards above the river’s center. Write an equation for the ellipse so that the x@axis coincides with the water level and the y@axis passes through the center of the arch.
hyperbola on which the ship is located. 6 yd
P
y
20 yd
13. A semielliptic archway has a height of 15 feet at the center and a width of 50 feet, as shown in the figure. The 50foot width consists of a twolane road. Can a truck that is 12 feet high and 14 feet wide drive under the archway without going into the other lane?
M2 (–100, 0)
200 miles
x M1 (100, 0)
10.3 50 feet
15 feet
14 feet 12 feet
14. An elliptical pool table has a ball placed at each focus. If one ball is hit toward the side of the table, explain what will occur.
In Exercises 27–33, find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. 27. y2 = 8x 28. x 2 + 16y = 0 29. (y  2)2 =  16x 30. (x  4)2 = 4(y + 1) 31. x 2 + 4y = 4 32. y2  4x  10y + 21 = 0 33. x 2  4x  2y = 0
Chapter 10 Summary, Review, and Test In Exercises 34–35, find the standard form of the equation of each parabola satisfying the given conditions. 34. Focus: (12, 0); Directrix: x = 12 35. Focus: (0, 11); Directrix: y = 11 36. An engineer is designing headlight units for automobiles. The unit has a parabolic surface with a diameter of 12 inches and a depth of 3 inches. The situation is illustrated in the figure, where a coordinate system has been superimposed. What is the equation of the parabola in this system? Where should the light source be placed? Describe this placement relative to the vertex. y
(6, 3)
3 inches x 6
–6 Vertex (0, 0) 12 inches
37. The George Washington Bridge spans the Hudson River from New York to New Jersey. Its two towers are 3500 feet apart and rise 316 feet above the road. As shown in the figure, the cable between the towers has the shape of a parabola and the cable just touches the sides of the road midway between the towers. What is the height of the cable 1000 feet from a tower? 2CTCDQNKE%CDNG
y
1093
41. 16x 2 + 64x + 9y2  54y + 1 = 0 42. 4x 2  9y2  8x + 12y  144 = 0
10.4 In Exercises 43–46, identify the conic represented by each equation without using a rotation of axes. 43. 44. 45. 46.
5x 2 + 223xy + 3y2  18 = 0 5x 2  8xy + 7y2  925x  9 = 0 x 2 + 6xy + 9y2  2y = 0 x 2  2xy + 3y2 + 2x + 4y  1 = 0
In Exercises 47–51, a. Rewrite the equation in a rotated x′y′@system without an x′y′@term. b. Express the equation involving x′ and y′ in the standard form of a conic section. c. Use the rotated system to graph the equation. 47. xy  4 = 0 48. x 2 + xy + y2  1 = 0 49. 4x 2 + 10xy + 4y2  9 = 0 50. 6x 2  6xy + 14y2  45 = 0 51. x 2 + 223xy + 3y2  1223x + 12y = 0
10.5 In Exercises 52–57, eliminate the parameter and graph the plane curve represented by the parametric equations. Use arrows to show the orientation of each plane curve. 52. x = 2t  1, y = 1  t;  ∞ 6 t 6 ∞
(1750, 316)
53. x = t 2, y = t  1; 1 … t … 3 –17 5
54. x = 4t 2, y = t + 1;  ∞ 6 t 6 ∞
0
55. x = 4 sin t, y = 3 cos t; 0 … t 6 p 350
0 fe
et
175
0 x
38. The giant satellite dish in the figure shown is in the shape of a parabolic surface. Signals strike the surface and are reflected to the focus, where the receiver is located. The diameter of the dish is 300 feet and its depth is 44 feet. How far, to the nearest foot, from the base of the dish should the receiver be placed? y
56. x = 3 + 2 cos t, y = 1 + 2 sin t; 0 … t 6 2p p 57. x = 3 sec t, y = 3 tan t; 0 … t … 4 58. Find two different sets of parametric equations for y = x 2 + 6. 59. The path of a projectile that is launched h feet above the ground with an initial velocity of 0 feet per second and at an angle u with the horizontal is given by the parametric equations x = (v0 cos u)t and y = h + (v0 sin u)t  16t 2,
300 feet
4GEGKXGT (150, 44) 44 feet
x
In Exercises 39–42, identify the conic represented by each equation without completing the square. 39. y2 + 4x + 2y  15 = 0 40. x 2 + 16y2  160y + 384 = 0
where t is the time, in seconds, after the projectile was launched. A football player throws a football with an initial velocity of 100 feet per second at an angle of 40° to the horizontal. The ball leaves the player’s hand at a height of 6 feet. a. Find the parametric equations that describe the position of the ball as a function of time. b. Describe the ball’s position after 1, 2, and 3 seconds. Round to the nearest tenth of a foot. c. How long, to the nearest tenth of a second, is the ball in flight? What is the total horizontal distance that it travels before it lands? d. Graph the parametric equations in part (a) using a graphing utility. Use the graph to determine when the ball is at its maximum height. What is its maximum height? Round answers to the nearest tenth.
1094
Chapter 10 Conic Sections and Analytic Geometry
10.6 In Exercises 60–65, a. If necessary, write the equation in one of the standard forms for a conic in polar coordinates. b. Determine values for e and p. Use the value of e to identify the conic section. c. Graph the given polar equation. 4 60. r = 1  sin u 61. r =
62. r =
6 2 + sin u
63. r =
2 3  2 cos u
64. r =
6 3 + 6 sin u
65. r =
8 4 + 16 cos u
6 1 + cos u
Chapter 10 Test You can check your answers against those at the back of the book. Stepbystep solutions are found in the Chapter Test Prep Videos available in MyLab Math and at youtube.com/user/ pearsonmathstats (playlist “Blitzer Algebra and Trigonometry 7e”).
10. An engineer is designing headlight units for cars. The unit shown in the figure below has a parabolic surface with a diameter of 6 inches and a depth of 3 inches. y
In Exercises 1–5, graph the conic section with the given equation. For ellipses, find the foci. For hyperbolas, find the foci and give the equations of the asymptotes. For parabolas, find the vertex, focus, and directrix. 1. 9x 2  4y2 = 36 3 inches
2. x 2 =  8y 3.
(x + 2)2 25
+
(y  5)2 9
= 1
x 3
–3
4. 4x 2  y2 + 8x + 2y + 7 = 0
Vertex (0, 0) 6 inches
5. (x + 5)2 = 8(y  1) In Exercises 6–8, find the standard form of the equation of the conic section satisfying the given conditions. 6. Ellipse; Foci: ( 7, 0), (7, 0); Vertices: (  10, 0), (10, 0) 7. Hyperbola; Foci: (0,  10), (0, 10); Vertices: (0,  7), (0, 7) 8. Parabola; Focus: (50, 0); Directrix: x = 50 9. A sound whispered at one focus of a whispering gallery can be heard at the other focus. The figure below shows a whispering gallery whose cross section is a semielliptical arch with a height of 24 feet and a width of 80 feet. How far from the room’s center should two people stand so that they can whisper back and forth and be heard?
a. Using the coordinate system that has been positioned on the unit, find the parabola’s equation. b. If the light source is located at the focus, describe its placement relative to the vertex. In Exercises 11–12, identify each equation without completing the square or using a rotation of axes. 11. x 2 + 9y2 + 10x  18y + 25 = 0 12. x 2 + y2 + xy + 3x  y  3 = 0 13. For the equation 7x 2  623xy + 13y2  16 = 0, determine what angle of rotation would eliminate the x′y′@term in a rotated x′y′@system. In Exercises 14–15, eliminate the parameter and graph the plane curve represented by the parametric equations. Use arrows to show the orientation of each plane curve.
24 feet
14. x = t 2, y = t  1;  ∞ 6 t 6 ∞ 15. x = 1 + 3 sin t, y = 2 cos t; 0 … t 6 2p
80 feet
In Exercises 16–17, identify the conic section and graph the polar equation. 2 4 17. r = 16. r = 1  cos u 2 + sin u
Chapter 10 Summary, Review, and Test
1095
Cumulative Review Exercises (Chapters 1–10) Solve each equation or inequality in Exercises 1–7. 1. 2(x  3) + 5x = 8(x  1) 2. 3(2x  4) 7 2(6x  12) 3. x  5 = 2x + 7 4. (x  2)2 = 20 5. 2x  1 Ú 7
a. b. c. d. e. f.
Find the domain and the range of f. What is the relative minimum and where does it occur? Find the interval on which f is increasing. Find f(  1)  f(0). Find ( f ∘ f )(1). Use arrow notation to complete this statement: f(x) S ∞ as ̱̱̱̱̱̱̱̱̱ or as ̱̱̱̱̱̱̱̱̱.
6. 3x 3 + 4x 2  7x + 2 = 0 7. log2(x + 1) + log2(x  1) = 3
g. Graph g(x) = f(x  2) + 1. h. Graph h(x) =  f(2x).
Solve each system in Exercises 8–10. 8. b
3x + 4y = 2 2x + 5y = 1
9. b
2x 2  y2 =  8 x  y = 6
16. If f(x) = x 2  4 and g(x) = x + 2, find (g ∘ f )(x). 17. Expand using logarithmic properties. Where possible, evaluate logarithmic expressions. log5 ¢
x  y + z = 17 10. c  4x + y + 5z = 2 2x + 3y + z = 8 In Exercises 11–13, graph each equation, function, or system in a rectangular coordinate system. 11. f(x) = (x  1)2  4 12.
y2 x2 + = 1 9 4
13. c
5x + y … 10 1 y Ú x + 2 4
14. a. List all possible rational roots of 32x 3  52x 2 + 17x + 3 = 0. b. The graph of f(x) = 32x 3  52x 2 + 17x + 3 is shown in a [ 1, 3, 1] by [  2, 6, 1] viewing rectangle. Use the graph of f and synthetic division to solve the equation in part (a).
x 3 1y ≤ 125
18. Write the slopeintercept form of the equation of the line passing through (1, 4) and (  5, 8). 19. RentaTruck charges a daily rental rate for a truck of $39 plus $0.76 a mile. A competing agency, Ace Truck Rentals, charges $25 a day plus $0.84 a mile for the same truck. How many miles must be driven in a day to make the daily cost of both agencies the same? What will be the cost? 20. The local cable television company offers two deals. Basic cable service with one movie channel costs $35 per month. Basic service with two movie channels cost $45 per month. Find the charge for the basic cable service and the charge for each movie channel. 21. The bar graph shows the number of interracial married couples in the United States, in thousands (for example, 651, the data for 1980, represents 651,000), for selected years from 1980 through 2018.
15. The figure shows the graph of y = f(x) and its two vertical asymptotes. y 5 4 3 2 1 –5 –4 –3 –2 –1–1 –2 –3 –4 –5
Number of Interracial Married Couples (thousands)
Interracial Married Couples in the United States 4000 3237 3000
2478
2000 1464 1000
964 651
1980
1990
y=f x
2000 Year
2010
2018
Source: U.S. Census Bureau 1 2 3 4 5
x
The linear function f(x) = 69x + 402 models the number of interracial married couples in the United States, in thousands, x years after 1980.
1096
Chapter 10 Conic Sections and Analytic Geometry
21. (continued) a. What is the slope of the model f (x) = 69x + 402 and what does it represent? b. Does this model overestimate or underestimate the data for 2010 in the bar graph on the previous page? By how much? c. If trends shown by the data continue, use the model to project the number of interracial married couples in the United States, in thousands, in 2030.
22. Verify the identity:
csc u  sin u = cot2 u. sin u
23. Graph one complete cycle of y = 2 cos(2x + p).
24. If v = 3i  6j and w = i + j, find (v # w)w. 25. Solve for u: sin 2u = sin u, 0 … u 6 2p.
26. In oblique triangle ABC, A = 64°, B = 72°, and a = 13.6. Solve the triangle. Round lengths to the nearest tenth.
11
Sequences, Induction, and Probability
Something incredible has happened. Your college roommate, a gifted athlete, has been given a sixyear contract with a professional baseball team. He will be playing against the likes of Mike Trout of the Los Angeles Angels and Kris Bryant of the Chicago Cubs. Management offers him three options. One is a beginning salary of $1,700,000 with annual increases of $70,000 per year starting in the second year. A second option is $1,700,000 the first year with an annual increase of 2% per year beginning in the second year. The third option involves less money the first year—$1,500,000—but there is an annual increase of 9% yearly after that. Which option offers the most money over the sixyear contract?
