polynomial and rational functions

ONE OF THE JOYS OF YOUR LIFE IS YOUR

dog, your very special buddy. Lately,however, you’ve noticed that your

companion is slowing down a bit. He’snow 8 years old and you wonder how thistranslates into human years. Youremember something about every year of

a dog’s life being equal to seven yearsfor a human. Is there a more accuratedescription?

This problem appears asExercises 63–64 in Exercise Set 2.5.

HERE IS A FUNCTION THATmodels the age in humanyears, of a dog that is xyears old:

The function contains variablesto powers that are wholenumbers and is an example of apolynomial function. In this

chapter, we study polynomialfunctions and functions that

consist of quotients of polynomials,called rational functions.

-1.2367x2+ 11.460x + 2.914.

H1x2 = -0.001618x4+ 0.077326x3

H1x2,

265

CHAPTER

2Polynomial and Rational Functions

T

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266 Chapter 2 • Polynomial and Rational Functions

Objectives❶ Add and subtract complex

numbers.

❷ Multiply complex numbers.

❸ Divide complex numbers.

❹ Perform operations withsquare roots of negativenumbers.

❺ Solve quadratic equationswith complex imaginarysolutions. Who is this kid warning us about our

eyeballs turning black if we attempt tofind the square root of Don’tbelieve what you hear on the street.Although square roots of negative num-bers are not real numbers, they do play asignificant role in algebra. In this sec-tion, we move beyond the real numbersand discuss square roots with negativeradicands.

The Imaginary Unit iIn the next section, we will study equations whose solutions may involve the squareroots of negative numbers. Because the square of a real number is never negative,there is no real number x such that To provide a setting in which such equa-tions have solutions, mathematicians invented an expanded system of numbers, thecomplex numbers. The imaginary number i, defined to be a solution of the equation

is the basis of this new set.

The Imaginary Unit iThe imaginary unit i is defined as

where

Using the imaginary unit i, we can express the square root of any negativenumber as a real multiple of i. For example,

We can check this result by squaring and obtaining

A new system of numbers, called complex numbers, is based on adding multi-ples of i, such as to the real numbers.

Complex Numbers and Imaginary NumbersThe set of all numbers in the form

with real numbers a and b, and i, the imaginary unit, is called the set of complexnumbers. The real number a is called the real part and the real number b iscalled the imaginary part of the complex number If then thecomplex number is called an imaginary number (Figure 2.1). An imaginarynumber in the form bi is called a pure imaginary number.

b Z 0,a + bi.

a + bi

5i,

15i22 = 52 i2

= 251-12 = -25

-25.5i

2-25 = 2-1 225 = i225 = 5i.

i2= -1.i = 2-1,

x2= -1,

x2= -1.

-9?

© 2005 Roz Chast from Cartoonbank.com.All rights reserved.

SECTION 2.1 Complex Numbers

Complex numbersa + bi

Realnumbers

a + bi with b = 0

Imaginarynumbers

a + bi with b ≠ 0

Figure 2.1 The complex numbersystem

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Section 2.1 • Complex Numbers 267

The following examples, usingthe same integers as in Example1, show how operations withcomplex numbers are just likeoperations with polynomials.

a.

b.

= 6 + 7x

= -5 + x + 11 + 6x

1-5 + x2 - 1-11 - 6x2= 12 - 7x

15 - 11x2 + 17 + 4x2

❶ Add and subtract complexnumbers.

Here are some examples of complex numbers. Each number can be written inthe form

Can you see that b, the imaginary part, is not zero in the first two complex numbers?Because these complex numbers are imaginary numbers. Furthermore, theimaginary number is a pure imaginary number. By contrast, the imaginary part ofthe complex number on the right is zero. This complex number is not an imaginarynumber. The number 3, or is a real number.

A complex number is said to be simplified if it is expressed in the standardform If b is a radical, we usually write i before b. For example, we write

rather than which could easily be confused with Expressed in standard form, two complex numbers are equal if and only if

their real parts are equal and their imaginary parts are equal.

Equality of Complex Numbersif and only if and

Operations with Complex NumbersThe form of a complex number is like the binomial Consequently, wecan add, subtract, and multiply complex numbers using the same methods we used forbinomials, remembering that

Adding and Subtracting Complex Numbers

1.In words, this says that you add complex numbers by adding their real parts,adding their imaginary parts, and expressing the sum as a complex number.

2.In words, this says that you subtract complex numbers by subtracting theirreal parts, subtracting their imaginary parts, and expressing the difference asa complex number.

EXAMPLE 1 Adding and Subtracting Complex Numbers

Perform the indicated operations, writing the result in standard form:

a. b.

Solution

a.

Remove the parentheses.

Group real and imaginary terms.

Add real parts and add imaginary parts.

Simplify.

b.

Remove the parentheses. Change signs of real and imag-inary parts in the complex number being subtracted.

Group real and imaginary terms.

Add real parts and add imaginary parts.

Simplify.= 6 + 7i

= 1-5 + 112 + 11 + 62i= -5 + 11 + i + 6i

= -5 + i + 11 + 6i

1-5 + i2 - 1-11 - 6i2= 12 - 7i

= 15 + 72 + 1-11 + 42i= 5 + 7 - 11i + 4i

= 5 - 11i + 7 + 4i

15 - 11i2 + 17 + 4i2

1-5 + i2 - 1-11 - 6i2.15 - 11i2 + 17 + 4i2

1a + bi2 - 1c + di2 = 1a - c2 + 1b - d2i

1a + bi2 + 1c + di2 = 1a + c2 + 1b + d2i

i2= -1.

a + bx.a + bi

b = d.a = ca + bi = c + di

7 + 15i .7 + 15i,7 + i15a + bi.

3 + 0i,

2ib Z 0,

–4+6i

a, the realpart, is −4.

b, the imaginarypart, is 6.

3=3+0i

a, the realpart, is 3.


2i=0+2i

a, the realpart, is 0.


a + bi.

Study Tip

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❷ Multiply complex numbers.

❸ Divide complex numbers.

Add or subtract as indicated:

a. b.

Multiplication of complex numbers is performed the same way as multiplicationof polynomials, using the distributive property and the FOIL method.After completingthe multiplication, we replace any occurrences of with This idea is illustrated inthe next example.

EXAMPLE 2 Multiplying Complex Numbers

Find the products:

a. b.

Solutiona.

Distribute 4i throughout the parentheses.Multiply.Replace with Simplify to and write in standard form.

b.

Use the FOIL method.

Group real and imaginary terms.Combine real and imaginary terms.

Find the products:

a. b.

Complex Conjugates and DivisionIt is possible to multiply complex numbers and obtain a real number. This occurswhen we multiply and

Use the FOIL method.

Notice that this product eliminates i.

For the complex number we define its complex conjugate to be The multiplication of complex conjugates results in a real number.

Conjugate of a Complex NumberThe complex conjugate of the number is and the complex conju-gate of is The multiplication of complex conjugates gives a realnumber.

Complex conjugates are used to divide complex numbers. By multiplying thenumerator and the denominator of the division by the complex conjugate of thedenominator, you will obtain a real number in the denominator.

1a - bi21a + bi2 = a2+ b2

1a + bi21a - bi2 = a2+ b2

a + bi.a - bia - bi,a + bi

a - bi.a + bi,

= a2+ b2

i2 � � 1 = a2- b21-12

=a2-abi+abi-b2i2

F O I L

1a + bi21a - bi2

a - bi.a + bi

15 + 4i216 - 7i2.7i12 - 9i2CheckPoint 2

= -29 - 29i= -14 - 15 - 35i + 6i

i2 � � 1= -14 - 35i + 6i + 151-12=–14-35i+6i+15i2

F O I L

17 - 3i21-2 - 5i212i � 20= 20 + 12i

� 1.i2= 12i - 201-12= 12i - 20i2= 4i # 3 - 4i # 5i

4i13 - 5i2

17 - 3i21-2 - 5i2.4i13 - 5i2

-1.i2

12 + 6i2 - 112 - i2.15 - 2i2 + 13 + 3i2CheckPoint 1

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❹ Perform operations withsquare roots of negativenumbers.

EXAMPLE 3 Using Complex Conjugates to Divide Complex Numbers

Divide and express the result in standard form:

Solution The complex conjugate of the denominator, is Multipli-cation of both the numerator and the denominator by will eliminate fromthe denominator.

Multiply the numerator and the denominator by thecomplex conjugate of the denominator.

14+35i+8i+20i2

22+52=

F O I L

7 + 4i

2 - 5i=

17 + 4i212 - 5i2 #

12 + 5i212 + 5i2

i2 + 5i2 + 5i.2 - 5i,

7 + 4i

2 - 5i.

Use the FOIL method in the numerator andin the denominator.

Combine imaginary terms and replace with

Combine real terms in the numerator:

Express the answer in standard form.

Observe that the quotient is expressed in the standard form with and

Divide and express the result in standard form:

Roots of Negative NumbersThe square of and the square of both result in :

Consequently, in the complex number system has two square roots, namely,and We call the principal square root of

Principal Square Root of a Negative NumberFor any positive real number b, the principal square root of the negative num-ber is defined by

Consider the multiplication problem

This problem can also be given in terms of principal square roots of negativenumbers:

Because the product rule for radicals only applies to real numbers, multiplyingradicands is incorrect. When performing operations with square roots of negativenumbers, begin by expressing all square roots in terms of i. Then perform theindicated operation.

2-25 # 2-4.

5i # 2i = 10i2= 10(-1) = -10.

2-b = i2b .

-b

-16.4i-4i.4i-16

14i22 = 16i2= 161-12 = -16 1-4i22 = 16i2

= 161-12 = -16.

-16-4i4i

5 + 4i

4 - i.Check

Point 3

b =4329 .

a = - 629a + bi,

= - 629

+

4329

i

14 � 201� 12 � 14 � 20 � �6. =

-6 + 43i

29

� 1.i2 =

14 + 43i + 201-1229

1a � bi21a � bi2 � a2 � b2

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Correct: Incorrect:

EXAMPLE 4 Operations Involving Square Rootsof Negative Numbers

Perform the indicated operations and write the result in standard form:

a. b. c.

Solution Begin by expressing all square roots of negative numbers in terms of i.

a.

b.

c.

Write the complex number in standard form.

Simplify.

Perform the indicated operations and write the result in standard form:CheckPoint 4

= - 53

+ i223

=

-2515

+

5i2215

250 � 225 # 2 � 522=

-25 + 5i2215

2�b � i2b=

-25 + i25015

-25 + 2-5015

= -4 - 2i25

= 1 - 2i25 + 51-12 = 1 - 2i25 + 5i2

(A + B)2 = A2 + 2 A B + B2

(–1+�–5B2=(–1+i�5B2=(–1)2+2(–1)(i�5B+(i�5B2

= 3i22 - 2i22 = i22

2-18 - 2-8 = i218 - i28 = i29 # 2 - i24 # 2

-25 + 2-5015

.A -1 + 2-5 B22-18 - 2-8

= 10 = 10i2= 101-12 = -10

= 2100 = 5i # 2i

2-25 # 2-4 = 2-251-42 2-25 # 2-4 = i225 # i24

a. b. c.

Quadratic Equations with Complex Imaginary SolutionsWe have seen that a quadratic equation can be expressed in the general form

All quadratic equations can be solved by the quadratic formula:

Recall that the quantity which appears under the radical sign in thequadratic formula, is called the discriminant. If the discriminant is negative, aquadratic equation has no real solutions. However, quadratic equations withnegative discriminants do have two solutions. These solutions are imaginarynumbers that are complex conjugates.

b2- 4ac,

x =

-b ; 3b2- 4ac

2a.

ax2+ bx + c = 0, a Z 0.

-14 + 2-122

.A -2 + 2-3 B22-27 + 2-48

This stamp honors the work doneby the German mathematicianCarl Friedrich Gauss (1777–1855)with complex numbers. Gaussrepresented complex numbers aspoints in the plane.

Complex Numbers ona Postage Stamp


If you need to review quadraticequations and how to solvethem, read Section P. 7, begin-ning on page 84.

❺ Solve quadratic equationswith complex imaginarysolutions.

Study Tip

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EXAMPLE 5 A Quadratic Equation with Imaginary Solutions

Solve using the quadratic formula:

Solution The given equation is in general form. Begin by identifying the valuesfor a, b, and c.

Use the quadratic formula.

Substitute the values for a, b, and c: and

Subtract under the radical. Because the numberunder the radical sign is negative, the solutions willnot be real numbers.

Factor 2 from the numerator.

Divide numerator and denominator by 2.

Write the complex numbers in standard form.

The solutions are complex conjugates, and the solution set is

Solve using the quadratic formula:

x2- 2x + 2 = 0.

CheckPoint 5

e 13

+ i 111

3,

13

- i 111

3f or e 1

3; i 111

3f .

=

13

; i 211

3

=

1 ; i2113

=

2 A1 ; i211 B6

� 2i211 2�44 � 4411121� 12 =

2 ; 2i2116

=

2 ; 2-446

�1�22 � 2 and 1�222 � 1�221�22 � 4. =

2 ; 24 - 486

c � 4,b � �2,a � 3,

=

-1-22 ; 41-222 - 41321422132

x =

-b ; 3b2- 4ac

2a

3x2-2x+4=0

a = 3 b = −2 c = 4

3x2- 2x + 4 = 0.

EXERCISE SET 2.1

Practice ExercisesIn Exercises 1–8, add or subtract as indicated and write

the result in standard form.

1. 2.

3. 4.

5.

6.

7. 8. 15i - 112 - 11i28i - 114 - 9i27 - 1-9 + 2i2 - 1-17 - i26 - 1-5 + 4i2 - 1-13 - i2

1-7 + 5i2 - 1-9 - 11i213 + 2i2 - 15 - 7i21-2 + 6i2 + 14 - i217 + 2i2 + 11 - 4i2

In Exercises 9–20, find each product and write the result instandard form.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20. 15 - 2i2212 + 3i221-7 - i21-7 + i21-5 + i21-5 - i212 + 7i212 - 7i213 + 5i213 - 5i218 - 4i21-3 + 9i217 - 5i21-2 - 3i21-4 - 8i213 + i21-5 + 4i213 + i2-8i12i - 72-3i17i - 52

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In Exercises 21–28, divide and express the result in standard form.

21. 22.

23. 24.

25. 26.

27. 28.

In Exercises 29–44, perform the indicated operations and writethe result in standard form.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

41. 42.

43. 44.

In Exercises 45–50, solve each quadratic equation using thequadratic formula. Express solutions in standard form.

45. 46.

47. 48.

49. 50.

Practice PlusIn Exercises 51–56, perform the indicated operation(s)

and write the result in standard form.

51.

52.

53.

54.

55.

56.

57. Evaluate

58. Evaluate

59. Evaluate for

60. Evaluate for

Application ExercisesComplex numbers are used in electronics to describe thecurrent in an electric circuit. Ohm’s law relates the cur-

rent in a circuit, I, in amperes, the voltage of the circuit, E, in volts,and the resistance of the circuit, R, in ohms, by the formula

Use this formula to solve Exercises 61–62.

61. Find E, the voltage of a circuit, if amperes andohms.R = 13 + 7i2

I = 14 - 5i2E = IR.

x = 4i.x2

+ 113 - x

x = 3i.x2

+ 192 - x

x2- 2x + 5 for x = 1 - 2i.

x2- 2x + 2 for x = 1 + i.

52-8 + 32-18

52-16 + 32-81

14 - i22 - 11 + 2i2212 + i22 - 13 - i2218 + 9i212 - i2 - 11 - i211 + i212 - 3i211 - i2 - 13 - i213 + i2

3x2= 4x - 63x2

= 8x - 7

2x2+ 2x + 3 = 04x2

+ 8x + 13 = 0

x2- 2x + 17 = 0x2

- 6x + 10 = 0

A32-7 B A22-8 BA32-5 B A -42-12 B2-12 A2-4 - 22 B2-8 A2-3 - 25 B-15 - 2-18

33-6 - 2-12

48

-12 + 2-2832

-8 + 2-3224

A -2 + 2-11 B2A -3 - 2-7 B2A -5 - 2-9 B2A -2 + 2-4 B252-8 + 32-1852-16 + 32-81

2-81 - 2-1442-64 - 2-25

3 - 4i

4 + 3i

2 + 3i

2 + i

-6i

3 + 2i

8i

4 - 3i

5i

2 - i

2i

1 + i

34 + i

23 - i

62. Find E, the voltage of a circuit, if amperes andohms.

63. The mathematician Girolamo Cardano is credited with thefirst use (in 1545) of negative square roots in solving thenow-famous problem, “Find two numbers whose sum is 10and whose product is 40.” Show that the complex numbers

and satisfy the conditions of the prob-lem. (Cardano did not use the symbolism or even

. He wrote R⁄ .m 15 for , meaning “radix minus 15.”He regarded the numbers R⁄ .m 15 and R⁄ .m 15 as“fictitious” or “ghost numbers,” and considered theproblem “manifestly impossible.” But in a mathematicallyadventurous spirit, he exclaimed, “Nevertheless, we willoperate.”)

Writing in Mathematics64. What is i?

65. Explain how to add complex numbers. Provide an examplewith your explanation.

66. Explain how to multiply complex numbers and give anexample.

67. What is the complex conjugate of ? What happenswhen you multiply this complex number by its complexconjugate?

68. Explain how to divide complex numbers. Provide an examplewith your explanation.

69. Explain each of the three jokes in the cartoon on page 266.

70. A stand-up comedian uses algebra in some jokes, includingone about a telephone recording that announces “You havejust reached an imaginary number. Please multiply by i anddial again.” Explain the joke.

Explain the error in Exercises 71–72.

71.

72.

Critical Thinking Exercises73. Which one of the following is true?

a. Some irrational numbers are not complex numbers.

b. is an imaginary number.

c.

d. In the complex number system, (the sum of twosquares) can be factored as

In Exercises 74–76, perform the indicated operations and writethe result in standard form.

74. 75.

76.8

1 +

2i

1 + i

1 + 2i+

1 - i

1 - 2i

4(2 + i)(3 - i)

1x + yi21x - yi2.x2

+ y2

7 + 3i

5 + 3i=

75

13 + 7i213 - 7i2

A2-9 B2 = 2-9 # 2-9 = 281 = 9

2-9 + 2-16 = 2-25 = i225 = 5i

2 + 3i

5 -5 +

1-151-15i115

5 - i1155 + i115

R = 13 + 5i2I = 12 - 3i2

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Section 2.2 • Quadratic Functions 273

Objectives❶ Recognize characteristics

of parabolas.

❷ Graph parabolas.

❸ Determine a quadraticfunction’s minimum ormaximum value.

❹ Solve problems involving aquadratic function’sminimum or maximumvalue.

The Food Stamp Program is the first line of defense against hunger for millions ofAmerican families. The program provides benefits for eligible participants to pur-chase approved food items at approved food stores. Over half of all participants arechildren; one out of six is a low-income older adult. The function

models the number of households, , in millions, participating in the program xyears after 1999. For example, to find the number of households receiving foodstamps in 2005, substitute 6 for x because 2005 is 6 years after 1999:

Thus, in 2005, 12.6 million households received food stamps.The function is an example of a quadratic func-

tion. We have seen that a quadratic function is any function of the form

where a, b, and c are real numbers, with A quadratic function is a polynomialfunction whose greatest exponent is 2. In this section, we study quadratic functionsand their graphs.

Graphs of Quadratic FunctionsThe graph of any quadratic function is called a parabola.Parabolas are shaped like cups,as shown in Figure 2.2. If the coefficient of (the value of a in ) is posi-tive, the parabola opens upward. If the coefficient of is negative, the graph opensdownward.The vertex (or turning point) of the parabola is the lowest point on the graphwhen it opens upward and the highest point on the graph when it opens downward.

x2ax2

+ bx + cx2

a Z 0.

f1x2 = ax2+ bx + c,

f1x2 = 0.22x2- 0.50x + 7.68

f162 = 0.221622 - 0.50162 + 7.68 = 12.6.

f1x2f1x2 = 0.22x2

- 0.50x + 7.68

x

y

a > 0: Parabola opens upward. a < 0: Parabola opens downward.

Vertex (minimum point)

f(x) = ax2 + bx + c (a > 0)

Axis of symmetry

Vertex (maximum point)

f(x) = ax2 + bx + c (a < 0)

Axis of symmetryx

y

Figure 2.2 Characteristics of graphs of quadratic functions

❶ Recognize characteristics ofparabolas.

SECTION 2.2 Quadratic Functions

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g(x) = a(x − h)2 + k

Vertex: (0, 0)

Vertex: (0, 0)

f(x) = ax2

f(x) = ax2

Axis of symmetry: x = h Axis of symmetry: x = h

Vertex: (h, k)x

x

y y

g(x) = a(x − h)2 + kVertex: (h, k)

Look at the unusual image of the word “mirror” shown below.The artist, ScottKim, has created the image so thatthe two halves of the whole aremirror images of each other. Aparabola shares this kind of sym-metry, in which a line through thevertex divides the figure in half.Parabolas are symmetric with re-spect to this line, called the axis ofsymmetry. If a parabola is foldedalong its axis of symmetry, the twohalves match exactly.

❷ Graph parabolas.

Graphing Quadratic Functions in Standard FormIn our earlier work with transformations, we applied a series of transformations to thegraph of The graph of this function is a parabola. The vertex for thisparabola is In Figure 2.3(a), the graph of for is shown inblack; it opens upward. In Figure 2.3(b), the graph of for is shownin black; it opens downward.

a 6 0f1x2 = ax2a 7 0f1x2 = ax210, 02.f1x2 = x2.

Figure 2.3(a) and 2.3(b) also show the graph of in blue.Compare these graphs to those of Observe that h determines the hori-zontal shift and k determines the vertical shift of the graph of

Consequently, the vertex on the black graph of moves to the pointon the blue graph of The axis of symmetry is the

vertical line whose equation is The form of the expression for g is convenient because it immediately identi-

fies the vertex of the parabola as This is the standard form of a quadraticfunction.

1h, k2.x = h.

g1x2 = a1x - h22 + k.1h, k2 f1x2 = ax210, 02

g(x)=a(x-h)2+k.

If h > 0, the graph off(x) = ax2 is shifted h

units to the right.

If k > 0, the graph ofy = a(x − h)2 is shifted

k units up.

f1x2 = ax2 :

f1x2 = ax2.g1x2 = a1x - h22 + k

Figure 2.3(b)Parabola opens downward.a 6 0:

Figure 2.3(a)Parabola opens upward.a 7 0:

Transformations of f1x2 = ax2

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The Standard Form of a Quadratic FunctionThe quadratic function

is in standard form. The graph of is a parabola whose vertex is the pointThe parabola is symmetric with respect to the line If the

parabola opens upward; if the parabola opens downward.

The sign of a in determines whether the parabolaopens upward or downward. Furthermore, if is small, the parabola opens moreflatly than if is large. Here is a general procedure for graphing parabolas whoseequations are in standard form:

Graphing Quadratic Functions with Equations in Standard FormTo graph

1. Determine whether the parabola opens upward or downward. If itopens upward. If it opens downward.

2. Determine the vertex of the parabola. The vertex is 3. Find any x-intercepts by solving The function’s real zeros are the

x-intercepts.4. Find the y-intercept by computing 5. Plot the intercepts, the vertex, and additional points as necessary. Connect

these points with a smooth curve that is shaped like a cup.

In the graphs that follow, we will show each axis of symmetry as a dashed ver-tical line. Because this vertical line passes through the vertex, its equation is

The line is dashed because it is not part of the parabola.

EXAMPLE 1 Graphing a Quadratic Function in Standard Form

Graph the quadratic function

Solution We can graph this function by following the steps in the preceding box.We begin by identifying values for a, h, and k.

Step 1 Determine how the parabola opens. Note that a, the coefficient of isThus, this negative value tells us that the parabola opens downward.

Step 2 Find the vertex. The vertex of the parabola is at Because and the parabola has its vertex at

Step 3 Find the x-intercepts by solving Replace with 0 in

Find x-intercepts, setting equal to 0.

Solve for x. Add to both sidesof the equation.Divide both sides by 2.

Apply the square root property.

Add 3 to both sides in each equation.

The x-intercepts are 1 and 5. The parabola passes through and 15, 02.11, 02. x = 5 x = 1

24 � 2 x - 3 = 2 x - 3 = -2

x - 3 = 24 or x - 3 = -24

1x - 322 = 4

21x � 322 21x - 322 = 8

f1x2 0 = -21x - 322 + 8

f1x2 = -21x - 322 + 8.f1x2f1x2 � 0.

13, 82.k = 8,h = 31h, k2.

a 6 0;-2.x2,

f(x)=a(x-h)2+k

f(x)=–2(x-3)2+8

a = −2 h = 3 k = 8

Standard form

Given function

f1x2 = -21x - 322 + 8.

x = h.1h, k2,

f102.f1x2 = 0.

1h, k2.a 6 0,

a 7 0,

f1x2 = a1x - h22 + k,

ƒa ƒ

ƒa ƒ

f1x2 = a1x - h22 + k

a 6 0,a 7 0,x = h.1h, k2. f

f1x2 = a1x - h22 + k, a Z 0

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−1

1234

1312111098765

1 2 3−1−3−4−6−7 −5

y

x

Vertex: (−3, 1)

Axis of symmetry: x = −3

y-intercept: 10

Figure 2.5 The graph off1x2 = 1x + 322 + 1

−2

68

10

−4−6−8

−10

1 2 3 4 8765−1−2

y

x

Axis of symmetry: x = 3y-intercept: −10

Vertex: (3, 8)

x-intercept: 5x-intercept: 1

Figure 2.4 The graph off1x2 = -21x - 322 + 8


Step 4 Find the y-intercept by computing . Replace x with 0 in

The y-intercept is The parabola passes through

Step 5 Graph the parabola. With a vertex at x-intercepts at 1 and 5, and ay-intercept at the graph of is shown in Figure 2.4.The axis of symmetry is thevertical line whose equation is


EXAMPLE 2 Graphing a Quadratic Function in Standard Form


Solution We begin by finding values for a, h, and k.

Step 1 Determine how the parabola opens. Note that a, the coefficient of is 1.Thus, this positive value tells us that the parabola opens upward.

Step 2 Find the vertex. The vertex of the parabola is at Because and the parabola has its vertex at

Step 3 Find the x-intercepts by solving Replace with 0 inBecause the vertex is at which lies above the x-axis,

and the parabola opens upward, it appears that this parabola has no x-intercepts.Wecan verify this observation algebraically.

Find possible x-intercepts, settingequal to 0.

Solve for x. Subtract 1 from bothsides.

Apply the square root property.

The solutions are

Because this equation has no real solutions, the parabola has no x-intercepts.


The y-intercept is 10. The parabola passes through

Step 5 Graph the parabola. With a vertex at no x-intercepts, and ay-intercept at 10, the graph of is shown in Figure 2.5. The axis of symmetry is thevertical line whose equation is

Graph the quadratic function f1x2 = 1x - 222 + 1.CheckPoint 2

x = -3.f

1-3, 12,10, 102.

f102 = 10 + 322 + 1 = 32+ 1 = 9 + 1 = 10

f1x2 = 1x + 322 + 1.f102

�3_i. x = -3 + i x = -3 - i

2� 1 � i x + 3 = i x + 3 = - i

x + 3 = 2-1 or x + 3 = -2-1

-1 = 1x + 322f1x2 0 = 1x + 322 + 1

1-3, 12,f1x2 = 1x + 322 + 1.f1x2f1x2 � 0.

1-3, 12.k = 1,h = -31h, k2.

a 7 0;x2,

f(x)=1(x-(–3))2+1

f(x)=(x+3)2+1

f(x)=a(x-h)2+k

a = 1 h = −3 k = 1

Standard form of quadratic function

Given function

f1x2 = 1x + 322 + 1.

f1x2 = -1x - 122 + 4.CheckPoint 1

x = 3.f-10,

13, 82,10, -102.-10.

f102 = -210 - 322 + 8 = -21-322 + 8 = -2192 + 8 = -10

f1x2 = -21x - 322 + 8.f102

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Graphing Quadratic Functions in the Form Quadratic functions are frequently expressed in the form Howcan we identify the vertex of a parabola whose equation is in this form? Completingthe square provides the answer to this question.

Factor out a from

Write the trinomial as thesquare of a binomial andsimplify the constant term.

Compare this form of the equation with a quadratic function’s standard form.

The important part of this observation is that h, the x-coordinate of the vertex, is

The y-coordinate can be found by evaluating the function at

The Vertex of a Parabola Whose Equation Is

Consider the parabola defined by the quadratic function

The parabola’s vertex is

We can apply our five-step procedure and graph parabolas in the formThe only step that is different is how we determine the vertex.

EXAMPLE 3 Graphing a Quadratic Function in the Form

Graph the quadratic function Use the graph to identify thefunction’s domain and its range.

SolutionStep 1 Determine how the parabola opens. Note that a, the coefficient of is

Thus, this negative value tells us that the parabola opens downward.

Step 2 Find the vertex. We know that the x-coordinate of the vertex is We identify a, b, and c in

f(x)=–x2-2x+1

a = −1 c = 1b = −2

f1x2 = ax2+ bx + c.

x = - b

2a.

a 6 0;-1.x2,

f1x2 = -x2- 2x + 1.

f1x2 � ax2 � bx � c

f1x2 = ax2+ bx + c.

a - b

2a, fa -

b

2ab b .

f1x2 = ax2+ bx + c.

f1x2 � ax2 � bx � c

- b

2a.-

b

2a.

f(x)=aax- a– bb2+c-b2a

b2

4a

b2a

b2

4a

f(x)=a(x-h)2+k

h = − k = c −

Standard form

Equation under discussion

= a ¢x +

b

2a≤2

+ c -

b2

4a

a �

=aax2+ x+ b+c-aa bba

b2

4a2

b2

4a2

Complete the square by adding the square of half

the coefficient of x.

By completing the square, we added

. To avoid changing the

function’s equation, we mustsubtract this term.

b2

4a2

ax2 � bx. = aax2+

ba

xb + c

f1x2 = ax2+ bx + c

f1x2 = ax2+ bx + c.

f1x2 � ax2 � bx � c

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The domain of any quadraticfunction includes all real num-bers. If the vertex is the graph’shighest point, the range includesall real numbers at or below they-coordinate of the vertex. If thevertex is the graph’s lowestpoint, the range includes all realnumbers at or above they-coordinate of the vertex.


Figure 2.6(a) The graph off1x2 = -x2

- 2x + 1Study Tip

Substitute for a and for b into the equation for the x-coordinate:

The x-coordinate of the vertex is and the vertex is at We substitute for x in the equation of the function, to find the y-coordinate:

The vertex is at

Step 3 Find the x-intercepts by solving Replace with 0 inWe obtain This equation cannot be

solved by factoring. We will use the quadratic formula to solve it.0 = -x2

- 2x + 1.f1x2 = -x2- 2x + 1.

f1x2f1x2 � 0.

1-1, 22.f1-12 = -1-122 - 21-12 + 1 = -1 + 2 + 1 = 2.

f1x2 = -x2- 2x + 1,

-11-1, f1-122.-1

x = - b

2a= -

-221-12 = - a -2

-2b = -1.

-2-1

x

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3−4−5

Range: Outputson y-axis fall at

or below 2.

Domain: Inputson x-axis includeall real numbers.

Figure 2.6(b) Determining the domainand range of f1x2 = -x2

- 2x + 1

x

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3−4−5

y-intercept: 1Vertex: (−1, 2)

x-intercept: 0.4x-intercept: −2.4

Axis of symmetry: x = −1

Now we are ready to determine the domain and range of Wecan use the parabola, shown again in Figure 2.6(b), to do so.To find the domain, lookfor all the inputs on the x-axis that correspond to points on the graph. As the graphwidens and continues to fall at both ends, can you see that these inputs include allreal numbers?

To find the range, look for all the outputs on the y-axis that correspond to points onthe graph. Figure 2.6(b) shows that the parabola’s vertex, is the highest1-1, 22,

Domain of f is 5x ƒx is a real number6 or 1- q , q2.

f1x2 = -x2- 2x + 1.

To locate the x-intercepts, weneed decimal approximations.

Thus, there is no need to simplifythe radical form of the solutions.

–b_�b2-4acx= =2a

–(–2)_�(–2)2-4(–1)(1)=

2(–1)2_�4-(–4)

–2

x= or≠–2.42+�8

–2x= ≠0.4

2-�8–2

–x2-2x+1=0

a = −1 c = 1b = −2

The x-intercepts are approximately and 0.4. The parabola passes throughand


The y-intercept is 1. The parabola passes through

Step 5 Graph the parabola. With a vertex at x-intercepts at approxi-mately and 0.4, and a y-intercept at 1, the graph of is shown in Figure 2.6(a).The axis of symmetry is the vertical line whose equation is x = -1.

f-2.41-1, 22,10, 12.

f102 = -02- 2 # 0 + 1 = 1

f1x2 = -x2- 2x + 1.

f10210.4, 02.1-2.4, 02 -2.4

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❸ Determine a quadraticfunction’s minimum ormaximum value.

point on the graph. Because the y-coordinate of the vertex is 2, outputs on the y-axisfall at or below 2.

Graph the quadratic function Use the graph toidentify the function’s domain and its range.

Minimum and Maximum Values of Quadratic FunctionsConsider the quadratic function If the parabola opensupward and the vertex is its lowest point. If the parabola opens downward

and the vertex is its highest point.The x-coordinate of the vertex is Thus, we can

find the minimum or maximum value of by evaluating the quadratic function at

Minimum and Maximum: Quadratic FunctionsConsider the quadratic function

1. If then has a minimum that occurs at This minimum

value is

2. If then has a maximum that occurs at This maximum

value is

In each case, the value of x gives the location of the minimum or maximum

value. The value of y, or gives that minimum or maximum value.

EXAMPLE 4 Obtaining Information about a Quadratic Functionfrom Its Equation

Consider the quadratic function

a. Determine, without graphing, whether the function has a minimum value or amaximum value.

b. Find the minimum or maximum value and determine where it occurs.c. Identify the function’s domain and its range.

Solution We begin by identifying a, b, and c in the function’s equation:

a. Because the function has a maximum value.

b. The maximum value occurs at

The maximum value occurs at and the maximum value ofis

We see that the maximum is at x = 1.-10

f112 = -3 # 12+ 6 # 1 - 13 = -3 + 6 - 13 = -10.

f1x2 = -3x2+ 6x - 13

x = 1

x = - b

2a= -

621-32 = -

6-6

= -1-12 = 1.

a 6 0,

f(x)=–3x2+6x-13.

b = 6a = −3 c = −13

f1x2 = -3x2+ 6x - 13.

fa - b

2ab ,

fa - b

2ab .

x = - b

2a.fa 6 0,

fa - b

2ab .

x = - b

2a.fa 7 0,

f1x2 = ax2+ bx + c.

x = - b

2a.

f

- b

2a.

a 6 0,a 7 0,f1x2 = ax2

+ bx + c.

f1x2 = -x2+ 4x + 1.Check

Point 3

Range of f is 5y ƒy … 26 or 1- q , 24.

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❹ Solve problems involving aquadratic function’sminimum or maximum value.

c. Like all quadratic functions, the domain is Because the function’smaximum value is the range includes all real numbers at or below The range is

We can use the graph of to visualize the results of Example 4. Figure 2.7shows the graph in a by viewing rectangle. The maximum function featureverifies that the function’s maximum is at

Notice that x gives the location of the maxi-mum and y gives the maximum value. Notice, too,that the maximum value is and not the orderedpair

Repeat parts (a) through (c) of Example 4 using the quadratic function

Applications of Quadratic FunctionsWhen did the minimum number of households participate in the food stampprogram? What is the age of a driver having the least number of car accidents? Ifyou throw a baseball vertically upward, after how many seconds will it reach itsmaximum height and what is that height? The answers to these questions involvefinding the maximum or minimum value of a quadratic function, as well as wherethis value occurs.

EXAMPLE 5 The Food Stamp Program

Figure 2.8 shows the number ofU.S. households, in millions,participating in the Food StampProgram from 1999 through2004. The function

models the number of house-holds, , in millions, partici-pating in the program x yearsafter 1999. According to thisfunction, in which year was thenumber of participants at a min-imum? How many householdsreceived food stamps for thatyear? How well does this modelthe data shown in Figure 2.8?

Solution We begin by identifying a, b, and c in the function’s equation:

Because the function has a minimum value. The minimum value occurs at

x = - b

2a= -

1-0.502210.222 =

0.500.44

L 1.

a 7 0,

f(x)=0.22x2-0.50x+7.68.

a = 0.22 c = 7.68b = −0.50

f1x2

f1x2 = 0.22x2- 0.50x + 7.68

f1x2 = 4x2- 16x + 1000.

CheckPoint 4

11, -102. -10

x = 1.-10

3-50, 20, 1043-6, 6, 14f1x2 = -3x2

+ 6x - 13

1- q , -104. -10.-10,1- q , q2.

[−6, 6, 1] by [−50, 20, 10]

Range is (−∞, −10].

Figure 2.7

2004

10.6

2003

9.2

2002

8.2

2001

7.5

2000

7.4

11

10

9

8

7

Num

ber

Rec

eivi

ngFo

od S

tam

ps (

mill

ions

)

U.S. Households on Food Stamps

Year1999

7.7

6

Figure 2.8Source: Food Stamp Program

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Because of the decreasing-increasing cuplike shape of the data in Figure 2.8, a quadraticfunction is an appropriate model. We entered the datausing

(number of years after 1999, millions of participants).

