polynomial and rational functions · polynomial and rational functions and their graphs. •...

y

x

175

If you ride the Giant Dipper roller coaster, you’ll climb to high points and descend to low points along a continuous track. If the track does not loop or overlap, then its shape can be modeled by a function. In this chapter you will study polynomial functions, and learn how the degree and the coe� cients of polynomial functions determine the shape of their graphs. � e graphs of polynomial functions o� en have hills and valleys, and sometimes resemble the track of a roller coaster.

Polynomial and Rational FunctionsPolynomial and Rational Functions

444444

175

CHAP TE R O B J EC TIV ES

• Discover some properties of polynomial and rational functions and their graphs.

• Review the properties, graphs, and techniques associated with quadratic functions.

• Given a polynomial function, determine from the graph what degree it might be, and vice versa.

• Given a polynomial function, fi nd the zeros from the equation or the graph.

• Given a set of points, fi nd the equation for the polynomial function that fi ts the data exactly or fi ts best for a given degree.

• Discover some properties of polynomial and rational functions and their graphs.

• Multiply, divide, add, and subtract rational expressions, and resolve proper algebraic fractions into the sum of two or more partial fractions.

• Given a rational algebraic function f, fi nd x for a given value of f (x).

175A Chapter 4 Interleaf: Polynomial and Rational Functions

OverviewIn this chapter students signifi cantly expand what they have learned about polynomial and rational functions. Factoring, completing the square, and the quadratic formula are reviewed. An examination of all zeros of the quadratic function leads to an investigation of real and nonreal complex zeros, the fundamental theorem of algebra, and graphs of higher-degree polynomial functions. Students study rational functions, asymptotes, and discontinuities. Finally, partial fractions and fractional equations complete the chapter.

A cumulative review of Chapters 1–4 appears in Section 4-9.

Using This ChapterTh is chapter focuses on polynomial and rational functions, the corresponding polynomial and rational expressions, and polynomial and rational equations. Th e study of functions includes an examination of the graphs, asymptotes, discontinuities, and fi tting of functions to data. Th e study of expressions involves factoring, completing the square, operations with rational expressions, and partial fractions. Th e study of equations includes the quadratic formula, complex numbers, fractional equations, and extraneous solutions. Th e relationships between the zeros and the graph of a function are emphasized throughout the chapter. Portions of this chapter review topics students learned in second-year algebra. You may prefer to select only a few problems from the familiar material and concentrate on the new content in the chapter.

Teaching ResourcesExplorationsExploration 4-3: Graphs and Zeros of Cubic FunctionsExploration 4-3a: Cubic Function GraphsExploration 4-3b: Synthetic SubstitutionExploration 4-3c: Sum and Product of the Zeros of a PolynomialExploration 4-4a: Fitting a Polynomial Function to PointsExploration 4-5: Transformations of the Parent Reciprocal

FunctionExploration 4-5a: Rational Functions and DiscontinuitiesExploration 4-6a: Introduction to Partial FractionsExploration 4-7a: Fractional Equations

Blackline MastersSections 4-8 and 4-9

Assessment ResourcesTest 8, Sections 4-1 to 4-4, Forms A and BTest 9, Sections 4-5 to 4-7, Forms A and BTest 10, Chapter 4, Forms A and BTest 11, Cumulative Test, Chapters 1–4, Forms A and B

Technology ResourcesSketchpad Presentation SketchesPolynomial Fit Present.gsp

ActivitiesCAS Activity 4-2a: Coeffi cients of Quadratic FunctionsCAS Activity 4-5a: End Behavior of a Rational Function

Calculator ProgramsSYNSUB

Polynomial and Rational FunctionsPolynomial and Rational FunctionsC h a p t e r 4

Chapter 4 Interleaf 175B

Standard Schedule Pacing Guide

Block Schedule Pacing Guide

Day Section Suggested Assignment

1 4-1 Introduction to Polynomial and Rational Functions 1–8

24-2 Quadratic Functions, Factoring, and Complex

NumbersRA, Q1–Q10, 1, 5, 7, 9, 13–19 odd, 23, 25, 28

3 33–37 odd, 41, 43, 47, 53, 55, 56, 57, 61, 63, 65b, 65d, 664

4-3 Graphs and Zeros of Higher-Degree FunctionsRA, Q1–Q10, 5–9, 11–19 odd

5 21, 25, 29, 30, 35–38

6 4-4 Fitting Polynomial Functions to Data RA, Q1–Q10, 1, (3), 4, 5, 7–9, 13, 15, (16)

74-5 Rational Functions: Asymptotes and

DiscontinuitiesRA, Q1–Q10, 1–5, 7, 9, 13–21 odd

8 23–31 odd, 35, 36

9 4-6 Partial Fractions and Operations with Rational Expressions RA, Q1–Q10, 7, 9, 13, 16, 21, 25, 29, 31, 41

10 4-7 Fractional Equations and Extraneous Solutions RA, Q1–Q10, 1–3, 5–11 odd, 15–27 odd

114-8 Chapter Review and Test

R0–R7, T1–T1912 Section 4-9 Problems 1–1313

4-9 Cumulative Review, Chapters 1–414–29

14 30–37

Chapter 4 Interleaf 175B

Day Section Suggested Assignment

1 4-2 Quadratic Functions, Factoring, and Complex Numbers

RA, Q1–Q10, 1, 5, 7, 9, 13–19 odd, 23, 25, 28, 33–37 odd, 41, 43, 47, 53

2 4-3 Graphs and Zeros of Higher-Degree Functions RA, Q1–Q10, 5–9, 11–21 odd, 25, 29, 30, 35–38

34-4 Fitting Polynomial Functions to Data RA, Q1–Q10, 1, 4, 5, 7–9, 13, 15

4-5 Rational Functions: Asymptotes and Discontinuities RA, Q1–Q10, 1–5, 7, 9, 13–21 odd

4 4-5 Rational Functions: Asymptotes and Discontinuities 23–31 odd, 35, 36

5 4-6 Partial Fractions and Operations with Rational Expressions RA, Q1–Q10, 7, 9, 13, 16, 21, 25, 29, 31, 41

6 4-7 Fractional Equations and Extraneous Solutions RA, Q1–Q10, 1–3, 5–11 odd

74-7 Fractional Equations and Extraneous Solutions 15–27 odd

4-8 Chapter Review and Test R0–R7, T1–T19

8 4-8 Chapter Review and Test Section 4-9 Problems 1–13

177Section 4-1: Introduction to Polynomial and Rational Functions

Introduction to Polynomial and Rational Functions Recall from Section 1-2 that a polynomial function has an equation such as f (x) x 2 2x 8 where no operations other than addition, subtraction, or multiplication are performed on the independent variable x. (Squaring is equivalent to multiplying.) Function f is a quadratic function because the highest power of x is the second power. A rational function has an equation such as g (x) x 3 5x 2 2x 25 _____________ x 3 , where the y-value is a ratio of two polynomials—in this case a cubic polynomial divided by a linear polynomial. Figures 4-1a and 4-1b show the graphs of functions f and g.

x422

10f(x)

10

x422

10g(x)

10Asymptote

Figure 4-1a Figure 4-1b

Discover some properties of polynomial and rational functions and their graphs.

Introduction to Polynomial and Rational Functions

4 -1


Objective

1. Function f appears to have x-intercepts at 2 and 4. Con� rm by direct substitution that f ( 2) and f (4) both equal zero. � en explain why 2 and 4 are also called zeros of function f.

2. � e x-value 2 appears to be a zero of function g. Con� rm or refute this observation numerically.

3. Function g has a vertical asymptote at x 3. Try to � nd g (3). � en explain why g (3) is unde� ned. Explain the behavior of g (x) when x is close to 3 by � nding g (3.001) and g (2.999).

4. Plot the graphs of functions f and g on the same screen. What do you notice about the two graphs when x is far from 3? When x is close to 3? How could you describe the relationship of the graph of function f to the graph of function g?

5. Let function h be the vertical translation of function f by 10 units. � at is, h(x) f (x) 10. Show by graphing that no zeros of function h are real numbers.

6. Explain why a quadratic function can have no more than two real numbers as zeros.

7. As a challenge, show that the complex number 1 i (where i

___ 1 ) is a zero of h (x).

Explain why a function can have a zero that is not an x-intercept.

8. What did you learn as a result of doing this problem set?

1. Function f appears to have f appears to have f x-intercepts at 2 5. Let function h be the vertical translation of

Exploratory Problem Set 4–1

Chapter 4: Polynomial and Rational Functions176

In this chapter you’ll learn about polynomial functions. You’ll � nd the zeros of these functions, a generalization of the concept of x-intercept. � e techniques you learn will allow you to analyze rational functions, in which f (x) is a ratio of two polynomials. � ese ratios can represent average rates of change. You’ll investigate these concepts in four ways.

� e graph of the polynomial function f (x) x 4 8 x 2 8x 15 has two x-intercepts.

f (x) x 4 0 x 3 8 x 2 8x 15 (x 1)(x 3)(x 2 i)(x 2 i)

f (1) 0, f (3) 0, f ( 2 i) 0, and f ( 2 i) 0

Zeros of the function are x 1, x 3, x 2 i, and x 2 i.

A fourth-degree function has exactly four zeros, counting multiple zeros, i f the domain of the function includes complex numbers. Only the real-number zeros appear as x-intercepts on the graph, but zeros can also include complex numbers.

GRAPHICALLY

40

x

f(x)

31

ALGEBRAICALLY

NUMERICALLY

VERBALLY

In this chapter you’ll learn about polynomial functions. You’ll � nd the

Mathematical Overview

176 Chapter 4: Polynomial and Rational Functions

When using a grapher, the Table feature is an invaluable tool. In Exploratory Problem Set 4-1, set the table to start at x 5 25 with an increment of 0.1. Students will readily see the unusual behavior of function g near x 5 3. Some of the built-in features of the grapher can also be helpful. Zoom Fit automatically adjusts the window so that it includes the minimum and maximum y-values of the function between the x-min and x-max values. Zoom Decimal

sets the x-window to 24.7 x 4.7. Th is creates the “friendly window” in which the cursor moves by increments of 0.1 and lands on “nice” domain values. Th e Table, Zoom Fit, and Zoom Decimal features do not always help the student to discover “weird behavior” of functions. In actuality, the grapher can be cumbersome for graphing polynomial and rational functions.

S e c t i o n 4 -1PL AN N I N G

Class Time 1 __ 2 day

Homework AssignmentProblems 1–8

TE ACH I N G

Important Terms and ConceptsRational functionZerosComplex zeros

Section Notes

A polynomial function was defi ned in Section 1-2 as

f(x) 5 a n x n 1 a n21 x n21 1 . . . 1 a 1 x 1 1 a 0

Th is defi nition, with all its coeffi cients, subscripts, terms, and powers, can be confusing to students. Remind students that polynomial functions are continuous and have a domain of all real numbers. Some examples may make the defi nition more concrete and easier to understand. Th ese functions are polynomial functions: y 5 5, y 5 27 1 x 14 , and y 5 2x. However, y 5 x, y 5 sin x, y 5 x 7/2 , y 5  

__ x and

y 5 x 23 are not polynomial functions.

A rational function was fi rst defi ned in Section 1-2 as the quotient of two polynomial functions.

Students oft en have trouble choosing an appropriate viewing window when graphing polynomial and rational functions. Th e most important thing students need to learn is the underlying mathematics associated with the graphs of these functions.


Introduction to Polynomial and Rational Functions Recall from Section 1-2 that a polynomial function has an equation such as f (x) � x 2 � 2x � 8 where no operations other than addition, subtraction, or multiplication are performed on the independent variable x. (Squaring is equivalent to multiplying.) Function f is a quadratic function because the highest power of x is the second power. A rational function has an equation such as g (x) � x 3 � 5x 2 � 2x � 25 _____________ x � 3 , where the y-value is a ratio of two polynomials—in this case a cubic polynomial divided by a linear polynomial. Figures 4-1a and 4-1b show the graphs of functions f and g.

x42−2

10f(x)

−10

x42�2

10g(x)

�10Asymptote

Figure 4-1a Figure 4-1b


IntrodRation

4 -1

Discotheir

Objective

1. Function f appears to have x-intercepts at �2 and 4. Con� rm by direct substitution that f (�2) and f (4) both equal zero. � en explain why �2 and 4 are also called zeros of function f.

2. � e x-value �2 appears to be a zero of function g. Con� rm or refute this observation numerically.

3. Function g has a vertical asymptote at x � 3. Try to � nd g (3). � en explain why g (3) is unde� ned. Explain the behavior of g (x) when x is close to 3 by � nding g (3.001) and g (2.999).

4. Plot the graphs of functions f and g on the same screen. What do you notice about the two graphs when x is far from 3? When x is close to 3? How could you describe the relationship of the graph of function f to the graph of function g?

5. Let function h be the vertical translation of function f by 10 units. � at is, h(x) � f (x) � 10. Show by graphing that no zeros of function h are real numbers.

6. Explain why a quadratic function can have no more than two real numbers as zeros.

7. As a challenge, show that the complex number 1 � i (where i � �

___ �1 ) is a zero of h (x).

Explain why a function can have a zero that is not an x-intercept.

8. What did you learn as a result of doing this problem set?

1. Function f appears to havef x-intercepts at �2 5. Let function h be the vertical translation of

Exploratory Problem Set 4–1

Chapter 4: Polynomial and Rational Functions176

In this chapter you’ll learn about polynomial functions. You’ll � nd the zeros of these functions, a generalization of the concept of x-intercept. � e techniques you learn will allow you to analyze rational functions, in which f (x) is a ratio of two polynomials. � ese ratios can represent average rates of change. You’ll investigate these concepts in four ways.

� e graph of the polynomial function f (x) x 4 8 x 2 8x 15 has two x-intercepts.

f (x) x 4 0 x 3 8 x 2 8x 15 (x 1)(x 3)(x 2 i)(x 2 i)

f (1) 0, f (3) 0, f ( 2 i) 0, and f ( 2 i) 0

Zeros of the function are x 1, x 3, x 2 i, and x 2 i.

A fourth-degree function has exactly four zeros, counting multiple zeros, i f the domain of the function includes complex numbers. Only the real-number zeros appear as x-intercepts on the graph, but zeros can also include complex numbers.

GRAPHICALLY

40

x

f(x)

31

ALGEBRAICALLY

NUMERICALLY

VERBALLY

In this chapter you’ll learn about polynomial functions. You’ll � nd the

Mathematical Overview


PRO B LE M N OTES

1. f (2) 0; f (4) 0; 2 and 4 are called zeros of the function because when substituted for x, they make the function value equal zero.2. g(2) 1 ___

5 ; 2 is not a zero of function g.

Problem 3 shows a rational function with a vertical asymptote. 3. g(3) takes the form 1 _ 0 , which is unde� ned because of division by zero; g(3.001) 995.004001, g(2.999) 1005.003999; g(x) is very large in the positive direction when x is close to 3 on the positive side, and very large in the negative direction when x is close to 3 on the negative side.

In Problem 4, be sure to discuss how functions f and g are related to each other when x is far from 3 and when x is close to 3.

Students who use TI-Nspire graphers can graph the functions in Problem 4 by � rst de� ning them using function notation. Note: While it may seem simpler to de� ne these functions graphically using their algebraic expressions, doing so adds considerable clutter to the screen, especially with the larger algebraic form of g(x).

In Problem 7 you may want to challenge your students to � nd the second complex zero of the function.

Additional CAS Problems

1. Do the graphs of f and g as de� ned in the Exploratory Problem Set ever intersect?

2. Do the graphs of g and h ever intersect?

CAS Suggestions

Have students de� ne functions on a CAS in order to algebraically manipulate both the function and its graph.6. A quadratic function has two “branches,” one coming downward and the other going upward. Each branch can cross the x-axis at most once, meaning that there can be at most two x-intercepts that are real numbers.

7. h(1 i ) (1 i ) 2 2(1 i ) 8 10 0; An x-intercept of a function is a real-number zero. A nonreal complex-number zero is not an x-intercept.8. Answers will vary.

See page 989 for answers to Problems 4 and 5, and CAS Problems 1 and 2.

� e answer in Example 1 is called a quadratic trinomial, quadratic meaning “second degree” and trinomial meaning “three terms.” With practice you can combine the two middle terms mentally and write just the given expression and the answer, like this:

(3x 5)(2x 11) 6x 2 23x 55

Example 2 shows how to multiply three binomials.

Multiply the binomials and simplify: (x 5)(x 2)(x 1)

(x 5)(x 2)(x 1) Write the given expression.

(x 5) (x 2)(x 1) Associate two of the factors.

(x 5)( x 2 3x 2) Multiply the associated factors.

x 3 3x 2 2x 5x 2 15x 10 Multiply each term in (x 5) by each term in ( x 2 3x 2).

x 3 8x 2 17x 10 Combine like terms.

� e process of transforming an expression in polynomial form to factored form is called factoring the expression. Example 3 reviews how to factor a quadratic trinomial.

Factor: x 2 6x 40

x 2 6x 40 (x )(x ) Write two pairs of parentheses. � e � rst term in each factor will be x.

(x 10)(x 4) Find two factors of the constant term, 40, that sum to 6. � ose factors must

be 10 and 4 in order to get x 2 6x 40 when you multiply the binomials.

You can check the answer to Example 3 by multiplying (x 10)(x 4) and showing that the result is the given expression, x 2 6x 40.

Solutions of Quadratic EquationsA quadratic equation such as x 2 6x 23 17 might arise from � nding the values of x for the function y x 2 6x 23 that give y 17 (Figure 4-2b).

y 17

x

40

40

55 10

y

Figure 4-2b

Such an equation can be solved with the help of factoring. � e idea is to get a product equal to zero and then set each factor equal to zero.

Multiply the binomials and simplify: (EXAMPLE 2 ➤

(xSOLUTION

➤

Factor: EXAMPLE 3 ➤

x 2SOLUTION

➤

179Section 4-2: Quadratic Functions, Factoring, and Complex Numbers178

Quadratic Functions, Factoring, and Complex NumbersFigure 4-2a shows the graphs of three quadratic functions, f, g, and h. � ese graphs are parabolas.

f (x) x 2 4x 5

g (x) x 2 4x 4

h(x) x 2 4x 13

Figure 4-2a

Each graph di� ers from the others by only a vertical translation. � e graph of function f has two x-intercepts, at x 1 and x 5, where the graph crosses the x-axis. Function g has only one x-intercept, at x 2, where the graph touches the x-axis but does not cross it. Function h has no x-intercepts because the graph never touches the x-axis. Note that at each x-intercept, y 0.

In this section you will learn that each of these functions has exactly two values of x that make y equal zero. For function f, these values are x 1 and x 5. For function g, they are x 2 and x 2, which are equal to each other. For function h, they are the complex numbers x 2 3i and x 2 3i, where i is the unit imaginary number,

___ 1 . � e term zero of a function is used to include nonreal

values of x that make y equal zero. So every quadratic function has exactly two zeros but may have two, one, or no x-intercepts.

In order to understand the algebra of higher-degree polynomial functions, it will be helpful to review various properties and techniques for quadratic functions.

Review the properties, graphs, and techniques associated with quadratic functions.

Algebraic Operations with Polynomials� e expression (3x 5)(2x 11) is said to be in factored form because it is written as a product of two (binomial) factors. To transform the expression to polynomial form you use the distributive property to multiply each term in the � rst binomial by each term in the second.

Multiply the binomials and simplify: (3x 5)(2x 11)

(3x 5)(2x 11) Write the given expression.

6x 2 33x 10x 55 Multiply each term in (3x 5) by each term in (2x 11).

6x 2 23x 55 Combine like terms.

Quadratic Functions, Factoring, and Complex Numbers

4 -2

gh

fx

52

25y

�1

�20

No x-intercepts

Two x-intercepts

Onex-intercept


Objective

Multiply the binomials and simplify: (3EXAMPLE 1 ➤

(3SOLUTION

➤

Chapter 4: Polynomial and Rational Functions



Class Time1–2 days

Homework AssignmentDay 1: RA, Q1–Q10, Problems 1, 5, 7, 9,

13–19 odd, 23, 25, 28Day 2: Problems 33–37 odd, 41, 43, 47, 53,

55, 56, 57, 61, 63, 65b, 65d, 66

Technology Resources

CAS Activity 4-2a: Coeffi cients of Quadratic Functions

TE ACH I N G

Important Terms and ConceptsUnit imaginary number i 5  

___ 21

Zero of a function Quadratic function Factored form of a function Polynomial form of a function Quadratic trinomial Quadratic equation Multiplication property of zero and its

converse Quadratic formula Discriminant Complex number Real part of a complex number Complex plane Complex conjugates Vertex form of a quadratic function Completing the square

� e answer in Example 1 is called a quadratic trinomial, quadratic meaning “second degree” and trinomial meaning “three terms.” With practice you can combine the two middle terms mentally and write just the given expression and the answer, like this:

(3x 5)(2x 11) 6x 2 23x 55

Example 2 shows how to multiply three binomials.

Multiply the binomials and simplify: (x 5)(x 2)(x 1)

(x 5)(x 2)(x 1) Write the given expression.

(x 5) (x 2)(x 1) Associate two of the factors.

(x 5)( x 2 3x 2) Multiply the associated factors.

x 3 3x 2 2x 5x 2 15x 10 Multiply each term in (x 5) by each term in ( x 2 3x 2).

x 3 8x 2 17x 10 Combine like terms.

� e process of transforming an expression in polynomial form to factored form is called factoring the expression. Example 3 reviews how to factor a quadratic trinomial.

Factor: x 2 6x 40

x 2 6x 40 (x )(x ) Write two pairs of parentheses. � e � rst term in each factor will be x.

(x 10)(x 4) Find two factors of the constant term, 40, that sum to 6. � ose factors must

be 10 and 4 in order to get x 2 6x 40 when you multiply the binomials.

You can check the answer to Example 3 by multiplying (x 10)(x 4) and showing that the result is the given expression, x 2 6x 40.

Solutions of Quadratic EquationsA quadratic equation such as x 2 6x 23 17 might arise from � nding the values of x for the function y x 2 6x 23 that give y 17 (Figure 4-2b).

y 17

x

40

40

55 10

y

Figure 4-2b

Such an equation can be solved with the help of factoring. � e idea is to get a product equal to zero and then set each factor equal to zero.

Multiply the binomials and simplify: (EXAMPLE 2 ➤

(xSOLUTION

➤

Factor: EXAMPLE 3 ➤

x 2SOLUTION

➤

179Section 4-2: Quadratic Functions, Factoring, and Complex Numbers178

Quadratic Functions, Factoring, and Complex NumbersFigure 4-2a shows the graphs of three quadratic functions, f, g, and h. � ese graphs are parabolas.

f (x) x 2 4x 5

g (x) x 2 4x 4

h(x) x 2 4x 13

Figure 4-2a

Each graph di� ers from the others by only a vertical translation. � e graph of function f has two x-intercepts, at x 1 and x 5, where the graph crosses the x-axis. Function g has only one x-intercept, at x 2, where the graph touches the x-axis but does not cross it. Function h has no x-intercepts because the graph never touches the x-axis. Note that at each x-intercept, y 0.

In this section you will learn that each of these functions has exactly two values of x that make y equal zero. For function f, these values are x 1 and x 5. For function g, they are x 2 and x 2, which are equal to each other. For function h, they are the complex numbers x 2 3i and x 2 3i, where i is the unit imaginary number,

___ 1 . � e term zero of a function is used to include nonreal

values of x that make y equal zero. So every quadratic function has exactly two zeros but may have two, one, or no x-intercepts.

In order to understand the algebra of higher-degree polynomial functions, it will be helpful to review various properties and techniques for quadratic functions.


Algebraic Operations with Polynomials� e expression (3x 5)(2x 11) is said to be in factored form because it is written as a product of two (binomial) factors. To transform the expression to polynomial form you use the distributive property to multiply each term in the � rst binomial by each term in the second.

Multiply the binomials and simplify: (3x 5)(2x 11)

(3x 5)(2x 11) Write the given expression.

6x 2 33x 10x 55 Multiply each term in (3x 5) by each term in (2x 11).

6x 2 23x 55 Combine like terms.

Quadratic Functions, Factoring, and Complex Numbers

4 -2

gh

fx

52

25y

�1

�20

No x-intercepts

Two x-intercepts

Onex-intercept


Objective

Multiply the binomials and simplify: (3EXAMPLE 1 ➤

(3SOLUTION

➤


179Section 4-2: Quadratic Functions, Factoring, and Complex Numbers

Section Notes

This section is primarily a review of topics found in a second-year algebra course. You may want to omit the portions your students have already mastered and choose problems at the end of the section for review. In general, students are more successful if review is incorporated into the context of new material. This section is included for students who have not mastered the material and also to provide a reference for students.

The vocabulary in this section is important. Discuss the difference between a quadratic expression such as x 2 1 7x 1 10, a quadratic equation such as x 2 1 7x 1 2 5 212, and a quadratic function such as y 5 x 2 1 7x 1 10. Students may not readily see the differences between these mathematical entities. Clarifying the language of mathematics can make an enormous difference in fostering student understanding.

Following Figure 4-2a, the text discusses the difference between the x-intercepts and the zeros of function f by including both real and nonreal values of c such that f(c) 5 0. Make sure your students understand this distinction.

In this section, students get an early glimpse of a corollary of the fundamental theorem of algebra that states that a polynomial function of degree n has n zeros if multiple zeros and nonreal complex zeros are included. This theorem will be discussed in Section 4-3 in some detail.

181

x 6 14 ______ 2 or x 6 14 ______ 2

x 10 or x 4

�is solution agrees with Example 4.

Although using the quadratic formula may take longer than factoring, it works even when the quadratic trinomial will not factor and even when the discriminant, b 2 4ac, is a negative number.

Imaginary and Complex Numbers Recall from Figure 4-2a that the function h(x) x 2 4x 13 has no x-intercepts. If you set h(x) equal to zero and apply the quadratic formula to the resulting quadratic equation, x 2 4x 13 0, you get

x 4 _______

16 52 _____________ 2 4 ____

36 _________ 2

�e square root of 36 is not a real number because squares of real numbers are never negative. In order for such a quadratic equation to have solutions, mathematicians de�ne square roots of negative numbers to be another kind of number:

___ 1 is de�ned to be the unit imaginary number, designated by the

letter i.

____

36 __________

(36) ( 1)

___

36 ___

1

6i

Similarly,

___

9 _________

(9) ( 1) __

9 ___

1 3i

____

53 7.2801...i

_____

100 10i

�e solutions of the quadratic equation x 2 4x 13 0 above become

x 4 6i ______ 2 2 3i 2 3i or 2 3i

�e complex numbers 2 3i and 2 3i are the zeros of the function h de�ned at the beginning of this section. �ese numbers are not x-intercepts because the graph of function h does not reach the x-axis. �e real number 2 is the real part of 2 3i, and the real number 3 is the imaginary part of 2 3i.

Complex numbers can be plotted on the complex-number plane, sometimes called simply the complex plane. �e real part is plotted horizontally, and the imaginary part is plotted vertically, as shown in Figure 4-2c.

➤

3i

1 42 5314 23Real

Imaginary4i

i2i

3i4i

i2i

0

2 3i

Figure 4-2c

Section 4-2: Quadratic Functions, Factoring, and Complex Numbers180

Solve by factoring: x 2 6x 23 17

x 2 6x 23 17

x 2 6x 40 0 Make one side of the given equation equal zero.

(x 10)(x 4) 0 Factor the polynomial side.

x 10 0 or x 4 0 Set each factor equal to zero.

x 10 or x 4 Solve the two simpler equations.

Note that the two solutions agree with the graph in Figure 4-2b.

� e basis for solving an equation by factoring is the multiplication property of zero and its converse.

PROPERTY: Multiplication Property of Zero and Its Conversea b 0 if and only if a 0 or b 0

It is for this reason that you must � rst make one side of the equation in Example 4 equal zero before you factor the expression on the other side.

A quadratic equation can also be solved using the quadratic formula. You may recall this formula from earlier courses. � e proof is presented in Problems 65 and 66 of this section.

PROPERTY: The Quadratic FormulaAlgebraically: If ax 2 bx c 0 and a 0, then x b

________ b 2 4ac _______________ 2a .

Verbally: If ax 2 bx c 0 and a 0, then x equals the opposite of b, plus or minus the square root of the quantity b 2 4ac, all divided by 2a.

� e quantity b 2 4ac under the radical sign is called the discriminant because it “discriminates” between quadratic equations that have real-number solutions and quadratic equations that don’t.

Example 5 shows how to solve the quadratic equation in Example 4 using this formula.

Solve using the quadratic formula: x 2 6x 23 17

To satisfy the “if” part of the quadratic formula, you must make one side of the equation equal zero.

x 2 6x 23 17


x 6 _______________

6 2 4 1 40 _____________________ 2 1 a 1, b 6, c 40

x 6 ____

196 _________ 2 Evaluate the discriminant � rst.

Solve by factoring: EXAMPLE 4 ➤

x 2SOLUTION

➤

Solve using the quadratic formula: EXAMPLE 5 ➤


SOLUTION



Section Notes (continued)

Some graphers allow the user to enter multiple variations of the same basic equation by enclosing in braces the values of the term that varies. For instance, in the beginning of the section, the functions f, g, and h can be entered as f 1 (x) 5 x 2 2 4x 1 {25, 4, 13}. Th e grapher will put all three variations on the same axes in the order listed within the braces. Another way to see the eff ect of changing just one term is to enter f 1 (x) 5 x 2 2 4x 1 A. Th e grapher will use whatever value is stored in A when it graphs f 1 (x). Changing the value of A is a quick way to see the results of varying a particular term. Th is second method is more intuitive, although the fi rst method displays all three functions at the same time.

It is very important for students to be able to do the material in this section using pencil-and-paper techniques. Encourage students to use their graphers to check their work.

In Examples 1–3, enter the given expression in f 1 (x) and the fi nal expression in f 2 (x) and check to see if the graphs are identical.

Example 4 can be checked by entering f 1 (x) 5 x 2 2 6x 2 23 and f 2 (x) 5 17 and fi nding the x-values of the points of intersection. Another way to check Example 4 is to set the equation equal to zero and see if the zeros are the same as the answers, reinforcing the need to set the equation equal to zero before factoring.

Th e complex plane is sometimes called the Argand plane because it is used in Argand diagrams. Th ese are named aft er Jean-Robert Argand (1768–1822).

181

x 6 14 ______ 2 or x 6 14 ______ 2

x 10 or x 4

�is solution agrees with Example 4.

Although using the quadratic formula may take longer than factoring, it works even when the quadratic trinomial will not factor and even when the discriminant, b 2 4ac, is a negative number.

Imaginary and Complex Numbers Recall from Figure 4-2a that the function h(x) x 2 4x 13 has no x-intercepts. If you set h(x) equal to zero and apply the quadratic formula to the resulting quadratic equation, x 2 4x 13 0, you get

x 4 _______

16 52 _____________ 2 4 ____

36 _________ 2

�e square root of 36 is not a real number because squares of real numbers are never negative. In order for such a quadratic equation to have solutions, mathematicians de�ne square roots of negative numbers to be another kind of number:

___ 1 is de�ned to be the unit imaginary number, designated by the

letter i.

____

36 __________

(36) ( 1)

___

36 ___

1

6i

Similarly,

___

9 _________

(9) ( 1) __

9 ___

1 3i

____

53 7.2801...i

_____

100 10i

�e solutions of the quadratic equation x 2 4x 13 0 above become

x 4 6i ______ 2 2 3i 2 3i or 2 3i

�e complex numbers 2 3i and 2 3i are the zeros of the function h de�ned at the beginning of this section. �ese numbers are not x-intercepts because the graph of function h does not reach the x-axis. �e real number 2 is the real part of 2 3i, and the real number 3 is the imaginary part of 2 3i.

Complex numbers can be plotted on the complex-number plane, sometimes called simply the complex plane. �e real part is plotted horizontally, and the imaginary part is plotted vertically, as shown in Figure 4-2c.

➤

3i

1 42 5314 23Real

Imaginary4i

i2i

3i4i

i2i

0

2 3i

Figure 4-2c


Solve by factoring: x 2 6x 23 17

x 2 6x 23 17


(x 10)(x 4) 0 Factor the polynomial side.

x 10 0 or x 4 0 Set each factor equal to zero.

x 10 or x 4 Solve the two simpler equations.

Note that the two solutions agree with the graph in Figure 4-2b.

� e basis for solving an equation by factoring is the multiplication property of zero and its converse.

PROPERTY: Multiplication Property of Zero and Its Conversea b 0 if and only if a 0 or b 0

It is for this reason that you must � rst make one side of the equation in Example 4 equal zero before you factor the expression on the other side.

A quadratic equation can also be solved using the quadratic formula. You may recall this formula from earlier courses. � e proof is presented in Problems 65 and 66 of this section.

PROPERTY: The Quadratic FormulaAlgebraically: If ax 2 bx c 0 and a 0, then x b

________ b 2 4ac _______________ 2a .

Verbally: If ax 2 bx c 0 and a 0, then x equals the opposite of b, plus or minus the square root of the quantity b 2 4ac, all divided by 2a.

� e quantity b 2 4ac under the radical sign is called the discriminant because it “discriminates” between quadratic equations that have real-number solutions and quadratic equations that don’t.

Example 5 shows how to solve the quadratic equation in Example 4 using this formula.

Solve using the quadratic formula: x 2 6x 23 17


x 2 6x 23 17


x 6 _______________

6 2 4 1 40 _____________________ 2 1 a 1, b 6, c 40

x 6 ____

196 _________ 2 Evaluate the discriminant � rst.

Solve by factoring: EXAMPLE 4 ➤

x 2SOLUTION

➤

Solve using the quadratic formula: EXAMPLE 5 ➤


SOLUTION


181

Students should be able to do Examples 6 and 7 using paper and pencil. However, a grapher can be used to check their answers by changing the mode to complex a 1 bi form.

In Example 9, some students may find it easier to move the constant term to the other side of the equation before completing the square. Ensure that students master completing the square on a quadratic equation to find the solutions or on a quadratic function to find the zeros. It will come up again when the students study conic sections.

Differentiating Instruction• Pass out the list of Chapter 4

vocabulary, available at www.keypress.com/keyonline, for students to look up and translate in their bilingual dictionaries.

• Complex numbers may be new to ELL students. Consider giving them a one-page summary to use as a reference.

• Have students write the definitions of imaginary and complex numbers and the rules for doing all four operations with them in their journals.

• FOIL has no meaning to speakers of other languages. The box represents a mnemonic that some students may find more meaningful.

x 24x x 2 24x3 3x 212

• In some countries, the algorithms used for solving quadratic equations are different from those we teach. Allow students to use different methods if they show the process they use.

• Clarify the meaning of iff, and have students write it in their journals.

• Allow ELL students to do the Reading Analysis in pairs and have them write a summary of the section in their journals.

Section 4-2: Quadratic Functions, Factoring, and Complex Numbers

183

Note that complex solutions of a quadratic equation with real coe� cients will always be complex conjugates of each other, such as 2 3i and 2 3i. You can see why this is true by writing the quadratic formula this way:

x b ___ 2a _______

b 2 4ac _________ 2a

� e expression b ___ 2a is a real number. � e radical expression will be an imaginary number if the discriminant, b 2 4ac, is negative.

Vertex Form and Completing the Square � e equation

y 3(x 5) 2 43

is said to be in vertex form. As you can see graphically in Figure 4-2d, the vertex is (5, 43). Algebraically, substituting 5 for x makes the expression 3(x 5) 2 equal zero, giving 43 for the value of y.

5 105

50

50

y

x

43

Figure 4-2d

Example 8 shows how to transform an equation in vertex form to polynomial form (y ax 2 bx c) by multiplying and combining like terms.

Transform the equation y 3(x 5) 2 43 to polynomial form.

y 3(x 5) 2 43

y 3( x 2 10x 25) 43 Multiply (x 5)(x 5).

y 3 x 2 30x 75 43 Distribute the 3.

y 3x 2 30x 32 Combine like terms.

If the equation is in polynomial form, like the answer in Example 8, it is easy to � nd the y-intercept. You can substitute zero for x and get y 32 (in this case), or simply look at the equation and write the constant term, 32. � e result agrees with Figure 4-2d. � e box on the next page summarizes polynomial form and vertex form.

Transform the equationEXAMPLE 8 ➤

y 3(SOLUTION

➤


If the coe� cient of i is zero, then only the real part is le� . So real numbers are a subset of the complex numbers. You can see this by covering up the regions of the complex plane that are above and below the real-number line.

� e de� nitions and properties of complex numbers appear in the box.

DEFINITIONS: Imaginary and Complex Numbers i

___ 1 so that i 2 1

If x is a nonnegative real number, then

___

x i __

x

A complex number is a number of the form a bi, where

a is called the real part of a bi.

b is called the imaginary part of a bi.

� e complex numbers a bi and a bi are called complex conjugates of each other.

Complex numbers can be added, subtracted, and multiplied the same way as binomials.

Multiply and simplify: (7 4i)(9 10i)

(7 4i)(9 10i) 63 70i 36i 40i 2 Distribute each term in one number to each term of the other.

63 70i 36i 40 Because i 2 1.

23 106i Combine like terms.

� e main thing to remember in operating with complex numbers is that i 2 is equal to 1. All the properties and techniques for real numbers apply to complex numbers, including squaring a binomial.

Check that x 2 3i is a solution to the equation x 2 4x 13 0 by substituting it into the le� side of the equation and showing that the result equals zero.

(2 3i ) 2 4(2 3i) 13 ? 0 Substitute 2 3i for x.

4 12i 9i 2 8 12i 13 ? 0 Do the squaring and distributing.

4 9 8 13 ? 0 Combine like terms; 9i 2 9.

0 0 ✓ � e result does equal zero.

Multiply and simplify: (7 EXAMPLE 6 ➤

(7 SOLUTION

➤

Check thatsubstituting it into the le� side of the equation and showing that the result

EXAMPLE 7 ➤

(2 SOLUTION

➤



Technology Notes

CAS Activity 4-2a: Coeffi cients of Quadratic Functions has students explore the eff ect of b on the graph of the equation y ax 2 bx c. Students will need access to a CAS that allows the use of sliders. Students using a TI-Nspire CAS can download QuadraticCoeffi cients.tns, available at www.keymath.com/precalc, and transfer it to their grapher. Allow 25–30 minutes.

CAS Suggestions

When introducing the term zeros, consider introducing the CAS command of the same name. Th e Zeros command is very useful in later applications and students will benefi t signifi cantly from learning how to use it. Th e fi gure shows the correlation between the Solve and Zeros commands.

Th e next fi gure shows the Solve and Zeros commands for function h. Notice that while the CAS is in real number mode, neither command produces a solution because there are no real solutions. Inserting a “c” before each command tells the CAS to fi nd all complex solutions.

183

Note that complex solutions of a quadratic equation with real coe� cients will always be complex conjugates of each other, such as 2 3i and 2 3i. You can see why this is true by writing the quadratic formula this way:

x b ___ 2a _______

b 2 4ac _________ 2a

� e expression b ___ 2a is a real number. � e radical expression will be an imaginary number if the discriminant, b 2 4ac, is negative.

Vertex Form and Completing the Square � e equation

y 3(x 5) 2 43

is said to be in vertex form. As you can see graphically in Figure 4-2d, the vertex is (5, 43). Algebraically, substituting 5 for x makes the expression 3(x 5) 2 equal zero, giving 43 for the value of y.

5 105

50

50

y

x

43

Figure 4-2d

Example 8 shows how to transform an equation in vertex form to polynomial form (y ax 2 bx c) by multiplying and combining like terms.

Transform the equation y 3(x 5) 2 43 to polynomial form.

y 3(x 5) 2 43

y 3( x 2 10x 25) 43 Multiply (x 5)(x 5).

y 3 x 2 30x 75 43 Distribute the 3.

y 3x 2 30x 32 Combine like terms.

If the equation is in polynomial form, like the answer in Example 8, it is easy to � nd the y-intercept. You can substitute zero for x and get y 32 (in this case), or simply look at the equation and write the constant term, 32. � e result agrees with Figure 4-2d. � e box on the next page summarizes polynomial form and vertex form.

Transform the equationEXAMPLE 8 ➤

y 3(SOLUTION

➤


If the coe� cient of i is zero, then only the real part is le� . So real numbers are a subset of the complex numbers. You can see this by covering up the regions of the complex plane that are above and below the real-number line.

� e de� nitions and properties of complex numbers appear in the box.

DEFINITIONS: Imaginary and Complex Numbers i

___ 1 so that i 2 1

If x is a nonnegative real number, then

___

x i __

x

A complex number is a number of the form a bi, where

a is called the real part of a bi.

b is called the imaginary part of a bi.

� e complex numbers a bi and a bi are called complex conjugates of each other.

Complex numbers can be added, subtracted, and multiplied the same way as binomials.

Multiply and simplify: (7 4i)(9 10i)

(7 4i)(9 10i) 63 70i 36i 40i 2 Distribute each term in one number to each term of the other.

63 70i 36i 40 Because i 2 1.

23 106i Combine like terms.

� e main thing to remember in operating with complex numbers is that i 2 is equal to 1. All the properties and techniques for real numbers apply to complex numbers, including squaring a binomial.

Check that x 2 3i is a solution to the equation x 2 4x 13 0 by substituting it into the le� side of the equation and showing that the result equals zero.

(2 3i ) 2 4(2 3i) 13 ? 0 Substitute 2 3i for x.

4 12i 9i 2 8 12i 13 ? 0 Do the squaring and distributing.

4 9 8 13 ? 0 Combine like terms; 9i 2 9.

0 0 ✓ � e result does equal zero.

Multiply and simplify: (7 EXAMPLE 6 ➤

(7 SOLUTION

➤

Check thatsubstituting it into the le� side of the equation and showing that the result

EXAMPLE 7 ➤

(2 SOLUTION

➤


183

The Factor and Expand commands are tremendously useful CAS commands for changing the form of an expression. These are particularly useful for solving problems for which the point is the use of the factors or expanded form and not the algebraic manipulation itself. Be sure to emphasize the appropriate use of a CAS when using these commands. The Factor command also has a complex-number command.

A CAS can also solve general equations. Although expressed in a different form, the figure shows the CAS version of the quadratic formula.


185

PROCEDURE: Completing the SquareTo write a quadratic trinomial x 2 bx c as the square of a linear binomial:

x 2 -term and the x-term.

x and square the answer.

parentheses.

Note: For this procedure to work, the coe� cient of x 2 must be 1.


Reading Analysis

From what you have read in this section, what do you consider to be the main idea? What is meant by a zero of a function, and how is it related to an x-intercept? What is the “If . . .” part of the quadratic formula? What are the di� erences among real numbers, imaginary numbers, and complex numbers? What is meant by the discriminant of a quadratic equation? Why does i 2 equal 1? What is meant by the vertex form of a quadratic function equation, and what procedure is used to transform a quadratic equation in polynomial form into vertex form?

Quick Review

For Problems Q1 Q4, let g (x) 2 3f (4x 5). Function g is Q1. A horizontal translation of function f by ?

units Q2. A horizontal dilation of function f by a factor

of ? Q3. A vertical translation of function f by ?

units Q4. A vertical dilation of function f by a factor

of ? Q5. f (x) 3 2 x is an example of a(n) ?

function.

Q6. f (x) 4 x 3 is an example of a(n) ? power function.


Q8. f (x) 6 2x is an example of a(n) ? function.

Q9. A function found by regression � ts the data well if the residual plot has a ? pattern.

Q10. (3x 5) 2 A. 9x 2 25 B. 9x 2 25 C. 9x 2 30x 25 D. 9x 2 30x 25 E. 9x 2 30x 25

For Problems 1 10, multiply the binomials and simplify. 1. (4x 5)(2x 3) 2. (5x 2)(6x 3) 3. (x 8)(2x 9) 4. (4x 1)(x 10) 5. (3x 7 ) 2 6. (6x 8 ) 2 7. (5x 7)(5x 7) 8. (3x 4)(3x 4) 9. (x 2)(x 5)(x 6) 10. (x 4)(x 3)(x 5)

5min

Reading Analysis Q6. f (f (f x) x) x 4 x 3 is an example of a(n) ? power

Problem Set 4-2

184

DEFINITION: Quadratic Function A quadratic function is a function with the general equation

y ax 2 bx c where a, b, and c stand for constants and a 0.

� is is known as the polynomial form of the equation. (If a 0, the function is linear, not quadratic.)

Vertex form: If the equation of a quadratic function is in the form

y a(x h) 2 k or y k a(x h) 2

where a, h, and k stand for constants and a 0, then the vertex is at the point (h, k).

Memory aid: h stands for the horizontal coordinate of the vertex. k comes a� er h in the alphabet, just as y comes a� er x.

In both forms, the constant a is the vertical dilation of the parent function y x 2 . � e x-coordinate, h, of the vertex is the average of the two x-values that result from the quadratic formula, speci� cally, h b ___ 2a .

A quadratic equation in polynomial form can be transformed to vertex form by completing the square, as shown in Example 9.

Write the equation y 5x 2 35x 17 in vertex form, and then sketch its graph.

y 5x 2 35x 17 Write the given equation.

y 5(x 2 7x ) 17 Factor 5 out of the � rst two terms to make the x 2 -coe� cient equal 1.

y 5(x 2 7x 3.5 2 ) 17 5( 3.5 2 ) Complete the square by taking half the x-coe� cient, squaring it, adding the resulting 3.5 2 inside the parentheses, and then subtracting 5( 3.5 2 ) outside the parentheses.

y 5(x 3.5) 2 44.25 Write x 2 7x 3.5 2 as (x 3.5) 2 , and combine the constants.

Vertex: ( 3.5, 44.25) 3.5 is the value of x that makes (x 3.5) 2 equal zero. 44.25 is the value of y when x 3.5.

y-intercept: 17 Set x 0 in the original polynomial form.

Sketch the graph (Figure 4-2e), showing the vertex, the axis of symmetry, and the y-intercept. (It is not necessary to use the same scales on the two axes.)

Write the equation EXAMPLE 9 ➤

ySOLUTION

x 3.5

17

44.25

y

x

( 3.5, 44.25)

Figure 4-2e


➤


CAS Suggestions (continued)

In Example 9, students can use a CAS to complete the square. The general procedure is to:

1. Move the constants to the y-side of the equation.

2. Take 1 _ 2 coefficient of x ____________ leading coefficient and square it. Add the result to both sides of the equation.

3. Factor the x-side of the resulting equation.

Note: The CAS notation is not traditional, but it makes sense from the perspective of operating on an equation as an object. Often, the CAS rearranges the factored expression to avoid internal fractions, but the CAS results can easily be shown equivalent to the traditional results by doing the algebraic manipulations with pencil and paper.

Completing the square is challenging for many students. Using a CAS helps students avoid algebraic manipulation mistakes and allows them to focus on the process.