Here’s where you’ll find these applications: A similar problem appears as Exercise 67 in Exercise Set 11.3, and this problem appears as the Group Exercise on page 1134.
1098
Chapter 11 Sequences, Induction, and Probability
SECTION 11.1
Sequences and Summation Notation
WHAT YOU’LL LEARN 1
Find particular terms of a sequence from the general term.
2 Use recursion formulas. 3 Use factorial notation. 4 Use summation notation.
Sequences Many creations in nature involve intricate mathematical designs, including a variety of spirals. For example, the arrangement of the individual florets in the head of a sunflower forms spirals. In some species, there are 21 spirals in the clockwise direction and 34 in the counterclockwise direction. The precise numbers depend on the species of sunflower: 21 and 34, or 34 and 55, or 55 and 89, or even 89 and 144. This observation becomes even more interesting when we consider a sequence of numbers investigated by Leonardo of Pisa, also known as Fibonacci, an Italian mathematician in the thirteenth century. The Fibonacci sequence of numbers is an infinite sequence that begins as follows: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, . . . .
BLITZER BONUS Fibonacci Numbers on the Piano Keyboard
The first two terms are 1. Every term thereafter is the sum of the two preceding terms. For example, the third term, 2, is the sum of the first and second terms: 1 + 1 = 2. The fourth term, 3, is the sum of the second and third terms: 1 + 2 = 3, and so on. Did you know that the number of spirals in a daisy or a sunflower, 21 and 34, are two Fibonacci numbers? The number of spirals in a pine cone, 8 and 13, and a pineapple, 8 and 13, are also Fibonacci numbers. We can think of the Fibonacci sequence as a function. The terms of the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, . . .
One Octave
Numbers in the Fibonacci sequence can be found in an octave on the piano keyboard. The octave contains 2 black keys in one cluster and 3 black keys in another cluster, for a total of 5 black keys. It also has 8 white keys, for a total of 13 keys. The numbers 2, 3, 5, 8, and 13 are the third through seventh terms of the Fibonacci sequence.
are the range values for a function f whose domain is the set of positive integers. Domain:
1, T
2, T
3, T
4, T
5, T
6, T
7, T
...
Range:
1,
1,
2,
3,
5,
8,
13,
...
Thus, f(1) = 1, f(2) = 1, f(3) = 2, f(4) = 3, f(5) = 5, f(6) = 8, f(7) = 13, and so on. The letter a with a subscript is used to represent function values of a sequence, rather than the usual function notation. The subscripts make up the domain of the sequence, and they identify the location of a term. Thus, a1 represents the first term of the sequence, a2 represents the second term, a3 the third term, and so on. This notation is shown for the first six terms of the Fibonacci sequence: 1, a=
1, a=
2, a=
3, a=
5, a=
8. a=
The notation an represents the nth term, or general term, of a sequence. The entire sequence is represented by {an}.
Section 11.1 Sequences and Summation Notation
1099
Definition of a Sequence An infinite sequence {an} is a function whose domain is the set of positive integers. The function values, or terms, of the sequence are represented by a1, a2, a3, a4, . . . , an, . . . . Sequences whose domains consist only of the first n positive integers are called finite sequences.
1 Find particular terms of a sequence from the general term.
Writing Terms of a Sequence from the General Term
EXAMPLE 1
Write the first four terms of the sequence whose nth term, or general term, is given: ( 1)n a. an = 3n + 4 b. an = n . 3  1 Solution a. We need to find the first four terms of the sequence whose general term is an = 3n + 4. To do so, we replace n in the formula with 1, 2, 3, and 4. aUV VGTO
3∙1+4=3+4=7
aPF VGTO
3 ∙ 2 + 4 = 6 + 4 = 10
aTF VGTO
3 ∙ 3 + 4 = 9 + 4 = 13
aVJ VGTO
3 ∙ 4 + 4 = 12 + 4 = 16
The first four terms are 7, 10, 13, and 16. The sequence defined by an = 3n + 4 can be written as 7, 10, 13, 16, . . . , 3n + 4, . . . . GREAT QUESTION What effect does ( −1)n have on the terms of a sequence? The factor (  1)n in the general term of a sequence causes the signs of the terms to alternate between positive and negative, depending on whether n is even or odd.
b. We need to find the first four terms of the sequence whose general term is ( 1)n . To do so, we replace each occurrence of n in the formula with an = n 3  1 1, 2, 3, and 4. aUV VGTO
(–1)1 1
3 −1 aTF VGTO
(–1)3 3
3 −1
=
–1 1 =– 3−1 2
=
–1 1 =– 27 − 1 26
aPF VGTO aVJ VGTO
(–1)2 32 − 1 (–1)4 4
3 −1
1 The first four terms are  12, 18,  26 , and can be written as
1 80 .
=
1 1 = 9−1 8
=
1 1 = 81 − 1 80
The sequence defined by
( 1)n 3n  1
( 1)n 1 1 1 1 , ....  , ,  , , ..., n 2 8 26 80 3  1 CHECK POINT 1
Write the first four terms of the sequence whose nth term, or
general term, is given: a. an = 2n + 5
b. an =
( 1)n . 2n + 1
Although sequences are usually named with the letter a, any lowercase letter can be used. For example, the first four terms of the sequence {bn} = 5 1 12 2 n 6 are 1 b1 = 12, b2 = 14, b3 = 18, and b4 = 16 . Because a sequence is a function whose domain is the set of positive integers, the graph of a sequence is a set of discrete points. For example, consider the sequence whose general term is an = n1 . How does the graph of this sequence differ from the
Chapter 11 Sequences, Induction, and Probability
1100
TECHNOLOGY Graphing utilities can write the terms of a sequence and graph them. For example, to find the first 1 six terms of {an} = e f, enter n )GPGTCN VGTO SEQ
5VCTV CVa
graph of the function f(x) = 1x ? The graph of f(x) = 1x is shown in Figure 11.1(a) for positive values of x. To obtain the graph of the sequence {an} = 5 n1 6 , remove all the points from the graph of f except those whose x@coordinates are positive integers. Thus, we remove all points except (1, 1), 1 2, 12 2 , 1 3, 13 2 , 1 4, 14 2 , and so on. The remaining points are the graph of the sequence {an} = 5 n1 6 , shown in Figure 11.1(b). Notice that the horizontal axis is labeled n and the vertical axis is labeled an.
5VQR CVa
y
an
3
3
2
2
( 1 ÷ x, x, 1, 6, 1). 8CTKCDNG WUGFKP IGPGTCN VGTO
6JG pUVGRq HTQO aVQa aVQa GVEKU
The first few terms of the sequence in decimal form are shown in the viewing rectangle. By pressing the right arrow key to scroll right, you can see the remaining terms. The first six terms are also shown in fraction form.
(1, 1) 1
1
Q2, 12R 2
3
4
(1, 1)
1
1
Q3, 3 R Q4, 1 R 4
x
Q2, 12R
1
1
Q3, 3 R Q4, 1 R 4
2
3
4
n
Figure 11.1(b) The graph of 1 {an} = e f n
Figure 11.1(a) The graph of 1 f(x) = , x 7 0 x
Comparing a continuous graph to the graph of a sequence
Recursion Formulas In Example 1, the formulas used for the nth term of a sequence expressed the term as a function of n, the number of the term. Sequences can also be defined using recursion formulas. A recursion formula defines the nth term of a sequence as a function of the previous term. Our next example illustrates that if the first term of a sequence is known, then the recursion formula can be used to determine the remaining terms.
Some graphing calculators require that you fill in the information following SEQ on a separate screen and then paste it to the home screen.
EXAMPLE 2
Using a Recursion Formula
Find the first four terms of the sequence in which a1 = 5 and an = 3an  1 + 2 for n Ú 2. Solution
Let’s be sure we understand what is given. a1 = 5
2 Use recursion formulas.
6JGƂTUV VGTOKU
and
an
'CEJVGTO CHVGTVJGƂTUV
3an−1 + 2
= KU
VKOGUVJG RTGXKQWUVGTO
RNWU
Now let’s write the first four terms of this sequence. a1 = 5
This is the given first term.
a2 = 3a1 + 2
Use an = 3an − 1 + 2, with n = 2. Thus, a2 = 3a2 − 1 + 2 = 3a1 + 2.
= 3(5) + 2 = 17 a3 = 3a2 + 2 = 3(17) + 2 = 53
Substitute 5 for a1.
a4 = 3a3 + 2
Notice that a4 is defined in terms of a3.
Again use an = 3an − 1 + 2, with n = 3. Substitute 17 for a2. We used an = 3an − 1 + 2, with n = 4.
= 3(53) + 2 = 161
Use the value of a3, the third term, obtained above.
The first four terms are 5, 17, 53, and 161.
Section 11.1 Sequences and Summation Notation
1101
CHECK POINT 2 Find the first four terms of the sequence in which a1 = 3 and an = 2an  1 + 5 for n Ú 2.
3 Use factorial notation.
Factorial Notation Products of consecutive positive integers occur quite often in sequences. These products can be expressed in a special notation, called factorial notation.
BLITZER BONUS Factorial Notation
Factorials from 0 through 20 0!
1
1!
1
2!
2
3!
6
4!
24
5!
120
6!
720
7!
5040
8!
40,320
9!
If n is a positive integer, the notation n! (read “n factorial”) is the product of all positive integers from n down through 1. n! = n(n  1)(n  2) g(3)(2)(1) 0! (zero factorial), by definition, is 1. 0! = 1 The values of n! for the first six positive integers are 1! 2! 3! 4! 5! 6!
362,880
10!
3,628,800
11!
39,916,800
12!
479,001,600
13!
6,227,020,800
14!
87,178,291,200
15!
1,307,674,368,000
16!
20,922,789,888,000
17!
355,687,428,096,000
18!
6,402,373,705,728,000
19!
121,645,100,408,832,000
20!
2,432,902,008,176,640,000
As n increases, n! grows very rapidly. Factorial growth is more explosive than exponential growth discussed in Chapter 4.
TECHNOLOGY Most calculators have factorial keys. To find 5!, most calculators use one of the following: Many Scientific Calculators 5 x! Many Graphing Calculators 5! ENTER . Because n! becomes quite large as n increases, your calculator will display these larger values in scientific notation.
= = = = = =
1 2#1 = 2 3#2#1 = 6 4 # 3 # 2 # 1 = 24 5 # 4 # 3 # 2 # 1 = 120 6 # 5 # 4 # 3 # 2 # 1 = 720.
Factorials affect only the number or variable that they follow unless grouping symbols appear. For example, 2 # 3! = 2(3 # 2 # 1) = 2 # 6 = 12
whereas
(2 # 3)! = 6! = 6 # 5 # 4 # 3 # 2 # 1 = 720.
In this sense, factorials are similar to exponents.
EXAMPLE 3
Finding Terms of a Sequence Involving Factorials
Write the first four terms of the sequence whose nth term is 2n an = . (n  1)! Solution
We need to find the first four terms of the sequence. To do so, we 2n replace each n in with 1, 2, 3, and 4. (n  1)! aUV VGTO
2 2 21 = = =2 0! 1 (1 − 1)!
aPF VGTO
4 4 22 = = =4 (2 − 1)! 1! 1
aTF VGTO
23 8 8 = = =4 2! 2∙1 (3 − 1)!
aVJ VGTO
16 16 16 8 24 = = = = 3! 3∙2∙1 6 3 (4 − 1)!
The first four terms are 2, 4, 4, and 83.
1102
Chapter 11 Sequences, Induction, and Probability CHECK POINT 3
Write the first four terms of the sequence whose nth term is 20 an = . (n + 1)!