Upon entering the QUADratic REGression program,we obtain the results shown in the screen on the right.Thus, the quadratic function of best fit is

where x represents the number of years after 1999 andrepresents the number of U.S. households, in mil-

lions, on food stamps.f1x2

f1x2 = 0.22x2- 0.50x + 7.68,

Technology

Data:(0, 7.7), (1,7.4), (2, 7.5),(3, 8.2), (4, 9.2), (5, 10.6)

This means that the number of households receiving food stamps was at a minimumapproximately 1 year after 1999, in 2000. Using the model

the number of households, in millions, for that year was

In 2000, the number of households receiving food stamps was at a minimum of 7.4million. Because this is precisely what is shown in Figure 2.8 on the previous page,the function models the data extremely well.

f112 = 0.221122 - 0.50112 + 7.68 = 7.4.

f1x2 = 0.22x2- 0.50x + 7.68,

The function models the number of acci-dents, , per 50 million miles driven, for drivers x years old, where

What is the age of a driver having the least number of caraccidents? What is the minimum number of car accidents per 50 millionmiles driven?

Quadratic functions can also be modeled from verbal conditions. Once wehave obtained a quadratic function, we can then use the x-coordinate of the vertexto determine its maximum or minimum value. Here is a step-by-step strategy forsolving these kinds of problems:

Strategy for Solving Problems Involving Maximizing or MinimizingQuadratic Functions

1. Read the problem carefully and decide which quantity is to be maximizedor minimized.

2. Use the conditions of the problem to express the quantity as a function inone variable.

3. Rewrite the function in the form

4. Calculate If has a minimum at This minimum

value is If has a maximum at This maximum

value is

5. Answer the question posed in the problem.

EXAMPLE 6 Minimizing a Product

Among all pairs of numbers whose difference is 10, find a pair whose product is assmall as possible. What is the minimum product?

fa - b

2ab .

x = - b

2a.fa 6 0,fa -

b

2ab .

x = - b

2a.fa 7 0,-

b

2a.

f1x2 = ax2+ bx + c.

16 … x … 74.f1x2 f1x2 = 0.4x2

- 36x + 1000CheckPoint 5

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The feature of a graph-ing utility can be used to verifyour work in Example 6.

Enter y1 = x2 − 10x, thefunction for the product,when one of the numbers

is x.

The product is aminimum, −25, whenone of the numbers

is 5.

� TABLE �Technology

Solution

Step 1 Decide what must be maximized or minimized. We must minimize theproduct of two numbers. Calling the numbers x and y, and calling the product P, wemust minimize

Step 2 Express this quantity as a function in one variable. In the formula P is expressed in terms of two variables, x and y. However, because the difference ofthe numbers is 10, we can write

We can solve this equation for y in terms of x (or vice versa), substitute the resultinto and obtain P as a function of one variable.

Subtract x from both sides of

Multiply both sides of the equa-tion by and solve for y.

Now we substitute for y in

Because P is now a function of x, we can write

Step 3 Write the function in the form We apply thedistributive property to obtain

Step 4 Calculate If the function has a minimum at this value. The

voice balloons show that and

This means that the product, P, of two numbers whose difference is 10 is a minimumwhen one of the numbers, x, is 5.

Step 5 Answer the question posed by the problem. The problem asks for the twonumbers and the minimum product. We found that one of the numbers, x, is 5. Nowwe must find the second number, y.

The number pair whose difference is 10 and whose product is as small as possible is5, The minimum product is or

Among all pairs of numbers whose difference is 8, find a pair whose prod-uct is as small as possible. What is the minimum product?

CheckPoint 6

-25.51-52,-5.

y = x - 10 = 5 - 10 = -5

x = - b

2a= -

-102112 = -1-52 = 5

b = -10.a = 1

a>0,� b

2a.

P(x)=x(x-10)=x2-10x.

a = 1 b = −10

f1x2 � ax2 � bx � c.

P1x2 = x1x - 102.

P = xy = x1x - 102.P = xy.x - 10

� 1 y = x - 10

x � y � 10. -y = -x + 10

P = xy,

x - y = 10.

P = xy,

P = xy.

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x50 − x

Figure 2.9 What value of x willmaximize the rectangle’s area?

The graph of the area function

was obtained with a graphingutility using a by

viewing rectangle.The maximum function featureverifies that a maximum area of625 square yards occurs whenone of the dimensions is 25yards.

30, 700, 25430, 50, 24

A1x2 = x150 - x2

EXAMPLE 7 Maximizing Area

You have 100 yards of fencing to enclose a rectangular region. Find the dimensionsof the rectangle that maximize the enclosed area. What is the maximum area?

SolutionStep 1 Decide what must be maximized or minimized. We must maximize area.What we do not know are the rectangle’s dimensions, x and y.

Step 2 Express this quantity as a function in one variable. Because we must max-imize area, we have We need to transform this into a function in which A isrepresented by one variable. Because you have 100 yards of fencing, the perimeterof the rectangle is 100 yards. This means that

We can solve this equation for y in terms of x, substitute the result into andobtain A as a function in one variable. We begin by solving for y.

Subtract 2x from both sides.

Divide both sides by 2.

Divide each term in the numerator by 2.

Now we substitute for y in

The rectangle and its dimensions are illustrated in Figure 2.9. Because A is now afunction of x, we can write

This function models the area, A(x), of any rectangle whose perimeter is 100 yardsin terms of one of its dimensions, x.

Step 3 Write the function in the form We apply thedistributive property to obtain

Step 4 Calculate If the function has a maximum at this value. The

voice balloons show that and

This means that the area, of a rectangle with perimeter 100 yards is a maxi-mum when one of the rectangle’s dimensions, x, is 25 yards.

Step 5 Answer the question posed by the problem. We found that Figure 2.9 shows that the rectangle’s other dimension is The dimensions of the rectangle that maximize the enclosed area are 25 yards by 25yards.The rectangle that gives the maximum area is actually a square with an area of

or 625 square yards.

You have 120 feet of fencing to enclose a rectangular region. Find thedimensions of the rectangle that maximize the enclosed area. What is themaximum area?

The ability to express a quantity to be maximized or minimized as a function inone variable plays a critical role in solving max-min problems. In calculus, you will learna technique for maximizing or minimizing all functions, not only quadratic functions.

CheckPoint 7

25 yards # 25 yards,

50 - x = 50 - 25 = 25.x = 25.

A1x2,x = -

b

2a= -

5021-12 = 25

b = 50.a = -1

a<0,� b

2a.

A(x)=x(50-x)=50x-x2=–x2+50x.

a = −1 b = 50

f1x2 � ax2 � bx � c.

A1x2 = x150 - x2.

A = xy = x150 - x2A = xy.50 - x

y = 50 - x

y =

100 - 2x

2

2y = 100 - 2x

A = xy,

2x + 2y = 100.

A = xy.

Technology

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EXERCISE SET 2.2In Exercises 9–16, find the coordinates of the vertex for theparabola defined by the given quadratic function.

9. 10.

11. 12.

13. 14.

15. 16.

In Exercises 17–38, use the vertex and intercepts to sketch thegraph of each quadratic function. Give the equation of theparabola’s axis of symmetry. Use the graph to determine thefunction’s domain and range.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

In Exercises 39–44, an equation of a quadratic function is given.

a. Determine, without graphing, whether the function has aminimum value or a maximum value.

b. Find the minimum or maximum value and determinewhere it occurs.

c. Identify the function’s domain and its range.

39. 40.

41. 42.

43. 44.

Practice PlusIn Exercises 45–48, give the domain and the range of

each quadratic function whose graph is described.

45. The vertex is and the parabola opens up.

46. The vertex is and the parabola opens down.

47. at

48. at

In Exercises 49–52, write an equation in standard form of theparabola that has the same shape as the graph of butwith the given point as the vertex.

49. 50.51. 52.

In Exercises 53–56, write an equation in standard form of theparabola that has the same shape as the graph of or

but with the given maximum or minimum.

53. at 54. at 55. at 56. at x = 9Minimum = 0

x = 11Minimum = 0x = 5Maximum = -7

x = -2Maximum = 4

g1x2 = -3x2,f1x2 = 3x2

1-8, -621-10, -5217, 4215, 32

f1x2 = 2x2,

x = -6Minimum = 18

x = 10Maximum = -6

1-3, -421-1, -22

f1x2 = 6x2- 6xf1x2 = 5x2

- 5x

f1x2 = -2x2- 12x + 3f1x2 = -4x2

+ 8x - 3

f1x2 = 2x2- 8x - 3f1x2 = 3x2

- 12x - 1

f1x2 = 6 - 4x + x2f1x2 = 2x - x2- 2

f1x2 = 3x2- 2x - 4f1x2 = 2x2

+ 4x - 3

f1x2 = x2+ 4x - 1f1x2 = x2

+ 6x + 3

f1x2 = 5 - 4x - x2f1x2 = 2x - x2+ 3

f1x2 = 2x2- 7x - 4f1x2 = x2

+ 3x - 10

f1x2 = x2- 2x - 15f1x2 = x2

- 2x - 3

f1x2 = 1 - 1x - 322f1x2 = 4 - 1x - 122f1x2 =

54 - Ax -

12 B2f1x2 = 21x + 222 - 1

y - 3 = 1x - 122y - 1 = 1x - 322f1x2 = 1x - 322 + 2f1x2 = 1x - 122 + 2

f1x2 = 1x - 122 - 2f1x2 = 1x - 422 - 1

f1x2 = -2x2+ 8x - 1f1x2 = -x2

- 2x + 8

f1x2 = 3x2- 12x + 1f1x2 = 2x2

- 8x + 3

f1x2 = -21x + 422 - 8f1x2 = -21x + 122 + 5

f1x2 = -31x - 222 + 12f1x2 = 21x - 322 + 1

Practice ExercisesIn Exercises 1–4, the graph of a quadratic function is

given. Write the function’s equation, selecting from the followingoptions.

1. 2.

3. 4.

In Exercises 5–8, the graph of a quadratic function is given. Writethe function’s equation, selecting from the following options.

5. 6.

7. 8.

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

y

x

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

y

x

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

y

x

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

y

x

j1x2 = -x2- 1 h1x2 = x2

- 1

g1x2 = x2- 2x + 1 f1x2 = x2

+ 2x + 1

−1

123

654

−2

1 2 3 4−1−2−3−4

y

x

−1

123

654

−2

1 2 3 4−1−2−3−4

y

x

−1

123

654

−2

1 2 3 4−1−2−3−4

y

x

−1

123

654

−2

1 2 3 4−1−2−3−4

y

x

j1x2 = 1x - 122 - 1 h1x2 = 1x - 122 + 1

g1x2 = 1x + 122 + 1 f1x2 = 1x + 122 - 1

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2003

2.7

2000

2.5

1995

2.1

1990

2.0

1985

2.3

2.7

2.6

2.5

2.4

2.3

2.2

2.1

Per

Cap

ita

Win

e C

onsu

mpt

ion

(gal

lons

per

per

son)

Wine Consumption per U.S. Adult

Year1980

2.6

2.0

Source: Adams Business Media

2002

1032

2001

980

2000

750

1999

692

1998

750

1997

910

1996

1000

1300

1200

1100

1000

900

800

700

600

Num

ber

of G

ang-

Rel

ated

Mur

ders

Gang-Related Homicides in the U.S.

Year1995

1220

500

Source: Professor James Alan Fox, Northeastern University

Application Exercises57. The graph shows per capita U.S. adult wine consumption (in

gallons per person) for selected years from 1980 through2003. The function

models U.S. wine consumption, , in gallons per person, xyears after 1980. According to this function, in which yearwas wine consumption at a minimum? Round to the nearestyear. What does the function give for per capita consump-tion, to the nearest tenth of a gallon, for that year? How welldoes this model the data shown in the graph?

f1x2f1x2 = 0.005x2

- 0.104x + 2.626

58. After declining in the late 1990s, the number of gang-relatedmurders across the United States has increased in recentyears. The graph shows the number of gang-related homi-cides in the United States. The function

models the number of gang-related homicides across thenation, , x years after 1995. According to this function, inwhich year was the number of homicides at a minimum?Round to the nearest year. What does the function give forthe number of gang-related murders for that year? How welldoes this model the data shown in the graph?

f1x2

f1x2 = 33x2- 255x + 1230

59. A person standing close to the edge on the top of a 200-footbuilding throws a baseball vertically upward. The quadraticfunction

models the ball’s height above the ground, in feet, tseconds after it was thrown.a. After how many seconds does the ball reach its maximum

height? What is the maximum height?

b. How many seconds does it take until the ball finally hitsthe ground? Round to the nearest tenth of a second.

c. Find and describe what this means.

d. Use your results from parts (a) through (c) to graph thequadratic function. Begin the graph with and endwith the value of t for which the ball hits the ground.

60. A person standing close to the edge on the top of a 160-footbuilding throws a baseball vertically upward. The quadraticfunction

models the ball’s height above the ground, in feet, tseconds after it was thrown.

a. After how many seconds does the ball reach its maximumheight? What is the maximum height?

b. How many seconds does it take until the ball finally hitsthe ground? Round to the nearest tenth of a second.

c. Find and describe what this means.

d. Use your results from parts (a) through (c) to graph thequadratic function. Begin the graph with and endwith the value of t for which the ball hits the ground.

61. Among all pairs of numbers whose sum is 16, find a pairwhose product is as large as possible. What is the maximumproduct?

62. Among all pairs of numbers whose sum is 20, find a pairwhose product is as large as possible. What is the maximumproduct?

63. Among all pairs of numbers whose difference is 16, find apair whose product is as small as possible. What is the mini-mum product?


65. You have 600 feet of fencing to enclose a rectangular plotthat borders on a river. If you do not fence the side along theriver, find the length and width of the plot that will maximizethe area. What is the largest area that can be enclosed?

66. You have 200 feet of fencing to enclose a rectangular plotthat borders on a river. If you do not fence the side along the

x

x

River

600 − 2x

t = 0

s102

s1t2,s1t2 = -16t2

+ 64t + 160

t = 0

s102

s1t2,s1t2 = -16t2

+ 64t + 200

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river, find the length and width of the plot that will maximizethe area. What is the largest area that can be enclosed?

67. You have 50 yards of fencing to enclose a rectangular region.Find the dimensions of the rectangle that maximize theenclosed area. What is the maximum area?

68. You have 80 yards of fencing to enclose a rectangular region.Find the dimensions of the rectangle that maximize theenclosed area. What is the maximum area?

69. A rectangular playground is to be fenced off and divided intwo by another fence parallel to one side of the playground.Six hundred feet of fencing is used. Find the dimensions ofthe playground that maximize the total enclosed area. Whatis the maximum area?

70. A rectangular playground is to be fenced off and divided intwo by another fence parallel to one side of the playground.Four hundred feet of fencing is used. Find the dimensions ofthe playground that maximize the total enclosed area. Whatis the maximum area?

71. A rain gutter is made from sheets of aluminum that are 20inches wide by turning up the edges to form right angles.Determine the depth of the gutter that will maximize itscross-sectional area and allow the greatest amount of waterto flow. What is the maximum cross-sectional area?

72. A rain gutter is made from sheets of aluminum that are 12inches wide by turning up the edges to form right angles.Determine the depth of the gutter that will maximize itscross-sectional area and allow the greatest amount of waterto flow. What is the maximum cross-sectional area?

If you have difficulty obtaining the functions to be maximized inExercises 73–76, read Example 2 in Section 1.10 on pages 244–245.

73. On a certain route, an airline carries 8000 passengers permonth, each paying $50. A market survey indicates that foreach $1 increase in the ticket price, the airline will lose 100passengers. Find the ticket price that will maximize theairline’s monthly revenue for the route. What is themaximum monthly revenue?

74. A car rental agency can rent every one of its 200 cars at $30per day. Far each $1 increase in rate, five fewer cars arerented. Find the rental amount that will maximize theagency’s daily revenue. What is the maximum daily revenue?

75. The annual yield per walnut tree is fairly constant at 60pounds per tree when the number of trees per acre is 20 orfewer. For each additional tree over 20, the annual yield pertree for all trees on the acre decreases by 2 pounds due toovercrowding. How many walnut trees should be planed peracre to maximize the annual yield for the acre? What is themaximum number of pounds of walnuts per acre?

76. The annual yield per cherry tree is fairly constant at 50pounds per tree when the number of trees per acre is 30 orfewer. For each additional tree aver 30, the annual yield pertree for all trees on the acre decreases by 1 pound due toovercrowding. How many cherry trees should be planted peracre to maximize the annual yield for the acre? What is themaximum number of pounds of cherries per acre?

x

x

200 − 2x

Writing in Mathematics77. What is a quadratic function?78. What is a parabola? Describe its shape.79. Explain how to decide whether a parabola opens upward or

downward.80. Describe how to find a parabola’s vertex if its equation is

expressed in standard form. Give an example.81. Describe how to find a parabola’s vertex if its equation is in

the form Use asan example.

82. A parabola that opens upward has its vertex at (1, 2).Describe as much as you can about the parabola based onthis information. Include in your discussion the number ofx-intercepts (if any) for the parabola.

Technology Exercises83. Use a graphing utility to verify any five of your hand-drawn

graphs in Exercises 17–38.84. a. Use a graphing utility to graph in a

standard viewing rectangle. What do you observe?b. Find the coordinates of the vertex for the given quadratic

function.c. The answer to part (b) is Because the lead-

ing coefficient, 2, of the given function is positive, the vertexis a minimum point on the graph. Use this fact to help find aviewing rectangle that will give a relatively complete pictureof the parabola. With an axis of symmetry at thesetting for x should extend past this, so try and

The setting for y should include (and probablygo below) the y-coordinate of the graph’s minimum y-value,so try Experiment with Ymax until yourutility shows the parabola’s major features.

d. In general, explain how knowing the coordinates of aparabola’s vertex can help determine a reasonable view-ing rectangle on a graphing utility for obtaining a com-plete picture of the parabola.

In Exercises 85–88, find the vertex for each parabola. Then deter-mine a reasonable viewing rectangle on your graphing utility anduse it to graph the quadratic function.

85. 86.

87. 88.

89. The following data show fuel efficiency, in miles per gallon,for all U.S. automobiles in the indicated year.

y (Average Numberx of Miles per Gallon

(Years after 1940) for U.S. Automobiles)1940: 0 14.8

1950: 10 13.9

1960: 20 13.4

1970: 30 13.5

1980: 40 15.9

1990: 50 20.2

2000: 60 22.0

Source: U.S. Department of Transportation

y = 0.01x2+ 0.6x + 100y = 5x2

+ 40x + 600

y = -4x2+ 20x + 160y = -0.25x2

+ 40x

Ymin = -130.

Xmax = 30.Xmin = 0

x = 20.5,

120.5, -120.52.

y = 2x2- 82x + 720

f1x2 = x2- 6x + 8f1x2 = ax2

+ bx + c.

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Section 2.3 • Polynomial Functions and Their Graphs 287

a. Use a graphing utility to draw a scatter plot of the data.Explain why a quadratic function is appropriate for mod-eling these data.

b. Use the quadratic regression feature to find the quadraticfunction that best fits the data.

c. Use the model in part (b) to determine the worst year forautomobile fuel efficiency. What was the average numberof miles per gallon for that year?

d. Use a graphing utility to draw a scatter plot of the data andgraph the quadratic function of best fit on the scatter plot.


a. No quadratic functions have a range of b. The vertex of the parabola described by

is at c. The graph of has one y-intercept

and two x-intercepts.d. The maximum value of y for the quadratic function

is 1.

In Exercises 91–92, find the axis of symmetry for each parabolawhose equation is given. Use the axis of symmetry to find asecond point on the parabola whose y-coordinate is the same asthe given point.

91.

92.

In Exercises 93–94, write the equation of each parabola instandard form.

93. Vertex: The graph passes through the point

94. Vertex: The graph passes through the point

95. Find the point on the line whose equation is that is closest to the origin. Hint: Minimize the distance func-tion by minimizing the expression under the square root.

2x + y - 2 = 0

1-2, -32.1-3, -12;11, 42.1-3, -42;

f1x2 = 1x - 322 + 2; 16, 112f1x2 = 31x + 222 - 5; 1-1, -22

f1x2 = -x2+ x + 1

f1x2 = -21x + 422 - 815, 12.f1x2 = 21x - 522 - 1

1- q , q2.

96. A 300-room hotel can rent every one of its rooms at $80 perroom. For each $1 increase in rent, three fewer roams arerented. Each rented room costs the hotel $10 to service perday. How much should the hotel charge for each room tomaximize its daily profit? What is the maximum daily profit?

97. A track and field area is to be constructed in the shape of a rec-tangle with semicircles at each end.The inside perimeter of thetrack is to be 440 yards. Find the dimensions of the rectanglethat maximize the area of the rectangular portion of the field.

Group Exercise98. Each group member should consult an almanac, newspaper,

magazine, or the Internet to find data that initially increaseand then decrease, or vice versa, and therefore can be mod-eled by a quadratic function. Group members should selectthe two sets of data that are most interesting and relevant.For each data set selected,a. Use the quadratic regression feature of a graphing utility

to find the quadratic function that best fits the data.b. Use the equation of the quadratic function to make a pre-

diction from the data. What circumstances might affectthe accuracy of your prediction?

c. Use the equation of the quadratic function to write and solvea problem involving maximizing or minimizing the function.

Objectives❶ Identify polynomial

functions.

❷ Recognize characteristicsof graphs of polynomialfunctions.

❸ Determine end behavior.

❹ Use factoring to find zerosof polynomial functions.

❺ Identify zeros and theirmultiplicities.

❻ Use the Intermediate ValueTheorem.

❼ Understand therelationship betweendegree and turning points.

❽ Graph polynomialfunctions.

In 1980, U.S. doctors diagnosed 41 cases of a rare form of cancer, Kaposi’s sarcoma,that involved skin lesions, pneumonia, and severe immunological deficiencies. Allcases involved gay men ranging in age from 26 to 51. By the end of 2002, approxi-mately 890,000 Americans, straight and gay, male and female, old and young, wereinfected with the HIV virus.

Magnified 6000 times, this color-scannedimage shows a T-lymphocyte blood cell(green) infected with the HIV virus (red).Depletion of the number of T-cells causesdestruction of the immune system.

SECTION 2.3 Polynomial Functions and Their Graphs

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80,000

70,000

60,000

50,000

40,000

30,000

20,000

2001 2002

41,227 42,136

2000

41,239

1999

41,314

1998

43,225

1997

49,379

1996

61,124

1995

69,984

1994

73,086

1993

79,879

1992

79,657

1991

60,573

1990

49,546

1989

43,499

1988

36,126

1987

29,105

1986

19,404

1985

12,044

1984

6368

Num

ber

of C

ases

Dia

gnos

ed

AIDS Cases Diagnosed in the U.S., 1983–2002

Year1983

315310,000

Changing circumstances and unforeseen events can result in models forAIDS-related data that are not particularly useful over long periods of time. Forexample, the function

models the number of AIDS cases diagnosed in the United States x years after1983. The model was obtained using a portion of the data shown in Figure 2.10,namely cases diagnosed from 1983 through 1991, inclusive. Figure 2.11 shows thegraph of from 1983 through 1991. This function is an example of a polynomialfunction of degree 3.

Definition of a Polynomial FunctionLet n be a nonnegative integer and let be real numbers,with The function defined by

is called a polynomial function of degree n. The number the coefficient ofthe variable to the highest power, is called the leading coefficient.

A constant function where is a polynomial function ofdegree 0. A linear function where is a polynomial func-tion of degree 1. A quadratic function where is apolynomial function of degree 2. In this section, we focus on polynomialfunctions of degree 3 or higher.

Smooth, Continuous GraphsPolynomial functions of degree 2 or higher have graphs that are smooth andcontinuous. By smooth, we mean that the graphs contain only rounded curves withno sharp corners. By continuous, we mean that the graphs have no breaks and can bedrawn without lifting your pencil from the rectangular coordinate system.These ideasare illustrated in Figure 2.12 on the next page.

a Z 0,f1x2 = ax2+ bx + c,

m Z 0,f1x2 = mx + b,c Z 0,f1x2 = c,

an ,

f1x2 = an xn+ an - 1 xn - 1

+Á

+ a2 x2+ a1 x + a0

an Z 0.an , an - 1 , Á , a2 , a1 , a0

f

f1x2 = -49x3+ 806x2

+ 3776x + 2503

❶ Identify polynomialfunctions.

5000

60,000

0 1 2 3 4 5 6 7 8

Years after 1983

[0, 8, 1] by [0, 60,000, 5000]

f(x) = −49x3 + 806x2 + 3776x + 2503

Cas

es D

iagn

osed

Figure 2.11 The graph of a function mod-eling the number of AIDS cases from 1983through 1991

Figure 2.10Source: Department of Health and Human Services

❷ Recognize characteristics ofgraphs of polynomialfunctions.

Modeling AIDS-related data and making predictions about the epidemic’shavoc is serious business. Figure 2.10 shows the number of AIDS cases diagnosed inthe United States from 1983 through 2002.

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y

x

y

x

y

x

Smoothroundedcorner

Smoothroundedcorner

Smoothroundedcorner

Smoothroundedcorners

Discontinuous;a break in thegraph

Sharpcorner

Sharpcorner

Graphs of Polynomial Functions Not Graphs of Polynomial Functions

y

x

Figure 2.12 Recognizing graphs of polynomial functions

End Behavior of Polynomial FunctionsFigure 2.13 shows the graph of the function

which models the number of U.S. AIDS cases from 1983 through 1991. Look whathappens to the graph when we extend the year up through 2005. By year 21 (2004),the values of y are negative and the function no longer models AIDS cases. We’veadded an arrow to the graph at the far right to emphasize that it continues todecrease without bound. It is this far-right end behavior of the graph that makes itinappropriate for modeling AIDS cases into the future.

f1x2 = -49x3+ 806x2

+ 3776x + 2503,

Cas

es D

iagn

osed

Years after 1983

[0, 22, 1] by [−10,000, 85,000, 5000]

5000

85,000

5 10 15 20

Graph falls to the right.

Figure 2.13 By extending the viewingrectangle, we see that y is eventually negativeand the function no longer models thenumber of AIDS cases. Model breakdownoccurs by 2004.

❸ Determine end behavior.

The behavior of a graph of a function to the far left or the far right is called itsend behavior. Although the graph of a polynomial function may have intervalswhere it increases or decreases, the graph will eventually rise or fall without boundas it moves far to the left or far to the right.

How can you determine whether the graph of a polynomial function goes upor down at each end? The end behavior of a polynomial function

depends upon the leading term because when is large, the other terms arerelatively insignificant in size. In particular, the sign of the leading coefficient,and the degree, n, of the polynomial function reveal its end behavior. In terms of endbehavior, only the term of highest degree counts, as summarized by the LeadingCoefficient Test.

an ,ƒx ƒan xn,

f1x2 = an xn+ an - 1 xn - 1

+Á

+ a1 x + a0

Odd-degree polynomial functions have graphs with opposite behavior at each end. Even-degree polynomial functions have graphs with the same behavior at each end.

Study Tip

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The Leading Coefficient TestAs x increases or decreases without bound, the graph of the polynomial function

eventually rises or falls. In particular,

1. For n odd: 2. For n even:

f1x2 = an xn+ an - 1 xn - 1

+ an - 2 xn - 2+

Á+ a1 x + a0 1an Z 02

If the leading coeffi-cient is positive, thegraph falls to the leftand rises to the right.

If the leading coeffi-cient is negative, thegraph rises to the leftand falls to the right.

If the leading coeffi-cient is positive, thegraph rises to the leftand to the right.

If the leading coeffi-cient is negative, thegraph falls to the leftand to the right.

y

x

y

x

y

x

y

x

Rises right Rises right

Falls right

Rises left

Rises left

Falls left

Odd degree; positiveleading coefficient

Odd degree; negativeleading coefficient

Even degree; positiveleading coefficient

Even degree; negativeleading coefficient

Falls left

Falls right

an > 0 an < 0 an > 0 an < 0

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

Rises right

Falls left

Figure 2.14 The graph off1x2 = x3

+ 3x2- x - 3

EXAMPLE 1 Using the Leading Coefficient Test

Use the Leading Coefficient Test to determine the end behavior of the graph of

Solution We begin by identifying the sign of the leading coefficient and thedegree of the polynomial.

The degree of the function is 3, which is odd. Odd-degree polynomial functionshave graphs with opposite behavior at each end. The leading coefficient, 1, is posi-tive. Thus, the graph falls to the left and rises to the right. The graph of is shown inFigure 2.14.

Use the Leading Coefficient Test to determine the end behavior of thegraph of


Use end behavior to explain why

is only an appropriate model for AIDS cases for a limited time period.

f1x2 = -49x3+ 806x2

+ 3776x + 2503

f1x2 = x4- 4x2.

CheckPoint 1

f

f

The leading coefficient,1, is positive.

The degree of thepolynomial, 3, is odd.

f(x)=x3+3x2-x-3

f1x2 = x3+ 3x2

- x - 3.

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The degree of is 3, which is odd. Odd-degree polynomial functions have graphswith opposite behavior at each end. The leading coefficient, is negative. Thus,the graph rises to the left and falls to the right. The fact that the graph falls to theright indicates that at some point the number of AIDS cases will be negative, animpossibility. If a function has a graph that decreases without bound over time, itwill not be capable of modeling nonnegative phenomena over long time periods.Model breakdown will eventually occur.

The polynomial function

models the ratio of students to computers in U.S. public schools x yearsafter 1980. Use end behavior to determine whether this function could bean appropriate model for computers in the classroom well into the twenty-first century. Explain your answer.

If you use a graphing utility to graph a polynomial function, it is important toselect a viewing rectangle that accurately reveals the graph’s end behavior. If theviewing rectangle, or window, is too small, it may not accurately show the end behavior.


The graph of was obtained with a graphing utilityusing a by viewing rectangle. The graph is shown inFigure 2.15(a). Does the graph show the end behavior of the function?

3-10, 10, 143-8, 8, 14f1x2 = -x4+ 8x3

+ 4x2+ 2

f1x2 = -0.27x3+ 9.2x2

- 102.9x + 400

CheckPoint 2

-49,f

The leading coefficient,−49, is negative.

The degree of thepolynomial, 3, is odd.

f(x)=–49x3+806x2+3776x+2503

[–8, 8, 1] by [–10, 10, 1] [–10, 10, 1] by [–1000, 750, 250]

Figure 2.15(a)


The degree of is 4, which is even. Even-degree polynomial functions have graphswith the same behavior at each end. The leading coefficient, is negative. Thus,the graph should fall to the left and fall to the right. The graph in Figure 2.15(a) isfalling to the left, but it is not falling to the right. Therefore, the graph is not com-plete enough to show end behavior.A more complete graph of the function is shownin a larger viewing rectangle in Figure 2.15(b).

-1,f

The leading coefficient,−1, is negative.

The degree of thepolynomial, 4, is even.

f(x)=–x4+8x3+4x2+2

Figure 2.15(b)

pr02-265-374.I-hr 1/26/06 5:13 PM Page 291

−1

12345

−2−3

−5

1 2 3 4 5−1−2−3−4−5

y

x

x-intercept: −3 x-intercept: 1

x-intercept: −1

Figure 2.17

Figure 2.16


❹ Use factoring to find zeros ofpolynomial functions.

The graph of is shown in a standard viewing rectangle inFigure 2.16. Use the Leading CoefficientTest to determine whether the graph showsthe end behavior of the function. Explainyour answer.

Zeros of Polynomial FunctionsIf is a polynomial function, then the values of x for which is equal to 0 arecalled the zeros of . These values of x are the roots, or solutions, of the polynomialequation Each real root of the polynomial equation appears as anx-intercept of the graph of the polynomial function.

EXAMPLE 4 Finding Zeros of a Polynomial Function

Find all zeros of

Solution By definition, the zeros are the values of x for which is equal to 0.Thus, we set equal to 0:

We solve the polynomial equation for x as follows:

This is the equation needed to find the function’s zeros.Factor from the first two terms and from thelast two terms.A common factor of is factored from theexpression.Set each factor equal to 0.Solve for x.Remember that if then

The zeros of are and 1. The graph of in Figure 2.17 shows that each zerois an x-intercept. The graph passes through the points and 11, 02.1-3, 02, 1-1, 02,f-3, -1,f

x2 � —2d .x 2 � d, x = ;1 x = -3 x2

= 1 x + 3 = 0 or x2

- 1 = 0

x � 3 1x + 321x2- 12 = 0

� 1x 2 x21x + 32 - 11x + 32 = 0 x3

+ 3x2- x - 3 = 0

x3+ 3x2

- x - 3 = 0

f1x2 = x3+ 3x2

- x - 3 = 0.

f1x2 f1x2f1x2 = x3

+ 3x2- x - 3.

f1x2 = 0.f

f1x2f

f1x2 = x3+ 13x2

+ 10x - 4CheckPoint 3

A graphing utility can be used to verify that and 1 are the three real zeros of

Numeric Check Graphic CheckDisplay a table for the function. Display a graph for the function. The x-intercepts indicate

that and 1 are the real zeros.

The utility’s feature on the graph of also verifies that and 1 are the function’s real zeros.-3, -1,f� ZERO �


x-intercept: −1

[–6, 6, 1] by [–6, 6, 1]

Enter y1 = x3 + 3x2 − x − 3.

y1 is equal to 0when x = −3,x = −1, andx = 1.

−3, −1, and1 are the realzeros.

-3, -1,

f1x2 = x3+ 3x2

- x - 3.-3, -1,

Technology

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❺ Identify zeros and theirmultiplicities.

−1

1

−2−3−4

−6

−8−7

−5

1 2 3 4 5−1−2−3−4−5

y

x

x-intercept: 0 x-intercept: 2

Figure 2.18 The zeros ofnamely 0

and 2, are the x-intercepts for thegraph of .f

f1x2 = -x4+ 4x3

- 4x2,

Find all zeros of


Find all zeros of

Solution We find the zeros of by setting equal to 0 and solving the result-ing equation.

We now have a polynomial equation.

Multiply both sides by This step is optional.

Factor out

Factor completely.

Set each factor equal to 0.Solve for x.

The zeros of are 0 and 2. The graph of , shown in Figure2.18, has x-intercepts at 0 and 2.The graph passes through the points and

Find all zeros of

Multiplicities of ZerosWe can use the results of factoring to express a polynomial as a product of factors. Forinstance, in Example 5, we can use our factoring to express the function’s equationas follows:

Notice that each factor occurs twice. In factoring the equation for the polynomialfunction , if the same factor occurs k times, but not times, we call r azero with multiplicity k. For the polynomial function

0 and 2 are both zeros with multiplicity 2.Multiplicity provides another connection between zeros and graphs. The

multiplicity of a zero tells us whether the graph of a polynomial function touchesthe x-axis at the zero and turns around, or if the graph crosses the x-axis at the zero.For example, look again at the graph of in Figure 2.18.Each zero, 0 and 2, is a zero with multiplicity 2. The graph of touches, but does notcross, the x-axis at each of these zeros of even multiplicity. By contrast, a graphcrosses the x-axis at zeros of odd multiplicity.

Multiplicity and x-InterceptsIf r is a zero of even multiplicity, then the graph touches the x-axis and turnsaround at r. If r is a zero of odd multiplicity, then the graph crosses the x-axis at r.Regardless of whether the multiplicity of a zero is even or odd, graphs tend toflatten out at zeros with multiplicity greater than one.

If a polynomial function’s equation is expressed as a product of linear factors,we can quickly identify zeros and their multiplicities.

ff1x2 = -x4

+ 4x3- 4x2

f1x2 = -x21x - 222,k + 1x - rf

The factor xoccurs twice:x2 = x � x.

The factor (x − 2)occurs twice:

(x − 2)2 = (x − 2)(x − 2).

f(x)=–x4+4x3-4x2=–(x4-4x3+4x2)=–x2(x-2)2.

f1x2 = x4- 4x2.Check

Point 5

12, 02.10, 02ff1x2 = -x4+ 4x3

- 4x2

x = 0 x = 2 x2

= 0 or 1x - 222 = 0

x21x - 222 = 0

x 2. x21x2- 4x + 42 = 0

� 1. x4- 4x3

+ 4x2= 0

-x4+ 4x3

- 4x2= 0

f1x2f

f1x2 = -x4+ 4x3

- 4x2.

f1x2 = x3+ 2x2

- 4x - 8.CheckPoint 4

pr02-265-374.I-hr 1/26/06 5:13 PM 3

y

x

(b, f(b))f(b) > 0

f(c) = 0

(a, f(a))f(a) < 0

a

c b

Figure 2.20 The graph must cross thex-axis at some value between a and b.


❻ Use the Intermediate ValueTheorem.

EXAMPLE 6 Finding Zeros and Their Multiplicities

Find the zeros of and give the multiplicity of each zero.State whether the graph crosses the x-axis or touches the x-axis and turns around ateach zero.

Solution We find the zeros of by setting equal to 0:

Set each factor equal to 0.