185

PROCEDURE: Completing the SquareTo write a quadratic trinomial x 2 bx c as the square of a linear binomial:

x 2 -term and the x-term.

x and square the answer.

parentheses.

Note: For this procedure to work, the coe� cient of x 2 must be 1.


Reading Analysis

From what you have read in this section, what do you consider to be the main idea? What is meant by a zero of a function, and how is it related to an x-intercept? What is the “If . . .” part of the quadratic formula? What are the di� erences among real numbers, imaginary numbers, and complex numbers? What is meant by the discriminant of a quadratic equation? Why does i 2 equal 1? What is meant by the vertex form of a quadratic function equation, and what procedure is used to transform a quadratic equation in polynomial form into vertex form?

Quick Review

For Problems Q1 Q4, let g (x) 2 3f (4x 5). Function g is Q1. A horizontal translation of function f by ?

units Q2. A horizontal dilation of function f by a factor

of ? Q3. A vertical translation of function f by ?

units Q4. A vertical dilation of function f by a factor

of ? Q5. f (x) 3 2 x is an example of a(n) ?

function.



Q8. f (x) 6 2x is an example of a(n) ? function.

Q9. A function found by regression � ts the data well if the residual plot has a ? pattern.

Q10. (3x 5) 2 A. 9x 2 25 B. 9x 2 25 C. 9x 2 30x 25 D. 9x 2 30x 25 E. 9x 2 30x 25

For Problems 1 10, multiply the binomials and simplify. 1. (4x 5)(2x 3) 2. (5x 2)(6x 3) 3. (x 8)(2x 9) 4. (4x 1)(x 10) 5. (3x 7 ) 2 6. (6x 8 ) 2 7. (5x 7)(5x 7) 8. (3x 4)(3x 4) 9. (x 2)(x 5)(x 6) 10. (x 4)(x 3)(x 5)

5min

Reading Analysis Q6. f (f (f x) x) x 4 x 3 is an example of a(n) ? power

Problem Set 4-2

184

DEFINITION: Quadratic Function A quadratic function is a function with the general equation

y ax 2 bx c where a, b, and c stand for constants and a 0.

� is is known as the polynomial form of the equation. (If a 0, the function is linear, not quadratic.)

Vertex form: If the equation of a quadratic function is in the form

y a(x h) 2 k or y k a(x h) 2

where a, h, and k stand for constants and a 0, then the vertex is at the point (h, k).

Memory aid: h stands for the horizontal coordinate of the vertex. k comes a� er h in the alphabet, just as y comes a� er x.

In both forms, the constant a is the vertical dilation of the parent function y x 2 . � e x-coordinate, h, of the vertex is the average of the two x-values that result from the quadratic formula, speci� cally, h b ___ 2a .

A quadratic equation in polynomial form can be transformed to vertex form by completing the square, as shown in Example 9.

Write the equation y 5x 2 35x 17 in vertex form, and then sketch its graph.

y 5x 2 35x 17 Write the given equation.

y 5(x 2 7x ) 17 Factor 5 out of the � rst two terms to make the x 2 -coe� cient equal 1.

y 5(x 2 7x 3.5 2 ) 17 5( 3.5 2 ) Complete the square by taking half the x-coe� cient, squaring it, adding the resulting 3.5 2 inside the parentheses, and then subtracting 5( 3.5 2 ) outside the parentheses.

y 5(x 3.5) 2 44.25 Write x 2 7x 3.5 2 as (x 3.5) 2 , and combine the constants.

Vertex: ( 3.5, 44.25) 3.5 is the value of x that makes (x 3.5) 2 equal zero. 44.25 is the value of y when x 3.5.

y-intercept: 17 Set x 0 in the original polynomial form.

Sketch the graph (Figure 4-2e), showing the vertex, the axis of symmetry, and the y-intercept. (It is not necessary to use the same scales on the two axes.)

Write the equation EXAMPLE 9 ➤

ySOLUTION

x 3.5

17

44.25

y

x

( 3.5, 44.25)

Figure 4-2e


➤

185

PRO B LE M N OTES

Q1. 25Q2. 1 __ 4 Q3. 2Q4. 3Q5. Exponential Q6. Direct cubeQ7. Inverse square Q8. Linear (polynomial)Q9. RandomQ10. D

Problems 1–54 can be solved in a single step when using a CAS.

Problems 1–20 involve quadratic expressions that require simplifying (multiplying out) and factoring (the reverse of simplifying).1. 8 x 2 2 2x 2 152. 30 x 2 1 3x 2 63. 2 x 2 2 25x 1 724. 4 x 2 2 41x 1 105. 9 x 2 1 42x 1 496. 36 x 2 2 96x 1 647. 25 x 2 2 498. 9 x 2 2 169. x 3 2 9 x 2 1 8x 1 6010. x 3 1 4 x 2 2 17x 2 60



For Problems 61 64, a. Transform the equation in polynomial form to

vertex form by completing the square. b. Write the coordinates of the vertex. c. Use the vertex, the y-intercept, and the

re�ection of the y-intercept across the axis of symmetry to sketch the graph.

61. f (x) x 2 6x 11 62. f (x) x 2 8x 21 63. f (x) 3 x 2 24x 17 64. f (x) 8 x 2 40x 37 65. Solving Quadratics by Completing the Square

Problem: If you did not know the quadratic formula, you could solve a quadratic equation by completing the square. �ese steps show how it can be done.

5 x 2 30x 7 0 Write the given equation.

x 2 6x 1.4 0 Divide by 5 to make the x 2 -coe�cient

equal 1.

x 2 6x 1.4 Move the constant term to the right

side, and leave a space to complete the square.

x 2 6x 9 9 1.4 Add 9 to both sides to complete the

square on the le�.

(x 3 ) 2 7.6 Factor on the le�, and simplify on

the right.

x 3 ___

7.6 Take the square root of both sides.

x 3 ___

7.6 0.2431... or 5.7568...

Simplify.

Solve these equations by completing the square. a. x 2 6x 4 0 b. x 2 10x 21 0 c. x 2 7x 4 0 d. 7 x 2 14x 3 0 e. 2 x 2 10x 20 9

66. Derivation of the Quadratic Formula Problem: In this problem you will see how the quadratic formula is derived by completing the square to solve the general quadratic equation as in Problem 65. Write each step, parts a–g, on your paper, and supply the reason for each step.

a. a x 2 bx c 0 ?

b. x 2 b __ a x c __ a 0 ?

c. x 2 b __ a x c __ a ?

d. x 2 b __ a x b 2 _____ (2 a) 2

b 2 _____ (2 a) 2

c __ a ?

e. x b ___ 2a 2 b 2 4ac ________

(2a ) 2 ?

f. x b ___ 2a ________

b 2 4ac ____________ 2a ?

g. x b ________

b 2 4ac _______________ 2a ?

186

For Problems 11 20, factor the polynomial. 11. x 2 10x 21 12. x 2 5x 36 13. x 2 6x 40 14. x 2 17x 60 15. x 2 36 16. x 2 121 17. x 2 20x 100 18. x 2 18x 81 19. x 2 7x 20. x 2 9x

For Problems 21 30, solve the equation by factoring. 21. x 2 10x 21 0 22. x 2 5x 36 0 23. x 2 6x 55 15 24. x 2 17x 70 10 25. x 2 36 0 26. x 2 121 0 27. x 2 20x 100 0 28. x 2 18x 81 0 29. x 2 7x 0 30. x 2 9x 0

For Problems 31 36, solve the equation by using the quadratic formula. 31. x 2 10x 21 0 32. x 2 5x 36 0 33. 3 x 2 11x 5 0 34. 5 x 2 3x 7 0 35. 6 x 2 5x 8 36. 10 x 2 1 8x

For Problems 37 42, multiply the complex numbers and simplify. 37. (3 2i)(4 5i) 38. (5 8i)(3 i) 39. (7 3i)(7 3i) 40. (4 6i)(4 6i) 41. (6 7i ) 2 42. (9 7i ) 2

For Problems 43 46, solve for complex values of x. Check one of the solutions by substituting it into the given equation. 43. x 2 6x 34 0 44. x 2 14x 58 0 45. 9 x 2 30x 26 0 46. 25 x 2 10x 101 0

For Problems 47 50, one complex solution of the equation is given. Write the other complex solution, and check this second solution by substituting it into the given equation. 47. x 2 22x 150 20; x 11 3i 48. x 2 16x 100 11; x 8 5i 49. 25 x 2 2 10x; x 0.2 0.2i 50. 0.04 x 2 1.2x 45; x 15 30i

For Problems 51 54, �nd the zeros of the given function. Show that your answer is reasonable by plotting the graph and sketching the result. 51. f (x) x 2 2x 15 52. f (x) x 2 7x 6 53. f (x) x 2 6x 10 54. f (x) x 2 10x 29 55. Write the de�nition of “zero of a function.” 56. What is the di�erence between a zero of a

function and an x-intercept of a function?

For Problems 57 60, a. Transform the equation to polynomial form. b. Write the coordinates of the vertex. c. Use the vertex, the y-intercept, and the


57. f (x) 0.2(x 3 ) 2 5 58. f (x) 3(x 2 ) 2 8 59. f (x) 5(x 4 ) 2 7 60. f (x) 0.5(x 1 ) 2 3



Problem Notes (continued)11. (x 1 3)(x 1 7)12. (x 2 9)(x 1 4)13. (x 2 4)(x 1 10)14. (x 2 12)(x 2 5)15. (x 2 6)(x 1 6)16. (x 2 11)(x 1 11)17. (x 2 10 ) 2 18. (x 1 9 ) 2 19. x(x 1 7) 20. x(x 2 9)

Problems 21–36 involve solving quadratic equations by factoring or by using the quadratic formula.21. x 5 23, 27 22. x 5 9, 2423. x 5 4, 210 24. x 5 12, 525. x 5 6, 26 26. x 5 11, 21127. x 5 10 28. x 5 29 29. x 5 0, 27 30. x 5 0, 9 31. x 5 23, 27 32. x 5 9, 2433. x 5 3.1350..., 0.5316...34. x 5 1.5206..., 20.9206...35. x 5 1.6442..., 20.8109... 36. x 5 20.1550..., 20.6449...

Problems 37–42 are like Problems 1–10 except they involve imaginary numbers.37. 2 2 23i 38. 7 1 29i 39. 58 40. 5241. 213 1 84i 42. 32 2 126i

Problems 43–46 involve using the quadratic formula to solve quadratic equations with nonreal solutions.43. x 5 3 5i 44. x 5 7 3i 45. x 5 2 5 __ 3 1 __ 3 i46. x 5 0.2 2i

In Problems 47–50, students need to use the property that complex solutions are conjugates of each other.47. x 5 211 1 3i48. x 5 8 1 5i 49. x 5 0.2 2 0.2i50. x 5 215 2 30i

Problems 51–54 focus on the connection between the real zeros of a function and the x-intercepts of the graph.

Problems 55 and 56 point out the important distinction between a zero and an x-intercept.

In Problems 57–60, students graph a quadratic function in vertex form and fi nd the vertex, y-intercept, and axis of symmetry.

Th e Expand command can be used to solve the algebraic part of Problems 57–60.

Completing the square using a CAS as described in the CAS suggestions can be used to solve Problems 61–66.

In Problems 61–64, students are asked to complete the square and fi nd the same information as in Problems 57–60. Completing the square is an important skill for students to develop.


For Problems 61 64, a. Transform the equation in polynomial form to

vertex form by completing the square. b. Write the coordinates of the vertex. c. Use the vertex, the y-intercept, and the


61. f (x) x 2 6x 11 62. f (x) x 2 8x 21 63. f (x) 3 x 2 24x 17 64. f (x) 8 x 2 40x 37 65. Solving Quadratics by Completing the Square

Problem: If you did not know the quadratic formula, you could solve a quadratic equation by completing the square. �ese steps show how it can be done.

5 x 2 30x 7 0 Write the given equation.

x 2 6x 1.4 0 Divide by 5 to make the x 2 -coe�cient

equal 1.

x 2 6x 1.4 Move the constant term to the right

side, and leave a space to complete the square.

x 2 6x 9 9 1.4 Add 9 to both sides to complete the

square on the le�.

(x 3 ) 2 7.6 Factor on the le�, and simplify on

the right.

x 3 ___

7.6 Take the square root of both sides.

x 3 ___

7.6 0.2431... or 5.7568...

Simplify.

Solve these equations by completing the square. a. x 2 6x 4 0 b. x 2 10x 21 0 c. x 2 7x 4 0 d. 7 x 2 14x 3 0 e. 2 x 2 10x 20 9

66. Derivation of the Quadratic Formula Problem: In this problem you will see how the quadratic formula is derived by completing the square to solve the general quadratic equation as in Problem 65. Write each step, parts a–g, on your paper, and supply the reason for each step.

a. a x 2 bx c 0 ?

b. x 2 b __ a x c __ a 0 ?

c. x 2 b __ a x c __ a ?

d. x 2 b __ a x b 2 _____ (2 a) 2

b 2 _____ (2 a) 2

c __ a ?

e. x b ___ 2a 2 b 2 4ac ________

(2a ) 2 ?

f. x b ___ 2a ________

b 2 4ac ____________ 2a ?

g. x b ________

b 2 4ac _______________ 2a ?

186

For Problems 11 20, factor the polynomial. 11. x 2 10x 21 12. x 2 5x 36 13. x 2 6x 40 14. x 2 17x 60 15. x 2 36 16. x 2 121 17. x 2 20x 100 18. x 2 18x 81 19. x 2 7x 20. x 2 9x

For Problems 21 30, solve the equation by factoring. 21. x 2 10x 21 0 22. x 2 5x 36 0 23. x 2 6x 55 15 24. x 2 17x 70 10 25. x 2 36 0 26. x 2 121 0 27. x 2 20x 100 0 28. x 2 18x 81 0 29. x 2 7x 0 30. x 2 9x 0

For Problems 31 36, solve the equation by using the quadratic formula. 31. x 2 10x 21 0 32. x 2 5x 36 0 33. 3 x 2 11x 5 0 34. 5 x 2 3x 7 0 35. 6 x 2 5x 8 36. 10 x 2 1 8x

For Problems 37 42, multiply the complex numbers and simplify. 37. (3 2i)(4 5i) 38. (5 8i)(3 i) 39. (7 3i)(7 3i) 40. (4 6i)(4 6i) 41. (6 7i ) 2 42. (9 7i ) 2

For Problems 43 46, solve for complex values of x. Check one of the solutions by substituting it into the given equation. 43. x 2 6x 34 0 44. x 2 14x 58 0 45. 9 x 2 30x 26 0 46. 25 x 2 10x 101 0

For Problems 47 50, one complex solution of the equation is given. Write the other complex solution, and check this second solution by substituting it into the given equation. 47. x 2 22x 150 20; x 11 3i 48. x 2 16x 100 11; x 8 5i 49. 25 x 2 2 10x; x 0.2 0.2i 50. 0.04 x 2 1.2x 45; x 15 30i

For Problems 51 54, �nd the zeros of the given function. Show that your answer is reasonable by plotting the graph and sketching the result. 51. f (x) x 2 2x 15 52. f (x) x 2 7x 6 53. f (x) x 2 6x 10 54. f (x) x 2 10x 29 55. Write the de�nition of “zero of a function.” 56. What is the di�erence between a zero of a

function and an x-intercept of a function?

For Problems 57 60, a. Transform the equation to polynomial form. b. Write the coordinates of the vertex. c. Use the vertex, the y-intercept, and the


57. f (x) 0.2(x 3 ) 2 5 58. f (x) 3(x 2 ) 2 8 59. f (x) 5(x 4 ) 2 7 60. f (x) 0.5(x 1 ) 2 3


187

Problem 65 involves solving a quadratic by completing the square. Be sure students understand the diff erences between completing the square on a quadratic equation and completing the square on quadratic expressions.65a. x 5 23  

__ 5

5 20.7639..., 25.2360...65b. x 5 7, 365c. x 5 3.5  

_____ 16.25

5 7.5311..., 20.5311...65d. x 5 21  

__ 4 __ 7

5 20.2440..., 21.7559...65e. x 5 2.5  

____ 0.75

5 3.3660..., 1.6339...

In Problem 66, the quadratic formula is derived by completing the square. Th is is a nice exercise for students with strong algebraic skills. 66a. Write the given equation.66b. Divide both sides by a.66c. Subtract c __ a from both sides.66d. Complete the square on the left and add to the right to balance the equation.66e. Write the left side as a perfect square and add fractions on the right side.66f. Take the square root of both sides.66g. Subtract b __ 2a from both sides. Combine the fractions on the right side.


1. For what value(s) of b does y 5 2 x 2 1 bx 1 1 have exactly one zero?

2. If the graph of y 5 2 x 2 1 bx 1 1 has exactly one zero, what are the coordinates of its vertex?

63a. f (x) 5 3(x 2 4 ) 2 2 3163b. (4, 231)63c.

64a. f (x) 5 8(x 1 2. 5) 2 21364b. (22.5, 213)64c.

(0, 17)x

f (x)

4

�31

x

f (x)

�13�2.5

(0, 37)


See page 990 for answers to Problems 51–62 and CAS Problems 1 and 2.

Graphs of Polynomial FunctionsRecall the general equation of a polynomial function from Chapter 1. Here are some examples of polynomial functions.

f (x) 4 x 2 7x 3 Quadratic function, 2nd degree

f (x) 2 x 3 5 x 2 4x 7 Cubic function, 3rd degree

f (x) x 4 6 x 3 3 x 2 5x 8 Quartic function, 4th degree

f (x) 6 x 5 x 3 2x Quintic function, 5th degree

f (x) 5 x 9 4 x 8 11 x 3 63 9th-degree function (no special name)

In each case, f (x) is a polynomial expression. � e only operations performed on the variable in a polynomial are the polynomial operations: addition, subtraction, and multiplication. Nonnegative integer powers are allowed in a polynomial because they are de� ned in terms of multiplication. Polynomials are continuous for all real-number values of x; that is, there are no discontinuities caused by roots of negative numbers or division by zero. � e graphs are also smooth, which means that they have no abrupt changes or sharp corners such as those caused by absolute values.

� e degree of a polynomial is the greatest number of variables multiplied together in any one term. For a one-variable polynomial, the degree is the same as the highest exponent on the variable. � e coe� cient of the term with the highest degree is called the leading coe� cient. Figure 4-3a on page 190 shows examples of a quadratic, a cubic, and a quartic function.

189Section 4-3: Graphs and Zeros of Higher-Degree Functions

7. � e graphs of f (x) x 3 and p(x) (x 2) 3 1 are shown in the � gure. Name the two transformations of function f that were done to get function p. Show that x 1 is a zero of function p.

2 1 321

y

1

2 pf

x

1

One factor of p(x) is (x 1). For a challenge, see if you can � nd the other factor, and use the result to � nd the two nonreal complex zeros of p(x).

8. From what you have seen in this exploration, how many zeros does a cubic function have? Give two reasons why a cubic function might have fewer than this number of x-intercepts.

9. What did you learn as a result of doing this exploration that you did not know before?

EXPLORATION, continued

188

Graphs and Zeros of Higher-Degree FunctionsIn Section 4-2, you reviewed quadratic functions—polynomial functions of degree 2. In this section you will learn to identify cubic, quartic, quintic, and higher-degree functions from their equations or graphs, to � nd their real and complex zeros, and to � nd their x-intercepts.

Given a polynomial function,

In this exploration you will explore the zeros of cubic functions and the behavior of their graphs.

Graphs and Zeros of Higher-Degree Functions

4 -3

Given a polynomial function,Objective


1. � e � gure shows the graphs of f (x) x 3 (dashed)—the parent cubic function—and g (x) x 3 x, a vertical translation of function f by the variable amount x. Set g (x) 0 and show that function g has the three distinct real zeros shown in the graph by factoring the polynomial on the right side of the resulting equation.

2 1 321

y

1

2g

f

x

1

2. � e parent function f (x) x 3 can be factored as f (x) x x x. Use this factorization to explain why function f also has three real zeros but not three distinct real zeros.

3. � e � gure shows the graphs of f (x) x 3 and function v, a vertical translation of function f by 1 unit.

v(x) x 3 1 (polynomial form)

v(x) (x 1)( x 2 x 1) (factored form)

Multiply the factors to show that the factored form is equivalent to the polynomial form.

2 1 321

y

1

2 v f

x

1

4. Use the factored form of function v from Problem 3 to show that function v also has three zeros, but two of the zeros are nonreal complex numbers.

5. � e � gure shows the graphs of f (x) x 3 and h(x) x 3 x 2 , a vertical translation of function f by the variable amount x 2 . By � rst factoring h(x), show that function h has three real zeros, not all of which are distinct.

21

y

1h

f

x 1

2

132

6. From the graphs of functions f and h above, describe the behavior of the graph at a single zero such as function h has at x 1, a double zero such as function h has at x 0, and a triple zero, such as function f has at x 0.

1. � e � gure shows the graphs of f � gure shows the graphs of f � gure shows the graphs of ( ) 3 y E X P L O R AT I O N 4 -3: G r a p h s a n d Z e r o s o f C u b i c Fu n c t i o n s

continued



Class Time1–2 days

Homework AssignmentDay 1: RA, Q1–Q10, Problems 5–9,

11–19 oddDay 2: Problems 21, 25, 29, 30, 35–38

Teaching ResourcesExploration 4-3: Graphs and Zeros of

Cubic FunctionsExploration 4-3a: Cubic Function GraphsExploration 4-3b: Synthetic SubstitutionExploration 4-3c: Sum and Product of

the Zeros of a Polynomial

Technology Resources Calculator Program: SYNSUB

TE ACH I N G

Important Terms and ConceptsPolynomial expressionPolynomial operationsContinuousSmoothDegree of a polynomialLeading coeffi cientEnd behaviorDominateExtreme pointVertexCritical pointMultiplicityDouble zeroSynthetic substitutionNested formRemainder theoremFactor theoremFundamental theorem of algebraSums and products of zeros of a cubic

function

Exploration Notes

Exploration 4-3 introduces students to the zeros and graphs of cubic functions. In Problem 1, g(x) 5 f(x) 1 (2x) 5 x 3 2 x.

Function f is shift ed vertically by an amount of 2x. Th is “translates” the function down when x 0 and up when x 0. Make sure students understand that function g is not a translation of f in the usual sense, since the shape of g is not the same as the shape of f. Instead, function f has been translated by the variable amount 2x. Th is same idea comes up again in Problem 5, where h(x) 5 f(x) 2 x 2 5 x 3 2 x 2 , so f is translated (shift ed) by the variable amount 2 x 2 .

See page 192 for notes on additional explorations.

1. 0 5 x 3 2 x; 0 5 x( x 2 2 1); 0 5 x(x 2 1)(x 1 1); x 5 0, x 5 1, or x 5 21.2. Each x can equal zero, so there are three zeros. Because they are equal to each other, they are not distinct.3. v(x) 5 (x 2 1)( x 2 1 x 1 1) 5 x 3 1 x 2 1 x 2 x 2 2 x 2 1 5 x 3 2 14. v(x) 5 0 if and only if x 2 1 5 0 or x 2 1 x 1 1 5 0 ⇒ x 5 1 or x 5 21 i  

__ 3 _______ 2 .

Graphs of Polynomial FunctionsRecall the general equation of a polynomial function from Chapter 1. Here are some examples of polynomial functions.

f (x) 4 x 2 7x 3 Quadratic function, 2nd degree

f (x) 2 x 3 5 x 2 4x 7 Cubic function, 3rd degree

f (x) x 4 6 x 3 3 x 2 5x 8 Quartic function, 4th degree

f (x) 6 x 5 x 3 2x Quintic function, 5th degree

f (x) 5 x 9 4 x 8 11 x 3 63 9th-degree function (no special name)

In each case, f (x) is a polynomial expression. � e only operations performed on the variable in a polynomial are the polynomial operations: addition, subtraction, and multiplication. Nonnegative integer powers are allowed in a polynomial because they are de� ned in terms of multiplication. Polynomials are continuous for all real-number values of x; that is, there are no discontinuities caused by roots of negative numbers or division by zero. � e graphs are also smooth, which means that they have no abrupt changes or sharp corners such as those caused by absolute values.

� e degree of a polynomial is the greatest number of variables multiplied together in any one term. For a one-variable polynomial, the degree is the same as the highest exponent on the variable. � e coe� cient of the term with the highest degree is called the leading coe� cient. Figure 4-3a on page 190 shows examples of a quadratic, a cubic, and a quartic function.


7. � e graphs of f (x) x 3 and p(x) (x 2) 3 1 are shown in the � gure. Name the two transformations of function f that were done to get function p. Show that x 1 is a zero of function p.

2 1 321

y

1

2 pf

x

1

One factor of p(x) is (x 1). For a challenge, see if you can � nd the other factor, and use the result to � nd the two nonreal complex zeros of p(x).

8. From what you have seen in this exploration, how many zeros does a cubic function have? Give two reasons why a cubic function might have fewer than this number of x-intercepts.



188

Graphs and Zeros of Higher-Degree FunctionsIn Section 4-2, you reviewed quadratic functions—polynomial functions of degree 2. In this section you will learn to identify cubic, quartic, quintic, and higher-degree functions from their equations or graphs, to � nd their real and complex zeros, and to � nd their x-intercepts.

Given a polynomial function,

In this exploration you will explore the zeros of cubic functions and the behavior of their graphs.

Graphs and Zeros of Higher-Degree Functions

4 -3

Given a polynomial function,Objective


1. � e � gure shows the graphs of f (x) x 3 (dashed)—the parent cubic function—and g (x) x 3 x, a vertical translation of function f by the variable amount x. Set g (x) 0 and show that function g has the three distinct real zeros shown in the graph by factoring the polynomial on the right side of the resulting equation.

2 1 321

y

1

2g

f

x

1

2. � e parent function f (x) x 3 can be factored as f (x) x x x. Use this factorization to explain why function f also has three real zeros but not three distinct real zeros.

3. � e � gure shows the graphs of f (x) x 3 and function v, a vertical translation of function f by 1 unit.

v(x) x 3 1 (polynomial form)

v(x) (x 1)( x 2 x 1) (factored form)

Multiply the factors to show that the factored form is equivalent to the polynomial form.

2 1 321

y

1

2 v f

x

1

4. Use the factored form of function v from Problem 3 to show that function v also has three zeros, but two of the zeros are nonreal complex numbers.

5. � e � gure shows the graphs of f (x) x 3 and h(x) x 3 x 2 , a vertical translation of function f by the variable amount x 2 . By � rst factoring h(x), show that function h has three real zeros, not all of which are distinct.

21

y

1h

f

x 1

2

132

6. From the graphs of functions f and h above, describe the behavior of the graph at a single zero such as function h has at x 1, a double zero such as function h has at x 0, and a triple zero, such as function f has at x 0.

1. � e � gure shows the graphs of f � gure shows the graphs of f � gure shows the graphs of ( ) 3 y E X P L O R AT I O N 4 -3: G r a p h s a n d Z e r o s o f C u b i c Fu n c t i o n s

continued


Section Notes

In this section, students explore the relationship between the graph of a polynomial function and its degree, and they learn to find the zeros of a polynomial function from its equation or its graph. Synthetic substitution, the remainder theorem, the factor theorem, and various other theorems were once necessary to find or approximate solutions and to draw graphs for polynomial functions. Now complex equations can be solved and graphed by using a grapher.

If time is an issue, you may want to omit the portions of this section on the sum and product of the zeros of a cubic function. The corresponding properties for the zeros of a quadratic function (see Problem 38) are easier to derive and more useful.

Polynomial functions are continuous and have a domain of all real numbers. In very simple terms, this means the functions can be drawn without lifting the pencil off the page. There are no discontinuities caused by roots of negative numbers or division by zero. Polynomial functions are smooth, a legitimate mathematical term. They contain no endpoints (they go on forever), have no sharp points (cusps), and have no jumps (step discontinuities). In addition, a polynomial function must approach positive infinity or negative infinity as x gets large in the positive direction and, likewise, as x gets large in the negative direction. This last condition means that a polynomial function cannot be periodic like a sine function. Polynomial functions are predictable, presenting no surprises. Students will learn in calculus that polynomial functions are so well behaved that they are everywhere differentiable.

Thus, v(x) has three zeros, two of which are nonreal complex zeros.5. h(x) 5 x 2 (x 2 1), so h(x) 5 0 if and only if x 5 0, x 5 0, or x 5 1. Thus, function h has three real zeros, two of which are equal to each other.6. At a single zero, the graph crosses the x-axis at an angle. At a double zero, the graph just touches the x-axis but does not cross it. At a triple zero, the graph crosses the x-axis, but is tangent to the x-axis.

7. Horizontal translation by 2 units, vertical translation by 1 unit. p(1) 5 (1 2 2) 3 1 1 5 2 1 1 1 5 0 so 1 is a zero of p(x). The nonreal complex zeros of p(x) are 2.5i 0.5i  

__ 3 .

8. A cubic function has exactly three zeros. Zeros might be complex numbers, or real zeros might occur at the same place.9. Answers will vary.


DEFINITIONS: Zeros and x-Intercepts of a FunctionA zero of a function f is a number c, real or complex, for which f (c) 0.

An x-intercept of a function f is a real zero of that function.

As you saw in Exploration 4-3, it is possible for two or more zeros to occur at the same value of x. �e function

f (x) (x 2)(x 3 ) 2 (x 1 ) 3 (x 2) (x 3)(x 3) (x 1)(x 1)(x 1)

in Figure 4-3c has a single zero at x 2, where the graph crosses the x-axis at an angle. �ere is a double zero at x 3, where the graph only touches the x-axis but does not cross it. And there is a triple zero at x 1, where the graph levels o� and crosses the x-axis tangent to it. Algebraically, you can see from the factored form of f (x) on the far right that the multiplicity of a zero is the number of identical linear factors that are zero for that particular value of x.

x

f(x)

2 31

Triplezero

Singlezero

Doublezero

Figure 4-3c

Finding Zeros of a Polynomial Function: Synthetic SubstitutionSynthetic substitution is a quick pencil-and-paper method for evaluating a polynomial function. Suppose that f (x) x 3 9 x 2 x 105 and that you want to �nd f (6) by synthetic substitution.

x-value, 6, and the coe�cients of f (x) like this:

6 1 9 1 105Leave space here.Leave space here.

writing the answer in the next column, under the 9.

6 1 9 1 1056

1

190

y

Downward

Upward

x

Quadratic(second degree)

Two zeros

y

Cubic(third degree)

�ree zeros

x

y

Quartic(fourth degree)

Four zeros

x

Figure 4-3a

�e end behavior of a function f refers to what happens to f (x) as x gets large in either the positive or the negative direction. As shown in Figure 4-3a, all three example functions become large and positive as x gets large in the positive direction. But the odd-degree function (cubic) becomes large and negative as x gets large in the negative direction. �is happens because both positive and negative numbers raised to even powers are positive, whereas negative numbers raised to odd powers are negative. If the leading coe�cient is negative, as in y 5 x 3 , then this end behavior is re�ected across the x-axis.

As x gets large, the highest-degree term eventually dominates the other terms, and the graph looks like that of the highest-degree term. Figure 4-3b shows the graph of the cubic function f (x) x 3 7 x 2 10x 2 and the graph of the cubic term, g (x) x 3 . In the narrow viewing window on the le�, the two graphs are clearly di�erent. But if you zoom out, as in the graph on the right, the graphs look quite similar.

yg f

x–5

5

5

y

x2020

500 g f

Figure 4-3b

Refer back to Figure 4-3a. A quadratic function graph has two branches, one downward and one upward. As a result there is one extreme point (or vertex, or critical point) and as many as two x-intercepts. Similarly, a cubic function graph can have three branches, a quartic function graph can have four branches, and a quintic function graph can have �ve branches. �e number of x-intercepts can be as high as the degree of the function, and the number of extreme points can be up to one less than the degree. As you saw in the exploration at the beginning of this section, the number of zeros of a polynomial function is equal to the degree of the function if you count multiple zeros and complex zeros.




Discuss the two different views of functions f and g in Figure 4-3b. The viewing window on the left shows the local behavior of the functions. The window on the right shows the global behavior of the same functions. The window has been significantly enlarged in the view on the right, and therefore the local behavior is obscured.

The definition box on page 191 states the definitions of zeros and x-intercepts introduced in Section 4-2 and leads the students into the fundamental theorem of algebra and its corollaries. Figure 4-3c shows a sixth-degree polynomial function with three zeros. Function f has a single zero at x 5 2, a double zero at x 5 3 where the function “bounces” off the x-axis, and a triple zero at x 5 1. Make sure students understand that the n zeros of an nth-degree polynomial include multiple zeros and nonreal complex zeros. Therefore, an nth-degree polynomial function does not necessarily have n x-intercepts.

The text introduces synthetic substitution as a method for evaluating a polynomial function for a specific value of x and for dividing a polynomial by a linear binomial. Students need to understand that when they use synthetic substitution to divide a polynomial by a linear binomial they must substitute the value of x for which the linear binomial is equal to zero. So, for instance, to divide by (x 1 3), substitute 23, not 3. When doing synthetic substitution, it is helpful for students if you “box off” the remainder. For example, write the bottom line of the solution in Example 1a as 1 25 2 | 0 . This will help students avoid putting the remainder in with the coefficients of the quotient.

Note: Some texts use the term synthetic division instead of synthetic substitution. The second term is used intentionally in this text to emphasize the distinction in how you find the remainder. When you divide, you subtract to find the remainder; when you use synthetic substitution, you can substitute to find the remainder.

Example 1 revisits the three cubic functions from Section 4-1. One zero of each function is found by using synthetic substitution, and the remaining zeros are found by factoring or using the quadratic formula. Each of the cubic functions has three zeros.

• Function f has three distinct real zeros. The graph crosses the x-axis at each zero.


DEFINITIONS: Zeros and x-Intercepts of a FunctionA zero of a function f is a number c, real or complex, for which f (c) 0.

An x-intercept of a function f is a real zero of that function.

As you saw in Exploration 4-3, it is possible for two or more zeros to occur at the same value of x. �e function

f (x) (x 2)(x 3 ) 2 (x 1 ) 3 (x 2) (x 3)(x 3) (x 1)(x 1)(x 1)

in Figure 4-3c has a single zero at x 2, where the graph crosses the x-axis at an angle. �ere is a double zero at x 3, where the graph only touches the x-axis but does not cross it. And there is a triple zero at x 1, where the graph levels o� and crosses the x-axis tangent to it. Algebraically, you can see from the factored form of f (x) on the far right that the multiplicity of a zero is the number of identical linear factors that are zero for that particular value of x.

x

f(x)

2 31

Triplezero

Singlezero

Doublezero

Figure 4-3c

Finding Zeros of a Polynomial Function: Synthetic SubstitutionSynthetic substitution is a quick pencil-and-paper method for evaluating a polynomial function. Suppose that f (x) x 3 9 x 2 x 105 and that you want to �nd f (6) by synthetic substitution.

x-value, 6, and the coe�cients of f (x) like this:

6 1 9 1 105Leave space here.Leave space here.

writing the answer in the next column, under the 9.

6 1 9 1 1056

1

190

y

Downward

Upward

x

Quadratic(second degree)

Two zeros

y

Cubic(third degree)

�ree zeros

x

y

Quartic(fourth degree)

Four zeros

x

Figure 4-3a

�e end behavior of a function f refers to what happens to f (x) as x gets large in either the positive or the negative direction. As shown in Figure 4-3a, all three example functions become large and positive as x gets large in the positive direction. But the odd-degree function (cubic) becomes large and negative as x gets large in the negative direction. �is happens because both positive and negative numbers raised to even powers are positive, whereas negative numbers raised to odd powers are negative. If the leading coe�cient is negative, as in y 5 x 3 , then this end behavior is re�ected across the x-axis.

As x gets large, the highest-degree term eventually dominates the other terms, and the graph looks like that of the highest-degree term. Figure 4-3b shows the graph of the cubic function f (x) x 3 7 x 2 10x 2 and the graph of the cubic term, g (x) x 3 . In the narrow viewing window on the le�, the two graphs are clearly di�erent. But if you zoom out, as in the graph on the right, the graphs look quite similar.

yg f

x–5

5

5

y

x2020

500 g f

Figure 4-3b

Refer back to Figure 4-3a. A quadratic function graph has two branches, one downward and one upward. As a result there is one extreme point (or vertex, or critical point) and as many as two x-intercepts. Similarly, a cubic function graph can have three branches, a quartic function graph can have four branches, and a quintic function graph can have �ve branches. �e number of x-intercepts can be as high as the degree of the function, and the number of extreme points can be up to one less than the degree. As you saw in the exploration at the beginning of this section, the number of zeros of a polynomial function is equal to the degree of the function if you count multiple zeros and complex zeros.


191

Make sure students understand that the term complex zero is used to mean a nonreal complex zero.

Example 2 presents the graph of a polynomial function and asks students to identify the degree of the function and the number of real and nonreal complex zeros that the function could have.

More examples involving finding the zeros of a polynomial can be found in the Additional Class Examples.

When the leading coefficient is factored from a cubic function, the coefficients of the resulting cubic are related to the sums and products of the zeros of the function. These relationships are illustrated using the function f(x) 5 5 x 3 2 33 x 2 1 58x 2 24 and summarized in the property box before Example 3. When discussing the text in the property box, you may want to rewrite the general cubic function as

p(x) 5 a x 3 1 b __ a x 2 1 c __ a x 1 d __ a so that students can better see how the property relates to the specific function they just investigated.

The sums and products of the zeros of a cubic function closely parallel a similar property for the zeros of a quadratic function. See Problem 38 in the Problem Set.

In Example 3, students find the zeros of a cubic function and then verify that the sum, the sum of pairwise products, and the product correspond to the coefficients of the equation.

Example 4 shows how to find an equation of a cubic function given the sum, sum of pairwise products, and product of its zeros.

• Function g has a double zero, meaning that two of the zeros are the same number. The graph touches but does not cross the x-axis at the double zero and crosses the x-axis at the other zero.

• Function h has one real-number zero and two nonreal complex zeros. The nonreal complex zeros are complex conjugates. The graph intersects the x-axis at the real-number zero.

Finding a double zero with the intersect feature on a grapher can be problematic on some graphers because there is no change of sign at that point. For example, if you graph y 5 x 3 2 4 x 2 2 3x 1 18 and y 5 0 in a nonfriendly graphing window, you will not be able to find the zero x 5 3 using the intersect feature. However, the zero feature will find the double zero x 5 3 as well as the zero x 5 22.

Section 4-3: Graphs and Zeros of Higher-Degree Functions


Notice that the coe� cients of the quotient, 1, 3, and 19, are the values below the line in the synthetic substitution process. � us synthetic substitution gives you a way to do long division of a polynomial by a linear binomial expression. Just substitute the value of x for which the linear binomial equals zero.

� e fact that the remainder a� er division by (x 6) equals the value of f (6) is an example of the remainder theorem.

PROPERTY: The Remainder TheoremIf p(x) is a polynomial, then p(c) equals the remainder when p(x) is divided by the quantity (x c).

COROLLARY: The Factor Theorem� e quantity (x c) is a factor of the polynomial p(x) if and only if p(c) 0.

� e corollary is true because if the remainder equals zero, then (x c) divides p(x) evenly, which means (x c) is a factor of p(x).

Let f (x) x 3 4 x 2 3x 2.

Let g (x) x 3 4 x 2 3x 18.

Let h(x) x 3 4 x 2 3x 54.

� e graphs are shown in Figure 4-3d.

a. Show that 1 is a zero of f (x). Find the other two zeros, and check by graphing.

b. Show that 2 is a zero of g (x). Find the other two zeros, and check by graphing.

c. Show that 3 is a zero of h(x). Find the other two zeros, and check by graphing.

For each part of the problem, use synthetic substitution to show that the remainder is zero.

a.

� erefore, 1 is a zero of f (x). f ( 1) 0 because f ( 1) equals the remainder.

Let fEXAMPLE 1 ➤

g

h

f

x5

60y

�3

�40Figure 4-3d


SOLUTION

1 1 4 3 2 Synthetically substitute 1 for x.

� e remainder is zero.1 5 2

1 5 2 0

192

9 and the 6, and write the answer, 3, below the line. Multiply 3 by 6, and write the answer, 18, above the line. Repeat the steps, adding and multiplying. � e � nal result is

6 1 9 1 1056 18 114

1 3 19 9

f (6) 9 � e value of the function f (6) is the last number below the line.

Check:

f (6) 6 3 9( 6 2 ) 1(6) 105 9

6–31Multiply

Writ

e prod

uct

Bringitdown

6 1 –9 –1 105 CoefficientshereAdd

To see why synthetic substitution works, factor the polynomial into nested form:

f (x) x 3 9 x 2 x 105

(1x 9) x 2 x 105

(1x 9)x 1 x 105 Nested form.

In this form you can evaluate the polynomial by repeating these steps:

Multiply by x.

Add the next coe� cient.

Synthetic substitution is closely related to long division of polynomials. To see why, divide f (x) by (x 6), a linear binomial that equals zero when x is 6. First, divide the x from the term (x 6) into the x 3 from the polynomial. Write the answer, x 2 , above the x 2 -term of the polynomial.

x 2 3x 19 Quotient x 6 )

_________________ x 3 9 x 2 x 105

x 3 6 x 2 Multiply x 2 and (x 6). 3 x 2 x Subtract and bring x down. 3 x 2 18x Multiply 3x and (x 6). 19x 105 Subtract and bring the next term down.

19x 114 Multiply 19 and (x 6). 9 Remainder.

x 3 9 x 2 x 105 _________________ x 6 x 2 3x 19 9 _____ x 6 “Mixed-number” form.

Note: � e term “mixed-number” form is used here because of the similarity of this form to the result of whole-number division when there is a remainder. For example, when you divide 13 by 4, the quotient is 3 and the remainder is 1, so you can write 13 __ 4 as 3 1 _ 4 .



Differentiating Instruction• You may need to spend three days on

this section to help ELL students fully grasp the concepts.

• Have students write in their own words the factor theorem, the remainder theorem, and the fundamental theorem of algebra in their journals.

• Clarify the difference between the meaning of could be and that of is/are.

• Division algorithms vary from country to country. You may need to spend extra time helping ELL students learn the algorithms for division of polynomials and synthetic substitution.

Additional Exploration Notes

Exploration 4-3a leads students to explore the shape and some properties of graphs of cubic functions by plotting the graphs of three cubic functions and comparing them. Allow about 20 minutes. Exploration 4-3b shows students how to use synthetic substitution to evaluate and factor a function. Allow about 20 minutes for this exploration.

Exploration 4-3c leads students to discover the relationship between the coefficients of a cubic polynomial function and the sums and products of its zeros. After discovering the patterns, students must apply them to write an equation for a cubic function given the zeros of the function. Allow 20–25 minutes.


Notice that the coe� cients of the quotient, 1, 3, and 19, are the values below the line in the synthetic substitution process. � us synthetic substitution gives you a way to do long division of a polynomial by a linear binomial expression. Just substitute the value of x for which the linear binomial equals zero.

� e fact that the remainder a� er division by (x 6) equals the value of f (6) is an example of the remainder theorem.

PROPERTY: The Remainder TheoremIf p(x) is a polynomial, then p(c) equals the remainder when p(x) is divided by the quantity (x c).

COROLLARY: The Factor Theorem� e quantity (x c) is a factor of the polynomial p(x) if and only if p(c) 0.

� e corollary is true because if the remainder equals zero, then (x c) divides p(x) evenly, which means (x c) is a factor of p(x).

Let f (x) x 3 4 x 2 3x 2.

Let g (x) x 3 4 x 2 3x 18.

Let h(x) x 3 4 x 2 3x 54.

� e graphs are shown in Figure 4-3d.

a. Show that 1 is a zero of f (x). Find the other two zeros, and check by graphing.

b. Show that 2 is a zero of g (x). Find the other two zeros, and check by graphing.

c. Show that 3 is a zero of h(x). Find the other two zeros, and check by graphing.


a.

� erefore, 1 is a zero of f (x). f ( 1) 0 because f ( 1) equals the remainder.

Let fEXAMPLE 1 ➤

g

h

f

x5

60y

�3

�40Figure 4-3d


SOLUTION


� e remainder is zero.1 5 2

1 5 2 0

192

9 and the 6, and write the answer, 3, below the line. Multiply 3 by 6, and write the answer, 18, above the line. Repeat the steps, adding and multiplying. � e � nal result is

6 1 9 1 1056 18 114

1 3 19 9

f (6) 9 � e value of the function f (6) is the last number below the line.

Check:

f (6) 6 3 9( 6 2 ) 1(6) 105 9

6–31Multiply

Writ

e prod

uct

Bringitdown

6 1 –9 –1 105 CoefficientshereAdd

To see why synthetic substitution works, factor the polynomial into nested form:

f (x) x 3 9 x 2 x 105

(1x 9) x 2 x 105

(1x 9)x 1 x 105 Nested form.

In this form you can evaluate the polynomial by repeating these steps:

Multiply by x.

Add the next coe� cient.

Synthetic substitution is closely related to long division of polynomials. To see why, divide f (x) by (x 6), a linear binomial that equals zero when x is 6. First, divide the x from the term (x 6) into the x 3 from the polynomial. Write the answer, x 2 , above the x 2 -term of the polynomial.

x 2 3x 19 Quotient x 6 )

_________________ x 3 9 x 2 x 105

x 3 6 x 2 Multiply x 2 and (x 6). 3 x 2 x Subtract and bring x down. 3 x 2 18x Multiply 3x and (x 6). 19x 105 Subtract and bring the next term down.

19x 114 Multiply 19 and (x 6). 9 Remainder.

x 3 9 x 2 x 105 _________________ x 6 x 2 3x 19 9 _____ x 6 “Mixed-number” form.

Note: � e term “mixed-number” form is used here because of the similarity of this form to the result of whole-number division when there is a remainder. For example, when you divide 13 by 4, the quotient is 3 and the remainder is 1, so you can write 13 __ 4 as 3 1 _ 4 .


193

Technology Notes

Calculator Program: SYNSUB

calculates the value of a function for a particular input value using the synthetic substitution technique, as asked for in Problem 33. Th e program is available at www.keymath.com/precalc for download onto a TI-Nspire, TI-83, or TI-84 graphing calculator.

Additional Class Examples

1. Find the zeros of g(x) 5 6 x 3 1 17 x 2 2 24x 2 35 algebraically. Write g(x) in factored form.Solution: Th e fi gure shows the graph of g. By tracing the graph, x 5 21 appears to be a zero. Confi rm this algebraically by synthetic substitution.

x

g(x)

−1

100

21 6 17 224 23526 211 35

6 11 235 0

Synthetically substitute 21 for x. Th e remainder is zero.

[ 21 is a zero of g(x). g (21) 5 0 because it equals the remainder.

g(x) 5 (x 1 1)(6x 2 1 11x 2 35) (x 1 1) is a factor of g (x ) because it is 0 when x 5 21.