When evaluating fractions with factorials in the numerator and the denominator, try to reduce the fraction before performing the multiplications. For example, 26! . Rather than write out 26! as the product of all integers from 26 down consider 21! to 1, we can express 26! as 26! = 26 # 25 # 24 # 23 # 22 # 21!.
In this way, we can divide both the numerator and the denominator by the common factor, 21!. 26! 26 # 25 # 24 # 23 # 22 # 21! = = 26 # 25 # 24 # 23 # 22 = 7,893,600 21! 21!
EXAMPLE 4
Evaluating Fractions with Factorials
Evaluate each factorial expression: (n + 1)! 10! a. b. . 2!8! n! Solution 10! 10 # 9 # 8! 90 a. = # # = = 45 2!8! 2 1 8! 2 (n + 1)! (n + 1) # n! b. = = n + 1 n! n! CHECK POINT 4
14! a. 2!12!
4 Use summation notation.
Evaluate each factorial expression: n! . b. (n  1)!
Summation Notation It is sometimes useful to find the sum of the first n terms of a sequence. For example, consider the cost of raising a child born in the United States to a middleincome family. Table 11.1 shows the cost from year 1 (child is under 1) to year 18 (child is 17). (Costs include housing, food, health care, and clothing but exclude college costs.)
Table 11.1 The Cost of Raising a Child Born in the United States to a MiddleIncome Family Year Average Cost
1
Average Cost
3
4
5
6
7
8
9
$10,600 $10,930 $11,270 $11,960 $12,330 $12,710 $12,950 $13,350 $13,760
%JKNF KUWPFGT
Year
2
10
%JKNF KU
%JKNF KU
%JKNF KU
%JKNF KU
%JKNF KU
%JKNF KU
%JKNF KU
%JKNF KU
11
12
13
14
15
16
17
18
$13,970 $14,400 $14,840 $16,360 $16,860 $17,390 $18,430 $19,000 $19,590 %JKNF KU
%JKNF KU
Source: U.S. Department of Agriculture
%JKNF KU
%JKNF KU
%JKNF KU
%JKNF KU
%JKNF KU
%JKNF KU
%JKNF KU
Section 11.1 Sequences and Summation Notation
1103
We can let an represent the cost of raising a child in year n. The terms of the finite sequence in Table 11.1 are given as follows: 10,600, 10,930, 11,270, 11,960, 12,330, 12,710, 12,950, 13,350, 13,760, a
a
a
a
a
a
a
a
a
13,970, 14,400, 14,840, 16,360, 16,860, 17,390, 18,430, 19,000, 19,590. a
a
a
a
a
a
a
a
a
Why might we want to add the terms of this sequence? We do this to find the total cost of raising a child from birth through age 17. Thus, a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 + a9 + a10 + a11 + a12 + a13 + a14 + a15 + a16 + a17 + a18 = 10,600 + 10,930 + 11,270 + 11,960 + 12,330 + 12,710 + 12,950 + 13,350 + 13,760 + 13,970 + 14,400 + 14,840 + 16,360 + 16,860 + 17,390 + 18,430 + 19,000 + 19,590 = 260,700. We see that the total cost of raising a child from birth through age 17 is $260,700. There is a compact notation for expressing the sum of the first n terms of a sequence. For example, rather than write a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 + a9 + a10 + a11 + a12 + a13 + a14 + a15 + a16 + a17 + a18, we can use summation notation to express the sum as 18
a ai.
i=1
We read this expression as “the sum as i goes from 1 to 18 of ai.” The letter i is called the index of summation and is not related to the use of i to represent 2 1. You can think of the symbol g (the uppercase Greek letter sigma) as an instruction to add up the terms of a sequence. Summation Notation The sum of the first n terms of a sequence is represented by the summation notation n
a ai = a1 + a2 + a3 + a4 + g + an,
i=1
where i is the index of summation, n is the upper limit of summation, and 1 is the lower limit of summation.
Any letter can be used for the index of summation. The letters i, j, and k are used commonly. Furthermore, the lower limit of summation can be an integer other than 1. When we write out a sum that is given in summation notation, we are expanding the summation notation. Example 5 shows how to do this.
EXAMPLE 5
Using Summation Notation
Expand and evaluate the sum: 6
a. a (i 2 + 1) i=1
7
b. a [( 2)k  5] k=4
5
c. a 3. i=1
1104
Chapter 11 Sequences, Induction, and Probability
TECHNOLOGY Graphing utilities can calculate the sum of a sequence. For example, to find the sum of the sequence in Example 5(a), enter
SUM SEQ (x
2
Solution 6
a. To find a (i 2 + 1), we must replace i in the expression i 2 + 1 with all i=1
consecutive integers from 1 to 6, inclusive. Then we add. 6
2 2 2 2 2 a (i + 1) = (1 + 1) + (2 + 1) + (3 + 1) + (4 + 1)
+ 1, x, 1, 6, 1).
i=1
+ (52 + 1) + (62 + 1)
Then press ENTER ; 97 should be displayed. Use this capability to verify Example 5(b).
= 2 + 5 + 10 + 17 + 26 + 37 = 97 7
b. The index of summation in a [( 2)k  5] is k. First we evaluate ( 2)k  5 k=4
for all consecutive integers from 4 through 7, inclusive. Then we add. 7
k 4 5 a [( 2)  5] = [( 2)  5] + [( 2)  5]
k=4
+ [( 2)6  5] + [( 2)7  5] = (16  5) + ( 32  5) + (64  5) + ( 128  5) = 11 + ( 37) + 59 + ( 133) = 100 5
c. To find a 3, we observe that every term of the sum is 3. The notation i = 1 i=1
through 5 indicates that we must add the first five terms of a sequence in which every term is 3. 5
a 3 = 3 + 3 + 3 + 3 + 3 = 15
i=1
Expand and evaluate the sum:
CHECK POINT 5 6
5
a. a 2i 2
b. a (2k  3)
i=1
k=3
5
c. a 4. i=1
Although the domain of a sequence is the set of positive integers, any integers can be used for the limits of summation. For a given sum, we can vary the upper and lower limits of summation, as well as the letter used for the index of summation. By doing so, we can produce differentlooking summation notations for the same sum. For example, the sum of the squares of the first four positive integers, 12 + 22 + 32 + 42, can be expressed in a number of equivalent ways: 4
2 2 2 2 2 a i = 1 + 2 + 3 + 4 = 30
i=1 3
2 2 2 2 2 a (i + 1) = (0 + 1) + (1 + 1) + (2 + 1) + (3 + 1)
i=0
= 12 + 22 + 32 + 42 = 30 5
2 2 2 2 2 a (k  1) = (2  1) + (3  1) + (4  1) + (5  1)
k=2
= 12 + 22 + 32 + 42 = 30.
Section 11.1 Sequences and Summation Notation
1105
Writing Sums in Summation Notation
EXAMPLE 6
Express each sum using summation notation: a. 13 + 23 + 33 + g + 73
b. 1 +
1 1 1 1 + + + g + n  1. 3 9 27 3
Solution In each case, we will use 1 as the lower limit of summation and i for the index of summation. a. The sum 13 + 23 + 33 + g + 73 has seven terms, each of the form i 3, starting at i = 1 and ending at i = 7. Thus, 7
13 + 23 + 33 + g + 73 = a i 3. i=1
b. The sum 1 +
1 1 1 1 + + + g+ n1 3 9 27 3
has n terms, each of the form 1 +
Instructor Resources for Section 11.1 in MyLab Math
1 , starting at i = 1 and ending at i = n. Thus, 3i  1
n 1 1 1 1 1 + + + g + n  1 = a i  1. 3 9 27 3 i=1 3
Express each sum using summation notation: 1 1 1 1 b. 1 + + + + g + n  1 . a. 12 + 22 + 32 + g + 9 2 2 4 8 2 CHECK POINT 6
Table 11.2 contains some important properties of sums expressed in summation notation. Table 11.2 Properties of Sums Property n
Example n
1. a cai = c a ai, c any real number i=1
i=1
2 # 2 # 2 # 2 # 2 a 3i = 3 1 + 3 2 + 3 3 + 3 4 4
i=1 4
3 a i 2 = 3(12 + 22 + 32 + 42) = 3 # 12 + 3 # 22 + 3 # 32 + 3 # 42 i=1
4
4
i=1
i=1
Conclusion: a 3i 2 = 3 a i 2 n
n
n
2. a (ai + bi) = a ai + a bi i=1
i=1
i=1
4
2 2 2 2 2 a (i + i ) = (1 + 1 ) + (2 + 2 ) + (3 + 3 ) + (4 + 4 )
i=1 4
4
2 2 2 2 2 a i + a i = (1 + 2 + 3 + 4) + (1 + 2 + 3 + 4 )
i=1
i=1
= (1 + 12) + (2 + 22) + (3 + 32) + (4 + 42) 4
4
4
i=1
i=1
i=1
Conclusion: a (i + i 2) = a i + a i 2 n
n
n
3. a (ai  bi) = a ai  a bi i=1
i=1
i=1
5
2 3 2 3 2 3 2 3 a (i  i ) = (3  3 ) + (4  4 ) + (5  5 )
i=3 5 2
5
3 2 2 2 3 3 3 a i  a i = (3 + 4 + 5 )  (3 + 4 + 5 )
i=3
i=3
= (32  33) + (42  43) + (52  53) 5
5
5
i=3
i=3
i=3
Conclusion: a (i 2  i 3) = a i 2  a i 3
Chapter 11 Sequences, Induction, and Probability
1106
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. {an} = a1, a2, a3, a4, . . . , an, . . . represents an infinite , a function whose domain is the set of positive . The function values a1, a2, a3, . . . are called the . C2. The nth term of a sequence, represented by an, is called the term.
C6. 5!, called 5 integers from 0! = . (n + 3)! = C7. (n + 2)! n
+
i=1
C3. an = 5n  6 C4. an =
+
C8. a ai =
Write the second term of each sequence.
, is the product of all positive down through . By definition,
+ g+
. In
this summation notation, i is called the of summation, n is the of summation, and 1 is the of summation.
(  1)n
4n  1 C5. an = 2an  1  4, a1 = 3
11.1 EXERCISE SET Practice Exercises
In Exercises 29–42, find each indicated sum.
In Exercises 1–12, write the first four terms of each sequence whose general term is given.
29. a 5i
1. an = 3n + 2 n
3. an = 3
5. an = ( 3)n 7. an = ( 1) (n + 3) 2n 9. an = n + 4 (  1)n + 1 11. an = n 2  1 n
1 n 4. an = a b 3 1 n 6. an = a  b 3
n+1
8. an = ( 1) (n + 4) 3n 10. an = n + 5 ( 1)n + 1 12. an = n 2 + 1
a1 a1 a1 a1 a1 a1
= = = = = =
7 and an = an  1 + 5 for n Ú 2 12 and an = an  1 + 4 for n Ú 2 3 and an = 4an  1 for n Ú 2 2 and an = 5an  1 for n Ú 2 4 and an = 2an  1 + 3 for n Ú 2 5 and an = 3an  1  1 for n Ú 2
20. an =
17! 15! 16! 25. 2!14! (n + 2)! 27. n!
32. a i 3
i=1
i=1
5
4
33. a k(k + 4)
34. a (k  3)(k + 2)
1 35. a a  b 2
4 1 i 36. a a  b 3
k=1 4
i
i=1 9
38. a 12
i=5
39. a
i=0
i=2 7
37. a 11 4
k=1
i=3
(  1)
i
i!
4 ( 1)i + 1 40. a i = 0 (i + 1)! 5
42. a
i=1
(i + 2)! i!
In Exercises 43–54, express each sum using summation notation. Use 1 as the lower limit of summation and i for the index of summation.
(n + 1)!
n2 22. an =  2(n  1)!
18! 16! 20! 26. 2!18! (2n + 1)! 28. (2n)! 24.