The zeros of are with multiplicity 1, and with multi-plicity 2. Because the multiplicity of is odd, the graph crosses the x-axis at this zero.Because the multiplicity of is even, the graph touches the x-axis and turns around atthis zero.These relationships are illustrated by the graph of in Figure 2.19.

Find the zeros of and give the multiplicity ofeach zero. State whether the graph crosses the x-axis or touches the x-axisand turns around at each zero.

The Intermediate Value TheoremThe Intermediate Value Theorem tells us of the existence of real zeros.The idea behindthe theorem is illustrated in Figure 2.20.The figure shows that if lies belowthe x-axis and lies above the x-axis, the smooth, continuous graph of a poly-nomial function must cross the x-axis at some value c between a and b. This valueis a real zero for the function.

These observations are summarized in the Intermediate Value Theorem.

The Intermediate Value Theorem for PolynomialsLet be a polynomial function with real coefficients. If and haveopposite signs, then there is at least one value of c between a and b for which

Equivalently, the equation has at least one real rootbetween a and b.

EXAMPLE 7 Using the Intermediate Value Theorem

Show that the polynomial function has a real zero between 2 and 3.

Solution Let us evaluate at 2 and at 3. If and have opposite signs, thenthere is at least one real zero between 2 and 3. Using weobtain

f(2)=23-2 � 2-5=8-4-5=–1

f(2) is negative.

f1x2 = x3- 2x - 5,

f132f122f

f1x2 = x3- 2x - 5

f1x2 = 0f1c2 = 0.

f1b2f1a2f

f1b, f1b22 1a, f1a22

f1x2 = -4 Ax +12 B21x - 523Check

Point 6

f

32

-1

32 ,-1,f1x2 = 1x + 1212x - 322

This exponent is 1.Thus, the multiplicity

of −1 is 1.

x + 1 = 0x = −1

2x − 3 = 0x =

This exponent is 2.Thus, the multiplicity

of is 2.

(x+1)1(2x-3)2=0

32

32

1x + 1212x - 322 = 0.

f1x2f

f1x2 = 1x + 1212x - 322

−1 is a zero of odd multiplicity.Graph crosses x-axis.

32 is a zero of even multiplicity.Graph touches x-axis, flattens,

and turns around.

[−3, 3, 1] by [−10, 10, 1]

Figure 2.19 The graph off1x2 = 1x + 1212x - 322

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❼ Understand the relationshipbetween degree and turningpoints.

and

Because and the sign change shows that the polynomialfunction has a real zero between 2 and 3. This zero is actually irrational and isapproximated using a graphing utility’s feature as 2.0945515 in Figure 2.21.

Show that the polynomial function has a real zerobetween and

Turning Points of Polynomial FunctionsThe graph of is shownin Figure 2.22. The graph has four smooth turningpoints. At each turning point, the graph changesdirection from increasing to decreasing or vice versa.The given equation has 5 as its greatest exponentand is therefore a polynomial function of degree 5.Notice that the graph has four turning points. In gen-eral, if is a polynomial function of degree n, thenthe graph of has at most turning points.

Figure 2.22 illustrates that the y-coordinate ofeach turning point is either a relative maximum ora relative minimum of . Without the aid of agraphing utility or a knowledge of calculus, it isdifficult and often impossible to locate turningpoints of polynomial functions with degrees greaterthan 2. If necessary, test values can be taken between the x-intercepts to get a generalidea of how high the graph rises or how low the graph falls. For the purpose of graph-ing in this section, a general estimate is sometimes appropriate and necessary.

A Strategy for Graphing Polynomial FunctionsHere’s a general strategy for graphing a polynomial function. A graphing utility is avaluable complement, but not a necessary component, to this strategy. If you are usinga graphing utility, some of the steps listed in the following box will help you to selecta viewing rectangle that shows the important parts of the graph.

Graphing a Polynomial Function

1. Use the Leading Coefficient Test to determine the graph’s end behavior.2. Find x-intercepts by setting and solving the resulting polynomial

equation. If there is an x-intercept at r as a result of in the com-plete factorization of , thena. If k is even, the graph touches the x-axis at r and turns around.b. If k is odd, the graph crosses the x-axis at r.c. If the graph flattens out at

3. Find the y-intercept by computing .4. Use symmetry, if applicable, to help draw the graph:

a. y-axis symmetry:b. Origin symmetry:

5. Use the fact that the maximum number of turning points of the graph isto check whether it is drawn correctly.n - 1

f1-x2 = -f1x2.f1-x2 = f1x2

f1021r, 02.k 7 1,

f1x2 1x - r2kf1x2 = 0

f1x2 = an xn+ an - 1 xn - 1

+ an - 2 xn - 2+

Á+ a1 x + a0 , an Z 0

f

n � 1ff

f1x2 = x5- 6x3

+ 8x + 1

-2.-3f1x2 = 3x3

- 10x + 9CheckPoint 7

� ZERO �

f132 = 16,f122 = -1

f(3)=33-2 � 3-5=27-6-5=16.

f(3) is positive.

❽ Graph polynomial functions.

Remember that, without calculus,it is often impossible to give theexact location of turning points.However, you can obtain addi-tional points satisfying the func-tion to estimate how high thegraph rises or how low it falls. Toobtain these points, use values of xbetween (and to the left and rightof) the x-intercepts.

Study Tip

−1

1234

−2−3−4−5

1 3 4 5−1−2−3−4−5

y

x

Turning points:from increasingto decreasing

Turning points:from decreasingto increasing

f(x) = x5 − 6x3 + 8x + 1

Figure 2.22 Graph with fourturning points

y = x3 − 2x − 5

[–3, 3, 1] by [–10, 10, 1]

Figure 2.21

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EXAMPLE 8 Graphing a Polynomial Function

Graph:

Solution

Step 1 Determine end behavior. Identify the sign of the leading coefficient,and the degree, n, of the polynomial function.

Because the degree, 4, is even, the graph has the same behavior at each end.The lead-ing coefficient, 1, is positive.Thus, the graph rises to the left and rises to the right.

Step 2 Find x-intercepts (zeros of the function) by setting

Factor.

Factor completely.

Express the factorization in a morecompact form.

Set each factorization equal to 0.

Solve for x.

We see that and 1 are both repeated zeros with multiplicity 2. Because of the evenmultiplicity, the graph touches the x-axis at and 1 and turns around. Furthermore,the graph tends to flatten out at these zeros with multiplicity greater than one.

Step 3 Find the y-intercept by computing . We use andcompute .

There is a y-intercept at 1, so the graph passes through

y

x11

1

–

It appears that 1 is arelative maximum, but weneed more information

to be certain.

10, 12.f102 = 04

- 2 # 02+ 1 = 1

f102 f1x2 = x4- 2x2

+ 1f102y

x11

Risesleft

Risesright

–

-1-1

x = -1 x = 1

1x + 122 = 0 or 1x - 122 = 0

1x + 1221x - 122 = 0

1x + 121x - 121x + 121x - 12 = 0

1x2- 121x2

- 12 = 0

x4- 2x2

+ 1 = 0

f1x2 � 0.

y

x

Risesleft

Risesright

The leadingcoefficient,

1, is positive.

The degree of thepolynomial function,

4, is even.

f(x)=x4-2x2+1

an ,

f1x2 = x4- 2x2

+ 1.

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−1

12345

1 2 3 4 5−1−2−3−4−5

y

x

Figure 2.23 The graph off1x2 = x4

- 2x2+ 1

Step 4 Use possible symmetry to help draw the graph. Our partial graph suggestsy-axis symmetry. Let’s verify this by finding

Because the graph of is symmetric with respect to the y-axis.Figure 2.23 shows the graph of

Step 5 Use the fact that the maximum number of turning points of the graph isto check whether it is drawn correctly. Because the maximum

number of turning points is or 3. Because the graph in Figure 2.23 has threeturning points, we have not violated the maximum number possible. Can you seehow this verifies that 1 is indeed a relative maximum and is a turning point? Ifthe graph rose above 1 on either side of it would have to rise above 1 on theother side as well because of symmetry.This would require additional turning pointsto smoothly curve back to the x-intercepts. The graph already has three turningpoints, which is the maximum number for a fourth-degree polynomial function.

Use the five-step strategy to graph f1x2 = x3- 3x2.Check

Point 8

x = 0,10, 12

4 - 1,n = 4,n � 1

f1x2 = x4- 2x2

+ 1.ff1-x2 = f1x2,

f(–x)=(–x)4-2(–x)2+1=x4-2x2+1

f(x) = x4 - 2x2 + 1

Replace x with −x.

f1-x2.

EXERCISE SET 2.3

Practice ExercisesIn Exercises 1–10, determine which functions are poly-

nomial functions. For those that are, identify the degree.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

In Exercises 11–14, identify which graphs are not those of polyno-mial functions.

11. 12.

13. 14. y

x

y

x

y

x

y

x

f1x2 =

x2+ 73

f1x2 =

x2+ 7

x3

f1x2 = x

13

- 4x2+ 7f1x2 = x

12

- 3x2+ 5

h1x2 = 8x3- x2

+

2x

h1x2 = 7x3+ 2x2

+

1x

g1x2 = 6x7+ px5

+

23

xg1x2 = 7x5- px3

+

15

x

f1x2 = 7x2+ 9x4f1x2 = 5x2

+ 6x3

In Exercises 15–18, use the Leading Coefficient Test to determinethe end behavior of the graph of the given polynomial function.Then use this end behavior to match the polynomial function withits graph. [The graphs are labeled (a) through ]

15. 16.

17.

18.

(a) (b)

(c) (d)

In Exercises 19–24, use the Leading Coefficient Test to determinethe end behavior of the graph of the polynomial function.

19.

20.

21. f1x2 = 5x4+ 7x2

- x + 9

f1x2 = 11x3- 6x2

+ x + 3

f1x2 = 5x3+ 7x2

- x + 9

x

y

�2

�4

�6

�8

�2�4 2x

y

�2

�4

�6

�8

�2 2 4

1

−2

2−2

y

x

2 4 6x

y

2

4

6

8

10

f1x2 = -x3- x2

+ 5x - 3

f1x2 = 1x - 322f1x2 = x3

- 4x2f1x2 = -x4+ x2

1d2.

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22.

23.

24.

In Exercises 25–32, find the zeros for each polynomial function andgive the multiplicity for each zero. State whether the graph crossesthe x-axis, or touches the x-axis and turns around, at each zero.

25.

26.

27.

28.

29.

30.

31.

32.

In Exercises 33–40, use the Intermediate Value Theorem to showthat each polynomial has a real zero between the given integers.

33. between 1 and 2

34. between 0 and 1

35. between and 0

36. between 2 and 3

37. between and

38. between 1 and 2

39. between and

40. between 2 and 3

In Exercises 41–64,

a. Use the Leading Coefficient Test to determine the graph’send behavior.

b. Find the x-intercepts. State whether the graph crosses thex-axis, or touches the x-axis and turns around, at each inter-cept.

c. Find the y-intercept.d. Determine whether the graph has y-axis symmetry, origin

symmetry, or neither.e. If necessary, find a few additional points and graph the

function. Use the maximum number of turning points tocheck whether it is drawn correctly.

41. 42.

43. 44.

45. 46.

47. 48.

49. 50.

51. 52.

53. 54.

55.

56.

57.

58.

59.

60.

61.

62. f1x2 = -3x31x - 1221x + 32f1x2 = -2x31x - 1221x + 52f1x2 = -x21x + 221x - 22f1x2 = -x21x - 121x + 32f1x2 = x31x + 2221x + 12f1x2 = x21x - 1231x + 22f1x2 = -21x - 4221x2

- 252f1x2 = -31x - 1221x2

- 42f1x2 =

12 -

12 x4f1x2 = 3x2

- x3

f1x2 = 6x - x3- x5f1x2 = 6x3

- 9x - x5

f1x2 = -2x4+ 2x3f1x2 = -2x4

+ 4x3

f1x2 = x4- 6x3

+ 9x2f1x2 = x4- 2x3

+ x2

f1x2 = -x4+ 4x2f1x2 = -x4

+ 16x2

f1x2 = x4- x2f1x2 = x4

- 9x2

f1x2 = x3+ x2

- 4x - 4f1x2 = x3+ 2x2

- x - 2

f1x2 = 3x3- 8x2

+ x + 2;

-2-3f1x2 = 3x3- 10x + 9;

f1x2 = x5- x3

- 1;

-2-3f1x2 = x3+ x2

- 2x + 1;

f1x2 = x4+ 6x3

- 18x2;

-1f1x2 = 2x4- 4x2

+ 1;

f1x2 = x3- 4x2

+ 2;

f1x2 = x3- x - 1;

f1x2 = x3+ 5x2

- 9x - 45

f1x2 = x3+ 7x2

- 4x - 28

f1x2 = x3+ 4x2

+ 4x

f1x2 = x3- 2x2

+ x

f1x2 = -3 Ax +12 B1x - 423

f1x2 = 41x - 321x + 623f1x2 = 31x + 521x + 222f1x2 = 21x - 521x + 422

f1x2 = -11x4- 6x2

+ x + 3

f1x2 = -5x4+ 7x2

- x + 9

f1x2 = 11x4- 6x2

+ x + 3 63.

64.

Practice PlusIn Exercises 65–72, complete graphs of polynomial

functions whose zeros are integers are shown.a. Find the zeros and state whether the multiplicity of each

zero is even or odd.b. Write an equation, expressed as the product of factors, of a

polynomial function that might have each graph. Use aleading coefficient of 1 or and make the degree of assmall as possible.

c. Use both the equation in part (b) and the graph to find they-intercept.

65.

66.

67.

68.

69.

[−4, 4, 1] by [−40, 4, 4]

[−3, 3, 1] by [−10, 10, 1]

[−3, 6, 1] by [−10, 10, 1]

[−6, 6, 1] by [−40, 40, 10]

[−5, 5, 1] by [−12, 12, 1]

f-1,

f1x2 = 1x + 321x + 1231x + 42f1x2 = 1x - 2221x + 421x - 12

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450,000

400,000

350,000

300,000

500,000

250,000

200,000

150,000

100,000

2002

2001

2000

1999

1998

1997

1996

1995

1994

1993

1992

1991

Cumulative Number of Deaths from AIDS in the U.S.

Year

1990

50,000

483,920458,551

Cum

ulat

ive

Num

ber

of D

eath

s

Source: Centers for Disease Control

70.

71.

72.

Application ExercisesThe bar graph shows the cumulative number of deathsfrom AIDS in the United States from 1990 through 2002.

[−3, 3, 1] by [−5, 10, 1]

[−3, 3, 1] by [−5, 10, 1]

[−2, 5, 1] by [−40, 4, 4]

74. Use both functions to find the cumulative number of AIDSdeaths in 2002. Which function provides a better descriptionfor the actual number shown in the bar graph?

75. Use the Leading Coefficient Test to determine the endbehavior to the right for the graph of . Will this functionbe useful in modeling the cumulative number of AIDSdeaths over an extended period of time? Explain youranswer.

76. Use the Leading Coefficient Test to determine the endbehavior to the right for the graph of g. Will this functionbe useful in modeling the cumulative number of AIDSdeaths over an extended period of time? Explain youranswer.

77. Although it has been more than 50 years since the SupremeCourt ruled against school segregation, data from the CivilRights Project at Harvard University indicate that integra-tion and academic equality remain elusive. The graph showsthe percentage of the average African-American student’sclassmates who were white for the period from 1970through 2002.

f

36

35

34

33

37

32

31

30

Per

cent

age

of W

hite

Cla

ssm

ates

Year

Percentage of the Average African-AmericanStudent’s Classmates Who Were White

1970 1975 1995 2000 2005

2002

1980 1985 1990

29

Source: Civil Rights Project, Harvard University

The data in the bar graph can be modeled by the followingsecond- and third-degree polynomial functions:

Use these functions to solve Exercises 73–76.

73. Use both functions to find the cumulative number of AIDSdeaths in 2000. Which function provides a better descriptionfor the actual number shown in the bar graph?

g(x)=–84x3-702x2+50,609x+113,435.

f(x)=–2212x2+57,575x+107,896Cumulativenumber of

AIDS deathsx years after

1990

a. For which years was the percentage of white classmatesincreasing?

b. For which years was the percentage of white classmatesdecreasing?

c. How many turning points (from increasing to decreasingor from decreasing to increasing) does the graph have forthe period shown?

d. Suppose that a polynomial function is used to model thedata shown in the graph using

(number of years after 1970, percentage of the averageAfrican-American student’s classmates who were white).

Use the number of turning points to determine the degreeof the polynomial function of best fit.

e. For the model in part (d), should the leading coefficient ofthe polynomial function be positive or negative? Explainyour answer.

pr02-265-374.I-hr 1/26/06 5:14 PM Page 299

a. Between which stages was marital satisfaction for wivesdecreasing?

b. Between which stages was marital satisfaction for wivesincreasing?

c. How many turning points (from decreasing to increasingor from increasing to decreasing) are shown in the graphfor wives?

d. Suppose that a polynomial function is used to model thedata shown in the graph for wives using

(stage in the relationship, percentage indicating that themarriage was going well all the time).

Use the number of turning points to determine the degreeof the polynomial function of best fit.

e. For the model in part (d), should the leading coefficient ofthe polynomial function be positive or negative? Explainyour answer.

Writing in Mathematics79. What is a polynomial function?

80. What do we mean when we describe the graph of a polyno-mial function as smooth and continuous?

81. What is meant by the end behavior of a polynomial function?

82. Explain how to use the Leading Coefficient Test to deter-mine the end behavior of a polynomial function.

83. Why is a third-degree polynomial function with a negativeleading coefficient not appropriate for modeling non-negative real-world phenomena over a long period of time?

84. What are the zeros of a polynomial function and how arethey found?

85. Explain the relationship between the multiplicity of a zeroand whether or not the graph crosses or touches the x-axis atthat zero.

86. If is a polynomial function, and and have oppo-site signs, what must occur between a and b? If and have the same sign, does it necessarily mean that this will notoccur? Explain your answer.

f1b2f1a2f1b2f1a2f

87. Explain the relationship between the degree of a polynomialfunction and the number of turning points on its graph.

88. Can the graph of a polynomial function have no x-intercepts?Explain.

89. Can the graph of a polynomial function have noy-intercept? Explain.

90. Describe a strategy for graphing a polynomial function. Inyour description, mention intercepts, the polynomial’sdegree, and turning points.

91. The graphs shown in Exercise 78 indicate that marital satis-faction tends to be greatest at the beginning and at the endof the stages in the relationship, with a decline occurring inthe middle. What explanations can you offer for this trend?

Technology Exercises92. Use a graphing utility to verify any five of the graphs that

you drew by hand in Exercises 41–64.

Write a polynomial function that imitates the end behavior ofeach graph in Exercises 93–96. The dashed portions of the graphsindicate that you should focus only on imitating the left and rightbehavior of the graph and can be flexible about what occursbetween the left and right ends. Then use your graphing utility tograph the polynomial function and verify that you imitated theend behavior shown in the given graph.

93. 94.

95. 96.

In Exercises 97–100, use a graphing utility with a viewingrectangle large enough to show end behavior to graph eachpolynomial function.

97.

98.

99.

100.

In Exercises 101–102, use a graphing utility to graph and g in

the same viewing rectangle. Then use the featureto show that and g have identical end behavior.

101.

102. f1x2 = -x4+ 2x3

- 6x, g1x2 = -x4

f1x2 = x3- 6x + 1, g1x2 = x3

f� ZOOM OUT �

f

f1x2 = -x5+ 5x4

- 6x3+ 2x + 20

f1x2 = -x4+ 8x3

+ 4x2+ 2

f1x2 = -2x3+ 6x2

+ 3x - 1

f1x2 = x3+ 13x2

+ 10x - 4

Per

cent

age

Indi

cati

ngM

arri

age

Goe

s Wel

lA

ll th

e T

ime

Marital Satisfaction for Families with Children

Stage of Relationship

VIIIVIIVIVIVIIIIII

StageStageStageStage

I:II:

III:IV:

Beginning familiesChild-bearing familiesFamilies with preschool childrenFamilies with school-age children

StageStageStageStage

V:VI:

VII:VIII:

Families with teenagersFamilies with adult children leaving homeFamilies in the middle yearsAging families

10

20

30

40

50

Wives

Husbands

Source: Rollins, B., & Feldman, H. (1970), Marital satisfaction over the family lifecycle. Journal of Marriage and the Family, 32, 20–28.

78. The graphs show the percentage of husbands and wives withone or more children who said their marriage was going well“all the time” at various stages in their relationships.


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Section 2.4 • Dividing Polynomials; Remainder and Factor Theorems 301


a. If then the graph of falls to the leftand falls to the right.

b. A mathematical model that is a polynomial of degree nwhose leading term is n odd and is ideallysuited to describe nonnegative phenomena overunlimited periods of time.

c. There is more than one third-degree polynomialfunction with the same three x-intercepts.

d. The graph of a function with origin symmetry can rise tothe left and to the right.

an 6 0,an xn,

ff1x2 = -x3+ 4x,

Use the descriptions in Exercises 104–105 to write an equation ofa polynomial function with the given characteristics. Use agraphing utility to graph your function to see if you are correct. Ifnot, modify the function’s equation and repeat this process.

104. Crosses the x-axis at 0, and 3; lies above the x-axisbetween and 0; lies below the x-axis between 0 and 3

105. Touches the x-axis at 0 and crosses the x-axis at 2; lies belowthe x-axis between 0 and 2

-4-4,

Objectives❶ Use long division to divide

polynomials.

❷ Use synthetic division todivide polynomials.

❸ Evaluate a polynomialusing the RemainderTheorem.

❹ Use the Factor Theorem tosolve a polynomialequation.

A moth has moved into your closet. She appeared in your bedroom at night, but some-how her relatively stout body escaped your clutches. Within a few weeks, swarms ofmoths in your tattered wardrobe suggest that Mama Moth was in the family way.Theremust be at least 200 critters nesting in every crevice of your clothing.

Two hundred plus moth-tykes from one female moth—is this possible? Indeedit is. The number of eggs, , in a female moth is a function of her abdominalwidth, x, in millimeters, modeled by

Because there are 200 moths feasting on your favorite sweaters, Mama’s abdominalwidth can be estimated by finding the solutions of the polynomial equation

How can we solve such an equation? You might begin by subtracting 200 from bothsides to obtain zero on one side. But then what? The factoring that we used in theprevious section will not work in this situation.

In Section 2.5, we will present techniques for solving certain kinds of polynomialequations. These techniques will further enhance your ability to manipulatealgebraically the polynomial functions that model your world. Because thesetechniques are based on understanding polynomial division, in this section we lookat two methods for dividing polynomials. (We’ll return to Mama Moth’s abdominalwidth in the exercise set.)

14x3- 17x2

- 16x + 34 = 200.

f1x2 = 14x3- 17x2

- 16x + 34, 1.5 … x … 3.5.

f1x2

SECTION 2.4 Dividing Polynomials; Remainder and Factor Theorems

pr02-265-374.I-hr 1/26/06 5:14 PM Page 301


❶ Use long division to dividepolynomials.

Long Division of Polynomials and the Division AlgorithmWe begin by looking at division by a polynomial containing more than one term, such as

When a divisor has more than one term, the four steps used to divide wholenumbers–divide, multiply, subtract, bring down the next term—form the repetitiveprocedure for polynomial long division.

EXAMPLE 1 Long Division of Polynomials

Divide by

Solution The following steps illustrate how polynomial division is very similar tonumerical division.

Arrange the terms of the dividendand the divisor

in descending powers of x.

Divide (the first term in thedividend) by x (the first term in the

divisor): Align like terms.

x(x + 3) = x2 + 3x �x2+10x+21

x2+ 3x

x

x+3

x 2

x� x.

x 2x

x + 3�x2+ 10x + 21

1x � 321x 2 � 10x � 212x + 3�x2+ 10x + 21

x + 3.x2+ 10x + 21

Divisor has two termsand is a binomial.

The polynomial dividend hasthree terms and is a trinomial.

�x2+10x+21.x+3

Multiply each term in the divisorby x, aligning terms of

the product under like terms inthe dividend.

�x2+10x+21

x2+ 3x7x

x+3

x

� �

Change signs of thepolynomial being subtracted.

1x � 32Subtract from by changing the sign of each term in the lower expression and adding.

x

x + 3�x2+ 10x + 21

x2+ 3x p

7x + 21

x 2 � 10xx 2 � 3x

Bring down 21 from the originaldividend and add algebraically toform a new dividend.

x + 7x + 3�x2

+ 10x + 21x2

+ 3x

7x + 21

Find the second term of thequotient. Divide the first term of

by x, the first term of

the divisor:

�x2+10x+21

x2+ 3x7x+21

7x+210

x+3

x+ 7

7(x + 3) = 7x + 21

Remainder

� �

7xx

� 7.

7x � 21

Multiply the divisor by 7,aligning under like terms in the newdividend. Then subtract to obtainthe remainder of 0.

1x � 32

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The quotient is Because the remainder is 0, we can conclude that is afactor of and

Divide by

Before considering additional examples, let’s summarize the general proce-dure for dividing one polynomial by another.

Long Division of Polynomials

1. Arrange the terms of both the dividend and the divisor in descendingpowers of the variable.

2. Divide the first term in the dividend by the first term in the divisor. Theresult is the first term of the quotient.

3. Multiply every term in the divisor by the first term in the quotient. Writethe resulting product beneath the dividend with like terms lined up.

4. Subtract the product from the dividend.5. Bring down the next term in the original dividend and write it next to the

remainder to form a new dividend.6. Use this new expression as the dividend and repeat this process until the

remainder can no longer be divided. This will occur when the degree of theremainder (the highest exponent on a variable in the remainder) is less thanthe degree of the divisor.

In our next long division, we will obtain a nonzero remainder.


Divide by

Solution We begin by writing the dividend in descending powers of x.

Now we divide by 3x to obtain x, multiply x and the divisor, and subtract.

Now we divide by 3x to obtain multiply and the divisor, and subtract.-1-1,-3x

�6x3 - x2-5x+46x3-4x2

–3x+4

3x2-2x

3x2-5x

3x-2

2x2+ xx(3x − 2) = 3x2 − 2x

� �

Change signs ofthe polynomial

being subtracted.

Divide: =x.

Multiply: x(3x-2)=3x2-2x.Subtract 3x2-2x from 3x2-5xand bring down 4.

3x2

3x

3x2

�6x3- x2-5x+46x3-4x2

3x2-5x

3x-2

2x2

2x2(3x − 2) = 6x3 − 4x2

� �


being subtracted.

Divide: =2x2.

Multiply: 2x2(3x-2)=6x3-4x2.Subtract 6x3-4x2 from 6x3-x2

and bring down –5x.

6x3

3x

4 - 5x - x2+ 6x3

= 6x3- x2

- 5x + 4

3x - 2.4 - 5x - x2+ 6x3

x + 9.x2+ 14x + 45Check

Point 1

x2+ 10x + 21

x + 3= x + 7.

x2+ 10x + 21

x + 3x + 7.

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The quotient is and the remainder is 2. When there is a nonzeroremainder, as in this example, list the quotient, plus the remainder above the divisor.Thus,

An important property of division can be illustrated by clearing fractions inthe equation that concluded Example 2. Multiplying both sides of this equation by

results in the following equation:

Polynomial long division is checked by multiplying the divisor with the quotient andthen adding the remainder.This should give the dividend.The process illustrates theDivision Algorithm.

The Division AlgorithmIf and are polynomials, with and the degree of is lessthan or equal to the degree of , then there exist unique polynomials and such that

The remainder, equals 0 or it is of degree less than the degree of Ifwe say that divides evenly into and that and are

factors of .

Divide by Express the result in the formquotient, plus remainder divided by divisor.

If a power of x is missing in either a dividend or a divisor, add that power of xwith a coefficient of 0 and then divide. In this way, like terms will be aligned as youcarry out the long division.

x - 3.7 - 11x - 3x2+ 2x3Check

Point 2

f1x2 q1x2d1x2f1x2d1x2r1x2 = 0,d1x2.r1x2,

f(x) = d(x) � q(x) + r(x).

Dividend Divisor RemainderQuotient

r1x2 q1x2f1x2 d1x2d1x2 Z 0,d1x2f1x2

6x3-x2-5x+4=(3x-2)(2x2+x-1)+2.

Dividend Divisor Quotient Remainder

3x - 2

6x3-x2-5x+43x-2

23x-2

=2x2+x-1+ .

Quotient

Remainderabove divisor

2x2+ x - 1

�6x3 - x2-5x+46x3-4x2

–3x+2

2

–3x+4

3x2-2x

3x2-5x

3x-2

2x2+ x-1−1(3x − 2) = −3x + 2

�


being subtracted.

Remainder

�

Divide: =–1.

Multiply: –1(3x-2)=–3x+2.Subtract –3x+2 from –3x+4,leaving a remainder of 2.

–3x3x

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❷ Use synthetic division todivide polynomials.


Divide by

Solution We write the dividend, as to keep all like terms aligned.

The division process is finished because the degree of which is 1, is less thanthe degree of the divisor which is 2. The answer is

Divide by

Dividing Polynomials Using Synthetic DivisionWe can use synthetic division to divide polynomials if the divisor is of the form This method provides a quotient more quickly than long division. Let’s compare thetwo methods showing divided by

Long Division Synthetic Division

Notice the relationship between the polynomials in the long division process andthe numbers that appear in synthetic division.

5

48 53

–5

21 16

4

3 7

1

1

These are the coefficients of thedividend x3 + 4x2 − 5x + 5.

These are the coefficients ofthe quotient x2 + 7x + 16.

This is theremainder.

The divisor is x − 3.This is 3, or c, in x − c. 3

3 � 1 4 -5 5 3 21 48 1 7 16 53

�x3 +4x2 - 5x+ 5x3 -3x2

16x-48

53

16x+ 5

7x2 -21x

7x2 - 5x

x-3

x2 + 7x+16

� �

� �

� �

Remainder

Quotient

DividendDivisorx − c;c = 3

x - 3.x3+ 4x2

- 5x + 5

x - c.

x2- 2x.2x4

+ 3x3- 7x - 10Check

Point 3

6x4+ 5x3

+ 3x - 5

3x2- 2x

= 2x2+ 3x + 2 +

7x - 5

3x2- 2x

.

3x2- 2x,

7x - 5,

�6x4 +5x3+0x2+3x-5

Multiply.

6x4 -4x3

6x2-4x

7x-5

6x2+3x

9x3-6x2

9x3+0x2

3x2-2x

2x2+3x+2

� �

� �

� �

2x2(3x2 − 2x) = 6x4 − 4x3

3x(3x2 − 2x) = 9x3 − 6x2

2(3x2 − 2x) = 6x2 − 4x

Remainder

6x4+ 5x3

+ 0x2+ 3x - 56x4

+ 5x3+ 3x - 5,

3x2- 2x.6x4

+ 5x3+ 3x - 5

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Now let’s look at the steps involved in synthetic division.

Synthetic DivisionTo divide a polynomial by

Example1. Arrange the polynomial in

descending powers, with a 0 coefficient for any missing term.

x - c:

x - 3�x3+ 4x2

- 5x + 5

2. Write c for the divisor, Tothe right, write the coefficients ofthe dividend.

x - c. 3 � 1 4 -5 5

3. Write the leading coefficient of thedividend on the bottom row.

3 � 1 4 -5 5 p

1 Bring down 1.

4. Multiply c (in this case, 3) times thevalue just written on the bottomrow. Write the product in the nextcolumn in the second row.

5

Add.

Add.

Add.

–54

3

1

1

Written from1 7 16 53

the last row of the synthetic division

3

Multiply by 3: 3 � 1 = 3.

Multiply by 3: 3 � 7 = 21.

Multiply by 3: 3 � 16 = 48.

5–54

3

1

1 7

3

5

48

–5

215316

4

37

1

1

3

5–5

2116

4

3

1

7 1

3

�x3+4x2-5x+ 5x-3

1x2+7x+16+53

x-3

5. Add the values in this new column,writing the sum in the bottom row.

6. Repeat this series of multiplicationsand additions until all columns arefilled in.

7. Use the numbers in the last row towrite the quotient, plus the remain-der above the divisor. The degreeof the first term of the quotient isone less than the degree of the firstterm of the dividend. The finalvalue in this row is the remainder.

EXAMPLE 4 Using Synthetic Division

Use synthetic division to divide by

Solution The divisor must be in the form Thus, we write asThis means that Writing a 0 coefficient for the missing in

the dividend, we can express the division as follows:

Now we are ready to set up the problem so that we can use synthetic division.

8605

Use the coefficients of the dividend5x3 + 0x2 + 6x + 8 in descending powers of x.

This is c inx − (−2). –2

x - 1-22�5x3+ 0x2

+ 6x + 8.

x2-termc = -2.x - 1-22. x + 2x - c.

x + 2.5x3+ 6x + 8

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❸ Evaluate a polynomial usingthe Remainder Theorem.

We begin the synthetic division process by bringing down 5. This is followed by aseries of multiplications and additions.

1. Bring down 5. 2. Multiply: 3. Add:

4. Multiply: 5. Add:

6. Multiply: 7. Add:

The numbers in the last row represent the coefficients of the quotient and theremainder. The degree of the first term of the quotient is one less than that ofthe dividend. Because the degree of the dividend, is 3, the degreeof the quotient is 2. This means that the 5 in the last row represents

Thus,

Use synthetic division to divide

The Remainder TheoremLet’s consider the Division Algorithm when the dividend, , is divided by In this case, the remainder must be a constant because its degree is less than one, thedegree of

f(x) = (x -c)q(x) + r

Dividend Divisor Quotient The remainder, r, is a constant whendividing by x − c.

x - c.

x - c.f1x2

x3- 7x - 6 by x + 2.Check

Point 4

5x2- 10x + 26 -

44x + 2

x + 2�5x3+ 6x + 8

860

–10 –5220

5

5 –10 26 –44

–2

The remainderis −44.

The quotient is5x2 − 10x + 26.

5x2.5x3

+ 6x + 8,

-2 � 5 0 6 8s -10 20 -52 ∂ Add. 5 -10 26 -44

Multiply by −2.

860

–10 –5220

5

5 –10 26

–2

8 � 1�522 � �44.�21262 � �52.

-2 � 5 0 6 s 8 -10 20 ∂ Add. 5 -10 26

Multiply by −2.

860

–10 20

5

5 –10

–2

6 � 20 � 26.�21�102 � 20.

-2 � 5 0s6 8 -10∂ Add. 5 -10

Multiply by −2.

860

–10

5

5

–2-2 � 5 0 6 8

∂

5

0 � 1�102 � �10.�2152 � �10.

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❹ Use the Factor Theorem tosolve a polynomial equation.

Now let’s evaluate at c.

Find by letting in This will give an expression for r.

and

What does this last equation mean? If a polynomial is divided by the remain-der is the value of the polynomial at c.This result is called the Remainder Theorem.

The Remainder TheoremIf the polynomial is divided by then the remainder is

Example 5 shows how we can use the Remainder Theorem to evaluate a poly-nomial function at 2. Rather than substituting 2 for x, we divide the function by

The remainder is .

EXAMPLE 5 Using the Remainder Theorem to Evaluate a Polynomial Function

Given use the Remainder Theorem to find .

Solution By the Remainder Theorem, if is divided by then theremainder is . We’ll use synthetic division to divide.

The remainder, 5, is the value of . Thus, We can verify that this iscorrect by evaluating directly. Using we obtain

Given use the Remainder Theorem to find

The Factor TheoremLet’s look again at the Division Algorithm when the divisor is of the form

By the Remainder Theorem, the remainder r is so we can substitute for r:

Notice that if then

so that is a factor of .This means that for the polynomial function , ifthen is a factor of .

Let’s reverse directions and see what happens if is a factor of . Thismeans that

f1x2 = 1x - c2q1x2.f1x2x - c

f1x2x - cf1c2 = 0,f1x2f1x2x - c

f1x2 = 1x - c2q1x2f1c2 = 0,

f1x2 = 1x - c2q1x2 + f1c2.f1c2f1c2,

Dividend Divisor Quotient Constant remainder

f(x) = (x -c)q(x) + r

x - c.

f1-42.f1x2 = 3x3+ 4x2

- 5x + 3,CheckPoint 5

f122 = 23- 4 # 22

+ 5 # 2 + 3 = 8 - 16 + 10 + 3 = 5.

f1x2 = x3- 4x2

+ 5x + 3,f122 f122 = 5.f122

Remainder

3

52

5

2–2

1

1

2 –4

–41

f122 x - 2,f1x2f122f1x2 = x3

- 4x2+ 5x + 3,

f122x - 2.

f1c2.x - c,f1x2

x - c,

O � r � r.O # q1c2 � O f1c2 = r

c - c � O f1c2 = 0 # q1c2 + r

f1x2 � 1x � c2q1x2 � r.x � cf1c2 f1c2 = 1c - c2q1c2 + r

f

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Equivalently,2x3

- 3x2- 11x + 6 = 1x - 3212x2

+ 3x - 22.

Now we can solve the polynomial equation.

This is the given equation.

Factor using the result from thesynthetic division.

Factor the trinomial.


Solve for x.

The solution set is

Based on the Factor Theorem, the following statements are useful in solvingpolynomial equations:

1. If is divided by and the remainder is zero, then c is a zero of andc is a root of the polynomial equation

2. If is divided by and the remainder is zero, then is a factor of.

Solve the equation given that is a zero off1x2 = 15x3

+ 14x2- 3x - 2.