5 (x 1 1)(2x 1 7)(3x 2 5)Factor the quadratic by inspection.

g(x) 5 0 ⇔ x 1 1 5 0 or 2x 1 7 5 0 or 3x 2 5 5 0

Zeros are x 5 21, x 5 2 7 _ 2 , and x 5 5 _ 3 .



of Example 1, there are three zeros of h(x) but only one x-intercept, as you can see in Figure 4-3d. � e other two zeros are nonreal complex numbers. As you saw in Section 4-2, the two complex zeros are complex conjugates of each other.

� e results of Example 1 illustrate the fundamental theorem of algebra and its corollaries.

PROPERTY: The Fundamental Theorem of Algebra and Its Corollaries

A polynomial function has at least one zero in the set of complex numbers.

CorollaryAn nth-degree polynomial function has exactly n zeros in the set of complex numbers, counting multiple zeros.

CorollaryIf a polynomial has only real coe� cients, then any nonreal complex zeros appear in conjugate pairs.

Be sure you understand what the theorem says. As you saw in Figure 4-2c, the real numbers are a subset of the complex numbers, so the zeros of the function could be nonreal or real complex numbers. � e x-intercepts correspond to the real-number zeros of the function. From now on, these real-number zeros will be referred to as real zeros and nonreal complex zeros wil be referred to as complex zeros.

Identify the degree and the number of real and nonreal complex zeros that the polynomial function in Figure 4-3e could have. State whether the leading coe� cient is positive or negative.

� e function could be 5th or higher degree because it has � ve branches—down, up, down, up, down—resulting in four extreme points in the domain shown. � e degree of the polynomial must be odd, because for large negative values of x, f (x) becomes larger in the positive direction, whereas for large positive values of x, f (x) becomes larger in the negative direction.

� ere are three real zeros because there are three x-intercepts, none of which is a repeated zero. If the degree is 5, then there are two complex zeros because the total number of zeros must be � ve.

� e leading coe� cient is negative because f (x) becomes very large in the negative direction as x becomes larger in the positive direction.

f(x)

x

Figure 4-3e

Identify the degree and the number of real and nonreal complex zeros that the polynomial function

EXAMPLE 2 ➤

� e function could be 5th or higher degree because it has � ve branches—down, up, down, up, down—resulting

SOLUTION

➤

194

f (x) (x 1)( x 2 5x 2) (x 1) is a factor of f(x) because f(x) equals zero when x 1; coe�cients of the other factor appear in the bottom row of the synthetic substitution.

x 2 5x 2 0 Set the other factor equal to zero.

x 4.5615... and x 0.4384... are also zeros of f (x). Solve using the quadratic formula.

�e graph of f in Figure 4-3d crosses the x-axis at about 1, 0.4, and 4.6, values that agree with the algebraic solutions.

b.

�erefore, 2 is a zero of g (x). g( 2) 0 because it equals the remainder.

g (x) (x 2)( x 2 6x 9) (x 2) is a factor of g(x) because g(x) equals zero when x 2.

g (x) (x 2)(x 3)(x 3) �e second factor can itself be factored.

2, 3, and 3 are zeros of g (x). 3 is called a double zero of g (x).

�e graph of g in Figure 4-3d crosses the x-axis at 2 and touches the x-axis at 3, in agreement with the algebraic solutions.

c.

�erefore, 3 is a zero of h(x). h( 3) 0 because it equals the remainder.

h(x) (x 3)( x 2 7x 18) (x 3) is a factor of h(x) because h(x) equals zero when x 3.


x 7 ____________

7 2 4(1)(18) _________________ 2(1) Use the quadratic formula.

3.5 0.5 ____

23

3.5 2.3979...i or 3.5 2.3979...i Complex solutions are a conjugate pair.

�e graph of h in Figure 4-3d crosses the x-axis only at 3, which agrees with the algebraic solutions.

Notes:

x 3 is called a double zero of f (x). It appears twice, once for each factor (x 3). As you can see from Figure 4-3d, the graph of g just touches the x-axis and does not cross it at x 3. So there are three zeros, 2, 3, and 3, although there are only two distinct x-intercepts.


�e remainder is zero.2 12 18

1 6 9 0


�e remainder is zero.

3 21 541 7 18 0

➤



Additional Class Examples (continued)

2. Show that x 2 5 is a factor of f(x) 5 x 3 2 9 x 2 2 x 1 105. Write f(x) in factored form.Solution:

5 1 29 21 1055 220 2105

1 24 221 0

Synthetically substitute 5 for x. The remainder is zero. So (x 2 5) is a factor of f(x).

[ f(x) 5 (x 2 5)( x 2 2 4x 2 21) (x 2 5) is a factor of f (x) because f(5) 5 0.

5 (x 2 5)(x 1 3)(x 2 7) Factor the quadratic by inspection.

3. Double Zeros: Plot the graph of the quartic function

P(x) 5 (x 1 2)(x 2 1)(x 2 3)(x 2 3)Describe the difference in behavior at the double zero x 5 3 compared to that at the single zeros x 5 22 and x 5 1. Does the number of zeros agree with the corollary to the fundamental theorem of algebra?Solution: The graph in the figure below shows that at the double zero x 5 3, the graph is tangent to the x-axis. That is, the graph touches the x-axis but does not cross it. The graph does cross the x-axis at the single zeros 22 and 1 because P(x) changes sign at those points.Although the graph touches the x-axis only three times, there are still four zeros, counting the double zero at x 5 3. This fact agrees with the corollary to the fundamental theorem of algebra.

3

Double zero40

x

P(x)

Note that a double zero is also called a zero of multiplicity 2. Similarly, a triple zero has multiplicity 3, and so on.


of Example 1, there are three zeros of h(x) but only one x-intercept, as you can see in Figure 4-3d. � e other two zeros are nonreal complex numbers. As you saw in Section 4-2, the two complex zeros are complex conjugates of each other.

� e results of Example 1 illustrate the fundamental theorem of algebra and its corollaries.

PROPERTY: The Fundamental Theorem of Algebra and Its Corollaries

A polynomial function has at least one zero in the set of complex numbers.

CorollaryAn nth-degree polynomial function has exactly n zeros in the set of complex numbers, counting multiple zeros.

CorollaryIf a polynomial has only real coe� cients, then any nonreal complex zeros appear in conjugate pairs.

Be sure you understand what the theorem says. As you saw in Figure 4-2c, the real numbers are a subset of the complex numbers, so the zeros of the function could be nonreal or real complex numbers. � e x-intercepts correspond to the real-number zeros of the function. From now on, these real-number zeros will be referred to as real zeros and nonreal complex zeros wil be referred to as complex zeros.

Identify the degree and the number of real and nonreal complex zeros that the polynomial function in Figure 4-3e could have. State whether the leading coe� cient is positive or negative.

� e function could be 5th or higher degree because it has � ve branches—down, up, down, up, down—resulting in four extreme points in the domain shown. � e degree of the polynomial must be odd, because for large negative values of x, f (x) becomes larger in the positive direction, whereas for large positive values of x, f (x) becomes larger in the negative direction.

� ere are three real zeros because there are three x-intercepts, none of which is a repeated zero. If the degree is 5, then there are two complex zeros because the total number of zeros must be � ve.

� e leading coe� cient is negative because f (x) becomes very large in the negative direction as x becomes larger in the positive direction.

f(x)

x

Figure 4-3e

Identify the degree and the number of real and nonreal complex zeros that the polynomial function

EXAMPLE 2 ➤

� e function could be 5th or higher degree because it has � ve branches—down, up, down, up, down—resulting

SOLUTION

➤

194

f (x) (x 1)( x 2 5x 2) (x 1) is a factor of f(x) because f(x) equals zero when x 1; coe�cients of the other factor appear in the bottom row of the synthetic substitution.


x 4.5615... and x 0.4384... are also zeros of f (x). Solve using the quadratic formula.

�e graph of f in Figure 4-3d crosses the x-axis at about 1, 0.4, and 4.6, values that agree with the algebraic solutions.

b.

�erefore, 2 is a zero of g (x). g( 2) 0 because it equals the remainder.

g (x) (x 2)( x 2 6x 9) (x 2) is a factor of g(x) because g(x) equals zero when x 2.

g (x) (x 2)(x 3)(x 3) �e second factor can itself be factored.

2, 3, and 3 are zeros of g (x). 3 is called a double zero of g (x).

�e graph of g in Figure 4-3d crosses the x-axis at 2 and touches the x-axis at 3, in agreement with the algebraic solutions.

c.

�erefore, 3 is a zero of h(x). h( 3) 0 because it equals the remainder.

h(x) (x 3)( x 2 7x 18) (x 3) is a factor of h(x) because h(x) equals zero when x 3.


x 7 ____________

7 2 4(1)(18) _________________ 2(1) Use the quadratic formula.

3.5 0.5 ____

23

3.5 2.3979...i or 3.5 2.3979...i Complex solutions are a conjugate pair.

�e graph of h in Figure 4-3d crosses the x-axis only at 3, which agrees with the algebraic solutions.

Notes:

x 3 is called a double zero of f (x). It appears twice, once for each factor (x 3). As you can see from Figure 4-3d, the graph of g just touches the x-axis and does not cross it at x 3. So there are three zeros, 2, 3, and 3, although there are only two distinct x-intercepts.


�e remainder is zero.2 12 18

1 6 9 0


�e remainder is zero.

3 21 541 7 18 0

➤


195

4. Find all zeros, real and nonreal, of f(x) 5 x 3 2 4 x 2 1 22x 1 68.Solution: The graph indicates that there is a real zero at x 5 22.

−2

200

x

f(x)

By synthetic substitution of x 5 22, write f(x) in factored form.

22 1 24 22 6822 12 268

1 26 34 0

f(x) 5 (x 1 2)(x2 2 6x 1 34)x 5 3 5i Use the quadratic formula.

Zeros are 22, 3 1 5i, and 3 2 5i.



Plot the graph to help you � nd that x 3 is a zero (Figure 4-3g).

Find the other factors by synthetic substitution.

3 1 13 59 873 30 87

1 10 29 0

f (x) (x 3)( x 2 10x 29)

x 2 10x 29 0 x 10 ____________

100 4 1 29 ___________________ 2 1

10 ____

16 ___________ 2 5 2i

Sum: 3 (5 2i) (5 2i) 13 � e opposite of the quadratic coe� cient.

Pairwise-product sum:

3(5 2i) 3(5 2i) (5 2i)(5 2i) Recall that i 2 1. 15 6i 15 6i 25 4 59 � e linear coe� cient.

Product: 3(5 2i)(5 2i) 3(25 4) 87 � e opposite of the constant term.

Find the particular equation for a cubic function with integer coe� cients if the function’s zeros have the given sum, product, and sum of pairwise products. Con� rm these properties a� er � nding the zeros of the function.

Sum: 5 __ 3 Sum of pairwise products: 58 ___ 3 Product: 40 ___ 3

� e particular equation for one possible function is

y x 3 5 __ 3 x 2 58 ___ 3 x 40 ___ 3

� e simplest particular equation for a function with integer coe� cients is

f (x) 3 x 3 5 x 2 58x 40

By the graph (Figure 4-3h), the zeros are x 5, x 2 __ 3 , and x 4.

Check:

Sum: ( 5) 2 __ 3 4 5 __ 3 (Correct)

Sum of the pairwise products:

( 5) 2 __ 3 ( 5)(4) 2 __ 3 (4) 58 ___ 3 (Correct)

Product: ( 5) 2 __ 3 (4) 40 ___ 3 (Correct)

Plot the graph to help you � nd that(Figure 4-3g).

SOLUTION

x3

50

f(x)

–50

Figure 4-3g

Sum = – b _ a Sum of pairwise products = c _ a Product = – d _ a

➤

Find the particular equation for a cubic function with integer coe� cients if the function’s zeros have the given sum, product, and sum of pairwise products.

EXAMPLE 4 ➤

� e particular equation for one possible function isSOLUTION

f(x)

x45

23

50

Figure 4-3h

➤

196

Sums and Products of ZerosFigure 4-3f shows the graph of the cubic function

f (x) 5 x 3 33 x 2 58x 24

� e three zeros are x z 1 0.6, x z 2 2, and x z 3 4. If you factor out the leading coe� cient, 5, the sum and product of these zeros appear in the equation.

f (x) 5 x 3 33 ___ 5 x 2 58 ___ 5 x 24 ___ 5 5( x 3 6.6 x 2 11.6x 4.8)

z 1 z 2 z 3 0.6 2 4 6.6 � e opposite of the quadratic coe� cient.

z 1 z 2 z 3 (0.6)(2)(4) 4.8 � e opposite of the constant term.

� e sum of the products of the zeros taken two at a time equals the linear coe� cient.

z 1 z 2 z 1 z 3 z 2 z 3 (0.6)(2) (0.6)(4) (2)(4) 11.6 Equal to the linear coe� cient.

To see why these properties hold, start with f (x) in factored form, expand it, and combine like terms without completing the calculations.

f (x) 5(x 0.6)(x 2)(x 4)

5 x 3 0.6 x 2 2 x 2 4 x 2 (0.6)(2)x (0.6)(4)x (2)(4)x (0.6)(2)(4)

5 x 3 (0.6 2 4) x 2 (0.6)(2) (0.6)(4) (2)(4) x (0.6)(2)(4) Opposite of sum Sum of pairwise products Opposite of product

� is property is true, in general, for any cubic function. In Problems 38 and 39, you will extend this property to quadratic functions and to higher-degree functions.

PROPERTY: Sums and Products of the Zeros of a Cubic FunctionIf the cubic function p(x) ax 3 bx 2 cx d has zeros z 1 , z 2 , and z 3 , then

z 1 z 2 z 3 b __ a Sum of the zeros.

z 1 z 2 z 1 z 3 z 2 z 3 c __ a Sum of the pairwise products of the zeros.

z 1 z 2 z 3 d __ a Product of the zeros.

� is property allows you to determine something about the zeros of the function without actually calculating them. It also gives you a way to � nd the particular equation of a cubic function when you know its zeros.

Find the zeros of the function f (x) x 3 13 x 2 59x 87.

� en show that the sum of the zeros, the sum of their pairwise products, and the product of all three zeros correspond to the coe� cients of the equation that de� nes function f.

f(x)

x5

20

20

Figure 4-3f

Find the zeros of the function EXAMPLE 3 ➤



CAS Suggestions

Note that the Solve and Zeros commands give the zeros but not the multiplicities of a function or equation. Only the Factor command reveals repeated zeros.

Th e fi gure shows two equivalent CAS solutions showing that x 5 1 is a zero of p(x). Th e cFactor and cSolve commands make fi nding the nonreal complex factors and zeros of p(x) no more diffi cult than fi nding the real factors and zeros.

Students can use the propFrac command to perform polynomial division. Again, it is important to use a CAS in a way that enhances student learning. Students should understand the structure of polynomial division, using a CAS whenever the algebraic manipulation is not the point of the problem.


Plot the graph to help you � nd that x 3 is a zero (Figure 4-3g).

Find the other factors by synthetic substitution.

3 1 13 59 873 30 87

1 10 29 0

f (x) (x 3)( x 2 10x 29)

x 2 10x 29 0 x 10 ____________

100 4 1 29 ___________________ 2 1

10 ____

16 ___________ 2 5 2i

Sum: 3 (5 2i) (5 2i) 13 � e opposite of the quadratic coe� cient.

Pairwise-product sum:

3(5 2i) 3(5 2i) (5 2i)(5 2i) Recall that i 2 1. 15 6i 15 6i 25 4 59 � e linear coe� cient.

Product: 3(5 2i)(5 2i) 3(25 4) 87 � e opposite of the constant term.

Find the particular equation for a cubic function with integer coe� cients if the function’s zeros have the given sum, product, and sum of pairwise products. Con� rm these properties a� er � nding the zeros of the function.

Sum: 5 __ 3 Sum of pairwise products: 58 ___ 3 Product: 40 ___ 3

� e particular equation for one possible function is

y x 3 5 __ 3 x 2 58 ___ 3 x 40 ___ 3

� e simplest particular equation for a function with integer coe� cients is

f (x) 3 x 3 5 x 2 58x 40

By the graph (Figure 4-3h), the zeros are x 5, x 2 __ 3 , and x 4.

Check:

Sum: ( 5) 2 __ 3 4 5 __ 3 (Correct)

Sum of the pairwise products:

( 5) 2 __ 3 ( 5)(4) 2 __ 3 (4) 58 ___ 3 (Correct)

Product: ( 5) 2 __ 3 (4) 40 ___ 3 (Correct)

Plot the graph to help you � nd that(Figure 4-3g).

SOLUTION

x3

50

f(x)

–50

Figure 4-3g

Sum = – b _ a Sum of pairwise products = c _ a Product = – d _ a

➤

Find the particular equation for a cubic function with integer coe� cients if the function’s zeros have the given sum, product, and sum of pairwise products.

EXAMPLE 4 ➤

� e particular equation for one possible function isSOLUTION

f(x)

x45

23

50

Figure 4-3h

➤

196

Sums and Products of ZerosFigure 4-3f shows the graph of the cubic function

f (x) 5 x 3 33 x 2 58x 24

� e three zeros are x z 1 0.6, x z 2 2, and x z 3 4. If you factor out the leading coe� cient, 5, the sum and product of these zeros appear in the equation.

f (x) 5 x 3 33 ___ 5 x 2 58 ___ 5 x 24 ___ 5 5( x 3 6.6 x 2 11.6x 4.8)

z 1 z 2 z 3 0.6 2 4 6.6 � e opposite of the quadratic coe� cient.

z 1 z 2 z 3 (0.6)(2)(4) 4.8 � e opposite of the constant term.

� e sum of the products of the zeros taken two at a time equals the linear coe� cient.

z 1 z 2 z 1 z 3 z 2 z 3 (0.6)(2) (0.6)(4) (2)(4) 11.6 Equal to the linear coe� cient.

To see why these properties hold, start with f (x) in factored form, expand it, and combine like terms without completing the calculations.

f (x) 5(x 0.6)(x 2)(x 4)

5 x 3 0.6 x 2 2 x 2 4 x 2 (0.6)(2)x (0.6)(4)x (2)(4)x (0.6)(2)(4)

5 x 3 (0.6 2 4) x 2 (0.6)(2) (0.6)(4) (2)(4) x (0.6)(2)(4) Opposite of sum Sum of pairwise products Opposite of product

� is property is true, in general, for any cubic function. In Problems 38 and 39, you will extend this property to quadratic functions and to higher-degree functions.

PROPERTY: Sums and Products of the Zeros of a Cubic FunctionIf the cubic function p(x) ax 3 bx 2 cx d has zeros z 1 , z 2 , and z 3 , then

z 1 z 2 z 3 b __ a Sum of the zeros.

z 1 z 2 z 1 z 3 z 2 z 3 c __ a Sum of the pairwise products of the zeros.

z 1 z 2 z 3 d __ a Product of the zeros.

� is property allows you to determine something about the zeros of the function without actually calculating them. It also gives you a way to � nd the particular equation of a cubic function when you know its zeros.

Find the zeros of the function f (x) x 3 13 x 2 59x 87.

� en show that the sum of the zeros, the sum of their pairwise products, and the product of all three zeros correspond to the coe� cients of the equation that de� nes function f.

f(x)

x5

20

20

Figure 4-3f

Find the zeros of the function EXAMPLE 3 ➤



The sums and products of the zeros of a cubic polynomial property on page 196 can be verified using the command Expand((x2c1)(x2c2)(x2c3)) where c 1 , c 2 , and c 3 are the zeros. The CAS returns the result x 3 2 c 1 x 2 2 c 2 x 2 2 c 3 x 2 1 c 1 c 2 x 1 c 1 c 3 x 1 c 2 c 3 x 2 c 1 c 2 c 3 .

Combining like terms and equating the coefficients gives b 5 2( c 1 1 c 2 1 c 3 ), c 5 ( c 1 c 2 1 c 1 c 3 1 c 2 c 3 ), and d 5 2 c 1 c 2 c 3 , the equivalents of the property for a 5 1. Here, the CAS is used to do the algebra, keeping students focused on the property without losing them in the computation or asking them to simply memorize properties.


6.

7.

8.

For Problems 9–20, sketch the graph of the polynomial function described, or explain why no such function can exist. 9. Cubic function with two distinct negative zeros,

one positive zero, and a positive y-intercept 10. Cubic function with a negative double zero, a

positive zero, and a negative leading coe�cient 11. Cubic function with one real zero, two complex

zeros, and a positive leading coe�cient 12. Cubic function with no real zeros 13. Cubic function with no extreme points 14. Quartic function with no extreme points 15. Quartic function with no real zeros 16. Quartic function with two distinct positive

zeros, two distinct negative zeros, and a negative y-intercept

17. Quartic function with two double zeros 18. Quartic function with two distinct real zeros and

two complex zeros 19. Quartic function with �ve distinct real zeros 20. Quintic function with �ve distinct real zeros

For Problems 21–24, �nd quickly the sum, the product, and the sum of the pairwise products of the zeros from the coe�cients, using the properties in this section. �en �nd the zeros and con�rm that your answers satisfy the properties. 21. f (x) x 3 x 2 22x 40 22. f (x) x 3 x 2 7x 15 23. f (x) 5 x 3 18 x 2 7x 156 24. f (x) 2 x 3 9 x 2 8x 15

For Problems 25–28, �nd a particular equation for the cubic function, with zeros as described and leading coe�cient equal to 1. �en �nd the zeros and con�rm that your answers satisfy the given properties. 25. Sum: 4; sum of the pairwise products: 11;

product: 30 26. Sum: 9; sum of the pairwise products: 26;



product: 10

For Problems 29 and 30, a. By synthetic substitution, �nd p(c). b. Write p(x)

____ x c in mixed-number form. 29. p(x) x 3 7 x 2 5x 4, c 2 and c 3 30. p(x) x 3 9 x 2 2x 5, c 3 and c 2 31. State the remainder theorem. 32. State the factor theorem. 33. State the fundamental theorem of algebra. 34. State the two corollaries of the fundamental

theorem of algebra. 35. Spaceship Problem: Ella Vader (Darth’s other

daughter) is approaching Alderaan in her spaceship. Her distance from the surface, d(x), in kilometers, at time x, in minutes, a�er she starts a maneuver, is given by the function

d(x) x 4 22 x 3 158 x 2 414x 405 a. State the degree of the function, and give the

number of terms.

x

y

x

y

x

y

198

Reading Analysis

From what you have read in this section, what do you consider to be the main idea? How can you determine the degree of a polynomial function by looking at its graph? What are the similarities and di� erences between zeros of a function and x-intercepts? How is the factor theorem related to the remainder theorem? How can you � nd the sum and the product of the zeros of a cubic function simply by looking at the equation?

Quick Review Q1. Sketch the graph of y x 2 . Q2. Sketch the graph of y x 3 . Q3. Sketch the graph of y x 4 . Q4. Sketch the graph of y 2 x . Q5. Multiply the complex numbers and simplify:

(3 2i)(5 6i) Q6. Solve the equation using the quadratic formula:

x 2 14x 54 0 Q7. Solve the equation using the quadratic formula:

x 2 14x 58 0 Q8. Multiply and simplify: (4 7i ) 2 Q9. What type of function has the add–multiply

property? Q10. � e vertex of the graph of the quadratic

function f (x) 3(x 4 ) 2 5 is at A. (3, 4) B. (4, 5) C. (4, 5) D. ( 4, 5) E. ( 4, 5)

1. Given the function p(x) x 3 5 x 2 2x 8, a. Plot the graph using an appropriate

domain. How many branches (increasing or decreasing) does the graph have? How is this number related to the degree of p(x)?

b. Find graphically the three real zeros of p(x). c. Show by synthetic substitution that (x 1)

is a factor of p(x). Use the result to write the other factor, and factor it further if you can.

d. Explain the relationship between the zeros of p(x) and the factors of p(x).

2. Given the function p(x) x 3 3 x 2 9x 13, a. Plot the graph using an appropriate domain.

From the graph, � nd the real zero of p(x). b. Show by synthetic substitution that (x 1) is

a factor of p(x). Write the other factor. c. Find the other two zeros. Substitute one of the

two complex zeros into the equation for p(x) to show that it really is a zero.

d. Explain the relationship between the zeros of p(x) and the graph of p(x).

For the polynomial functions in Problems 3–8, give the minimum possible degree, the number of real zeros the function could have (counting multiple zeros), the number of complex zeros the function could have, and the sign of the leading coe� cient. 3.

4.

5.

5min

x

y

x

y

x

y


Reading Analysis b. Find graphically the three real zeros of p(x).

Problem Set 4-3


PRO B LE M N OTES

Q5. 3 1 28i Q6. x 5 7 i  __

5 Q7. x 5 7 3i Q8. 233 2 56iQ9. Exponential Q10. C

Problems 1–6 model and reinforce the concepts covered in Examples 1 and 2.

Th e Factor and Zeros commands, along with defi ning and evaluating functions, could be used to answer Problems 1–2.1a.

1

8

x

y

Th ree (two up and one down), the same as the degree of the polynomial1b. x 21, 2, 41c. p(x) 5 (x 1 1)(x 2 2)(x 2 4)1d. If c is a zero, then x 2 c is a factor.2a. x 21

2b. p(x) 5 (x 1 1)( x 2 2 4x 1 13) 2c. x 5 2 3i 2d. Th e one real zero corresponds to the one x-intercept.

In Problems 3–8 students are given a graph of a polynomial function and are asked about the leading coeffi cient and the nature of the zeros.3. Sixth-degree; four real distinct zeros; two nonreal complex zeros; positive leading coeffi cient4. Seventh-degree, three real zeros (including a double zero); four nonreal complex zeros; negative leading coeffi cient

1

13

x

y

5. Eighth-degree; six real zeros (including three double zeros); two nonreal complex zeros; positive leading coeffi cient6. Ninth-degree; seven real zeros (including a double zero); two nonreal complex zeros; positive leading coeffi cient

7. Fourth-degree; four real zeros (including a triple zero); no nonreal complex zeros; negative leading coeffi cient

Problems 9–20 ask students to sketch graphs of polynomial functions given information about their zeros, leading coeffi cients, and extreme points. Make sure students understand that the term complex zero is used to mean a nonreal complex zero.

Problems 21–28 provide practice with the properties of the sums and products of zeros demonstrated in Examples 3 and 4.


6.

7.

8.

For Problems 9–20, sketch the graph of the polynomial function described, or explain why no such function can exist. 9. Cubic function with two distinct negative zeros,

one positive zero, and a positive y-intercept 10. Cubic function with a negative double zero, a

positive zero, and a negative leading coe�cient 11. Cubic function with one real zero, two complex

zeros, and a positive leading coe�cient 12. Cubic function with no real zeros 13. Cubic function with no extreme points 14. Quartic function with no extreme points 15. Quartic function with no real zeros 16. Quartic function with two distinct positive

zeros, two distinct negative zeros, and a negative y-intercept

17. Quartic function with two double zeros 18. Quartic function with two distinct real zeros and

two complex zeros 19. Quartic function with �ve distinct real zeros 20. Quintic function with �ve distinct real zeros

For Problems 21–24, �nd quickly the sum, the product, and the sum of the pairwise products of the zeros from the coe�cients, using the properties in this section. �en �nd the zeros and con�rm that your answers satisfy the properties. 21. f (x) x 3 x 2 22x 40 22. f (x) x 3 x 2 7x 15 23. f (x) 5 x 3 18 x 2 7x 156 24. f (x) 2 x 3 9 x 2 8x 15

For Problems 25–28, �nd a particular equation for the cubic function, with zeros as described and leading coe�cient equal to 1. �en �nd the zeros and con�rm that your answers satisfy the given properties. 25. Sum: 4; sum of the pairwise products: 11;




product: 10

For Problems 29 and 30, a. By synthetic substitution, �nd p(c). b. Write p(x)

____ x c in mixed-number form. 29. p(x) x 3 7 x 2 5x 4, c 2 and c 3 30. p(x) x 3 9 x 2 2x 5, c 3 and c 2 31. State the remainder theorem. 32. State the factor theorem. 33. State the fundamental theorem of algebra. 34. State the two corollaries of the fundamental

theorem of algebra. 35. Spaceship Problem: Ella Vader (Darth’s other

daughter) is approaching Alderaan in her spaceship. Her distance from the surface, d(x), in kilometers, at time x, in minutes, a�er she starts a maneuver, is given by the function

d(x) x 4 22 x 3 158 x 2 414x 405 a. State the degree of the function, and give the

number of terms.

x

y

x

y

x

y

198

Reading Analysis

From what you have read in this section, what do you consider to be the main idea? How can you determine the degree of a polynomial function by looking at its graph? What are the similarities and di� erences between zeros of a function and x-intercepts? How is the factor theorem related to the remainder theorem? How can you � nd the sum and the product of the zeros of a cubic function simply by looking at the equation?

Quick Review Q1. Sketch the graph of y x 2 . Q2. Sketch the graph of y x 3 . Q3. Sketch the graph of y x 4 . Q4. Sketch the graph of y 2 x . Q5. Multiply the complex numbers and simplify:

(3 2i)(5 6i) Q6. Solve the equation using the quadratic formula:

x 2 14x 54 0 Q7. Solve the equation using the quadratic formula:

x 2 14x 58 0 Q8. Multiply and simplify: (4 7i ) 2 Q9. What type of function has the add–multiply

property? Q10. � e vertex of the graph of the quadratic

function f (x) 3(x 4 ) 2 5 is at A. (3, 4) B. (4, 5) C. (4, 5) D. ( 4, 5) E. ( 4, 5)

1. Given the function p(x) x 3 5 x 2 2x 8, a. Plot the graph using an appropriate

domain. How many branches (increasing or decreasing) does the graph have? How is this number related to the degree of p(x)?

b. Find graphically the three real zeros of p(x). c. Show by synthetic substitution that (x 1)

is a factor of p(x). Use the result to write the other factor, and factor it further if you can.

d. Explain the relationship between the zeros of p(x) and the factors of p(x).

2. Given the function p(x) x 3 3 x 2 9x 13, a. Plot the graph using an appropriate domain.

From the graph, � nd the real zero of p(x). b. Show by synthetic substitution that (x 1) is

a factor of p(x). Write the other factor. c. Find the other two zeros. Substitute one of the

two complex zeros into the equation for p(x) to show that it really is a zero.

d. Explain the relationship between the zeros of p(x) and the graph of p(x).

For the polynomial functions in Problems 3–8, give the minimum possible degree, the number of real zeros the function could have (counting multiple zeros), the number of complex zeros the function could have, and the sign of the leading coe� cient. 3.

4.

5.

5min

x

y

x

y

x

y


Reading Analysis b. Find graphically the three real zeros of p(x).

Problem Set 4-3

199

27. f (x) 5 x 3 2 8x 2 1 29x 2 52 5 (x 2 4)(x 2 2 2 3i )(x 2 2 1 3i )28. f (x) 5 x 3 1 5x 2 1 4x 2 10 5 (x 2 1)(x 1 3 2 i )(x 1 3 1 i )

Problems 29 and 30 require students to use synthetic substitution to evaluate a polynomial function and to write the quotient of the polynomial and a linear binomial in mixed-number form.

Th e propFrac command will convert the rational functions in Problems 29–30 into mixed-number form.

29a. p(2) 5 26; p(23) 5 2101

29b. p(x)

_____ x 2 2 5 x 2 2 5x 2 5 2 6 _____ x 2 2 ;

p(x)

_____ x 1 3 5 x 2 2 10x 1 35 2 101 _____ x 1 3

30a. p(3) 5 253; p(22) 5 253

30b. p(x)

_____ x 2 3 5 x 2 2 6x 2 16 2 53 _____ x 2 3 ;

p(x)

_____ x 1 2 5 x 2 2 11x 1 24 2 53 _____ x 1 2

Problems 31–34 require students to state the properties studied in this section.31. If p(x) is a polynomial, then p(c) equals the remainder when p(x) is divided by the quantity (x 2 c).32. (x 2 c) is a factor of polynomial p(x) if and only if p(c) 5 0.33. A polynomial function has at least one zero in the set of complex numbers.34. An nth-degree polynomial function has exactly n zeros in the set of complex numbers, counting multiple zeros. If a polynomial has only real coeffi cients, then any nonreal complex zeros appear in conjugate pairs.

Problems 35–37 are real-world applications of polynomial functions. Th ey can be used as group problems or included in the Day 2 homework.35a. Quartic function; fi ve terms

Problems 21–24 can be solved by fi nding the zeros and then manipulating the zeros to fi nd their various combination values.21. Sum 5 1, product 5 240; sum of pairwise products 5 222, x 5 25, 2, 422. Sum 5 21, product 5 15; sum of pairwise products 5 27, x 5 3, 22 i

23. Sum 5 2 18 __ 5 , product 5 156 ___ 5 ; sum of pairwise products 5 2 7 _ 5 , x 5 12 __ 5 , 23 2i

24. Sum 5 9 _ 2 , product 5 2 15 __ 2 ; sum of pairwise products 5 24, x 5 2 3 _ 2 , 1, 5

Problems 25–28 could be solved using a system of equations with the unknown zeros as variables.25. f (x) 5 x 3 2 4x 2 2 11x 1 30 5 (x 1 3)(x 2 2)(x 2 5)26. f (x) 5 x 3 2 9x 2 1 26x 2 24 5 (x 2 2)(x 2 3)(x 2 4)


See pages 991–992 for answers to Problems Q1–Q4 and 8–20.


38. Quadratic Function Sum and Product of Zeros Problem: By the quadratic formula, the zeros of the general quadratic function f (x) a x 2 bx c are

z 1 b ________

b 2 4ac _______________ 2a and

z 2 b ________

b 2 4ac _______________ 2a

By �nding z 1 z 2 and z 1 z 2 , show that there is a property of the sum and product of zeros of a quadratic function that is similar to the corresponding property for cubic functions.

39. Quartic Function Sum and Product of Zeros Problem: By repeated synthetic substitution or using your grapher, �nd the zeros of the function

f (x) 2 x 4 3 x 3 14 x 2 9x 18 �en �nd these quantities:

Sum of the zeros (“products” of the zeros taken one at a time)Sum of all possible products of zeros taken two at a timeSum of all possible products of zeros taken three at a timeProduct of the zeros (“sum” of the products taken all four at a time)

From the results of your calculations, make a conjecture about how the property of the sums and products of the zeros can be extended to functions of degree higher than 3.

40. Reciprocals of the Zeros Problem: Prove that if the function

p(x) a x 3 b x 2 cx d has zeros z 1 , z 2 , and z 3 , then the function

q(x) d x 3 c x 2 bx a has zeros 1 __ z 1 ,

1 __ z 2 , and 1 __ z 3 . 41. Horizontal Translation and Zeros Problem:

Let f (x) x 3 5 x 2 7x 12. Let g (x) be a horizontal translation of f (x) by 1 unit in the positive direction. Find the equation for g (x). Show algebraically that each zero of g (x) is 1 unit larger than the corresponding zero of f (x).

200

b. Plot the graph of function d and sketch the result. Use a window with 0 x 12 and an appropriate range for y.

c. How far was Ella from Alderaan when she started the maneuver? What part of the function tells you this?

d. Function d has an extreme point at a time soon a�er Ella started the maneuver. Find Ella’s distance from Alderaan at this point and the time at which she reached this point. Describe the method you use.

e. Function d has a double zero. Find this zero. Describe what happens to Ella’s spaceship at this zero.

f. Function d has two complex zeros. One of these zeros is x 2 i. What is the other complex zero? Show that 2 i really is a zero of function d.

g. Find the sum of the zeros of d(x), counting the double zero twice. Where does this number appear in the equation of the function?

36. Bent Board Problem: A group of students conducts a project in which they construct a bridge out of a board 10 � long. �ey nail down the board at distances x 0 �, x 4 �, and x 10 � from the le� end of the board. �en one of the group members stands on the board at distance x 3 �. As a result, the board bends into the shape shown in Figure 4-3i. �e vertical scale has been expanded for clarity.

0 4 10x

p(x)

Load here

Figure 4-3i

a. Assume that the board bends into the shape of the graph of a cubic function with zeros at the three support points. Find the equation for p(x), a cubic polynomial function with these three zeros and a leading coe�cient equal to 1.

b. �e actual de�ection, d(x), in inches, of the board from its rest position is a vertical dilation of function p by a constant factor of a. �at is, d(x) a p(x). If d(3) 0.7, �nd the value of a.

c. Explain why the equation for d(x) does not give meaningful answers for values of x less than 0 or greater than 10, and thus why the domain of x is 0 x 10.

d. At what value of x within the domain does the board reach its maximum positive de�ection? How far above the board’s rest position (the x-axis) is the board at this value of x?

37. Bungee Jumping Problem: Lucy Lastic bungee jumps from a high tower. On the way down she passes an elevated walkway on which her friends are standing. Her distance, f (x), in feet above the walkway, is given by the cubic function f (x) x 3 13 x 2 50x 56 where x is the number of seconds that have elapsed since she jumped. Part of the graph of function f is shown in Figure 4-3j.

x

f(x)

Figure 4-3j

a. How high above the platform was Lucy when she jumped? What part of the mathematical model tells you this?

b. By synthetic substitution, show that 2 is a zero of f (x) and thus that Lucy was passing the walkway at time x 2 s. Was she going down or coming up at this time?

c. Find the other zeros of f (x). Show that the sum, the product, and the sum of the pairwise products of these zeros agree with the properties you have learned in this section.

d. Explain why f (x) has the wrong end behavior for large positive values of x.



Problem Notes (continued)

35b.

35c. 405 km as shown by the y-intercept.35d. Th e extreme point is (2.1492..., 47.9782...). So Ella turned around and started moving away from Alderaan at about 2.15 minutes aft er she started the maneuver, when she was at about 48.0 km from the surface.35e. x 5 9 minutes; at this point, the spaceship just touches the surface of Alderaan, and then starts moving away.35f. x 5 2 2 i35g. 22; the opposite of the coeffi cient of the second term.36a. p(x) 5 x(x 2 4)(x 2 10) 5 x 3 2 14 x 2 1 40x 36b. a 5 20.0333... 36c. Th e board begins at x 5 0 and ends at x 5 10, so values of x outside of the domain [0, 10] would give meaningless values of d(x).36d. 2.1890... in. at x 5 7.5725... ft 37a. 56 ft as indicated by the y-intercept.37b. Synthetic substitution shows x 5 2 is a zero of f (x). Lucy was going down at this time.37c. Zeros are x 5 2, x 5 4, and x 5 7; the sum of the zeros is 13; 2 b _ a 5 2 13 ___

21 5 13. Th e product of the zeros is 56; 2 d _ a 5 2 56 ___

21 5 56. Th e sum of the pairwise products is 50; c _ a 5 250 ___

21 5 50.37d. For large positive values of x, f(x) keeps going downward as x increases, whereas Lucy actually comes up again and continues bounces of decreasing amplitude as time goes on.

3 6 12

500

x

d(x)

9

Minimum Doublezero

Problem 38 asks students to derive the sum and product properties of zeros of a quadratic function. Th e results are useful and reinforce the fact that the zeros of a quadratic function are symmetric to the vertical line through the vertex. Th ese properties parallel the corresponding properties for the zeros of a cubic function.

In Problem 38, students can use a CAS to do some of the cumbersome multiplication and addition of the generic forms of roots.


38. Quadratic Function Sum and Product of Zeros Problem: By the quadratic formula, the zeros of the general quadratic function f (x) a x 2 bx c are

z 1 b ________

b 2 4ac _______________ 2a and

z 2 b ________

b 2 4ac _______________ 2a

By �nding z 1 z 2 and z 1 z 2 , show that there is a property of the sum and product of zeros of a quadratic function that is similar to the corresponding property for cubic functions.

39. Quartic Function Sum and Product of Zeros Problem: By repeated synthetic substitution or using your grapher, �nd the zeros of the function

f (x) 2 x 4 3 x 3 14 x 2 9x 18 �en �nd these quantities:

Sum of the zeros (“products” of the zeros taken one at a time)Sum of all possible products of zeros taken two at a timeSum of all possible products of zeros taken three at a timeProduct of the zeros (“sum” of the products taken all four at a time)

From the results of your calculations, make a conjecture about how the property of the sums and products of the zeros can be extended to functions of degree higher than 3.

40. Reciprocals of the Zeros Problem: Prove that if the function

p(x) a x 3 b x 2 cx d has zeros z 1 , z 2 , and z 3 , then the function

q(x) d x 3 c x 2 bx a has zeros 1 __ z 1 ,

1 __ z 2 , and 1 __ z 3 . 41. Horizontal Translation and Zeros Problem:

Let f (x) x 3 5 x 2 7x 12. Let g (x) be a horizontal translation of f (x) by 1 unit in the positive direction. Find the equation for g (x). Show algebraically that each zero of g (x) is 1 unit larger than the corresponding zero of f (x).

200

b. Plot the graph of function d and sketch the result. Use a window with 0 x 12 and an appropriate range for y.

c. How far was Ella from Alderaan when she started the maneuver? What part of the function tells you this?

d. Function d has an extreme point at a time soon a�er Ella started the maneuver. Find Ella’s distance from Alderaan at this point and the time at which she reached this point. Describe the method you use.

e. Function d has a double zero. Find this zero. Describe what happens to Ella’s spaceship at this zero.

f. Function d has two complex zeros. One of these zeros is x 2 i. What is the other complex zero? Show that 2 i really is a zero of function d.

g. Find the sum of the zeros of d(x), counting the double zero twice. Where does this number appear in the equation of the function?

36. Bent Board Problem: A group of students conducts a project in which they construct a bridge out of a board 10 � long. �ey nail down the board at distances x 0 �, x 4 �, and x 10 � from the le� end of the board. �en one of the group members stands on the board at distance x 3 �. As a result, the board bends into the shape shown in Figure 4-3i. �e vertical scale has been expanded for clarity.

0 4 10x

p(x)

Load here

Figure 4-3i

a. Assume that the board bends into the shape of the graph of a cubic function with zeros at the three support points. Find the equation for p(x), a cubic polynomial function with these three zeros and a leading coe�cient equal to 1.

b. �e actual de�ection, d(x), in inches, of the board from its rest position is a vertical dilation of function p by a constant factor of a. �at is, d(x) a p(x). If d(3) 0.7, �nd the value of a.

c. Explain why the equation for d(x) does not give meaningful answers for values of x less than 0 or greater than 10, and thus why the domain of x is 0 x 10.

d. At what value of x within the domain does the board reach its maximum positive de�ection? How far above the board’s rest position (the x-axis) is the board at this value of x?

37. Bungee Jumping Problem: Lucy Lastic bungee jumps from a high tower. On the way down she passes an elevated walkway on which her friends are standing. Her distance, f (x), in feet above the walkway, is given by the cubic function f (x) x 3 13 x 2 50x 56 where x is the number of seconds that have elapsed since she jumped. Part of the graph of function f is shown in Figure 4-3j.

x

f(x)

Figure 4-3j

a. How high above the platform was Lucy when she jumped? What part of the mathematical model tells you this?

b. By synthetic substitution, show that 2 is a zero of f (x) and thus that Lucy was passing the walkway at time x 2 s. Was she going down or coming up at this time?

c. Find the other zeros of f (x). Show that the sum, the product, and the sum of the pairwise products of these zeros agree with the properties you have learned in this section.

d. Explain why f (x) has the wrong end behavior for large positive values of x.


201


1. Let the five zeros of a quintic function f(x) 5 a x 5 1 b x 4 1 c x 3 1 d x 2 1 ex 1 f be z 1 , z 2 , z 3 , z 4 , and z 5 . a. Write an equation for f in factored

form.b. Expand your answer to part a and

use that to determine the values of the sum and product of the zeros of f (x) in terms of a, b, c, d, e, and/or f.

2. Let g(x) 5 c n x n 1 c n21 x n21 1

c n22 x n22 1 . . . 1 c 2 x 2 1 c 1 x 1 c 0 be any nth-degree polynomial. Use your knowledge of the sums and products of the roots of polynomials with degrees 2–5 and use a CAS to determine the following values in terms of the coefficients of g.a. What is the sum of the zeros of g?b. What is the product of the zeros

of g?

3. An old mathematical manuscript about quartic functions was discovered. It listed 5 and 27 as two zeros of a quartic function whose expanded form began y 5 x 4 1 2 x 3 2 31 x 2 1 . . . . Unfortunately, damage to the manuscript obscured the values of the other two zeros as well as the last two terms of the expanded form of the function.a. Determine the values of the other

two zeros.b. What are the coefficients of the last

two terms in the expanded form of the polynomial? 38. z 1 1 z 2 5 2 b __ a ; z 1 z 2 5 c __ a

39. x 5 23, 2 3 _ 2 , 1, 2; (one at a time) 2 3 _ 2 ; (two at a time) 214 ___ 2 ; (three at a time) 2 29 ___ 2 ; (four at a time) 18 __ 2 . Conjecture: For a degree n polynomial a n x n 1 a n21 x n21 1 . . . 1 a 1 x 1 a 0 , the coefficient a k of the degree k term is a n (21) n2k (the sum of the product of roots taken k at a time).

40. For each nonzero root z n of P(x),

z n 3 Q   1 __ z n

5 z n 3 d __ z n

3 1 c __ z n 2 1 b __ z n

1 a

5 d 1 c z n 1 b z n 2 1 a z n

3 5 P   z n 5 0 Because z n is nonzero and z n

3 Q   1 __ z n 5 0, it

follows that Q   1 __ z n 5 0; in other words,   1 __ z n

is a zero of Q(x).41. g(x) 5 x 3 2 8 x 2 1 20x 2 25; zeros of f(x): x 5 4, 1 _ 2 1 _ 2 i  

___ 11 ;

zeros of g(x): x 5 5, 3 _ 2 1 _ 2 i  ___

11


See page 992 for answers to CAS Problems 1–3.

203Section 4-4: Fitting Polynomial Functions to Data

x

y

x

y

Figure 4-4b

Comparing the graphs on the le� in Figure 4-4b and Figure 4-4a, notice that if a function is increasing and concave up, then y is increasing at an increasing rate. If the function is increasing but concave down, then y is still increasing, but at a decreasing rate.

A cubic function has exactly one point of in� ection. Its x-coordinate is given by the formula in the box, a property you will prove in subsequent calculus courses.

PROPERTY: Horizontal Coordinate of a Cubic Function’s Inflection Point

For the cubic function f (x) ax 3 bx 2 cx d, the horizontal coordinate of the point of in� ection is

x b ___ 3a

Note: � e vertex of the quadratic function f (x) ax 2 bx c is at x b __ 2a .

A cubic function f contains the points (6, 38), (5, 74), (2, 50), and ( 1, 80).

a. Find the equation for f algebraically. Check by cubic regression.

b. Verify the answer in part a by plotting and tracing on the graph.

c. Find the x-coordinate of the point of in� ection and show that it agrees with the graph.

a. f (x) a x 3 b x 2 cx d Write the general equation.