5
31. a 2i 2
i! 41. a (i 1)! i=1
In Exercises 23–28, evaluate each factorial expression. 23.
i=1
4
5
In Exercises 19–22, the general term of a sequence is given and involves a factorial. Write the first four terms of each sequence. n2 19. an = n! 21. an = 2(n + 1)!
6
30. a 7i
i=1
2. an = 4n  1
The sequences in Exercises 13–18 are defined using recursion formulas. Write the first four terms of each sequence. 13. 14. 15. 16. 17. 18.
6
43. 12 + 22 + 32 + g + 152 44. 14 + 24 + 34 + g + 124 45. 2 + 22 + 23 + g + 211 46. 5 + 52 + 53 + g + 514 47. 1 + 2 + 3 + g + 30 48. 1 + 2 + 3 + g + 34 49.
1 2 3 14 + + + g+ 2 3 4 14 + 1
Section 11.1 Sequences and Summation Notation 1 2 3 16 + + + g+ 3 4 5 16 + 2
51. 4 + 52.
Hours per Day Spent by U.S. Adults on Digital Media
42 43 4n + + g+ 2 3 n
1 2 3 n + 2 + 3 + g+ n 9 9 9 9
53. 1 + 3 + 5 + g + (2n  1) 2
54. a + ar + ar + g + ar
n1
In Exercises 55–60, express each sum using summation notation. Use a lower limit of summation of your choice and k for the index of summation. 55. 5 + 7 + 9 + 11 + g + 31
Hours per Day on Digital Media
50.
7 6 5
58. a + ar + ar 2 + g + ar 14 59. a + (a + d) + (a + 2d) + g + (a + nd)
4
The Graph of {an} an
The Graph of {bn} bn
5 4 3 2 1
5 4 3 2 1
+ 1)
1 2011
2012
n
1 2 3 4 5
–1 –2 –3 –4 –5 5
62. a
i=1 5
(b2i
ai 2 65. a ¢ ≤ i = 4 bi
ai 3 66. a ¢ ≤ i = 4 bi
5
5
i=1
i=1
where n = 1, 2, 3, 4, 5, 6, 7, models the average number of hours per day that U.S. adult users spent on digital media 1 7 n years after 2010. Use this model to find a ai. Does 7 i=1 this underestimate or overestimate the rounded value you obtained in part (a)? By how much?
n
 1)
64. a (ai + 3bi)
67. a a2i + a bi2
2017
70. The bar graph shows the number of Americans who renounced their U.S. citizenship, many over tax laws, from 2013 through 2017.
63. a (2ai + bi) i=1 5
2016
2
i=1 5
Giving Up U.S. Citizenship 6000 Number of People Renouncing Their U.S. Citizenship
i=1 5
2015
3
an = 0.3n + 3.6,
In Exercises 61–68, use the graphs of {an} and {bn} to find each indicated sum.
61. a
2014 Year
3.7
b. The finite sequence whose general term is
Practice PLUS
(a2i
2013
5.9
4.3
60. (a + d) + (a + d 2) + g + (a + d n)
5
5.1
5.6
Let an represent the average number of hours per day that U.S. adult users spent on digital media n years after 2010. a. Use the numbers given in the graph to find and interpret 1 7 ai. Round to the nearest hour. 7 ia =1
57. a + ar + ar 2 + g + ar 12
1 2 3 4 5
4.9
5.4
Source: 2018 Internet Trends Report
56. 6 + 8 + 10 + 12 + g + 32
–1 –2 –3 –4 –5
1107
5411
5000 4004
4000 3000
3415 2999
2000 1000 2013
5
5
i=1
i=3
68. a a2i  a b2i
Application Exercises 69. The bar graph at the top of the next column shows the average number of hours per day that U.S. adult users spent on digital media (desktop/laptop, mobile, and other devices) from 2011 through 2017.
5133
2014
2015 Year
2016
2017
Source: Internal Revenue Service
Let an represent the number of Americans who gave up their U.S. citizenship n years after 2012. a. Use the numbers given in the graph to find and interpret 5
a ai.
i=1
1108
Chapter 11 Sequences, Induction, and Probability
70. (continued) b. The finite sequence whose general term is
88. As n increases, the terms of the sequence an = a1 +
1 n b n get closer and closer to the number e (where e ≈ 2.7183). Use a calculator to find a10, a100, a1000, a10,000, and a100,000, comparing these terms to your calculator’s decimal approximation for e.
an = 626n + 2313, where n = 1, 2, 3, 4, 5, models the number of Americans who gave up their U.S. citizenship n years after 2012. 5
Use this model to find a ai. Does this underestimate or i=1
overestimate the actual sum in part (a) on the previous page? By how much? 71. A deposit of $6000 is made in an account that earns 6% interest compounded quarterly. The balance in the account after n quarters is given by the sequence 0.06 n b , n = 1, 2, 3, . . . . 4 Find the balance in the account after five years. Round to the nearest cent. 72. A deposit of $10,000 is made in an account that earns 8% interest compounded quarterly. The balance in the account after n quarters is given by the sequence an = 6000a1 +
0.08 n n = 1, 2, 3, . . . . b , 4 Find the balance in the account after six years. Round to the nearest cent. an = 10,000a1 +
Explaining the Concepts 73. What is a sequence? Give an example with your description. 74. Explain how to write terms of a sequence if the formula for the general term is given. 75. What does the graph of a sequence look like? How is it obtained? 76. What is a recursion formula? 77. Explain how to find n! if n is a positive integer. 900! 78. Explain the best way to evaluate without a calculator. 899! 79. What is the meaning of the symbol g? Give an example with your description. 80. You buy a new car for $24,000. At the end of n years, the value of your car is given by the sequence 3 n an = 24,000a b , n = 1, 2, 3, . . . . 4 Find a5 and write a sentence explaining what this value represents. Describe the nth term of the sequence in terms of the value of your car at the end of each year.
Technology Exercises In Exercises 81–85, use a calculator’s factorial key to evaluate each expression. 200! 300 20! 81. 82. a b! 83. 198! 20 300 20! 54! 84. 85. (20  3)! (54  3)!3! 86. Use the SEQ (sequence) capability of a graphing utility to verify the terms of the sequences you obtained for any five sequences from Exercises 1–12 or 19–22. 87. Use the SUM SEQ (sum of the sequence) capability of a graphing utility to verify any five of the sums you obtained in Exercises 29–42.
Many graphing utilities have a sequencegraphing mode that plots the terms of a sequence as points on a rectangular coordinate system. Consult your manual; if your graphing utility has this capability, use it to graph each of the sequences in Exercises 89–92. What appears to be happening to the terms of each sequence as n gets larger? n n: [0, 10, 1] by an : [0, 1, 0.1] n + 1 100 90. an = n: [0, 1000, 100] by an : [0, 1, 0.1] n 2n2 + 5n  7 91. an = n: [0, 10, 1] by an : [0, 2, 0.2] n3 4 3n + n  1 92. an = n: [0, 10, 1] by an : [0, 1, 0.1] 5n4 + 2n2 + 1 89. an =
Critical Thinking Exercises Make Sense? In Exercises 93–96, determine whether each statement makes sense or does not make sense, and explain your reasoning. 93. Now that I’ve studied sequences, I realize that the joke in this cartoon is based on the fact that you can’t have a negative number of sheep.
94. By writing a1, a2, a3, a4, . . . , an, . . . , I can see that the range of a sequence is the set of positive integers. 95. It makes a difference whether or not I use parentheses around the expression following the summation symbol, 8
8
because the value of a (i + 7) is 92, but the value of a i + 7 i=1 i=1 is 43. 96. Without writing out the terms, I can see that ( 1)2n in (  1)2n an = causes the terms to alternate in sign. 3n In Exercises 97–100, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. 97.
n! 1 = (n  1)! n  1
Section 11.2 Arithmetic Sequences 98. The Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, . . . can be defined recursively using a0 = 1, a1 = 1; an = an  2 + an  1, where n Ú 2. 2
2
99. a (  1)i2i = 0
2
104. Use the graph of y = f(x) to graph y = f(x  2)  1. y 4 3 2 1
2
100. a aibi = a ai a bi
i=1
i=1
i=1
i=1
101. Write the first five terms of the sequence whose first term is 9 and whose general term is an  1 an = c 2 3an  1 + 5
1109
–5 –4 –3 –2 –1–1
y=f x
1 2 3 4 5
x
–2 –3 –4
if an  1 is even if an  1 is odd
(Section 2.5, Example 8)
for n Ú 2.
105. Solve: x 4  6x 3 + 4x 2 + 15x + 4 = 0. (Section 3.4, Example 5)
Group Exercise 102. Enough curiosities involving the Fibonacci sequence exist to warrant a flourishing Fibonacci Association, which publishes a quarterly journal. Do some research on the Fibonacci sequence by consulting the Internet or the research department of your library, and find one property that interests you. After doing this research, get together with your group to share these intriguing properties.
x 2x 5 =  . x  3 x  3 3
(Section 1.2, Example 5)
SECTION 11.2
Exercises 107–109 will help you prepare for the material covered in the next section.
108. Consider the sequence whose nth term is an = 4n  3. Find a2  a1, a3  a2, a4  a3, and a5  a4. What do you observe? 109. Use the formula an = 4 + (n  1)( 7) to find the eighth term of the sequence 4, 3,  10, . . . .
Arithmetic Sequences
WHAT YOU’LL LEARN 1 Find the common difference for an arithmetic sequence.
2 Write terms of an arithmetic sequence.
3 Use the formula for
the general term of an arithmetic sequence.
4 Use the formula for the
Preview Exercises
107. Consider the sequence 8, 3, 2, 7,  12, . . . . Find a2  a1, a3  a2, a4  a3, and a5  a4. What do you observe?
Retaining the Concepts 103. Solve:
106. Determine the amplitude, period, and phase shift of y = 12 cos(3x + p2 ). Then graph one period of the function. (Section 5.5, Example 6)
sum of the first n terms of an arithmetic sequence.
Your grandmother and her financial counselor are looking at options in case an adult residential facility is needed in the future. The good news is that your grandmother’s total assets are $500,000. The bad news is that adult residential community costs average $64,130 annually, increasing by $1800 each year. In this section, we will see how sequences can be used to model your grandmother’s situation and help her to identify realistic options.
Arithmetic Sequences The bar graph in Figure 11.2 at the top of the next page shows how much Americans spent on their pets, rounded to the nearest billion dollars, each year from 2005 through 2016.
1110
Chapter 11 Sequences, Induction, and Probability
Figure 11.2 Source: American Pet Products Manufacturers Association
Spending on Pets in the United States 59
Spending (billions of dollars)
60 57
57
55 53
54 51
51
49 47
48 45
45
43 41
42 39
39 37
36 33 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 Year
The graph illustrates that each year spending increased by $2 billion. The sequence of annual spending 37, 39, 41, 43, 45, 47, 49, c shows that each term after the first, 37, differs from the preceding term by a constant amount, namely, 2. This sequence is an example of an arithmetic sequence. Definition of an Arithmetic Sequence An arithmetic sequence is a sequence in which each term after the first differs from the preceding term by a constant amount. The difference between consecutive terms is called the common difference of the sequence.
1 Find the common difference for an arithmetic sequence.
The common difference, d, is found by subtracting any term from the term that directly follows it. In the following examples, the common difference is found by subtracting the first term from the second term, a2  a1. Arithmetic Sequence
Common Difference
142, 146, 150, 154, 158, c
d = 146  142 = 4
5,  2, 1, 4, 7, c
d = 2  (  5) = 2 + 5 = 3
8, 3,  2,  7,  12, c
d = 3  8 = 5
Figure 11.3 shows the graphs of the last two arithmetic sequences in our list. The common difference for the increasing sequence in Figure 11.3(a) is 3. The common difference for the decreasing sequence in Figure 11.3(b) is 5. an 8 6 4 2 –2 –4 –6 –8 –10 –12
bn %QPUVCPV VGTOVQVGTO EJCPIG KU 2 4 6 8 10
8 6 4 2
n
(KTUVVGTOKU–
Figure 11.3(a) The graph of {an} =  5,  2, 1, 4, 7, c
–2 –4 –6 –8 –10 –12
(KTUVVGTOKU
2 4 6 8 10
n
%QPUVCPV VGTOVQVGTO EJCPIG KU–
Figure 11.3(b) The graph of {bn} = 8, 3,  2,  7,  12, c
The graph of each arithmetic sequence in Figure 11.3 forms a set of discrete points lying on a straight line. This illustrates that an arithmetic sequence is a linear function whose domain is the set of positive integers.