-115x3+ 14x2

- 3x - 2 = 0CheckPoint 6

f1x2 x - cx - cf1x2f1x2 = 0.

fx - cf1x2

E -2, 12 , 3F . x = 3 x =

12 x = -2

x - 3 = 0 or 2x - 1 = 0 or x + 2 = 0

1x - 3212x - 121x + 22 = 0

1x - 3212x2+ 3x - 22 = 0

2x3- 3x2

- 11x + 6 = 0

Because the solution set of

is this implies that thepolynomial function

has x-intercepts (or zeros) atand 3. This is verified by

the graph of .


x-intercept: 12

[–10, 10, 1] by [–15, 15, 1]

f-2, 12 ,

f1x2 = 2x3- 3x2

- 11x + 6

E -2, 12 , 3F ,2x3

- 3x2- 11x + 6 = 0

Technology

If we replace x in with c, we obtain

Thus, if is a factor of , then We have proved a result known as the Factor Theorem.

The Factor TheoremLet be a polynomial.

a. If then is a factor of .

b. If is a factor of , then

The example that follows shows how the Factor Theorem can be used to solvea polynomial equation.

EXAMPLE 6 Using the Factor Theorem

Solve the equation given that 3 is a zero of

Solution We are given that 3 is a zero of Thismeans that Because the Factor Theorem tells us that is afactor of . We’ll use synthetic division to divide by

2x2+ 3x - 2

x - 3�2x3- 3x2

- 11x + 66

0–6

–11

63

2

2

3 –3

9–2

The remainder, 0,verifies that x − 3 is

a factor of2x3 − 3x2 − 11x + 6.

x - 3.f1x2f1x2 x - 3f132 = 0,f132 = 0.f1x2 = 2x3

- 3x2- 11x + 6.

f1x2 = 2x3- 3x2

- 11x + 6.2x3

- 3x2- 11x + 6 = 0

f1c2 = 0.f1x2x - c

f1x2x - cf1c2 = 0,

f1x2

f1c2 = 0.f1x2x - c

f1c2 = 1c - c2q1c2 = 0 # q1c2 = 0.

f1x2 = 1x - c2q1x2

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EXERCISE SET 2.4

Practice ExercisesIn Exercises 1–16, divide using long division. State the

quotient, and the remainder,

1.

2.

3.

4.

5.

6.

7.

8.

9. 10.

11. 12.

13. 14.

15. 16.

In Exercises 17–32, divide using synthetic division.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27. 28.

29. 30.

31.

32.

In Exercises 33–40, use synthetic division and the RemainderTheorem to find the indicated function value.

33.

34.

35.

36.

37.

38. f1x2 = x4- 5x3

+ 5x2+ 5x - 6; f122

f1x2 = x4+ 5x3

+ 5x2- 5x - 6; f132

f1x2 = 4x3+ 5x2

- 6x - 4; f1-22f1x2 = 3x3

- 7x2- 2x + 5; f1-32

f1x2 = x3- 7x2

+ 5x - 6; f132f1x2 = 2x3

- 11x2+ 7x - 5; f142

x5- 2x4

- x3+ 3x2

- x + 1x - 2

2x5- 3x4

+ x3- x2

+ 2x - 1x + 2

x7- 128

x - 2x4

- 256x - 4

x7+ x5

- 10x3+ 12

x + 2x5

+ x3- 2

x - 1

1x2- 6x - 6x3

+ x42 , 16 + x21x2

- 5x - 5x3+ x42 , 15 + x2

1x5+ 4x4

- 3x2+ 2x + 32 , 1x - 32

16x5- 2x3

+ 4x2- 3x + 12 , 1x - 22

15x3- 6x2

+ 3x + 112 , 1x - 2214x3

- 3x2+ 3x - 12 , 1x - 12

15x2- 12x - 82 , 1x + 32

13x2+ 7x - 202 , 1x + 52

1x2+ x - 22 , 1x - 12

12x2+ x - 102 , 1x - 22

2x5- 8x4

+ 2x3+ x2

2x3+ 1

18x4+ 9x3

+ 3x2

3x2+ 1

x4+ 2x3

- 4x2- 5x - 6

x2+ x - 2

6x3+ 13x2

- 11x - 15

3x2- x - 3

x4- 81

x - 34x4

- 4x2+ 6x

x - 4

3x2- 2x + 5x - 3

2x3+ 7x2

+ 9x - 20x + 3

14x2- 8x + 62 , 12x - 12

112x2+ x - 42 , 13x - 22

16x3+ 17x2

+ 27x + 202 , 13x + 4216x3

+ 7x2+ 12x - 52 , 13x - 12

1x3- 2x2

- 5x + 62 , 1x - 321x3

+ 5x2+ 7x + 22 , 1x + 22

1x2+ 3x - 102 , 1x - 22

1x2+ 8x + 152 , 1x + 52

r1x2.q1x2,39.

40.

41. Use synthetic division to divide

Use the result to find all zeros of 42. Use synthetic division to divide

Use the result to find all zeros of 43. Solve the equation given that 2 is a

zero of 44. Solve the equation given that

is a zero of 45. Solve the equation given that

is a root.46. Solve the equation given that

is a root.

Practice PlusIn Exercises 47–50, use the graph or the table to deter-

mine a solution of each equation. Use synthetic division to verifythat this number is a solution of the equation. Then solve the poly-nomial equation.

47.

48.

49.

y1 = 6x3 − 11x2 + 6x − 1

6x3- 11x2

+ 6x - 1 = 0

[−4, 0, 1] by [−25, 25, 5]

y = 2x3 + x2 − 13x + 6

0−4

−25

25

2x3+ x2

- 13x + 6 = 0

[0, 4, 1] by [−25, 25, 5]

y = x3 + 2x2 − 5x − 6

0 4

−25

25

x3+ 2x2

- 5x - 6 = 0

- 133x3

+ 7x2- 22x - 8 = 0

- 3212x3

+ 16x2- 5x - 3 = 0

f1x2 = 2x3- 3x2

- 11x + 6.-22x3

- 3x2- 11x + 6 = 0

f1x2 = 2x3- 5x2

+ x + 2.2x3

- 5x2+ x + 2 = 0

f.

f1x2 = x3- 2x2

- x + 2 by x + 1.

f.

f1x2 = x3- 4x2

+ x + 6 by x + 1.

f1x2 = 6x4+ 10x3

+ 5x2+ x + 1; fa -

23b

f1x2 = 2x4- 5x3

- x2+ 3x + 2; fa -

12b

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50.

Application Exercises51. a. Use synthetic division to show that 3 is a solution of the

polynomial equation

b. Use the solution from part (a) to solve this problem. Thenumber of eggs, , in a female moth is a function of herabdominal width, x, in millimeters, modeled by

What is the abdominal width when there are 211 eggs?

52. a. Use synthetic division to show that 2 is a solution of thepolynomial equation

b. Use the solution from part (a) to solve this problem. Thewidth of a rectangular box is twice the height and thelength is 7 inches more than the height. If the volume is 72cubic inches, find the dimensions of the box.

In Exercises 53–54, write a polynomial that represents the lengthof each rectangle.

53.

54.The width isx + units.

The area is8x3 − 6x2 − 5x + 3

square units.

34

The width isx + 0.2 units.

The area is0.5x3 − 0.3x2 + 0.22x + 0.06

square units.

h + 72h

h

2h3+ 14h2

- 72 = 0.

f1x2 = 14x3- 17x2

- 16x + 34.

f1x2

14x3- 17x2

- 16x - 177 = 0.

y1 = 2x3 + 11x2 − 7x − 6

2x3+ 11x2

- 7x - 6 = 0 During the 1980s, the controversial economist Arthur Lafferpromoted the idea that tax increases lead to a reduction ingovernment revenue. Called supply-side economics, the theoryuses functions such as

This function models the government tax revenue, , in tens ofbillions of dollars, in terms of the tax rate, x.The graph of thefunction is shown. It illustrates tax revenue decreasing quite dramati-cally as the tax rate increases.At a tax rate of (gasp) 100%, the gov-ernment takes all our money and no one has an incentive to work.With no income earned, zero dollars in tax revenue is generated.

Use function and its graph to solve Exercises 55–56.

55. a. Find and interpret (30). Identify the solution as a pointon the graph of the function.

b. Rewrite the function by using long division to perform

Then use this new form of the function to find (30). Doyou obtain the same answer as you did in part (a)?

c. Is a polynomial function? Explain your answer.

56. a. Find and interpret Identify the solution as a pointon the graph of the function.

b. Rewrite the function by using long division to perform

Then use this new form of the function to find Doyou obtain the same answer as you did in part (a)?

c. Is a polynomial function? Explain your answer.

Writing in Mathematics57. Explain how to perform long division of polynomials. Use

divided by in your explanation.

58. In your own words, state the Division Algorithm.

59. How can the Division Algorithm be used to check thequotient and remainder in a long division problem?

60. Explain how to perform synthetic division. Use the divisionproblem in Exercise 57 to support your explanation.

61. State the Remainder Theorem.

62. Explain how the Remainder Theorem can be used to findif What

advantage is there to using the Remainder Theorem in thissituation rather than evaluating directly?

63. How can the Factor Theorem be used to determine if is a factor of x3

- 2x2- 11x + 12?

x - 1

f1-62f1x2 = x4

+ 7x3+ 8x2

+ 11x + 5.f1-62

x - 32x3- 3x2

- 11x + 7

f

f1402.180x - 80002 , 1x - 1102.

f1402.f

f

180x - 80002 , 1x - 1102.

f

f

Gov

ernm

ent T

ax R

even

ue(t

ens

of b

illio

ns o

f dol

lars

)

y

x

Tax Rate10080604020

80

60

40

20

80x − 8000x − 110f(x) =

At a 100% tax rate,$0 in tax revenue isgenerated.

f1x2f1x2 =

80x - 8000x - 110

, 30 … x … 100.

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64. If you know that is a zero of

explain how to solve the equation

Technology Exercise65. For each equation that you solved in Exercises 43–46, use a

graphing utility to graph the polynomial function defined bythe left side of the equation. Use end behavior to obtain acomplete graph. Then use the graph’s x-intercepts to verifyyour solutions.


a. If a trinomial in x of degree 6 is divided by a trinomial in xof degree 3, the degree of the quotient is 2.

x3+ 7x2

+ 4x - 12 = 0.

f1x2 = x3+ 7x2

+ 4x - 12,

-2 b. Synthetic division could not be used to find the quotientof and

c. Any problem that can be done by synthetic division can alsobe done by the method for long division of polynomials.

d. If a polynomial long-division problem results in a remain-der that is a whole number, then the divisor is a factor ofthe dividend.

67. Find k so that is a factor of

68. When is divided by a polynomial, the quotientis and the remainder is 3. Find the polynomial.

69. Find the quotient of and

70. Synthetic division is a process for dividing a polynomial byThe coefficient of x is 1. How might synthetic division

be used if you are dividing by

71. Use synthetic division to show that 5 is a solution of

Then solve the polynomial equation.

x4- 4x3

- 9x2+ 16x + 20 = 0.

2x - 4?x - c.

xn+ 1.x3n

+ 1

2x - 32x2

- 7x + 9

20x3+ 23x2

- 10x + k.

4x + 3

x -12 .10x3

- 6x2+ 4x - 1

Objectives❶ Use the Rational Zero

Theorem to find possiblerational zeros.

❷ Find zeros of a polynomialfunction.

❸ Solve polynomialequations.

❹ Use the LinearFactorization Theoremto find polynomialswith given zeros.

❺ Use Descartes’s Ruleof Signs.

Tartaglia’s Secret Formula for One Solution of

- C3 B an

2b2

+ am

3b3

-

n

2

x = C3 B an

2b2

+ am

3b3

+

n

2

x3 � mx � n

You stole my formula!

SECTION 2.5 Zeros of Polynomial Functions

Popularizers of mathematics are sharing bizarre stories that are giving math asecure place in popular culture. One episode, able to compete with the wildest fareserved up by television talk shows and the tabloids, involves three Italianmathematicians and, of all things, zeros of polynomial functions.

Tartaglia (1499–1557), poor and starving, has found a formula that gives a rootfor a third-degree polynomial equation. Cardano (1501–1576) begs Tartaglia toreveal the secret formula, wheedling it from him with the promise he will find theimpoverished Tartaglia a patron. Then Cardano publishes his famous work ArsMagna, in which he presents Tartaglia’s formula as his own. Cardano uses his mosttalented student, Ferrari (1522–1565), who derived a formula for a root of a fourth-degree polynomial equation, to falsely accuse Tartaglia of plagiarism. The disputebecomes violent and Tartaglia is fortunate to escape alive.

The noise from this “You Stole My Formula” episode is quieted by the work ofFrench mathematician Evariste Galois (1811–1832). Galois proved that there is nogeneral formula for finding roots of polynomial equations of degree 5 or higher.There are, however, methods for finding roots. In this section, we study methods forfinding zeros of polynomial functions. We begin with a theorem that plays animportant role in this process.

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Section 2.5 • Zeros of Polynomial Functions 313

❶ Use the Rational ZeroTheorem to find possiblerational zeros.

Be sure you are familiar with the various kinds of zeros of polynomial functions. Here’s aquick example:

–3, , –�2, �2, 4-5i, 4+5i

f(x)=(x+3)(2x-1)(x+�2)(x-�2)(x-4+5i)(x-4-5i).

12

Rational zeros

Real zeros Nonreal zeros

Irrational zeros Complex imaginary zeros

Zeros:

Study Tip

The Rational Zero TheoremThe Rational Zero Theorem provides us with a tool that we can use to make a list ofall possible rational zeros of a polynomial function. Equivalently, the theorem givesall possible rational roots of a polynomial equation. Not every number in the list willbe a zero of the function, but every rational zero of the polynomial function willappear somewhere in the list.

The Rational Zero Theorem

If has integer coefficients and

(where is reduced to lowest terms) is a rational zero of , then p is a

factor of the constant term, and q is a factor of the leading coefficient,

You can explore the “why” behind the Rational Zero Theorem in Exercise 90of Exercise Set 2.5. For now, let’s see if we can figure out what the theorem tells usabout possible rational zeros.To use the theorem, list all the integers that are factorsof the constant term, Then list all the integers that are factors of the leadingcoefficient, Finally list all possible rational zeros:

EXAMPLE 1 Using the Rational Zero Theorem

List all possible rational zeros of

Solution The constant term is 4. We list all of its factors: The leadingcoefficient is Its factors are

Because

we must take each number in the first row, and divide by each numberin the second row,

Factors of 4Factors of –1

—1, —2, —4—1

=Possible rational zeros= =—1, —2, —4

Divide ±1by ±1.

Divide ±2by ±1.

Divide ±4by ±1.

;1.;1, ;2, ;4,

Possible rational zeros =

Factors of the constant termFactors of the leading coefficient

,

Factors of the constant term, 4:Factors of the leading coefficient, -1:

;1,;1

;2,

;4

;1.-1.;1, ;2, ;4.

f1x2 = -x4+ 3x2

+ 4.



.

an .a0 .

an .a0 ,

fp

q

p

qf1x2 = an xn

+ an - 1 xn - 1+

Á+ a1 x + a0

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Always keep in mind therelationship among zeros, roots,and x-intercepts. The zeros of afunction are the roots, or solu-tions, of the equation Furthermore, the real zeros, orreal roots, are the x-intercepts ofthe graph of .f

f1x2 = 0.f

Study Tip

❷ Find zeros of a polynomialfunction.

There are six possible rational zeros,and The graph of is shown in Figure 2.24. The x-intercepts are

and 2. Thus, and 2 are the actualrational zeros.


EXAMPLE 2 Using the Rational Zero Theorem


Solution The constant term is and the leading coefficient is 15.

There are 16 possible rational zeros. The actual solution set of

is which contains three of the 16 possible zeros.


How do we determine which (if any) of the possible rational zeros arerational zeros of the polynomial function? To find the first rational zero, we can usea trial-and-error process involving synthetic division: If is divided by and the remainder is zero, then c is a zero of . After we identify the first rationalzero, we use the result of the synthetic division to factor the original polynomial.Then we set each factor equal to zero to identify any additional rational zeros.


Find all zeros of

Solution We begin by listing all possible rational zeros.

Now we will use synthetic division to see if we can find a rational zero among thepossible rational zeros Keep in mind that if is divided byf1x2;1, ;2, ;3, ;6.

Divide the eight numbersin the numerator by ±1.

Factors of the constant term, –6Factors of the leading coefficient, 1

—1, —2, —3, —6—1

== =—1, —2, —3, —6

Possible rational zeros

f1x2 = x3+ 2x2

- 5x - 6.

fx - cf1x2

f1x2 = 4x5+ 12x4

- x - 3.

CheckPoint 2

E -1, - 13 , 25 F ,

15x3+ 14x2

- 3x - 2 = 0

Divide ±1and ±2by ±1.

=—1, —2,


13

— ,23

— ,


15

— ,25

— ,


115

— ,215

—


Factors of the constant term, -2Factors of the leading coefficient, 15

=

;1, ;2;1, ;3, ;5, ;15

-2

f1x2 = 15x3+ 14x2

- 3x - 2.

f1x2 = x3+ 2x2

- 5x - 6.

CheckPoint 1

-2-2

f1x2 = -x4+ 3x2

+ 4;4.;1, ;2,

−2 is a rational zero. 2 is a rational zero.

Figure 2.24 The graph ofshows that

and 2 are rational zeros.-2f1x2 = -x4

+ 3x2+ 4

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and the remainder is zero, then c is a zero of . Let’s start by testing 1. If 1 isnot a rational zero, then we will test other possible rational zeros.

The zero remainder tells us that 2 is a zero of the polynomial functionEquivalently, 2 is a solution, or root, of the polynomial

equation Thus, is a factor of the polynomial. Thefirst three numbers in the bottom row of the synthetic division on the right, 1, 4, and3, give the coefficients of the other factor. This factor is

Finding the zeros of isthe same as finding the roots of this equation.

Factor using the result from the synthetic division.

Factor completely.

or or Set each factor equal to zero.

Solve for x.

The solution set is The zeros of are and 2.

Find all zeros of

Our work in Example 3 involved finding zeros of a third-degree polynomialfunction.The Rational Zero Theorem is a tool that allows us to rewrite such functionsas products of two factors, one linear and one quadratic. Zeros of the quadratic factorare found by factoring, the quadratic formula, or the square root property.


Find all zeros of

Solution We begin by listing all possible rational zeros.

Now we will use synthetic division to see if we can find a rational zero among thefour possible rational zeros.


Factors of the constant term, -3Factors of the leading coefficient, 1

=

;1, ;3;1

= ;1, ;3

f1x2 = x3+ 7x2

+ 11x - 3.

f1x2 = x3+ 8x2

+ 11x - 20.

CheckPoint 3

-3, -1,f5-3, -1, 26. x = -1 x = -3 x = 2

x + 1 = 0 x + 3 = 0 x - 2 = 0

1x - 221x + 321x + 12 = 0

1x - 221x2+ 4x + 32 = 0

f1x2 � x3 � 2x 2 � 5x � 6 x3+ 2x2

- 5x - 6 = 0

x2+ 4x + 3.

x - 2x3+ 2x2

- 5x - 6 = 0.f1x2 = x3

+ 2x2- 5x - 6.

Coefficients off (x) = x3 + 2x2 − 5x − 6

–6

–2 –8

–5

3 –2

2

1 3

1

1

1Possible

rational zero

The nonzero remaindershows that 1 is not a zero.

Test 1.

Coefficients off (x) = x3 + 2x2 − 5x − 6

–6

6 0

–5

8 3

2

2 4

1

1

2Possible

rational zero

The zero remaindershows that 2 is a zero.

Test 2.

fx - c

Test 1. Test . Test 3. Test .

-3 � 1 7 11 -3-3 -12 3

1 4 -1 0

3 � 1 7 11 -33 30 123

1 10 41 120

-1 � 1 7 11 -3-1 -6 -5

1 6 5 -8

1 � 1 7 11 -31 8 19

1 8 19 16

�3�1

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The zero remainder in the final synthetic division on the previous page tells us thatis a zero of the polynomial function To find all

zeros of , we proceed as follows:

Finding the zeros of is the same thing asfinding the roots of

This result is from the last synthetic divisionon the previous page. The first three numbersin the bottom row, 1, 4, and give thecoefficients of the second factor.

or Set each factor equal to 0.Solve the linear equation.

We can use the quadratic formula to solve

We use the quadratic formula because cannot be factored.

Let and

Multiply and subtract under the radical:

Divide the numerator and the denominator by 2.

The solution set is The zeros of are and Among these three real zeros, one zero is rational and two areirrational.

Find all zeros of

If the degree of a polynomial function or equation is 4 or higher, it is oftennecessary to find more than one linear factor by synthetic division.

One way to speed up the process of finding the first zero is to graph thefunction. Any x-intercept is a zero.

EXAMPLE 5 Solving a Polynomial Equation

Solve:

Solution Recall that we refer to the zeros of a polynomial function and the rootsof a polynomial equation. Because we are given an equation, we will use the word“roots,” rather than “zeros,” in the solution process. We begin by listing all possiblerational roots.

Possible rational roots

The graph of is shown in Figure 2.25. Becausethe x-intercept is 2, we will test 2 by synthetic division and show that it is a rootof the given equation. Without the graph, the procedure would be to start the

f1x2 = x4- 6x2

- 8x + 24

= ;1, ;2, ;3, ;4, ;6, ;8, ;12, ;24

=

;1, ;2, ;3, ;4, ;6, ;8, ;12, ;24;1

=

Factors of the constant term, 24Factors of the leading coefficient, 1

x4- 6x2

- 8x + 24 = 0.

f1x2 = x3+ x2

- 5x - 2.CheckPoint 4

-2 + 25.-3, -2 - 25,fE -3, -2 - 25, -2 + 25F .

= -2 ; 25

220 � 24 # 5 � 225 =

-4 ; 2252

42 � 41121� 12 � 16 � 1�42 � 16 � 4 � 20. =

-4 ; 2202

c � � 1.a � 1, b � 4, =

-4 ; 342- 41121-12

2112

x2 � 4x � 1 x =

-b ; 3b2- 4ac

2a

x2+ 4x - 1 = 0.

x = -3 x2

+ 4x - 1 = 0 x + 3 = 0

� 1,

1x + 321x2+ 4x - 12 = 0

f1x2 � 0.f x3

+ 7x2+ 11x - 3 = 0

ff1x2 = x3

+ 7x2+ 11x - 3.-3

❸ Solve polynomial equations.

x-intercept: 2

Figure 2.25 The graph ofin a

by viewingrectangle

[-2, 10, 1][-1, 5, 1]f1x2 = x4

- 6x2- 8x + 24

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trial-and-error synthetic division with 1 and proceed until a zero remainder is found,as we did in Example 4.

Now we can rewrite the given equation in factored form.


This is the result obtained from thesynthetic division. The first four numbersin the bottom row, 1, 2, and give the coefficients of the second factor.


We can use the same approach to look for rational roots of the polynomial equationlisting all possible rational roots. Without the graph in

Figure 2.25, the procedure would be to start testing possible rational roots by trial-and-error synthetic division. However, take a second look at the graph inFigure 2.25. Because the graph turns around at 2, this means that 2 is a root of evenmultiplicity. Thus, 2 must also be a root of confirmed bythe following synthetic division.

Now we can solve the original equation as follows:


This factorization wasobtained from the firstsynthetic division.

This factorization wasobtained from the secondsynthetic division. The firstthree numbers in the bottomrow, 1, 4, and 6, give thecoefficients of the third factor.

or or Set each factor equal to 0.Solve the linear equations.

We can use the quadratic formula to solve

We use the quadratic formula becausecannot be factored.

Let and

Multiply and subtract under the radical:

Simplify. = -2 ; i22

2�8 � 441221� 12 � 2i22 =

-4 ; 2i222

42- 4112162 � 16 - 24 � �8. =

-4 ; 2-82

c � 6.a � 1, b � 4, =

-4 ; 342- 4112162

2112

x2 � 4x � 6 x =

-b ; 3b2- 4ac

2a

x2+ 4x + 6 = 0.

x = 2 x = 2 x2

+ 4x + 6 = 0 x - 2 = 0 x - 2 = 0

1x - 221x - 221x2+ 4x + 62 = 0

1x - 221x3+ 2x2

- 2x - 122 = 0

x4- 6x2

- 8x + 24 = 0

These are the coefficients ofx3 + 2x2 − 2x − 12 = 0.

The zero remainder indicates that 2 is aroot of x3 + 2x2 − 2x − 12 = 0.

–12

120

–2

8 6

2

2 4

1

1

2

x3+ 2x2

- 2x - 12 = 0,

x3+ 2x2

- 2x - 12 = 0,

x - 2 = 0 or x3+ 2x2

- 2x - 12 = 0

� 12,�2,

1x - 221x3+ 2x2

- 2x - 122 = 0

x4- 6x2

- 8x + 24 = 0

Careful!x4 − 6x2 − 8x + 24 = x4 + 0x3 − 6x2 − 8x + 24

The zero remainder indicates that 2 is a root ofx4 − 6x2 − 8x + 24 = 0.

24

–24 0

–8

–4 –12

–6

4 –2

0

2 2

1

1

2

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The solution set of the original equation, isThe graph of in

Figure 2.25 illustrates that a graphing utility does not reveal the two imaginary roots.

In Example 5, 2 is a repeated root of the equation with multiplicity 2. Theexample illustrates two general properties:

Properties of Polynomial Equations

1. If a polynomial equation is of degree n, then counting multiple rootsseparately, the equation has n roots.

2. If is a root of a polynomial equation with real coefficients then the complex imaginary number is also a root. Complex imagi-nary roots, if they exist, occur in conjugate pairs.

Solve:

The Fundamental Theorem of AlgebraThe fact that a polynomial equation of degree n has n roots is a consequence of atheorem proved in 1799 by a 22-year-old student named Carl Friedrich Gauss in hisdoctoral dissertation. His result is called the Fundamental Theorem of Algebra.

The Fundamental Theorem of AlgebraIf is a polynomial of degree n, where then the equation has at least one complex root.

Suppose, for example, that represents a polynomial equation ofdegree n. By the Fundamental Theorem of Algebra, we know that this equation hasat least one complex root; we’ll call it By the Factor Theorem, we know that

is a factor of . Therefore, we obtain

The degree of the polynomial is


If the degree of is at least 1, by the Fundamental Theorem of Algebra, theequation has at least one complex root. We’ll call it The FactorTheorem gives us

The degree of is

The degree of is


Let’s see what we have up to this point and then continue the process.

This is the original polynomial equation ofdegree n.

This is the result from our first application ofthe Fundamental Theorem.

This is the result from our second applicationof the Fundamental Theorem.

By continuing this process, we will obtain the product of n linear factors. Settingeach of these linear factors equal to zero results in n complex roots.Thus, if is apolynomial of degree n, where then has exactly n roots, whereroots are counted according to their multiplicity.

f1x2 = 0n Ú 1,f1x2

1x - c121x - c22q21x2 = 0

1x - c12q11x2 = 0

f1x2 = 0

x - c2 = 0 or q21x2 = 0.

n � 2.q21x2 1x - c22q21x2 = 0

n � 1.q11x2 q11x2 = 0

c2 .q11x2 = 0q11x2

x - c1 = 0 or q11x2 = 0.

n � 1.q11x2 1x - c12q11x2 = 0

f1x2x - c1

c1 .

f1x2 = 0

f1x2 = 0n Ú 1,f1x2

x4- 6x3

+ 22x2- 30x + 13 = 0.Check

Point 5

a - bi1b Z 02,a + bi

f1x2 = x4- 6x2

- 8x + 2452, -2 - i12, -2 + i126. x4- 6x2

- 8x + 24 = 0,

x-intercept: 2

Figure 2.25 (repeated)

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The graph ofshown

in a by viewing rectangle, verifies that and 2 are real zeros. By tracingalong the curve, we can check that

−2 is a zero. 2 is a zero.

f132 = -150.

-2[-200, 20, 20][-3, 3, 1]

f1x2 = -3x4+ 9x2

+ 12,

The Linear Factorization TheoremIn Example 5, we found that has as asolution set, where 2 is a repeated root with multiplicity 2. The polynomial can befactored over the complex nonreal numbers as follows:

This fourth-degree polynomial has four linear factors. Just as an nth-degreepolynomial equation has n roots, an nth-degree polynomial has n linear factors.This is formally stated as the Linear Factorization Theorem.

The Linear Factorization TheoremIf where and then

where are complex numbers (possibly real and not necessarilydistinct). In words: An nth-degree polynomial can be expressed as the productof a nonzero constant and n linear factors.

Many of our problems involving polynomial functions and polynomialequations dealt with the process of finding zeros and roots.The Linear FactorizationTheorem enables us to reverse this process, finding a polynomial function when thezeros are given.

EXAMPLE 6 Finding a Polynomial Function with Given Zeros

Find a fourth-degree polynomial function with real coefficients that has 2,and i as zeros and such that

Solution Because i is a zero and the polynomial has real coefficients, the conjugate,must also be a zero.We can now use the Linear Factorization Theorem.

This is the linear factorization for afourth-degree polynomial.

Use the given zeros:and,

from above,

Multiply:

Complete the multiplication.

To find use the fact that

Solve for

Simplify: Divide both sides by 50.

Substituting for in the formula for , we obtain

Equivalently,

f1x2 = -3x4+ 9x2

+ 12.

f1x2 = -31x4- 3x2

- 42.f1x2an-3

an = -381 � 27 � 4 � 50. 50an = -150

an . an181 - 27 - 42 = -150

f132 � � 150.an , f132 = an134

- 3 # 32- 42 = -150

f1x2 = an1x4- 3x2

- 42x2 � 1� 12 � x2 � 1.

1x � i21x � i2 � x2- i2 � = an1x2

- 421x2+ 12

c4 � � i.c1 � �2, c2 � 2, c3 � i,

= an1x + 221x - 221x - i21x + i2 f1x2 = an1x - c121x - c221x - c321x - c42

- i,

f132 = -150.-2,f1x2

c1 , c2 , Á , cn

f1x2 = an1x - c121x - c22Á 1x - cn2,an Z 0,n Ú 1f1x2 = an xn

+ an - 1 xn - 1+

Á+ a1 x + a0 ,

f(x)=x4-6x2-8x+24

=[x-(–2+i�2)][x-(–2-i�2)](x-2)(x-2).

These are the four zeros.

These are the linear factors.

E2, -2 ; i22Fx4- 6x2

- 8x + 24 = 0❹ Use the Linear FactorizationTheorem to find polynomialswith given zeros.

Technology

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The number of real zeros given by Descartes’s Rule of Signs includes rational zeros froma list of possible rational zeros, as well as irrational zeros not on the list. It does not includeany imaginary zeros.

“An equation can have as many true[positive] roots as it contains changesof sign, from plus to minus or from minusto plus.” René Descartes (1596–1650)in La Géométrie (1637)

Study Tip

❺ Use Descartes’s Rule of Sign.

Find a third-degree polynomial function with real coefficients thathas and i as zeros and such that

Descartes’s Rule of SignsBecause an nth-degree polynomial equation might have roots that are imaginary num-bers,we should note that such an equation can have at most n real roots.Descartes’s Ruleof Signs provides even more specific information about the number of real zeros that apolynomial can have. The rule is based on considering variations in sign between con-secutive coefficients.For example, the function has three sign changes:

Descartes’s Rule of SignsLet be a polynomial withreal coefficients.

1. The number of positive real zeros of is eithera. the same as the number of sign changes of

orb. less than the number of sign changes of by a positive even integer.If has only one variation in sign, then has exactly one positive real zero.

2. The number of negative real zeros of is eithera. the same as the number of sign changes of

orb. less than the number of sign changes of by a positive even integer.If has only one variation in sign, then has exactly one negative real zero.ff1-x2

f1-x2f1-x2f

ff1x2f1x2

f1x2f

f1x2 = an xn+ an - 1 xn - 1

+Á

+ a2 x2+ a1 x + a0

f(x)=3x7-2x5-x4+7x2+x-3.

sign change sign change sign change

f1x2 = 3x7- 2x5

- x4+ 7x2

+ x - 3

f112 = 8.-3f1x2Check

Point 6

Polynomial Function Sign Changes Conclusion

3 There are 3 positivereal zeros.

orThere is positive real zero.

2 There are 2 positivereal zeros.

orThere are positive real zeros.

1 There is 1 positivereal zero.f(x)=–7x6-5x4+x+9

sign change

2 - 2 = 0

f(x)=4x5+2x4-3x2+x+5

sign change sign change

3 - 2 = 1

f(x)=3x7-2x5-x4+7x2+x-3.

sign change sign change sign change

Table 2.1 Descartes’s Rule of Signs and Positive Real Zeros

Table 2.1 illustrates what Descartes’s Rule of Signs tells us about the positivereal zeros of various polynomial functions.

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EXAMPLE 7 Using Descartes’s Rule of Signs

Determine the possible numbers of positive and negative real zeros of

Solution1. To find possibilities for positive real zeros, count the number of sign changes in

the equation for Because all the coefficients are positive, there are novariations in sign. Thus, there are no positive real zeros.

2. To find possibilities for negative real zeros, count the number of sign changesin the equation for We obtain this equation by replacing x with inthe given function.

Now count the sign changes.

There are three variations in sign. The number of negative real zeros of iseither equal to the number of sign changes, 3, or is less than this number by aneven integer. This means that either there are 3 negative real zeros or there is

negative real zero.

What do the results of Example 7 mean in terms of solving

Without using Descartes’s Rule of Signs, we list the possible rational roots as follows:

However, Descartes’s Rule of Signs informed us that has no positive real zeros. Thus, the polynomial equation has no positive real roots. This means that we can eliminate the positive numbersfrom our list of possible rational roots. Possible rational roots include only and We can use synthetic division and test two of the three possible rationalroots of as follows:

By solving the equation you will find that this equation ofdegree 3 has three roots. One root is and the other two roots are imaginarynumbers in a conjugate pair. Verify this by completing the solution process.

Determine the possible numbers of positive and negative real zeros off1x2 = x4

- 14x3+ 71x2

- 154x + 120.CheckPoint 7

-1x3

+ 2x2+ 5x + 4 = 0,

4

–10 –6

5

0 5

2

–2 0

1

1

–2Test−2.

The nonzero remaindershows that −2 is not a root.

4

–4 0

5

–1 4

2

–1 1

1

1

–1Test−1.

The zero remainder showsthat −1 is a root.

x3+ 2x2

+ 5x + 4 = 0-4.

-1, -2,

x3+ 2x2

+ 5x + 4 = 0f1x2 = x3

+ 2x2+ 5x + 4

=

Factors of the constant term, 4Factors of the leading coefficient, 1

=

;1, ;2, ;4;1

= ;1, ;2, ;4

Possible rational roots

x3+ 2x2

+ 5x + 4 = 0?

3 - 2 = 1

f

f(x)=–x3+2x2-5x+4x

sign change sign changesign change

=–x3+2x2-5x+4

f(–x)=(–x)3+2(–x)2+5(–x)+4

f(x) = x3 + 2x2 + 5x+4

Replace x with −x.

-xf1-x2.

f1x2.

f1x2 = x3+ 2x2

+ 5x + 4.

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EXERCISE SET 2.5

Practice ExercisesIn Exercises 1–8, use the Rational Zero Theorem to list

all possible rational zeros for each given function.

1.2.3.4.5.6.7.8.


a. List all possible rational zeros.b. Use synthetic division to test the possible rational zeros and

find an actual zero.c. Use the quotient from part (b) to find the remaining zeros

of the polynomial function.

9.10.11.12.13.14.15.16.


a. List all possible rational roots.

b. Use synthetic division to test the possible rational roots andfind an actual root.

c. Use the quotient from part (b) to find the remaining rootsand solve the equation.

17. 18.19. 20.21.22.23.24.

In Exercises 25–32, find an nth-degree polynomial function withreal coefficients satisfying the given conditions. If you are using agraphing utility, use it to graph the function and verify the realzeros and the given function value.

25. 1 and 5i are zeros;

26. 4 and 2i are zeros;

27. and are zeros;

28. 6 and are zeros;

29. i and 3i are zeros;

30. and i are zeros;

31. 5, and are zeros;

32. and are zeros; f112 = 1002 + 3i-4, 13 ,n = 4;

f112 = -963 + 2i-2,n = 4;

f112 = 18-2, - 12 ,n = 4;

f1-12 = 20n = 4;

f122 = -636-5 + 2in = 3;

f122 = 914 + 3i-5n = 3;

f1-12 = -50n = 3;

f1-12 = -104n = 3;

x4- 2x2

- 16x - 15 = 0x4

- 2x3- 5x2

+ 8x + 4 = 02x3

- 5x2- 6x + 4 = 0

6x3+ 25x2

- 24x + 5 = 0x3

- 5x2+ 17x - 13 = 0x3

- 10x - 12 = 0x3

- 2x2- 7x - 4 = 0x3

- 2x2- 11x + 12 = 0

f1x2 = x3- 4x2

+ 8x - 5f1x2 = 2x3

+ 6x2+ 5x + 2

f1x2 = 2x3+ x2

- 3x + 1f1x2 = x3

+ 4x2- 3x - 6

f1x2 = 2x3- 5x2

+ x + 2f1x2 = 2x3

- 3x2- 11x + 6

f1x2 = x3- 2x2

- 11x + 12f1x2 = x3

+ x2- 4x - 4

f1x2 = 4x5- 8x4

- x + 2f1x2 = x5

- x4- 7x3

+ 7x2- 12x - 12

f1x2 = 3x4- 11x3

- 3x2- 6x + 8

f1x2 = 4x4- x3

+ 5x2- 2x - 6

f1x2 = 2x4+ 3x3

- 11x2- 9x + 15

f1x2 = 3x4- 11x3

- x2+ 19x + 6

f1x2 = x3+ 3x2

- 6x - 8f1x2 = x3

+ x2- 4x - 4

In Exercises 33–38, use Descartes’s Rule of Signs to determine thepossible number of positive and negative real zeros for each givenfunction.