216a 36b 6c d 38 Substitute 6 for x and 38 for f(x), and so forth, to get a system of equations. 125a 25b 5c d 74

8a 4b 2c d 50 a b c d 80

216

125 8

1

36

25 4

1

6

5 2

1

1

1 1

1

1

38

74 50

80

2

15

19

44 Solve the system using matrices.

f (x) 2 x 3 15 x 2 19x 44 Cubic regression gives the same equation, with R 2 1.

A cubic functionEXAMPLE 1 ➤

a. ff (fSOLUTION


Fitting Polynomial Functions to Data Figure 4-4a shows four cubic functions that could model the position of a moving object as a function of time. From le� to right in the � gure, the graphs show an object that

y 0) while slowing down (decreasing slope) and then speeds up again (increasing slope)

y 0) going in the negative direction

In all four cases, there is only one x-intercept, indicating that the other two zeros are nonreal complex numbers.

x

y

Stopsmomentarily

x

y

Going back

x

y Coming toward thereference point

Passing thereference point

Figure 4-4a

In this section you will apply curve-� tting techniques to polynomial functions.

Given a set of points, � nd the equation for the polynomial function that � ts

Concavity and Points of In� ectionA smoothly curved graph can have a concave (hollowed out) side and a convex (bulging) side. For reasons you will learn when you study calculus, mathematicians usually refer to the concave side. Figure 4-4b shows regions where the graph is concave up or concave down. � e point at which the direction of concavity changes is called a point of in� ection. � e word in� ection originates from the British spelling, in� exion, meaning “not � exed.”

Fitting Polynomial Functions to Data Figure 4-4a shows four cubic functions that could model the position of a

4 - 4

U.S. speed skater Shani Davis competed in the men’s 1500 m at the 2006 Winter Olympic Games in Torino, Italy, winning the silver medal.

x

y

Slowingdown

Speeding up


Given a set of points, � nd the equation for the polynomial function that � ts Objective


S e c t i o n 4 - 4PL AN N I N G

Class Time1 day

Homework AssignmentRA, Q1–Q10, Problems 1, (3), 4, 5, 7–9,

13, 15, (16)

Teaching ResourcesExploration 4-4a: Fitting a Polynomial

Function to PointsTest 8, Sections 4-1 to 4-4, Forms A and B


Presentation Sketch: Polynomial Fit Present.gsp

TE ACH I N G

Important Terms and ConceptsPoint of infl ectionConcave upConcave downTh ird diff erencesConstant-nth-diff erences propertyDominant highest-degree term

Section Notes

In this section, students fi t polynomial functions to data. If you did not cover Chapter 3, you will need to show students how to enter data in lists, create scatter plots, and run regression analyses.

Shortly aft er Figure 4-4b, there is a paragraph that discusses concavity. A function that increases at an increasing rate is concave up, and a function that increases at a decreasing rate is concave down. Although this language may seem cumbersome, it is an important concept in calculus. Students usually understand it readily, and it is important to include it in your explanation of this material.


x

y

x

y

Figure 4-4b

Comparing the graphs on the le� in Figure 4-4b and Figure 4-4a, notice that if a function is increasing and concave up, then y is increasing at an increasing rate. If the function is increasing but concave down, then y is still increasing, but at a decreasing rate.

A cubic function has exactly one point of in� ection. Its x-coordinate is given by the formula in the box, a property you will prove in subsequent calculus courses.

PROPERTY: Horizontal Coordinate of a Cubic Function’s Inflection Point

For the cubic function f (x) ax 3 bx 2 cx d, the horizontal coordinate of the point of in� ection is

x b ___ 3a

Note: � e vertex of the quadratic function f (x) ax 2 bx c is at x b __ 2a .

A cubic function f contains the points (6, 38), (5, 74), (2, 50), and ( 1, 80).

a. Find the equation for f algebraically. Check by cubic regression.

b. Verify the answer in part a by plotting and tracing on the graph.

c. Find the x-coordinate of the point of in� ection and show that it agrees with the graph.

a. f (x) a x 3 b x 2 cx d Write the general equation.

216a 36b 6c d 38 Substitute 6 for x and 38 for f(x), and so forth, to get a system of equations. 125a 25b 5c d 74

8a 4b 2c d 50 a b c d 80

216

125 8

1

36

25 4

1

6

5 2

1

1

1 1

1

1

38

74 50

80

2

15

19

44 Solve the system using matrices.

f (x) 2 x 3 15 x 2 19x 44 Cubic regression gives the same equation, with R 2 1.

A cubic functionEXAMPLE 1 ➤

a. ff (fSOLUTION


Fitting Polynomial Functions to Data Figure 4-4a shows four cubic functions that could model the position of a moving object as a function of time. From le� to right in the � gure, the graphs show an object that

y 0) while slowing down (decreasing slope) and then speeds up again (increasing slope)

y 0) going in the negative direction

In all four cases, there is only one x-intercept, indicating that the other two zeros are nonreal complex numbers.

x

y

Stopsmomentarily

x

y

Going back

x

y Coming toward thereference point


Figure 4-4a

In this section you will apply curve-� tting techniques to polynomial functions.

Given a set of points, � nd the equation for the polynomial function that � ts

Concavity and Points of In� ectionA smoothly curved graph can have a concave (hollowed out) side and a convex (bulging) side. For reasons you will learn when you study calculus, mathematicians usually refer to the concave side. Figure 4-4b shows regions where the graph is concave up or concave down. � e point at which the direction of concavity changes is called a point of in� ection. � e word in� ection originates from the British spelling, in� exion, meaning “not � exed.”

Fitting Polynomial Functions to Data Figure 4-4a shows four cubic functions that could model the position of a

4 - 4

U.S. speed skater Shani Davis competed in the men’s 1500 m at the 2006 Winter Olympic Games in Torino, Italy, winning the silver medal.

x

y

Slowingdown

Speeding up


Given a set of points, � nd the equation for the polynomial function that � ts Objective

203

Th e property box about the horizontal coordinate of the infl ection point of a cubic function off ers an interesting parallel to the horizontal coordinate of a vertex of a quadratic function. Students should understand that the point of infl ection of a cubic function provides information about the concavity of the function.

Example 1 goes through the steps for fi nding a cubic function containing a given set of points. (In this case, we are looking for a function that fi ts the points exactly.) You may need to review or instruct your students on how to solve a system of equations using matrices. (See Section 13-2.)

Remind students that they can calculate diff erence lists by entering the list into their grapher and using List or

deltaList to calculate the diff erences.

Th e property box aft er Example 1 extends the constant-second-diff erences property of quadratics to all polynomial functions.

Th e data in Example 2 are not as “nice” as those in Example 1. In other words, the points do not fall exactly along the graph of a cubic function. Students will need to use the built-in regression feature of their graphers to fi nd the polynomial that best fi ts the data. Th e example illustrates that looking at the coeffi cient of determination is not suffi cient for choosing the best model. In this case, the quartic model has a greater R 2 -value than the cubic model. However, the quartic function has the wrong endpoint behavior, so the cubic model is the better choice. Use this example to emphasize to students that they should not apply the regression cap abilities of their graphers indiscriminately.

Section 4-4: Fitting Polynomial Functions to Data


a. Figure 4-4d shows the scatter plot. A cubic function is a reasonable mathematical model because it can reverse direction twice (it has two extreme points), as shown by the scatter plot.

b. By cubic regression, the equation is

f (x) 1.5782... x 3 25.0119... x 2 117.2669...x 145.4761...

R 2 0.9611..., indicating a reasonably good � t because it is close to 1. � e graph of f in Figure 4-4e shows that it is a reasonably good � t.

c. From the graph or the table, the x-value when the object passes the reference point going backward is close to 6. Use the zeros, intersect, or solver feature on your grapher.

x 5.8908... 5.9 s

d. Quartic regression gives a coe� cient of determination R 2 0.9861..., which is closer to 1 than the R 2 0.961... from the cubic regression. However, the quartic function would have a third extreme point and would cross the x-axis a fourth time, which wasn’t mentioned in the statement of the problem.

a. Figure 4-4d shows the scatter plot. A cubic function is a reasonable mathematical model because it can reverse direction twice (it has two

SOLUTION

➤

Reading Analysis

From what you have read in this section, what do you consider to be the main idea? How is the constant-second-di� erences property of quadratic functions extended to higher-degree functions? How would you go about � nding the equation for the quartic function that best � ts a given set of data points algebraically? Numerically?

Quick Review Q1. Multiply: (x 3)(x 5) Q2. Multiply and simplify: (3 2i)(5 4i) Q3. Expand: (x 7 ) 2 Q4. Expand and simplify: (5 3i ) 2 Q5. What is the maximum number of extreme

points a quintic function graph can have? Q6. One zero of a particular cubic function with

real-number coe� cients is 7 4i. What is another zero?

Q7. Sketch the graph of a cubic function with a positive double zero and x 3 -coe� cient 2.

Q8. Find the sum of the zeros of the function

f (x) 2 x 3 7 x 2 5x 13 Q9. If polynomial p(x) has remainder 7 when

divided by (x 5), � nd p(5). Q10. If polynomial p(x) has p 3 ___ 2 0, then a factor

of p(x) is A. x 3 B. x 3 C. x 2 D. x 2 E. 2x 3

1. Given p(x) x 3 5 x 2 2x 10, a. Plot the graph using an appropriate domain.

Sketch the result. b. How many zeros does the function have?

How many extreme points does the graph have? How are these numbers related to the degree of p(x)?

c. Make a table of values of p(x) for each integer value of x from 4 to 9. Show that the third di� erences between the p(x)-values are constant.

d. Find the x-coordinate of the point of in� ection, and mark this point on the graph.

5min

Reading Analysis Q8. Find the sum of the zeros of the function

Problem Set 4-4

x

40

2

f(x)

Figure 4-4d

x2

40f(x)

Figure 4-4e

204

b. � e graph is shown in Figure 4-4c. Tracing to x 6, x 5, x 2, and x 1 con� rms that the given points are on the graph. Note that for large positive values of x, the values of f (x) get larger in the negative direction, consistent with the fact that the leading coe� cient is negative.

c. x b ___ 3a 15 ______ 3( 2) 2.5, which agrees with the graph in Figure 4-4c

Note: If you do not recall the matrix solution of linear systems from previous courses, see Section 13-2 for a refresher.

Recall the constant-second-di� erences property of quadratic functions. For a cubic function, the third di� erences between the y-values are constant. � e table shows the result for the function f (x) 2 x 3 15 x 2 19x 44 from Example 1.

x f (x)

1 800 44

3630

121 38

618

122 50

126

123 68

186

124 80

1218

125 74

630

6 3836

PROPERTY: Constant-nth-Differences PropertyFor an nth-degree polynomial function, if the x-values are equally spaced, then the f (x)-values have constant nth di� erences.

You will prove this property in Problem 14.

An object moving in a straight line passes a reference point at time x 2 s. It slows down, stops, reverses direction, and passes the reference point going backward. � en it stops and reverses direction again, passing the reference point a third time. � e table shows its displacements, f (x), in meters, at various times.

a. Make a scatter plot of the data. Explain why a cubic function would be a reasonable mathematical model for displacement as a function of time.

b. Write the equation for the best-� tting cubic function. Plot the graph of the function on the scatter plot in part a.

c. Use the equation in part b to calculate the approximate time the object passes the reference point going backward.

d. Show that a quartic function gives a coe� cient of determination closer to 1 but has the wrong behavior for the given information.

x21

150

5 6

150f(x)

x-2.5

In�ectionpoint

Figure 4-4c➤

An object moving in a straight line passes a reference point at timedown, stops, reverses direction, and passes the reference point going backward.

EXAMPLE 2 ➤

x (s) f(x) (m)

2 03 274 245 136 47 118 69 32



Diff erentiating Instruction• Make sure ELL students understand

the language in Example 2—especially part d.

• Show students how to use the solve feature on their graphers. Th is will help alleviate some of the language diffi culty.

• Students should write in their journals about constant-nth-diff erences in their own words.

• Some students may not have learned matrices to solve systems of equations. Consider giving an overview at this time, or waiting until Chapter 13 to solve equations using matrices.

• In Problem 8, you may need to explain the cultural signifi cance of the Pilgrims. Be aware that Problem 13 is language heavy even though it is review. Take the time to make sure ELL students understand the question.

• Have ELL students do Problem 14 in pairs and write the problem and solution in their journals.

Exploration Notes

Exploration 4-4a requires students to fi nd the equation for a cubic function that fi ts a given set of points, both by using matrices and by using the regression feature of their graphers. Students then factor the equation and show that the zeros of the equation agree with the graph. You can assign this exploration as a group activity to be completed in class or as a homework assignment or review sheet. Allow about 20 minutes to complete this exploration.

Technology Notes

Presentation Sketch: Polynomial Fit Present.gsp at www.keypress.com/keyonline demonstrates how any n points can be fi t to a polynomial of degree n 2 1.

CAS Suggestions

Students can use a CAS to solve Example 1. Consider having them use a system of equations. Some students fi nd it more straightforward to record ordered pairs in lists and use the equation to be solved in its generic form. Solve(y = a x 3 + b x 2 + cx + d, a) | x = {6, 5, 2, –1}

and y = {38, 74, 50, 80} yields a 5 22, b 5 15, c 5 219, and d 5 44.

Alternatively, students can run a cubic regression using the lists of ordered pairs.


a. Figure 4-4d shows the scatter plot. A cubic function is a reasonable mathematical model because it can reverse direction twice (it has two extreme points), as shown by the scatter plot.

b. By cubic regression, the equation is

f (x) 1.5782... x 3 25.0119... x 2 117.2669...x 145.4761...

R 2 0.9611..., indicating a reasonably good � t because it is close to 1. � e graph of f in Figure 4-4e shows that it is a reasonably good � t.

c. From the graph or the table, the x-value when the object passes the reference point going backward is close to 6. Use the zeros, intersect, or solver feature on your grapher.

x 5.8908... 5.9 s

d. Quartic regression gives a coe� cient of determination R 2 0.9861..., which is closer to 1 than the R 2 0.961... from the cubic regression. However, the quartic function would have a third extreme point and would cross the x-axis a fourth time, which wasn’t mentioned in the statement of the problem.

a. Figure 4-4d shows the scatter plot. A cubic function is a reasonable mathematical model because it can reverse direction twice (it has two

SOLUTION

➤

Reading Analysis

From what you have read in this section, what do you consider to be the main idea? How is the constant-second-di� erences property of quadratic functions extended to higher-degree functions? How would you go about � nding the equation for the quartic function that best � ts a given set of data points algebraically? Numerically?

Quick Review Q1. Multiply: (x 3)(x 5) Q2. Multiply and simplify: (3 2i)(5 4i) Q3. Expand: (x 7 ) 2 Q4. Expand and simplify: (5 3i ) 2 Q5. What is the maximum number of extreme

points a quintic function graph can have? Q6. One zero of a particular cubic function with

real-number coe� cients is 7 4i. What is another zero?

Q7. Sketch the graph of a cubic function with a positive double zero and x 3 -coe� cient 2.

Q8. Find the sum of the zeros of the function

f (x) 2 x 3 7 x 2 5x 13 Q9. If polynomial p(x) has remainder 7 when

divided by (x 5), � nd p(5). Q10. If polynomial p(x) has p 3 ___ 2 0, then a factor

of p(x) is A. x 3 B. x 3 C. x 2 D. x 2 E. 2x 3

1. Given p(x) x 3 5 x 2 2x 10, a. Plot the graph using an appropriate domain.

Sketch the result. b. How many zeros does the function have?

How many extreme points does the graph have? How are these numbers related to the degree of p(x)?

c. Make a table of values of p(x) for each integer value of x from 4 to 9. Show that the third di� erences between the p(x)-values are constant.

d. Find the x-coordinate of the point of in� ection, and mark this point on the graph.

5min

Reading Analysis Q8. Find the sum of the zeros of the function

Problem Set 4-4

x

40

2

f(x)

Figure 4-4d

x2

40f(x)

Figure 4-4e

204

b. � e graph is shown in Figure 4-4c. Tracing to x 6, x 5, x 2, and x 1 con� rms that the given points are on the graph. Note that for large positive values of x, the values of f (x) get larger in the negative direction, consistent with the fact that the leading coe� cient is negative.

c. x b ___ 3a 15 ______ 3( 2) 2.5, which agrees with the graph in Figure 4-4c

Note: If you do not recall the matrix solution of linear systems from previous courses, see Section 13-2 for a refresher.

Recall the constant-second-di� erences property of quadratic functions. For a cubic function, the third di� erences between the y-values are constant. � e table shows the result for the function f (x) 2 x 3 15 x 2 19x 44 from Example 1.

x f (x)

1 800 44

3630

121 38

618

122 50

126

123 68

186

124 80

1218

125 74

630

6 3836

PROPERTY: Constant-nth-Differences PropertyFor an nth-degree polynomial function, if the x-values are equally spaced, then the f (x)-values have constant nth di� erences.

You will prove this property in Problem 14.

An object moving in a straight line passes a reference point at time x 2 s. It slows down, stops, reverses direction, and passes the reference point going backward. � en it stops and reverses direction again, passing the reference point a third time. � e table shows its displacements, f (x), in meters, at various times.

a. Make a scatter plot of the data. Explain why a cubic function would be a reasonable mathematical model for displacement as a function of time.

b. Write the equation for the best-� tting cubic function. Plot the graph of the function on the scatter plot in part a.

c. Use the equation in part b to calculate the approximate time the object passes the reference point going backward.

d. Show that a quartic function gives a coe� cient of determination closer to 1 but has the wrong behavior for the given information.

x21

150

5 6

150f(x)

x-2.5

In�ectionpoint

Figure 4-4c➤

An object moving in a straight line passes a reference point at timedown, stops, reverses direction, and passes the reference point going backward.

EXAMPLE 2 ➤

x (s) f(x) (m)

2 03 274 245 136 47 118 69 32


205

PRO B LE M N OTES

Q1. x 2 2 8x 1 15Q2. 7 1 22iQ3. x 2 2 14x 1 49Q4. 16 1 30i Q5. FourQ6. 27 2 4i

Q7.

Q8. 2 7 __ 2 Q9. 7Q10. E

Problems 1 and 2 ask students to explore the zeros, extreme points, constant diff erences of polynomial functions, x-coordinate of a point of infl ection, and concavity of the function.

Th e zeros in Problem 1 are easily found using the cSolve command. 1a., 1d.

1b. Th e function has three zeros (equaling the degree of the function) and two extreme points (one less than the degree of the function).1c. Th e third diff erences all equal 6.1d. x 5 1.6666...

y

x

1

10

�10

x

p(x)Point of in�ection



x (ft)f (x)

(0.001 in.)

0 0

1 116

2 448

3 972

a. Write the equation for the cubic function that �ts the data in the table. Show that the linear and constant coe�cients are zero.

b. How far does the diving board de�ect when the person is standing at its end, x 10 �?

c. Show that the function has another zero but that it is out of the domain of the function.

d. Sketch the graph of the function, showing both vertices. Darken the part of the graph that is in the domain determined by the 10-�-long diving board.

6. Two-Stage Rocket Problem: A two-stage rocket is �red straight up. A�er the �rst stage �nishes �ring, the rocket slows down until the second stage starts �ring. Its altitudes, h(x), in feet above the ground, at each 10 s a�er �ring, are given in the table.

x (s) h(x) (ft)

10 1750

20 3060

30 3510

40 3700

50 4230

60 5700

a. Show that the h(x)-values have constant third di�erences.

b. What type of function will �t the data exactly? Write its equation.

c. Plot the graph of h. Based on the graph, does the rocket start coming back down before the second stage �res? How can you tell?

d. �e �rst stage of the rocket was �red at time x 0. How do you explain the fact that the function in part b has a zero at time x 3 s?

7. Cylinder in a Paraboloid Problem: Figure 4-4g shows the paraboloid formed by rotating about the y-axis the parabola portion of the graph y 16 x 2 that is above the x-axis. A cylinder is inscribed in the paraboloid with its upper rim touching the parabola at the sample point (x, y) in the �rst quadrant. Both x and y are measured in centimeters.

y

x

(x, y)

Figure 4-4g

a. Find the volume of the cylinder as a function of x and y. (Recall that the volume of a cylinder equals the area of the base multiplied by the altitude.) �en substitute 16 x 2 for y to get an equation for v(x), the volume of the cylinder as a function of x alone. Explain why function v is a quartic function of x.

b. Show that function v has four real zeros, two of which are a double zero. Explain why one of the real zeros is outside the real-world domain of this function. A graph might help.

c. Find the value of x that gives the cylinder its maximum volume. What is the maximum volume?

206

2. Given p(x) x 4 6 x 3 6 x 2 12x 11, a. Plot the graph using an appropriate domain.

Sketch the result. b. How many zeros does the function have? How

many extreme points does the graph have? How are these numbers related to the degree of p(x)?

c. Make a table of values of p(x) for each integer value of x from 2 to 4. Show that the fourth di�erences between the p(x)-values are constant.

d. Darken the parts of the graph that are concave down.

3. �e values in the table give the coordinates of points that are on the graph of function f:

x f (x)

2 25.4

3 13.1

4 3.8

5 23.5

6 44.2

7 64.1

a. Make a scatter plot of the points. b. Show that the third di�erences between the

f (x)-values are constant. c. Find algebraically the equation for the cubic

function that �ts the �rst four points. Show that running a cubic regression on all six points gives the same equation. Plot the equation for the function on the same screen as the scatter plot from part a.

d. Find the x-coordinate of the point of in�ection, and mark this point on the graph.

4. �e values in the table give the coordinates of points that are on the graph of function g:

x g(x)

2 33 254 315 276 2217 6478 1425

a. Make a scatter plot of the points. b. Show that the fourth di�erences between the

g (x)-values are constant. c. Find algebraically the equation for the quartic

function that �ts the �rst �ve points. Show that running a quartic regression on all seven points gives the same equation. Plot the equation for the function on the same screen as the scatter plot from part a.

d. Darken the parts of the graph that are concave up.

5. Diving Board Problem: Figure 4-4f shows a diving board de�ected by a person standing on it. �eoretical results on strength of materials indicate that the de�ection of such a cantilever beam below its horizontal rest position at any point x from the built-in end of the beam is a cubic function of x. Suppose that these de�ections, f (x), are measured in thousandths of an inch, where x is measured in feet.

f xx

Figure 4-4f



Problem Notes (continued)2a.

2b. Two zeros (less than the degree of the function) and three extreme points (one less than the degree).2c. The fourth differences all equal 224.2d. Concave down below x 20.3 and above x 3.3.

1

50x

p(x)

Problems 3 and 4 require students to find algebraically the equations of polynomials that fit given data points.3a.

3b. The third differences all equal 1.8.3c. f (x) 5 0.3 x 3 2 5 x 2 1 7x 1 29; Cubic regression gives the same result, with R 2 5 1.

2 7x

f(x)

20

�20 Point of in�ection

3d. The point of inflection is at x 5 5.5555...

1

50x

p(x)

2x

f (x)

20

�207

4a.

4b. The fourth differences all equal 24.

4c. g(x) 5 x 4 2 6 x 3 1 7 x 2 2 8 x 1 17; Quartic regression gives the same answer, with R 2 5 1.

x

g(x)

200

2 7

x

g(x)

2200

7


x (ft)f (x)

(0.001 in.)

0 0

1 116

2 448

3 972

a. Write the equation for the cubic function that �ts the data in the table. Show that the linear and constant coe�cients are zero.

b. How far does the diving board de�ect when the person is standing at its end, x 10 �?

c. Show that the function has another zero but that it is out of the domain of the function.

d. Sketch the graph of the function, showing both vertices. Darken the part of the graph that is in the domain determined by the 10-�-long diving board.

6. Two-Stage Rocket Problem: A two-stage rocket is �red straight up. A�er the �rst stage �nishes �ring, the rocket slows down until the second stage starts �ring. Its altitudes, h(x), in feet above the ground, at each 10 s a�er �ring, are given in the table.

x (s) h(x) (ft)

10 1750

20 3060

30 3510

40 3700

50 4230

60 5700

a. Show that the h(x)-values have constant third di�erences.

b. What type of function will �t the data exactly? Write its equation.

c. Plot the graph of h. Based on the graph, does the rocket start coming back down before the second stage �res? How can you tell?

d. �e �rst stage of the rocket was �red at time x 0. How do you explain the fact that the function in part b has a zero at time x 3 s?

7. Cylinder in a Paraboloid Problem: Figure 4-4g shows the paraboloid formed by rotating about the y-axis the parabola portion of the graph y 16 x 2 that is above the x-axis. A cylinder is inscribed in the paraboloid with its upper rim touching the parabola at the sample point (x, y) in the �rst quadrant. Both x and y are measured in centimeters.

y

x

(x, y)

Figure 4-4g

a. Find the volume of the cylinder as a function of x and y. (Recall that the volume of a cylinder equals the area of the base multiplied by the altitude.) �en substitute 16 x 2 for y to get an equation for v(x), the volume of the cylinder as a function of x alone. Explain why function v is a quartic function of x.

b. Show that function v has four real zeros, two of which are a double zero. Explain why one of the real zeros is outside the real-world domain of this function. A graph might help.

c. Find the value of x that gives the cylinder its maximum volume. What is the maximum volume?

206

2. Given p(x) x 4 6 x 3 6 x 2 12x 11, a. Plot the graph using an appropriate domain.

Sketch the result. b. How many zeros does the function have? How

many extreme points does the graph have? How are these numbers related to the degree of p(x)?

c. Make a table of values of p(x) for each integer value of x from 2 to 4. Show that the fourth di�erences between the p(x)-values are constant.

d. Darken the parts of the graph that are concave down.

3. �e values in the table give the coordinates of points that are on the graph of function f:

x f (x)

2 25.4

3 13.1

4 3.8

5 23.5

6 44.2

7 64.1

a. Make a scatter plot of the points. b. Show that the third di�erences between the

f (x)-values are constant. c. Find algebraically the equation for the cubic

function that �ts the �rst four points. Show that running a cubic regression on all six points gives the same equation. Plot the equation for the function on the same screen as the scatter plot from part a.

d. Find the x-coordinate of the point of in�ection, and mark this point on the graph.

4. �e values in the table give the coordinates of points that are on the graph of function g:

x g(x)

2 33 254 315 276 2217 6478 1425

a. Make a scatter plot of the points. b. Show that the fourth di�erences between the

g (x)-values are constant. c. Find algebraically the equation for the quartic

function that �ts the �rst �ve points. Show that running a quartic regression on all seven points gives the same equation. Plot the equation for the function on the same screen as the scatter plot from part a.

d. Darken the parts of the graph that are concave up.

5. Diving Board Problem: Figure 4-4f shows a diving board de�ected by a person standing on it. �eoretical results on strength of materials indicate that the de�ection of such a cantilever beam below its horizontal rest position at any point x from the built-in end of the beam is a cubic function of x. Suppose that these de�ections, f (x), are measured in thousandths of an inch, where x is measured in feet.

f xx

Figure 4-4f


207

Problem 6 models the position of a moving object and is similar to problems students will encounter in calculus.6a. The third differences all equal 600.6b. Cubic function: h(x) 5 0.1 x 3 2 10.3 x 2 1 370x 2 1020 6c.

There is no vertex, so the rocket does not start descending before the second stage fires.6d. There are at least two possibilities: Either the rocket engine ran for three seconds before the release mechanism was activated or the rocket was launched from the bottom of a deep hole or from a submarine.

Problem 7 asks students to work in three dimensions, finding the radius that produces a cylinder of maximum volume inscribed in a paraboloid. 7a. Volume 5 x 2 y, so v(x) 5 (16 x 2 2 x 4 ); The function is quartic because of the x 4 .7b. v(x) 5 x x (4 2 x)(4 1 x), so zeros are x 5 0, 0, 4, or 24. x 5 0 is a double zero, x 5 24 is a zero out of the domain.7c. About 201.06 c m 3 at x < 2.83 cm.

10

1000 x

h(x)

4d. Concave up below x 0.46 and above x 2.54.

x

g(x)

2

200

7 5a. f(x) 5 24 x 3 1 120x 2 ; Cubic regression gives the same result, with R 2 5 1.

5b. 8 in.5c. Zeros are x 5 0 ft or 30 ft5d.

10 20 30x

f(x)

10,000



Squanto Teaching Pilgrims by Charles Je�erys (�e Granger Collection, New York)

a. Write the equation for the best-�tting cubic function. Plot the equation and the data on the same screen. Sketch the result.

b. According to the cubic model, what is the maximum number of bean plants the Pilgrims had in the �rst year? A�er how many weeks did the number of bean plants reach this maximum? Did any plants survive through the next winter? If so, what was the smallest number of plants, and when did the number reach this minimum? If not, when did the last plant die, and when did the �rst plant emerge the next spring?

c. Show that if B(8) had been 273 instead of 272, the conclusions of part b would be much di�erent. (�is phenomenon is called sensitive dependence on initial conditions.)

d. In what year did the �rst Pilgrims arrive in America? What did they name the place where they landed?

11. River Bend Problem: A river meanders back and forth across Route 66. �ree crossings are 1.7 mi, 3.8 mi, and 5.5 mi east of the intersection of Route 66 and Farm Road 13, or FM 13 (Figure 4-4h).

a. Use the sum and product of zeros properties to �nd quickly an equation of the cubic function y x 3 b x 2 cx d that has 1.7, 3.8, and 5.5 as zeros. What is the y-intercept?

1.7mi

IntersectionRoute 66

FM 13North

3.8 mi5.5 mi

Figure 4-4h

b. Let f (x) be the number of miles north of Route 66 for a point on the river that is distance x, in miles, east of FM 13. Suppose that the river crosses FM 13 at a point 4.1 mi north of the intersection. What negative vertical dilation of the equation in part a would give a cubic function with the correct f (x)-intercept as well as the correct x-intercepts? Write the particular equation for f (x).

c. Plot function f. Sketch the result. What is the graphical signi�cance of the fact that the leading coe�cient is negative?

d. Based on the cubic model, what is the farthest south of Route 66 that the river goes between the 1.7-mi crossing and the 3.8-mi crossing? What is the farthest north of Route 66 that the river goes between the 3.8-mi crossing and the 5.5-mi crossing? How far east of the 5.5-mi crossing would you have to go for the river to be 10 mi south of Route 66?

12. Airplane Payload Problem: �e number of kilograms of “payload” an airplane can carry equals the number of kilograms the wings can li� minus the mass of the airplane, minus the mass of the crew and their equipment. Use these facts to write an equation for the payload as a function of the airplane’s length:

�e plane’s mass is directly proportional to the cube of the plane’s length.�e plane’s li� is directly proportional to the square of the plane’s length.

a. Assume that a plane of a particular design and length L 20 m can li� 2000 kg and has mass 800 kg. Write an equation for the li� and an equation for the mass as functions of L.

208

8. Christmas Tree Problem: A charity organization sells Christmas trees of varying heights. Some prices they charge are shown in the table.

x (ft) p(x) (dollars)

4 58

5 78

6 100

7 130

8 174

a. Use the �rst four data points to �nd algebraically the equation for the cubic function that �ts these data. �en show that the price of an 8-� tree �ts the mathematical model.

b. Plot the graph of function p. Use a window with 0 x 10 and 50 y 300. Sketch the result.

c. How much would you expect to pay for a 20-� tree? Is this surprising?

d. According to your cubic model, trees below a certain height are worthless. At what height does this happen? What part of the mathematical model tells you this?

e. Find the three zeros of this cubic function. Use the result to explain how you know there are no x-intercepts other than the one you found in part d.

f. �e price of relatively short trees increases at a decreasing rate. �e price of relatively tall trees increases at an increasing rate. At what height does this behavior change? Name the feature the graph has at this point.

9. Television Set Pricing: A retail store has various sizes of television sets made by the same manufacturer. �e price of a television set is a function of the screen size, measured in inches along the diagonal. �e prices are listed in the table.

x (in.) p(x) (dollars)

2 160

5 100

7 120

12 250

17 220

21 200

27 340

32 680

35 1100

a. Make a scatter plot of the data. Based on the scatter plot, explain why a quartic function is a more appropriate mathematical model than a cubic function.

b. Write the equation for the best-�tting quartic function, p(x). Plot function p on the same screen as the scatter plot.

c. Based on the quartic model, which size television set is most overpriced?

d. What real-world reason can you think of to explain why the 17-in. and 21-in. sets are less expensive than the smaller, 12-in. set?

10. Pilgrim’s Bean Crop Problem: When the Pilgrims arrived in America, they brought along seeds from which to grow crops. If they had planted beans, the number of bean plants would have increased rapidly, leveled o�, and then decreased with the approach of winter. �e next spring, a new crop would have come up. Suppose that the number of bean plants, B(x), as a function of x, in weeks since the planting, is given by this table:

x B(x)

3 59

4 113

5 160

6 203

7 240

8 272




Problem 8 integrates several skills in a multistep real-world application.

In Problem 8, students can defi ne the function in part a on a CAS. Th en function notation can be used in combination with the cZeros command to solve the rest of the problem.8a. p (x) 5 x 3 2 14 x 2 1 85x 2 122; p (8) 5 174, so the model fi ts the last data point.8b.

5 10

300

x

p (x)

Decreasingrate

Point ofin�ection

Increasingrate

8c. $3978; Surprising!8d. Trees less than 2 ft tall are predicted to be worthless. p(2) 5 0; x-intercept.8e. x 5 2, 6 5i; Cubic functions have only three zeros, so there are no other zeros besides 2 and 6 5i. So there can be no x-intercepts other than x 5 2.8f. Th e point of infl ection, x 5 4.6666..., is where the price behavior changes.9a.

Th e function appears to have three vertices, so it must have degree at least four.

10

500

x

p(x)

9b. p(x) 5 0.0058... x 4 2 0.3289... x 3 1 6.1803... x 2 2 36.6237...x 1 193.0963... with R 2 5 0.9931...

10

500

x

p(x)

9c. Th e 12-in.-model is the most overpriced because its residual is greatest.9d. Answers will vary.

In Problem 10, part c demonstrates the signifi cance of initial conditions in real-world problems.10a. B(x) 5 0.0370... x 3 2 3.2896... x 2 1 75.1812...x 2 137.8333... with R 2 5 0.9999...


Squanto Teaching Pilgrims by Charles Je�erys (�e Granger Collection, New York)

a. Write the equation for the best-�tting cubic function. Plot the equation and the data on the same screen. Sketch the result.

b. According to the cubic model, what is the maximum number of bean plants the Pilgrims had in the �rst year? A�er how many weeks did the number of bean plants reach this maximum? Did any plants survive through the next winter? If so, what was the smallest number of plants, and when did the number reach this minimum? If not, when did the last plant die, and when did the �rst plant emerge the next spring?

c. Show that if B(8) had been 273 instead of 272, the conclusions of part b would be much di�erent. (�is phenomenon is called sensitive dependence on initial conditions.)

d. In what year did the �rst Pilgrims arrive in America? What did they name the place where they landed?

11. River Bend Problem: A river meanders back and forth across Route 66. �ree crossings are 1.7 mi, 3.8 mi, and 5.5 mi east of the intersection of Route 66 and Farm Road 13, or FM 13 (Figure 4-4h).

a. Use the sum and product of zeros properties to �nd quickly an equation of the cubic function y x 3 b x 2 cx d that has 1.7, 3.8, and 5.5 as zeros. What is the y-intercept?

1.7mi

IntersectionRoute 66

FM 13North

3.8 mi5.5 mi

Figure 4-4h

b. Let f (x) be the number of miles north of Route 66 for a point on the river that is distance x, in miles, east of FM 13. Suppose that the river crosses FM 13 at a point 4.1 mi north of the intersection. What negative vertical dilation of the equation in part a would give a cubic function with the correct f (x)-intercept as well as the correct x-intercepts? Write the particular equation for f (x).

c. Plot function f. Sketch the result. What is the graphical signi�cance of the fact that the leading coe�cient is negative?

d. Based on the cubic model, what is the farthest south of Route 66 that the river goes between the 1.7-mi crossing and the 3.8-mi crossing? What is the farthest north of Route 66 that the river goes between the 3.8-mi crossing and the 5.5-mi crossing? How far east of the 5.5-mi crossing would you have to go for the river to be 10 mi south of Route 66?

12. Airplane Payload Problem: �e number of kilograms of “payload” an airplane can carry equals the number of kilograms the wings can li� minus the mass of the airplane, minus the mass of the crew and their equipment. Use these facts to write an equation for the payload as a function of the airplane’s length:

�e plane’s mass is directly proportional to the cube of the plane’s length.�e plane’s li� is directly proportional to the square of the plane’s length.

a. Assume that a plane of a particular design and length L 20 m can li� 2000 kg and has mass 800 kg. Write an equation for the li� and an equation for the mass as functions of L.

208

8. Christmas Tree Problem: A charity organization sells Christmas trees of varying heights. Some prices they charge are shown in the table.

x (ft) p(x) (dollars)

4 58

5 78

6 100

7 130

8 174

a. Use the �rst four data points to �nd algebraically the equation for the cubic function that �ts these data. �en show that the price of an 8-� tree �ts the mathematical model.

b. Plot the graph of function p. Use a window with 0 x 10 and 50 y 300. Sketch the result.

c. How much would you expect to pay for a 20-� tree? Is this surprising?

d. According to your cubic model, trees below a certain height are worthless. At what height does this happen? What part of the mathematical model tells you this?

e. Find the three zeros of this cubic function. Use the result to explain how you know there are no x-intercepts other than the one you found in part d.

f. �e price of relatively short trees increases at a decreasing rate. �e price of relatively tall trees increases at an increasing rate. At what height does this behavior change? Name the feature the graph has at this point.

9. Television Set Pricing: A retail store has various sizes of television sets made by the same manufacturer. �e price of a television set is a function of the screen size, measured in inches along the diagonal. �e prices are listed in the table.

x (in.) p(x) (dollars)

2 160

5 100

7 120

12 250

17 220

21 200

27 340

32 680

35 1100

a. Make a scatter plot of the data. Based on the scatter plot, explain why a quartic function is a more appropriate mathematical model than a cubic function.

b. Write the equation for the best-�tting quartic function, p(x). Plot function p on the same screen as the scatter plot.

c. Based on the quartic model, which size television set is most overpriced?

d. What real-world reason can you think of to explain why the 17-in. and 21-in. sets are less expensive than the smaller, 12-in. set?

10. Pilgrim’s Bean Crop Problem: When the Pilgrims arrived in America, they brought along seeds from which to grow crops. If they had planted beans, the number of bean plants would have increased rapidly, leveled o�, and then decreased with the approach of winter. �e next spring, a new crop would have come up. Suppose that the number of bean plants, B(x), as a function of x, in weeks since the planting, is given by this table:

x B(x)

3 59

4 113

5 160

6 203

7 240

8 272



Th e Expand command can quickly convert the factored form of the cubic function in Problem 11 without resorting to the sum and product of zeros properties of polynomials. For part b, the fi gure shows how the leading coeffi cient of f can be determined on a CAS without requiring the hint. From here, a CAS can easily give the equation in factored or expanded form.

Students using a TI-Nspire can solve Problem 11 part d by placing a point on the graph and dragging it to the maximum, minimum, or zero. Th e TI-Nspire will lock in the points on the graph unless they are dragged off the graph.11a. y 5 x 3 2 11 x 2 1 36.71x 2 35.53; the y-intercept is 235.53 mi.

11b. 2 4.1 mi ______ 35.53 mi ;

y 5 4.1 ____ 35.53 x 3 1 45.1 ____ 35.53 x 2 2 150.511 _____ 35.53 x 1 4.111c.

Th e river goes south before the fi rst crossing and aft er the last.11d. Farthest south is 0.3618... mi (y 5 2 0.3618...) at 2.5676... mi. Farthest north is 0.2508... mi at 4.7656... mi. 22.8568... mi east of the zero at 5.5 mi.

12a. lift 5 5L 2 ; mass 5 0.1 L 3

5.53.8

4.1Farm Road 13

Route 661.7

10

300

x

B(x)

10b. Approximately 375 plants aft er approximately 15.5 weeks. Th e last plant died at about week 38, and the fi rst plant of the next spring sprouted at about week 49.

10c. If B(8) 5 273, then the new best-fi tting cubic function is B(x) 5 0.0833... x 3 2 3.9642... x 2 1 78.3095... x 2 2 142.4285... ( R 2 5 0.9999...). Th e graph no longer has any vertices; the number of beans still keeps increasing.10d. 1620; Plymouth Colony (Plymouth Rock)

211Section 4-5: Rational Functions: Asymptotes and Discontinuities

Rational Functions:Asymptotes and Discontinuities In Section 4–1, you were introduced to rational algebraic functions such as f (x) x 3 5 x 2 8x 6 ____________ x 3 . A rational algebraic function is de� ned to be a function that is a ratio of two polynomials. In this section you will investigate the behavior of such functions at values of x that make the denominator equal zero, such as at x 3 for function f above.


In this exploration you will analyze transformations of the parent reciprocal function.

Rational Functions:Asymptotes and Discontinuities

4 -5


Objective

1. � e � gure shows the graph of the parent reciprocal function, f (x) 1 _ x . Find f (1), f (2), f (3), f 1 _ 2 , and f ( 2). Are these values consistent with the graph? Find f (1000) and f (0.001). Based on the answers, explain why the y-axis is a vertical asymptote and the x-axis is a horizontal asymptote.

5

x

5

10

10

10 5 5

f(x)

10

2. � e � gure shows the graph of function g, a translation of the parent reciprocal function in both the x- and y-directions. Where are the vertical and horizontal asymptotes for function g? What is the magnitude of each translation? Write the particular equation for g (x). Show that the value of g (4) given by the equation agrees with the graph.

x

5

10

10

10 5

g(x)

10

5

5

3. � e � gure shows the graph of function h, a dilation of the parent reciprocal function. Considering the dilation to be vertical, write the equation for h(x). Considering the dilation to be horizontal, write another equation for h(x). Show that the value of h( 6) given by each equation agrees with the graph.

10 5

5

x

h(x)

10

10

5

10

5

1. � e � gure shows the graph of the parent ( )

E X P L O R AT I O N 4 -5: Tr a n s f o r m a t i o n s o f t h e P a r e n t R e c i p r o c a l Fu n c t i o n

continued

210

b. Assume that the crew and their equipment have mass 400 kg. Write the equation for P(L), the payload the plane can carry in kilograms.

c. Make a table of values of P(L) for each 10 m from 0 m to 50 m.

d. Function P is cubic and thus has three zeros. Find these three zeros, and explain what each represents in the real world.

13. Behavior of Polynomial Functions for Large Values of x: Figure 4-4i shows

f (x) x 3 7 x 2 10x 2 (solid) and g (x) x 3 (dashed)

� e graph on the right is zoomed out by a factor of 4 in the x-direction and by a factor of 64 (equal to 4 3 ) in the y-direction.

yg f

x–5

5

5

Figure 4-4i

a. Plot the two graphs on your grapher with a window as shown in the graph on the le� . � en change the window to widen the domain by a factor of 4 and the range by a factor of 64. Does the result resemble the graph on the right in Figure 4-4i?

b. Zoom out again by the same factors. Sketch the resulting graphs.

c. What do you notice about the shapes of the two graphs as you zoom out farther and farther? Can you still see the intercepts and vertices of the graph of f ? What do you think is the reason for saying that the highest-degree term dominates the function for large positive and negative values of x?

14. Constant-nth-Di� erences Proof Project: Let f (x) a x 3 b x 2 cx d.

a. Find algebraically four consecutive values of f (x) for which the x-values are k units apart. � at is, � nd f (x), f (x k), f (x 2k), and f (x 3k). Expand the powers.

b. Show that the third di� erences between the values in part a are independent of x and are equal to 6ak 3 .

c. Let g (x) 5 x 3 11 x 2 13x 19. Find g (3), g (10), g (17), g (24), and g (31). By � nding the third di� erences between consecutive values, show numerically that the conclusion in part b is correct.

15. Coe� cient of Determination Review Problem: a. Enter the data from Example 2 into your

grapher and run the cubic regression. Con� rm that the coe� cient of determination is 0.9611..., as shown in the example.

b. Using the appropriate list features on your grapher, � nd SS res , the sum of the squares of the residual deviations of each data point from the regression curve.

c. Find the mean of the given f (x)-values. Find SS dev , the sum of the squares of the deviations of each data point from this mean value.

d. Recall that the coe� cient of determination is de� ned to be the fraction of SS dev that is removed by the regression. � at is,

R 2 SS dev SS res __________ SS dev

Con� rm that this formula gives 0.9611..., the value found by regression.

16. Journal Problem: Enter into your journal what you have learned so far about higher-degree polynomial functions. Include things such as the shapes of the graphs and constant di� erences and their relationship to the degree of the polynomial.

y

x2020

500 g f




12b. P(L) 5 20.1L 3 1 5L 2 2 400

In Problem 12 part d, students can use a Zeros or Solve command to calculate the zeros of function P.

12d. L 5 10, 20 20  __

2 5 28.2842... m, 10 m, and 48.2842... m; At L 5 10 m, the plane becomes large enough to lift itself and the crew. At L 5 48.2842... m, the plane becomes so large it cannot lift itself. L 5 28.2842... m has no real-world meaning.

Problem 13 illustrates why most of the fl uctuations that create zeros and vertices in the graph of a polynomial function occur relatively near the origin. Knowing that fl uctuations occur near the origin applies to setting the window on a grapher. Th e x-minimum and x-maximum usually can be between 210 and 10, even when the y-minimum and y-maximum are extremely large or small.13a. Th e graphs match.13b.

13c. Both graphs look similar to y 5 x3. Th e vertices and inter cepts of the f graph are hard to see. Th e terms of lower degree do not signifi cantly aff ect the graph for large x.

Problem 14 requires good algebraic skills and is appropriate for group work or as a graded take-home quiz. 14b. Th ird diff erence 5 6a k 3 ; this expression does not involve x.14c. g(3) 5 56, g(10) 5 4,011, g(17) 5 21,588, g(24) 5 63,077, g(31) 5 138,768. Th ird diff erences all equal 10,290. Th is agrees with 6a k 3 5 6(5)(7 ) 3 5 10,290.

20,000

50

y

x

Problem 15 provides a refresher on the coeffi cient of determination.15a. R 2 5 0.9611...15b. SS res 5 67.0324...15c.

_ y 5 10.875; SS dev 5 1724.875

15d. R 2 5 1724.875 2 67.0324... __________________ 1724.875 5 0.9611...16. Answers will vary.


1. Use two diff erent procedures to determine an equation for a quintic function whose graph contains the points (22, 9), (21, 4), (1, 6), (2, 13), (3, 64) and (4, 219). What do you notice about the solution?

2. What is the smallest value of x for which the function in CAS Problem 1 has a y-coordinate of 100?See page 992 for answers to Problems 12c

and 14a, and CAS Problems 1 and 2.