Section 11.2 Arithmetic Sequences
1111
If the first term of an arithmetic sequence is a1, each term after the first is obtained by adding d, the common difference, to the previous term. This can be expressed recursively as follows: an = an−1 + d. #FFdVQVJGVGTOKPCP[ RQUKVKQPVQIGVVJGPGZVVGTO
To use this recursion formula, we must be given the first term.
2 Write terms of an arithmetic sequence.
Writing the Terms of an Arithmetic Sequence
EXAMPLE 1
Write the first six terms of the arithmetic sequence in which a1 = 6 and an = an  1  2. Solution The recursion formula a1 = 6 and an = an  1  2 indicates that each term after the first, 6, is obtained by adding 2 to the previous term. a1 a2 a3 a4 a5 a6
= = = = = =
6 a1 a2 a3 a4 a5
This is given.

2 2 2 2 2
= = = = =
6  2 4  2 2  2 0  2 2  2
= = = = =
4 2 0 2 4
Use an = an − 1 − 2 with n = 2. Use an = an − 1 − 2 with n = 3. Use an = an − 1 − 2 with n = 4. Use an = an − 1 − 2 with n = 5. Use an = an − 1 − 2 with n = 6.
The first six terms are 6, 4, 2, 0, 2, and 4. Write the first six terms of the arithmetic sequence in which a1 = 100 and an = an  1  30. CHECK POINT 1
3 Use the formula for the
general term of an arithmetic sequence.
The General Term of an Arithmetic Sequence Consider an arithmetic sequence whose first term is a1 and whose common difference is d. We are looking for a formula for the general term, an. Let’s begin by writing the first six terms. The first term is a1. The second term is a1 + d. The third term is a1 + d + d, or a1 + 2d. Thus, we start with a1 and add d to each successive term. The first six terms are a1, aƂTUV VGTO
a1 + d,
a1 + 2d,
a1 + 3d,
a1 + 4d,
a1 + 5d.
aUGEQPF VGTO
aVJKTF VGTO
aHQWTVJ VGTO
aƂHVJ VGTO
aUKZVJ VGTO
Compare the coefficient of d and the subscript of a denoting the term number. Can you see that the coefficient of d is 1 less than the subscript of a denoting the term number? a3: third term = a1 + 2d
a4: fourth term = a1 + 3d
1PGNGUUVJCPQTKU VJGEQGHƂEKGPVQHd
1PGNGUUVJCPQTKU VJGEQGHƂEKGPVQHd
1112
Chapter 11 Sequences, Induction, and Probability
Thus, the formula for the nth term is an: nth term = a1 + (n − 1)d. 1PGNGUUVJCPnQTn−KU VJGEQGHƂEKGPVQHd
General Term of an Arithmetic Sequence The nth term (the general term) of an arithmetic sequence with first term a1 and common difference d is an = a1 + (n  1)d.
EXAMPLE 2
Using the Formula for the General Term of an Arithmetic Sequence
Find the eighth term of the arithmetic sequence whose first term is 4 and whose common difference is 7. Solution To find the eighth term, a8, we replace n in the formula with 8, a1 with 4, and d with 7. an = a1 + (n  1)d a8 = 4 + (8  1)( 7) = 4 + 7( 7) = 4 + ( 49) = 45 The eighth term is 45. We can check this result by writing the first eight terms of the sequence: 4, 3, 10, 17, 24, 31, 38, 45. CHECK POINT 2 Find the ninth term of the arithmetic sequence whose first term is 6 and whose common difference is 5.
EXAMPLE 3
Using the Formula for the General Term of an Arithmetic Sequence
Write a formula for the general term (the nth term) of the arithmetic sequence: 2, 8, 18, 28, c. Solution an = a1 + (n  1)d
Use the formula for the general term of an arithmetic sequence.
an = 2 + (n  1)d an = 2 + (n  1)10
a1, the first term, is −2. d, the common difference, is found by subtracting the first term, −2, from the second term, 8. d = 8 − (−2) = 8 + 2 = 10
an = 2 + 10n  10 an = 10n  12 CHECK POINT 3
Distribute 10 to each term in parentheses. Simplify. This is the formula for the general term.
Write a formula for the general term (the nth term) of the
arithmetic sequence: 4, 2, 8, 14, c.
Section 11.2 Arithmetic Sequences
1113
Using an Arithmetic Sequence to Model Changes in the U.S. Population
EXAMPLE 4
The graph in Figure 11.4 shows the percentage of the U.S. population by race/ ethnicity for 2010, with projections by the U.S. Census Bureau for 2050. U.S. Population by Race/Ethnicity 2010 Census
2050 Projections
Percentage of U.S. Population
80% 70%
64
60% 46
50% 40%
30 30% 16
20%
12
15
10%
5 White
Latino
African American
9
Asian
Figure 11.4 Source: U.S. Census Bureau
The data show that in 2010, 64% of the U.S. population was white. On average, this is projected to decrease by approximately 0.45% per year. a. Write a formula for the nth term of the arithmetic sequence that describes the percentage of the U.S. population that will be white n years after 2009. b. What percentage of the U.S. population is projected to be white in 2030? Solution a. With a yearly decrease of 0.45%, we can express the percentage of the white population by the following arithmetic sequence: 64,
64 − 0.45 = 63.55,
aRGTEGPVCIGQHYJKVGUKP VJGRQRWNCVKQPKP [GCTCHVGT
63.55 − 0.45 = 63.10, … .
aRGTEGPVCIGQHYJKVGUKP VJGRQRWNCVKQPKP [GCTUCHVGT
aRGTEGPVCIGQHYJKVGUKP VJGRQRWNCVKQPKP [GCTUCHVGT
In the sequence 64, 63.55, 63.10, c, the first term, a1, represents the percentage of the population that was white in 2010. Each subsequent year this amount decreases by 0.45%, so d = 0.45. We use the formula for the general term of an arithmetic sequence to write the nth term of the sequence that describes the percentage of whites in the population n years after 2009. an = a1 + (n  1)d
This is the formula for the general term of an arithmetic sequence.
an = 64 + (n  1)( 0.45)
a1 = 64 and d = − 0.45.
an = 64  0.45n + 0.45
Distribute − 0.45 to each term in parentheses.
an = 0.45n + 64.45
Simplify.
Thus, the percentage of the U.S. population that will be white n years after 2009 can be described by an = 0.45n + 64.45.
1114
Chapter 11 Sequences, Induction, and Probability
b. Now we need to project the percentage of the population that will be white in 2030. The year 2030 is 21 years after 2009. Thus, n = 21. We substitute 21 for n in an = 0.45n + 64.45. a21 = 0.45(21) + 64.45 = 55 The 21st term of the sequence is 55. Thus, 55% of the U.S. population is projected to be white in 2030. CHECK POINT 4 The data in Figure 11.4 on the previous page show that in 2010, 16% of the U.S. population was Latino. On average, this is projected to increase by approximately 0.35% per year. a. Write a formula for the nth term of the arithmetic sequence that describes the percentage of the U.S. population that will be Latino n years after 2009. b. What percentage of the U.S. population is projected to be Latino in 2030?
4 Use the formula for the sum of the first n terms of an arithmetic sequence.
The Sum of the First n Terms of an Arithmetic Sequence The sum of the first n terms of an arithmetic sequence, denoted by Sn, and called the nth partial sum, can be found without having to add up all the terms. Let Sn = a1 + a2 + a3 + g + an be the sum of the first n terms of an arithmetic sequence. Because d is the common difference between terms, Sn can be written forward and backward as follows:
(QTYCTF5VCTVYKVJ VJGƂTUVVGTOa GGRCFFKPId $CEMYCTF5VCTVYKVJ VJGNCUVVGTOan GGRUWDVTCEVKPId
+ (a1 + d) + (a1 + 2d) + … + an + (an − d) + (an − 2d) + … + a1 2Sn = (a1 + an) + (a1 + an) + (a1 + an) + … + (a1 + an). Sn = a1 Sn = an
Add the two equations.
Consider the last equation 2Sn = (a1 + an) + (a1 + an) + (a1 + an) + g + (a1 + an). Because there are n sums of (a1 + an) on the right side, we can express this side as n(a1 + an). Thus, the last equation can be written as follows: 2Sn = n(a1 + an) Sn =
n (a1 + an). 2
Solve for Sn , dividing both sides by 2.
We have proved the following result: The Sum of the First n Terms of an Arithmetic Sequence The sum, Sn, of the first n terms of an arithmetic sequence is given by n Sn = (a1 + an), 2 in which a1 is the first term and an is the nth term. n (a + an), we 2 1 need to know the first term, a1, the last term, an, and the number of terms, n. The following examples illustrate how to use this formula.
To find the sum of the terms of an arithmetic sequence using Sn =
Section 11.2 Arithmetic Sequences
EXAMPLE 5
1115
Finding the Sum of n Terms of an Arithmetic Sequence
Find the sum of the first 100 terms of the arithmetic sequence: 1, 3, 5, 7, c. Solution By finding the sum of the first 100 terms of 1, 3, 5, 7, c, we are finding the sum of the first 100 odd numbers. To find the sum of the first 100 terms, S100, we replace n in the formula with 100. n (a + an) 2 1 100 S100 = (a + a100) 2 1 Sn =
6JGƂTUVVGTO aKU
9GOWUVƂPFa VJGVJVGTO
We use the formula for the general term of a sequence to find a100. The common difference, d, of 1, 3, 5, 7, c, is 2. an = a1 + (n  1)d
This is the formula for the nth term of an arithmetic
a100 = 1 + (100  1) # 2
sequence. Use it to find the 100th term. Substitute 100 for n, 2 for d, and 1 (the first term) for a1.
= 1 + 99 # 2
Perform the subtraction in parentheses.
= 1 + 198 = 199
Multiply (99 # 2 = 198) and then add.
Now we are ready to find the sum of the 100 terms 1, 3, 5, 7, c, 199. Sn = S100 =
n (a + an) 2 1
Use the formula for the sum of the first n terms of an arithmetic sequence. Let n = 100, a1 = 1, and a100 = 199.
100 (1 + 199) = 50(200) = 10,000 2
The sum of the first 100 odd numbers is 10,000. Equivalently, the 100th partial sum of the sequence 1, 3, 5, 7, cis 10,000. CHECK POINT 5
Find the sum of the first 15 terms of the arithmetic sequence:
3, 6, 9, 12, c.
TECHNOLOGY
EXAMPLE 6
Using Sn to Evaluate a Summation 25
To find
Find the following sum: a (5i  9)
i=1
on a graphing utility, enter
SUM SEQ (5x  9, x, 1, 25, 1). Then press ENTER .
a (5i  9).
i=1
25
Solution
# # # # a (5i  9) = (5 1  9) + (5 2  9) + (5 3  9) + g + (5 25  9) 25
i=1
= 4
+ 1
+ 6
+ g + 116
By evaluating the first three terms and the last term, we see that a1 = 4; d, the common difference, is 1  ( 4), or 5; and a25, the last term, is 116. Sn = S25 =
n (a + an) 2 1
Use the formula for the sum of the first n terms of an arithmetic sequence. Let n = 25, a1 = − 4, and a25 = 116.