33.34.35.36.37.38.

In Exercises 39–52, find all zeros of the polynomial function orsolve the given polynomial equation. Use the Rational ZeroTheorem, Descartes’s Rule of Signs, and possibly the graph of thepolynomial function shown by a graphing utility as an aid inobtaining the first zero or the first root.

39.40.41.42.43.44.45.46.47.48.49.50.51.52.

Practice PlusExercises 53–60, show incomplete graphs of given

polynomial functions.a. Find all the zeros of each function.b. Without using a graphing utility, draw a complete graph of

the function.53.

54.

[−2, 0, 1] by [−10, 10, 1]

f1x2 = -x3+ 3x2

- 4

[−5, 0, 1] by [−40, 25, 5]

f1x2 = -x3+ x2

+ 16x - 16

4x5+ 12x4

- 41x3- 99x2

+ 10x + 24 = 02x5

+ 7x4- 18x2

- 8x + 8 = 03x4

- 11x3- 3x2

- 6x + 8 = 04x4

- x3+ 5x2

- 2x - 6 = 0f1x2 = 2x4

+ 3x3- 11x2

- 9x + 15f1x2 = 3x4

- 11x3- x2

+ 19x + 6x4

- x3+ 2x2

- 4x - 8 = 0x4

- 3x3- 20x2

- 24x - 8 = 0f1x2 = x4

- 4x3- x2

+ 14x + 10f1x2 = x4

- 2x3+ x2

+ 12x + 83x3

- 8x2- 8x + 8 = 0

2x3- x2

- 9x - 4 = 0f1x2 = x3

+ 12x2+ 21x + 10

f1x2 = x3- 4x2

- 7x + 10

f1x2 = 4x4- x3

+ 5x2- 2x - 6

f1x2 = 2x4- 5x3

- x2- 6x + 4

f1x2 = -2x3+ x2

- x + 7f1x2 = 5x3

- 3x2+ 3x - 1

f1x2 = x3+ 7x2

+ x + 7f1x2 = x3

+ 2x2+ 5x + 4

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55.

56.

57.

58.

59.

60.

[0, 2, 1] by [−10, 10, 1]

f1x2 = -5x4+ 4x3

- 19x2+ 16x + 4

[0, 4, 1] by [−20, 25, 5]

f1x2 = 3x5+ 2x4

- 15x3- 10x2

+ 12x + 8

[0, 4, 1] by [−50, 50, 10]

f1x2 = 2x4+ 2x3

- 22x2- 18x + 36

[0, 1, ~] by [−10, 10, 1]

f1x2 = 2x4- 3x3

- 7x2- 8x + 6

[0, 2, Z] by [−3, 15, 1]

f1x2 = 3x3+ 2x2

+ 2x - 1

[−2, 0, 1] by [−10, 10, 1]

f1x2 = 4x3- 8x2

- 3x + 9 Application ExercisesThe graphs are based on a study of the percentage ofprofessional works completed in each age decade of life

by 738 people who lived to be at least 79. Use the graphs to solveExercises 61–62.

Per

cent

age

of W

orks

Com

plet

ed

Age Trends in Professional Productivity

Age Decade70s60s50s40s30s20s

30

25

20

15

10

5

ArtsSciences

Source : Dennis, W. (1966), Creative productivity between the ages of20 and 80 years. Journal of Gerontology, 21, 1–8.

61. Suppose that a polynomial function is used to model thedata shown in the graph for the arts using

(age decade, percentage of works completed).a. Use the graph to solve the polynomial equation

Describe what this means in terms of an age decade andproductivity.

b. Describe the degree and the leading coefficient of afunction that can be used to model the data in the graph.

62. Suppose that a polynomial function g is used to model thedata shown in the graph for the sciences using

(age decade, percentage of works completed).a. Use the graph to solve the polynomial equation

Find only the meaningful value of x and then describe whatthis means in terms of an age decade and productivity.

b. Describe the degree and the leading coefficient of afunction g that can be used to model the data in the graph.

The polynomial function

models the age in human years, of a dog that is x years old,where Although the coefficients make it difficult to solveequations algebraically using this function, a graph of the function makes approximate solutions possible.

Use the graph shown to solve Exercises 63–64. Round allanswers to the nearest year.

x Ú 1.H1x2,

H1x2 = -0.001618x4+ 0.077326x3

- 1.2367x2+ 11.460x + 2.914

g1x2 = 20.

f

f1x2 = 27.

f

Age of Dog (years)2220181614121086420

H(x)

x0102030405060708090

100

Age

in H

uman

Yea

rs

Dog’s Age in Human Years

Source : U.C. Davis

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(Be sure to turn back a page and refer to the graph to solveExercises 63–64.)63. If you are 25, what is the equivalent age for dogs?64. If you are 90, what is the equivalent age for dogs?65. Set up an equation to answer the question in either Exercise

63 or 64. Bring all terms to one side and obtain zero on theother side. What are some of the difficulties involved insolving this equation? Explain how the Intermediate ValueTheorem can be used to verify the approximate solution thatyou obtained from the graph.

66. The concentration of a drug, in parts per million, in apatient’s blood x hours after the drug is administered is givenby the function

How many hours after the drug is administered will it beeliminated from the bloodstream?

67. A box with an open top is formed by cutting squares out ofthe corners of a rectangular piece of cardboard 10 inches by 8inches and then folding up the sides. If x represents thelength of the side of the square cut from each corner ofthe rectangle, what size square must be cut if the volumeof the box is to be 48 cubic inches?

Writing in Mathematics68. Describe how to find the possible rational zeros of a polynomial

function.69. How does the linear factorization of , that is,

show that a polynomial equation of degree n has n roots?70. Describe how to use Descartes’s Rule of Signs to determine the

possible number of positive real zeros of a polynomial function.71. Describe how to use Descartes’s Rule of Signs to determine the

possible number of negative roots of a polynomial equation.72. Why must every polynomial equation of degree 3 have at

least one real root?73. Explain why the equation has no rational

roots.74. Suppose is a root of a polynomial equation. What does this

tell us about the leading coefficient and the constant term inthe equation?

75. Use the graphs for Exercises 61–62 to describe one similarityand one difference between age trends in professionalproductivity in the arts and the sciences.

Technology ExercisesThe equations in Exercises 76–79 have real roots that are

rational. Use the Rational Zero Theorem to list all possible ratio-nal roots. Then graph the polynomial function in the given view-ing rectangle to determine which possible rational roots are actualroots of the equation.

76. by 77. by 78. by 79. by 3-5, 5, 143-2, 2, 144x4

+ 4x3+ 7x2

- x - 2 = 0;3-45, 45, 1543-4, 3, 142x4

+ 7x3- 4x2

- 27x - 18 = 0;3-3, 2, 1430, 2, 146x3

- 19x2+ 16x - 4 = 0;

3-50, 50, 1043-1, 6, 142x3- 15x2

+ 22x + 15 = 0;

34

x4+ 6x2

+ 2 = 0

f1x2 = an1x - c121x - c22Á 1x - cn2,f1x2

8 − 2x 10 − 2x

x

f1x2 = -x4+ 12x3

- 58x2+ 132x.

80. Use Descartes’s Rule of Signs to determine the possible numberof positive and negative real zeros of What does this mean in terms of the graph of ? Verify yourresult by using a graphing utility to graph .

81. Use Descartes’s Rule of Signs to determine the possiblenumber of positive and negative real zeros of

Verify your result byusing a graphing utility to graph .

82. Write equations for several polynomial functions of odddegree and graph each function. Is it possible for the graph tohave no real zeros? Explain. Try doing the same thing forpolynomial functions of even degree. Now is it possible tohave no real zeros?

Use a graphing utility to obtain a complete graph for eachpolynomial function in Exercises 83–86.Then determine the numberof real zeros and the number of imaginary zeros for each function.

83.

84.

85.

86.


a. The equation has one positivereal root.

b. Descartes’s Rule of Signs gives the exact number of positiveand negative real roots for a polynomial equation.

c. Every polynomial equation of degree 3 has at least onerational root.

d. None of the above is true.88. Give an example of a polynomial equation that has no real

roots. Describe how you obtained the equation.89. If the volume of the solid shown in the figure is 208 cubic

inches, find the value of x.

90. In this exercise, we lead you through the steps involved in theproof of the Rational Zero Theorem. Consider the polynomialequation

where is a rational root reduced to lowest terms.

a. Substitute for x in the equation and show that the

equation can be written as

b. Why is p a factor of the left side of the equation?c. Because p divides the left side, it must also divide the right

side. However, because is reduced to lowest terms, p

cannot divide q. Thus, p and q have no common factorsother than and 1. Because p does divide the right sideand it is not a factor of what can you conclude?qn,

-1

p

q

+ an - 2 pn - 2 q2

+Á

+ a1 pqn - 1= -a0 qn.an pn

+ an - 1 pn - 1q

p

q

p

q

an xn+ an - 1 xn - 1

+ an - 2 xn - 2+

Á+ a1 x + a0 = 0,

x + 2

x

2x + 1x + 5

33

x3+ 5x2

+ 6x + 1 = 0

f1x2 = x6- 64

f1x2 = 3x4+ 4x3

- 7x2- 2x - 3

f1x2 = 3x5- 2x4

+ 6x3- 4x2

- 24x + 16

f1x2 = x3- 6x - 9

ff1x2 = x5

- x4+ x3

- x2+ x - 8.

ff

f1x2 = 3x4+ 5x2

+ 2.

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Mid-Chapter Check Point 325

d. Rewrite the equation from part (a) with all terms contain-ing q on the left and the term that does not have a factorof q on the right. Use an argument that parallels parts (b)and (c) to conclude that q is a factor of

In Exercises 91–94, the graph of a polynomial function is given.What is the smallest degree that each polynomial could have?

91. 92. y

x

y

x

an .

93. 94.

95. Explain why a polynomial function of degree 20 cannot crossthe x-axis exactly once.

y

x

y

x

CHAPTER 2MID-CHAPTER CHECK POINTWhat You Know: We performed operations with com-plex numbers and used the imaginary unit

to represent solutions of quadratic equationswith negative discriminants. Only real solutions correspondto x-intercepts. We graphed quadratic functions using ver-tices, intercepts, and additional points, as necessary. Welearned that the vertex of is and the vertex of is We used the vertex to solve problems that involved minimiz-ing or maximizing quadratic functions. We learned a numberof techniques for finding the zeros of a polynomial function of degree 3 or higher or, equivalently, finding the roots, orsolutions, of the equation For some functions, thezeros were found by factoring . For other functions, welisted possible rational zeros and used synthetic division andthe Factor Theorem to determine the zeros. We saw thatgraphs cross the x-axis at zeros of odd multiplicity and touchthe x-axis and turn around at zeros of even multiplicity. Welearned to graph polynomial functions using zeros, the Lead-ing Coefficient Test, intercepts, and symmetry. We checkedgraphs using the fact that a polynomial function of degree nhas a graph with at most turning points. After findingzeros of polynomial functions, we reversed directions byusing the Linear Factorization Theorem to find functions withgiven zeros.

In Exercises 1–6, perform the indicated operations and write theresult in standard form.

1. 2.

3. 4.

5. 6.

7. Solve and express solutions in standard form:

In Exercises 8–11, graph the given quadratic function. Give eachfunction’s domain and range.

8. 9.10. 11 f1x2 = 3x2

- 6x + 1f1x2 = -x2- 4x + 5

f1x2 = 5 - 1x + 222f1x2 = 1x - 322 - 4

x12x - 32 = -4.

A2 - 2-3 B22-75 - 2-12

1 + i

1 - i11 + i214 - 3i2

3i12 + i216 - 2i2 - 17 - i2

n - 1

f1x2f1x2 = 0.

f

A - b2a , f A -

b2a B B .f1x2 = ax2

+ bx + c1h, k2f1x2 = a1x - h22 + k

where i2= -12 i 1i = 2-1,

In Exercises 12–20, find all zeros of each polynomial function.Then graph the function.

12. 13.

14. 15.

16. 17.

18. 19.

20.

In Exercises 21–26, solve each polynomial equation.

21.22.23.24.25.26.27. A company manufactures and sells bath cabinets.The function

models the company’s daily profit, when x cabinets aremanufactured and sold per day. How many cabinets shouldbe manufactured and sold per day to maximize thecompany’s profit? What is the maximum daily profit?

28. Among all pairs of numbers whose sum is find a pairwhose product is as large as possible. What is the maximumproduct?

29. The base of a triangle measures 40 inches minus twice themeasure of its height. For what measure of the height does thetriangle have a maximum area? What is the maximum area?

In Exercises 30–31, divide, using synthetic division if possible.

30.

31.

In Exercises 32–33, find an nth-degree polynomial function withreal coefficients satisfying the given conditions.

32. 1 and i are zeros;

33. 2 (with multiplicity 2) and 3i are zeros;

34. Does have a real zero between 1 and 2?f1x2 = x3- x - 5

f102 = 36n = 4;

f1-12 = 8n = 3;

12x4- 13x3

+ 17x2+ 18x - 242 , 1x - 42

16x4- 3x3

- 11x2+ 2x + 42 , 13x2

- 12

-18,

P1x2,P1x2 = -x2

+ 150x - 4425

2x4+ x3

- 17x2- 4x + 6 = 0

x4- x3

- 11x2= x + 12

2x3+ 5x2

- 200x - 500 = 012x + 1213x - 22312x - 72 = 06x3

- 11x2+ 6x - 1 = 0

x3- 3x + 2 = 0

f1x2 = -x3+ 5x2

- 5x - 3

f1x2 = x3- 2x2

+ 26xf1x2 = 2x3- 2x

f1x2 = -6x3+ 7x2

- 1f1x2 = -1x + 126f1x2 = x4

- 5x2+ 4f1x2 = x3

- x2- 4x + 4

f1x2 = -1x - 2221x + 122f1x2 = 1x - 2221x + 123

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Objectives❶ Find the domain of rational

functions.

❷ Use arrow notation.

❸ Identify verticalasymptotes.

❹ Identify horizontalasymptotes.

❺ Use transformations tograph rational functions.

❻ Graph rational functions.

❼ Identify slant asymptotes.

❽ Solve applied problemsinvolving rationalfunctions.

Technology is now promising to bring light, fast, and beautiful wheelchairs to millionsof disabled people. The cost of manufacturing these radically different wheelchairscan be modeled by rational functions. In this section, we will see how graphs of thesefunctions illustrate that low prices are possible with high production levels, which areurgently needed in this situation. There are more than half a billion people with dis-abilities in developing countries; an estimated 20 million need wheelchairs right now.

Rational FunctionsRational functions are quotients of polynomial functions. This means that rationalfunctions can be expressed as

where p and q are polynomial functions and The domain of a rationalfunction is the set of all real numbers except the x-values that make the denominatorzero. For example, the domain of the rational function

is the set of all real numbers except 0, 2, and

EXAMPLE 1 Finding the Domain of a Rational Function

Find the domain of each rational function:

a. b. c.

Solution Rational functions contain division. Because division by 0 is undefined,we must exclude from the domain of each function values of x that cause thepolynomial function in the denominator to be 0.

a. The denominator of is 0 if Thus, x cannot equal 3.

The domain of consists of all real numbers except 3. We can express thedomain in set-builder or interval notation:

Domain of f = 1- q , 32 ´ 13, q2. Domain of f = 5x ƒx Z 36

f

x = 3.f1x2 =

x2- 9

x - 3

h1x2 =

x + 3

x2+ 9

.g1x2 =

x

x2- 9

f1x2 =

x2- 9

x - 3

-5.

x2+7x+9x(x-2)(x+5)

This is p(x).

This is q(x).f(x)=

q1x2 Z 0.

f1x2 =

p1x2q1x2 ,

❶ Find the domain of rationalfunctions.

SECTION 2.6 Rational Functions and Their Graphs

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Section 2.6 • Rational Functions and Their Graphs 327

Because the domain of a rationalfunction is the set of all realnumbers except those for whichthe denominator is 0, you canidentify such numbers by settingthe denominator equal to 0 andsolving for x. Exclude theresulting real values of x from thedomain.

Study Tip

❷ Use arrow notation.

b. The denominator of is 0 if or Thus, the domain

of g consists of all real numbers except and 3.We can express the domain inset-builder or interval notation:

c. No real numbers cause the denominator of to equal 0. Thedomain of h consists of all real numbers.

Find the domain of each rational function:CheckPoint 1

Domain of h = 1- q , q2h1x2 =

x + 3

x2+ 9

Domain of g = 1- q , -32 ´ 1-3, 32 ´ 13, q2. Domain of g = 5x ƒx Z -3, x Z 36

-3

x = 3.x = -3g1x2 =

x

x2- 9

a. b. c.

The most basic rational function is the reciprocal function, defined by

The denominator of the reciprocal function is zero when so the

domain of is the set of all real numbers except 0.Let’s look at the behavior of near the excluded value 0.We start by evaluatingto the left of 0.f1x2 f

f

x = 0,f1x2 =

1x

.

h1x2 =

x + 5

x2+ 25

.g1x2 =

x

x2- 25

f1x2 =

x2- 25

x - 5

x

-1000-100-10-2-1f1x2 �1x

-0.001-0.01-0.1-0.5-1

x approaches 0 from the left.

y

x

Mathematically, we say that “x approaches 0 from the left.” From the table and theaccompanying graph, it appears that as x approaches 0 from the left, the functionvalues, , decrease without bound.We say that “ approaches negative infinity.”We use a special arrow notation to describe this situation symbolically:

Observe that the minus superscript on the 0 is read “from the left.”Next, we evaluate to the right of 0.f1x2 1x : 0-21-2

As x S 0–, f(x) S – q.As x approaches 0

from the left, f(x) approachesnegative infinity (that is,

the graph falls).

f1x2f1x2

x 0.001 0.01 0.1 0.5 1

1000 100 10 2 1f1x2 �1x

x approaches 0 from the right.

y

x

Mathematically, we say that “x approaches 0 from the right.” From the table and theaccompanying graph, it appears that as x approaches 0 from the right, the functionvalues, , increase without bound. We say that “ approaches infinity.” Weagain use a special arrow notation to describe this situation symbolically:

f1x2f1x2

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If x is far from 0, then is close

to 0. By contrast, if x is close to 0,

then is far from 0.1x

1x

Study Tip


Observe that the plus superscript on the 0 is read “from the right.”Now let’s see what happens to the function values, , as x gets farther away

from the origin. The following tables suggest what happens to as x increases ordecreases without bound.

f1x2f1x21x : 0+21+2

As x S 0±, f(x) S q.As x approaches 0

from the right, f(x) approachesinfinity (that is, the graph rises).

y

x

Figure 2.26approaches 0 as x increasesor decreases withoutbound.

f1x2

x increases without bound:

x 1 10 100 1000

1 0.1 0.01 0.001f1x2 �1x

x decreases without bound:

x

-0.001-0.01-0.1-1f1x2 �1x

-1000-100-10-1

It appears that as x increases or decreases without bound, the function values, ,are getting progressively closer to 0.

Figure 2.26 illustrates the end behavior of as x increases or decreases

without bound. The graph shows that the function values, , are approaching 0.This means that as x increases or decreases without bound, the graph of isapproaching the horizontal line (that is, the x-axis). We use arrow notation todescribe this situation:

Thus, as x approaches infinity or as xapproaches negative infinity the func-tion values are approaching zero:

The graph of the reciprocal function

is shown in Figure 2.27. Unlike the graph

of a polynomial function, the graph of thereciprocal function has a break and is composed oftwo distinct branches.

The arrow notation used throughout ourdiscussion of the reciprocal function is summarizedin the following box:

f1x2 =

1x

f1x2: 0.1x : - q2,1x : q2

As x S q, f(x) S 0 and as x S – q, f(x) S 0.

As x approaches negative infinity(that is, decreases without bound),

f(x) approaches 0.

As x approaches infinity(that is, increases without bound),

f(x) approaches 0.

y = 0f

f1x2f1x2 =

1x

f1x2

Arrow NotationSymbol Meaning

x approaches a from the right.x approaches a from the left.x approaches infinity; that is, x increases without bound.x approaches negative infinity; that is, x decreases without bound.

In calculus, you will use limits to convey ideas involving a function’s endbehavior or its possible asymptotic behavior. For example, examine the graph of

in Figure 2.27 and its end behavior to the right. As the values of

approach 0: In calculus, this is symbolized by

This is read “the limit of as xapproaches infinity equals zero.”

f1x2limx: q

f1x2 = 0.

f1x2: 0.f1x2x : q ,f1x2 =

1x

x : - q

x : q

x : a-

x : a+

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

Figure 2.27 The graph of the

reciprocal function f1x2 =

1x

pr02-265-374.I-hr 1/26/06 5:14 PM Page 328


1

2

3

x

y

1 2 3 4 5

4

−1−2−3−4−5

As x S 0−, f (x) S q.Function values increase

without bound.

As x S 0+, f (x) S q.Function values increase

without bound.

As x S q (increaseswithout bound),

f (x) S 0.

As x S −q (decreaseswithout bound),

f (x) S 0. Figure 2.28 The graph of f1x2 =

1

x2

❸ Identify vertical asymptotes.

Another basic rational function is The graph of this even function,

with y-axis symmetry and positive function values, is shown in Figure 2.28. Like thereciprocal function, the graph has a break and is composed of two distinct branches.

f1x2 =

1

x2 .

Vertical Asymptotes of Rational Functions

Look again at the graph of in Figure 2.28.The curve approaches, but does

not touch,the y-axis. The y-axis, or is said to be a vertical asymptote of thegraph. A rational function may have no vertical asymptotes, one vertical asymptote,or several vertical asymptotes. The graph of a rational function never intersects avertical asymptote. We will use dashed lines to show asymptotes.

Definition of a Vertical AsymptoteThe line is a vertical asymptote of the graph of a function if increases or decreases without bound as x approaches a.

f1x2fx = a

x = 0,

f1x2 =

1

x2

y

x

y

x

y

x

y

x

As x → a+, f(x) → q . As x → a−, f(x) → q . As x → a+, f(x) → −q . As x → a−, f(x) → −q .

x = af

f

f

fa

x = a

a

x = a

a

x = a

a

lim f(x)=q xSa±

lim f(x)=−q xSa±

lim f(x)=−q xSa–

lim f(x)=q xSa–

Thus, as x approaches a from either the left or the right, or

If the graph of a rational function has vertical asymptotes, they can be locatedusing the following theorem:

Locating Vertical Asymptotes

If is a rational function in which and have no common

factors and a is a zero of the denominator, then is a verticalasymptote of the graph of .f

x = aq1x2,

q1x2p1x2f1x2 =

p1x2

q1x2

f1x2: - q .f1x2: q

pr02-265-374.I-hr 1/27/06 2:25 PM Page 329

x

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3−4−5

Verticalasymptote: x = −3

Verticalasymptote: x = 3


EXAMPLE 2 Finding the Vertical Asymptotes of a Rational Function

Find the vertical asymptotes, if any, of the graph of each rational function:

a. b. c.

Solution Factoring is usually helpful in identifying zeros of denominators.

a.

There are no common factors in the numerator and the denominator. Thezeros of the denominator are and 3. Thus, the lines and arethe vertical asymptotes for the graph of . [See Figure 2.29(a).]

b. We will use factoring to see if there are common factors.

The only zero of the denominator of in simplified form is 3. Thus, the lineis the only vertical asymptote of the graph of g. [See Figure 2.29(b).]

c. We cannot factor the denominator of over the real numbers.

The denominator has no real zeros. Thus, the graph of h has no verticalasymptotes. [See Figure 2.29(c).]

x+3x2+9

h(x)=

No real numbers make this denominator 0.

h1x2x = 3

g1x2

x+3x2-9

(x+3)(x+3)(x-3)

g(x)= =1

x-3=

This denominatoris 0 if x = 3.

There is a common factor,x + 3, so simplify.

fx = 3x = -3-3

xx2-9

x(x+3)(x-3)

f(x)= =

This factor is0 if x = −3.

This factor is0 if x = 3.

h1x2 =

x + 3

x2+ 9

.g1x2 =

x + 3

x2- 9

f1x2 =

x

x2- 9

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3−4−5x


There is ahole in the graph

correspondingto x = −3.

x

y

1 2 3 4 5−0.1

0.10.20.30.40.5

−0.2−0.3−0.4−0.5

−2−3−4−5

Find the vertical asymptotes, if any, of the graph of each rational function:CheckPoint 2

a. b. c. h1x2 =

x - 1

x2+ 1

.g1x2 =

x - 1

x2- 1

f1x2 =

x

x2- 1

Figure 2.29(b)

The graph of has one

vertical asymptote.

g1x2 =

x + 3

x2- 9

Figure 2.29(c)

The graph of has no

vertical asymptotes.

h1x2 =

x + 3

x2+ 9

Figure 2.29(a)

The graph of has two

vertical asymptotes.

f1x2 =

x

x2- 9

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x

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3−4−5

Hole correspondingto x = 2

f (x) =x − 2x2 − 4

Figure 2.30 A graph with a holecorresponding to the denominator’s zero

The graph of the rational function drawn by hand in Figure 2.29(a), is

graphed below in a by viewing rectangle. The graph is shown inconnected mode and in dot mode. In connected mode, the graphing utility plots manypoints and connects the points with curves. In dot mode, the utility plots the same points,but does not connect them.

The steep lines in connected mode that are “almost” the vertical asymptotes andare not part of the graph and do not represent the vertical asymptotes.The graphing

utility has incorrectly connected the last point to the left of with the first point tothe right of It has also incorrectly connected the last point to the left of with the first point to the right of The effect is to create two near-vertical segmentsthat look like asymptotes.This erroneous effect does not appear using dot mode.

x = 3.x = 3x = -3.

x = -3x = 3

x = -3

Dot Mode

This might appear to be thevertical asymptote x = −3,but it is neither vertical

nor an asymptote.

This might appear to be thevertical asymptote x = 3,but it is neither vertical

nor an asymptote.

Connected Mode

3-4, 4, 143-5, 5, 14f1x2 =

x

x2- 9

,

Technology

A value where the denominator of a rational function is zero does not neces-sarily result in a vertical asymptote. There is a hole corresponding to and nota vertical asymptote, in the graph of a rational function under the following condi-tions: The value a causes the denominator to be zero, but there is a reduced form ofthe function’s equation in which a does not cause the denominator to be zero.

Consider, for example, the function

Because the denominator is zero when the function’s domain is all realnumbers except 2. However, there is a reduced form of the equation in which 2 doesnot cause the denominator to be zero:

Figure 2.30 shows that the graph has a hole corresponding to Graphingutilities do not show this feature of the graph.

x = 2.

Denominator iszero at x = 2.

In this reduced form, 2 does notresult in a zero denominator.

x2-4x-2

(x+2)(x-2)x-2f(x)= =x+2, x � 2.=

x = 2,

f1x2 =

x2- 4

x - 2.

x = a,

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x

y

1 2 3 4 5

−1

1

−1−2−3−4−5

y = 0

f (x) =2x2 + 1

4x

Figure 2.31(a) The horizontalasymptote of the graph is y = 0.


❹ Identify horizontalasymptotes.

Horizontal Asymptotes of Rational Functions

Figure 2.27, repeated at the left, shows the graph of the reciprocal function

As and as the function values are approaching 0: Theline (that is, the x-axis) is a horizontal asymptote of the graph. Many, but notall, rational functions have horizontal asymptotes.

Definition of a Horizontal AsymptoteThe line is a horizontal asymptote of the graph of a function if approaches b as x increases or decreases without bound.

Recall that a rational function may have several vertical asymptotes. Bycontrast, it can have at most one horizontal asymptote. Although a graph can neverintersect a vertical asymptote, it may cross its horizontal asymptote.

If the graph of a rational function has a horizontal asymptote, it can be locatedusing the following theorem:

Locating Horizontal AsymptotesLet be the rational function given by

The degree of the numerator is n. The degree of the denominator is m.

1. If the x-axis, or is the horizontal asymptote of the graph of .

2. If the line is the horizontal asymptote of the graph of .

3. If the graph of has no horizontal asymptote.

EXAMPLE 3 Finding the Horizontal Asymptote of a Rational Function

Find the horizontal asymptote, if any, of the graph of each rational function:

a. b. c.

Solution

a.

The degree of the numerator, 1, is less than the degree of the denominator, 2.Thus, the graph of has the x-axis as a horizontal asymptote. [See Figure2.31(a).] The equation of the horizontal asymptote is y = 0.

f

f1x2 =

4x

2x2+ 1

h1x2 =

4x3

2x2+ 1

.g1x2 =

4x2

2x2+ 1

f1x2 =

4x

2x2+ 1

fn 7 m,

fy =

an

bmn = m,

fy = 0,n 6 m,

f1x2 =

an xn+ an - 1 xn - 1

+Á

+ a1 x + a0

bm xm+ bm - 1 xm - 1

+Á

+ b1 x + b0, an Z 0, bm Z 0.

f

y = b

y

x

y

x

y

x

As x → q, f(x) → b. As x → q, f(x) → b. As x → q, f(x) → b.

f

ff

y = b

y = b

lim f(x)=b xSq

lim f(x)=b xSq

lim f(x)=b xSq

f1x2fy = b

y = 0f1x2: 0.x : - q ,x : q

f1x2 =

1x

.

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

Figure 2.27 The graph of

(repeated)

f1x2 =

1x

pr02-265-374.I-hr 1/27/06 2:27 PM Page 332


(1, 1)(−1, 1)

(q, 4)(− q, 4)

(2, ~)(−2, ~)

3

4

x

y

1 2

1

2

−1−2

x = 0

y = 0y = 0

Even function: f(−x) = f(x)•• y-axis symmetry

f(x) = 1x2

x

y

1 2

1

2

−1

−2

−1−2

y = 0(1, 1)

(−1, −1)

(− q, −2)

(−2, − q)

(q, 2)

(2, q)

y = 0

x = 0

x = 0

Odd function: f(−x) = −f(x)•• Origin symmetry

f(x) = 1x

Table 2.2 Graphs of Common Rational Functions

❺ Use transformations to graphrational functions.

b.

The degree of the numerator, 2, is equal to the degree of the denominator, 2.The leading coefficients of the numerator and denominator, 4 and 2, are usedto obtain the equation of the horizontal asymptote. The equation of the hori-zontal asymptote is or [See Figure 2.31(b).]

c.

The degree of the numerator, 3, is greater than the degree of the denominator,2. Thus, the graph of h has no horizontal asymptote. [See Figure 2.31(c).]

h1x2 =

4x3

2x2+ 1

y = 2.y =42

g1x2 =

4x2

2x2+ 1

x

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3−4−5

f (x) =2x2 + 1

4x3

Figure 2.31(c) The graph has nohorizontal asymptote.

2

x

y

1 2 3 4 5

1

−1−2−3−5 −4

y = 2f (x) =

2x2 + 14x2

Figure 2.31(b) The horizontalasymptote of the graph is y = 2.

Find the horizontal asymptote, if any, of the graph of each rational function:CheckPoint 3

a. b. c.

Using Transformations to Graph Rational Functions

Table 2.2 shows the graphs of two rational functions, and Thedashed green lines indicate the asymptotes.

f1x2 =

1

x2 .f1x2 =

1x

h1x2 =

9x3

3x2+ 1

.g1x2 =

9x

3x2+ 1

f1x2 =

9x2

3x2+ 1

Some rational functions can be graphed using transformations (horizontal shift-ing, stretching or shrinking, reflecting, vertical shifting) of these two common graphs.

pr02-265-374.I-hr 1/26/06 5:14 PM Page 333


❻ Graph rational functions.

EXAMPLE 4 Using Transformations to Graph a Rational Function

Use the graph of to graph

Solution

g1x2 =

1

1x - 222 + 1.f1x2 =

1

x2

Graph y = .

Shift 2 units to theright. Add 2 to each

x-coordinate.

We’ve identified twopoints and the asymptotes.

Begin with f(x) = .1x2

The graph of y = 1(x − 2)2

showing two points andthe asymptotes

The graph of g(x) = + 11(x − 2)2

showing two points and theasymptotes

(1, 1)(−1, 1)

3

4

x

y

1 2

1

2

−1−2

(3, 1)(1, 1)

3

4

x

y

1 2 3 4

1

2(3, 2)(1, 2)

3

4

x

y

1 2 3 4

1

2

1(x − 2)2 Graph g(x) = + 1.

Shift 1 unit up. Add 1 to each y-coordinate.

1(x − 2)2

y = 0y = 0 y = 0y = 0

y = 1

x = 2 x = 2x = 0

Use the graph of to graph

Graphing Rational FunctionsRational functions that are not transformations of or can begraphed using the following suggestions:

Strategy for Graphing a Rational FunctionThe following strategy can be used to graph

where p and q are polynomial functions with no common factors.

1. Determine whether the graph of has symmetry.

2. Find the y-intercept (if there is one) by evaluating .3. Find the x-intercepts (if there are any) by solving the equation 4. Find any vertical asymptote(s) by solving the equation 5. Find the horizontal asymptote (if there is one) using the rule for determining

the horizontal asymptote of a rational function.6. Plot at least one point between and beyond each x-intercept and vertical

asymptote.7. Use the information obtained previously to graph the function between and

beyond the vertical asymptotes.

q1x2 = 0.p1x2 = 0.

f102 f1-x2 = -f1x2: origin symmetry

f1-x2 = f1x2: y-axis symmetry

f

f1x2 =

p1x2q1x2 ,

f1x2 =

1

x2f1x2 =

1x

g1x2 =

1x + 2

- 1.f1x2 =

1x

CheckPoint 4

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EXAMPLE 5 Graphing a Rational Function

Graph:

Solution

Step 1 Determine symmetry.

Because does not equal or the graph has neither y-axis nororigin symmetry.

Step 2 Find the y-intercept. Evaluate

The y-intercept is 0, so the graph passes through the origin.

Step 3 Find x-intercept(s). This is done by solving

Set the numerator equal to 0.


There is only one x-intercept. This verifies that the graph passes through the origin.

Step 4 Find the vertical asymptote(s). Solve thereby finding zeros ofthe denominator.

Set the denominator equal to 0.

Add 1 to both sides.

The equation of the vertical asymptote is

Step 5 Find the horizontal asymptote. Because the numerator and denominator

of have the same degree, 1, the leading coefficients of the numerator

and denominator, 2 and 1, respectively, are used to obtain the equation of thehorizontal asymptote. The equation is

The equation of the horizontal asymptote is

Step 6 Plot points between and beyond each x-intercept and vertical asymptote.With an x-intercept at 0 and a vertical asymptote at we evaluate the functionat 2, and 4.-2, -1, 12 ,

x = 1,

y = 2.

y =

21

= 2.

f1x2 =

2x

x - 1

x = 1.

x = 1

x - 1 = 0

q1x2 = 0,

x = 0

2x = 0

p1x2 = 0.

f102 =

2 # 00 - 1

=

0-1

= 0

f102.-f1x2,f1x2f1-x2

f1-x2 =

21-x2-x - 1

=

-2x

-x - 1=

2x

x + 1

f1x2 =

2x

x - 1.

x 2 4

1 483

-243

f1x2 �2x

x � 1

12

-1-2

Figure 2.32 shows these points, the y-intercept, the x-intercept, and the asymptotes.

1234

765

−2−3

1 2 3 4 5−1−2−3−4−5

y

x

Horizontalasymptote: y = 2


x-intercept andy-intercept

Figure 2.32 Preparing to graph the

rational function f1x2 =

2x

x - 1

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Because the graph has y-axissymmetry, it is not necessary toevaluate the even function at and again at 3.

This also applies to evaluation atand 1.-1

f1-32 = f132 =275

-3

1234

765

−2−1

−3

1 2 3 4 5−1−2−3−4−5

y

x

y = 2

x = 1


f1x2 =

2x

x - 1


The graph of

obtained using the dot mode in aby viewing

rectangle, verifies that our hand-drawn graph is correct.

3-6, 6 143-6, 6, 14

y =

2x

x - 1,

Technology

Study Tip

Graph:


Graph:

Solution

Step 1 Determine symmetry. The graph

of is symmetric with respect to the y-axis.

Step 2 Find the y-intercept. The y-intercept is 0, so

the graph passes through the origin.

Step 3 Find the x-intercept(s). so The x-intercept is 0, verifyingthat the graph passes through the origin.

Step 4 Find the vertical asymptote(s). Set


Add 4 to both sides.

Use the square root property.