Rational Functions:Asymptotes and Discontinuities In Section 4–1, you were introduced to rational algebraic functions such as f (x) x 3 5 x 2 8x 6 ____________ x 3 . A rational algebraic function is de� ned to be a function that is a ratio of two polynomials. In this section you will investigate the behavior of such functions at values of x that make the denominator equal zero, such as at x 3 for function f above.


In this exploration you will analyze transformations of the parent reciprocal function.

Rational Functions:Asymptotes and Discontinuities

4 -5


Objective

1. � e � gure shows the graph of the parent reciprocal function, f (x) 1 _ x . Find f (1), f (2), f (3), f 1 _ 2 , and f ( 2). Are these values consistent with the graph? Find f (1000) and f (0.001). Based on the answers, explain why the y-axis is a vertical asymptote and the x-axis is a horizontal asymptote.

5

x

5

10

10

10 5 5

f(x)

10

2. � e � gure shows the graph of function g, a translation of the parent reciprocal function in both the x- and y-directions. Where are the vertical and horizontal asymptotes for function g? What is the magnitude of each translation? Write the particular equation for g (x). Show that the value of g (4) given by the equation agrees with the graph.

x

5

10

10

10 5

g(x)

10

5

5

3. � e � gure shows the graph of function h, a dilation of the parent reciprocal function. Considering the dilation to be vertical, write the equation for h(x). Considering the dilation to be horizontal, write another equation for h(x). Show that the value of h( 6) given by each equation agrees with the graph.

10 5

5

x

h(x)

10

10

5

10

5

1. � e � gure shows the graph of the parent ( )

E X P L O R AT I O N 4 -5: Tr a n s f o r m a t i o n s o f t h e P a r e n t R e c i p r o c a l Fu n c t i o n

continued

210

b. Assume that the crew and their equipment have mass 400 kg. Write the equation for P(L), the payload the plane can carry in kilograms.

c. Make a table of values of P(L) for each 10 m from 0 m to 50 m.

d. Function P is cubic and thus has three zeros. Find these three zeros, and explain what each represents in the real world.

13. Behavior of Polynomial Functions for Large Values of x: Figure 4-4i shows

f (x) x 3 7 x 2 10x 2 (solid) and g (x) x 3 (dashed)

� e graph on the right is zoomed out by a factor of 4 in the x-direction and by a factor of 64 (equal to 4 3 ) in the y-direction.

yg f

x–5

5

5

Figure 4-4i

a. Plot the two graphs on your grapher with a window as shown in the graph on the le� . � en change the window to widen the domain by a factor of 4 and the range by a factor of 64. Does the result resemble the graph on the right in Figure 4-4i?

b. Zoom out again by the same factors. Sketch the resulting graphs.

c. What do you notice about the shapes of the two graphs as you zoom out farther and farther? Can you still see the intercepts and vertices of the graph of f ? What do you think is the reason for saying that the highest-degree term dominates the function for large positive and negative values of x?

14. Constant-nth-Di� erences Proof Project: Let f (x) a x 3 b x 2 cx d.

a. Find algebraically four consecutive values of f (x) for which the x-values are k units apart. � at is, � nd f (x), f (x k), f (x 2k), and f (x 3k). Expand the powers.

b. Show that the third di� erences between the values in part a are independent of x and are equal to 6ak 3 .

c. Let g (x) 5 x 3 11 x 2 13x 19. Find g (3), g (10), g (17), g (24), and g (31). By � nding the third di� erences between consecutive values, show numerically that the conclusion in part b is correct.

15. Coe� cient of Determination Review Problem: a. Enter the data from Example 2 into your

grapher and run the cubic regression. Con� rm that the coe� cient of determination is 0.9611..., as shown in the example.

b. Using the appropriate list features on your grapher, � nd SS res , the sum of the squares of the residual deviations of each data point from the regression curve.

c. Find the mean of the given f (x)-values. Find SS dev , the sum of the squares of the deviations of each data point from this mean value.

d. Recall that the coe� cient of determination is de� ned to be the fraction of SS dev that is removed by the regression. � at is,

R 2 SS dev SS res __________ SS dev

Con� rm that this formula gives 0.9611..., the value found by regression.

16. Journal Problem: Enter into your journal what you have learned so far about higher-degree polynomial functions. Include things such as the shapes of the graphs and constant di� erences and their relationship to the degree of the polynomial.

y

x2020

500 g f


211


Class Time2 days

Homework AssignmentDay 1: RA, Q1–Q10, Problems 1–5, 7, 9,

13–21 oddDay 2: Problems 23–31 odd, 35, 36

Teaching ResourcesExploration 4-5: Transformations of the

Parent Reciprocal FunctionExploration 4-5a: Rational Functions and

Discontinuities


CAS Activity 4-5a: End Behavior ofa Rational Function

TE ACH I N G

Important Terms and ConceptsParent reciprocal functionRational algebraic function (rational

function)Rational algebraic fractionProper rational algebraic fractionDiscontinuityRemovable discontinuityVertical asymptoteImproper rational algebraic fractionEnd behavior

Exploration Notes

Exploration 4-5 presents a series of graphs which are transformations of the parent reciprocal function f(x) 5 1 _ x . Students are asked to fi nd the equations of the transformed graph. Problem 4 is the most challenging in the exploration but if students have done Problems 1–3 thoughtfully then Problem 4 should not be too diffi cult. Allow 20–25 minutes to complete this exploration.

See page 215 for notes on additional explorations.

1. f(1) 5 1, f(2) 5 1 _ 2 , f(3) 5 1 _ 3 , f   1 _ 2 5 2, f(22) 5 2 1 _ 2 . Th ese values agree with the graph. f(1000) 5 0.001, so the x-axis is a horizontal asymptote. f(0.001) 5 1000, so the y-axis is a vertical asymptote.2. Th e vertical asymptote is at x 5 5, for a horizontal translation by 5 units. Th e horizontal asymptote is at y 5 3, for a vertical translation by 3 units.g(x) 5 3 1 1 ____ x 2 5 ; g(4) 5 2, which agrees with the graph.

3. h(x) 5 4 _ x because h(1) 5 4, which is 4 times f(1). h(x) 5 1 ___  x/4 , a horizontal dilation by a factor of 4, which simplifi es to h(x) 5 4 _ x , agreeing with the vertical dilation equation.

h(26) 5 4 _ x 5 4 ___ 26 5 2 2 _ 3 ; h(26) 5 1 ____ (x/4)

5 1 _____ (26/4) 5 1 _____ 2(3/2) 5 2 2 _ 3 , which agrees with

the graph.

Section 4-5: Rational Functions: Asymptotes and Discontinuities


Example 1 shows two interesting things that can happen to a function whose equation contains a proper rational algebraic fraction.

For the function f (x) 3x 9 ___________ x 2 7x 12

2,

a. Simplify the fraction. Describe what happens at values of x that make the denominator equal zero.

b. Use the result from part a to sketch the graph.

a. First, factor the numerator and the denominator:

f (x) 3(x 3) ____________ (x 3)(x 4) 2

If x 3 or x 4, the denominator is zero, so there is no value for f ( 3) or f ( 4). Function f is said to have a discontinuity at these values of x.

If you cancel the (x 3) factors, you get

f (x) 3 ______ (x 4) 2, provided x 3 x 3 because f ( 3) is unde� ned.

b. � e graph in Figure 4-5a shows a horizontal asymptote at y 2, a vertical asymptote at x 4, and a gap (circled) in the graph at x 3.

1010

5

x

5

5

10

5

Removablediscontinuity

10 f(x)

Figure 4-5a

Example 1 shows that two kinds of discontinuity can occur in a function graph at a value of x that makes the denominator zero. � e box gives the de� nition of these kinds of discontinuity.

DEFINITION: DiscontinuitiesIf the denominator of the equation for a function f contains the factor (x c), then f (c) is unde� ned because of division by zero, and the graph has a discontinuity at x c.

If the factor (x c) is canceled out in simplifying the equation, then the discontinuity is a removable discontinuity, and the graph has a gap at that one point.

If the factor (x c) is still present in the denominator a� er all possible simplifying has been done, then f (x) becomes in� nitely large (in the positive or negative direction, or both) as x gets close to c, and the graph has a vertical asymptote at x c.

For the function EXAMPLE 1 ➤

a. First, factor the numerator and SOLUTION

➤


4. �e �gure shows the graph of function c, which is a vertical dilation and a vertical and horizontal translation of the parent reciprocal function. Use the asymptotes to �nd the magnitude of each translation. Find the vertical dilation factor, and explain how you calculated it. Write the equation for c(x), and show that the value of c( 1) agrees with the graph.

5

x

5

10

10

10 5

c(x)

10

5



DEFINITION: Rational Algebraic FunctionA rational algebraic function (or, more simply, a “rational function”) is a function for which f (x) can be expressed as a rational algebraic fraction, that is, as the ratio of two polynomials. �e general equation is

f (x) n(x) ____ d(x) t

where the numerator, n(x), and the denominator, d(x), are both polynomial expressions.

�e de�nition of a rational algebraic fraction is similar to the de�nition of a rational number: “a number that can be expressed as a ratio of two integers.” Just as not all ratios of two numbers are rational numbers, not all ratios of expressions are rational algebraic fractions. For example,

__ 7 ____ log 13 is not a rational number, and

x___ x 1.5 is not a rational algebraic fraction.

Proper Rational Algebraic FractionsRecall that a fraction like 3 _ 7 is called a proper fraction because the numerator is less than the denominator and therefore its value is less than 1. Similarly, if a rational algebraic fraction has a numerator of degree less than the degree of the denominator, then the expression is called a proper rational algebraic fraction. For example, 3x 9 _________ x 2 7x 12 is a proper rational algebraic fraction. If the equation for a rational function f contains an expression of the form constant expression

____________ linear expression , then function f may be a transformation of the parent reciprocal function, as you saw in Problem 4 of Exploration 4-5.


Section Notes

In this section, students work with rational functions, asymptotes, and discontinuities. Th e asymptotes can be vertical, horizontal, oblique, or non-linear. Th e concept of limit is not explicitly discussed until Chapter 16, but it is implied whenever asymptotes are involved. A discontinuity can be either removable or infi nite. Because there are so many subtleties and variations to the graphs of rational algebraic functions, students need to understand rather than memorize this material.

Because division by zero is undefi ned, rational functions are discontinuous at values of x for which the denominator is equal to zero. Example 1 illustrates both an infi nite and a removable discontinuity. Th e name removable discontinuity comes from the fact that the discontinuity could be “removed” by defi ning f(23) 5 5. In this example, be very clear with your students that f(x) 3 ____ x 1 4 1 2. Th e function isf(x) 5 3 ____ x 1 4 1 2 only under the condition that x 3.

Th e behavior of a function near a discontinuity can be described using the idea of a limit. If you mention limits at all, you should keep the discussion very informal and intuitive; the topic will be covered more formally in Chapter 16, and in calculus.

Be alert to students thinking that is a number. Th e language increasing without bound and decreasing without bound creates mental images of vertical asymptotes and helps to avoid the misconception that is a number.

When using a grapher to demonstrate rational functions it is helpful to choose a “friendly window” on some graphers so that removable discontinuities appear as gaps in the function.

4. Th e asymptotes are at x 5 24 and y 5 2. Moving 1 unit to the right, x 5 23, the point on the graph is 3 units above the horizontal asymptote, so the vertical dilation is by a factor of 3. c(x) 5 2 1 3 ____ x + 4 . c(21) 5 2 1 3 _____

21 1 4 5 3, which agrees with the graph.5. Answers will vary.


Example 1 shows two interesting things that can happen to a function whose equation contains a proper rational algebraic fraction.

For the function f (x) 3x 9 ___________ x 2 7x 12

2,

a. Simplify the fraction. Describe what happens at values of x that make the denominator equal zero.

b. Use the result from part a to sketch the graph.

a. First, factor the numerator and the denominator:

f (x) 3(x 3) ____________ (x 3)(x 4) 2

If x 3 or x 4, the denominator is zero, so there is no value for f ( 3) or f ( 4). Function f is said to have a discontinuity at these values of x.

If you cancel the (x 3) factors, you get

f (x) 3 ______ (x 4) 2, provided x 3 x 3 because f ( 3) is unde� ned.

b. � e graph in Figure 4-5a shows a horizontal asymptote at y 2, a vertical asymptote at x 4, and a gap (circled) in the graph at x 3.

1010

5

x

5

5

10

5


10 f(x)

Figure 4-5a

Example 1 shows that two kinds of discontinuity can occur in a function graph at a value of x that makes the denominator zero. � e box gives the de� nition of these kinds of discontinuity.

DEFINITION: DiscontinuitiesIf the denominator of the equation for a function f contains the factor (x c), then f (c) is unde� ned because of division by zero, and the graph has a discontinuity at x c.

If the factor (x c) is canceled out in simplifying the equation, then the discontinuity is a removable discontinuity, and the graph has a gap at that one point.

If the factor (x c) is still present in the denominator a� er all possible simplifying has been done, then f (x) becomes in� nitely large (in the positive or negative direction, or both) as x gets close to c, and the graph has a vertical asymptote at x c.

For the function EXAMPLE 1 ➤

a. First, factor the numerator and SOLUTION

➤


4. �e �gure shows the graph of function c, which is a vertical dilation and a vertical and horizontal translation of the parent reciprocal function. Use the asymptotes to �nd the magnitude of each translation. Find the vertical dilation factor, and explain how you calculated it. Write the equation for c(x), and show that the value of c( 1) agrees with the graph.

5

x

5

10

10

10 5

c(x)

10

5



DEFINITION: Rational Algebraic FunctionA rational algebraic function (or, more simply, a “rational function”) is a function for which f (x) can be expressed as a rational algebraic fraction, that is, as the ratio of two polynomials. �e general equation is

f (x) n(x) ____ d(x) t

where the numerator, n(x), and the denominator, d(x), are both polynomial expressions.

�e de�nition of a rational algebraic fraction is similar to the de�nition of a rational number: “a number that can be expressed as a ratio of two integers.” Just as not all ratios of two numbers are rational numbers, not all ratios of expressions are rational algebraic fractions. For example,

__ 7 ____ log 13 is not a rational number, and

x___ x 1.5 is not a rational algebraic fraction.

Proper Rational Algebraic FractionsRecall that a fraction like 3 _ 7 is called a proper fraction because the numerator is less than the denominator and therefore its value is less than 1. Similarly, if a rational algebraic fraction has a numerator of degree less than the degree of the denominator, then the expression is called a proper rational algebraic fraction. For example, 3x 9 _________ x 2 7x 12 is a proper rational algebraic fraction. If the equation for a rational function f contains an expression of the form constant expression

____________ linear expression , then function f may be a transformation of the parent reciprocal function, as you saw in Problem 4 of Exploration 4-5.

213

The function in Example 2 has neither a removable nor an infinite discontinuity, yet it is a rational algebraic function. The denominator is never equal to 0. Therefore, the function is never undefined and the domain is all real numbers.

In Example 3, synthetic substitution is used to transform functions to mixed-number form. Refer to the Section Notes in Section 4-3 for hints on doing synthetic substitution. In part c the most expedient way to find the values of f(x) and g(x) is to use the table feature. Function g has a quadratic asymptote. Students have become accustomed to seeing only linear asymptotes, so this example provides a new twist.

Emphasize the difference between a proper rational algebraic fraction and an improper rational algebraic fraction. In a rational function f(x) 5 n(x)

___ d(x) , if the degree of n(x) is less than the degree of d(x), then the function has a horizontal asymptote. If the degree of n(x) is greater than or equal to the degree of d(x), the function can be transformed to mixed-number form using synthetic substitution or long division. Transforming a rational function to mixed-number form is helpful when learning integration techniques in calculus.

Nonvertical asymptotes for the graph of a rational function can be linear (oblique), quadratic (Example 3) or any other polynomial function. A rational function can have only one nonvertical asymptote. Otherwise, the graph would fail the vertical line test and, therefore, would not be a function.

The x-intercepts show up in the numerator of the reduced form of the rational function. In Example 4, there is an x-intercept at 2 5 _ 3 .



Improper Rational Algebraic FractionsFunctions f and g are examples of improper rational algebraic fractions:

f (x) x 3 5 x 2 8x 6 ________________ x 3 g (x) x 3 5 x 2 8x 5 ________________ x 3

In an improper rational algebraic fraction, the numerator is of equal or higher degree than the denominator. � e equations of f (x) and g (x) are the same except that the numerators di� er by 1. However, as shown in Figure 4-5c, the graphs of the two functions are quite di� erent. Both functions have a discontinuity at x 3, but the discontinuity for function f is a removable discontinuity while the discontinuity for function g is a vertical asymptote.

x3

10

f(x)


g(x)

Verticalasymptote

x3

10

Figure 4-5c

Example 3 shows you how to analyze functions f and g algebraically to see why their graphs have such di� erent behavior at the discontinuity.

Refer to functions f and g above.

a. Transform functions f and g to mixed-number form using synthetic substitution.

b. Use the results from part a to explain why function f has a removable discontinuity at x 3 but function g has a vertical asymptote at x 3.

c. Make a table showing the values of f (x) and g (x) for x 2, 2.9, 2.99, 3, 3.01, 3.1, and 4. From the results, describe the di� erence in behavior of the two functions for values of x close to 3.

d. Find the exact coordinates of the removable discontinuity in function f.

a. f (x) x 3 5 x 2 8x 6 ________________ x 3 g (x) x 3 5 x 2 8x 5 ________________ x 3

3 1 5 8 6 3 1 5 8 53 6 6 3 6 6

1 2 2 0 1 2 2 1

f (x) x 2 2x 2, provided x 3 g (x) x 2 2x 2 1 _____ x 3

Refer to functions EXAMPLE 3 ➤

SOLUTION


Some proper rational algebraic fractions have denominators that never equal zero. In this case there are no discontinuities in the corresponding function graph, as shown in Example 2.

For the rational function f (x) 5 ________ (x 3) 2 1 2,

a. Explain why there are no discontinuities in the graph.

b. Explain why the graph has a horizontal asymptote at y 2.

c. What feature will the graph of function f have at x 3?

d. Sketch the graph of function f. Verify that your sketch is reasonable by plotting the graph of function f on your grapher. Is the graph concave up or concave down at x 3?

a. Function f has no discontinuities because the denominator (x 3) 2 1 is never less than 1, so it never equals zero.

b. � e graph of function f has a horizontal asymptote at y 2 because the denominator gets very large as x gets large in either the positive or the negative direction. � e algebraic fraction gets very small, causing values of f (x) to get closer and closer to 2.

c. � e denominator has its minimum value, 1, at x 3. At this value of x, f (x) 3. All other values of x give a value of f (x) less than 3, so the graph has a high point at (3, 3).

d. Sketch the graph as in Figure 4-5b. Plotting the graph on your grapher con� rms that your sketch is reasonable. � e graph is concave down at x 3.

10

5

x

5

5 5

AsymptoteConcave down

High point

f(x)

Figure 4-5b

For the rational function EXAMPLE 2 ➤

SOLUTION

➤


Differentiating Instruction• This section is language heavy.

Consider providing an outline or worksheet to help ELL students take notes.

• Go over the mathematical meaning of undefined.

• Discuss the answers to Exploration 4-5, Problem 5 as a class.

• The term proper fraction is not used in all countries and so is not always a part of an ELL student’s vocabulary in their primary language. Make sure they understand what it means.

• Enlarge and pass out Figure 4-5a so students can trace it with their fingers. This will help students understand the concepts being demonstrated.

• Have students write in their journals the definitions of rational function, proper rational function, improper rational function, discontinuity, and removable discontinuity in their own words.

• Make sure students know how to find values of functions on their graphers.

• Have students summarize the properties of this section in their journals.

• Have ELL students do the Reading Analysis by themselves. Consider giving an announced homework quiz over a portion of the Reading Analysis to encourage students to answer the questions thoroughly.

• Consider introducing the word oblique, as slant has common non-mathematical meanings.


Improper Rational Algebraic FractionsFunctions f and g are examples of improper rational algebraic fractions:

f (x) x 3 5 x 2 8x 6 ________________ x 3 g (x) x 3 5 x 2 8x 5 ________________ x 3

In an improper rational algebraic fraction, the numerator is of equal or higher degree than the denominator. � e equations of f (x) and g (x) are the same except that the numerators di� er by 1. However, as shown in Figure 4-5c, the graphs of the two functions are quite di� erent. Both functions have a discontinuity at x 3, but the discontinuity for function f is a removable discontinuity while the discontinuity for function g is a vertical asymptote.

x3

10

f(x)


g(x)

Verticalasymptote

x3

10

Figure 4-5c

Example 3 shows you how to analyze functions f and g algebraically to see why their graphs have such di� erent behavior at the discontinuity.

Refer to functions f and g above.

a. Transform functions f and g to mixed-number form using synthetic substitution.

b. Use the results from part a to explain why function f has a removable discontinuity at x 3 but function g has a vertical asymptote at x 3.

c. Make a table showing the values of f (x) and g (x) for x 2, 2.9, 2.99, 3, 3.01, 3.1, and 4. From the results, describe the di� erence in behavior of the two functions for values of x close to 3.

d. Find the exact coordinates of the removable discontinuity in function f.

a. f (x) x 3 5 x 2 8x 6 ________________ x 3 g (x) x 3 5 x 2 8x 5 ________________ x 3

3 1 5 8 6 3 1 5 8 53 6 6 3 6 6

1 2 2 0 1 2 2 1

f (x) x 2 2x 2, provided x 3 g (x) x 2 2x 2 1 _____ x 3

Refer to functions EXAMPLE 3 ➤

SOLUTION


Some proper rational algebraic fractions have denominators that never equal zero. In this case there are no discontinuities in the corresponding function graph, as shown in Example 2.

For the rational function f (x) 5 ________ (x 3) 2 1 2,

a. Explain why there are no discontinuities in the graph.

b. Explain why the graph has a horizontal asymptote at y 2.

c. What feature will the graph of function f have at x 3?

d. Sketch the graph of function f. Verify that your sketch is reasonable by plotting the graph of function f on your grapher. Is the graph concave up or concave down at x 3?

a. Function f has no discontinuities because the denominator (x 3) 2 1 is never less than 1, so it never equals zero.

b. � e graph of function f has a horizontal asymptote at y 2 because the denominator gets very large as x gets large in either the positive or the negative direction. � e algebraic fraction gets very small, causing values of f (x) to get closer and closer to 2.

c. � e denominator has its minimum value, 1, at x 3. At this value of x, f (x) 3. All other values of x give a value of f (x) less than 3, so the graph has a high point at (3, 3).

d. Sketch the graph as in Figure 4-5b. Plotting the graph on your grapher con� rms that your sketch is reasonable. � e graph is concave down at x 3.

10

5

x

5

5 5

AsymptoteConcave down

High point

f(x)

Figure 4-5b

For the rational function EXAMPLE 2 ➤

SOLUTION

➤

215

Additional Exploration Notes

Exploration 4-5a introduces the concepts of asymptotes and removable discontinuities. Th e function in Problem 5 has both an asymptote and a removable discontinuity. Allow 20–25 minutes to complete this exploration.

Technology Notes

CAS Activity 4-5a: End Behavior of a Rational Function has students explore the information that can be gathered from the mixed-number and improper rational algebraic form of a rational function, reinforcing the idea that diff erent algebraic forms of a function give diff erent information about the behavior of a function. Allow 25–30 minutes.



Example 4 shows the behavior of a rational function in which the numerator and the denominator are polynomials of the same degree.

Plot the graph of the rational function f (x), showing any discontinuities and stating which kind they are.

f (x) 3 x 2 25x 50 _____________ x 2 16x 60

Describe the end behavior (the behavior when x is far away from zero) and explain how this behavior is related to the coe� cients in the rational expression.

f (x) 3 x 2 25x 50 _____________ x 2 16x 60

f (x) (3x 5)(x 10) ______________ (x 10)(x 6) Factor the numerator and denominator.

f (x) 3x 5 ______ x 6 , provided x 10 x 10 because f (10) is unde� ned.

Figure 4-5e shows the graph, with a vertical asymptote at x 6 and a removable discontinuity at x 10.

x


y 310

f(x)

10

10

10x 6

Figure 4-5e

� e end behavior shows the graph approaching the horizontal asymptote y 3 as x gets large in both the positive and the negative directions. � e 3 is equal to the ratio of the leading coe� cients in the numerator and denominator, in both the simpli� ed and the unsimpli� ed fractions.

� e end behavior of a function refers to the behavior of a function near the positive and negative ends of its domain. For the rational functions in this section, the domain extends to in� nity in both directions. � e box on the next page summarizes the end behavior of rational functions.

Plot the graph of the rational functionstating

EXAMPLE 4 ➤

f (f (f x) x) xSOLUTION

➤


b. Function f has a removable discontinuity at x 3 because (x 3) is a factor of both the numerator and the denominator. (Because there is still no value for f (3), you must state x 3.)

Function g has a vertical asymptote at x 3 because (x 3) is still in the denominator a�er the fraction has been simpli�ed.

c. �e table shows that f (x) stays close to 5 when x is close to 3 but that g (x) becomes very large in the negative direction when x is close to 3 on the negative side and very large in the positive direction when x is close to 3 on the positive side.

x f (x) g (x)

2 2 1

2.9 4.61 5.39

2.99 4.9601 95.0399

3 (none) (none)

3.01 5.0401 105.0401

3.1 5.41 15.41

4 10 11

d. Substitute 3 for x in the simpli�ed form of f (x):

y (3 ) 2 2(3) 2 5 You are not allowed to say f (3) 5 because function f is unde�ned at x 3.

�e removable discontinuity is at (3, 5). �e table in part c shows that values of f (x) are close to 5 if x is close to 3.

Note that an interesting feature of function g shows up if you plot the graph of g and the graph of y x 2 2x 2, the polynomial part of the mixed-number form, on the same screen. As shown in Figure 4-5d, the graph of y x 2 2x 2 is a curved asymptote for the graph of g.

y

g(x)

x

10

3

Curvedasymptote

Figure 4-5d

➤


CAS Suggestions

In Example 3, students can use the propFrac command on a CAS to convert functions f and g into mixed-number form without the labor of synthetic substitution.

Th ere are two important points to notice about this result. First, the order of the terms is not the same as that in the text. Second, notice the warning at the bottom of the screen. Whenever possible, a CAS attempts to automatically simplify any expressions. In the case of f, the CAS gives students a warning that the domain of the function might be diff erent (and it actually is!).


Example 4 shows the behavior of a rational function in which the numerator and the denominator are polynomials of the same degree.

Plot the graph of the rational function f (x), showing any discontinuities and stating which kind they are.

f (x) 3 x 2 25x 50 _____________ x 2 16x 60

Describe the end behavior (the behavior when x is far away from zero) and explain how this behavior is related to the coe� cients in the rational expression.

f (x) 3 x 2 25x 50 _____________ x 2 16x 60

f (x) (3x 5)(x 10) ______________ (x 10)(x 6) Factor the numerator and denominator.

f (x) 3x 5 ______ x 6 , provided x 10 x 10 because f (10) is unde� ned.

Figure 4-5e shows the graph, with a vertical asymptote at x 6 and a removable discontinuity at x 10.

x


y 310

f(x)

10

10

10x 6

Figure 4-5e

� e end behavior shows the graph approaching the horizontal asymptote y 3 as x gets large in both the positive and the negative directions. � e 3 is equal to the ratio of the leading coe� cients in the numerator and denominator, in both the simpli� ed and the unsimpli� ed fractions.

� e end behavior of a function refers to the behavior of a function near the positive and negative ends of its domain. For the rational functions in this section, the domain extends to in� nity in both directions. � e box on the next page summarizes the end behavior of rational functions.

Plot the graph of the rational functionstating

EXAMPLE 4 ➤

f (f (f x) x) xSOLUTION

➤


b. Function f has a removable discontinuity at x 3 because (x 3) is a factor of both the numerator and the denominator. (Because there is still no value for f (3), you must state x 3.)

Function g has a vertical asymptote at x 3 because (x 3) is still in the denominator a�er the fraction has been simpli�ed.

c. �e table shows that f (x) stays close to 5 when x is close to 3 but that g (x) becomes very large in the negative direction when x is close to 3 on the negative side and very large in the positive direction when x is close to 3 on the positive side.

x f (x) g (x)

2 2 1

2.9 4.61 5.39

2.99 4.9601 95.0399

3 (none) (none)

3.01 5.0401 105.0401

3.1 5.41 15.41

4 10 11

d. Substitute 3 for x in the simpli�ed form of f (x):

y (3 ) 2 2(3) 2 5 You are not allowed to say f (3) 5 because function f is unde�ned at x 3.

�e removable discontinuity is at (3, 5). �e table in part c shows that values of f (x) are close to 5 if x is close to 3.

Note that an interesting feature of function g shows up if you plot the graph of g and the graph of y x 2 2x 2, the polynomial part of the mixed-number form, on the same screen. As shown in Figure 4-5d, the graph of y x 2 2x 2 is a curved asymptote for the graph of g.

y

g(x)

x

10

3

Curvedasymptote

Figure 4-5d

➤

217

To avoid having the CAS automatically simplify an expression, first define the expression as a function. Even though the function appears to reduce when f(x) is typed, the figure shows that the CAS clearly retains the original expression indicating that f(3) is undefined.

Using the propFrac command to convert a rational function to mixed-number form, students can quickly identify any vertical asymptotes (but not removable discontinuities), and, using the property in the box on page 218, any horizontal asymptote as well. Students might use this command to reveal the features of the graph of the function in Example 4, and to explore various other improper rational functions.



3. What causes a rational algebraic function to have a discontinuity at a particular value of x? What two kinds of discontinuities have you encountered so far?

4. Write the equation for g (x), a transformation of the parent reciprocal function that is a vertical dilation by a factor of 5, a vertical translation by

4 units, and a horizontal translation by 3 units.

For Problems 5–8, function f is a transformation of the parent reciprocal function. Write the vertical dilation and translation and the horizontal translation. Calculate the values of f (x) 1 unit to the le� of the vertical asymptote and 1 unit to the right. Sketch the graph of function f, showing these features. 5. f (x) 3 2 _____ x 4

6. f (x) 2 3 _____ x 4

7. f (x) 4 _____ x 1 3

8. f (x) 1 _____ x 2 3

For Problems 9–22, f (x) equals an improper rational algebraic expression with numerator and denominator of the same degree.

a. Write f (x) in the form p(x) r(x) ___ d(x) .

b. Find the x- and y-coordinates of any removable discontinuities and the location of any vertical asymptotes.

c. Show that the horizontal asymptote is located at the value of y equal to the ratio of the leading coe�cients of the numerator and the denominator.

d. Find the x- and y-intercepts. e. Sketch the graph.

9. f (x) x 3 _____ x 2 10. f (x) x 1 _____ x 3

11. f (x) 2 x _____ x 4 12. f (x) 3 x _____ x 5

13. f (x) 3x 2 ______ x 1 14. f (x) 1 2x ______ x 3

15. f (x) (x 2)(x 7) _____________ (x 7)(x 5)

16. f (x) (x 2)(x 3) ____________ (x 3)(x 4)

17. f (x) x 2 3x 4 ___________ x 2 2x 3

18. f (x) x 2 6x 8 ____________ x 2 9x 20

19. f (x) 3 x 2 5x 28 ____________ x 2 x 20

20. f (x) 2 x 2 3x 20 ______________ x 2 16

21. f (x) x 2 4 ______ x 2 4

22. f (x) x 2 9 ___________ x 2 4x 10

For Problems 23–34, function f is an improper rational algebraic expression with a numerator of higher degree than the denominator.

a. Write f (x) in mixed-number form or polynomial form.


c. Write the equation of any nonvertical asymptotes.


23. f (x) x 2 2x 8 __________ x 1

24. f (x) x 2 6x 8 __________ x 1

25. f (x) x 2 4 ______ x 3

26. f (x) x 2 2x 3 __________ x 2

27. f (x) 2 x 2 11x 12 _____________ x 4

28. f (x) 8 2x x 2 __________ x 2

29. f (x) x 3 x 2 7x 13 ________________ x 2

30. f (x) x 3 5 x 2 2x 10 _________________ x 3

31. f (x) x 3 5 x 2 4x 16 ________________ x 3

32. f (x) x 3 x 2 13x 18 _________________ x 2

33. f (x) x 3 10 x 2 17x 28 ___________________ x 4

34. f (x) x 3 4 x 2 16x 11 __________________ x 1


Reading Analysis

From what you have read in this section, what do you consider to be the main idea? What causes a function to have a discontinuity at a particular value of x? What makes a discontinuity removable, and what makes a discontinuity a vertical asymptote? What operation can you perform on the equation of a rational function to reveal a nonvertical asymptote?

Quick Review Q1. Sketch the graph of a quadratic function. Q2. Sketch the graph of a cubic function with a

double zero at x 3. Q3. Sketch the graph of a cubic function with two

nonreal zeros. Q4. Sketch the graph of a quartic function with

four distinct real zeros and a negative leading coe� cient.

Q5. How many extreme points can the graph of a 7th-degree function have?

Q6. Write the zeros of the function f (x) (3x 5)(2x 7)(x 4).

Q7. How many times does the graph of g (x) (3x 5)( x 2 5x 2) cross the x-axis?

Q8. What transformation of function f gives function g if g (x) f 1 _ 3 x ?

Q9. Multiply and simplify: (3 5i ) 2 Q10. Which of these is not a rational number?

A. 7 B. 3 __ 4 C. ___

25

D. 2 5 E. All are rational numbers.

1. Explain the similarities and di� erences between a rational number and a rational algebraic expression.

2. Explain the similarities and di� erences between a numerical improper fraction and an improper algebraic expression.

5min

Reading Analysis Q5. How many extreme points can the graph of a

Problem Set 4-5

PROPERTY: End Behavior and Nonvertical Asymptotes

A rational function with an improper rational algebraic expression f (x) n(x) ___ d(x) ,

where the degree of n(x) is greater than or equal to the degree of d(x),

can be transformed to the form f (x) p(x) r(x) ___ d(x) , where p(x), remainder

r(x), and denominator d(x) are polynomials and r(x) ___ d(x) is a proper rational

algebraic fraction (that is, r(x) is of lower degree than d(x)). � e remainder

fraction, r(x) ___ d(x) , approaches zero as x gets large in the positive or negative

direction.

� e graph of y p(x) forms a non-vertical asymptote for the graph of

function f. If y p(x) is a nonconstant linear equation, it is called a slant

asymptote. � e graph of function f approaches the graph of this asymptote as

x gets large in the positive or negative direction.

If the degrees of n(x) and d(x) are the same, then f (x) can be transformed to

f (x) k r(x) ___ d(x) , where k is a constant. � en k is a vertical translation and the

line y k is a horizontal asymptote for the graph of function f.


PRO B LE M N OTES

Q1. Q2.

Q3. Q4.

Q5. Six or fewerQ6. x 5 27 ___ 2 , 5 __ 3 , 4Q7. Th ree Q8. Horizontal dilation by a factor of 3Q9. 216 1 30i Q10. E

Th e Reading Analysis questions and Problems 1–4 are all designed to help students understand the concepts associated with rational algebraic functions. 1. A rational number is a number that can be written as a ratio of two integers. A rational algebraic expression is an expression that can be written as a ratio of two polynomials.2. A numerical improper fraction is a fraction with numerator greater than or equal to the denominator. An improper algebraic fraction is a rational algebraic expression with numerator of equal or higher degree than the denominator.3. A rational algebraic function has a discontinuity at any value of x that makes the denominator equal zero. It can be a removable discontinuity or a vertical asymptote.4. g(x) 5 24 1 5 _____ x 2 3

Problems 5–8 involve rational functions where the translations, dilations, and refl ections do not involve additional algebra.

Th e propFrac command converts every function in Problems 9–36 into the form f (x) 5 p(x) 1 r(x)

___ d(x) . It also

y

x

y

x3

y

x

y

x

quickly identifi es the horizontal and slant asymptotes in Problems 23–36.

In Problems 9–22, the rational algebraic functions each have numerator and denominator of the same degree. Encourage your students to do these using paper and pencil fi rst, and then check them on their grapher.

9a. f (x) 5 1 1 25 _____ x 1 2 9b. No removable discontinuities, vertical asymptote: x 5 22.9c. Horizontal asymptote: y 5 1, which equals the ratio of the leading coeffi cients9d. x-intercept: x 5 3; y-intercept: f (0) 5 1.5


3. What causes a rational algebraic function to have a discontinuity at a particular value of x? What two kinds of discontinuities have you encountered so far?

4. Write the equation for g (x), a transformation of the parent reciprocal function that is a vertical dilation by a factor of 5, a vertical translation by

4 units, and a horizontal translation by 3 units.

For Problems 5–8, function f is a transformation of the parent reciprocal function. Write the vertical dilation and translation and the horizontal translation. Calculate the values of f (x) 1 unit to the le� of the vertical asymptote and 1 unit to the right. Sketch the graph of function f, showing these features. 5. f (x) 3 2 _____ x 4

6. f (x) 2 3 _____ x 4

7. f (x) 4 _____ x 1 3

8. f (x) 1 _____ x 2 3

For Problems 9–22, f (x) equals an improper rational algebraic expression with numerator and denominator of the same degree.

a. Write f (x) in the form p(x) r(x) ___ d(x) .


c. Show that the horizontal asymptote is located at the value of y equal to the ratio of the leading coe�cients of the numerator and the denominator.


9. f (x) x 3 _____ x 2 10. f (x) x 1 _____ x 3

11. f (x) 2 x _____ x 4 12. f (x) 3 x _____ x 5

13. f (x) 3x 2 ______ x 1 14. f (x) 1 2x ______ x 3

15. f (x) (x 2)(x 7) _____________ (x 7)(x 5)

16. f (x) (x 2)(x 3) ____________ (x 3)(x 4)

17. f (x) x 2 3x 4 ___________ x 2 2x 3

18. f (x) x 2 6x 8 ____________ x 2 9x 20

19. f (x) 3 x 2 5x 28 ____________ x 2 x 20

20. f (x) 2 x 2 3x 20 ______________ x 2 16

21. f (x) x 2 4 ______ x 2 4

22. f (x) x 2 9 ___________ x 2 4x 10

For Problems 23–34, function f is an improper rational algebraic expression with a numerator of higher degree than the denominator.

a. Write f (x) in mixed-number form or polynomial form.


c. Write the equation of any nonvertical asymptotes.


23. f (x) x 2 2x 8 __________ x 1

24. f (x) x 2 6x 8 __________ x 1

25. f (x) x 2 4 ______ x 3

26. f (x) x 2 2x 3 __________ x 2

27. f (x) 2 x 2 11x 12 _____________ x 4

28. f (x) 8 2x x 2 __________ x 2

29. f (x) x 3 x 2 7x 13 ________________ x 2

30. f (x) x 3 5 x 2 2x 10 _________________ x 3

31. f (x) x 3 5 x 2 4x 16 ________________ x 3

32. f (x) x 3 x 2 13x 18 _________________ x 2

33. f (x) x 3 10 x 2 17x 28 ___________________ x 4

34. f (x) x 3 4 x 2 16x 11 __________________ x 1


Reading Analysis

From what you have read in this section, what do you consider to be the main idea? What causes a function to have a discontinuity at a particular value of x? What makes a discontinuity removable, and what makes a discontinuity a vertical asymptote? What operation can you perform on the equation of a rational function to reveal a nonvertical asymptote?

Quick Review Q1. Sketch the graph of a quadratic function. Q2. Sketch the graph of a cubic function with a

double zero at x 3. Q3. Sketch the graph of a cubic function with two

nonreal zeros. Q4. Sketch the graph of a quartic function with

four distinct real zeros and a negative leading coe� cient.

Q5. How many extreme points can the graph of a 7th-degree function have?

Q6. Write the zeros of the function f (x) (3x 5)(2x 7)(x 4).

Q7. How many times does the graph of g (x) (3x 5)( x 2 5x 2) cross the x-axis?

Q8. What transformation of function f gives function g if g (x) f 1 _ 3 x ?

Q9. Multiply and simplify: (3 5i ) 2 Q10. Which of these is not a rational number?

A. 7 B. 3 __ 4 C. ___

25

D. 2 5 E. All are rational numbers.

1. Explain the similarities and di� erences between a rational number and a rational algebraic expression.

2. Explain the similarities and di� erences between a numerical improper fraction and an improper algebraic expression.

5min

Reading Analysis Q5. How many extreme points can the graph of a

Problem Set 4-5

PROPERTY: End Behavior and Nonvertical Asymptotes

A rational function with an improper rational algebraic expression f (x) n(x) ___ d(x) ,

where the degree of n(x) is greater than or equal to the degree of d(x),

can be transformed to the form f (x) p(x) r(x) ___ d(x) , where p(x), remainder

r(x), and denominator d(x) are polynomials and r(x) ___ d(x) is a proper rational

algebraic fraction (that is, r(x) is of lower degree than d(x)). � e remainder

fraction, r(x) ___ d(x) , approaches zero as x gets large in the positive or negative

direction.

� e graph of y p(x) forms a non-vertical asymptote for the graph of

function f. If y p(x) is a nonconstant linear equation, it is called a slant

asymptote. � e graph of function f approaches the graph of this asymptote as

x gets large in the positive or negative direction.

If the degrees of n(x) and d(x) are the same, then f (x) can be transformed to

f (x) k r(x) ___ d(x) , where k is a constant. � en k is a vertical translation and the

line y k is a horizontal asymptote for the graph of function f.

219

10e.

11a. f (x) 5 21 1 22 _____ x 2 4

11b. No removable discontinuities, vertical asymptote: x 5 4

11c. Horizontal asymptote: y 5 21, which equals the ratio of the leading coeffi cients

11d. x-intercept: x 5 2; y-intercept: f (0) 5 20.5

11e.

12a. f (x) 5 21 1 8 _____ x 1 5

12b. No removable discontinuities, vertical asymptote: x 5 25.

12c. Horizontal asymptote: y 5 21, which equals the ratio of the leading coeffi cients

12d. x-intercept: x 5 3; y-intercept: f (0) 5 0.6

Th e Factor command could be helpful in Problems 17–34.

In Problems 23–34, some of the functions have nonvertical asymptotes.

�5 5x

f (x)

5

10�10

�5

�10

10

�5 5

10

x

f (x)

5

10�10

�5

�10

9e.

10a. f (x) 5 1 1 4 _____ x 2 3 10b. No removable discontinuities, vertical asymptote: x 5 3.10c. Horizontal asymptote: y 5 1, which equals the ratio of the leading coeffi cients10d. x-intercept: x 5 21; y-intercept: f (0) 5 2 1 __ 3

�5 5

10

x

f (x)

5

10�10

�5

�10


See pages 993–997 for answers to Problems 5–8 and 12e–34.

221Section 4-6: Partial Fractions and Operations with Rational Expressions

Partial Fractions and Operations with Rational ExpressionsFrom previous courses, recall how to multiply, divide, add, and subtract fractions:

Multiply: 2 __ 3 5 __ 7 2 5 ____ 3 7 10 ___ 21 Top times top, over bottom times bottom.

Divide: 2 __ 3 5 __ 7 2 __ 3 7 __ 5 14 ___ 15 Multiply by the reciprocal of the second fraction.

Add: 2 __ 3 5 __ 7 2 __ 3 7 __ 7 5 __ 7 3 __ 3 14 ___ 21 15 ___ 21 29 ___ 21 Get a common denominator.

Subtract: 2 __ 3 5 __ 7 14 ___ 21 15 ___ 21 1 ___ 21 Get a common denominator.

In this section you will perform these operations on algebraic fractions. You will also learn about partial fractions, in which you start with the answer to an addition problem and � nd the fractions that were added. You will use these techniques in the next section to solve fractional equations.

Multiplication and DivisionTwo properties of real numbers and a de� nition allow you to multiply and divide rational expressions and simplify the result. � ese properties, stated in the box, can be proved from the � eld axioms in Appendix A.

PROPERTIES: Products of Fractions

Reciprocal of a Product PropertyAlgebraically: If x 0 and y 0, then 1 __ xy 1 __ x 1 __ y .

Verbally: � e reciprocal of a product of two nonzero numbers equals the product of the two reciprocals.

Multiplication Property of Fractions Algebraically: If x 0 and y 0, then ab ___ xy a __ x b __ y .

Verbally: A quotient of two products can be split into a product of two quotients. Or, the other way around: To multiply two fractions, multiply their numerators and multiply their denominators.

Partial Fractions and Operations with Rational Expressions

4 - 6

Objective

4 - 64 - 6


35. Functions f, g, and h have the same numerator but di� erent denominators. Use appropriate algebra to describe the features of each graph in Figures 4-5f to 4-5h, including the location and type of each discontinuity, the x- and y-intercepts, and the location of any slant asymptotes.

a. f (x) (x 4)(x 1)(x 3) __________________ (x 4)(x 3)

10

10

10

y

5

x

5

f

5 5

10

Figure 4-5f

b. g (x) (x 4)(x 1)(x 3) __________________ (x 3)(x 2)

10

10

10

y

5

x

5

g

5 5

10Figure 4-5g

c. h(x) (x 4)(x 1)(x 3) __________________ (x 4)(x 2)

10

10

10

y

5

x

5

h

5 5

10Figure 4-5h

36. Functions f, g, and h have the same numerator but di� erent denominators. Use appropriate algebra to describe the features of each graph in Figures 4-5i to 4-5k, including the location and type of each discontinuity, the x- and y-intercepts, and the location of any slant asymptotes.

a. f (x) x 3 3 x 2 13x 15 __________________ x 2 6x 5

10

10

10

y

5

x

5

f

5 5

10

Figure 4-5i

b. g (x) x 3 3 x 2 13x 15 __________________ x 2 16

y

x

10

10

g

10

10

55

5

5

20

15

Figure 4-5j

c. h(x) x 3 3 x 2 13x 15 _________________ x 2 16

10

10

10

y

5

x

5

h

5 5

10

Figure 4-5k



Problems 35 and 36 are nice comparison problems that show how one seemingly small change in the function results in a huge change in the graph.35a. Th e graph of f is the line y 5 x 2 1 with removable discontinuities at x 5 4 and x 5 23, x-intercept 1, and y-intercept 21.35b. Th ere are vertical asymptotes at x 5 3 and x 5 22, and x-intercepts 4, 1, and 23. Th e y-intercept is g(0) 5 22. g(x) 5 x 2 1 1 26x 2 6 _______ x 2 2 x 2 6 , so the diagonal asymptote has equation y 5 x 2 1.35c. Th ere is a vertical asymptote at x 5 22, a removable discontinuity at x 5 4, and x-intercepts 1 and 23. Th e y-intercept is h(0) 5 21.5. h(x) 5 x 2 3 ____ x 1 2 , if x 4, so the diagonal asymptote has equation y 5 x.36a. Th e graph of f is the line y 5 x 2 3 with removable discontinuities at x 5 21 and x 5 25, and x-intercept 3. Th e y-intercept is f (0) 5 23.36b. Th ere are vertical asymptotes at x 5 24 and x 5 4, and x-intercepts 3, 21, and 25. Th e y-intercept is g(0) 5 15 __ 16 . g(x) 5 x 1 3 1 3x 1 33 _____ x 2 2 16 , so the diagonal asymptote has equation y 5 x 1 3.36c. Th ere are no vertical asymptotes.Th e x-intercepts are 3, 21, and 25. Th e y-intercept is h(0) 5 2 15 __ 16 . h(x) 5 x 1 3 1 229x 2 63 _______ x 2 1 16 , so the diagonal asymptote has equation y 5 x 1 3.