25 25 ( 4 + 116) = (112) = 1400 2 2
1116
Chapter 11 Sequences, Induction, and Probability
Thus, 25
a (5i  9) = 1400.
i=1
30
CHECK POINT 6
Find the following sum:
a (6i  11).
i=1
EXAMPLE 7
Modeling Total Residential Community Costs over a SixYear Period
Your grandmother has assets of $500,000. One option that she is considering involves an adult residential community for a sixyear period beginning in 2022. The model an = 1800n + 64,130 describes yearly adult residential community costs n years after 2021. Does your grandmother have enough to pay for the facility? Solution We must find the sum of an arithmetic sequence whose general term is an = 1800n + 64,130. The first term of the sequence corresponds to the facility’s costs in the year 2022. The last term corresponds to costs in the year 2027. Because the model describes costs n years after 2021, n = 1 describes the year 2022 and n = 6 describes the year 2027. an = 1800n + 64,130
This is the given formula for the
a1 = 1800 # 1 + 64,130 = 65,930 a6 = 1800 # 6 + 64,130 = 74,930
general term of the sequence. Find a1 by replacing n with 1. Find a6 by replacing n with 6.
The first year the facility will cost $65,930. By year 6, the facility will cost $74,930. Now we must find the sum of the costs for all six years. We focus on the sum of the first six terms of the arithmetic sequence 65,930, 67,730, … , 74,930. a
a
a
We find this sum using the formula for the sum of the first n terms of an arithmetic sequence. We are adding six terms: n = 6. The first term is 65,930: a1 = 65,930. The last term—that is, the sixth term—is 74,930: a6 = 74,930. n (a + an) 2 1 6 S6 = (65,930 + 74,930) = 3(140,860) = 422,580 2
Sn =
Instructor Resources for Section 11.2 in MyLab Math
Total adult residential community costs for your grandmother are predicted to be $422,580. Because your grandmother’s assets are $500,000, she has enough to pay for the facility for the sixyear period.
CHECK POINT 7 In Example 7, how much would it cost for the adult residential community for a tenyear period beginning in 2022?
Section 11.2 Arithmetic Sequences
1117
CONCEPT AND VOCABULARY CHECK Fill in each blank so that the resulting statement is true. C1. A sequence in which each term after the first differs from the preceding term by a constant amount is called a/an sequence. The difference between consecutive terms is called the of the sequence.
C3. The sum, Sn, of the first n terms of the sequence described in Exercise C1 is given by the formula
C2. The nth term of the sequence described in Exercise C1 is given by the formula an = , where a1 is the and d is the of the sequence.
C4. The first term of a (6i  4) is i=1 is .
, where a1 is the .
Sn = is the
and an
20
and the last term
17
C5. The first three terms of a (5i + 3) are
,
,
i=1
and
. The common difference is
.
11.2 EXERCISE SET Practice Exercises
29. a1 = 20, d =  4
In Exercises 1–14, write the first six terms of each arithmetic sequence.
30. a1 = 70, d =  5 31. an = an  1 + 3, a1 = 4
2. a1 = 300, d = 50
32. an = an  1 + 5, a1 = 6
3. a1 =  7, d = 4
4. a1 = 8, d = 5
33. an = an  1  10, a1 = 30
5. a1 = 300, d = 90
6. a1 = 200, d =  60
34. an = an  1  10, a1 = 33
1. a1 = 200, d = 20
7. a1 =
5 2,
d = 
1 2
8. a1 =
3 4,
d = 
1 4
9. an = an  1 + 6, a1 = 9
10. an = an  1 + 4, a1 =  7
11. an = an  1  10, a1 = 30
12. an = an  1  20, a1 = 50
13. an = an  1  0.4, a1 = 1.6
35. Find the sum of the first 20 terms of the arithmetic sequence: 4, 10, 16, 22, . . . . 36. Find the sum of the first 25 terms of the arithmetic sequence: 7, 19, 31, 43, . . . .
14. an = an  1  0.3, a1 = 1.7
37. Find the sum of the first 50 terms of the arithmetic sequence: 10,  6,  2, 2, . . . .
In Exercises 15–22, find the indicated term of the arithmetic sequence with first term, a1, and common difference, d.
38. Find the sum of the first 50 terms of the arithmetic sequence: 15,  9,  3, 3, . . . .
15. Find a6 when a1 = 13, d = 4. 16. Find a16 when a1 = 9, d = 2.
39. Find 1 + 2 + 3 + 4 + g + 100, the sum of the first 100 natural numbers.
17. Find a50 when a1 = 7, d = 5.
40. Find 2 + 4 + 6 + 8 + g + 200, the sum of the first 100 positive even integers.
18. Find a60 when a1 = 8, d = 6.
41. Find the sum of the first 60 positive even integers.
19. Find a200 when a1 = 40, d = 5. 20. Find a150 when a1 = 60, d = 5. 21. Find a60 when a1 = 35, d =  3. 22. Find a70 when a1 = 32, d = 4. In Exercises 23–34, write a formula for the general term (the nth term) of each arithmetic sequence. Do not use a recursion formula. Then use the formula for an to find a20, the 20th term of the sequence. 23. 1, 5, 9, 13, c 24. 2, 7, 12, 17, c 25. 7, 3, 1, 5, c 26. 5, 2, 1, 4, c 27. a1 = 9, d = 2 28. a1 = 6, d = 3
42. Find the sum of the first 80 positive even integers. 43. Find the sum of the even integers between 21 and 45. 44. Find the sum of the odd integers between 30 and 54. For Exercises 45–50, write out the first three terms and the last term. Then use the formula for the sum of the first n terms of an arithmetic sequence to find the indicated sum. 17
45. a (5i + 3) i=1 30
47. a (  3i + 5) i=1 100
49. a 4i i=1
20
46. a (6i  4) i=1 40
48. a ( 2i + 6) i=1 50
50. a ( 4i) i=1
Chapter 11 Sequences, Induction, and Probability
n
51. Find a14 + b12. 52. Find a16 + b18. 53. If {an} is a finite sequence whose last term is 83, how many terms does {an} contain? 54. If {bn} is a finite sequence whose last term is 93, how many terms does {bn} contain? 55. Find the difference between the sum of the first 14 terms of {bn} and the sum of the first 14 terms of {an}. 56. Find the difference between the sum of the first 15 terms of {bn} and the sum of the first 15 terms of {an}. 57. Write a linear function f(x) = mx + b, whose domain is the set of positive integers, that represents {an}. 58. Write a linear function g(x) = mx + b, whose domain is the set of positive integers, that represents {bn}. Use a system of two equations in two variables, a1 and d, to solve Exercises 59–60. 59. Write a formula for the general term (the nth term) of the arithmetic sequence whose second term, a2, is 4 and whose sixth term, a6, is 16. 60. Write a formula for the general term (the nth term) of the arithmetic sequence whose third term, a3, is 7 and whose eighth term, a8, is 17.
Application Exercises The bar graph shows two probable majors of U.S. college freshmen in 1980 and in 2017. Exercises 61–62 involve developing arithmetic sequences that model the data. Probable Majors of U.S. College Freshmen Business
Biology
Percentage of College Freshmen
24%
21.2
20% 15.5
16%
13.8
12% 8% 4%
The bar graph shows the average dormitory charges at public and private fouryear U.S. colleges. Use the information displayed by the graph to solve Exercises 65–66. Average Dormitory Charges at FourYear U.S. Colleges $8000
Public Institutions Private Institutions
$7000 $6000 $5000 $4000 $3000 $2000
6717
1 2 3 4 5
6017
–2 –4 –6
6464
n
1 2 3 4 5
5850
–2 –4 –6
6228
14 12 10 8 6 4 2
14 12 10 8 6 4 2
5677
bn
an
6026
Use the graphs of the arithmetic sequences {an} and {bn} to solve Exercises 51–58.
61. In 1980, 4.5% of college freshmen declared biology as their probable major. On average, this percentage has increased by approximately 0.3 each year. a. Write a formula for the nth term of the arithmetic sequence that models the percentage of college freshmen who declared biology as their probable major n years after 1979. b. Use the formula in part (a) to project the percentage of college freshmen who will declare biology as their probable major in 2023. 62. In 1980, 21.2% of college freshmen declared business as their probable major. On average, this percentage has decreased by approximately 0.2 each year. a. Write a formula for the nth term of the arithmetic sequence that models the percentage of college freshmen who declared business as their probable major n years after 1979. b. Use the formula in part (a) to project the percentage of college freshmen who will declare business as their probable major in 2023. 63. Company A pays $44,000 yearly with raises of $1600 per year. Company B pays $48,000 yearly with raises of $1000 per year. Which company will pay more in year 10? How much more? 64. Company A pays $53,000 yearly with raises of $1200 per year. Company B pays $56,000 yearly with raises of $800 per year. Which company will pay more in year 10? How much more?
5479
Practice PLUS
Dormitory Charges
1118
$1000 2014
2015 2016 2017 Ending Year in the School Year
Source: U.S. Department of Education
65. a. Use the numbers shown by the bar graph to find the total dormitory charges at public colleges for a fouryear period from the school year ending in 2014 through the school year ending in 2017. b. The model an = 179n + 5309
4.5 1980
2017 Year
Source: The Higher Education Research Institute
describes dormitory charges at public colleges in academic year n, where n = 1 corresponds to the school year ending in 2014, n = 2 to the school year ending in 2015, and so on. Use this model and the formula for Sn to find
Section 11.2 Arithmetic Sequences the total dormitory charges at public colleges for a fouryear period from the school year ending in 2014 through the school year ending in 2017. Does this underestimate or overestimate the actual sum that you obtained in part (a)? By how much? 66. a. Use the numbers shown by the bar graph to find the total dormitory charges at private colleges for a fouryear period from the school year ending in 2014 through the school year ending in 2017. b. The model an = 231n + 5782
67.
68.
69.
70.
71.
describes the dormitory charges at private colleges in academic year n, where n = 1 corresponds to the school year ending in 2014, n = 2 to the school year ending in 2015, and so on. Use this model and the formula for Sn to find the total dormitory charges at private colleges for a fouryear period from the school year ending in 2014 through the school year ending in 2017. How does this compare with the actual sum you obtained in part (a)? Use one of the models in Exercises 65–66 and the formula for Sn to find the total dormitory charges for your undergraduate education. How well does the model describe your anticipated costs? A company offers a starting yearly salary of $53,000 with raises of $2500 per year. Find the total salary over a tenyear period. You are considering two job offers. Company A will start you at $49,000 a year and guarantee a raise of $2600 per year. Company B will start you at a higher salary, $57,000 a year, but will only guarantee a raise of $1200 per year. Find the total salary that each company will pay over a tenyear period. Which company pays the greater total amount? A theater has 30 seats in the first row, 32 seats in the second row, increasing by 2 seats per row for a total of 26 rows. How many seats are there in the theater? A section in a stadium has 20 seats in the first row, 23 seats in the second row, increasing by 3 seats each row for a total of 38 rows. How many seats are in this section of the stadium?
Explaining the Concepts 72. What is an arithmetic sequence? Give an example with your explanation. 73. What is the common difference in an arithmetic sequence? 74. Explain how to find the general term of an arithmetic sequence. 75. Explain how to find the sum of the first n terms of an arithmetic sequence without having to add up all the terms.
1119
Critical Thinking Exercises Make Sense? In Exercises 78–81, determine whether each statement makes sense or does not make sense, and explain your reasoning. 78. Rather than performing the addition, I used the formula Sn = n2 (a1 + an) to find the sum of the first 30 terms of the sequence 2, 4, 8, 16, 32, . . . . 79. I was able to find the sum of the first 50 terms of an arithmetic sequence even though I did not identify every term. 80. The sequence for the number of seats per row in our movie theater as the rows move toward the back is arithmetic with d = 1 so that people don’t block the view of those in the row behind them. 81. Beginning at 6:45 a.m., a bus stops on my block every 23 minutes, so I used the formula for the nth term of an arithmetic sequence to describe the stopping time for the nth bus of the day. 82. In the sequence 21,700, 23,172, 24,644, 26,116, . . . , which term is 314,628? 83. A degreeday is a unit used to measure the fuel requirements of buildings. By definition, each degree that the average daily temperature is below 65°F is 1 degreeday. For example, an average daily temperature of 42°F constitutes 23 degreedays. If the average temperature on January 1 was 42°F and fell 2°F for each subsequent day up to and including January 10, how many degreedays are included from January 1 to January 10? 84. Show that the sum of the first n positive odd integers, 1 + 3 + 5 + g + (2n  1), is n2.