The vertical asymptotes are and

Step 5 Find the horizontal asymptote. Because the numerator and denominator

of have the same degree, 2, their leading coefficients, 3 and 1, are

used to determine the equation of the horizontal asymptote. The equation is

Step 6 Plot points between and beyond each x-intercept and vertical asymptote.With an x-intercept at 0 and vertical asymptotes at and we evaluatethe function at and 4.-3, -1, 1, 3,

x = 2,x = -2

y =31 = 3.

f1x2 =

3x2

x2- 4

x = 2.x = -2

x = ;2

x2= 4

x2- 4 = 0

q1x2 = 0.

x = 0:3x2= 0,

f102 =

3 # 02

02- 4

=

0-4

= 0:

f

f1-x2 =

31-x221-x22 - 4

=

3x2

x2- 4

= f1x2:

f1x2 =

3x2

x2- 4

.

f1x2 =

3x

x - 2.Check

Point 5

x 1 3 4

4275

-1-1275

f1x2 �3x2

x2 � 4

-1-3

Figure 2.34 at the top of the next page shows these points, the y-intercept, thex-intercept, and the asymptotes.

1234

765

−2−3

1 2 3 4 5−1−2−3−4−5

y

x




Figure 2.32 (repeated) Preparing to

graph the rational function f1x2 =

2x

x - 1

Step 7 Graph the function. The graph of is shown in Figure 2.33.f1x2 =

2x

x - 1

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Step 7 Graph the function. The graph of is shown in Figure 2.35.The y-axis symmetry is now obvious.

f1x2 =

3x2

x2- 4

−1

1234

765

−2−3

1 2 3 4 5−1−2−3−4−5

y

x

x = 2x = −2

y = 3


f1x2 =

3x2

x2- 4

−1

1234

765

−2−3

1 2 3 4 5−1−2−3−4−5

y

x


Verticalasymptote: x = −2



Figure 2.34 (repeated) Preparing to graph

f1x2 =

3x2

x2- 4

The graph of

generated by a graphing utility,verifies that our hand-drawngraph is correct.

[–6, 6, 1] by [–6, 6, 1]

y =

3x2

x2- 4

,

Technology

Graph:

Example 7 illustrates that not every rational function has vertical and horizontalasymptotes.


Graph:

Solution

Step 1 Determine symmetry. The graph

of is symmetric with respect to the y-axis.

Step 2 Find the y-intercept. The y-intercept is 0.

Step 3 Find the x-intercept(s). so The x-intercept is 0.

Step 4 Find the vertical asymptote. Set


Subtract 1 from both sides.

Although this equation has imaginary roots there are no real roots. Thus,the graph of has no vertical asymptotes.

Step 5 Find the horizontal asymptote. Because the degree of the numerator, 4, isgreater than the degree of the denominator, 2, there is no horizontal asymptote.

Step 6 Plot points between and beyond each x-intercept and vertical asymptote.With an x-intercept at 0 and no vertical asymptotes, let’s look at function values at

f1x = ; i2,

x2= -1

x2+ 1 = 0

q1x2 = 0.

x = 0:x4= 0,

f102 =

04

02+ 1

=

01

= 0:

f

f1-x2 =

1-x241-x22 + 1

=

x4

x2+ 1

= f1x2:

f1x2 =

x4

x2+ 1

.

f1x2 =

2x2

x2- 9

.CheckPoint 6

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−1

1234

67

5

−2−3

1 2 3 4 65−1−2−3−4

y

x

Vertical asymptote:x = 1

Slant asymptote:y = x + 1


with a slant asymptotef1x2 =

x2+ 1

x - 1


−1

1234

8765

−2

1 2 3 4 5−1−2−3−4−5

y

x


f1x2 =

x4

x2+ 1

❼ Identify slant asymptotes.

1, and 2. You can evaluate the function at 1 and 2. Use y-axis symmetry toobtain function values at and

f1-12 = f112 and f1-22 = f122.-2:-1

-2, -1,

x 1 2

165

12

12

165

f1x2 �x4

x2 � 1

-1-2

Step 7 Graph the function. Figure 2.36 shows the graph of using the pointsobtained from the table and y-axis symmetry. Notice that as x approaches infinity ornegative infinity ( or ), the function values, , are getting largerwithout bound

Graph:

Slant AsymptotesExamine the graph of

shown in Figure 2.37. Note that the degree of the numerator, 2, is greater than thedegree of the denominator, 1. Thus, the graph of this function has no horizontalasymptote. However, the graph has a slant asymptote,

The graph of a rational function has a slant asymptote if the degree of thenumerator is one more than the degree of the denominator. The equation of the slantasymptote can be found by division. For example, to find the slant asymptote for the

graph of divide into

Observe that

As the value of is approximately 0. Thus, when is large, the

function is very close to This means that as or as the graph of gets closer and closer to the line whose equation is Theline is a slant asymptote of the graph.

In general, if p and q have no common factors, and the degree of

p is one greater than the degree of q, find the slant asymptote by dividing into. The division will take the form

The equation of the slant asymptote is obtained by dropping the term with theremainder. Thus, the equation of the slant asymptote is y = mx + b.

p(x)q(x)

remainderq(x)

.=mx+b+

Slant asymptote:y = mx + b

p1x2 q1x2f1x2 =

p1x2q1x2 ,

y = x + 1y = x + 1.f

x : - q ,x : qy = x + 1 + 0.

ƒx ƒ

2x - 1

ƒx ƒ : q ,

x2+1x-1

2x-1

=x+1+f(x)= .

The equation of the slant asymptote is y = x + 1.

1x + 1 +

2x - 1

x - 1�x2+ 0x + 1

Remainder

1

21

0

1

1 1

11

x2+ 1:x - 1f1x2 =

x2+ 1

x - 1,

y = x + 1.

f1x2 =

x2+ 1

x - 1,

f1x2 =

x4

x2+ 2

.CheckPoint 7

3f1x2: q4. f1x2x : - qx : q

f

pr02-265-374.I-hr 1/26/06 5:14 PM Page 338


−1

1234

765

−2−3

1 2 3 4 8765−1−2

y

x


Slant asymptote:y = x − 1


f1x2 =

x2- 4x - 5x - 3

EXAMPLE 8 Finding the Slant Asymptote of a Rational Function

Find the slant asymptote of

Solution Because the degree of the numerator, 2, is exactly one more than thedegree of the denominator, 1, and is not a factor of the graph of

has a slant asymptote. To find the equation of the slant asymptote, divide into

The equation of the slant asymptote is Using our strategy for graphing

rational functions, the graph of is shown in Figure 2.38.

Find the slant asymptote of

ApplicationsThere are numerous examples of asymptotic behavior in functions that describe real-world phenomena. Let’s consider an example from the business world.The costfunction, C, for a business is the sum of its fixed and variable costs:

The average cost per unit for a company to produce x units is the sum of itsfixed and variable costs divided by the number of units produced. The average costfunction is a rational function that is denoted by Thus,

EXAMPLE 9 Average Cost of Producing a Wheelchair

A company is planning to manufacture wheelchairs that are light, fast, andbeautiful. The fixed monthly cost will be $500,000 and it will cost $400 to produceeach radically innovative chair.

a. Write the cost function, C, of producing x wheelchairs.

b. Write the average cost function, of producing x wheelchairs.

c. Find and interpret and

d. What is the horizontal asymptote for the graph of the average cost function,Describe what this represents for the company.C?

C1100,0002.C110002, C110,0002,C,

Cost of producing x units:fixed plus variable costs

Number of units produced

(fixed cost)+cxxC(x)= .

C.

Cost per unit times thenumber of units produced, x.

C(x)=(fixed cost)+cx.

f1x2 =

2x2- 5x + 7x - 2

.CheckPoint 8

f1x2 =

x2- 4x - 5x - 3

y = x - 1.

Drop the remainderterm and you'll have

the equation ofthe slant asymptote.�x2-4x-5x-3

1x-1-8

x-3Remainder

–5

–8–3

–4

1

3 1

3–1

x2- 4x - 5:

x - 3fx2

- 4x - 5,x - 3

f1x2 =

x2- 4x - 5x - 3

.

❽ Solve applied problemsinvolving rational functions.

pr02-265-374.I-hr 1/26/06 5:14 PM Page 339

Ave

rage

Cos

t per

Whe

elch

air

for

the

Com

pany

10,000 100,00050,000

C(x)

x

Number of WheelchairsProduced per Month

400405410415420425430435440445450455

y = 400

Figure 2.39 As production levelincreases, the average cost per wheel-chair approaches $400:

limx: q

C1x2 = 400.


Solution

a. The cost function of producing x wheelchairs, C, is the sum of the fixed costand the variable costs.

b. The average cost function of producing x wheelchairs, is the sum of thefixed and variable costs divided by the number of wheelchairs produced.

c. We evaluate at 1000, 10,000, and 100,000, interpreting the results.

The average cost per wheelchair of producing 1000 wheelchairs per month is $900.

The average cost per wheelchair of producing 10,000 wheelchairs per month is$450.

The average cost per wheelchair of producing 100,000 wheelchairs per monthis $405. Notice that with higher production levels, the cost of producing eachwheelchair decreases.

d. We developed the average cost function

in which the degree of the numerator, 1, is equal to the degree of the denomi-nator, 1. The leading coefficients of the numerator and denominator, 400 and1, are used to obtain the equation of the horizontal asymptote.The equation ofthe horizontal asymptote is

The horizontal asymptote is shown in Figure 2.39. This means that the morewheelchairs produced per month, the closer the average cost per wheelchairfor the company comes to $400. The least possible cost per wheelchair isapproaching $400. Competitively low prices take place with high productionlevels, posing a major problem for small businesses.

The time: the not-too-distant future. A new company is hoping to replacetraditional computers and two-dimensional monitors with its virtualreality system.The fixed monthly cost will be $600,000 and it will cost $500to produce each system.

a. Write the cost function, C, of producing x virtual reality systems.

b. Write the average cost function, of producing x virtual reality systems.

c. Find and interpret and

d. What is the horizontal asymptote for the graph of the average costfunction, Describe what this represents for the company.C?

C1100,0002.C110002, C110,0002,C,

CheckPoint 9

y =

4001

or y = 400.

C1x2 =

400x + 500,000x

C1100,0002 =

4001100,0002 + 500,000

100,000= 405

C110,0002 =

400110,0002 + 500,000

10,000= 450

C110002 =

400110002 + 500,000

1000= 900

C

C1x2 =

500,000 + 400xx

or C1x2 =

400x + 500,000x

C,

C(x)=500,000+400x

Variable cost: $400 foreach wheelchair produced

Fixed cost is$500,000.

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If an object moves at an average velocity the distance, s, covered in time t isgiven by the formula

Thus, Objects that move in accordance with this formulaare said to be in uniform motion. In Example 10, we use a rational function to modeltime, t, in uniform motion. Solving the uniform motion formula for t, we obtain

Thus, time is the quotient of distance and average velocity.

EXAMPLE 10 Time Involved in Uniform Motion

Two commuters drove to work a distance of 40 miles and then returned again on thesame route.The average velocity on the return trip was 30 miles per hour faster thanthe average velocity on the outgoing trip. Express the total time required to com-plete the round trip, T, as a function of the average velocity on the outgoing trip, x.

Solution As specified, the average velocity on the outgoing trip is represented byx. Because the average velocity on the return trip was 30 miles per hour faster thanthe average velocity on the outgoing trip, let

The sentence that we use as a verbal model to write our rational function is

The function that expresses the total time required to complete the round trip is

Once you have modeled a problem’s conditions with a function, you can use agraphing utility to explore the function’s behavior. For example let’s graph the functionin Example 10. Because it seems unlikely that an average outgoing velocity exceeds 60miles per hour with an average return velocity that is 30 miles per hour faster, we graph

the function for Figure 2.40 shows the graph of in

a by viewing rectangle. Notice that the function is decreasing.This shows decreasing times with increasing average velocities. Can you see that thevertical asymptote is or the y-axis? This indicates that close to an outgoingaverage velocity of zero miles per hour, the round trip will take nearly forever:

Two commuters drove to work a distance of 20 miles and then returnedagain on the same route. The average velocity on the return trip was 10miles per hour slower than the average velocity on the outgoing trip.Express the total time required to complete the round trip, T, as a func-tion of the average velocity on the outgoing trip, x.

CheckPoint 10

limx:0

T1x2 = q .

x = 0,

30, 10, 1430, 60, 34T1x2 =

40x

+

40x + 30

0 … x … 60.

T1x2 =

40x

+

40x + 30

.

40x

40x+30

±T(x) .=

Total time onthe round trip

This is outgoingdistance, 40 miles,divided by outgoing

velocity, x.

This is returndistance, 40 miles,divided by returnvelocity, x + 30.

equalstime on theoutgoing trip plus

time on thereturn trip.

x + 30 = the average velocity on the return trip.

t =

sv

.

distance = velocity # time.

s = vt.

v,


As average

velocity increases, time for the tripdecreases: lim

x: q

T1x2 = 0.

T1x2 =

40x

+

40x + 30

.

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EXERCISE SET 2.619. As _____.

20. As _____.

In Exercises 21–28, find the vertical asymptotes, if any, of thegraph of each rational function.

21. 22.

23. 24.

25. 26.

27. 28.

In Exercises 29–36, find the horizontal asymptote, if any, of thegraph of each rational function.

29. 30.

31. 32.

33. 34.

35. 36.

In Exercises 37–48, use transformations of or

to graph each rational function.

37. 38.

39. 40.

41. 42.

43. 44.

45. 46.

47. 48.

In Exercises 49–70, follow the seven steps on page 334 to grapheach rational function.

49. 50.

51. 52.

53. 54. f1x2 =

4x2

x2- 9

f1x2 =

2x2

x2- 1

f1x2 =

4x

x2- 1

f1x2 =

2x

x2- 4

f1x2 =

3x

x - 1f1x2 =

4x

x - 2

h1x2 =

1

1x - 322 + 2h1x2 =

1

1x - 322 + 1

h1x2 =

1

x2 - 3h1x2 =

1

x2 - 4

g1x2 =

1

1x + 122g1x2 =

1

1x + 222

g1x2 =

1x + 2

- 2g1x2 =

1x + 1

- 2

h1x2 =

1x

+ 1h1x2 =

1x

+ 2

g1x2 =

1x - 2

g1x2 =

1x - 1

f1x2 =

1

x2

f1x2 =

1x

f1x2 =

-3x + 75x - 2

f1x2 =

-2x + 13x + 5

h1x2 =

15x3

3x2+ 1

h1x2 =

12x3

3x2+ 1

g1x2 =

15x2

3x2+ 1

g1x2 =

12x2

3x2+ 1

f1x2 =

15x

3x2+ 1

f1x2 =

12x

3x2+ 1

r1x2 =

x

x2+ 3

r1x2 =

x

x2+ 4

h1x2 =

x

x1x - 32h1x2 =

x

x1x + 42

g1x2 =

x + 3x1x - 32g1x2 =

x + 3x1x + 42

f1x2 =

x

x - 3f1x2 =

x

x + 4

x : - q , f1x2:x : q , f1x2:Practice Exercises

In Exercises 1–8, find the domain of each rationalfunction.

1. 2.

3. 4.

5. 6.

7. 8.

Use the graph of the rational function in the figure shown tocomplete each statement in Exercises 9–14.

9. As _____.

10. As _____.

11. As _____.

12. As _____.

13. As _____.

14. As _____.

Use the graph of the rational function in the figure shown tocomplete each statement in Exercises 15–20.

15. As _____.

16. As _____.

17. As _____.

18. As _____.x : -2-, f1x2:x : -2+, f1x2:x : 1-, f1x2:x : 1+, f1x2:

x

1

−1

2−5

2

3 4 51−4 −3 −2 −1

y


Horizontal asymptote:y = 1

Vertical asymptote:x = −2

x : q , f1x2:x : - q , f1x2:x : 1+, f1x2:x : 1-, f1x2:x : -3+, f1x2:x : -3-, f1x2:

x

1

−2

2−5−1

−3

2

3

31−4 −3 −2 −1

y


Horizontal asymptote:y = 0

Vertical asymptote:x = −3

f1x2 =

x + 8

x2+ 64

f1x2 =

x + 7

x2+ 49

h1x2 =

x + 8

x2- 64

h1x2 =

x + 7

x2- 49

g1x2 =

2x2

1x - 221x + 62g1x2 =

3x2

1x - 521x + 42

f1x2 =

7x

x - 8f1x2 =

5x

x - 4

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55. 56.

57. 58.

59. 60.

61. 62.

63. 64.

65. 66.

67. 68.

69. 70.

In Exercises 71–78, a. Find the slant asymptote of the graph ofeach rational function and b. Follow the seven-step strategy anduse the slant asymptote to graph each rational function.

71. 72.

73. 74.

75. 76.

77. 78.

Practice PlusIn Exercises 79–84, the equation for is given by the

simplified expression that results after performing the indicatedoperation. Write the equation for and then graph the function.

79. 80.

81. 82.

83. 84.

In Exercises 85–88, use long division to rewrite the equation for in the form quotient, plus remainder divided by divisor.Thenuse this form of the function’s equation and transformations of

to graph g.

85. 86.

87. 88. g1x2 =

2x - 9x - 4

g1x2 =

3x - 7x - 2

g1x2 =

3x + 7x + 2

g1x2 =

2x + 7x + 3

f1x2 =

1x

g

x -

1x

x +

1x

1 -

3x + 2

1 +

1x - 2

2

x2+ 3x + 2

-

4

x2+ 4x + 3

x

2x + 6-

9

x2- 9

x - 510x - 2

,

x2- 10x + 25

25x2- 1

5x2

x2- 4

#x2

+ 4x + 4

10x3

f

f

f1x2 =

x3- 1

x2- 9

f1x2 =

x3+ 1

x2+ 2x

f1x2 =

x2- x + 1x - 1

f1x2 =

x2+ x - 6x - 3

f1x2 =

x2+ 4x

f1x2 =

x2+ 1x

f1x2 =

x2- 4x

f1x2 =

x2- 1x

f1x2 =

x2- 4x + 3

1x + 122f1x2 =

3x2+ x - 4

2x2- 5x

f1x2 =

x2

x2+ x - 6

f1x2 =

x2+ x - 12

x2- 4

f1x2 =

2x4

x2+ 1

f1x2 =

x4

x2+ 2

f1x2 =

x - 4

x2- x - 6

f1x2 =

x + 2

x2+ x - 6

f1x2 =

4x2

x2+ 1

f1x2 =

2x2

x2+ 4

f1x2 =

-2

x2- x - 2

f1x2 =

2

x2+ x - 2

f1x2 = - 2

x2- 1

f1x2 = - 1

x2- 4

f1x2 =

-3x

x + 2f1x2 =

-x

x + 1Application Exercises

89. A company is planning to manufacture mountain bikes. Thefixed monthly cost will be $100,000 and it will cost $100 toproduce each bicycle.a. Write the cost function, C, of producing x mountain bikes.b. Write the average cost function, of producing x

mountain bikes.c. Find and interpret and d. What is the horizontal asymptote for the graph of the

average cost function, Describe what this means inpractical terms.

90. A company that manufactures running shoes has a fixedmonthly cost of $300,000. It costs $30 to produce each pair ofshoes.a. Write the cost function, C, of producing x pairs of shoes.b. Write the average cost function, of producing x pairs of

shoes.c. Find and interpret and d. What is the horizontal asymptote for the graph of the aver-

age cost function, Describe what this represents for thecompany.

91. The function

models the pH level, , of the human mouth x minutesafter a person eats food containing sugar. The graph of thisfunction is shown in the figure.

a. Use the graph to obtain a reasonable estimate, to thenearest tenth, of the pH level of the human mouth 42minutes after a person eats food containing sugar.

b. After eating sugar, when is the pH level the lowest? Usethe function’s equation to determine the pH level, to thenearest tenth, at this time.

c. According to the graph, what is the normal pH level of thehuman mouth?

d. What is the equation of the horizontal asymptote associatedwith this function? Describe what this means in terms of themouth’s pH level over time.

e. Use the graph to describe what happens to the pH levelduring the first hour.

92. A drug is injected into a patient and the concentration of thedrug in the bloodstream is monitored. The drug’s concentra-tion, in milligrams per liter, after t hours is modeled by

C1t2 =

5t

t2+ 1

.

C1t2,

6.5

6.0

7.0

5.5

5.0

4.5

605430 36 42 48

pH L

evel

of t

heH

uman

Mou

th

Number of Minutes afterEating Food Containing Sugar

6 12 18 24 66

4.0

f (x) =

y

x

6.5x2 − 20.4x + 234x2 + 36

f1x2f1x2 =

6.5x2- 20.4x + 234

x2+ 36

C?

C1100,0002.C110002, C110,0002,C,

C?

C140002.C15002, C110002, C120002,C,

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20022001200019951990198019701960

Popu

lati

on (

mill

ions

)

U.S. Population, by Gender

Year1950

Male Female1501401301201101009080

146.

714

1.7

145.

0

140.

0

143.

4

138.

1

134.

5

128.

3

127.

9

122.

0

116.

5

110.

1

104.

398

.9

91.0

88.3

75.8

74.8

70

Source: U.S. Census Bureau

The graph of this rational function, obtained with a graphingutility, is shown in the figure.

a. Use the graph to obtain a reasonable estimate of thedrug’s concentration after 3 hours. Then verify thisestimate algebraically.

b. Use the function’s equation, to find the

horizontal asymptote for the graph. Describe what thismeans about the drug’s concentration in the patient’sbloodstream as time increases.

Among all deaths from a particular disease, the percentage thatare smoking related (21–39 cigarettes per day) is a function of thedisease’s incidence ratio. The incidence ratio describes thenumber of times more likely smokers are than nonsmokers to diefrom the disease. The following table shows the incidence ratiosfor heart disease and lung cancer for two age groups.

Incidence Ratios

Heart Disease Lung CancerAges 55–64 1.9 10

Ages 65–74 1.7 9

Source: Alexander M. Walker, Observations and Inference, EpidemiologyResources Inc., 1991.

For example, the incidence ratio of 9 in the table means thatsmokers between the ages of 65 and 74 are 9 times more likelythan nonsmokers in the same age group to die from lung cancer.The rational function

models the percentage of smoking-related deaths among alldeaths from a disease, in terms of the disease’s incidenceratio, x. The graph of the rational function is shown. Use thisfunction to solve Exercises 93–96.

93. Find Describe what this means in terms of theincidence ratio, 10, given in the table. Identify your solutionas a point on the graph.

P1102.

Per

cent

age

of D

eath

s fr

om th

eD

isea

se T

hat A

re S

mok

ing

Rel

ated

y

x

The Disease’s Incidence Ratio:The number of times more likely smokers are

than nonsmokers to die from the disease

10987654321

100

80

60

40

20

100(x − 1)xP(x) =

P1x2,P1x2 =

1001x - 12x

C1t2 =

5t

t2+ 1

,

5xx2 + 1

y =

[0, 10, 1] by [0, 3, 1]

94. Find Round to the nearest percent. Describe what thismeans in terms of the incidence ratio, 9, given in the table.Identify your solution as a point on the graph.

95. What is the horizontal asymptote of the graph? Describewhat this means about the percentage of deaths caused bysmoking with increasing incidence ratios.

96. According to the model and its graph, is there a disease forwhich all deaths are caused by smoking? Explain your answer.

97. The graph shows the U.S. population, by gender, for selectedyears from 1950 through 2002.

P192.

a. Write a fraction that shows the ratio of males to femalesin 1995.Then express the fraction as a decimal, rounded tothe nearest thousandth. How many males per 1000females were there in 1995?

b. How many males per 1000 females were there in 2002?

c. The function models the male U.S.population, in millions, x years after 1950. Thefunction models the female U.S.population, in millions, x years after 1950. Write afunction that models the ratio of males to females x yearsafter 1950.

d. Use the function that you wrote in part (c) to find thenumber of males per 1000 females in 1995. How well doesthe function model the actual number that youdetermined in part (a)?

e. Use the function that you wrote in part (c) to find thenumber of males per 1000 females in 2002. How well doesthe function model the actual number that youdetermined in part (b)?

f. What is the equation of the horizontal asymptote associatedwith the function in part (c)? Round to the nearestthousandth. What does this mean about the number ofmales per 1000 females over time?

Exercises 98–101 involve writing a rational function that models aproblem’s conditions.

98. You drive from your home to a vacation resort 600 milesaway. You return on the same highway. The average velocityon the return trip is 10 miles per hour slower than theaverage velocity on the outgoing trip. Express the total timerequired to complete the round trip, T, as a function of theaverage velocity on the outgoing trip, x.

99. A tourist drives 90 miles along a scenic highway and thentakes a 5-mile walk along a hiking trail. The average velocitydriving is nine times that while hiking. Express the total timefor driving and hiking, T, as a function of the average velocityon the hike, x.

q1x2,q1x2 = 1.324x + 76.71

p1x2,p1x2 = 1.256x + 74.2

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100. A contractor is constructing the house shown in the figure.The cross section up to the roof is in the shape of a rectan-gle. The area of the rectangular floor of the house is 2500square feet. Express the perimeter of the rectangular floor,P, as a function of the width of the rectangle, x.

101. The figure shows a page with 1-inch margins at the top andthe bottom and half-inch side margins.A publishing companyis willing to vary the page dimensions subject to the conditionthat the printed area of the page is 50 square inches.Express the total area of the page, A, as a function of thewidth of the rectangle containing the print, x.

Writing in Mathematics102. What is a rational function?103. Use everyday language to describe the graph of a rational

function such that as 104. Use everyday language to describe the behavior of a graph

near its vertical asymptote if as andas

105. If you are given the equation of a rational function,explain howto find the vertical asymptotes, if any, of the function’s graph.

106. If you are given the equation of a rational function, explainhow to find the horizontal asymptote, if any, of thefunction’s graph.

107. Describe how to graph a rational function.108. If you are given the equation of a rational function, how can

you tell if the graph has a slant asymptote? If it does, howdo you find its equation?

109. Is every rational function a polynomial function? Why orwhy not? Does a true statement result if the two adjectivesrational and polynomial are reversed? Explain.

110. Although your friend has a family history of heart disease,he smokes, on average, 25 cigarettes per day. He sees thetable showing incidence ratios for heart disease (see Exer-cises 93–96) and feels comfortable that they are less than 2,compared to 9 and 10 for lung cancer. He claims that allfamily deaths have been from heart disease and decides notto give up smoking. Use the given function and its graph todescribe some additional information not given in the tablethat might influence his decision.

x : -2+.f1x2: - q

x : -2-f1x2: q

x : - q , f1x2: 3.f

1 in.ere I encounter the most popular fallacy of our times. It is not considered sufficient that the law should be just; it

must be philanthropic. Nor is it sufficient that the law should guar-antee to every citizen the free and inoffensive use of his faculties for physical, intellectual, and moral self-improvement. Instead, it is demanded that the law should directly extend welfare, education, and morality throughout the nation.

This is the seductive lure of socialism. And I repeat again: These two uses of the law are in direct contradiction to each other. We must choose between them. A citizen cannot at the same time be free and not free.

Enforced Fraternity Destroys Liberty

Mr. de Lamartine once wrote to me thusly: "Your doctrine is only the half of my program. You have stopped at liberty; I go on to fraternity." I answered him: "The second half of your program will destroy the first." In fact, it is impossible for me to separate the word fraternity from the word voluntary. I cannot possibly under-stand how fraternity can be legally enforced without liberty being legally destroyed, and thus justice being legally trampled under-foot. Legal plunder has two roots: One of them, as I have said be-fore, is in human greed; the other is in false philanthropy.

At this point, I think that I should explain exactly what I mean by the word plunder.

Plunder Violates Ownership

I do not, as is often done, use the word in any vague, uncertain, ap-proximate, or metaphorical sense. I use it in its scientific accept-ance as expressing the idea opposite to that of property [wages, land, money, or whatever]. When a portion of wealth is transfer-red from the person who owns it without his consent and

x

y

1 in.

q in.q in.

Length

Width: x

Technology Exercises111. Use a graphing utility to verify any five of your hand-drawn

graphs in Exercises 37–78.

112. Use a graphing utility to graph and in

the same viewing rectangle. For odd values of n, how does

changing n affect the graph of

113. Use a graphing utility to graph and

in the same viewing rectangle. For even values of n, how

does changing n affect the graph of

114. Use a graphing utility to graph

What differences do you observe between the graph of andthe graph of g? How do you account for these differences?

115. The rational function

models the number of arrests, , per 100,000 drivers, fordriving under the influence of alcohol, as a function of adriver’s age, x.a. Graph the function in a by viewing

rectangle.b. Describe the trend shown by the graph.c. Use the and features or the

maximum function feature of your graphing utility tofind the age that corresponds to the greatest number ofarrests. How many arrests, per 100,000 drivers, are therefor this age group?


a. The graph of a rational function cannot have both avertical asymptote and a horizontal asymptote.

b. It is not possible to have a rational function whose graphhas no y-intercept.

c. The graph of a rational function can have three horizontalasymptotes.

d. The graph of a rational function can never cross a verticalasymptote.

117. Which one of the following is true?

a. The function is a rational function.

b. The x-axis is a horizontal asymptote for the graph of

c. The number of televisions that a company can produceper week after t weeks of production is given by

Using this model, the company will eventually be able toproduce 30,000 televisions in a single week.

d. None of the given statements is true.

N1t2 =

3000t2+ 30,000t

t2+ 10t + 25

.

f1x2 =

4x - 1x + 3

.

f1x2 =

11x - 3

� TRACE �� ZOOM �

30, 400, 20430, 70, 54

f1x2f1x2 =

27,7251x - 142x2

+ 9- 5x

f

f1x2 =

x2- 4x + 3x - 2

and g1x2 =

x2- 5x + 6x - 2

.

y =

1xn?

y =

1

x6y =

1

x2 , y =

1

x4 ,

y =

1xn?

1

x5y =

1x

, y =

1

x3 ,

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Objectives❶ Solve polynomial

inequalities.

❷ Solve rational inequalities.

❸ Solve problems modeledby polynomial or rationalinequalities.

People are going to live longer in the twenty-first century. This will put addedpressure on the Social Security and Medicare systems. The bar graph in Figure 2.41shows the cost of Medicare, in billions of dollars, through 2005.

Medicare Spending

Spen

ding

(bi

llion

s of

dol

lars

)

100

200

300

400

$500

01995 1996 1997 1998

Year1999 2000 2001 2002 2003 2004 2005

$178$199

$219$240

$263$288

$315$345

$379$416

$458

Figure 2.41Source: Congressional Budget Office

SECTION 2.7 Polynomial and Rational Inequalities

Medicare spending, , in billions of dollars, x years after 1995 can be modeled bythe quadratic function

To determine in which years Medicare spending will exceed $500 billion, we mustsolve the inequality

We begin by subtracting 500 from both sides. This will give us zero on the right:

1.2x2+ 15.2x - 318.6 7 0.

1.2x2+ 15.2x + 181.4 - 500 7 500 - 500

1.2x2+15.2x+181.4>500.

Medicare spendingexceeds $500 billion.

f1x2 = 1.2x2+ 15.2x + 181.4.

f1x2

In Exercises 118–121, write the equation of a rational function

having the indicated properties, in which the degrees

of p and q are as small as possible. More than one correctfunction may be possible. Graph your function using a graphingutility to verify that it has the required properties.

118. has a vertical asymptote given by a horizontalasymptote y-intercept at and no x-intercept.-1,y = 0,

x = 3,f

f1x2 =

p1x2q1x2

119. has vertical asymptotes given by and ahorizontal asymptote y-intercept at x-intercepts

at and 3, and y-axis symmetry.

120. has a vertical asymptote given by a slant asymptotewhose equation is y-intercept at 2, and x-interceptsat and 2.

121. has no vertical, horizontal, or slant asymptotes, and nox-intercepts.f

-1y = x,

x = 1,f

-3

92 ,y = 2,

x = 2,x = -2f

pr02-265-374.I-hr 1/26/06 5:14 PM Page 346

x

y

1 2 3 4 5 6 7−1

1234567

−2−3

−1−2−3

x2 − 7x + 10 < 0

x2 − 7x + 10 > 0

Boundary points

Figure 2.42

Section 2.7 • Polynomial and Rational Inequalities 347

❶ Solve polynomial inequalities.

The form of is Such an inequality iscalled a polynomial inequality.

Definition of a Polynomial InequalityA polynomial inequality is any inequality that can be put into one of the forms

where is a polynomial function.

In this section, we establish the basic techniques for solving polynomial inequal-ities.We will use these techniques to solve inequalities involving rational functions.

Solving Polynomial InequalitiesGraphs can help us visualize the solutions of polynomial inequalities. For example,the graph of is shown in Figure 2.42.The x-intercepts, 2 and 5,are boundary points between where the graph lies above the x-axis, shown in blue, andwhere the graph lies below the x-axis, shown in red.

Locating the x-intercepts of a polynomial function, , is an important step in find-ing the solution set for polynomial inequalities in the form or Weuse the x-intercepts of as boundary points that divide the real number line intointervals.On each interval, the graph of is either above the x-axis or belowthe x-axis For this reason, x-intercepts play a fundamental role in solvingpolynomial inequalities. The x-intercepts are found by solving the equation

Procedure for Solving Polynomial Inequalities

1. Express the inequality in the form

where is a polynomial function.2. Solve the equation The real solutions are the boundary points.3. Locate these boundary points on a number line, thereby dividing the

number line into intervals.4. Choose one representative number, called a test value, within each interval

and evaluate at that number.a. If the value of is positive, then for all numbers, x, in the interval.b. If the value of is negative, then for all numbers, x, in the interval.

5. Write the solution set, selecting the interval or intervals that satisfy thegiven inequality.

This procedure is valid if is replaced by or is replaced by However, ifthe inequality involves or include the boundary points [the solutions of

] in the solution set.

EXAMPLE 1 Solving a Polynomial Inequality

Solve and graph the solution set on a real number line:

Solution

Step 1 Express the inequality in the form or We begin byrewriting the inequality so that 0 is on the right side.

This is the given inequality.

Subtract 15 from both sides.

Simplify.

This inequality is equivalent to the one we wish to solve. It is in the form where f1x2 = 2x2

+ x - 15.f1x2 7 0,

2x2+ x - 15 7 0

2x2+ x - 15 7 15 - 15

2x2+ x 7 15

f1x2>0.f1x2<0

2x2+ x 7 15.

f1x2 = 0Ú ,…

Ú .7…6

f1x2 6 0f

f1x2 7 0f

f

f1x2 = 0.f

f1x2 6 0 or f1x2 7 0,

f1x2 = 0.3f1x2 6 04. 3f1x2 7 04f

ff1x2 7 0.f1x2 6 0

f

f1x2 = x2- 7x + 10

f

f1x2 6 0, f1x2 7 0, f1x2 … 0, or f1x2 Ú 0,

ax2+ bx + c 7 0.1.2x2

+ 15.2x - 318.6 7 0

pr02-265-374.I-hr 1/26/06 5:14 PM Page 347


The solution set for

or, equivalently,

can be verified with a graphingutility. The graph of

wasobtained using a by

viewing rectangle.The graph lies above the x-axis,representing for all x in

or

52x =x = −3

above x-axisabove x-axis

A52 , q B .1- q , -327 ,

3-16, 6, 143-10, 10, 14

f1x2 = 2x2+ x - 15

2x2+ x - 15 7 0

2x2+ x 7 15

Technology

Step 2 Solve the equation We find the x-intercepts of by solving the equation

This polynomial equation is aquadratic equation.

Factor.


Solve for x.

The x-intercepts of are and We will use these x-intercepts as boundary pointson a number line.

Step 3 Locate the boundary points on a number line and separate the line intointervals. The number line with the boundary points is shown as follows:

The boundary points divide the number line into three intervals:

Step 4 Choose one test value within each interval and evaluate at that number.f

q B .A52 ,52 BA -3,-321- q ,

1 2 3 4 50−1−2−3

−3

−4−5x

e

52 .-3f

x =52 x = -3

2x - 5 = 0 or x + 3 = 0

12x - 521x + 32 = 0

2x2+ x - 15 = 0

2x2+ x - 15 = 0.

f1x2 = 2x2+ x - 15f1x2� 0.

Substitute intoInterval Test Value Conclusion

for all x in

0 for all x in

3 for all x in

= 6, positive

a52

, q b .f1x2 7 0 f132 = 2 # 32+ 3 - 15a5

2, q b

= -15, negative

a -3, 52b .f1x2 6 0 f102 = 2 # 02

+ 0 - 15a -3, 52b

= 13, positive1- q , -32.f1x2 7 0 f1-42 = 21-422 + 1-42 - 15-41- q , -32

f1x2 � 2x2 � x � 15

Step 5 Write the solution set, selecting the interval or intervals that satisfy thegiven inequality. We are interested in solving where

Based on our work in step 4, we see that for all x in

or Thus, the solution set of the given inequality,or, equivalently, is

The graph of the solution set on a number line is shown as follows:

Solve and graph the solution set:

EXAMPLE 2 Solving a Polynomial Inequality


Solution

Step 1 Express the inequality in the form or We begin byrewriting the inequality so that 0 is on the right side.

f1x2 » 0.f1x2 ◊ 0

x3+ x2

… 4x + 4.

x2- x 7 20.Check

Point 1

1 2 3 4 50−1−2−3

−3

−4−5

ex

1- q , -32 ´ A52 , q B or Ex ƒx 6 -3 or x 752 F .