1. When the propFrac command is applied to the rational function in Problem 36b, the result is g(x) 5 x 1 3 1 3(x 1 11)

_______ x 2 2 16 , so y 5 x 1 3 is the end behavior asymptote for g(x). Solve the system of equations

{ y 5 g(x) y 5 x 1 3

What happens to the graph of y 5 g(x) at the point defi ned by the solution to the

system? How could you have identifi ed this point without any additional work by using the results of the propFrac command?

2. Use the results of Problems 35 and 36 to explain how you could have known without solving a system that functions f(x) 5 x 2 2 2x 2 8 and g(x) 5 x 3 2 5 x 2 2 2x 1 25 ____________ x 2 3 never intersect.

See page 997 for answers toCAS Problems 1 and 2.


Partial Fractions and Operations with Rational ExpressionsFrom previous courses, recall how to multiply, divide, add, and subtract fractions:

Multiply: 2 __ 3 5 __ 7 2 5 ____ 3 7 10 ___ 21 Top times top, over bottom times bottom.

Divide: 2 __ 3 5 __ 7 2 __ 3 7 __ 5 14 ___ 15 Multiply by the reciprocal of the second fraction.

Add: 2 __ 3 5 __ 7 2 __ 3 7 __ 7 5 __ 7 3 __ 3 14 ___ 21 15 ___ 21 29 ___ 21 Get a common denominator.

Subtract: 2 __ 3 5 __ 7 14 ___ 21 15 ___ 21 1 ___ 21 Get a common denominator.

In this section you will perform these operations on algebraic fractions. You will also learn about partial fractions, in which you start with the answer to an addition problem and � nd the fractions that were added. You will use these techniques in the next section to solve fractional equations.

Multiplication and DivisionTwo properties of real numbers and a de� nition allow you to multiply and divide rational expressions and simplify the result. � ese properties, stated in the box, can be proved from the � eld axioms in Appendix A.

PROPERTIES: Products of Fractions

Reciprocal of a Product PropertyAlgebraically: If x 0 and y 0, then 1 __ xy 1 __ x 1 __ y .

Verbally: � e reciprocal of a product of two nonzero numbers equals the product of the two reciprocals.

Multiplication Property of Fractions Algebraically: If x 0 and y 0, then ab ___ xy a __ x b __ y .

Verbally: A quotient of two products can be split into a product of two quotients. Or, the other way around: To multiply two fractions, multiply their numerators and multiply their denominators.

Partial Fractions and Operations with Rational Expressions

4 - 6

Objective

4 - 64 - 6


35. Functions f, g, and h have the same numerator but di� erent denominators. Use appropriate algebra to describe the features of each graph in Figures 4-5f to 4-5h, including the location and type of each discontinuity, the x- and y-intercepts, and the location of any slant asymptotes.

a. f (x) (x 4)(x 1)(x 3) __________________ (x 4)(x 3)

10

10

10

y

5

x

5

f

5 5

10

Figure 4-5f

b. g (x) (x 4)(x 1)(x 3) __________________ (x 3)(x 2)

10

10

10

y

5

x

5

g

5 5

10Figure 4-5g

c. h(x) (x 4)(x 1)(x 3) __________________ (x 4)(x 2)

10

10

10

y

5

x

5

h

5 5

10Figure 4-5h

36. Functions f, g, and h have the same numerator but di� erent denominators. Use appropriate algebra to describe the features of each graph in Figures 4-5i to 4-5k, including the location and type of each discontinuity, the x- and y-intercepts, and the location of any slant asymptotes.

a. f (x) x 3 3 x 2 13x 15 __________________ x 2 6x 5

10

10

10

y

5

x

5

f

5 5

10

Figure 4-5i

b. g (x) x 3 3 x 2 13x 15 __________________ x 2 16

y

x

10

10

g

10

10

55

5

5

20

15

Figure 4-5j

c. h(x) x 3 3 x 2 13x 15 _________________ x 2 16

10

10

10

y

5

x

5

h

5 5

10

Figure 4-5k



Class Time1 day

Homework AssignmentRA, Q1–10, Problems 7, 9, 13, 16, 21, 25,

29, 31, 41

Teaching ResourcesExploration 4-6a: Introduction to Partial

Fractions

TE ACH I N G

Important Terms and ConceptsReciprocal of a product propertyMultiplication property of fractionsDefi nition of division of fractionsCanceling in a fractionResolving into partial fractionsHeaviside’s methodHeaviside shortcut

Section Notes

Th is section begins with a quick review of operations with rational expressions and then discusses partial fractions. Partial fractions “undo” the process of adding or subtracting rational expressions. If your students are profi cient at working with rational expressions, then omit the fi rst part of the section and go directly to partial fractions. Include a few rational-expression problems to give your students an opportunity to revisit the material.

223

Addition and SubtractionTo add or subtract two or more rational expressions, such as 3 ____ x 4 5 ____ x 7 , you must � rst � nd a common denominator. � en add the numerators. Example 2 shows how to do this.

Add by � rst � nding a common denominator: 3 _____ x 4 5 _____ x 7

3 _____ x 4 5 _____ x 7

x 7 ______ x 7 3 _____ x 4 5 _____ x 7 x 4 ______ x 4 Multiply each fraction by a “clever” form of 1.

x 7 3 ____________ x 7 x 4 5 x 4 ____________ x 7 x 4 Multiply the numerators. Multiply the denominators.

x 7 3 5 x 4 __________________ x 7 x 4 Add the numerators; keep the common denominator.

3x 21 5x 20 ________________ x 7 x 4 Multiply in the numerator.

8x 1 ___________ x 2 3x 28

Combine like terms. Multiply in the denominator.

Partial Fractions� e process of adding or subtracting two rational expressions can be reversed to � nd the two simpler proper fractions that were added or subtracted. Finding these two fractions from the answer is called resolving into partial fractions.

To resolve 8x 1 _________ x 2 3x 28 into partial fractions, � rst factor the denominator:

8x 1 ___________ x 2 3x 28

8x 1 ____________ x 4 x 7

So the two target fractions will have denominators (x 4) and (x 7). Observe that the � rst fraction is a proper rational algebraic expression—the numerator is of lower degree than the denominator. So the two target fractions will also be proper fractions, meaning that their numerators will be of degree zero—that is, constant. Picking letters such as A and B to stand for these constant numerators, write:

8x 1 ____________ (x 4)(x 7) A _____ x 4 B _____ x 7

Next follow these steps:

8x 1 ______ (x 7) A B _____ x 7 (x 4) Multiply both sides of the equation by (x 4) to remove the discontinuity on the le� and leave the constant A by itself on the right, with no denominator.

8( 4) 1 _________ 4 7 A Substitute 4 for x to make the second term on the right equal zero.

A 33 ____ 11 3 Simplify.

Add by � rst � nding a common denominator: EXAMPLE 2 ➤

3_____x 4SOLUTION

➤

Section 4-6: Partial Fractions and Operations with Rational Expressions222 Chapter 4: Polynomial and Rational Functions

DEFINITION: Division of Fractions Algebraically: If y 0, then x __ y x 1 __ y .

Verbally: Dividing a number by y means multiplying that number by the reciprocal of y.

Applied to two fractions, this de� nition becomes

a __ x b __ y a __ x y __ b , provided x 0, y 0, and b 0

Example 1 shows how to apply these properties to a product of two rational expressions and simplify the result.

Multiply and simplify: x 2 5x 6 __________ x 2 x 20

x 2 3x 4 __________ x 2 x 2

x 2 5x 6 __________ x 2 x 20

x 2 3x 4 __________ x 2 x 2

x 2 5x 6 x 2 3x 4 ________________________ x 2 x 20 x 2 x 2

Numerator times numerator over denominator times denominator.

x 2 x 3 x 1 x 4 _______________________ x 4 x 5 x 1 x 2 Factor the numerator and the denominator.

x 3 x 1 x 2 x 4 _______________________ x 5 x 1 x 2 x 4 Commute factors appropriately.

x 3 _____ x 5 x 1 x 2 x 4 _________________ x 1 x 2 x 4 Use the multiplication property of fractions.

x 3 _____ x 5 , x 1, x 2, x 4 Because the second fraction equals 1.

Note that the restrictions on x must be stated for the factors that cancel, as they were for removable discontinuities. Of course, x cannot equal 5 either, but this is clear from the fact that the term (x 5) still appears in the denominator.

� e process can be simpli� ed if the numerator and denominator involve only multiplication of factors. If you realize this, you can cancel by striking out the common factors in the numerator and denominator without doing the commuting and breaking the rational expression into two fractions.

x 2 x 3 x 1 x 4 _______________________ x 4 x 5 x 1 x 2 x 3 _____ x 5 , x 1, x 2, x 4

For this purpose it is helpful to de� ne canceling in a fraction.

DEFINITION: Canceling in a FractionCanceling in a fraction means dividing the numerator and denominator by the same common factor.

� is de� nition will help you avoid mistakes such as canceling the x’s in a fraction

like x 3 ____ x 5 . Because canceling means division, which distributes over addition and

subtraction, if you cancel the x’s the result is 1 3 __ x

____ 1 5 __ x

, which is by no means simpler!

Multiply and simplify: EXAMPLE 1 ➤

x__________x 2SOLUTION

➤



Emphasize these points in your discussion:

• The definition of division of fractions provides the justification for how two fractions are divided. Students may have memorized a _ x b _ y 5 a _ x y _ b provided x 0, y 0, and b 0 without understanding that this is true because y _ b is the reciprocal of b _ y .

• Remind students that properties are “two-way streets.” Many students know that 1 _ x 1 _ y 5 1 __ xy , but many do not recognize that 1 __ xy 5 1 _ x 1 _ y . Make a point of writing the properties both ways.

• In Example 1, keep stressing to your students that x 2 1 ____ x 2 1 is a form of 1. Likewise, x 1 2 ____ x 1 2 and x 1 4 ____ x 1 4 are other forms of 1 dressed up in different costumes. We can “cancel” these forms of 1 because of the multiplication identity property.

• In Example 2, the justification for adding fractions once you find a common denominator is the distributive property, where the common factor is 1 _________  x 2 7  x 1 4 .

Note: The reciprocal of a product property is provable from the field axioms, and the multiplication property of fractions is provable from the reciprocal of a product property.

If the denominator of a rational expression is the product of linear factors, then you can rewrite the expression as the sum of partial fractions with linear denominators. Partial fractions are introduced in this chapter to help with the graphing of rational algebraic functions. This skill is also helpful in calculus when integrating rational algebraic expressions with a product in the denominator. In addition, they appear in some differential equation problems.

Example 3 shows two ways to resolve a fraction into its partial fractions. Heaviside’s method of resolving partial fractions is more tedious than the Heaviside shortcut but it is well worth doing. Students will remember the “long” way much more easily than the shortcut. Make sure your students can do partial fraction problems both the long way and using Heaviside’s shortcut.

The last four problems in Problem Set 4-6 show how to resolve a fractional expression when the denominator is not a product of linear factors, and when there is a repeated linear factor in the denominator. Note that the Heaviside shortcut does not work for these problems.

223

Addition and SubtractionTo add or subtract two or more rational expressions, such as 3 ____ x 4 5 ____ x 7 , you must � rst � nd a common denominator. � en add the numerators. Example 2 shows how to do this.

Add by � rst � nding a common denominator: 3 _____ x 4 5 _____ x 7

3 _____ x 4 5 _____ x 7

x 7 ______ x 7 3 _____ x 4 5 _____ x 7 x 4 ______ x 4 Multiply each fraction by a “clever” form of 1.

x 7 3 ____________ x 7 x 4 5 x 4 ____________ x 7 x 4 Multiply the numerators. Multiply the denominators.

x 7 3 5 x 4 __________________ x 7 x 4 Add the numerators; keep the common denominator.

3x 21 5x 20 ________________ x 7 x 4 Multiply in the numerator.

8x 1 ___________ x 2 3x 28

Combine like terms. Multiply in the denominator.

Partial Fractions� e process of adding or subtracting two rational expressions can be reversed to � nd the two simpler proper fractions that were added or subtracted. Finding these two fractions from the answer is called resolving into partial fractions.

To resolve 8x 1 _________ x 2 3x 28 into partial fractions, � rst factor the denominator:

8x 1 ___________ x 2 3x 28

8x 1 ____________ x 4 x 7

So the two target fractions will have denominators (x 4) and (x 7). Observe that the � rst fraction is a proper rational algebraic expression—the numerator is of lower degree than the denominator. So the two target fractions will also be proper fractions, meaning that their numerators will be of degree zero—that is, constant. Picking letters such as A and B to stand for these constant numerators, write:

8x 1 ____________ (x 4)(x 7) A _____ x 4 B _____ x 7

Next follow these steps:

8x 1 ______ (x 7) A B _____ x 7 (x 4) Multiply both sides of the equation by (x 4) to remove the discontinuity on the le� and leave the constant A by itself on the right, with no denominator.

8( 4) 1 _________ 4 7 A Substitute 4 for x to make the second term on the right equal zero.

A 33 ____ 11 3 Simplify.

Add by � rst � nding a common denominator: EXAMPLE 2 ➤

3_____x 4SOLUTION

➤


DEFINITION: Division of Fractions Algebraically: If y 0, then x __ y x 1 __ y .

Verbally: Dividing a number by y means multiplying that number by the reciprocal of y.

Applied to two fractions, this de� nition becomes

a __ x b __ y a __ x y __ b , provided x 0, y 0, and b 0

Example 1 shows how to apply these properties to a product of two rational expressions and simplify the result.

Multiply and simplify: x 2 5x 6 __________ x 2 x 20

x 2 3x 4 __________ x 2 x 2

x 2 5x 6 __________ x 2 x 20

x 2 3x 4 __________ x 2 x 2

x 2 5x 6 x 2 3x 4 ________________________ x 2 x 20 x 2 x 2

Numerator times numerator over denominator times denominator.

x 2 x 3 x 1 x 4 _______________________ x 4 x 5 x 1 x 2 Factor the numerator and the denominator.

x 3 x 1 x 2 x 4 _______________________ x 5 x 1 x 2 x 4 Commute factors appropriately.

x 3 _____ x 5 x 1 x 2 x 4 _________________ x 1 x 2 x 4 Use the multiplication property of fractions.

x 3 _____ x 5 , x 1, x 2, x 4 Because the second fraction equals 1.

Note that the restrictions on x must be stated for the factors that cancel, as they were for removable discontinuities. Of course, x cannot equal 5 either, but this is clear from the fact that the term (x 5) still appears in the denominator.

� e process can be simpli� ed if the numerator and denominator involve only multiplication of factors. If you realize this, you can cancel by striking out the common factors in the numerator and denominator without doing the commuting and breaking the rational expression into two fractions.

x 2 x 3 x 1 x 4 _______________________ x 4 x 5 x 1 x 2 x 3 _____ x 5 , x 1, x 2, x 4

For this purpose it is helpful to de� ne canceling in a fraction.

DEFINITION: Canceling in a FractionCanceling in a fraction means dividing the numerator and denominator by the same common factor.

� is de� nition will help you avoid mistakes such as canceling the x’s in a fraction

like x 3 ____ x 5 . Because canceling means division, which distributes over addition and

subtraction, if you cancel the x’s the result is 1 3 __ x

____ 1 5 __ x

, which is by no means simpler!

Multiply and simplify: EXAMPLE 1 ➤

x__________x 2SOLUTION

➤


• Resolving into partial fractions may be diffi cult to understand for ELL students. Consider working through extra examples.

• Consider allowing ELL students to do the Reading Analysis in pairs.

• For Problems 13–16, students may fi nd common denominator of the denominators easier to deal with than clever form of one.

• If you assign Problems 39–42, have ELL students do them in pairs to help them overcome the challenging language.

Exploration Notes

Exploration 4-6a introduces students to partial fractions. Students are fi rst asked to add algebraic fractions, and then deconstruct the process using a diff erent algebraic fraction. Allow 15–20 minutes for this exploration.

CAS Suggestions

When learning the procedure in Example 1, students should not enter the entire original problem into a CAS. Th e machine will immediately reduce it to the fi nal form of the answer. However, students could use the Factor command on the expressions in each numerator and denominator. Students can use a CAS to check their work on these types of problems. Having immediate feedback on the accuracy of algebraic work is a tremendously valuable learning tool for students, as is the reminder at the bottom of the screen for students to add back in their domain restrictions.

Keep in mind that a signifi cant role of a CAS is as a support for algebraic manipulation when the point of the problem is something other than the manipulation.

Diff erentiating Instruction• Algorithms are not standard from

country to country. Consider separating algebra algorithms and arithmetic algorithms when you teach.

• Phrases such as “ invert and multiply” may not be familiar to ELL students.

• Have students write in their journals the algorithms and properties for algebraic fractions in their own words.

• Point out that the partial fraction algorithms do not give the only possible answers; they give the simplest. Th is will help those students who come from diff erent algorithmic traditions.

225

Q9. Where does the graph of f (x) 5 x 2 3x 7 _________ 2 x 2 11x 1 have a horizontal asymptote?

Q10. �e x-coordinate of the vertex of y 5 x 2 30x 17 is

A. 6 B. 5 C. 3 D. 3 E. 5

For Problems 1–12, multiply or divide and simplify. 1. 2 _____ x 2 x 2 4 ______ 4

2. x 2 7x 12 ___________ 12 4 _____ x 4

3. x 2 4x 3 __________ 5x x 1 _____ x 5

4. x 2 64 _______ x 2 16

x 8 _____ x 4

5. x 2 6x _______ 6 x 2 6 ________ x 3 6 x 2

6. x 2 4 ______ 2x 4 2 _____ x 2

7. x 2 x 2 ___________ x 2 4x 12

x 2 5x 6 __________ x 2 2x 1

8. x 2 3x 10 ___________ x 2 7x 6

x 2 2x 3 __________ x 2 x 6

9. x 2 7x 12 ___________ x 2 x 6

x 2 16 __________ x 2 x 2

10. x 2 6x 8 __________ x 2 5x 6

x 2 7x 12 ___________ x 2 4x 4

11. x 3 _____ x 7 x 7 _____ x 5 x 5 _____ x 3

12. x 9 ______ x 10 x 8 _____ x 9 x 10 ______ x 8

Problems 13–16 involve complex fractions, which are fractions with one or more fractions in the numerator, the denominator, or both. Multiply the fraction by a clever form of 1 that eliminates the “minor” denominators. �en do any other possible simplifying.

13. x 3 2 _ x

_________ 1 4 __ x 2

Multiply by x 2 __ x 2

.

14. x 5 6 _ x

_________ 1 9 __ x 2

15. x 3 12 ____ x 5

____________ x 8 42 ____ x 5

Multiply by x 5 _____ x 5 .

16. x 3 5 ____ x 3

___________ x 2 4 ____ x 3

For Problems 17–28, add or subtract and simplify. 17. 1 _____ x 1 1 _____ x 1

18. 3 _____ x 1 1 _____ 1 x

19. 2x 1 ______ x 1 2x 1 ______ x 1

20. x 3 _____ x 3 x 3 _____ x 3

21. x 2 x 2 x 6 _________ x 3

22. 2x 5 x 2 2x 15 ___________ x 3

23. 1 _____ x 3 1 _____ x 3 2x ______ x 2 9

24. 3 _____ x 6 4x _______ x 2 36

2 _____ x 6

25. 3x 13 ___________ x 2 3x 10

16 __________ x 2 6x 5

26. 6 ___________ x 2 7x 12

5x 9 __________ x 2 2x 3

27. x 2 _________ x 2 x 2

x 4 __________ x 2 5x 4

28. x 4 ___________ x 2 3x 28

x 5 ___________ x 2 2x 35

For Problems 29–38, resolve the algebraic fraction into partial fractions.

29. 11x 15 __________ x 2 3x 2

30. 7x 25 __________ x 2 7x 8

31. 5x 11 __________ x 2 2x 8

32. 3x 12 ___________ x 2 5x 50

33. 21 ___________ x 2 7x 10

34. 10x ___________ x 2 9x 36

35. 9 x 2 25x 50 __________________ (x 1)(x 7)(x 2)

36. 7 x 2 22x 54 __________________ (x 2)(x 4)(x 1)

37. 4 x 2 15x 1 ________________ x 3 2 x 2 5x 6

38. 3 x 2 22x 31 __________________ x 3 8 x 2 19x 12


Reading Analysis

From what you have read in this section, what do you consider to be the main idea? What property of fractions allows you to multiply two fractions by multiplying the numerators and multiplying the denominators? Why can’t you add two fractions by adding the numerators and adding the denominators? What process allows you to take the answer to the sum of two algebraic fractions and break it into the original fractions that were added?

Quick Review Q1. Multiply: (x 3)(x 7) Q2. Factor: x 2 7x 8

For Problems Q3–Q8, f (x) (x 1)(x 2) ______________ (x 1)(x 2)(x 3) .

Q3. Find the zeros of the denominator. Q4. Find the zeros of the numerator. Q5. Find the x-coordinate(s) of any removable

discontinuities. Q6. Find the x-coordinate(s) of any vertical

asymptotes. Q7. Find any x-intercepts. Q8. Find the y-intercept.

Following similar steps but multiplying both sides by (x 7) instead of (x 4) allows you to conclude that B 5. So

8x 1 ____________ (x 4)(x 7) 3 _____ x 4 5 _____ x 7

� is process is known as Heaviside’s method a� er Oliver Heaviside (1850 1925). A verbal description of a shortcut to this method is “To � nd the numerator of the fraction containing (x 4), cover up the (x 4) in the original fraction with your � nger and substitute 4 for x (the value of x that makes (x 4) equal zero) in what remains.” � e result is the numerator of the partial fraction with denominator (x 4). Example 3 shows how to apply this shortcut.

Resolve into partial fractions: 8x 1 ____________ (x 4)(x 7)

Write 8x 1 _________ (x 4)(x 7) ____ x 4 ____ x 7 .

Cover the (x 4) with your � nger and substitute 4 for x in what is le� .

8x 1 3x 4 x 7( )(x 7)

8( 4) 1 /( 4 7) 3


8x 1(x 4)( )

3x 4 x 7

5 8(7) 1 /(7 4) 5

8x 1 ____________ (x 4)(x 7) 3 _____ x 4 5 _____ x 7

Resolve into partial fractions: EXAMPLE 3 ➤

Write 8_________(x

SOLUTION

➤

Reading Analysis Quick Review

Problem Set 4-6


CAS Suggestions (continued)

Th e common denominator command, comDenom, achieves the results of Example 2. Notice, however, that the CAS automatically expands the denominator and warns about potential domain issues from factor simplifi cation.

Th e Expand command decomposes a rational algebraic fraction into its partial fractions. Note: When applied to an improper rational algebraic fraction, both the Expand and propFrac commands determine the same expression for p(x) in the p(x) 1 r(x)

___ d(x) form, but the Expand command also attempts to decompose the r(x)

___ d(x) into partial fractions, if they exist. See the fi gure below for a comparison of the results of the two commands using the expression in Example 3.

PRO B LE M N OTES

Note: Domain restrictions are noted only for those values of x whose denominator factors have canceled out. Other domain restrictions can be seen by inspection in the fi nal answer.Q1. x 2 1 4x 2 21Q2. (x 2 8)(x 1 1)Q3. 1, 2, and 23Q4. 1 and 22Q5. x 5 1Q6. x 5 2 and x 5 23Q7. 22 Q8. 2 1 __ 3 Q9. y 5 2.5 Q10. C

A CAS will automatically simplify each of the expressions in Problems 1–28 to its fi nal simplifi ed form. Encourage students to use a CAS to confi rm their work.

Problems 1–12 involve multiplication and division of variable rational expressions. Th ese should be review for your students.

1. x 1 2 _____ 2 , x 2

2. x 1 3 _____ 3 , x 24

3. (x 1 3)(x 1 5) ___________ 5x , or x 2 1 8x 1 15 ___________ 5x , x 21, x 25

4. x 2 8 _____ x 2 4 , x 24, x 28

5. x 2 1 6 _____ 6x , x 26

6. 1, x 2, x 22

225

Q9. Where does the graph of f (x) 5 x 2 3x 7 _________ 2 x 2 11x 1 have a horizontal asymptote?

Q10. �e x-coordinate of the vertex of y 5 x 2 30x 17 is

A. 6 B. 5 C. 3 D. 3 E. 5

For Problems 1–12, multiply or divide and simplify. 1. 2 _____ x 2 x 2 4 ______ 4

2. x 2 7x 12 ___________ 12 4 _____ x 4

3. x 2 4x 3 __________ 5x x 1 _____ x 5

4. x 2 64 _______ x 2 16

x 8 _____ x 4

5. x 2 6x _______ 6 x 2 6 ________ x 3 6 x 2

6. x 2 4 ______ 2x 4 2 _____ x 2

7. x 2 x 2 ___________ x 2 4x 12

x 2 5x 6 __________ x 2 2x 1

8. x 2 3x 10 ___________ x 2 7x 6

x 2 2x 3 __________ x 2 x 6

9. x 2 7x 12 ___________ x 2 x 6

x 2 16 __________ x 2 x 2

10. x 2 6x 8 __________ x 2 5x 6

x 2 7x 12 ___________ x 2 4x 4

11. x 3 _____ x 7 x 7 _____ x 5 x 5 _____ x 3

12. x 9 ______ x 10 x 8 _____ x 9 x 10 ______ x 8

Problems 13–16 involve complex fractions, which are fractions with one or more fractions in the numerator, the denominator, or both. Multiply the fraction by a clever form of 1 that eliminates the “minor” denominators. �en do any other possible simplifying.

13. x 3 2 _ x

_________ 1 4 __ x 2

Multiply by x 2 __ x 2

.

14. x 5 6 _ x

_________ 1 9 __ x 2

15. x 3 12 ____ x 5

____________ x 8 42 ____ x 5

Multiply by x 5 _____ x 5 .

16. x 3 5 ____ x 3

___________ x 2 4 ____ x 3

For Problems 17–28, add or subtract and simplify. 17. 1 _____ x 1 1 _____ x 1

18. 3 _____ x 1 1 _____ 1 x

19. 2x 1 ______ x 1 2x 1 ______ x 1

20. x 3 _____ x 3 x 3 _____ x 3

21. x 2 x 2 x 6 _________ x 3

22. 2x 5 x 2 2x 15 ___________ x 3

23. 1 _____ x 3 1 _____ x 3 2x ______ x 2 9

24. 3 _____ x 6 4x _______ x 2 36

2 _____ x 6

25. 3x 13 ___________ x 2 3x 10

16 __________ x 2 6x 5

26. 6 ___________ x 2 7x 12

5x 9 __________ x 2 2x 3

27. x 2 _________ x 2 x 2

x 4 __________ x 2 5x 4

28. x 4 ___________ x 2 3x 28

x 5 ___________ x 2 2x 35

For Problems 29–38, resolve the algebraic fraction into partial fractions.

29. 11x 15 __________ x 2 3x 2

30. 7x 25 __________ x 2 7x 8

31. 5x 11 __________ x 2 2x 8

32. 3x 12 ___________ x 2 5x 50

33. 21 ___________ x 2 7x 10

34. 10x ___________ x 2 9x 36

35. 9 x 2 25x 50 __________________ (x 1)(x 7)(x 2)

36. 7 x 2 22x 54 __________________ (x 2)(x 4)(x 1)

37. 4 x 2 15x 1 ________________ x 3 2 x 2 5x 6

38. 3 x 2 22x 31 __________________ x 3 8 x 2 19x 12


Reading Analysis

From what you have read in this section, what do you consider to be the main idea? What property of fractions allows you to multiply two fractions by multiplying the numerators and multiplying the denominators? Why can’t you add two fractions by adding the numerators and adding the denominators? What process allows you to take the answer to the sum of two algebraic fractions and break it into the original fractions that were added?

Quick Review Q1. Multiply: (x 3)(x 7) Q2. Factor: x 2 7x 8

For Problems Q3–Q8, f (x) (x 1)(x 2) ______________ (x 1)(x 2)(x 3) .

Q3. Find the zeros of the denominator. Q4. Find the zeros of the numerator. Q5. Find the x-coordinate(s) of any removable

discontinuities. Q6. Find the x-coordinate(s) of any vertical

asymptotes. Q7. Find any x-intercepts. Q8. Find the y-intercept.

Following similar steps but multiplying both sides by (x 7) instead of (x 4) allows you to conclude that B 5. So

8x 1 ____________ (x 4)(x 7) 3 _____ x 4 5 _____ x 7

� is process is known as Heaviside’s method a� er Oliver Heaviside (1850 1925). A verbal description of a shortcut to this method is “To � nd the numerator of the fraction containing (x 4), cover up the (x 4) in the original fraction with your � nger and substitute 4 for x (the value of x that makes (x 4) equal zero) in what remains.” � e result is the numerator of the partial fraction with denominator (x 4). Example 3 shows how to apply this shortcut.

Resolve into partial fractions: 8x 1 ____________ (x 4)(x 7)

Write 8x 1 _________ (x 4)(x 7) ____ x 4 ____ x 7 .


8x 1 3x 4 x 7( )(x 7)

8( 4) 1 /( 4 7) 3


8x 1(x 4)( )

3x 4 x 7

5 8(7) 1 /(7 4) 5

8x 1 ____________ (x 4)(x 7) 3 _____ x 4 5 _____ x 7

Resolve into partial fractions: EXAMPLE 3 ➤

Write 8_________(x

SOLUTION

➤

Reading Analysis Quick Review

Problem Set 4-6


15. x 1 3 _____ x 2 2 , x 25, x 1

16. x 1 2 _____ x 1 1 , x 3, x 2

In Problems 17–28, students add and subtract variable rational expressions that require fi nding a common denominator.

17. 2x ____________ (x 2 1)(x 1 1)

18. 2 _____ x 2 1

19. 2 2 4x ______ x 2 2 1

20. 12x ______ x 2 2 9

21. 22x _____ x 2 3

22. x, x 3

23. 2 _____ x 1 3 , x 3

24. 1 _____ x 1 6 , x 6

25. 3x 1 9 ____________ (x 1 2)(x 2 1) , x 5

26. 5x 1 10 ____________ (x 2 4)(x 1 1) , x 3

27. 2x ______ x 2 2 1 , x 2, x 4

28. 14 _______ x 2 2 49 , x 24, x 5

Problems 29–38 require resolving into partial fractions.

Problems 29–38 are solved quickly using the Expand command.

29. 4 _____ x 2 1 1 7 _____ x 2 2

30. 22 _____ x 1 1 1 9 _____ x 2 8

31. 3.5 _____ x 1 2 1 1.5 _____ x 2 4

32. 1.8 _____ x 1 5 1 1.2 ______ x 2 10

33. 7 _____ x 1 2 1 27 _____ x 1 5

34. 2 _____ x 1 3 1 8 ______ x 2 12

35. 2 _____ x 1 1 1 3 _____ x 2 7 1 4 _____ x 1 2

36. 3 _____ x 2 2 1 21 _____ x 1 4 1 5 _____ x 2 1

37. 3 _____ x 2 2 1 2 _____ x 1 1 1 21 _____ x 1 3

38. 3 _____ x 2 4 1 24 _____ x 2 3 1 22 _____ x 2 1

7. x 1 1 _____ x 2 1 , x 6, x 22

8. x 1 5 _____ x 2 6 , x 2, x 1, x 23

9. x 2 1 _____ x 1 4 , x 3, x 22, x 4

10. (x 2 2)(x 2 2) ____________ (x 2 3)(x 2 3) , or x 2 2 4x 1 4 __________ x 2 2 6x 1 9 ,

x 2, x 4

11. (x 1 5 ) 2 _______ (x 2 7 ) 2 , or x 2 1 10x 1 25 ____________ x 2 2 14x 1 49 , x 25, x 23

12. (x 1 8 ) 2 ________ (x 2 10 ) 2 , or x 2 1 16x 1 64 _____________ x 2 2 20x 1 100 ,

x 29, x 28

Problems 13–16 involve complex fractions. Consider assigning a few of these as review problems.

13. x 2 1 x ______ x 2 2 , x 0, x 22

14. x 2 2 2x _______ x 1 3 , x 0, x 3

227Section 4-7: Fractional Equations and Extraneous Solutions

Fractional Equations and Extraneous Solutions For each kind of function you have studied so far, you have evaluated f (x) for various given values of x and solved for x if a value of f (x) is given. � at is,

x, � nd y.

y, � nd x.

In this section you will solve for x if function f is a rational algebraic function.

Given a rational function f, � nd x for a given value of f (x).

Suppose you are given the rational algebraic function

f (x) x 2 5x 10 ___________ x 3

and you are to � nd x if f (x) 6. Figure 4-7a shows the graph of f and a line plotted at y 6.

y

5

x

5

f

5 5

x 7x 4

y 6

Figure 4-7a

It is relatively easy to � nd that x 4 or x 7 numerically, by making a table, or graphically, using the intersect or solver feature of your grapher. Example 1 shows how to � nd these values algebraically.

For f (x) x 2 5x 10 ___________ x 3 , � nd x algebraically if f (x) 6.

x 3 Write any domain restrictions.

6 x 2 5x 10 ___________ x 3 Substitute 6 for f(x).

6(x 3) x 2 5x 10 Multiply both sides by (x 3) to eliminate the fraction.

6x 18 x 2 5x 10

0 x 2 11x 28 Make one side equal zero.

0 (x 4)(x 7) Factor the quadratic trinomial (or solve using the quadratic formula).

x 4 or x 7 � ese values of x make the right side equal zero.

Neither value of x is excluded from the domain, so both 4 and 7 are solutions.

Fractional Equations and Extraneous Solutions

4 -7

Given a rationalObjective

For EXAMPLE 1 ➤

x 3SOLUTION

➤

4 -7


Unfactorable Quadratics

Heaviside’s method does not work for an algebraic fraction such as

7 x 2 4x _____________ ( x 2 1)(x 2)

which has an unfactorable quadratic factor, ( x 2 1), in the denominator (unless you are willing to use imaginary numbers). Because the numerators of the corresponding partial fractions need be only one degree lower than the degree of their denominators, the quadratic term can have a linear numerator. � us the partial fractions would be

7 x 2 4x _____________ ( x 2 1)(x 2)

Ax B _______ x 2 1

C _____ x 2

where A, B, and C stand for constants. Adding the fractions on the right, then simplifying, gives

7 x 2 4x _____________ ( x 2 1)(x 2)

Ax B _______ x 2 1

C _____ x 2

(Ax B)(x 2) C( x 2 1) ________________________ ( x 2 1)(x 2)

A x 2 2Ax Bx 2B C x 2 C ____________________________ ( x 2 1)(x 2)

So A x 2 C x 2 7 x 2 , 2Ax Bx 4x, and 2B C 0. Solving the system

7 A C

4 2A B

0 2B C

gives A 3, B 2, and C 4. � erefore,

7 x 2 4x _____________ ( x 2 1)(x 2)

3x 2 ______ x 2 1

4 _____ x 2

For Problems 39 and 40, resolve the rational expression into partial fractions.

39. 4 x 2 6x 11 _____________ ( x 2 1)(x 4)

40. 4 x 2 15x 1 ________________ x 3 5 x 2 3x 1

Repeated Linear Factors

If a power of a linear factor appears in the denominator, the fraction could have come from adding partial fractions with that power or any lower power. For example, the algebraic fraction

x 2 4x 18 _____________ (x 4)(x 1 ) 2

has a repeated linear factor, (x 1), in the denominator. In this case the partial fractions will include one fraction with denominator (x 1 ) 2 . But they could also include a fraction with denominator (x 1). So the fractions would be

A _____ x 4

B _____ x 1

C Dx _______ (x 1) 2

Because the numerator has only three coe� cients, one of the four constants, A, B, C, or D, is arbitrary. To get the simplest partial fractions, it is helpful to let D be the arbitrary constant, and set D equal to zero. � e partial fractions thus have the form

x 2 4x 18 _____________ (x 4)(x 1 ) 2

A _____ x 4

B _____ x 1

C _______ (x 1 ) 2

Adding the fractions on the right, then simplifying, gives

A(x 1 ) 2 B(x 4)(x 1) C(x 4) __________________________________ (x 4)(x 1 ) 2

A x 2 2Ax A B x 2 3Bx 4B Cx 4C _______________________________________ (x 4)(x 1 ) 2

So A x 2 B x 2 x 2 , 2Ax 3Bx Cx 4x, and A 4B 4C 18. Solving the system

1 A B

4 2A 3B C

18 A 4B 4C


x 2 4x 18 _____________ (x 4)(x 1 ) 2

2 _____ x 4

1 _____ x 1

3 _______ (x 1 ) 2


41. 4 x 2 18x 6 _____________ (x 5)(x 1 ) 2

42. 3 x 2 53x 245 ______________ x 3 14 x 2 49x



Th e Expand command also works on the unfactorable quadratic and repeated linear terms described in the paragraphs preceding Problems 39 and 41. Note, however, that in an attempt to minimize the numerator of each term, the fi gure shows the CAS splitting the quadratic term of 3x 1 2 _____ x 2 1 1 into two fractions, establishing again the principle that use of a CAS develops and encourages a deeper sense of mathematical equivalence. Answers to Problems 39–42 therefore can be quickly confi rmed with the Expand command under the conditions just noted.

Problems 39 and 40 are challenging problems for which the denominators cannot be factored into linear factors. Note that Heaviside’s method does not work.

39. x 1 2 ______ x 2 1 1 1 3 _____ x 1 4

40. 3 _____ x 2 1 1 x 2 2 __________ x 2 2 4x 2 1

Problems 41 and 42 present repeated linear factors. Again, Heaviside’s method does not work.

41. 1 _____ x 1 5 1 3 _____ x 1 1 1 22 _______ (x 1 1) 2

42. 5 __ x 1 22 _____ x 2 7 1 3 _______ (x 2 7 ) 2 Additional CAS Problems

1. Because x 2 1 1 5 (x 1 i)(x 2 i), the rational expression 3x 1 2 _____ x 2 1 1 can be decomposed to A ____ x 1 i 1 B ____ x 2 i for some constants A and B. Use the methods in Problems 29–38 to determine the values of A and B by hand, and use your CAS to confi rm that your answer is correct.

2. Is there a value of C for which a function defi ned by f(x) 5 3 ____ x 1 2 1 C ____ x 2 5 has a vertical asymptote at x 5 5 only?

See page 997 for answers to CAS Problems 1 and 2.


Fractional Equations and Extraneous Solutions For each kind of function you have studied so far, you have evaluated f (x) for various given values of x and solved for x if a value of f (x) is given. Th at is,

• Given x, fi nd y.

• Given y, fi nd x.

In this section you will solve for x if function f is a rational algebraic function.

Given a rational algebraic function f, fi nd x for a given value of f (x).

Suppose you are given the rational algebraic function

f (x) � x 2 � 5x � 10 ___________ x � 3

and you are to fi nd x if f (x) � 6. Figure 4-7a shows the graph of f and a line plotted at y � 6.

y

5

x

−5

f

−5 5

x = 7x = 4

y = 6

Figure 4-7a

It is relatively easy to fi nd that x � 4 or x � 7 numerically, by making a table, or graphically, using the intersect or solver feature of your grapher. Example 1 shows how to fi nd these values algebraically.

For f (x) � x 2 � 5x � 10 ___________ x � 3 , fi nd x algebraically if f (x) � 6.

x � 3 Write any domain restrictions.

6 � x 2 � 5x � 10 ___________ x � 3 Substitute 6 for f(x).

6(x � 3) � x 2 � 5x � 10 Multiply both sides by (x � 3) to eliminate the fraction.

6x � 18 � x 2 � 5x � 10

0 � x 2 � 11x � 28 Make one side equal zero.

0 � (x � 4)(x � 7) Factor the quadratic trinomial (or solve using the quadratic formula).

x � 4 or x � 7 Th ese values of x make the right side equal zero.

Neither value of x is excluded from the domain, so both 4 and 7 are solutions.

FractioSolutio

4 -7

GiveObjective

ForEXAMPLE 1

x �SOLUTION


Unfactorable Quadratics

Heaviside’s method does not work for an algebraic fraction such as

7 x 2 4x _____________ ( x 2 1)(x 2)

which has an unfactorable quadratic factor, ( x 2 1), in the denominator (unless you are willing to use imaginary numbers). Because the numerators of the corresponding partial fractions need be only one degree lower than the degree of their denominators, the quadratic term can have a linear numerator. � us the partial fractions would be

7 x 2 4x _____________ ( x 2 1)(x 2)

Ax B _______ x 2 1

C _____ x 2

where A, B, and C stand for constants. Adding the fractions on the right, then simplifying, gives

7 x 2 4x _____________ ( x 2 1)(x 2)

Ax B _______ x 2 1

C _____ x 2

(Ax B)(x 2) C( x 2 1) ________________________ ( x 2 1)(x 2)

A x 2 2Ax Bx 2B C x 2 C ____________________________ ( x 2 1)(x 2)

So A x 2 C x 2 7 x 2 , 2Ax Bx 4x, and 2B C 0. Solving the system

7 A C

4 2A B

0 2B C


7 x 2 4x _____________ ( x 2 1)(x 2)

3x 2 ______ x 2 1

4 _____ x 2


39. 4 x 2 6x 11 _____________ ( x 2 1)(x 4)

40. 4 x 2 15x 1 ________________ x 3 5 x 2 3x 1

Repeated Linear Factors

If a power of a linear factor appears in the denominator, the fraction could have come from adding partial fractions with that power or any lower power. For example, the algebraic fraction

x 2 4x 18 _____________ (x 4)(x 1 ) 2

has a repeated linear factor, (x 1), in the denominator. In this case the partial fractions will include one fraction with denominator (x 1 ) 2 . But they could also include a fraction with denominator (x 1). So the fractions would be

A _____ x 4

B _____ x 1

C Dx _______ (x 1) 2

Because the numerator has only three coe� cients, one of the four constants, A, B, C, or D, is arbitrary. To get the simplest partial fractions, it is helpful to let D be the arbitrary constant, and set D equal to zero. � e partial fractions thus have the form

x 2 4x 18 _____________ (x 4)(x 1 ) 2

A _____ x 4

B _____ x 1

C _______ (x 1 ) 2

Adding the fractions on the right, then simplifying, gives

A(x 1 ) 2 B(x 4)(x 1) C(x 4) __________________________________ (x 4)(x 1 ) 2

A x 2 2Ax A B x 2 3Bx 4B Cx 4C _______________________________________ (x 4)(x 1 ) 2

So A x 2 B x 2 x 2 , 2Ax 3Bx Cx 4x, and A 4B 4C 18. Solving the system

1 A B

4 2A 3B C

18 A 4B 4C


x 2 4x 18 _____________ (x 4)(x 1 ) 2

2 _____ x 4

1 _____ x 1

3 _______ (x 1 ) 2


41. 4 x 2 18x 6 _____________ (x 5)(x 1 ) 2

42. 3 x 2 53x 245 ______________ x 3 14 x 2 49x



Class Time1 day

Homework AssignmentRA, Q1–10, Problems 1–3, 5–11 odd,

15–27 odd

Teaching ResourcesExploration 4-7a: Fractional EquationsTest 9, Sections 4-5 to 4-7, Forms A and B

TE ACH I N G

Important Terms and ConceptsFractional equationExtraneous solutionIrreversible stepDepressed equation

Section Notes

� e goal of this section is to solve fractional or rational algebraic equations for x, which leads to the possibility of extraneous solutions. Students previously encountered extraneous solutions when solving radical equations, exponential equations, and logarithmic equations. Extraneous solutions arise when an irreversible operation is performed, such as squaring both sides of an equation.

Students can verify numerically that particular x-values produce the given function values by using a grapher. It may be di� cult to � nd points of intersection graphically when they occur near asymptotes. An algebraic solution may be more e� cient in these instances.

229

Both graphs have asymptotes at x 3 and x 2. Although it is a bit di� cult to see, the graphs intersect at x 7. It is not clear whether the graphs intersect where they come together near the asymptote at x 2. � e algebraic solution allows you to resolve this ambiguity.

Find the intersections of the graphs of f (x) x ____ x 2 2 ____ x 3 and g (x) 10 _______ x 2 x 6 algebraically.

x 2, x 3 Write any domain restrictions.

x _____ x 2

2 _____ x 3

10 _________ x 2 x 6

Set f (x) equal to g(x).

x _____ x 2 2 _____ x 3 10 ____________ (x 2)(x 3) Factor the denominator on the right side.

To get rid of the fractions, you can multiply both sides of the equation by (x 2)(x 3), which is the least common multiple of the three denominators.

(x 2)(x 3) x _____ x 2

2 _____ x 3

10 ____________ (x 2)(x 3) (x 2)(x 3)

On the right side, the denominator cancels immediately, leaving 10. On the le� side, the (x 2)(x 3) must be distributed to both fractions before canceling.

(x 2)(x 3) x _____ x 2 (x 2)(x 3) 2 _____ x 3 10

(x 3)(x) (x 2)(2) 10 Cancel.

x 2 5x 14 0 Distribute, and combine like terms.

(x 2)(x 7) 0 Factor.

x 2 or x 7

� e value x 2 is one of the values excluded from the domain. A solution of the transformed equation x 2 5x 14 0 that does not work in the original equation is called an extraneous solution. Strike through x 2 and write the word “extraneous.”

extraneousx 2 or x 7

x 7 is the only solution. Write the solution.

Note: � e extraneous solution in Example 3 occurs where both graphs have a common asymptote. It is possible for the graphs of two rational functions to cross each other near a common asymptote. � e most reliable way to be sure of the behavior of the two functions is to � nd an algebraic solution.

Find the intersections of the graphs of g (g (g x) x) x

EXAMPLE 3 ➤

x 2,x

SOLUTION

➤

Section 4-7: Fractional Equations and Extraneous Solutions228 Chapter 4: Polynomial and Rational Functions

� e equation 6 x 2 5x 10 _________ x 3 in the solution to Example 1 is called a fractional equation because there is a variable in a denominator. An equation such as 1 _ 2 x 8 13 is not called a fractional equation, even though it has a fraction in it, because there are no variables in a denominator.

Suppose you are asked to � nd x for function f in Example 1 if f (x) 4. Figure 4-7b shows that a horizontal line drawn at y 4 does not intersect the graph of function f. Example 2 shows how to verify this algebraically.

y

5

x

5

f

5 5

Doesn’t intersect!y 4

Figure 4-7b

For f (x) x 2 5x 10 _________ x 3 , � nd x algebraically if f (x) 4.