Retaining the Concepts 85. Write an equation in pointslope form and slopeintercept form for the line passing through ( 2, 6) and perpendicular to the line whose equation is x  3y + 9 = 0. (Section 2.4, Example 2) 86. Among all pairs of numbers whose sum is 24, find a pair whose product is as large as possible. What is the maximum product? (Section 3.1, Example 6) 87. Solve: log2(x + 9)  log2 x = 1. (Section 4.4, Example 7) x 88. Graph y = 3 tan for p 6 x 6 3p. 2 (Section 5.6, Example 1)
Preview Exercises
Technology Exercises
Exercises 89–91 will help you prepare for the material covered in the next section. a2 a3 a4 89. Consider the sequence 1, 2, 4,  8, 16, . . . . Find , , , a1 a2 a3 a5 and . What do you observe? a4
76. Use the SEQ (sequence) capability of a graphing utility and the formula you obtained for an to verify the value you found for a20 in any five exercises from Exercises 23–34. 77. Use the capability of a graphing utility to calculate the sum of a sequence to verify any five of your answers to Exercises 45–50.
91. Use the formula an = a13n  1 to find the seventh term of the sequence 11, 33, 99, 297, . . . .
90. Consider the sequence whose nth term is an = 3 # 5n. Find a5 a2 a3 a4 , , , and . What do you observe? a1 a2 a3 a4
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Chapter 11 Sequences, Induction, and Probability
SECTION 11.3
Geometric Sequences and Series
WHAT YOU’LL LEARN 1 Find the common ratio of a geometric sequence.
2 Write terms of a geometric sequence.
3 Use the formula for the
general term of a geometric sequence.
4 Use the formula for the
sum of the first n terms of a geometric sequence.
5 Find the value of an annuity.
6 Use the formula for the
sum of an infinite geometric series.
Here we are at the closing moments of an online job interview. You managed to answer all the tough questions without losing your poise, and now you’ve been offered a job. As a matter of fact, your qualifications are so terrific that you’ve been offered two jobs—one just the day before, with a rival company in the same field! One company offers $60,000 the first year, with increases of 6% per year for four years after that. The other offers $62,000 the first year, with annual increases of 3% per year after that. Over a fiveyear period, which is the better offer? If salary raises amount to a certain percent each year, the yearly salaries over time form a geometric sequence. In this section, we investigate geometric sequences and their properties. After studying the section, you will be in a position to decide which job offer to accept: You will know which company will pay you more over five years.
Geometric Sequences Figure 11.5 shows a sequence in which the number of squares is increasing. From left to right, the number of squares is 1, 5, 25, 125, and 625. In this sequence, each term after the first, 1, is obtained by multiplying the preceding term by a constant amount, namely, 5. This sequence of increasing numbers of squares is an example of a geometric sequence.
Figure 11.5 A geometric sequence of squares
1 Find the common ratio of a geometric sequence.
Definition of a Geometric Sequence A geometric sequence is a sequence in which each term after the first is obtained by multiplying the preceding term by a fixed nonzero constant. The amount by which we multiply each time is called the common ratio of the sequence. The common ratio, r, is found by dividing any term after the first term by the term that directly precedes it. In the following examples, the common ratio is found by a2 dividing the second term by the first term, . a1
GREAT QUESTION What happens to the terms of a geometric sequence when the common ratio is negative? When the common ratio is negative, the signs of the terms alternate.
Geometric sequence
Common ratio
1, 5, 25, 125, 625, . . .
r =
4, 8, 16, 32, 64, . . . 6, 12, 24, 48, 96, . . . 1 1 9, 3, 1,  , , . . . 3 9
5 = 5 1 8 r = = 2 4 12 r = = 2 6 3 1 r = = 9 3
Section 11.3 Geometric Sequences and Series an 125 100 75 50 25 1 2 3 4 5
n
Figure 11.6 The graph of {an} = 1, 5, 25, 125, c
2 Write terms of a geometric sequence.
1121
Figure 11.6 shows a partial graph of the first geometric sequence in our list. The graph forms a set of discrete points lying on the exponential function f(x) = 5x  1. This illustrates that a geometric sequence with a positive common ratio other than 1 is an exponential function whose domain is the set of positive integers. How do we write out the terms of a geometric sequence when the first term and the common ratio are known? We multiply the first term by the common ratio to get the second term, multiply the second term by the common ratio to get the third term, and so on.
EXAMPLE 1
Writing the Terms of a Geometric Sequence
Write the first six terms of the geometric sequence with first term 6 and common ratio 13. Solution The first term is 6. The second term is 6 # 31, or 2. The third term is 2 # 13, or 23. The fourth term is 23 # 31, or 29, and so on. The first six terms are 2 2 2 2 6, 2, , , , and . 3 9 27 81 Write the first six terms of the geometric sequence with first term 12 and common ratio 12. CHECK POINT 1
3 Use the formula for the
general term of a geometric sequence.
The General Term of a Geometric Sequence Consider a geometric sequence whose first term is a1 and whose common ratio is r. We are looking for a formula for the general term, an. Let’s begin by writing the first six terms. The first term is a1. The second term is a1r. The third term is a1r # r, or a1r 2. The fourth term is a1r 2 # r, or a1r 3, and so on. Starting with a1 and multiplying each successive term by r, the first six terms are a 1,
a1r,
aƂTUV VGTO
aUGEQPF VGTO
a 1r 2, aVJKTF VGTO
a 1r 3, aHQWTVJ VGTO
a 1r 4, aƂHVJ VGTO
a 1r 5. aUKZVJ VGTO
Can you see that the exponent on r is 1 less than the subscript of a denoting the term number? a3: third term = a1r2
a4: fourth term = a1r3
1PGNGUUVJCPQTKU VJGGZRQPGPVQPr
1PGNGUUVJCPQTKU VJGGZRQPGPVQPr
Thus, the formula for the nth term is GREAT QUESTION When using a1r n  1 to find the nth term of a geometric sequence, what should I do first? Be careful with the order of operations when evaluating a1r n  1. First subtract 1 in the exponent and then raise r to that power. Finally, multiply the result by a1.
an = a1r n−1. 1PGNGUUVJCPnQTn− KUVJGGZRQPGPVQPr
General Term of a Geometric Sequence The nth term (the general term) of a geometric sequence with first term a1 and common ratio r is an = a1r n  1.
Chapter 11 Sequences, Induction, and Probability
1122
Using the Formula for the General Term of a Geometric Sequence
EXAMPLE 2
BLITZER BONUS Geometric Population Growth
Find the eighth term of the geometric sequence whose first term is 4 and whose common ratio is 2. Solution To find the eighth term, a8, we replace n in the formula with 8, a1 with 4, and r with 2. an = a1r n  1 a8 = 4( 2)8  1 = 4( 2)7 = 4( 128) = 512 The eighth term is 512. We can check this result by writing the first eight terms of the sequence: 4, 8, 16, 32, 64, 128, 256, 512. Find the seventh term of the geometric sequence whose first term is 5 and whose common ratio is 3. CHECK POINT 2
Economist Thomas Malthus (1766–1834) predicted that population would increase as a geometric sequence and food production would increase as an arithmetic sequence. He concluded that eventually population would exceed food production. If two sequences, one geometric and one arithmetic, are increasing, the geometric sequence will eventually overtake the arithmetic sequence, regardless of any head start that the arithmetic sequence might initially have.
In Chapter 4, we studied exponential functions of the form f(x) = bx and used an exponential function to model the growth of the U.S. population from 1970 through 2020 (Example 1 on page 550). From 2010 through 2020, the nation’s population grew at the slowest rate since the 1930s. Consequently, in our next example, we consider the country’s population growth from 2010 through 2020. Because a geometric sequence is an exponential function whose domain is the set of positive integers, geometric growth and exponential growth mean the same thing.
Geometric Population Growth
EXAMPLE 3
The accompanying table shows the population of the United States in 2010, with estimates for 2011 through 2020.
Year
2010
2011
2012
2013
2014
2015
2016
2017
2018
2019
2020
Population (millions)
308.7
310.9
313.0
315.2
317.4
319.7
321.9
324.2
326.4
328.7
331.0
GREAT QUESTION Why are estimates given for 2011 through 2020, but an actual value for 2010? The United States conducts a census every ten years. The first census, in 1790, counted almost four million people. The 2010 census counted approximately 309 million people. At the time of this printing, the data for the 2020 census were not yet available. Therefore, we have used estimated populations for the years after 2010.
a. Show that the population is increasing geometrically. b. Write the general term for the geometric sequence modeling the population of the United States, in millions, n years after 2009. c. Project the U.S. population, in millions, for the year 2030. Solution a. First, we use the sequence of population growth, 308.7, 310.9, 313.0, 315.2, and so on, to divide the population for each year by the population in the preceding year. 310.9 ≈ 1.007, 308.7
313.0 ≈ 1.007, 310.9
315.2 ≈ 1.007 313.0
Continuing in this manner, we will keep getting approximately 1.007. This means that the population is increasing geometrically with r ≈ 1.007. The population of the United States in any year shown in the sequence is approximately 1.007 times the population the year before.
Section 11.3 Geometric Sequences and Series
1123
b. The sequence of the U.S. population growth is 308.7, 310.9, 313.0, 315.2, 317.4, 319.7, . . . . Because the population is increasing geometrically, we can find the general term of this sequence using an = a1r n  1. In this sequence, a1 = 308.7 and [from part (a)] r ≈ 1.007. We substitute these values into the formula for the general term. This gives the general term for the geometric sequence modeling the U.S. population, in millions, n years after 2009. an = 308.7(1.007)n  1 c. We can use the formula for the general term, an, in part (b) to project the U.S. population for the year 2030. The year 2030 is 21 years after 2009—that is, 2030  2009 = 21. Thus, n = 21. We substitute 21 for n in an = 308.7(1.007)n 1. a21 = 308.7(1.007)21  1 = 308.7(1.007)20 ≈ 354.9 The model projects that the United States will have a population of approximately 354.9 million in the year 2030. Write the general term for the geometric sequence
CHECK POINT 3
3, 6, 12, 24, 48, . . . . Then use the formula for the general term to find the eighth term.
BLITZER BONUS
Ponzi Schemes and Geometric Sequences
A Ponzi scheme is an investment fraud that pays returns to existing investors from funds contributed by new investors rather than from legitimate investment activity. Here’s a simplified example: a1 Round 1
6JGUEJGOGT VCMGU GCEJHTQOVJG ƂTUV KPXGUVQTU
,
a2 Round 2
6JGUEJGOGTVCMGU GCEJHTQO PGYKPXGUVQTU RQEMGVKPI CPFRC[KPIVJG ƂTUVKPXGUVQTU GCEJ
,
a3 Round 3
6JGUEJGOGTVCMGU GCEJHTQO PGYKPXGUVQTU RQEMGVKPI CPFRC[KPIVJG RTGXKQWUKPXGUVQTU GCEJ
,
a4 Round 4
, …
6JGUEJGOGTVCMGU GCEJHTQO PGYKPXGUVQTU RQEMGVKPI CPFRC[KPIVJG RTGXKQWUKPXGUVQTU GCEJ
The number of investors needed to continue this Ponzi scheme, 2, 4, 8, 16, . . . , and the money collected in each round, $200, $400, $800, $1600, . . . , form rapidly growing geometric sequences. With no legitimate earnings, the scheme requires a consistent geometric flow of money from new investors to continue. Ponzi schemes tend to collapse when it becomes difficult to recruit new investors or when a large number of investors ask to cash out.
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Chapter 11 Sequences, Induction, and Probability
4 Use the formula for the
sum of the first n terms of a geometric sequence.