2x2+ x - 15 7 0,

2x2+ x 7 15,A52 , q B .1- q , -32

f1x2 7 0f1x2 = 2x2+ x - 15.

2x2+ x - 15 7 0,

pr02-265-374.I-hr 1/26/06 5:14 PM Page 348



or, equivalently,

can be verified with a graphingutility. The graph of

lieson or below the x-axis, represent-ing for all x in or

x = 2x = −1

x = −2

Belowx-axis

Belowx-axis

[−4, 4, 1] by [−7, 3, 1]

[-1, 24.1- q , -24… ,

f1x2 = x3+ x2

- 4x - 4

x3+ x2

- 4x - 4 … 0

x3+ x2

… 4x + 4

Technology


Subtract from both sides.

Simplify.

This inequality is equivalent to the one we wish to solve. It is in the form where

Step 2 Solve the equation We find the x-intercepts ofby solving the equation

This polynomial equation is of degree 3.

Factor from the first two terms and from the last two terms.

A common factor of is factored fromthe expression.


Solve for x.

Use the square root property.

The x-intercepts of are and 2. We will use these x-intercepts as boundarypoints on a number line.


The boundary points divide the number line into four intervals:

Step 4 Choose one test value within each interval and evaluate at that number.f

1- q , -22 1-2, -12 1-1, 22 12, q2.

4 531 20−1−2

−1−2 2

−3−4−5x

-2, -1,f

x = ;2

x = -1 x2= 4

x + 1 = 0 or x2- 4 = 0

x � 1 1x + 121x2- 42 = 0

�4x 2 x21x + 12 - 41x + 12 = 0

x3+ x2

- 4x - 4 = 0

x3+ x2

- 4x - 4 = 0.f1x2 = x3+ x2

- 4x - 4f1x2� 0.

f1x2 = x3+ x2

- 4x - 4.f1x2 … 0,

x3+ x2

- 4x - 4 … 0

4x � 4 x3+ x2

- 4x - 4 … 4x + 4 - 4x - 4

x3+ x2

… 4x + 4


forall x in

forall x in

0 for all x in

3 forall x in 12, q2. = 20, positivef1x2 7 0 f132 = 33

+ 32- 4 # 3 - 412, q2

1-1, 22. = -4, negativef1x2 6 0 f102 = 03

+ 02- 4 # 0 - 41-1, 22

1-2, -12. = 0.875, positivef1x2 7 0 f1-1.52 = 1-1.523 + 1-1.522 - 41-1.52 - 4-1.51-2, -12

1- q , -22. = -10, negativef1x2 6 0 f1-32 = 1-323 + 1-322 - 41-32 - 4-31- q , -22

f1x2 � x3 � x2 � 4x � 4

Step 5 Write the solution set, selecting the interval or intervals that satisfy thegiven inequality. We are interested in solving where

Based on our work in step 4, we see that forall x in or However, because the inequality involves (lessthan or equal to), we must also include the solutions of namely and 2, in the solution set. Thus, the solution set of the given inequal-ity, or, equivalently, is


4 531 20−1−2−3−4−5x

or 5x ƒx … -2 or -1 … x … 26.1- q , -24 ´ 3-1, 24

x3+ x2

- 4x - 4 … 0,x3+ x2

… 4x + 4,-2, -1,

x3+ x2

- 4x - 4 = 0,…1-1, 22.1- q , -22 f1x2 6 0f1x2 = x3

+ x2- 4x - 4.

x3+ x2

- 4x - 4 … 0,

pr02-265-374.I-hr 1/26/06 5:14 PM Page 349

Do not begin solving

by multiplying both sides byWe do not know if

is positive or negative.Thus, wedo not know whether or not tochange the sense of theinequality.

x + 3x + 3.

x + 1x + 3

Ú 2


Study Tip


Solving Rational InequalitiesA rational inequality is any inequality that can be put into one of the forms

where is a rational function. An example of a rational inequality is

This inequality is in the form where is the rational function given by

The graph of is shown in Figure 2.43.We can find the x-intercept of by setting the numerator equal to 0:

We can determine where is undefined by setting the denominator equal to 0:

By setting both the numerator and the denominator of equal to 0, weobtained and These numbers separate the x-axis into three intervals:

and On each interval, the graph of is either abovethe x-axis or below the x-axis

Examine the graph in Figure 2.43 carefully. Can you see that it is above thex-axis for all x in or shown in blue? Thus, the solution set of

is By contrast, the graph of lies below the

x-axis for all x in shown in red. Thus, the solution set of is

The first step in solving a rational inequality is to bring all terms to one side,obtaining zero on the other side. Then express the rational function on the nonzeroside as a single quotient. The second step is to set the numerator and the denomina-tor of equal to zero.The solutions of these equations serve as boundary points thatseparate the real number line into intervals. At this point, the procedure is the sameas the one we used for solving polynomial inequalities.

EXAMPLE 3 Solving a Rational Inequality


Solution

Step 1 Express the inequality so that one side is zero and the other side is a singlequotient. We subtract 2 from both sides to obtain zero on the right.

x + 1x + 3

Ú 2.

f

1-2, -12.

3x + 32x + 4

6 01-2, -12,

f1- q , -22 ´ 1-1, q2.3x + 32x + 4

7 0

1-1, q2,1- q , -22

3f1x2 6 04.3f1x2 7 04f1-1, q2.1- q , -22, 1-2, -12,

-1.-2f

x=–2.

2x=–4

2x+4=0

f is undefined at −2.Figure 2.43 shows thatthe function’s verticalasymptote is x = −2.

f

x=–1.

3x=–3

3x+3=0

f has an x-interceptat −1 and passesthrough (−1, 0).

ff

f1x2 =

3x + 32x + 4

.

ff1x2 7 0,

3x + 32x + 4

7 0.

f

f1x2 6 0, f1x2 7 0, f1x2 … 0, or f1x2 Ú 0,

x3+ 3x2

… x + 3.CheckPoint 2

❷ Solve rational inequalities.

x

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3−4−5

Abovex-axis Above

x-axis

Belowx-axis


f1x2 =

3x + 32x + 4

pr02-265-374.I-hr 1/27/06 2:28 PM Page 350

Never include the value thatcauses a rational function’sdenominator to equal zero in thesolution set of a rationalinequality. Division by zero isundefined.


Study Tip


Subtract 2 from both sides, obtaining 0on the right.

The least common denominator is Express 2 in terms of this denominator.

Subtract rational expressions.

Apply the distributive property.

Simplify.

This inequality is equivalent to the one we wish to solve. It is in the form

where

Step 2 Set the numerator and the denominator of equal to zero. The realsolutions are the boundary points.

Set the numerator and denominator equalto 0. These are the values that make theprevious quotient zero or undefined.

Solve for x.

We will use these solutions as boundary points on a number line.


The boundary points divide the number line into three intervals:

Step 4 Choose one test value within each interval and evaluate at that number.

Substitute into

Interval Test Value Conclusion

for all

x in

for all

x in

0 for all

x in

Step 5 Write the solution set, selecting the interval or intervals that satisfy the

given inequality. We are interested in solving where

Based on our work in step 4, we see that for all x in However,1-5, -32.f1x2 7 0

f1x2 =

-x - 5x + 3

.-x - 5x + 3

Ú 0,

1-3, q2. = - 53 , negative

f1x2 6 0 f102 =

-0 - 50 + 3

1-3, q21-5, -32. = 1, positive

f1x2 7 0 f1-42 =

-1-42 - 5

-4 + 3-41-5, -32

1- q , -52. = - 13 , negative

f1x2 6 0 f1-62 =

-1-62 - 5

-6 + 3-61- q , -52

f1x2 ��x � 5x � 3

f

1- q , -52 1-5, -32 1-3, q2.

31 20−1−2

−3−5

−3−6−7 −4−5x

x = -5 x = -3

-x - 5 = 0 x + 3 = 0

f

f1x2 =

-x - 5x + 3

.

f1x2 Ú 0,

-x - 5x + 3

Ú 0

x + 1 - 2x - 6

x + 3Ú 0

x + 1 - 21x + 32

x + 3Ú 0

x � 3. x + 1x + 3

-

21x + 32x + 3

Ú 0

x + 1x + 3

- 2 Ú 0

x + 1x + 3

Ú 2

pr02-265-374.I-hr 1/26/06 5:14 PM Page 351

Because is positive, it ispossible so solve

by first multiplying both sides by(where ). This

will not change the sense of theinequality and will clear thefraction. Try using this solutionmethod and compare it to thesolution on pages 350–352.

x Z -31x + 322

x + 1x + 3

Ú 2

1x + 322Discovery



or, equivalently,

can be verified with a graphing utility. The graph of

lies on or above the x-axis,

representing for all x in 3-5, -32.Ú ,

f1x2 =

-x - 5x + 3

-x - 5x + 3

Ú 0

x + 1x + 3

Ú 2

Technology

x = −3x = −5

Graph lieson or abovethe x-axis.

[−8, 8, 1] by [−3, 3, 1]

❸ Solve problems modeled bypolynomial or rationalinequalities.

because the inequality involves (greater than or equal to), we must also includethe solution of namely the value that we obtained when we set the nu-merator of equal to zero.Thus, we must include in the solution set.The solutionset of the given inequality is


31 20−1−2−3−6−7 −4−5x

3-5, -32 or 5x ƒ -5 … x 6 -32.

-5ff1x2 = 0,

Ú


ApplicationsWe are surrounded by evidence that the world is profoundly mathematical. Forexample, did you know that every time you throw an object vertically upward, itschanging height above the ground can be described by a quadratic function? Thesame function can be used to describe objects that are falling, such as sky divers.

The Position Function for a Free-Falling Object Near Earth’s SurfaceAn object that is falling or vertically projected into the air has its height abovethe ground, in feet, given by

where is the original velocity (initial velocity) of the object, in feet persecond, t is the time that the object is in motion, in seconds, and is the originalheight (initial height) of the object, in feet.

In Example 4, we solve a polynomial inequality in a problem about theposition of a free-falling object.

EXAMPLE 4 Using the Position Function

A ball is thrown vertically upward from the top of the Leaning Tower of Pisa (190feet high) with an initial velocity of 96 feet per second (Figure 2.44). During whichtime period will the ball’s height exceed that of the tower?

s0

v0

s1t2 = -16t2+ v0t + s0 ,

s1t2,

2x

x + 1Ú 1.Check

Point 3

pr02-265-374.I-hr 1/26/06 5:14 PM Page 352

190 feet

t = 0s0 = 190v0 = 96

Figure 2.44 Throwing a ball from 190feet with a velocity of 96 feet per second


Solution

This is the position function for a free-falling object.

Because andsubstitute these

values into the formula.

This is the inequality that models the problem’squestion. We must find t.

Subtract 190 from both sides. This inequalityis in the form where

Solve the equation

Factor.


Solve for t. The boundary points are 0 and 6.

Locate these values on a number line, with

The intervals are and For our purposes, the mathematicalmodel is useful only from until the ball hits the ground. (By setting

equal to zero, we find the ball hits the ground afterapproximately 7.57 seconds.) Thus, we use and for our intervals.


1 for allt in

7 for allt in

We are interested in solving where We seethat for all t in This means that the ball’s height exceeds that of thetower between 0 and 6 seconds.

10, 62.f1t2 7 0f1t2 = -16t2

+ 96t.-16t2+ 96t 7 0,

16, 7.572. = -112, negativef1t2 6 0 f172 = -16 # 72

+ 96 # 716, 7.57210, 62. = 80, positive

f1t2 7 0 f112 = -16 # 12+ 96 # 110, 62

f1t2 � �16t2 � 96t

16, 7.57210, 62t L 7.57;-16t2+ 96t + 190

t = 016, q2.10, 62,1- q , 02,

t » 0.

5 6 7 81 2 3 40−1−2t

t = 0 t = 6

-16t = 0 or t - 6 = 0

-16t1t - 62 = 0

f1t2 � 0. -16t2+ 96t = 0

f1t2 � � 16t 2 � 96t.f1t2>0,

-16t2+ 96t 7 0

-16t2+ 96t + 190 7 190

–16t2 + 96t +190 > 190

When will s(t), the ball’s height, exceed that of the tower?

s0 1initial position2 � 190,v0 1initial velocity2 � 96 s1t2 = -16t2

+ 96t + 190

s1t2 = -16t2+ v0 t + s0

The graphs of

and

are shown in a

viewing rectangle. The graphsshow that the ball’s heightexceeds that of the towerbetween 0 and 6 seconds.

[0, 8, 1] by [0, 360, 36]

secondsin motion

height,in feet

y2 = 190

y1 = -16x2+ 96x + 190 Height of Ball:

y1 = −16x2 + 96x + 190

Height of Tower:y2 = 190 Ball hits ground

after 7.57 seconds.

360

0 6 8

Seconds in Motion[0, 8, 1] by [0, 360, 36]

Hei

ght (

feet

)

Technology

pr02-265-374.I-hr 1/26/06 5:14 PM Page 353


An object is propelled straight up from ground level with an initial velocityof 80 feet per second. Its height at time t is modeled by

where the height, is measured in feet and the time, t, is measured inseconds. In which time interval will the object be more than 64 feet abovethe ground?

s1t2,s1t2 = -16t2

+ 80t,

CheckPoint 4

EXERCISE SET 2.7

Practice ExercisesSolve each polynomial inequality in Exercises 1–38

and graph the solution set on a real number line. Express eachsolution set in interval notation.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29.

30.

31. 32.

33. 34.

35. 36.

37. 38.

Solve each rational inequality in Exercises 39–56 and graph thesolution set on a real number line. Express each solution set ininterval notation.

39. 40.

41. 42.

43. 44.

45. 46.3x + 56 - 2x

Ú 04 - 2x

3x + 4… 0

-x - 3x + 2

… 0-x + 2x - 4

Ú 0

x + 5x + 2

6 0x + 3x + 4

6 0

x + 5x - 2

7 0x - 4x + 3

7 0

x3… 4x2x3

Ú 9x2

x3- x2

+ 9x - 9 7 0x3+ x2

+ 4x + 4 7 0

x3+ 7x2

- x - 7 6 0x3- 3x2

- 9x + 27 6 0

x3+ 2x2

- 4x - 8 Ú 0x3+ 2x2

- x - 2 Ú 0

1x + 121x + 221x + 32 Ú 0

1x - 121x - 221x - 32 Ú 0

4x2- 4x + 1 Ú 0x2

- 6x + 9 6 0

x2… 2x + 2x2

… 4x - 2

-x2+ 2x Ú 0-x2

+ x Ú 0

3x2- 5x … 02x2

+ 3x 7 0

x2+ 2x 6 0x2

- 4x Ú 0

4x2+ 1 Ú 4x5x … 2 - 3x2

3x2+ 16x 6 -54x2

+ 7x 6 -3

6x2+ x 7 12x2

+ x 6 15

9x2+ 3x - 2 Ú 03x2

+ 10x - 8 … 0

x2- 2x + 1 7 0x2

- 6x + 9 6 0

x2+ x - 6 7 0x2

+ 5x + 4 7 0

x2- 4x + 3 6 0x2

- 5x + 4 7 0

1x + 121x - 72 … 01x - 721x + 32 … 0

1x + 321x - 52 7 01x - 421x + 22 7 0

47. 48.

49. 50.

51. 52.

53. 54.

55. 56.

Practice PlusIn Exercises 57–60, find the domain of each function.

57. 58.

59. 60.

Solve each inequality in Exercises 61–66 and graph the solutionset on a real number line.

61. 62.

63. 64.

65. 66.

In Exercises 67–68, use the graph of the polynomial function tosolve each inequality.

67. 68. 2x3+ 11x2

6 7x + 62x3+ 11x2

Ú 7x + 6

f (x) = 2x3 + 11x2 − 7x − 6

−7 3

−10

70

x2- 3x + 2

x2- 2x - 3

7 0x2

- x - 2

x2- 4x + 3

7 0

1x + 1

7

2x - 1

3x + 3

7

3x - 2

ƒx2+ 6x + 1 ƒ 7 8ƒx2

+ 2x - 36 ƒ 7 12

f1x2 = A x

2x - 1- 1f1x2 = A 2x

x + 1- 1

f1x2 =

134x2- 9x + 2

f1x2 = 32x2- 5x + 2

x

x + 2Ú 2

x - 2x + 2

… 2

1x - 3

6 1x + 4

2x - 1… 3

x

x - 17 2

x + 1x + 3

6 2

1x + 321x - 22x + 1

… 01x + 421x - 12

x + 2… 0

x + 4x

7 0x

x - 37 0

pr02-265-374.I-hr 1/26/06 5:14 PM Page 354


In Exercises 69–70, use the graph of the rational function to solveeach inequality.

69.

70.

Application ExercisesUse the position function

to answer Exercises 71–72.

71. Divers in Acapulco, Mexico, dive headfirst at 8 feet persecond from the top of a cliff 87 feet above the Pacific Ocean.During which time period will the diver’s height exceed thatof the cliff?

72. You throw a ball straight up from a rooftop 160 feet highwith an initial velocity of 48 feet per second. During whichtime period will the ball’s height exceed that of the rooftop?

The bar graph in Figure 2.41 on page 346 shows the cost ofMedicare, in billions of dollars, through 2005. Using the regressionfeature of a graphing utility, these data can be modeled by

In each function, x represents the number of years after 1995. Usethese functions to solve Exercises 73–76.

73. The graph indicates that Medicare spending reached $379billion in 2003. Find the amount predicted by each of thefunctions, and g, for that year. How well do the functionsmodel the value in the graph?

74. The graph indicates that Medicare spending reached $458billion in 2005. Find the amount predicted by each of thefunctions, and g, for that year. How well do the functionsmodel the value in the graph? Which function serves as abetter model for that year?

75. For which years does the quadratic model indicate thatMedicare spending will exceed $536.6 billion?

76. For which years does the quadratic model indicate thatMedicare spending will exceed $629.4 billion?

f

f

a quadratic function, g1x2 = 1.2x2+ 15.2x + 181.4.

a linear function, f1x2 = 27x + 163;

1v0 = initial velocity, s0 = initial position, t = time2s1t2 = -16t2

+ v0t + s0

141x + 22 7 -

341x - 22

141x + 22 … -

341x - 22

f (x) = x + 1x2 − 4

−4 4

−4

4

It’s vacation time. You drive 90 miles along a scenic highway andthen take a 5-mile run along a hiking trail. Your driving rate isnine times that of your running rate. The graph shows the totaltime you spend driving and running, as a function of yourrunning rate, x. Use the rational function and its graph to solveExercises 77–81.

77. Describe your running rate if you have no more than a totalof 3 hours for driving and running. Use a rational inequalityto solve the problem. Then explain how your solution isshown on the graph.

78. Describe your running rate if you have no more than a totalof 5 hours for driving and running. Use a rational inequalityto solve the problem. Then explain how your solution isshown on the graph.

79. Describe the behavior of the graph as What does thisshow about the time driving and running as a function ofyour running rate?

80. Describe the behavior of the graph as What does thisshow about the time driving and running as a function ofyour running rate?

81. Describe how to use the formula and the problem’s

verbal conditions to obtain the function’s equation displayedin the voice balloon.

82. The perimeter of a rectangle is 50 feet. Describe the possiblelengths of a side if the area of the rectangle is not to exceed114 square feet.

83. The perimeter of a rectangle is 180 feet. Describe thepossible lengths of a side if the area of the rectangle is not toexceed 800 square feet.

Writing in Mathematics

84. What is a polynomial inequality?

85. What is a rational inequality?

86. If is a polynomial or rational function, explain howthe graph of can be used to visualize the solution set of theinequality f1x2 6 0.

ff

t =

s

v

x : 0+.

x : q .

Tim

e D

rivi

ng a

nd R

unni

ng (

hour

s)

y

x

Running Rate (miles per hour)1211109876543210

1112

910

78

56

34

12

909x

5x

15xf (x) = = +

f1x2,

pr02-265-374.I-hr 1/26/06 5:14 PM Page 355


Technology Exercises

87. Use a graphing utility to verify your solution sets to anythree of the polynomial inequalities that you solved alge-braically in Exercises 1–38.

88. Use a graphing utility to verify your solution sets to anythree of the rational inequalities that you solvedalgebraically in Exercises 39–56.

Solve each inequality in Exercises 89–94 using a graphing utility.

89.

90.

91.

92.

93.

94.

Critical Thinking Exercises

95. Which one of the following is true?

a. The solution set of is

b. The inequality can be solved by multiplying

both sides by resulting in the equivalent inequality

c. and have the samesolution set.

d. None of these statements is true.

96. Write a polynomial inequality whose solution set is

97. Write a rational inequality whose solution set is

In Exercises 98–101, use inspection to describe each inequality’ssolution set. Do not solve any of the inequalities.

98. 99.

100. 101.

102. The graphing utility screen shows the graph of

y = 4x2 − 8x + 7

[–2, 6, 1] by [–2, 8, 1]

y = 4x2- 8x + 7.

1

1x - 222 7 01x - 222 6 -1

1x - 222 … 01x - 222 7 0

1- q , -42 ´ 33, q2.3-3, 54.

x + 3x - 1

Ú 01x + 321x - 12 Ú 0

x - 2 6 21x + 32.x + 3,

x - 2x + 3

6 2

15, q2.x27 25

1x + 1

…

2x + 4

x + 2x - 3

… 2

x - 4x - 1

… 0

x3+ x2

- 4x - 4 7 0

2x2+ 5x - 3 … 0

x2+ 3x - 10 7 0

a. Use the graph to describe the solution set of

b. Use the graph to describe the solution set of

c. Use an algebraic approach to verify each of yourdescriptions in parts (a) and (b).

103. The graphing utility screen shows the graph of

Write and solve a quadratic inequalitythat explains why the graph only appears for

Group Exercise

104. This exercise is intended as a group learning experience andis appropriate for groups of three to five people. Beforeworking on the various parts of the problem, reread thedescription of the position function on page 352.

a. Drop a ball from a height of 3 feet, 6 feet, and 12 feet.Record the number of seconds it takes for the ball to hitthe ground.

b. For each of the three initial positions, use the positionfunction to determine the time required for the ball tohit the ground.

c. What factors might result in differences between thetimes that you recorded and the times indicated by thefunction?

d. What appears to be happening to the time required for afree-falling object to hit the ground as its initial height isdoubled? Verify this observation algebraically and witha graphing utility.

e. Repeat part (a) using a sheet of paper rather than a ball.What differences do you observe? What factor seems tobe ignored in the position function?

f. What is meant by the acceleration of gravity and howdoes this number appear in the position function for afree-falling object?

y = �27 − 3x2

[–5, 5, 1] by [0, 6, 1]

-3 … x … 3.y = 227 - 3x2 .

4x2- 8x + 7 6 0.

4x2- 8x + 7 7 0.

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Section 2.8 • Modeling Using Variation 357

Objectives❶ Solve direct variation

problems.

❷ Solve inverse variationproblems.

❸ Solve combined variationproblems.

❹ Solve problems involvingjoint variation.

Have you ever wondered how telecommunication companies estimate the numberof phone calls expected per day between two cities? The formula

shows that the daily number of phone calls, C, increases as the populations of thecities, and in thousands, increase and decreases as the distance, d, between thecities increases.

Certain formulas occur so frequently in applied situations that they are givenspecial names. Variation formulas show how one quantity changes in relation toother quantities. Quantities can vary directly, inversely, or jointly. In this section, welook at situations that can be modeled by each of these kinds of variation.And thinkof this: The next time you get one of those “all-circuits-are-busy” messages, you willbe able to use a variation formula to estimate how many other callers you’recompeting with for those precious 5-cent minutes.

Direct VariationWhen you swim underwater, the pressure in your ears depends on the depth at whichyou are swimming. The formula

describes the water pressure, p, in pounds per square inch, at a depth of d feet. Wecan use this linear function to determine the pressure in your ears at various depths:

If d = 20, p=0.43(20)= 8.6.

If d = 40, p=0.43(40)=17.2.

Doubling the depth doubles the pressure.

If d = 80, p=0.43(80)=34.4.

Doubling the depth doubles the pressure.

p = 0.43d

P2 ,P1

C =

0.02P1 P2

d2

❶ Solve direct variationproblems.

SECTION 2.8 Modeling Using Variation

At a depth of 20 feet, water pressure is 8.6pounds per square inch.



The formula illustrates that water pressure is a constant multipleof your underwater depth. If your depth is doubled, the pressure is doubled; ifyour depth is tripled, the pressure is tripled; and so on. Because of this, the pres-sure in your ears is said to vary directly as your underwater depth. The equationof variation is

p = 0.43d.

p = 0.43d

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70

60

50

40

30

20

160140120

Pre

ssur

e in

Div

er's

Ear

s(p

ound

s pe

r sq

uare

inch

)

Depth (feet)20 40 60 80 100

10

Unsafe foramateur divers

p = 0.43d

p

dFigure 2.45 Water pressure at variousdepths

Generalizing our discussion of pressure and depth on the previous page, weobtain the following statement:

Direct VariationIf a situation is described by an equation in the form

where k is a nonzero constant, we say that y varies directly as x or y is directlyproportional to x. The number k is called the constant of variation or theconstant of proportionality.

Can you see that the direct variation equation, is a special case of thelinear function When and becomes

Thus, the slope of a direct variation equation is k, the constant of variation.Because b, the y-intercept, is 0, the graph of a direct variation equation is a linepassing through the origin. This is illustrated in Figure 2.45, which shows the graphof Water pressure varies directly as depth.p = 0.43d:

y = kx.y = mx + bb = 0,m = ky � mx � b ?

y � kx,

y = kx,

Problems involving direct variation can be solved using the following proce-dure. This procedure applies to direct variation problems, as well as to the otherkinds of variation problems that we will discuss.

Solving Variation Problems

1. Write an equation that describes the given English statement.2. Substitute the given pair of values into the equation in step 1 and solve for

k, the constant of variation.3. Substitute the value of k into the equation in step 1.4. Use the equation from step 3 to answer the problem’s question.

EXAMPLE 1 Solving a Direct Variation Problem

Many areas of Northern California depend on the snowpack of the Sierra Nevadamountain range for their water supply. The volume of water produced from meltingsnow varies directly as the volume of snow. Meteorologists have determined that250 cubic centimeters of snow will melt to 28 cubic centimeters of water. How muchwater does 1200 cubic centimeters of melting snow produce?

Solution

Step 1 Write an equation. We know that y varies directly as x is expressed as

By changing letters, we can write an equation that describes the following Englishstatement: Volume of water, W, varies directly as volume of snow, S.

W = kS

y = kx.

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Step 2 Use the given values to find k. We are told that 250 cubic centimeters ofsnow will melt to 28 cubic centimeters of water. Substitute 250 for S and 28 for W inthe direct variation equation. Then solve for k.

Volume of water varies directly asvolume of melting snow.

250 cubic centimeters of snow meltto 28 cubic centimeters of water.


Simplify.

Step 3 Substitute the value of k into the equation.

This is the equation from step 1.

Replace k, the constant ofvariation, with 0.112.

Step 4 Answer the problem’s question. How much water does 1200 cubiccentimeters of melting snow produce? Substitute 1200 for S in andsolve for W.

Use the equation from step 3.

Substitute 1200 for S.

Multiply.

A snowpack measuring 1200 cubic centimeters will produce 134.4 cubic centimetersof water.

The number of gallons of water, W, used when taking a shower variesdirectly as the time, t, in minutes, in the shower.A shower lasting 5 minutesuses 30 gallons of water. How much water is used in a shower lasting 11minutes?

The direct variation equation is a linear function. If then theslope of the line is positive. Consequently, as x increases, y also increases.

A direct variation situation can involve variables to higher powers. Forexample, y can vary directly as or as

Direct Variation with Powersy varies directly as the nth power of x if there exists some nonzero constant ksuch that

We also say that y is directly proportional to the nth power of x.

Direct variation with whole number powers is modeled by polynomialfunctions. In our next example, the graph of the variation equation is the familiarparabola.

EXAMPLE 2 Solving a Direct Variation Problem

The distance, s, that a body falls from rest varies directly as the square of the time, t,of the fall. If skydivers fall 64 feet in 2 seconds, how far will they fall in 4.5 seconds?

Solution

Step 1 Write an equation. We know that y varies directly as the square of x isexpressed as

y = kx2.

y = kxn.

x31y = kx32.x21y = kx22

k>0,y � kx

CheckPoint 1

W = 134.4

W = 0.112112002 W = 0.112S

W = 0.112S

W = 0.112S

W = kS

0.112 = k

28

250=

k12502250

28 = k12502 W = kS

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By changing letters, we can write an equation that describes the following Englishstatement: Distance, s, varies directly as the square of time, t, of the fall.

Step 2 Use the given values to find k. Skydivers fall 64 feet in 2 seconds. Substi-tute 64 for s and 2 for t in the direct variation equation. Then solve for k.

Distance varies directly as the square of time.

Skydivers fall 64 feet in 2 seconds.

Simplify:


Simplify.



Replace k, the constant of variation, with 16.

Step 4 Answer the problem’s question. How far will the skydivers fall in 4.5seconds? Substitute 4.5 for t in and solve for s.

Thus, in 4.5 seconds, the skydivers will fall 324 feet.

We can express the variation equation from Example 2 in function notation,writing

The distance that a body falls from rest is a function of the time, t, of the fall. Theparabola that is the graph of this quadratic function is shown in Figure 2.46.The graphincreases rapidly from left to right, showing the effects of the acceleration of gravity.

The distance required to stop a car varies directly as the square of itsspeed. If 200 feet are required to stop a car traveling 60 miles per hour,how many feet are required to stop a car traveling 100 miles per hour?

Inverse VariationThe distance from San Francisco to Los Angeles is 420 miles. The time that it takesto drive from San Francisco to Los Angeles depends on the rate at which one drivesand is given by

For example, if you average 30 miles per hour, the time for the drive is

or 14 hours. If you average 50 miles per hour,the time for the drive is

or 8.4 hours. As your rate (or speed) increas-es, the time for the trip decreases and viceversa. This is illustrated by the graph inFigure 2.47.

Time =

42050

= 8.4,

Time =

42030

= 14,

Time =

420Rate

.

CheckPoint 2

s1t2 = 16t2.

s = 1614.522 = 16120.252 = 324

s = 16t2

s = 16t2

s = kt2

16 = k

644

=

4k

4

22 � 4. 64 = 4k

64 = k # 22

s = kt2

s = kt2

Tim

e fo

r Tri

p (h

ours

)

t

r

5

10

15

20

25

Driving Rate (miles per hour)20 40 60 80 100 120

Averaging 30 mph,the trip takes 14 hours.

Averaging 50 mph,the trip takes 8.4 hours.

420r t =

Figure 2.47

Dis

tanc

e th

e Sk

ydiv

ers

Fall

(fee

t)

s(t)

t

Time the Skydivers Fall (seconds)654321

100

200

300

400

Distance Fallen bySkydivers over Time

Figure 2.46 The graph of s1t2 = 16t2


❷ Solve inverse variationproblems.

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2P

2V

P

V

Doubling the pressurehalves the volume.

y

x

kxy = , k > 0 and x > 0

Figure 2.48 The graph of the inversevariation equation


We can express the time for the San Francisco–Los Angeles trip using t fortime and r for rate:

This equation is an example of an inverse variation equation. Time, t, varies inverselyas rate, r. When two quantities vary inversely, one quantity increases as the otherdecreases and vice versa.

Generalizing, we obtain the following statement:

Inverse VariationIf a situation is described by an equation in the form

where k is a nonzero constant, we say that y varies inversely as x or y is inverselyproportional to x. The number k is called the constant of variation.

Notice that the inverse variation equation

is a rational function. For and the graph of the function takes on theshape shown in Figure 2.48.

We use the same procedure to solve inverse variation problems as we did tosolve direct variation problems. Example 3 illustrates this procedure.

EXAMPLE 3 Solving an Inverse Variation Problem

When you use a spray can and press the valve at the top, you decrease the pressureof the gas in the can. This decrease of pressure causes the volume of the gas in thecan to increase. Because the gas needs more room than is provided in the can, itexpands in spray form through the small hole near the valve. In general, if thetemperature is constant, the pressure, P, of a gas in a container varies inversely asthe volume, V, of the container. The pressure of a gas sample in a container whosevolume is 8 cubic inches is 12 pounds per square inch. If the sample expands to avolume of 22 cubic inches, what is the new pressure of the gas?

Solution

Step 1 Write an equation. We know that y varies inversely as x is expressed as

By changing letters, we can write an equation that describes the following Englishstatement: The pressure, P, of a gas in a container varies inversely as the volume, V.

Step 2 Use the given values to find k. The pressure of a gas sample in a containerwhose volume is 8 cubic inches is 12 pounds per square inch. Substitute 12 for P and8 for V in the inverse variation equation. Then solve for k.

Pressure varies inversely as volume.

The pressure in an 8 cubic-inchcontainer is 12 pounds per square inch.

Multiply both sides by 8.

Simplify. 96 = k

12 # 8 =

k

8# 8

12 =

k

8

P =

k

V

P =

k

V.

y =

kx

.

x 7 0,k 7 0

y =

kx

, or f1x2 =

kx

,

y =

kx

,

t =

420r

.

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❸ Solve combined variationproblems.



Replace k, the constant of variation,with 96.

Step 4 Answer the problem’s question. We need to find the pressure when thevolume expands to 22 cubic inches. Substitute 22 for V and solve for P.

When the volume is 22 cubic inches, the pressure of the gas is pounds per squareinch.

The length of a violin string varies inversely as the frequency of itsvibrations. A violin string 8 inches long vibrates at a frequency of 640cycles per second. What is the frequency of a 10-inch string?

Combined VariationIn combined variation, direct and inverse variation occur at the same time. Forexample, as the advertising budget, A, of a company increases, its monthly sales, S, alsoincrease. Monthly sales vary directly as the advertising budget:

By contrast, as the price of the company’s product, P, increases, its monthly sales, S,decrease. Monthly sales vary inversely as the price of the product:

We can combine these two variation equations into one combined equation:

The following example illustrates an application of combined variation.

EXAMPLE 4 Solving a Combined Variation Problem

The owners of Rollerblades Plus determine that the monthly sales, S, of its skatesvary directly as its advertising budget, A, and inversely as the price of the skates, P.When $60,000 is spent on advertising and the price of the skates is $40, the monthlysales are 12,000 pairs of rollerblades.

a. Write an equation of variation that describes this situation.

b. Determine monthly sales if the amount of the advertising budget is increasedto $70,000.

Solution

a. Write an equation.

kAP

S= .

Translate “sales vary directly asthe advertising budget and

inversely as the skates’ price.”

kAP

S= .

Monthly sales , S, vary directlyas the advertising budget, A,and inversely as the price of

the product, P.

S =

k

P.

S = kA.

CheckPoint 3

4 4

11

P =

96V

=

9622

= 4 411

P =

96V

P =

k

V

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❹ Solve problems involvingjoint variation.

Use the given values to find k.

When $60,000 is spent on advertisingand the price is $40

monthly sales are 12,000units

Divide 60,000 by 40.

Divide both sides of the equation by 1500.

Simplify.

Therefore, the equation of variation that describes monthly sales is

Substitute 8 for k in

b. The advertising budget is increased to $70,000, so The skates’price is still $40, so

This is the equation from part (a).

Substitute 70,000 for A and 40 for P.

Simplify.

With a $70,000 advertising budget and $40 price, the company can expect tosell 14,000 pairs of rollerblades in a month (up from 12,000).

The number of minutes needed to solve an exercise set of variationproblems varies directly as the number of problems and inversely as thenumber of people working to solve the problems. It takes 4 people 32minutes to solve 16 problems. How many minutes will it take 8 people tosolve 24 problems?

Joint VariationJoint variation is a variation in which a variable varies directly as the product of two ormore other variables.Thus, the equation is read “y varies jointly as x and z.”

Joint variation plays a critical role in Isaac Newton’s formula for gravitation:

The formula states that the force of gravitation, , between two bodies varies jointlyas the product of their masses, and and inversely as the square of thedistance between them, d. (G is the gravitational constant.) The formula indicatesthat gravitational force exists between any two objects in the universe, increasing asthe distance between the bodies decreases. One practical result is that the pull of themoon on the oceans is greater on the side of Earth closer to the moon. Thisgravitational imbalance is what produces tides.

EXAMPLE 5 Modeling Centrifugal Force

The centrifugal force, C, of a body moving in a circle varies jointly with the radius ofthe circular path, r, and the body’s mass, m, and inversely with the square of the time,t, it takes to move about one full circle.A 6-gram body moving in a circle with radius100 centimeters at a rate of 1 revolution in 2 seconds has a centrifugal force of 6000dynes. Find the centrifugal force of an 18-gram body moving in a circle with radius100 centimeters at a rate of 1 revolution in 3 seconds.

m2 ,m1

F

F = G

m1 m2

d2 .

y = kxz

CheckPoint 4

S = 14,000

S =

8170,000240

S =

8A

P

P = 40.A = 70,000.

S �kAP

.S =

8A

P.

8 = k

12,0001500

=

k # 15001500

12,000 = k # 1500

1S � 12,0002.1P � 402,1A � 60,0002 12,000 =

k160,000240

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Solution

Translate “Centrifugal force, C, varies jointly withradius, r, and mass, m, and inversely with thesquare of time, t.”