4 x 2 5x 10 ___________ x 3

4(x 3) x 2 5x 10

4x 12 x 2 5x 10

0 x 2 9x 22

b 2 4ac ( 9 ) 2 4(1)(22) 7 Find the discriminant.

No real solutions. � e discriminant is negative.

Interesting things sometimes happen if you try to � nd the points at which the graphs of two rational algebraic functions intersect. Suppose

f (x) x _____ x 2 2 _____ x 3 and g(x) 10 _________ x 2 x 6

Figure 4-7c shows the graphs of f (x) and g(x) plotted on the same axes.

1010

5

y

5

5 5x

Graphs intersect atx 7

f (x)

g(x)

Figure 4-7c

For EXAMPLE 2 ➤

x 3SOLUTION

➤



In Example 3, the extraneous solutions were introduced in the fourth step of the solution when both sides of the equation were multiplied by the variable expression  x 2 2  x 1 3 . Multiplying by this expression is not reversible, because you cannot divide by  x 2 2  x 1 3 without encountering domain problems.

Students can check their algebraic work by plotting the equations f 1 (x) 5 x ____ x 2 2 1 2 ____ x 1 3 and f 2 (x) 5 10 _______ x 2 1 x 2 6 on their grapher and then finding the x-coordinates of the points of intersection. Or students can enter f 1 (x) and f 2 (x), turn off f 1 (x) and f 2 (x), then enter f 3 (x) 5 f 1 (x) 2 f 2 (x). The zeros are the x-coordinates of the fractional equation.

The only zero of f 3 (x) for Example 3 is 27, the solution to the given fractional equation. The extraneous solution, x 5 2, is excluded from the domain of both f(x) and g(x). This method of finding the zeros of the difference of two functions is often easier than finding the x-coordinates of the points of intersection of the functions.

Emphasize to your students that when solving equations they cannot multiply or divide by a variable expression without introducing the possibility of creating extraneous solutions. This is also the case if they square both sides of an equation. To solve a fractional equation, they will often need to multiply by a variable expression. It is critical to keep the domain restrictions in mind. Differentiating Instruction

• Enlarging Figures 4-7a and 4-7c will make it easier for students to follow along with the discussion.

• Have ELL students do the Reading Analysis by themselves. Consider giving an announced homework quiz over a portion of the Reading Analysis to encourage students to answer the questions thoroughly.

• Clarify the meaning of surprising in the Reading Analysis for ELL students. The word different might be clearer.

• Discussing Problems 1 and 2 as a class will help students understand why solutions can be lost or gained when solving an equation.

• Students may be tempted to skip problems that involve writing. Make sure that they answer the writing questions, as this is an important skill to develop.

229

Both graphs have asymptotes at x 3 and x 2. Although it is a bit di� cult to see, the graphs intersect at x 7. It is not clear whether the graphs intersect where they come together near the asymptote at x 2. � e algebraic solution allows you to resolve this ambiguity.

Find the intersections of the graphs of f (x) x ____ x 2 2 ____ x 3 and g (x) 10 _______ x 2 x 6 algebraically.

x 2, x 3 Write any domain restrictions.

x _____ x 2

2 _____ x 3

10 _________ x 2 x 6

Set f (x) equal to g(x).

x _____ x 2 2 _____ x 3 10 ____________ (x 2)(x 3) Factor the denominator on the right side.

To get rid of the fractions, you can multiply both sides of the equation by (x 2)(x 3), which is the least common multiple of the three denominators.

(x 2)(x 3) x _____ x 2

2 _____ x 3

10 ____________ (x 2)(x 3) (x 2)(x 3)

On the right side, the denominator cancels immediately, leaving 10. On the le� side, the (x 2)(x 3) must be distributed to both fractions before canceling.

(x 2)(x 3) x _____ x 2 (x 2)(x 3) 2 _____ x 3 10

(x 3)(x) (x 2)(2) 10 Cancel.

x 2 5x 14 0 Distribute, and combine like terms.

(x 2)(x 7) 0 Factor.

x 2 or x 7

� e value x 2 is one of the values excluded from the domain. A solution of the transformed equation x 2 5x 14 0 that does not work in the original equation is called an extraneous solution. Strike through x 2 and write the word “extraneous.”

extraneousx 2 or x 7

x 7 is the only solution. Write the solution.

Note: � e extraneous solution in Example 3 occurs where both graphs have a common asymptote. It is possible for the graphs of two rational functions to cross each other near a common asymptote. � e most reliable way to be sure of the behavior of the two functions is to � nd an algebraic solution.

Find the intersections of the graphs of g (g (g x) x) x

EXAMPLE 3 ➤

x 2,x

SOLUTION

➤


� e equation 6 x 2 5x 10 _________ x 3 in the solution to Example 1 is called a fractional equation because there is a variable in a denominator. An equation such as 1 _ 2 x 8 13 is not called a fractional equation, even though it has a fraction in it, because there are no variables in a denominator.

Suppose you are asked to � nd x for function f in Example 1 if f (x) 4. Figure 4-7b shows that a horizontal line drawn at y 4 does not intersect the graph of function f. Example 2 shows how to verify this algebraically.

y

5

x

5

f

5 5

Doesn’t intersect!y 4

Figure 4-7b

For f (x) x 2 5x 10 _________ x 3 , � nd x algebraically if f (x) 4.


4 x 2 5x 10 ___________ x 3

4(x 3) x 2 5x 10

4x 12 x 2 5x 10

0 x 2 9x 22

b 2 4ac ( 9 ) 2 4(1)(22) 7 Find the discriminant.

No real solutions. � e discriminant is negative.

Interesting things sometimes happen if you try to � nd the points at which the graphs of two rational algebraic functions intersect. Suppose

f (x) x _____ x 2 2 _____ x 3 and g(x) 10 _________ x 2 x 6

Figure 4-7c shows the graphs of f (x) and g(x) plotted on the same axes.

1010

5

y

5

5 5x

Graphs intersect atx 7

f (x)

g(x)

Figure 4-7c

For EXAMPLE 2 ➤

x 3SOLUTION

➤


Exploration Notes

Exploration 4-7a introduces fractional equations, alerting students to extraneous solutions and the importance of confi rming that their solutions are in the domain of the original equation. Allow 20–25 minutes for this exploration.

CAS Suggestions

Th e Solve command is a central CAS command for this section. Th e fi gure shows the solutions to Examples 1 and 2. If you want to fi nd the complex solutions, use the cSolve command.

Th e fi gure shows that a CAS oft en catches domain restrictions such as those in Example 3. Evidence that some information may be missing from this solution is suggested by the domain warning at the bottom of the screen, a warning that was not issued by the CAS for either Example 1 or 2.

231

6. Given f (x) x 2 6x 3 _________ x 4 , a. Find algebraically the values of x for which

f (x) 6. b. Find graphically the values of x for which

f (x) 3. c. Sketch the graph, showing the answers to

parts a and b. d. Is the range of function f the set of all real

numbers? Explain. 7. Given f (x) x x ____ x 2 and g (x) 2 ____ x 2 ,

a. Find algebraically all values of x for which f (x) g (x). Discard any extraneous solutions.

b. Plot the graphs of functions f and g on your grapher. Sketch the result. Show that the graphs con�rm your answer to part a.

8. Given f (x) x 2x ____ x 1 and g (x) 3 x ____ x 1 , a. Find algebraically all values of x for which

f (x) g (x). Discard any extraneous solutions. b. Plot the graphs of functions f and g on your

grapher. Sketch the result. Show that the graphs con�rm your answer to part a.

For Problems 9–28, state any domain restrictions and solve the equation algebraically. Show the step where you discard any extraneous solutions. 9. x _____

x 3 7 _____

x 5 24 ___________

x 2 2x 15

10. x _____ x 2

7 _____ x 5

14 ___________ x 2 3x 10

11. 3x _____ x 4

4x _____ x 3

84 __________ x 2 x 12

12. 4x _____ x 1

5x _____ x 2

2 __________ x 2 3x 2

13. 3 _____ x 3

4 _____ x 4

25 ___________ x 2 7x 12

14. 11x _______ x 20 24 ___ x 11 88 ________ x(x 20)

15. x 2 _____ x 3 x 2 _____ x 6 2

16. 3x 2 ______ x 1 2x 4 ______ x 2 5

17. 2 _____ x 2

x _____ x 2

x 2 4 ______ x 2 4

18. x _____ x 4

4 _____ x 4

x 2 16 _______ x 2 16

19. 1 _____ 1 x 1 x _____ x 1

20. x _____ x 1

2 ______ x 2 1

8 _____ x 1

21. x 3 ______ 2 x 3

18x _______ 4 x 2 9

22. 3 22 _____ x 5 6x 1 ______ 2x 7

23. 4x ______ x 2 9

x 1 __________ x 2 6x 9

2 _____ x 3

24. x __________ x 2 2x 1

2 _____ x 1

4 ______ x 2 1

25. 3x _____ x 2

2x _____ x 3

30 _________ x 2 x 6

26. 5 _____ x 6

4 _____ x 3

x 39 ___________ x 2 3x 18

27. River Barge Problem: A tugboat pushes a string of barges 50 mi up the Mississippi River against a current of 4 mi/h, stops for an hour to pick up a string of empty barges, and then returns to the starting point.

a. If x is the number of miles per hour the tugboat travels through the water, explain why the total time, f (x), for the round trip is given by the function f (x) 50 ____ x 4 1 50 ____ x 4 .

b. How long will the round trip take if the tugboat goes 20 mi/h? 10 mi/h? 5 mi/h? Does the round trip take 4 times as long at 5 mi/h than it does at 20 mi/h, more than 4 times as long, or less than 4 times as long? Do you �nd this surprising?

c. �e tugboat owner wants to complete the round trip in 8 h so that she will not have to pay the crew for overtime. Find, algebraically, the minimum speed at which the tugboat can travel through the water to accomplish this.


Reading Analysis

From what you have read in this section, what do you consider to be the main idea? How does this main idea correspond to operations you have performed on other kinds of functions? Why is not every equation with fractions in it called a fractional equation? What surprising type of solution might you � nd when you solve a fractional equation?

Quick Review Q1. Find x if 5x 7 13. Q2. Find x if (2x 5)(x 3) 0. Q3. Find x if x 2 5x 14 0. Q4. Find x if x 8 10. Q5. Find x if 2 _ 3 x 24. Q6. Find x if x 1/3 8. Q7. Find x if 3 x 200. Q8. Find x if 3 _ x 12. Q9. Multiply: (3x 7 ) 2 Q10. “(x a) is a factor of P(x) if and only if

P(a) 0” is a statement of A. � e multiplication property of zero B. � e factor theorem C. � e multiplicative identity axiom D. � e synthetic substitution property E. � e converse of the multiplication property

of zero

1. Extraneous Solutions Problem: Start with the equation x 3.

a. Multiply each side by (x 4). Explain why the transformed equation is true when x 4 but the original equation is not.

b. What name is given to the solution x 4 for the transformed equation?

c. Multiplying both sides of an equation by an expression such as (x 4) that can equal zero is called an irreversible step. Why do you suppose this name is used?

2. Depressed Equation Problem: Start with the equation x 2 3x.

a. Explain why x 0 and x 3 are both solutions of the equation.

b. Divide each side of the equation by x. � e transformed equation is called a “depressed equation” because its degree is lower than the degree of the original equation. Write all solutions of the depressed equation.

c. Why is it more dangerous to divide both members by an expression that can equal zero than it is to multiply by such an expression?

3. Given f (x) x 2 6x 11 __________ x 2 , a. Find algebraically the values of x for which

f (x) 2. b. Show algebraically that f (x) never equals 4. c. Does f (x) ever equal 7? Justify your answer. d. Con� rm the results of parts a, b, and c by

plotting the graph of function f on your grapher and sketching the result.

4. Given f (x) x 2 10x 32 _________ x 5 ,

a. Find algebraically the values of x for which f (x) 8.

b. Show algebraically that f (x) never equals 5. c. Does f (x) ever equal 5? Justify your answer. d. Con� rm the results of parts a, b, and c by


5. Given f (x) x 2 4x 1 ________ x 3 , a. Find algebraically the values of x for which




numbers? Explain.

5min

Reading Analysis 2. Depressed Equation Problem: Start with the

Problem Set 4-7


PRO B LE M N OTES

Q1. x 5 4 Q2. x 5 2.5, 23Q3. x 5 22, 7 Q4. x 5 18, 22Q5. x 5 36 Q6. x5 512Q7. x 5 4.8227... Q8. x 5 1 __ 4 Q9. 9 x 2 2 42x 1 49Q10. B

Problem 1 is a simple example of how multiplying by a variable expression increased the degree of the resulting equation and introduced an extraneous solution.1a. x(x 2 4) 5 3(x 2 4); Th e transformed equation is true for x 5 4 because both sides equal 0 if x 5 4.1b. Extraneous solution1c. Multiplying by (x 2 4) is an irreversible step because you cannot divide by (x 2 4) without risking division by 0.

In Problem 2 dividing by a variable expression creates a depressed question with a decreased degree, and a solution is lost in the process.2a. x 2 5 3x ⇒ x (x 2 3) 5 0 ⇒ x 5 0, 32b. Divide both sides by x : x 5 3. Only x 5 3 is a solution of the depressed equation.2c. Dividing by a variable expression is more dangerous because the lost solution is hard to recover. Multiplying by a variable expression may give extra solutions, but these can be discarded by checking to see if they satisfy the original equation.

Problems 3–8 walk students through the algebraic and graphical process of fi nding x for a given value of f(x) in a fractional equation.

Most parts of Problems 3–27 can be solved quickly on a CAS. Because a CAS issues domain restrictions, students can use this capability to check their work carefully, even though it will not return the extraneous solutions directly.

7a. x 5 21 or x 5 2 ⇒ x 5 21

7b. Th e graph shows that f (x) 5 g(x) only at x 5 21, and there is an asymptote at x 5 2, where the extraneous solution is.

f

g

10

5

�5

�5 5x

y

In Problems 9–26, students are asked to solve fractional equations algebraically. Th ere are interesting variations in the solutions to these problems. Some have no solutions, others have one, two, three, or an infi nite number of solutions.

9. x 3, x 25;x 5 21 or x 5 3 ⇒ x 5 21

10. x 22, x 5; x 5 0 or x 5 22 ⇒ x 5 0

extraneous

extraneous

extraneous

231

6. Given f (x) x 2 6x 3 _________ x 4 , a. Find algebraically the values of x for which




numbers? Explain. 7. Given f (x) x x ____ x 2 and g (x) 2 ____ x 2 ,

a. Find algebraically all values of x for which f (x) g (x). Discard any extraneous solutions.

b. Plot the graphs of functions f and g on your grapher. Sketch the result. Show that the graphs con�rm your answer to part a.

8. Given f (x) x 2x ____ x 1 and g (x) 3 x ____ x 1 , a. Find algebraically all values of x for which

f (x) g (x). Discard any extraneous solutions. b. Plot the graphs of functions f and g on your

grapher. Sketch the result. Show that the graphs con�rm your answer to part a.

For Problems 9–28, state any domain restrictions and solve the equation algebraically. Show the step where you discard any extraneous solutions. 9. x _____

x 3 7 _____

x 5 24 ___________

x 2 2x 15

10. x _____ x 2

7 _____ x 5

14 ___________ x 2 3x 10

11. 3x _____ x 4

4x _____ x 3

84 __________ x 2 x 12

12. 4x _____ x 1

5x _____ x 2

2 __________ x 2 3x 2

13. 3 _____ x 3

4 _____ x 4

25 ___________ x 2 7x 12

14. 11x _______ x 20 24 ___ x 11 88 ________ x(x 20)

15. x 2 _____ x 3 x 2 _____ x 6 2

16. 3x 2 ______ x 1 2x 4 ______ x 2 5

17. 2 _____ x 2

x _____ x 2

x 2 4 ______ x 2 4

18. x _____ x 4

4 _____ x 4

x 2 16 _______ x 2 16

19. 1 _____ 1 x 1 x _____ x 1

20. x _____ x 1

2 ______ x 2 1

8 _____ x 1

21. x 3 ______ 2 x 3

18x _______ 4 x 2 9

22. 3 22 _____ x 5 6x 1 ______ 2x 7

23. 4x ______ x 2 9

x 1 __________ x 2 6x 9

2 _____ x 3

24. x __________ x 2 2x 1

2 _____ x 1

4 ______ x 2 1

25. 3x _____ x 2

2x _____ x 3

30 _________ x 2 x 6

26. 5 _____ x 6

4 _____ x 3

x 39 ___________ x 2 3x 18

27. River Barge Problem: A tugboat pushes a string of barges 50 mi up the Mississippi River against a current of 4 mi/h, stops for an hour to pick up a string of empty barges, and then returns to the starting point.

a. If x is the number of miles per hour the tugboat travels through the water, explain why the total time, f (x), for the round trip is given by the function f (x) 50 ____ x 4 1 50 ____ x 4 .

b. How long will the round trip take if the tugboat goes 20 mi/h? 10 mi/h? 5 mi/h? Does the round trip take 4 times as long at 5 mi/h than it does at 20 mi/h, more than 4 times as long, or less than 4 times as long? Do you �nd this surprising?

c. �e tugboat owner wants to complete the round trip in 8 h so that she will not have to pay the crew for overtime. Find, algebraically, the minimum speed at which the tugboat can travel through the water to accomplish this.


Reading Analysis

From what you have read in this section, what do you consider to be the main idea? How does this main idea correspond to operations you have performed on other kinds of functions? Why is not every equation with fractions in it called a fractional equation? What surprising type of solution might you � nd when you solve a fractional equation?

Quick Review Q1. Find x if 5x 7 13. Q2. Find x if (2x 5)(x 3) 0. Q3. Find x if x 2 5x 14 0. Q4. Find x if x 8 10. Q5. Find x if 2 _ 3 x 24. Q6. Find x if x 1/3 8. Q7. Find x if 3 x 200. Q8. Find x if 3 _ x 12. Q9. Multiply: (3x 7 ) 2 Q10. “(x a) is a factor of P(x) if and only if

P(a) 0” is a statement of A. � e multiplication property of zero B. � e factor theorem C. � e multiplicative identity axiom D. � e synthetic substitution property E. � e converse of the multiplication property

of zero

1. Extraneous Solutions Problem: Start with the equation x 3.

a. Multiply each side by (x 4). Explain why the transformed equation is true when x 4 but the original equation is not.

b. What name is given to the solution x 4 for the transformed equation?

c. Multiplying both sides of an equation by an expression such as (x 4) that can equal zero is called an irreversible step. Why do you suppose this name is used?

2. Depressed Equation Problem: Start with the equation x 2 3x.

a. Explain why x 0 and x 3 are both solutions of the equation.

b. Divide each side of the equation by x. � e transformed equation is called a “depressed equation” because its degree is lower than the degree of the original equation. Write all solutions of the depressed equation.

c. Why is it more dangerous to divide both members by an expression that can equal zero than it is to multiply by such an expression?

3. Given f (x) x 2 6x 11 __________ x 2 , a. Find algebraically the values of x for which

f (x) 2. b. Show algebraically that f (x) never equals 4. c. Does f (x) ever equal 7? Justify your answer. d. Con� rm the results of parts a, b, and c by


4. Given f (x) x 2 10x 32 _________ x 5 ,

a. Find algebraically the values of x for which f (x) 8.

b. Show algebraically that f (x) never equals 5. c. Does f (x) ever equal 5? Justify your answer. d. Con� rm the results of parts a, b, and c by


5. Given f (x) x 2 4x 1 ________ x 3 , a. Find algebraically the values of x for which




numbers? Explain.

5min

Reading Analysis 2. Depressed Equation Problem: Start with the

Problem Set 4-7


23. x 3, x 23;x 5 5 or x 5 23 ⇒ x 5 5

24. x 1, x 21;x 5 2 or x 5 21 ⇒ x 5 2

25. x 2, x 23 x 5 2 or x 5 23 ⇒ No solutions

26. x 6, x 23All real numbers are solutions except 6 and 23.

Problem 27 is a real-world problem involving rational fractions.27a. distance 5 rate time, so time 5 distance ____ rate . Th e rate upstream is (x 2 4) mi/h, the rate downstream is (x 1 4) mi/h, and the round trip distance is 50 mi. Th e waiting time is 1 hour. [ f (x) 5 50 _____ x 2 4 1 1 1 50 _____ x 1 4 27b. f (20) 6.2 hours; f (10) 12.9 hours; f (5) 56.6 hoursAt 5 mi/h, which is 1 _ 4 of 20 mi/h, the round trip takes more than 4 times as long; in fact, it takes almost 10 times as long! 27c. Th e minimum speed is about 15.3 mi/h for a less-than-8-hour round trip.


1. Examples 1 and 2 showed how diff erent values of k could make k 5 x 2 2 5x 1 10 _________ x 2 3 have two or no real solutions. Determine any value(s) of k for which k 5 x 2 2 5x 1 10 _________ x 2 3 has exactly one solution.

2. In Example 3, x ____ x 2 2 1 C ____ x 1 3 5 10 _______ x 2 1 x 2 6 was shown to have only one solution when C 5 2. Are there any other values of C for which x ____ x 2 2 1 C ____ x 1 3 5 10 _______ x 2 1 x 2 6 has only one solution?

extraneous

extraneous

extraneous extraneous

11. x 24, x 3; x 5 24 or x 5 3 ⇒ No solutions

12. x 1, x 2; x 5 21 or x 5 22

13. x 3, x 4; x 5 7

14. x 220, x 0; x 5 2

15. x 3, x 6; x 5 14 ___ 3 5 4 2 __ 3

16. x 1, x 22; x 5 6

17. x 22, x 2 x 5 2 ⇒ No solutions

18. x 24, x 4All real numbers are solutions except 24 and 4.

19. x 1All real numbers are solutions except 1.

20. x 1, x 21; x 5 2, 5

21. x 3 __ 2 , x 2 3 __ 2 ;

x 5 3 or x 5 3 __ 2 ⇒ x 5 3

22. x 25, x 23.5; x 5 22

See pages 997–998 for answers to Problems 3–6 and 8, and CAS Problems 1 and 2.

extraneous

extraneous

extraneous

extraneous

233

R2. a. Multiply and simplify: (4x 7)(2x 5) b. Expand and simplify: (3x 10 ) 2

c. Factor the polynomial: x 2 9x 10

d. Factor the polynomial: x 2 121 e. Solve the equation by factoring:

x 2 7x 12 f. Solve the equation using the quadratic

formula: 5 x 2 7x 13 0 g. Solve the equation using the quadratic

formula: x 2 10x 34 0 h. Show by direct substitution that x 2 3i

is a solution to the equation x 2 4x 13. i. Quick! Write the other solution to the

equation in part h. j. Write the coordinates of the vertex:

y 3(x 5 ) 2 7 k. Transform into polynomial form:

y 3(x 5 ) 2 7 l. Complete the square to transform to vertex

form: y 5 x 2 80x 57 R3. a. Figure 4-8b shows the graph of a polynomial

function f.

f(x)

1 22 3 4 5x

1

Figure 4-8b

i. Write the single zero. ii. Write the double zero. iii. Write the triple zero. iv. Write a possible equation for f (x) in

factored form. v. Write the smallest possible degree of

function f. vi. Give the sign of the leading coe� cient

of f (x).

b. Given the polynomial function f (x) 2 x 3 43 x 2 271x 440, i. Find the sum of the zeros (without

actually calculating the zeros). ii. Find the sum of the pairwise products of

the zeros. iii. Find the product of the zeros. iv. Con� rm your answers in parts i–iii by

� nding the zeros using your grapher. c. Sketch the graph of a quintic function

that has three distinct real zeros and two complex zeros.

d. i. State the remainder theorem. ii. Use the remainder theorem to � nd the

remainder if x 10 723 is divided by x 2.

iii. Explain why the factor theorem is a special case of the remainder theorem.

R4. a. Figure 4-8c shows the graph of the cubic function f (x) x 3 8 x 2 29x 52.

f(x)

1 2 3 4 5x

1

Figure 4-8c

i. From the graph, at approximately what value of x is the point of in� ection?

ii. Find algebraically the exact x-coordinate of the point of in� ection.

iii. For what values of x is the graph concave up?

iv. Find the three zeros of function f.

Section 4-8: Chapter Review and Test232 Chapter 4: Polynomial and Rational Functions

Chapter Review and TestIn this chapter you extended your knowledge of quadratic functions to include higher-degree polynomial functions. A quadratic function may have two, one, or no x-intercepts and has exactly two zeros. Zeros that have real-number values are x-intercepts, while complex zeros are not x-intercepts. By the fundamental theorem of algebra and its corollaries, an nth-degree polynomial function has exactly n zeros, counting complex values and repeated zeros. You learned that while polynomial functions are de� ned for all real values of x, rational algebraic functions are unde� ned at values of x that make a denominator zero. � e graph of a rational function may have a vertical asymptote or a removable discontinuity at any such x-value and may have horizontal, slant, or curved asymptotes if the numerator is of equal or higher degree than the denominator. You have also reviewed operations with algebraic fractions and solutions to fractional equations.

Chapter Review and TestIn this chapter you extended your knowledge of quadratic functions to include

4 - 8

R0. Update your journal with what you have learned in this chapter. Include things such as � e most important thing you have learned

as a result of studying this chapter � e new terms you have learned and what

they mean Signi� cant features of polynomial and

rational function graphs � e types of real-world situations

polynomial and rational functions can model

� e similarities and di� erences between multiplying and adding rational expressions, and between adding rational expressions and resolving a rational expression into partial fractions

How to solve a fractional equation and why a fractional equation may have extraneous solutions

R1. Figure 4-8a shows the graphs of

f (x) x 2 x 6 (solid) and

g (x) x 3 5 x 2 2x 29 _________________ x 4 (dashed).

21−1−2−3−4−5

10

20

30

−10

y

x

g

f

5 43 6

Figure 4-8a

a. Explain why there is a value of f (4) but no value of g (4).

b. What feature does the graph of function g have at x 4?

c. Show that x 2 and x 3 are zeros of function f.

d. Show numerically that g (6) is close to f (6) but g (4.01) is not close to f (4.01).

e. Based on your answer to part d, explain why the graph of function f is a curved asymptote for function g.

R0. Update your journal with what you have R1. Figure 4-8a shows the graphs of

Review Problems



Class Time2 days (including 1 day for testing)

Homework AssignmentDay 1: Problems R0–R7, T1–T19Day 2: Section 4-9, Problems 1–13

Teaching ResourcesBlackline Master

Problem C1dTest 10, Chapter 4, Forms A and B

TE ACH I N G

Important Terms and ConceptsRational root theorem

Section Notes

Section 4-8 contains a set of review problems, a set of concept problems, and a chapter test. Th e review problems include one problem for each section in the chapter. You may wish to use the chapter test as an additional set of review problems.

Encourage students to practice the no calculator problems without a calculator so that they are prepared for the test problems for which they cannot use a calculator.

Diff erentiating Instruction• Go over the review problems in class,

perhaps by having students present their solutions. You might assign students to write up their solutions before class starts.

• Because many cultures’ norms highly value helping peers, ELL students oft en help each other on tests. You can limit this tendency by making multiple versions of the test.

• Consider giving a group test the day before the individual test so that students can learn from each other as

they review, and they can identify what they don’t know prior to the individual test. Give a copy of the test to each group member, have them work together, and then randomly choose one paper from the group to grade. Grade the test on the spot, so students know what they need to review further. Make this test worth 1 _ 3 the value of the individual test, or less.

• ELL students may need more time to take the test.

• ELL students will benefi t from having access to their bilingual dictionaries while taking the test.

• In Problem R3a, clarify the meaning of possible equation.

• Allow ELL students to answer Problem R6 in their primary language, in English, or in both.

• ELL students may fi nd Problem C1f diffi cult. Consider discussing this in class instead of assigning it as homework.

233

R2. a. Multiply and simplify: (4x 7)(2x 5) b. Expand and simplify: (3x 10 ) 2

c. Factor the polynomial: x 2 9x 10

d. Factor the polynomial: x 2 121 e. Solve the equation by factoring:

x 2 7x 12 f. Solve the equation using the quadratic

formula: 5 x 2 7x 13 0 g. Solve the equation using the quadratic

formula: x 2 10x 34 0 h. Show by direct substitution that x 2 3i

is a solution to the equation x 2 4x 13. i. Quick! Write the other solution to the

equation in part h. j. Write the coordinates of the vertex:

y 3(x 5 ) 2 7 k. Transform into polynomial form:

y 3(x 5 ) 2 7 l. Complete the square to transform to vertex

form: y 5 x 2 80x 57 R3. a. Figure 4-8b shows the graph of a polynomial

function f.

f(x)

1 22 3 4 5x

1

Figure 4-8b

i. Write the single zero. ii. Write the double zero. iii. Write the triple zero. iv. Write a possible equation for f (x) in

factored form. v. Write the smallest possible degree of

function f. vi. Give the sign of the leading coe� cient

of f (x).

b. Given the polynomial function f (x) 2 x 3 43 x 2 271x 440, i. Find the sum of the zeros (without

actually calculating the zeros). ii. Find the sum of the pairwise products of

the zeros. iii. Find the product of the zeros. iv. Con� rm your answers in parts i–iii by

� nding the zeros using your grapher. c. Sketch the graph of a quintic function

that has three distinct real zeros and two complex zeros.

d. i. State the remainder theorem. ii. Use the remainder theorem to � nd the

remainder if x 10 723 is divided by x 2.

iii. Explain why the factor theorem is a special case of the remainder theorem.

R4. a. Figure 4-8c shows the graph of the cubic function f (x) x 3 8 x 2 29x 52.

f(x)

1 2 3 4 5x

1

Figure 4-8c

i. From the graph, at approximately what value of x is the point of in� ection?

ii. Find algebraically the exact x-coordinate of the point of in� ection.

iii. For what values of x is the graph concave up?

iv. Find the three zeros of function f.


Chapter Review and TestIn this chapter you extended your knowledge of quadratic functions to include higher-degree polynomial functions. A quadratic function may have two, one, or no x-intercepts and has exactly two zeros. Zeros that have real-number values are x-intercepts, while complex zeros are not x-intercepts. By the fundamental theorem of algebra and its corollaries, an nth-degree polynomial function has exactly n zeros, counting complex values and repeated zeros. You learned that while polynomial functions are de� ned for all real values of x, rational algebraic functions are unde� ned at values of x that make a denominator zero. � e graph of a rational function may have a vertical asymptote or a removable discontinuity at any such x-value and may have horizontal, slant, or curved asymptotes if the numerator is of equal or higher degree than the denominator. You have also reviewed operations with algebraic fractions and solutions to fractional equations.

Chapter Review and TestIn this chapter you extended your knowledge of quadratic functions to include

4 - 8

R0. Update your journal with what you have learned in this chapter. Include things such as � e most important thing you have learned

as a result of studying this chapter � e new terms you have learned and what

they mean Signi� cant features of polynomial and

rational function graphs � e types of real-world situations

polynomial and rational functions can model

� e similarities and di� erences between multiplying and adding rational expressions, and between adding rational expressions and resolving a rational expression into partial fractions

How to solve a fractional equation and why a fractional equation may have extraneous solutions

R1. Figure 4-8a shows the graphs of

f (x) x 2 x 6 (solid) and

g (x) x 3 5 x 2 2x 29 _________________ x 4 (dashed).

21−1−2−3−4−5

10

20

30

−10

y

x

g

f

5 43 6

Figure 4-8a

a. Explain why there is a value of f (4) but no value of g (4).

b. What feature does the graph of function g have at x 4?

c. Show that x 2 and x 3 are zeros of function f.

d. Show numerically that g (6) is close to f (6) but g (4.01) is not close to f (4.01).

e. Based on your answer to part d, explain why the graph of function f is a curved asymptote for function g.

R0. Update your journal with what you have R1. Figure 4-8a shows the graphs of

Review Problems

233Section 4-8: Chapter Review and Test

PRO B LE M N OTES

Th e Factor and Solve commands can be used to solve Problems R1 and R2. Th e CAS variation on completing the square gives an alternate approach to Problem R2 part l. R0. Journal entries will vary.R1a. f (4) 5 6, which is a real number. g(4) would involve division by zero, so there is no value for g(4).R1b. A vertical asymptote at x 5 4R1c. f (3) 5 0 and f (22) 5 0R1d. f (6) 5 24 and g(6) 5 26.5 (Close!); f (4.01) 5 6.0701 and g(4.01) 5 506.0701 (Not close!!)R1e. Th e graph of function g gets arbitrarily close to the graph of function f as x gets farther away from zero, showing that the graph of f is a curved asymptote for the graph of g.R2a. 8 x 2 2 6x 2 35R2b. 9 x 2 2 60x 1 100R2c. (x 2 10)(x 1 1)R2d. (x 2 11)(x 1 11)R2e. x 5 3, 4R2f. x 5 7   

____ 309 _________ 10

⇒ x 5 2.4578..., 21.0578...R2g. x 5 25 3iR2h. (2 1 3 i ) 2 2 4(2 1 3i) 5 4 1 12i 1 9 i 2 2 8 2 12i 5 4 2 9 2 8 5 213R2i. Th e other solution is x 5 2 2 3i, the complex conjugate.R2j. (5, 27)R2k. y 5 3 x 2 2 30x 1 68R2l. y 5 5(x 1 8 ) 2 2 263R3a. i. x 5 1R3a. ii. x 5 21R3a. iii. x 5 4R3a. iv. f (x) 5 (x 1 1 ) 2 (x 2 1)(x 2 4) 3

R3a. v. 6th degree

R3a. vi. PositiveR3b. i. 21.5R3b. ii. 135.5R3b. iii. 220R3b. iv. Zeros are 2.5, 8, and 11;2.5 1 8 1 11 5 21.5;(2.5)(8) 1 (2.5)(11) 1 (8)(11) 5 135.5;(2.5)(8)(11) 5 220

Problem R4b requires data regression analysis.

R4a. i. x 2 or 3.

R4a. ii. x 5 2.6666...

R4a. iii. Th e graph is concave up for x 2.6666... .

R4a. iv. Th e zeros are x 5 4, 2 1 3i, and 2 2 3i.

See page 998 for answers to Problems R3c–R3d.

235

i. Simplify the fraction by factoring and canceling.

ii. Explain algebraically why there is a removable discontinuity at x 3.

iii. Explain algebraically why the x-axis is a horizontal asymptote but there are no vertical asymptotes.

R6. a. State the multiplication property of fractions algebraically and verbally.

b. State the de�nition of division algebraically and verbally.

c. Divide and simplify. State any domain restrictions.

x 2 5x 6 __________ x 2 4x 5

x 2 2x 3 ___________ x 2 9x 20

d. Add and simplify. State any domain restrictions.

16 __________ x 2 2x 3

3x 13 __________ x 2 5x 6

e. Resolve into partial fractions:

6 x 2 x 31 __________________ (x 2)(x 1)(x 3)

R7. a. Starting with the equation x 4, multiply both sides by (x 3). What value of x is a solution of the transformed equation that was not a solution of the original equation? What special name is given to this new solution? What name is given to the step where you multiplied both sides of the equation by (x 3), an expression that can equal zero?

b. Solve by factoring: x 2 5x 14 0 c. Given f (x) x 2 2x 15 _________ x 5 ,

i. Substitute 2 for f (x), and solve for x by �rst multiplying both sides by (x 5). Show that one solution is valid and one solution is extraneous.

ii. Substitute 8 for f (x), and solve for x as in part i.

iii. Sketch the graph of function f. Show the results of parts i and ii.

d. Each of these equations illustrates a di�erent situation that could occur when you solve a fractional equation. Solve each equation by multiplying both sides by the least common multiple of the denominators. Describe what happens.

i. x _____ x 2

1 _____ x 4

12 __________ x 2 2x 8

ii. 5 _____ x 2

5 _____ x 4

30 __________ x 2 2x 8

iii. x _____ x 2

2 _____ x 4

12 __________ x 2 2x 8

iv. x _____ x 2

2 _____ x 4

19 __________ x 2 2x 8

e. Airplane Flight Time Problem: Suppose you are a route planner for an airline company that is planning a daily shuttle service between Chicago and San Francisco. You want to know how fast an airplane would have to travel in order to complete a round trip in less than 24 hours, so that only one plane is needed to provide the shuttle service. Let x be the plane’s airspeed in miles per hour, and let f (x) be the number of hours it takes to travel round trip. i. You �nd that the two airports are about

2000 mi apart. �e prevailing wind is from west to east at about 50 mi/h, so the plane’s ground speed from Chicago to San Francisco is (x 50) mi/h, and its ground speed for the return trip is (x 50) mi/h. Recalling that time distance _____ rate , write an equation for f (x), including 2 h “turn-around” time at each airport.

ii. Assuming that no intermediate stops are needed for refueling or maintenance, how long would it take a Boeing 737 �ying at 500 mi/h to travel round trip? A blimp �ying at 60 mi/h? Is this surprising?

iii. Let f (x) 24 h, and solve the resulting equation to �nd the minimum speed a plane would have to �y in order to travel round trip in no more than 24 h.


b. Stock Market Problem: �e stock for a new biotechnology company goes on the market. It is an immediate success, and the price of the stock rises dramatically. �e price per share of the stock is listed in the table.

Weeks Dollars per Share

1 14.00

2 26.00

3 60.00

4 110.00

5 170.00

6 234.00

i. Find algebraically the equation for the cubic function that �ts the �rst four data points. Show the steps leading to your answer.

ii. Show that the ��h and sixth data points agree with the cubic function in part i.

iii. Run a cubic regression on the data. Explain how the result con�rms that the cubic function in part i �ts all six data points.

iv. Plot the graph of the cubic function in part i. Sketch the result.

v. If you use this mathematical model to extrapolate a few months into the future, what does it predict will eventually happen to the price of the stock? �ink of a real-world reason why this end behavior might occur.

R5. a. Write the de�nition of a rational algebraic function.

b. What is the di�erence between a proper algebraic fraction and an improper algebraic fraction?

c. For the rational algebraic function f (x) 3x 2 _____ x 2 , i. Write the equation in the form f (x) p(x) r(x)

___ d(x) . ii. What are the three transformations of

the parent reciprocal function, y 1 _ x , that appear in function f ?

iii. Sketch the graph. Show the horizontal and vertical asymptotes.

iv. Explain how to �nd the horizontal asymptote without transforming the equation to the form f (x) p(x) r(x)

___ d(x) . d. Given the rational functions

f (x) x 3 13 x 2 57x 81 ___________________ x 3 and

g (x) x 3 13 x 2 57x 80 ___________________ x 3

i. Plot both graphs on your grapher. Sketch the result, identifying which function is which.

ii. Function f has a removable discontinuity at x 3. Remove the discontinuity by canceling. Find the y-coordinate of the discontinuity.

iii. Function g has a vertical asymptote at x 3. Write an equation for g (x) in the form p(x) r(x)

___ d(x) , and explain why the discontinuity at x 3 cannot be removed.

iv. Function g also has a curved asymptote. Write the equation for this asymptote and describe how the asymptote relates to the graphs of f (x) and g (x).

e. Figure 4-8d shows the graph of h(x) 10x 30 _____________ x 3 3 x 2 5x 15 .

1010

h(x)2

5 5x

Figure 4-8d



R4b. i. f (x) 5 2 x 3 1 17x 2 2 32x 1 30R4b. ii. f (5) 5 170; f (6) 5 234R4b. iii. y 5 2 x 3 1 17 x 2 2 32x 1 30;

R 2 5 1 indicates perfect fit.R4b. iv.

R4b. v. The model shows the share price dipping at first, then soaring up above $400, then plummeting to zero at 15 weeks. Reasons will vary.R5a. A rational algebraic function is a function with a general equation that can be expressed as f (x) 5 n (x)

___ d (x) , where n(x) and d(x) are polynomials.R5b. A proper algebraic fraction has a numerator of lower degree than the denominator. An improper algebraic fraction has a numerator of equal or greater degree than the denominator.R5c. i. f (x) 5 3 1 4 _____ x 2 2 R5c. ii. Vertical translation by 3 units, vertical dilation by a factor of 4, and horizontal translation by 2 units.R5c. iii.

R5c. iv. The ratio of the leading coefficients of the numerator to denominator, 3 to 1, gives the horizontal asymptote: y 5 3.

10

500

x

f (x)

400

300

200

100

155

�5

f (x)

�10

5

�5

�10 10

5

x

R5d. i. The graph of f (x) is dashed; the graph of g(x) is solid.

x

y

fg

5

5

R5d. ii. f (x) 5 x 2 2 10x 1 27, x 3; y-coordinate 6R5d. iii. g(x) 5 x 2 2 10x 1 27 1 1 ____ x 2 3 ; The term 1 ____ x 2 3 is infinite if x 5 3, so the discontinuity is a vertical asymptote and cannot be removed.R5d. iv. The curved asymptote is y 5 x 2 2 10x 1 27, the polynomial part of the equation for function g, and the same graph as function f.

235

i. Simplify the fraction by factoring and canceling.

ii. Explain algebraically why there is a removable discontinuity at x 3.

iii. Explain algebraically why the x-axis is a horizontal asymptote but there are no vertical asymptotes.

R6. a. State the multiplication property of fractions algebraically and verbally.

b. State the de�nition of division algebraically and verbally.

c. Divide and simplify. State any domain restrictions.

x 2 5x 6 __________ x 2 4x 5

x 2 2x 3 ___________ x 2 9x 20

d. Add and simplify. State any domain restrictions.

16 __________ x 2 2x 3

3x 13 __________ x 2 5x 6

e. Resolve into partial fractions:

6 x 2 x 31 __________________ (x 2)(x 1)(x 3)

R7. a. Starting with the equation x 4, multiply both sides by (x 3). What value of x is a solution of the transformed equation that was not a solution of the original equation? What special name is given to this new solution? What name is given to the step where you multiplied both sides of the equation by (x 3), an expression that can equal zero?

b. Solve by factoring: x 2 5x 14 0 c. Given f (x) x 2 2x 15 _________ x 5 ,

i. Substitute 2 for f (x), and solve for x by �rst multiplying both sides by (x 5). Show that one solution is valid and one solution is extraneous.

ii. Substitute 8 for f (x), and solve for x as in part i.

iii. Sketch the graph of function f. Show the results of parts i and ii.

d. Each of these equations illustrates a di�erent situation that could occur when you solve a fractional equation. Solve each equation by multiplying both sides by the least common multiple of the denominators. Describe what happens.

i. x _____ x 2

1 _____ x 4

12 __________ x 2 2x 8

ii. 5 _____ x 2

5 _____ x 4

30 __________ x 2 2x 8

iii. x _____ x 2

2 _____ x 4

12 __________ x 2 2x 8

iv. x _____ x 2

2 _____ x 4

19 __________ x 2 2x 8

e. Airplane Flight Time Problem: Suppose you are a route planner for an airline company that is planning a daily shuttle service between Chicago and San Francisco. You want to know how fast an airplane would have to travel in order to complete a round trip in less than 24 hours, so that only one plane is needed to provide the shuttle service. Let x be the plane’s airspeed in miles per hour, and let f (x) be the number of hours it takes to travel round trip. i. You �nd that the two airports are about

2000 mi apart. �e prevailing wind is from west to east at about 50 mi/h, so the plane’s ground speed from Chicago to San Francisco is (x 50) mi/h, and its ground speed for the return trip is (x 50) mi/h. Recalling that time distance _____ rate , write an equation for f (x), including 2 h “turn-around” time at each airport.

ii. Assuming that no intermediate stops are needed for refueling or maintenance, how long would it take a Boeing 737 �ying at 500 mi/h to travel round trip? A blimp �ying at 60 mi/h? Is this surprising?

iii. Let f (x) 24 h, and solve the resulting equation to �nd the minimum speed a plane would have to �y in order to travel round trip in no more than 24 h.


b. Stock Market Problem: �e stock for a new biotechnology company goes on the market. It is an immediate success, and the price of the stock rises dramatically. �e price per share of the stock is listed in the table.

Weeks Dollars per Share

1 14.00

2 26.00

3 60.00

4 110.00

5 170.00

6 234.00

i. Find algebraically the equation for the cubic function that �ts the �rst four data points. Show the steps leading to your answer.

ii. Show that the ��h and sixth data points agree with the cubic function in part i.

iii. Run a cubic regression on the data. Explain how the result con�rms that the cubic function in part i �ts all six data points.

iv. Plot the graph of the cubic function in part i. Sketch the result.

v. If you use this mathematical model to extrapolate a few months into the future, what does it predict will eventually happen to the price of the stock? �ink of a real-world reason why this end behavior might occur.

R5. a. Write the de�nition of a rational algebraic function.

b. What is the di�erence between a proper algebraic fraction and an improper algebraic fraction?

c. For the rational algebraic function f (x) 3x 2 _____ x 2 , i. Write the equation in the form f (x) p(x) r(x)

___ d(x) . ii. What are the three transformations of

the parent reciprocal function, y 1 _ x , that appear in function f ?

iii. Sketch the graph. Show the horizontal and vertical asymptotes.

iv. Explain how to �nd the horizontal asymptote without transforming the equation to the form f (x) p(x) r(x)

___ d(x) . d. Given the rational functions

f (x) x 3 13 x 2 57x 81 ___________________ x 3 and

g (x) x 3 13 x 2 57x 80 ___________________ x 3

i. Plot both graphs on your grapher. Sketch the result, identifying which function is which.

ii. Function f has a removable discontinuity at x 3. Remove the discontinuity by canceling. Find the y-coordinate of the discontinuity.

iii. Function g has a vertical asymptote at x 3. Write an equation for g (x) in the form p(x) r(x)

___ d(x) , and explain why the discontinuity at x 3 cannot be removed.

iv. Function g also has a curved asymptote. Write the equation for this asymptote and describe how the asymptote relates to the graphs of f (x) and g (x).

e. Figure 4-8d shows the graph of h(x) 10x 30 _____________ x 3 3 x 2 5x 15 .

1010

h(x)2

5 5x

Figure 4-8d


R6b. Algebraically: If y 0, then x _ y 5 x 1 _ y . Verbally: “Dividing a number by y means multiplying that number by the reciprocal of y.”

R6c. x 2 2 6x 1 8 __________ x 2 1 2x 1 1 , x 3, x 4, x 5

R6d. 3x 1 15 _________ x 2 2 x 2 2 , x 3

R6e. 3 _____ x 2 2 1 22 _____ x 1 1 1 5 _____ x 2 3

R7a. x(x 2 3) 5 4(x 2 3); The value x 5 3 is now a solution. The new solution is an extraneous solution. The step is an irreversible step. (You can’t divide by (x 2 3) without risking division by zero.)R7b. x 5 7, 22R7c. i. x 5 21 or x 5 5 The valid solution is x 5 21.R7c. ii. x 5 5 No valid solutionsR7c. iii. The graph shows that y 5 2 intersects the graph of f at x 5 21 only, and the graph of y 5 8 intersects the graph of f at x 5 5, where there is a removable discontinuity.