The Sum of the First n Terms of a Geometric Sequence The sum of the first n terms of a geometric sequence, denoted by Sn and called the nth partial sum, can be found without having to add up all the terms. Recall that the first n terms of a geometric sequence are a1, a1r, a1r 2, . . . , a1r n  2, a1r n  1. We find the sum of a1, a1r, a1r 2, . . . , a1r n  2, a1r n  1 as follows: Sn = a1 + a1r + a1r 2 + g + a1r n  2 + a1r n  1
Sn is the sum of the first n terms of the sequence.
rSn = a1r + a1r 2 + a1r 3 + g + a1r n  1 + a1r n
Multiply both sides of the equation by r.
Sn  rSn = a1  a1r n
Subtract the second equation from the first equation.
Sn(1  r) = a1(1  r n)
Factor out Sn on the left and
GREAT QUESTION What is the sum of the first n terms of a geometric sequence if the common ratio is 1? If the common ratio is 1, the geometric sequence is a1, a1, a1, a1, . . . . The sum of the first n terms of this sequence is na1 :
a1 on the right.
a1(1  r n) Sn = . 1  r
Solve for Sn by dividing both sides by 1 − r (assuming that r 3 1).
We have proved the following result: The Sum of the First n Terms of a Geometric Sequence The sum, Sn, of the first n terms of a geometric sequence is given by
S n = a 1 + a 1 + a 1 + … + a 1. There are n terms = na1.
Sn =
a1(1  r n) , 1  r
in which a1 is the first term and r is the common ratio (r ≠ 1). a1(1  r n) , we 1  r need to know the first term, a1, the common ratio, r, and the number of terms, n. The following examples illustrate how to use this formula. To find the sum of the terms of a geometric sequence using Sn =
EXAMPLE 4
Finding the Sum of the First n Terms of a Geometric Sequence
Find the sum of the first 18 terms of the geometric sequence: 2, 8, 32, 128, . . . . Solution with 18.
To find the sum of the first 18 terms, S18, we replace n in the formula
Sn = S18 =
a1(1 − r n) 1−r a1(1 − r18) 1−r
6JGƂTUVVGTO aKU
9GOWUVƂPFr VJGEQOOQPTCVKQ
We can find the common ratio by dividing the second term of 2, 8, 32, 128, . . . by the first term. r =
a2 8 = = 4 a1 2
Section 11.3 Geometric Sequences and Series
1125
Now we are ready to find the sum of the first 18 terms of 2, 8, 32, 128, . . . . Sn =
S18 =
a1(1  r n) 1  r
Use the formula for the sum of the first n terms of a geometric sequence.
2[1  ( 4)18] 1  ( 4)
a1 (the first term) = 2, r = 4, and n = 18 because we want the sum of the first 18 terms.
= 27,487,790,694
Use a calculator.
The sum of the first 18 terms is 27,487,790,694. Equivalently, this number is the 18th partial sum of the sequence 2, 8, 32, 128, . . . . CHECK POINT 4
Find the sum of the first nine terms of the geometric sequence:
2, 6, 18, 54, . . . .
Using Sn to Evaluate a Summation
EXAMPLE 5
# i a 6 2. 10
Find the following sum:
i=1
Solution
Let’s write out a few terms in the sum.
# i # # 2 # 3 # 10 a 6 2 = 6 2 + 6 2 + 6 2 + g+ 6 2 10
i=1
TECHNOLOGY To find
# i a6 2 10
Do you see that each term after the first is obtained by multiplying the preceding term by 2? To find the sum of the 10 terms (n = 10), we need to know the first term, a1, and the common ratio, r. The first term is 6 # 2 or 12: a1 = 12. The common ratio is 2.
i=1
on a graphing utility, enter
SUM SEQ (6
Sn =
a1(1  r n) 1  r
Use the formula for the sum of the first n terms of a geometric sequence.
S10 =
12(1  210) 1  2
a1 (the first term) = 12, r = 2, and n = 10 because we are adding 10 terms.
* 2x, x, 1, 10, 1).
Then press ENTER.
= 12,276
Use a calculator.
Thus,
# i a 6 2 = 12,276. 10
i=1
# i a 2 3. 8
CHECK POINT 5
Find the following sum:
i=1
Some of the exercises in the previous Exercise Set involved situations in which salaries increased by a fixed amount each year. A more realistic situation is one in which salary raises increase by a certain percent each year. Example 6 shows how such a situation can be modeled using a geometric sequence.
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Chapter 11 Sequences, Induction, and Probability
EXAMPLE 6
Computing a Lifetime Salary
A union contract specifies that each worker will receive a 5% pay increase each year for the next 30 years. One worker is paid $55,000 the first year. What is this person’s total lifetime salary over a 30year period? Solution The salary for the first year is $55,000. With a 5% raise, the secondyear salary is computed as follows: Salary for year 2 = 55,000 + 55,000(0.05) = 55,000(1 + 0.05) = 55,000(1.05). Each year, the salary is 1.05 times what it was in the previous year. Thus, the salary for year 3 is 1.05 times 55,000(1.05), or 55,000(1.05)2. The salaries for the first five years are given in the table. Yearly Salaries Year 1 55,000
Year 2
Year 3
55,000(1.05)
Year 4 2
55,000(1.05)
Year 5 3
55,000(1.05)
N 4
55,000(1.05)
c
The numbers in the bottom row form a geometric sequence with a1 = 55,000 and r = 1.05. To find the total salary over 30 years, we use the formula for the sum of the first n terms of a geometric sequence, with n = 30. Sn =
S30 =
a1(1  r n) 1  r 55,000[1 − (1.05)30] 1 − 1.05
6QVCNUCNCT[ QXGT[GCTU
Use Sn =
a1(1 − rn)
with 1 − r a1 = 55,000, r = 1.05, and n = 30.
≈ 3,654,137
Use a calculator.
The total salary over the 30year period is approximately $3,654,137 . CHECK POINT 6 A job pays a salary of $65,000 the first year. During the next 29 years, the salary increases by 6% each year. What is the total lifetime salary over the 30year period?
5 Find the value of an annuity.
Annuities The compound interest formula A = P(1 + r)t gives the future value, A, after t years, when a fixed amount of money, P, the principal, is deposited in an account that pays an annual interest rate r (in decimal form) compounded once a year. However, money is often invested in small amounts at periodic intervals. For example, to save for retirement, you might decide to place $1000 into an Individual Retirement Account (IRA) at the end of each year until you retire. An annuity is a sequence of equal payments made at equal time periods. An IRA is an example of an annuity. Suppose P dollars are deposited into an account at the end of each year. The account pays an annual interest rate, r, compounded annually. At the end of the first year, the account contains P dollars. At the end of the second year, P dollars is deposited again. At the time of this deposit, the first deposit has received interest
Section 11.3 Geometric Sequences and Series
1127
earned during the second year. The value of the annuity is the sum of all deposits made plus all interest paid. Thus, the value of the annuity after two years is P + P(1 + r). &GRQUKVQHP FQNNCTUCVGPFQH UGEQPF[GCT
(KTUV[GCTFGRQUKV QHPFQNNCTUYKVJ KPVGTGUVGCTPGFHQT C[GCT
The value of the annuity after three years is P
+
&GRQUKVQHP FQNNCTUCVGPFQH VJKTF[GCT
P(1 + r)
+
5GEQPF[GCTFGRQUKV QHPFQNNCTUYKVJ KPVGTGUVGCTPGFHQT C[GCT
P(1 + r)2. (KTUV[GCTFGRQUKV QHPFQNNCTUYKVJ KPVGTGUVGCTPGF QXGTVYQ[GCTU
The value of the annuity after t years is P + P(1 + r) + P(1 + r)2 + P(1 + r)3 + … + P(1 + r)t−1. &GRQUKVQHP FQNNCTUCVGPFQH [GCTt
(KTUV[GCTFGRQUKV QHPFQNNCTUYKVJ KPVGTGUVGCTPGF QXGTt−[GCTU
This is the sum of the terms of a geometric sequence with first term P and common ratio 1 + r. We use the formula Sn =
a1(1  r n) 1  r
to find the sum of the terms: St =
P[1  (1 + r)t ] P[1  (1 + r)t ] P[(1 + r)t  1] = = . r r 1  (1 + r)
This formula gives the value of an annuity after t years if interest is compounded once a year. We can adjust the formula to find the value of an annuity if equal payments are made at the end of each of n yearly compounding periods. Value of an Annuity: Interest Compounded n Times per Year If P is the deposit made at the end of each compounding period for an annuity at r percent annual interest compounded n times per year, the value, A, of the annuity after t years is PJa1 + A =
EXAMPLE 7
r nt b  1R n r n
.
Determining the Value of an Annuity
At age 25, to save for retirement, you decide to deposit $200 at the end of each month into an IRA with a historical return equivalent to 7.5% compounded monthly. a. Using the historical rate of return, how much will you have from the IRA when you retire at age 65? b. Find the amount earned on your investment.
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Chapter 11 Sequences, Induction, and Probability
BLITZER BONUS
Solution a. Because you are 25, the amount that you will have from the IRA when you retire at 65 is its value after 40 years.
Stashing Cash and Making Taxes Less Taxing As you prepare for your future career, retirement probably seems very far away. Making regular deposits into an IRA may not be fun, but there is a special incentive from Uncle Sam that makes it far more appealing. Traditional IRAs are taxdeferred savings plans. This means that you do not pay taxes on deposits and interest until you begin withdrawals, typically at retirement. Before then, yearly deposits count as adjustments to gross income and are not part of your taxable income. Not only do you get a tax break now, but also you ultimately earn more. This is because you do not pay taxes on interest from year to year, allowing earnings to accumulate until you start withdrawals. With a tax code that encourages longterm savings, opening an IRA early in your career is a smart way to gain more control over how you will spend a large part of your life.
PJa1 + A =
r nt b  1R n r n
# 0.075 12 40 b  1R 12 0.075 12 200[(1 + 0.00625)480  1] 0.00625 200[(1.00625)480  1] 0.00625 200(19.8989  1) 0.00625 604,765
200Ja1 + A =
= = ≈ ≈
Use the formula for the value of an annuity.
The annuity involves monthend deposits of $200: P = 200. The interest rate is 7.5%: r = 0.075. The interest is compounded monthly: n = 12. The number of years is 40: t = 40. Using parentheses keys, this can be performed in a single step on a graphing calculator.
Use a calculator to find (1.00625)480: 1.00625 yx 480 5 .
After 40 years, you will have approximately $604,765 when retiring at age 65. b. Earnings = Value of the IRA − Total deposits ≈ $604,765 − $200 ∙ 12 ∙ 40 RGTOQPVJ×OQPVJU RGT[GCT×[GCTU
= $604,765 − $96,000 = $508,765 You earned approximately $508,765, more than five times the amount of your contributions to the IRA. CHECK POINT 7 At age 30, to save for retirement, you decide to deposit $100 at the end of each month into an IRA with a historical return equivalent to 9.5% compounded monthly. a. Assuming the historical rate of return, how much will you have from the IRA when you retire at age 65? b. Find the amount earned on your investment.
6 Use the formula for the
sum of an infinite geometric series.
Geometric Series An infinite sum of the form a1 + a1r + a1r 2 + a1r 3 + g + a1r n  1 + g with first term a1 and common ratio r is called an infinite geometric series. How can we determine which infinite geometric series have sums and which do not? We look at what happens to r n as n gets larger in the formula for the sum of the first n terms of this series, namely, Sn =
a1(1  r n) . 1  r
If r is any number between 1 and 1, that is, 1 6 r 6 1, the term r n approaches 0 as n gets larger. For example, consider what happens to r n for r = 12 : 1 1 1 a b = 2 2
1 2 1 a b = 2 4
1 3 1 a b = 2 8
1 4 1 a b = 2 16
1 5 1 a b = 2 32
6JGUGPWODGTUCTGCRRTQCEJKPICUnIGVUNCTIGT
1 6 1 . a b = 2 64
Section 11.3 Geometric Sequences and Series
1129
Take another look at the formula for the sum of the first n terms of a geometric sequence. Sn =
a1(1 − r n) 1−r
+H–