�1

8

x

f (x)

5

2Valid solution

Ext.sol.

R7d. i. x 5 25 or x 5 2 ⇒ x 5 25 (One solution is extraneous.) R7d. ii. All real numbers are solutions except x 5 2 and x 5 24.R7d. iii. x 5 2 or x 5 24 ⇒ No valid solutions R7d. iv. x 5 3 or x 5 25 (Both solutions are valid.)R7e. i. f (x) 5 2000 ______ x 2 50 1 2000 ______ x 1 50 1 4

R7e. ii. 12.1 h for 737; 222.2 h for the blimp!R7e. iii. The plane must fly at least about 212 mi/h.

R5e. i. h(x) 5 10 _____ x 2 1 5

, x 3R5e. ii. There is a removable discontinuity at x 5 3 because the factor (x 2 3) cancels from the denominator.R5e. iii. The denominator is of higher degree than the numerator, so the values of h(x) approach zero as x gets far from the origin, making the x-axis a horizontal asymptote. There are no vertical asymptotes because the denominator ( x 2 1 5) never equals zero.

R6a. Algebraically: If x 0 and y 0, then ab __ xy 5 a _ x b _ y . Verbally: “A quotient of two products can be split into a product of two quotients,” or “To multiply two fractions, multiply their numerators and multiply their denominators.”

extraneous

extraneous

extraneous

extraneous extraneous

237

Part 1: No calculators allowed (T1–T11) T1. Multiply and simplify: (x 5)(3x 7) T2. Factor: x 2 7x 60 T3. Solve by factoring: x 2 7x 60 0 T4. Expand and simplify: (5 3i ) 2 T5. Write the zeros of the function

f (x) (x 5)(x 2)(x 1). T6. Write the x-intercepts of function f in

Problem T5. T7. Figure 4-8h shows the graph of the function

g (x) x 3 x 2 7x 15.

g(x)

x

50

－50

5－5

Figure 4-8h

a. Explain graphically why function g has two nonreal zeros.

b. Use synthetic substitution to show that x 3 is really a zero of function g.

c. Find the nonreal zeros of function g. d. Show that the sum of the pairwise products

of the zeros equals the coe� cient of the x-term in the given equation.

T8. For the polynomial function f in Figure 4-8i, which has no nonreal zeros, a. Write the zeros. b. Write the degree of the function. c. If the leading coe� cient equals 1, write the

equation for f (x) in factored form.

f(x)

x55

Figure 4-8i

T9. Function h in Figure 4-8j is a transformation of the parent reciprocal function, y 1 _ x .

10

h(x)

5

5x

5

10

5 Figure 4-8j

a. Name the three transformations. b. Write an equation for h(x).

T10. Resolve f (x) 4 _________ (x 1)(x 3) into partial fractions.

C3. Rational Function Surprise Problem: Figure 4-8g shows the graph of the rational function

f (x) x 3 6 x 2 9x 46 _________________ x 2 2x 15

1010

f(x)

10

x5 5

Figure 4-8g

a. Transform the equation for f (x) to the form p(x) r(x)

___ d(x) , and use the result to write equations for the three asymptotes. Explain why there are no removable discontinuities.

b. � e diagonal asymptote appears to intersect the graph of function f somewhere between x 10 and x 10. Using appropriate algebra, either � nd the point(s) where the asymptote intersects the graph or show that the asymptote does not intersect the graph.

Part 1: No calculators allowed (T1–T11) T8. For the polynomial function f in Figure 4-8i,

Chapter Test

Section 4-8: Chapter Review and Test236

14. Xxxxxxxxxxxxx


C1. Graphs of Complex Zeros Problem: Figure 4-8eshows the graph of the cubic function f (x) x 3 18 x 2 105x 146 and the line containing the points (2, 0) and 8, f (8) . You will learn a way to use this line to � nd the complex zeros of a cubic function.

105

f(x)100

50

x

Figure 4-8e

a. Find the two nonreal zeros of f (x). b. Show algebraically that the line is tangent to

the graph of function f at the point 8, f (8) by � nding the equation for the line and setting it equal to f (x). Show that x 8 is a double zero of the resulting equation and thus the graph touches the line but does not cross it at x 8.

c. Show that the real part, a, of the nonreal zeros equals the x-coordinate of the point of tangency. Show that the imaginary part, b, is equal to the square root of the absolute value of the slope of the tangent line.

d. Figure 4-8f shows the graph of another cubic function, g, but the equation for g (x) is not given. On a copy of the � gure, construct a tangent line to the graph of function g at the appropriate point (x, g (x)) that also contains the x-intercept. Use the technique of part c to � nd graphically the nonreal complex zeros.

2 4 6

10

5

x

g(x)

Figure 4-8f

e. For integer values of x, the values of g (x) in Figure 4-8f are also integers. Use four of these integer points to � nd an equation for g (x). � en calculate the three zeros and con� rm that your graphical solution is correct.

f. Find Paul J. Nahin’s book An Imaginary Tale: � e Story of

___ 1 , published by

Princeton University Press in 1998. Consult pages 27 30 for an in-depth analysis of the property you have learned in this problem.

C2. � e Rational Root � eorem: a. By synthetic substitution, show that 1 is a

zero of the function f (x) 6 x 3 17 x 2 24x 35. Use the results to show that f (x) factors into three linear factors:

f (x) (x 1)(2x 7)(3x 5) What are the other two zeros of f (x)? b. � e box contains a statement of the rational

root theorem from algebra. Show that the zeros of f (x) in part a agree with the conclusion of this theorem.

PROPERTY: Rational Root Theorem

If the rational number n _ d is a zero of the polynomial function p with integer coe� cients (or a root of the polynomial equation p(x) 0), then n is a factor of the constant term of the polynomial and d is a factor of the leading coe� cient.

c. Show that these two polynomial functions agree with the rational root theorem:

g (x) 3 x 3 19 x 2 13x 35

h(x) 6 x 3 35 x 2 31x 280

C1. Graphs of Complex Zeros Problem: Figure 4-8e e. For integer values of x, the values of g (g (g x)

Concept Problems



Problem C1 presents a fascinating and little-known method for finding complex zeros of a cubic function graphically. Students draw a line containing the one line through the x-intercept that is tangent to the graph at some other point. The real part of each complex zero equals the x-coordinate of the point of tangency. The imaginary part equals the positive square root of the slope of the tangent line for one of the zeros, and the negative square root for the other zero. This is a good problem for your stronger students.C1a. x 5 8 3iC1b. f (8) 5 54; The equation of the line is y 5 9x 2 18. x 3 2 18 x 2 1 105x 2 146 5 9x 2 18 ⇒ (x 2 2)(x 2 8)(x 2 8) 5 0 ⇒ x 5 2, 8, 8. So 8 is a double zero, and the line just touches the graph at x 5 8 without crossing, meaning that the line is tangent to the graph.C1c. The complex zeros are a bi 5 8 3i. a 5 8 is the x-coordinate of the point of tangency. The slope of the tangent line is 9, so b 5 3 is the square root of the absolute value of the slope of the tangent line.

A blackline master for Problem C1d is available in the Instructor’s Resource Book.C1d. The graph shows the tangent line through (2, 4) and (6, 0), with slope 21. Nonreal complex zeros are 2 i.

2x

g(x)

5

10

4 6 C1e. The integer points on the graph are shown in the table.

x g(x)

1 10

2 4

3 6

4 10

5 10

6 0

g(x) 5 2 x 3 1 10 x 2 2 29x 1 30. The three zeros are x 5 6, 2 i, which agrees with the graphical solution.C1f. Student research problem.C2a.

f (x) 5 (x 1 1)(6 x 2 1 11x 2 35) 5 (x 1 1)(2x 1 7)(3x 2 5); Zeros are 21, 2 7 _ 2 , 5 _ 3 .

21 6 17 224 23526 211 35

6 11 235 0

21 6 17 224 23526 211 35

6 11 235 0

237

Part 1: No calculators allowed (T1–T11) T1. Multiply and simplify: (x 5)(3x 7) T2. Factor: x 2 7x 60 T3. Solve by factoring: x 2 7x 60 0 T4. Expand and simplify: (5 3i ) 2 T5. Write the zeros of the function

f (x) (x 5)(x 2)(x 1). T6. Write the x-intercepts of function f in

Problem T5. T7. Figure 4-8h shows the graph of the function

g (x) x 3 x 2 7x 15.

g(x)

x

50

－50

5－5

Figure 4-8h

a. Explain graphically why function g has two nonreal zeros.

b. Use synthetic substitution to show that x 3 is really a zero of function g.

c. Find the nonreal zeros of function g. d. Show that the sum of the pairwise products

of the zeros equals the coe� cient of the x-term in the given equation.

T8. For the polynomial function f in Figure 4-8i, which has no nonreal zeros, a. Write the zeros. b. Write the degree of the function. c. If the leading coe� cient equals 1, write the

equation for f (x) in factored form.

f(x)

x55

Figure 4-8i

T9. Function h in Figure 4-8j is a transformation of the parent reciprocal function, y 1 _ x .

10

h(x)

5

5x

5

10

5 Figure 4-8j

a. Name the three transformations. b. Write an equation for h(x).

T10. Resolve f (x) 4 _________ (x 1)(x 3) into partial fractions.

C3. Rational Function Surprise Problem: Figure 4-8g shows the graph of the rational function

f (x) x 3 6 x 2 9x 46 _________________ x 2 2x 15

1010

f(x)

10

x5 5

Figure 4-8g

a. Transform the equation for f (x) to the form p(x) r(x)

___ d(x) , and use the result to write equations for the three asymptotes. Explain why there are no removable discontinuities.

b. � e diagonal asymptote appears to intersect the graph of function f somewhere between x 10 and x 10. Using appropriate algebra, either � nd the point(s) where the asymptote intersects the graph or show that the asymptote does not intersect the graph.

Part 1: No calculators allowed (T1–T11) T8. For the polynomial function f in Figure 4-8i,

Chapter Test

Section 4-8: Chapter Review and Test236

14. Xxxxxxxxxxxxx


C1. Graphs of Complex Zeros Problem: Figure 4-8eshows the graph of the cubic function f (x) x 3 18 x 2 105x 146 and the line containing the points (2, 0) and 8, f (8) . You will learn a way to use this line to � nd the complex zeros of a cubic function.

105

f(x)100

50

x

Figure 4-8e

a. Find the two nonreal zeros of f (x). b. Show algebraically that the line is tangent to

the graph of function f at the point 8, f (8) by � nding the equation for the line and setting it equal to f (x). Show that x 8 is a double zero of the resulting equation and thus the graph touches the line but does not cross it at x 8.

c. Show that the real part, a, of the nonreal zeros equals the x-coordinate of the point of tangency. Show that the imaginary part, b, is equal to the square root of the absolute value of the slope of the tangent line.

d. Figure 4-8f shows the graph of another cubic function, g, but the equation for g (x) is not given. On a copy of the � gure, construct a tangent line to the graph of function g at the appropriate point (x, g (x)) that also contains the x-intercept. Use the technique of part c to � nd graphically the nonreal complex zeros.

2 4 6

10

5

x

g(x)

Figure 4-8f

e. For integer values of x, the values of g (x) in Figure 4-8f are also integers. Use four of these integer points to � nd an equation for g (x). � en calculate the three zeros and con� rm that your graphical solution is correct.

f. Find Paul J. Nahin’s book An Imaginary Tale: � e Story of

___ 1 , published by

Princeton University Press in 1998. Consult pages 27 30 for an in-depth analysis of the property you have learned in this problem.

C2. � e Rational Root � eorem: a. By synthetic substitution, show that 1 is a

zero of the function f (x) 6 x 3 17 x 2 24x 35. Use the results to show that f (x) factors into three linear factors:

f (x) (x 1)(2x 7)(3x 5) What are the other two zeros of f (x)? b. � e box contains a statement of the rational

root theorem from algebra. Show that the zeros of f (x) in part a agree with the conclusion of this theorem.

PROPERTY: Rational Root Theorem

If the rational number n _ d is a zero of the polynomial function p with integer coe� cients (or a root of the polynomial equation p(x) 0), then n is a factor of the constant term of the polynomial and d is a factor of the leading coe� cient.

c. Show that these two polynomial functions agree with the rational root theorem:

g (x) 3 x 3 19 x 2 13x 35

h(x) 6 x 3 35 x 2 31x 280

C1. Graphs of Complex Zeros Problem: Figure 4-8e e. For integer values of x, the values of g (g (g x)

Concept Problems


C3b. The asymptote intersects the graph at the point (7, 11).

�5

10

f (x)

5�10 10

(7, 11)

x

T1. x 2 2 22x 1 35T2. (x 2 12)(x 1 5)T3. x 5 12, 25T4. 16 2 30i T5. x 5 5, 2, and 21T6. The x-intercepts are 5, 2, and 21.T7a. There are two nonreal complex zeros because the vertex on the left approaches but does not cross the x-axis.

T7b.

Therefore x 5 3 is a real-number zero of g(x).T7c. x 5 22 i T7d. The coefficient of the x-term is 27. 3(22 1 i ) 1 3(22 2 i ) 1 (22 1 i )(22 2 i ) 5 27T8a. 22, 1, 1, and 3T8b. 4T8c. f (x) 5 2(x 1 2) (x 2 1) 2 (x 2 3)T9a. Vertical translation by 3 units, vertical dilation by a factor of 2, and horizontal translation by 4 units

T9b. h(x) 5 2 _____ x 2 4 1 3

T10. f (x) 5 22 _____ x 2 1 1 2 _____ x 2 3

3 1 1 27 2153 12 15

1 4 5 0

C2b. In 21 5 21 __ 1 , 21 is a factor of 235 and 1 is a factor of 6. In 27 __ 2 , 27 is a factor of 235 and 2 is a factor of 6. In 5 _ 3 , 5 is a factor of 235 and 3 is a factor of 6.

C2c. For g(x) the possible rational roots are

1, 5, 7, 35 ______ 1, 3 , that is, 1 _ 1 , 1 _ 3 , 5 _ 1 , 5 _ 3 , and so

on. By synthetic substitution, g(x) 5 (x 1 1)(x 2 5)(3x 2 7), so the roots are 21, 5, 7 _ 3 .

For h(x) the possible rational roots are 1, 2, 4, 5, 7, 8, 10, 14, 20, 28, 35, 40, 56, 70, 112, 140, 280

__________________________________ 1, 2, 3, 6 By synthetic substitution, h(x) 5 (x 2 5)(2x 2 7)(3x 1 8), so the roots are 5, 7 _ 2 , 2 8 _ 3 .

C3a. f (x) 5 x 1 4 1 22x 1 14 ____________ (x 1 5)(x 2 3) .Vertical

asymptotes: x 5 25 and x 5 3, diagonal asymptote: y 5 x 1 4. There are no removable discontinuities because the factors (x 1 5) and (x 2 3) cannot be factored out.

239

Part 1: No calculators allowed (1–29)

In Chapter 1, you studied functions and transformations that can be performed on functions. 1. What is the di� erence between a relation that is a

function and a relation that is not a function? 2. Name the transformations applied to function f

to get function g, if a. g (x) 5f (3x) b. g (x) 4 f (x 7)

3. In Figure 4-9a, name the transformations applied to function f to get function h. Write an equation for h(x) in terms of function f.

yf

h

x

5

5

Figure 4-9a

4. On a copy of Figure 4-9b, sketch the graph of y f 1 (x) and the line y x. Explain how the graphs of functions f and f 1 are related to the line.

f(x)

x1

1

f

Figure 4-9b

5. If f ( x) f (x) for all values of x in the domain, then function f is a(n) ? function.

6. Write an equation for g (x) in terms of f (x) that has all these features:

A horizontal dilation by a factor of 2 A vertical dilation by a factor of 3 A horizontal translation by 4 units A vertical translation by 5 units

7. Name the kind of function speci� ed by the equations

x t 3 7ty 9 2 t 2

8. Figure 4-9c shows the graph of a function f. On a copy of the � gure, draw the graph of

a. g (x) f (x)

b. h(x) f x Indicate which graph is which.

10

y

5

x

5

5

10

5

f

10

Figure 4-9c

In Chapter 2, you studied properties of elementary functions. 9. Write the general equation for a(n)

a. Linear function b. Quadratic function c. Logarithmic function d. Exponential function e. Power function

Cumulative Review, Chapters 1–4� ese problems constitute a 2- to 3-hour “rehearsal” for your examination on Chapters 1–4.

Cumulative Review, Chapters 1–4� ese problems constitute a 2

4 - 9

Section 4-9: Cumulative Review, Chapters 1–4

Part 1: No calculators allowed (1–29) 6. Write an equation for g (g (g x) in terms of f) in terms of f) in terms of (f (f x) that

Review Problems


T11. � e equation x 3 5 has x 8 as its only solution. If you multiply both sides by (x 2), the resulting equation has another solution. What is this other solution? What name is given to this new solution?

Part 2: Graphing calculators allowed (T12–T19) T12. Factor x 2 22x 720 by � rst � nding the zeros

either graphically or numerically. T13. Find the complex-number solutions of

x 2 10x 44 10 using the quadratic formula. Substitute one of the solutions into the given equation and show that the quadratic expression on the le� really equals 10 for this value of x.

T14. Transform y 3 x 2 24x 37 into vertex form by completing the square; then state the coordinates of the vertex. Show that the value of h, the horizontal coordinate of the vertex, agrees with the formula h b ___ (2a) .

T15. Driving Problem: Hezzy Tate slows down as her car approaches an intersection, then speeds up again as she passes through it. Figure 4-8k and the table show her distance, d(t) in feet, from the beginning of the intersection at time t, in seconds.

t d(t)3 134 205 276 407 65

Figure 4-8k

5t

d(t)

50

100

a. Run a cubic regression to � nd an equation for d(t). How can you tell from the result that the equation � ts all � ve data points?

b. At what time, t, did Hezzy � rst enter the intersection? Show that the cubic model has no other real zeros by � nding the other zeros of function d.

c. � e slope of the graph indicates Hezzy’s speed. At what time, t, was she going the slowest? What special name is given to this point on the graph of function d?

d. At time t 3 s, is the graph concave down or concave up?

T16. Let f (x) x 2 ____ x 3 and g (x) x 3 ____ x 2 .

a. Find f (x) g (x). b. Find f (x) g (x). c. Find f (x) g (x).

T17. Figure 4-8l shows the graph of the rational function f (x) x 3 3 x 2 25x 75 _____________ x 2 3x 10 . Simplify the algebraic fraction, and then transform it to the form p(x) r(x)

___ d(x) . Explain how the simpli� ed fraction allows you to � nd the two asymptotes, the removable discontinuity, and the x-intercepts.

5−5−10

10

10

−10

x

f (x)

Figure 4-8l

T18. Figure 4-8m shows f (x) le� side of fractional equation (dashed) and g (x) right side (solid). Solve this equation by multiplying both sides of the equation by the least common multiple of the denominators to eliminate the fractions and solving the resulting equation. How does the valid solution agree with the graphs of functions f and g? What feature do the graphs have at the extraneous solution of x?

x _____ x 3

7 _____ x 5

24 ___________ x 2 2x 15

1010

y5

x

5

g

f

5 5

Figure 4-8m

T19. What did you learn as a result of taking this test that you did not know before?


Problem Notes (continued)T11. 2; extraneous solutionT12. (x 1 18)(x 2 40)

Students can use the cSolve command to solve Problem T13.T13. 25 3iCheck x 5 25 1 3i: (25 1 3i ) 2 1 10(25 1 3i ) 1 44 5 25 2 30i 1 9 i 2 2 50 1 30i 1 44 5 25 2 9 2 50 1 44 5 10

See the CAS Suggestions in Section 4-2 for help completing the square in Problem T14.T14. y 5 3(x 2 4 ) 2 2 11; Vertex at (4, 211), which agrees with the value found by the formula: h 5 2(224)

______ 2(3) 5 4

See the CAS Suggestions in Section 4-4 for ways to solve Problem T15.T15a. d(t) 5 t 3 2 12 t 2 1 54t 2 68; R 2 5 1.T15b. t 5 2 s; the other zeros are t 5 5 3i, so t 5 2 is the only real zeroT15c. She was going the slowest att 5 4 s; the point of infl ection.T15d. Concave down

In Problem T16, students can quickly confi rm their answers by entering the given expressions once they have defi ned functions f and g.

T16a. f (x) g(x) 5 x 2 2 _____ x 1 2 , x 3

T16b. f (x) g(x) 5 x 2 2 4 __________ x 2 2 6x 1 9 , x 22

T16c. f (x) 2 g(x) 5 6x 2 13 _________ x 2 2 x 2 6

T17. f (x) 5 (x 2 3)(x 2 5) ____________ (x 2 2) , x 25

By synthetic substitution, f (x) 5 x 2 6 1 3 ____ x 2 2 . Th ere is a removable discontinuity at x 5 25 because the (x 1 5) factor in the denominator cancels. Th ere is a vertical asymptote at x 5 2 because the (x 2 2) factor in the denominator does not cancel.

Th e diagonal asymptote is y 5 x 2 6, the polynomial part of the mixed-number form of the equation.Th e graph has x-intercepts x 5 3 and x 5 5 because the factors (x 2 3) and (x 2 5) in the numerator do not cancel.T18. x 5 21 or x 5 3 Th e valid solution is x 5 21.Th e graphs show the valid solution at x 5 21 where the graphs cross, and the

asymptote at x 5 3, corresponding to the extraneous solution.T19. Answers will vary.

extraneous

239

Part 1: No calculators allowed (1–29)

In Chapter 1, you studied functions and transformations that can be performed on functions. 1. What is the diff erence between a relation that is a

function and a relation that is not a function? 2. Name the transformations applied to function f

to get function g, if a. g (x) � 5f (3x) b. g (x) � 4 � f (x � 7)

3. In Figure 4-9a, name the transformations applied to function f to get function h. Write an equation for h(x) in terms of function f.

yf

h

x

5

5

Figure 4-9a

4. On a copy of Figure 4-9b, sketch the graph of y � f �1 (x) and the line y � x. Explain how the graphs of functions f and f �1 are related to the line.

f(x)

x1

1

f

Figure 4-9b

5. If f (�x) � �f (x) for all values of x in the domain, then function f is a(n) ? function.

6. Write an equation for g (x) in terms of f (x) that has all these features:

• A horizontal dilation by a factor of 2 • A vertical dilation by a factor of 3 • A horizontal translation by 4 units • A vertical translation by 5 units

7. Name the kind of function specifi ed by the equations

x � t 3 � 7ty � 9 � 2 t 2

8. Figure 4-9c shows the graph of a function f. On a copy of the fi gure, draw the graph of

a. g (x) � f (x)

b. h(x) � f � x � Indicate which graph is which.

−10

y

5

x

−5

−5

−10

5

f

10

Figure 4-9c

In Chapter 2, you studied properties of elementary functions. 9. Write the general equation for a(n)

a. Linear function b. Quadratic function c. Logarithmic function d. Exponential function e. Power function

Cumulative Review, Chapters 1–4Th ese problems constitute a 2- to 3-hour “rehearsal” for your examination on Chapters 1–4.

CumuTh ese problem

4 - 9


Part 1: No calculators allowed (1–29) 6. Write an equation for g(g x) in terms of ff (f x) that

Review Problems


T11. � e equation x 3 5 has x 8 as its only solution. If you multiply both sides by (x 2), the resulting equation has another solution. What is this other solution? What name is given to this new solution?

Part 2: Graphing calculators allowed (T12–T19) T12. Factor x 2 22x 720 by � rst � nding the zeros

either graphically or numerically. T13. Find the complex-number solutions of

x 2 10x 44 10 using the quadratic formula. Substitute one of the solutions into the given equation and show that the quadratic expression on the le� really equals 10 for this value of x.

T14. Transform y 3 x 2 24x 37 into vertex form by completing the square; then state the coordinates of the vertex. Show that the value of h, the horizontal coordinate of the vertex, agrees with the formula h b ___ (2a) .

T15. Driving Problem: Hezzy Tate slows down as her car approaches an intersection, then speeds up again as she passes through it. Figure 4-8k and the table show her distance, d(t) in feet, from the beginning of the intersection at time t, in seconds.

t d(t)3 134 205 276 407 65

Figure 4-8k

5t

d(t)

50

100

a. Run a cubic regression to � nd an equation for d(t). How can you tell from the result that the equation � ts all � ve data points?

b. At what time, t, did Hezzy � rst enter the intersection? Show that the cubic model has no other real zeros by � nding the other zeros of function d.

c. � e slope of the graph indicates Hezzy’s speed. At what time, t, was she going the slowest? What special name is given to this point on the graph of function d?

d. At time t 3 s, is the graph concave down or concave up?

T16. Let f (x) x 2 ____ x 3 and g (x) x 3 ____ x 2 .

a. Find f (x) g (x). b. Find f (x) g (x). c. Find f (x) g (x).

T17. Figure 4-8l shows the graph of the rational function f (x) x 3 3 x 2 25x 75 _____________ x 2 3x 10 . Simplify the algebraic fraction, and then transform it to the form p(x) r(x)

___ d(x) . Explain how the simpli� ed fraction allows you to � nd the two asymptotes, the removable discontinuity, and the x-intercepts.

5−5−10

10

10

−10

x

f (x)

Figure 4-8l

T18. Figure 4-8m shows f (x) le� side of fractional equation (dashed) and g (x) right side (solid). Solve this equation by multiplying both sides of the equation by the least common multiple of the denominators to eliminate the fractions and solving the resulting equation. How does the valid solution agree with the graphs of functions f and g? What feature do the graphs have at the extraneous solution of x?

x _____ x 3

7 _____ x 5

24 ___________ x 2 2x 15

1010

y5

x

5

g

f

5 5

Figure 4-8m

T19. What did you learn as a result of taking this test that you did not know before?

239Section 4-9: Cumulative Review, Chapters 1–4


Class Time2 days

Homework AssignmentDay 1: Problems 14–29Day 2: Problems 30–37

Teaching ResourcesBlackline Master

Problems 4, 8, 14–16, 30, 33b, 36gTes t 11, Cumulative Test, Chapters 1–4,

Forms A and B

TE ACH I N G

Section Notes

� ese problems review Chapters 1–4. If you are giving a cumulative examination at this point, they provide a good pre-exam review of these chapters. If you do not give an examination at this time, consider giving this as an extended assignment over two to three weeks along with the usual homework problems. One option is to consider using the problems as a take-home group assignment (or take-home test), in which groups of two or three students submit one paper per group for a grade. Allot time periodically for the groups to discuss their work in class.

Before assigning the problem set, have a class discussion about time management and how to break a large project into small parts with periodic self-imposed deadlines. If you feel your students are not ready for self-imposed deadlines, create a time schedule for groups to submit designated problems periodically over the three- to four-week period.

1. A function is a relation for which there is never a value of x that has more than one value of y.2a. Vertical dilation by a factor of 5 and horizontal dilation by a factor of 1 _ 3 2b. Vertical translation by 4 units and horizontal translation by 7 units3. Horizontal translation by 3 units and vertical translation by 2 units; h(x) 2 f (x 3)

5. Odd6. g(x) 3f 1 __ 2 (x 4) 5 7. Parametric function9a. y mx b9b. y a x 2 bx c, a 09c. y a b ln x or y a b log x, b 09d. y a b x , a 0, b 09e. y a x b , a 0, b 0

See pages 998–999 for answers to Problems 4 and 8.

241

20. Suppose you run a power regression on a set of data. �e correlation coe�cient, r, is 0.99, and the residual plot shows a de�nite pattern. What do these two facts tell you about how well the power function �ts the data?

For Problems 21–23, Figure 4-9e shows the graph of y 5000(0.7 2 x ) on a special kind of graph paper. 21. What kind of graph paper is this? Why is this

kind of graph paper useful if there is a wide range of y-values for the function being graphed?

22. Find the approximate value of y for x 8. Find the approximate value of x for y 80.

y 5000(0.72x)

0 5 10 1510

20

50

100

200

500

1000

2000

5000

10000y

x

Figure 4-9e

23. Take the logarithm of both sides of the given equation. Use the result to explain why the graph of this function is a straight line on this kind of graph paper.

In Chapter 4, you studied polynomial and rational functions and how imaginary numbers are useful in analyzing the behavior of these functions. 24. Multiply: (3x 2)(5x 7) 25. Factor: x 2 11x 12 26. Expand and simplify: (7 3i ) 2 27. For the function

f (x) (x 2)(x 6)(x 6),

a. Write the degree of f (x). b. Write the x-intercepts of function f. c. Write the zeros of function f. d. Find quickly the coe�cient of the x 2 -term if

f (x) is expanded into polynomial form. 28. For the rational function

f (x) (x 3)(x 2) __________________ (x 2)(x 5)(x 7) ,

a. Write the x-coordinate(s) of any removable discontinuities.

b. Write the equation(s) of any vertical asymptotes.

c. Write any x-intercept(s).

29. Resolve 18 _________ (x 4)(x 2) into partial fractions.

Section 4-9: Cumulative Review, Chapters 1–4240 Chapter 4: Polynomial and Rational Functions

10. Sketch graphs of these functions. a. f (x) 2(x 3 ) 2 5 b. g (x) e x 2 c. h(x) log 10 x 1

11. Name the pattern followed by the y-values of functions with regularly spaced x-values for

a. Logarithmic functions b. Power functions c. Quadratic functions

12. Name the type of function shown by each graph. a.

b.

c.

d.

e.

f.

13. Use the appropriate property or de�nition to complete the statement.

a. p log c m if and only if ? b. log 7 log 8 log ? c. log 32 ? log 2

In Chapter 3, you �t equations of functions to scattered data.

For Problems 14–16, Figure 4-9d shows a set of points with a dashed horizontal line drawn at y _ y , where _ y is the mean of the y-values.

y

y y

x

Figure 4-9d

14. On a copy of the �gure, sketch the deviation from the mean for the rightmost point.

15. Sketch what you think is the best-�tting linear function. Show the residual for the rightmost point. How does the size of the residual compare with the size of the deviation?

16. Based on what you know about the mean-mean point, ( __

x , _ y ), show where __ x would be on the

graph in Problem 14. 17. Suppose the regression equation for a set of data

is y 3x 5. What does the residual equal for the data point (4, 15)?

18. For a regression line, what is true about the sum of the squares of the residuals, SS res ?

19. Suppose the sum of the squares of the deviations is SS dev 100 and SS res 36. What does the coe�cient of determination, r 2 , equal?

y

x

y

x

y

x

y

x

y

x

x

y


PRO B LE M N OTESTh e problems are written in chronological order according to the chapters.

Blackline masters for Problems 4, 8, and 14–16 are available in the Instructor’s Resource Book.10.

11a. Multiply–add11b. Multiply–multiply11c. Constant-second-diff erences12a. Logarithmic12b. Exponential12c. Logistic12d. Quadratic12e. Power12f. Linear13a. p 5 lo g c m if and only if c p 5 m 13b. log 7 1 log 8 5 log 5613c. log 32 5 5 log 214.–16.

15. Th e residual is smaller than the deviation.16. See graph above.17. 2218. SS res is a minimum.

19. r 2 5 0.64

�2�2

g f

h x

y

y

x

y � yRegression line

ResidualDev

Average-average point

241

20. Suppose you run a power regression on a set of data. �e correlation coe�cient, r, is 0.99, and the residual plot shows a de�nite pattern. What do these two facts tell you about how well the power function �ts the data?

For Problems 21–23, Figure 4-9e shows the graph of y 5000(0.7 2 x ) on a special kind of graph paper. 21. What kind of graph paper is this? Why is this

kind of graph paper useful if there is a wide range of y-values for the function being graphed?

22. Find the approximate value of y for x 8. Find the approximate value of x for y 80.

y 5000(0.72x)

0 5 10 1510

20

50

100

200

500

1000

2000

5000

10000y

x

Figure 4-9e

23. Take the logarithm of both sides of the given equation. Use the result to explain why the graph of this function is a straight line on this kind of graph paper.

In Chapter 4, you studied polynomial and rational functions and how imaginary numbers are useful in analyzing the behavior of these functions. 24. Multiply: (3x 2)(5x 7) 25. Factor: x 2 11x 12 26. Expand and simplify: (7 3i ) 2 27. For the function

f (x) (x 2)(x 6)(x 6),

a. Write the degree of f (x). b. Write the x-intercepts of function f. c. Write the zeros of function f. d. Find quickly the coe�cient of the x 2 -term if

f (x) is expanded into polynomial form. 28. For the rational function

f (x) (x 3)(x 2) __________________ (x 2)(x 5)(x 7) ,

a. Write the x-coordinate(s) of any removable discontinuities.

b. Write the equation(s) of any vertical asymptotes.

c. Write any x-intercept(s).

29. Resolve 18 _________ (x 4)(x 2) into partial fractions.

Section 4-9: Cumulative Review, Chapters 1–4240 Chapter 4: Polynomial and Rational Functions

10. Sketch graphs of these functions. a. f (x) 2(x 3 ) 2 5 b. g (x) e x 2 c. h(x) log 10 x 1

11. Name the pattern followed by the y-values of functions with regularly spaced x-values for

a. Logarithmic functions b. Power functions c. Quadratic functions

12. Name the type of function shown by each graph. a.

b.

c.

d.

e.

f.

13. Use the appropriate property or de�nition to complete the statement.

a. p log c m if and only if ? b. log 7 log 8 log ? c. log 32 ? log 2

In Chapter 3, you �t equations of functions to scattered data.

For Problems 14–16, Figure 4-9d shows a set of points with a dashed horizontal line drawn at y _ y , where _ y is the mean of the y-values.

y

y y

x

Figure 4-9d

14. On a copy of the �gure, sketch the deviation from the mean for the rightmost point.

15. Sketch what you think is the best-�tting linear function. Show the residual for the rightmost point. How does the size of the residual compare with the size of the deviation?

16. Based on what you know about the mean-mean point, ( __

x , _ y ), show where __ x would be on the

graph in Problem 14. 17. Suppose the regression equation for a set of data

is y 3x 5. What does the residual equal for the data point (4, 15)?

18. For a regression line, what is true about the sum of the squares of the residuals, SS res ?

19. Suppose the sum of the squares of the deviations is SS dev 100 and SS res 36. What does the coe�cient of determination, r 2 , equal?

y

x

y

x

y

x

y

x

y

x

x

y


20. The power function fits the data very well, but there is still some unexplained pattern in the data. (The negative r indicates that there is a downward trend to the data.) 21. Add–multiply semilog. Taking the logarithm of the y-values compresses them into a narrower range.22. y (8) 360; x 12.6 23. log y 5 (log 5000) 1 (log 0.72)x, which shows that log y is linear in x.24. 15 x 2 1 11x 2 1425. (x 2 12)(x 1 1)26. 49 1 42i 1 9 i 2 5 40 1 42i 27a. Cubic (third degree)27b. x 5 2, 2627c. 2, 26, and 26 (a double zero)27d. 1028a. x 5 2228b. x 5 25 and x 5 728c. x-intercept: x 5 329. 3 _____ x 2 4 1 23 _____ x 1 2

243

33. Spindletop Problem: On January 10, 1901, the �rst oil-well gusher in Texas happened at Spindletop near Beaumont. In the following months, many more wells were drilled. Assume that the data in the table are number of wells, y, as a function of number of months, x, a�er January 10, 1901.

x (months) y (wells)

0 1

1 3

2 10

3 27

4 75

5 150

a. Run a regression to �nd the particular equation for the best-�tting logistic function for these data.

b. Plot the logistic function on your grapher. Sketch the result on a copy of Figure 4-9h. Show the point of in�ection.

y (wells)

x (months)

400

300

200

100

15105Figure 4-9h

c. How many wells does your model predict were ultimately drilled at Spindletop?

d. Why is a logistic model more reasonable than an exponential model for this problem?

34. Cricket Problem: �e frequency at which crickets chirp increases as the temperature increases. Let x be the temperature, in degrees Fahrenheit, and let y be the number of chirps per second. Suppose the data in the table have been measured. �e data are graphed in Figure 4-9i.

x y

50 3555 5560 7465 9370 11275 13080 14785 16590 18295 200

y

10080604020

200

150

100

50

x

a. A linear function appears to �t the data. Write the linear regression equation, and give numerical evidence from the regression result that a linear function �ts very well.

b. If you extrapolate to temperatures below 50°F, what does the linear function indicate will eventually happen to the crickets? What, then, would be a reasonable lower bound for the domain of the linear function?

c. If you extrapolate to temperatures above 95°F, what does the linear function indicate will eventually happen to the crickets? Is this a reasonable endpoint behavior?

d. Enter a list into your grapher to calculate the residuals, y y. Use the results to make a residual plot. Sketch the residual plot. What information do you get from the residual plot concerning how well the linear function �ts the data?

35. Logarithmic Function Problem: a. Use the de�nition of logarithm to evaluate

log 9 53. b. Use the log-of-a-power property to solve the

exponential equation 3 4x 93. c. Use the change-of-base property to evaluate

log 5 47 using natural logarithms. d. Plot the graph of f (x) ln x. Sketch the

result. Explain why f (1) 0. Give numerical evidence to show how f (5) and f (7) are related to f (35).


Figure 4-9i


Part 2: Graphing calculators allowed (30–37) 30. Figure 4-9f shows the graph of the function

f (x) x 2 7. a. With your grapher in parametric mode, plot

the graph of function f and its inverse relation. Sketch the graph of the inverse relation on a copy of the given �gure.

b. Find an equation for the inverse of function f. Explain why you are not permitted to use the symbol f 1 (x) for this inverse relation.

f

1010

y

x

5

5

10

5

10

5

Figure 4-9f

31. Figure 4-9g shows the graph of the parametric function

x t 3 7t y 9 2 t 2

Find the coordinates of the points (x, y) for t 1, t 2, and t 3. Do these points lie on the given graph?

1010

y

x

5

5

10

5

10

5

Figure 4-9g

32. Light Intensity Problem: �e table shows the intensity, y, of light beneath the water surface as a function of distance, x, in meters, below the surface.

x y

0 1003 506 258 16

11 817 2

a. What pattern do the �rst three data points follow? What type of function has this pattern?

b. Find the particular equation for the function in part a algebraically by substituting the second and third points into the general equation. Show that the particular equation gives values for the last three points that are close to the values in the table.

c. Run the appropriate kind of regression to �nd the function of the type in part a that best �ts all six data points. Write the correlation coe�cient, and explain how it indicates that the function �ts the data quite well.

d. Use the regression equation from part c to predict the light intensity at a depth of 14 m. Which do you use, interpolation or extrapolation, to �nd this intensity? How do you decide?



A blackline master for Problem 30 is available in the Instructor’s Resource Book.30a.

30b. y 5  _____

x 1 7 ; The inverse relation is not a function.31. When t 5 21, the coordinates are (6, 7). When t 5 2, the coordinates are (26, 1). When t 5 3, the coordinates are (6, 29). The three points lie on the graph.

�5 10�10 5

10

5

�5

�10

t � 2

t � �1

y

x

t � 3

32a. Add–multiply; Exponential function32b. y 5 100 (0.7937...) x 32c. y 5 99.9085...(0.794 6...) x r 5 20.999993537..., indicating a good fit because it is close to 21.32d. y 5 3.9989...  4.0 units Interpolation, because 14 m is within the given data.33a. y 5 263.8637... ____________________ 1 1 314.7597... e 21.2056...x

10

5

�5

�10

�5 10�10 5

y

x

243

33. Spindletop Problem: On January 10, 1901, the �rst oil-well gusher in Texas happened at Spindletop near Beaumont. In the following months, many more wells were drilled. Assume that the data in the table are number of wells, y, as a function of number of months, x, a�er January 10, 1901.

x (months) y (wells)

0 1

1 3

2 10

3 27

4 75

5 150

a. Run a regression to �nd the particular equation for the best-�tting logistic function for these data.

b. Plot the logistic function on your grapher. Sketch the result on a copy of Figure 4-9h. Show the point of in�ection.

y (wells)

x (months)

400

300

200

100

15105Figure 4-9h

c. How many wells does your model predict were ultimately drilled at Spindletop?

d. Why is a logistic model more reasonable than an exponential model for this problem?

34. Cricket Problem: �e frequency at which crickets chirp increases as the temperature increases. Let x be the temperature, in degrees Fahrenheit, and let y be the number of chirps per second. Suppose the data in the table have been measured. �e data are graphed in Figure 4-9i.

x y

50 3555 5560 7465 9370 11275 13080 14785 16590 18295 200

y

10080604020

200

150

100

50

x

a. A linear function appears to �t the data. Write the linear regression equation, and give numerical evidence from the regression result that a linear function �ts very well.

b. If you extrapolate to temperatures below 50°F, what does the linear function indicate will eventually happen to the crickets? What, then, would be a reasonable lower bound for the domain of the linear function?

c. If you extrapolate to temperatures above 95°F, what does the linear function indicate will eventually happen to the crickets? Is this a reasonable endpoint behavior?

d. Enter a list into your grapher to calculate the residuals, y y. Use the results to make a residual plot. Sketch the residual plot. What information do you get from the residual plot concerning how well the linear function �ts the data?

35. Logarithmic Function Problem: a. Use the de�nition of logarithm to evaluate

log 9 53. b. Use the log-of-a-power property to solve the

exponential equation 3 4x 93. c. Use the change-of-base property to evaluate

log 5 47 using natural logarithms. d. Plot the graph of f (x) ln x. Sketch the

result. Explain why f (1) 0. Give numerical evidence to show how f (5) and f (7) are related to f (35).


Figure 4-9i


Part 2: Graphing calculators allowed (30–37) 30. Figure 4-9f shows the graph of the function

f (x) x 2 7. a. With your grapher in parametric mode, plot

the graph of function f and its inverse relation. Sketch the graph of the inverse relation on a copy of the given �gure.

b. Find an equation for the inverse of function f. Explain why you are not permitted to use the symbol f 1 (x) for this inverse relation.

f

1010

y

x

5

5

10

5

10

5

Figure 4-9f

31. Figure 4-9g shows the graph of the parametric function

x t 3 7t y 9 2 t 2

Find the coordinates of the points (x, y) for t 1, t 2, and t 3. Do these points lie on the given graph?

1010

y

x

5

5

10

5

10

5

Figure 4-9g

32. Light Intensity Problem: �e table shows the intensity, y, of light beneath the water surface as a function of distance, x, in meters, below the surface.

x y

0 1003 506 258 16

11 817 2

a. What pattern do the �rst three data points follow? What type of function has this pattern?

b. Find the particular equation for the function in part a algebraically by substituting the second and third points into the general equation. Show that the particular equation gives values for the last three points that are close to the values in the table.

c. Run the appropriate kind of regression to �nd the function of the type in part a that best �ts all six data points. Write the correlation coe�cient, and explain how it indicates that the function �ts the data quite well.

d. Use the regression equation from part c to predict the light intensity at a depth of 14 m. Which do you use, interpolation or extrapolation, to �nd this intensity? How do you decide?


34a. y 5 3.6472...x 2 145.1272... r 5 0.999717..., indicating a good fit.34b. Extrapolating to lower values of x indicates that the number of chirps per minute becomes negative, which is unreasonable. The domain should be about x 40.34c. Extrapolation to higher values of x indicates that the chirp rate keeps increasing. Actually, the crickets will die if the temperature gets too high.34d. The graph shows a definite pattern. So there is something in the data that is not accounted for by the linear function.

20 40 60 80 100

1

2

�1

�2

x

Residual

35a. 1.8069...

35b. x 5 1.0314...35c. 2.3922...35d.

f (1) 5 0 because ln 1 5 0, because ln 1 5 lo g e 1, and e 0 5 1. f(5) 5 ln 5 5 1.6094...; f(7) 5 ln 7 5 1.9459... f(35) 5 ln 35 5 3.5553... 5 1.6094... 1 1.9459... 5 f(5) 1 f(7)

2

1

�2

f(x) � ln x

x

A blackline master for Problem 33b is available in the Instructor’s Resource Book.33b. Actual point of inflection is (4.7707..., 131.9318...).

5 10 15

100

200

300

400y (oil wells)

Point of inflectionx (months)

33c. About 264 wells33d. An exponential model predicts that the number of oil wells would grow without bound. The logistic model shows that the number of wells will level off because of overcrowding.


36. Tree Problem: Ann R. Burr has a tree nursery in which she stocks various sizes of live oak trees. Figure 4-9j shows the price, f (x), in dollars, she charges for a tree of height x, in feet, planted on a customer’s property. For short trees, the price per foot decreases as the height increases. For taller trees, the price per foot increases because the trees are harder to move and plant.

f x

x

Figure 4-9j

a. �e particular equation for this function is the cubic function

f (x) x 3 25 x 2 249x 225 Show by synthetic substitution that x 1 is a

zero of this function. What is the real-world meaning of this fact?

b. Find the other two zeros of function f. How does the fact that these are nonreal zeros agree with the given graph?

c. Add the three zeros of function f. Where does this sum appear in the equation for f (x)?

d. �e graph goes o� the scale. If the vertical scale were extended far enough, how much would a 20-� tree cost?

e. �e slope of the graph changes from decreasing to increasing at the point of in�ection. Find the x-coordinate of this point. Does this value agree with the graph?

f. �e average rate of change in price from 10 � to x � is given by the rational function r(x) f (x) f (10)

________ x 10 . Find an equation for r(x) explicitly in terms of x. Using appropriate algebra, show that the graph of function r has a removable discontinuity at x 10. Find the y-value of this discontinuity.

g. �e y-value you found in part f is the instantaneous rate of change in price at x 10 �. On a copy of the given graph, plot a line through the point 10, f (10) with slope equal to this instantaneous rate. Describe the relationship of this line to the given graph.

37. What is the most important thing you have learned as a result of completing this Cumulative Review?



36a.

f (x) 5 (x 2 1)( x2 2 24x 1 225), so x 5 1 is a zero of function f, which means 1-foot trees are free!36b. x 5 12 9i, which agrees with the fact that there is only one real zero (one x-intercept).36c. 25, which equals the opposite of the x 2 -coefficient.36d. $275536e. x 5 8.3333..., which agrees with the graph, as shown in part g.36f. r (x) 5 x 2 2 15x 1 99, x 10 There is a removable discontinuity in the graph of function r at x 5 10, with y-coordinate 49.

A blackline master for Problem 36g is available in the Instructor’s Resource Book.36g. The graph shows that the line through (10, 765) with slope 49 is tangent to the graph of function f at that point. The instantaneous rate of change at x 5 10 is $49/ft.

5x ft

10 15 20

500

1000

1500

2000

Point of inflection

Tangent to graph

f (x) dollars

37. Answers will vary.

1 1 225 249 22251 224 225

1 224 225 0

polynomial and rational functions · polynomial and rational functions and their graphs. •...

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