6000 =

k1100216222

C =

krm

t2

EXERCISE SET 2.810. y varies jointly as m and the square of n and inversely as p.

when and Find y whenand

Practice PlusIn Exercises 11–20, write an equation that expresses

each relationship. Then solve the equation for y.

11. x varies jointly as y and z.

12. x varies jointly as y and the square of z.

13. x varies directly as the cube of z and inversely as y.

14. x varies directly as the cube root of z and inversely as y.

15. x varies jointly as y and z and inversely as the square rootof

16. x varies jointly as y and z and inversely as the square of

17. x varies jointly as z and the sum of y and

18. x varies jointly as z and the difference between y and

19. x varies directly as z and inversely as the difference betweeny and

20. x varies directly as z and inversely as the sum of y and w.

w.

w.

w.

w.

w.

p = 10.m = 3, n = 4,p = 6.m = 2, n = 1,y = 15

Practice ExercisesUse the four-step procedure for solving variation

problems given on page 358 to solve Exercises 1–10.

1. y varies directly as x. when Find y when

2. y varies directly as x. when Find y when

3. y varies inversely as x. when Find y when

4. y varies inversely as x. when Find y when

5. y varies directly as x and inversely as the square of z.when and Find y when and

6. a varies directly as b and inversely as the square of c.when and Find a when and

7. y varies jointly as x and z. when and Find y when and

8. C varies jointly as A and T. when andFind C when and

9. y varies jointly as a and b and inversely as the square root ofc. when and Find y when

and c = 9.a = 5, b = 3,c = 25.a = 3, b = 2,y = 12

T = 6.A = 2400T = 4.A = 2100C = 175

z = 12.x = 8z = 5.x = 2y = 25

c = 8.b = 4c = 6.b = 9a = 7

z = 6.x = 3z = 5.x = 50y = 20

x = 9.x = 3.y = 6

x = 2.x = 5.y = 12

x = 13.x = 5.y = 45

x = 12.x = 5.y = 65

A 6-gram body moving in a circle withradius 100 centimeters at 1 revolution in2 seconds has a centifugal force of 6000dynes

Simplify.

Divide both sides by 150 and solve for k.

Substitute 40 for k in the model for centrifugal force.

C =

4011002118232

C =

40rm

t2

40 = k

6000 = 150k

1C � 60002.1t � 22

1r � 10021m � 62

Find centifugal force, C, of an 18-gram bodymoving in a circle with radius 100

centimeters at 1 revolution in 3 seconds

Simplify.

The centrifugal force is 8000 dynes.

The volume of a cone, V, varies jointly as its height, h, and the square of itsradius, r. A cone with a radius measuring 6 feet and a height measuring 10feet has a volume of cubic feet. Find the volume of a cone having aradius of 12 feet and a height of 2 feet.

120p

CheckPoint 5

= 8000

1t � 32.1r � 1002

1m � 182

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Application ExercisesUse the four-step procedure for solving variation prob-lems given on page 358 to solve Exercises 21–36.

21. An alligator’s tail length, T, varies directly as its body length,B.An alligator with a body length of 4 feet has a tail length of3.6 feet. What is the tail length of an alligator whose bodylength is 6 feet?

22. An object’s weight on the moon, M, varies directly as itsweight on Earth, E. Neil Armstrong, the first person to stepon the moon on July 20, 1969, weighed 360 pounds on Earth(with all of his equipment on) and 60 pounds on the moon.What is the moon weight of a person who weighs 186 poundson Earth?

23. The height that a ball bounces varies directly as the heightfrom which it was dropped. A tennis ball dropped from 12inches bounces 8.4 inches. From what height was the tennisball dropped if it bounces 56 inches?

24. The distance that a spring will stretch varies directly as theforce applied to the spring. A force of 12 pounds is needed tostretch a spring 9 inches. What force is required to stretch thespring 15 inches?

25. If all men had identical body types, their weight would varydirectly as the cube of their height. Shown below is RobertWadlow, who reached a record height of 8 feet 11 inches (107inches) before his death at age 22. If a man who is 5 feet 10inches tall (70 inches) with the same body type as Mr. Wad-low weighs 170 pounds, what was Robert Wadlow’s weightshortly before his death?

26. On a dry asphalt road, a car’s stopping distance varies directlyas the square of its speed.A car traveling at 45 miles per hourcan stop in 67.5 feet. What is the stopping distance for a cartraveling at 60 miles per hour?

Tail length, TBody length, B

27. The figure shows that a bicyclist tips the cycle when making aturn. The angle B, formed by the vertical direction and thebicycle, is called the banking angle. The banking angle variesinversely as the cycle’s turning radius. When the turningradius is 4 feet, the banking angle is 28°. What is the bankingangle when the turning radius is 3.5 feet?

28. The water temperature of the Pacific Ocean varies inverselyas the water’s depth. At a depth of 1000 meters, the watertemperature is 4.4° Celsius. What is the water temperature ata depth of 5000 meters?

29. Radiation machines, used to treat tumors, produce an intensityof radiation that varies inversely as the square of the distancefrom the machine. At 3 meters, the radiation intensity is 62.5milliroentgens per hour. What is the intensity at a distance of2.5 meters?

30. The illumination provided by a car’s headlight varies inverselyas the square of the distance from the headlight. A car’sheadlight produces an illumination of 3.75 footcandles at adistance of 40 feet. What is the illumination when thedistance is 50 feet?

31. Body-mass index, or BMI, takes both weight and height intoaccount when assessing whether an individual is underweightor overweight. BMI varies directly as one’s weight, in pounds,and inversely as the square of one’s height, in inches. Inadults, normal values for the BMI are between 20 and 25,inclusive. Values below 20 indicate that an individual isunderweight and values above 30 indicate that an individualis obese. A person who weighs 180 pounds and is 5 feet, or 60inches, tall has a BMI of 35.15. What is the BMI, to the near-est tenth, for a 170 pound person who is 5 feet 10 inches tall.Is this person overweight?

32. One’s intelligence quotient, or IQ, varies directly as a person’smental age and inversely as that person’s chronological age. Aperson with a mental age of 25 and a chronological age of 20has an IQ of 125. What is the chronological age of a personwith a mental age of 40 and an IQ of 80?

33. The heat loss of a glass window varies jointly as the window’sarea and the difference between the outside and inside tem-peratures.A window 3 feet wide by 6 feet long loses 1200 Btuper hour when the temperature outside is 20° colder than thetemperature inside. Find the heat loss through a glass windowthat is 6 feet wide by 9 feet long when the temperature outsideis 10° colder than the temperature inside.

B°B°

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34. Kinetic energy varies jointly as the mass and the square of thevelocity. A mass of 8 grams and velocity of 3 centimeters persecond has a kinetic energy of 36 ergs. Find the kinetic energyfor a mass of 4 grams and velocity of 6 centimeters per second.

35. Sound intensity varies inversely as the square of the distancefrom the sound source. If you are in a movie theater and youchange your seat to one that is twice as far from the speakers,how does the new sound intensity compare to that of youroriginal seat?

36. Many people claim that as they get older, time seems to passmore quickly. Suppose that the perceived length of a period oftime is inversely proportional to your age. How long will a yearseem to be when you are three times as old as you are now?

37. The average number of daily phone calls, C, between two citiesvaries jointly as the product of their populations, and and inversely as the square of the distance, d, between them.

a. Write an equation that expresses this relationship.

b. The distance between San Francisco (population: 777,000)and Los Angeles (population: 3,695,000) is 420 miles. Ifthe average number of daily phone calls between thecities is 326,000, find the value of k to two decimal placesand write the equation of variation.

c. Memphis (population: 650,000) is 400 miles from New Or-leans (population: 490,000). Find the average number ofdaily phone calls, to the nearest whole number, betweenthese cities.

38. The force of wind blowing on a window positioned at a rightangle to the direction of the wind varies jointly as the area ofthe window and the square of the wind’s speed. It is knownthat a wind of 30 miles per hour blowing on a window mea-suring 4 feet by 5 feet exerts a force of 150 pounds. During astorm with winds of 60 miles per hour, should hurricane shut-ters be placed on a window that measures 3 feet by 4 feet andis capable of withstanding 300 pounds of force?

39. The table shows the values for the current, I, in an electriccircuit and the resistance, R, of the circuit.

I (amperes) 0.5 1.0 1.5 2.0 2.5 3.0 4.0 5.0

R (ohms) 12.0 6.0 4.0 3.0 2.4 2.0 1.5 1.2

a. Graph the ordered pairs in the table of values, with valuesof I along the x-axis and values of R along the y-axis.Connect the eight points with a smooth curve.

b. Does current vary directly or inversely as resistance? Useyour graph and explain how you arrived at your answer.

c. Write an equation of variation for I and R, using one ofthe ordered pairs in the table to find the constant of varia-tion. Then use your variation equation to verify the otherseven ordered pairs in the table.

Writing in Mathematics40. What does it mean if two quantities vary directly?41. In your own words, explain how to solve a variation problem.42. What does it mean if two quantities vary inversely?43. Explain what is meant by combined variation. Give an

example with your explanation.44. Explain what is meant by joint variation. Give an example

with your explanation.

P2 ,P1

In Exercises 45–46, describe in words the variation shown by thegiven equation.

45. 46.

47. We have seen that the daily number of phone calls betweentwo cities varies jointly as their populations and inversely asthe square of the distance between them.This model, used bytelecommunication companies to estimate the line capacitiesneeded among various cities, is called the gravity model.Compare the model to Newton’s formula for gravitationon page 363 and describe why the name gravity model isappropriate.

Technology Exercise48. Use a graphing utility to graph any three of the variation

equations in Exercises 21–30. Then along eachcurve and identify the point that corresponds to the prob-lem’s solution.

Critical Thinking Exercises49. In a hurricane, the wind pressure varies directly as the square

of the wind velocity. If wind pressure is a measure of ahurricane’s destructive capacity, what happens to thisdestructive power when the wind speed doubles?

50. The illumination from a light source varies inversely as thesquare of the distance from the light source. If you raise alamp from 15 inches to 30 inches over your desk, what hap-pens to the illumination?

51. The heat generated by a stove element varies directly as thesquare of the voltage and inversely as the resistance. If thevoltage remains constant, what needs to be done to triple theamount of heat generated?

52. Galileo’s telescope brought about revolutionary changes inastronomy. A comparable leap in our ability to observe theuniverse took place as a result of the Hubble Space Tele-scope. The space telescope was able to see stars and galaxieswhose brightness is of the faintest objects observable usingground-based telescopes. Use the fact that the brightness of apoint source, such as a star, varies inversely as the square ofits distance from an observer to show that the space tele-scope was able to see about seven times farther than aground-based telescope.

Group Exercise53. Begin by deciding on a product that interests the group

because you are now in charge of advertising this product.Members were told that the demand for the product variesdirectly as the amount spent on advertising and inversely asthe price of the product. However, as more money is spent onadvertising, the price of your product rises. Under whatconditions would members recommend an increased expensein advertising? Once you’ve determined what your product is,write formulas for the given conditions and experiment withhypothetical numbers. What other factors might you take intoconsideration in terms of your recommendation? How dothese factor affect the demand for your product?

150

� TRACE �

z = kx21yz =

k1x

y2

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Summary, Review, and Test 367

Chapter 2Summary, Review, and TestSummary

DEFINITIONS AND CONCEPTS EXAMPLES

c. The graph of a quadratic function is a parabola. The vertex is or A procedure for

graphing a quadratic function is given in the box on page 275.

a - b

2a, fa -

b

2ab b .1h, k2 Ex. 1, p. 275;

Ex. 2, p. 276;Ex. 3, p. 277

d. See the box on page 279 for minimum or maximum values of quadratic functions. Ex. 4, p. 279;Ex. 5, p. 280

e. A strategy for solving problems involving maximizing or minimizing quadratic functions is given in the boxon page 281.

Ex. 6, p. 281;Ex. 7, p. 283

2.3 Polynomial Functions and Their Graphs

a. Polynomial Function of Degree n:

b. The graphs of polynomial functions are smooth and continuous.

an Z 0f1x2 = an xn+ an - 1 xn - 1

+Á

+ a2 x2+ a1 x + a0 ,

Fig. 2.12, p. 289

c. The end behavior of the graph of a polynomial function depends on the leading term, given by the LeadingCoefficient Test in the box on page 290.

Ex. 1, p. 290;Ex. 2, p. 290;Ex. 3, p. 291

d. The values of x for which is equal to 0 are the zeros of the polynomial function . These values are theroots, or solutions, of the polynomial equation f1x2 = 0.

ff1x2 Ex. 4, p. 292;Ex. 5, p. 293

e. If occurs k times in a polynomial function’s factorization, r is a repeated zero with multiplicity k. If kis even, the graph touches the x-axis and turns around at r. If k is odd, the graph crosses the x-axis at r.

x - r Ex. 6, p. 294

f. The Intermediate Value Theorem: If is a polynomial function and and have opposite signs,there is at least one value of c between a and b for which f1c2 = 0.

f1b2f1a2f Ex. 7, p. 294

g. If is a polynomial of degree n, the graph of has at most turning points.n - 1ff Fig. 2.22, p. 295

h. A strategy for graphing a polynomial function is given in the box on page 295. Ex. 8, p. 296

2.1 Complex Numbers

a. The imaginary unit i is defined as

The set of numbers in the form is called the set of complex numbers; a is the real part and b is theimaginary part. If the complex number is a real number. If the complex number is an imagi-nary number. Complex numbers in the form bi are called pure imaginary numbers.

b Z 0,b = 0,a + bi

i = 2-1, where i2= -1.

b. Rules for adding and subtracting complex numbers are given in the box on page 267. Ex. 1, p. 267

c. To multiply complex numbers, multiply as if they are polynomials. After completing the multiplication,replace with and simplify.-1i2

Ex. 2, p. 268

d. The complex conjugate of is and vice versa. The multiplication of complex conjugates gives areal number: 1a + bi21a - bi2 = a2

+ b2.

a - bia + bi

2.2 Quadratic Functions

a. A quadratic function is of the form

b. The standard form of a quadratic function is f1x2 = a1x - h22 + k, a Z 0.

f1x2 = ax2+ bx + c, a Z 0.

e. To divide complex numbers, multiply the numerator and the denominator by the complex conjugate of thedenominator.

Ex. 3, p. 269

f. When performing operations with square roots of negative numbers, begin by expressing all square roots interms of i. The principal square root of is defined by2-b = i2b .

-b

g. Quadratic equations with negative discriminants have imaginarysolutions that are complex conjugates.

1b2- 4ac 6 021ax2

+ bx + c = 0, a Z 02 Ex. 5, p. 271

Ex. 4, p. 270

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2.6 Rational Functions and Their Graphs

a. Rational function: p and q are polynomial functions and The domain of is the set

of all real numbers excluding values of x that make zero.q1x2fq1x2 Z 0.f1x2 =

p1x2q1x2 ; Ex. 1, p. 326

b. Arrow notation is summarized in the box on page 328.

c. The line is a vertical asymptote of the graph of if increases or decreases without bound as xapproaches a. Vertical asymptotes are identified using the location theorem in the box on page 329.

f1x2fx = a Ex. 2, p. 330

d. The line is a horizontal asymptote of the graph of if approaches b as x increases or decreaseswithout bound. Horizontal asymptotes are identified using the location theorem in the box on page 332.

f1x2fy = b Ex. 3, p. 332

e. Table 2.2 on page 333 shows the graphs of and Some rational functions can be

graphed using transformations of these common graphs.

f1x2 =

1

x2 .f1x2 =

1x

Ex. 4, p. 334

f. A strategy for graphing rational functions is given in the box on page 334. Ex. 5, p. 335;Ex. 6, p. 336;Ex. 7, p. 337

g. The graph of a rational function has a slant asymptote when the degree of the numerator is one more thanthe degree of the denominator. The equation of the slant asymptote is found using division and droppingthe remainder term.

Ex. 8, p. 339

2.7 Polynomial and Rational Inequalities

a. A polynomial inequality can be expressed as or where is apolynomial function. A procedure for solving polynomial inequalities is given in the box on page 347.

ff1x2 Ú 0,f1x2 6 0, f1x2 7 0, f1x2 … 0, Ex. 1, p. 347;Ex. 2, p. 348

b. Number of roots: If is a polynomial of degree then, counting multiple roots separately, theequation has n roots.

c. If is a root of then is also a root.

d. The Linear Factorization Theorem: An nth-degree polynomial can be expressed as the product of n linearfactors. Thus, f1x2 = an1x - c121x - c22Á 1x - cn2.

a - bif1x2 = 0,a + bi

f1x2 = 0n Ú 1,f1x2

Ex. 6, p. 319

e. Descartes’s Rule of Signs:The number of positive real zeros of equals the number of sign changes of or is less than that number by an even integer. The number of negative real zeros of applies a similarstatement to f1-x2.

ff1x2f Table 2.1, p. 320;

Ex. 7, p. 321


2.4 Dividing Polynomials; Remainder and Factor Theorems

a. Long division of polynomials is performed by dividing, multiplying, subtracting, bringing down the nextterm, and repeating this process until the degree of the remainder is less than the degree of the divisor. Thedetails are given in the box on page 303.

Ex. 1, p. 302;Ex. 2, p. 303;Ex. 3, p. 305

b. The Division Algorithm: The dividend is the product of the divisor and thequotient plus the remainder.

c. Synthetic division is used to divide a polynomial by The details are given in the box on page 306.x - c.

f1x2 = d1x2q1x2 + r1x2.Ex. 4, p. 306

d. The Remainder Theorem: If a polynomial is divided by then the remainder is f1c2.x - c,f1x2 Ex. 5, p. 308

e. The Factor Theorem: If is a factor of a polynomial function , then c is a zero of and a root ofIf c is a zero of or a root of then is a factor of .f1x2x - cf1x2 = 0,ff1x2 = 0.

ff1x2x - c Ex. 6, p. 309

2.5 Zeros of Polynomial Functions

a. The Rational Zero Theorem states that the possible rational zeros of a polynomial

The theorem is stated in the box on page 313.function =


.

Ex. 1, p. 313;Ex. 2, p. 314;Ex. 3, p. 314;Ex. 4, p. 315;Ex. 5, p. 316

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Review Exercises 369

2.8 Modeling Using Variation

a. A procedure for solving variation problems is given in the lower box on page 358.

b. English Statement Equation

y varies directly as x. Ex. 1, p. 358y is directly proportional to x.

y varies directly as Ex. 2, p. 359y is directly proportional to

y varies inversely as x. Ex. 3, p. 361;y =

k

x

xn.y = kxnxn.

y = kx

Review Exercises

5

6

4

3

2

Div

orce

Rat

e pe

r10

00 P

opul

atio

n

Year

U.S. Divorce Rate

1960 1970 1980 1990

2003

2000

1

Source: National Center for Health Statistics

b. A rational inequality can be expressed as or where is a rationalfunction. The procedure for solving such inequalities begins with expressing them so that one side is zeroand the other side is a single quotient. Find boundary points by setting the numerator and denominatorequal to zero. Then follow a procedure similar to that for solving polynomial inequalities.

ff1x2 Ú 0,f1x2 6 0, f1x2 7 0, f1x2 … 0, Ex. 3, p. 350

2.1

In Exercises 1–10 perform the indicated operations and write theresult in standard form.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

Ire Exercises 11–12, solve each quadratic equation using thequadratic formula. Express solutions in standard form.

11. 12.

2.2In Exercises 13–16, use the vertex and intercepts to sketch thegraph of each quadratic function. Give the equation for theparabola’s axis of symmetry. Use the graph to determine the func-tion’s domain and range.

13. 14.

15. 16.

In Exercises 17–18, use the function’s equation, and not its graph, tofind

a. the minimum or maximum value and where it occurs.

b. the function’s domain and its range.

f1x2 = 2x2- 4x - 6f1x2 = -x2

+ 2x + 3

f1x2 = 1x + 422 - 2f1x2 = -1x + 122 + 4

2x2- 6x + 5 = 0x2

- 2x + 4 = 0

4 + 2-82

A -2 + 2-100 B22-32 - 2-18

3 + 4i

4 - 2i

65 + i

17 + 8i217 - 8i213 - 4i2217 - i212 + 3i24i13i - 2218 - 3i2 - 117 - 7i2

17. 18.

19. The function

models the yearly growth of a young redwood tree, , ininches, with x inches of rainfall per year. How many inches ofrainfall per year result in maximum tree growth? What is themaximum yearly growth?

20. Suppose that a quadratic function is used to model the datashown in the graph using

f1x2f1x2 = -0.02x2

+ x + 1

f1x2 = 2x2+ 12x + 703f1x2 = -x2

+ 14x - 106

y is inversely proportional to x. Ex. 4, p. 362

y varies inversely as y =

k

xn

xn.y is inversely proportional to

y varies jointly as x and z. Ex. 5, p. 363y = kxz

xn.

(number of years after 1960, divorce rate per 1000 population).

Determine, without obtaining an actual quadratic functionthat models the data, the approximate coordinates of the ver-tex for the function’s graph. Describe what this means inpractical terms.

pr02-265-374.I-hr 1/26/06 5:14 PM Page 369


21. A field bordering a straight stream is to be enclosed.The sidebordering the stream is not to be fenced. If 1000 yards offencing material is to be used, what are the dimensions of thelargest rectangular field that can be fenced? What is the max-imum area?


23. You have 1000 feet of fencing to construct six corrals, asshown in the figure. Find the dimensions that maximize theenclosed area. What is the maximum area?

24. The annual yield per fruit tree is fairly constant at 150pounds per tree when the number of trees per acre is 35 orfewer. For each additional tree over 35, the annual yield pertree for all trees on the acre decreases by 4 pounds due toovercrowding. How many fruit trees should be planted peracre to maximize the annual yield for the acre? What is themaximum number of pounds of fruit per acre?

2.3In Exercises 25–28, use the Leading Coefficient Test to determinethe end behavior of the graph of the given polynomial function.Then use this end behavior to match the polynomial function withits graph. [The graphs are labeled (a) through (d).]

25. 26.

27. 28.

a. b.

c. d.

29. The polynomial function

models the number of thefts, , in thousands, in the UnitedStates x years after 1987. Will this function be useful inmodeling the number of thefts over an extended period oftime? Explain your answer.

f1x2f1x2 = -0.87x3

+ 0.35x2+ 81.62x + 7684.94

y

x

y

x

y

x

y

x

f1x2 = -x4+ 1f1x2 = x5

- 5x3+ 4x

f1x2 = x6- 6x4

+ 9x2f1x2 = -x3+ x2

+ 2x

x

y

30 A herd of 100 elk is introduced to a small island.The number ofelk, , after x years is modeled by the polynomial function

Use the Leading Coefficient Test to determine the graph’send behavior to the right. What does this mean about whatwill eventually happen to the elk population?

In Exercises 31–32, find the zeros for each polynomial function andgive the multiplicity of each zero. State whether the graph crosses thex-axis, or touches the x-axis and turns around, at each zero.

31.

32.

33. Show that has a real zero between 1 and 2.

In Exercises 34–39,a. Use the Leading Coefficient Test to determine the graph’s

end behavior.b. Determine whether the graph has y-axis symmetry, origin

symmetry, or neither.c. Graph the function.

34.

35.

36.

37.

38.

39.

In Exercises 40–41, graph each polynomial function.

40.

41.

2.4In Exercises 42–44, divide using long division.

42.

43.

44.

In Exercises 45–46, divide using synthetic division.

45.

46.

47. Given use the RemainderTheorem to find

48. Use synthetic division to divide by Use the result to find all zeros of .

49. Solve the equation given that 4 is a root.

2.5In Exercises 50–51, use the Rational Zero Theorem to list all pos-sible rational zeros for each given function.

50.

51. f1x2 = 3x5- 2x4

- 15x3+ 10x2

+ 12x - 8

f1x2 = x4- 6x3

+ 14x2- 14x + 5

x3- 17x + 4 = 0

fx - 2.f1x2 = 2x3

+ x2- 13x + 6

f1-132.f1x2 = 2x3

- 7x2+ 9x - 3,

13x4- 2x2

- 10x2 , 1x - 2213x4

+ 11x3- 20x2

+ 7x + 352 , 1x + 52

14x4+ 6x3

+ 3x - 12 , 12x2+ 12

110x3- 26x2

+ 17x - 132 , 15x - 3214x3

- 3x2- 2x + 12 , 1x + 12

f1x2 = -x31x + 4221x - 12f1x2 = 2x21x - 1231x + 22

f1x2 = 3x4- 15x3

f1x2 = -x4+ 6x3

- 9x2

f1x2 = -x4+ 25x2

f1x2 = 2x3+ 3x2

- 8x - 12

f1x2 = 4x - x3

f1x2 = x3- x2

- 9x + 9

f1x2 = x3- 2x - 1

f1x2 = x3- 5x2

- 25x + 125

f1x2 = -21x - 121x + 2221x + 523

f1x2 = -x4+ 21x2

+ 100.

f1x2

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Review Exercises 371

In Exercises 52–53, use Descartes’s Rule of Signs to determine thepossible number of positive and negative real zeros for each givenfunction.

52.

53.

54. Use Descartes’s Rule of Signs to explain whyhas no real roots.

For Exercises 55–61,

a. List all possible rational roots or rational zeros.b. Use Descartes’s Rule of Signs to determine the possible

number of positive and negative real roots or real zeros.c. Use synthetic division to test the possible rational roots or

zeros and find an actual root or zero.d. Use the quotient from part (c) to find all the remaining

zeros or roots.

55.

56.

57.

58.

59.

60.

61.

In Exercises 62–63, find an nth-degree polynomial function withreal coefficients satisfying the given conditions. If you are using agraphing utility, graph the function and verify the real zeros andthe given function value.

62. 2 and are zeros;

63. i is a zero; is a zero of multiplicity 2;

In Exercises 64–65, find all the zeros of each polynomial functionand write the polynomial as a product of linear factors.

64.

65.

In Exercises 66–69, graphs of fifth-degree polynomial functionsare shown. In each case, specify the number of real zeros and thenumber of imaginary zeros. Indicate whether there are any realzeros with multiplicity other than 1.

66. 67.

68. 69. y

x

y

x

y

x

y

x

g1x2 = x4- 6x3

+ x2+ 24x + 16

f1x2 = 2x4+ 3x3

+ 3x - 2

f1-12 = 16-3n = 4;

f112 = -102 - 3in = 3;

f1x2 = 2x4+ x3

- 9x2- 4x + 4

4x4+ 7x2

- 2 = 0

x4- x3

- 7x2+ x + 6 = 0

2x3+ 9x2

- 7x + 1 = 0

8x3- 36x2

+ 46x - 15 = 0

f1x2 = 6x3+ x2

- 4x + 1

f1x2 = x3+ 3x2

- 4

2x4+ 6x2

+ 8 = 0

f1x2 = 2x5- 3x3

- 5x2+ 3x - 1

f1x2 = 3x4- 2x3

- 8x + 5

2.6

In Exercises 70–71, use transformations of or

to graph each rational function.

70. 71.

In Exercises 72–79, find the vertical asymptotes, if any, thehorizontal asymptote, if one exists, and the slant asymptote, ifthere is one, of the graph of each rational function. Then graphthe rational function.

72. 73.

74. 75.

76. 77.

78. 79.

80. A company is planning to manufacture affordable graphingcalculators. The fixed monthly cost will be $50,000 and it willcost $25 to produce each calculator.a. Write the cost function, C, of producing x graphing

calculators.b. Write the average cost function, of producing x graph-

ing calculators.c. Find and interpret and

d. What is the horizontal asymptote for the graph of thisfunction and what does it represent?

81. In Palo Alto, California, a government agency orderedcomputer-related companies to contribute to a monetary poolto clean up underground water supplies. (The companies hadstored toxic chemicals in leaking underground containers.)The rational function

models the cost, in tens of thousands of dollars, forremoving x percent of the contaminants.a. Find and interpret b. What is the equation for the vertical asymptote? What

does this mean in terms of the variables given by thefunction?

Exercises 82–83 involve rational functions that model the givensituations. In each case, find the horizontal asymptote as and then describe what this means in practical terms.

82. the number of bass, , after x

months in a lake that was stocked with 120 bass

83. the percentage, of people in the

United States with x years of education who are unemployed

84. The function models the number of non-violent prisoners, in thousands, in New York State prisonsx years after 1980. The function modelsthe total number of prisoners, in thousands, in New YorkState prisons x years after 1980.

q1x2,q1x2 = 3.04x + 21.79

p1x2,p1x2 = 1.96x + 3.14

P1x2,P1x2 =

72,900

100x2+ 729

;

f1x2f1x2 =

150x + 1200.05x + 1

;

x : q

C1902 - C1502.C1x2,

C1x2 =

200x

100 - x

C1100,0002.C1502, C11002, C110002,

C,

g1x2 =

4x2- 16x + 162x - 3

f1x2 =

-2x3

x2+ 1

y =

x2+ 2x - 3x - 3

y =

x2

x + 1

r1x2 =

x2+ 4x + 3

1x + 222h1x2 =

x2- 3x - 4

x2- x - 6

g1x2 =

2x - 4x + 3

f1x2 =

2x

x2- 9

h1x2 =

1x - 1

+ 3g1x2 =

1

1x + 222 - 1

f1x2 =

1

x2

f1x2 =

1x

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Chapter 2 Test

In Exercises 1–3, perform the indicated operations and write theresult in standard form.

1. 2.

3.4. Solve and express solutions in standard form:

In Exercises 5–6, use the vertex and intercepts to sketch the graphof each quadratic function. Give the equation for the parabola’saxis of symmetry. Use the graph to determine the function’sdomain and range.

5. 6.

7. Determine, without graphing, whether the quadratic functionhas a minimum value or a maxi-

mum value. Then finda. the minimum or maximum value and where it occurs.b. the function’s domain and its range.

8. The function models the dailyprofit, , in hundreds of dollars, for a company that manu-factures x computers daily. How many computers should bemanufactured each day to maximize profit? What is the max-imum daily profit?

f1x2f1x2 = -x2

+ 46x - 360

f1x2 = -2x2+ 12x - 16

f1x2 = x2- 2x - 3f1x2 = 1x + 122 + 4

x2= 4x - 8.

21-49 + 31-64

52 - i

16 - 7i212 + 5i2

9. Among all pairs of numbers whose sum is 14, find a pair whoseproduct is as large as possible. What is the maximum product?

10. Consider the function

a. Use factoring to find all zeros of .

b. Use the Leading Coefficient Test and the zeros of tograph the function.

11. Use end behavior to explain why the graph shown belowcannot be the graph of Then use intercepts toexplain why the graph cannot represent

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

y

x

f1x2 = x5- x.

f1x2 = x5- x.

f

f

f1x2 = x3- 5x2

- 4x + 20.

a. Write a function that models the fraction of nonviolentprisoners in New York State prisons x years after 1980.

b. What is the equation of the horizontal asymptote associatedwith the function in part (a)? Describe what this meansabout the percentage, to the nearest tenth of a percent, ofnonviolent prisoners in New York State prisons over time.

c. Use your equation in part (b) to explain why, in 1998, NewYork State implemented a strategy where more nonviolentoffenders are granted parole and more violent offendersare denied parole.

85. A jogger ran 4 miles and then walked 2 miles. The averagevelocity running was 3 miles per hour faster than the averagevelocity walking. Express the total time for running andwalking, T, as a function of the average velocity walking, x.

86. The area of a rectangular floor is 1000 square feet. Expressthe perimeter of the floor, P, as a function of the width of therectangle, x.

2.7In Exercises 87–92, solve each inequality and graph the solutionset on a real number line.

87. 88.

89. 90.

91. 92.

93. Use the position function

s1t2 = -16t2+ v0 t + s0

x + 3x - 4

… 51x + 121x - 22

x - 1Ú 0

x - 6x + 2

7 0x3+ 2x2

7 3x

2x2+ 9x + 4 Ú 02x2

+ 5x - 3 6 0

to solve this problem. A projectile is fired vertically upwardfrom ground level with an initial velocity of 48 feet persecond. During which time period will the projectile’s heightexceed 32 feet?

2.8Solve the variation problems in Exercises 94–99.

94. An electric bill varies directly as the amount of electricityused. The bill for 1400 kilowatts of electricity is $98. What isthe bill for 2200 kilowatts of electricity?

95. The distance that a body falls from rest is directly propor-tional to the square of the time of the fall. If skydivers fall 144feet in 3 seconds, how far will they fall in 10 seconds?

96. The time it takes to drive a certain distance is inverselyproportional to the rate of travel. If it takes 4 hours at 50miles per hour to drive the distance, how long will it take at40 miles per hour?

97. The loudness of a stereo speaker, measured in decibels,varies inversely as the square of your distance from thespeaker. When you are 8 feet from the speaker, the loudnessis 28 decibels. What is the loudness when you are 4 feet fromthe speaker?

98. The time required to assemble computers varies directly asthe number of computers assembled and inversely as thenumber of workers. If 30 computers can be assembled by 6workers in 10 hours, how long would it take 5 workers toassemble 40 computers?

99. The volume of a pyramid varies jointly as its height and thearea of its base.A pyramid with a height of 15 feet and a basewith an area of 35 square feet has a volume of 175 cubic feet.Find the volume of a pyramid with a height of 20 feet and abase with an area of 120 square feet.

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Chapter 2 Test 373

12. The graph of is shown in thefigure.

a. Based on the graph of , find the root of the equationthat is an integer.

b. Use synthetic division to find the other two roots of

13. Use the Rational Zero Theorem to list all possible rationalzeros of

14. Use Descartes’s Rule of Signs to determine the possiblenumber of positive and negative real zeros of

15. Solve:

16. Consider the function whose equation is given by

a. List all possible rational zeros.

b. Use the graph of in the figure shown and syntheticdivision to find all zeros of the function.

17. Use the graph of in the figure shownto factor

18. Find a fourth-degree polynomial function with realcoefficients that has 1, and i as zeros and such thatf132 = 160.

-1,f1x2

−1

123

−2−3

−5−4

1 2 3 4−1−2−3−4

y

x

f(x) = x3 + 3x2 − 4

x3+ 3x2

- 4.f1x2 = x3

+ 3x2- 4

−5

5101520

−10−15−20

1 2 3 4−1−2−3−4

y

x

f(x) = 2x4 − x3 − 13x2 + 5x + 15

f

f1x2 = 2x4- x3

- 13x2+ 5x + 15.

x3+ 9x2

+ 16x - 6 = 0.

f1x2 = 3x5- 2x4

- 2x2+ x - 1.

f1x2 = 2x3+ 11x2

- 7x - 6.

−3

−2

−1

1

4321

y

x

f(x) = 6x3 − 19x2 + 16x − 4

6x3- 19x2

+ 16x - 4 = 0.

6x3- 19x2

+ 16x - 4 = 0f

f1x2 = 6x3- 19x2

+ 16x - 4 19. The figure shows an incomplete graph ofFind all the zeros of the

function. Then draw a complete graph.

In Exercises 20–25, find the domain of each rational function andgraph the function.

20. 21.

22. 23.

24. 25.

26. A company is planning to manufacture pocket-sized televi-sions. The fixed monthly cost will be $300,000 and it will cost$10 to produce each television.

a. Write the average cost function, of producing xtelevisions.

b. What is the horizontal asymptote for the graph of thisfunction and what does it represent?

27. Rational functions can be used to model learning. Many ofthese functions model the proportion of correct responses asa function of the number of trials of a particular task. Onesuch model, called a learning curve, is

where is the proportion of correct responses after xtrials. If there are no correct responses. If

all responses are correct. The graph of the rationalfunction is shown.

Pro

port

ion

of C

orre

ct R

espo

nses

A Learning Curvey

x

Number of Learning Trials151413121110987654321

1.21.11.00.90.80.70.60.50.40.30.20.1

0.9x − 0.40.9x + 0.1f(x) =

f1x2 = 1,f1x2 = 0,

f1x2f1x2 =

0.9x - 0.40.9x + 0.1

,

C,

f1x2 =

4x2

x2+ 3

f1x2 =

x + 1

x2+ 2x - 3

f1x2 =

x2- 9

x - 2f1x2 =

x

x2- 16

f1x2 =

1x - 1

+ 2f1x2 =

1

1x + 322

x

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3−4−5

f1x2 = -3x3- 4x2

+ x + 2.

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Cumulative Review Exercises (Chapters P–2)Use the graph of to solve Exercises 1–6.

1. Find the domain and the range of .

2. Find the zeros and the least possible multiplicity of each zero.

3. Where does the relative maximum occur?

4. Find

5. Use arrow notation to complete this statement:as _____ or as _____.

6. Graph

In Exercises 7–12, solve each equation or inequality.

7. ƒ2x - 1 ƒ = 3

g1x2 = f1x + 22 + 1.

f1x2: q

1f � f21-12.

f

3

4

x

y

1 2

1

2

−1−2

y = f (x)

y = f1x2 8.

9.

10.

11.

12.

In Exercises 13–18, graph each equation in a rectangular coordi-nate system. If two functions are given, graph both in the samesystem.

13.

14.

15.

16.

17. and

18.

In Exercises 19–20, let and

19. Find

20. Find f1x + h2 - f1x2

h.

1f � g21x2.g1x2 = 4x - 1.f1x2 = 2x2

- x - 1

x2+ y2

- 2x + 4y - 4 = 0

g1x2 = - ƒx ƒ - 1f1x2 = ƒx ƒ

f1x2 =

x - 1x - 2

f1x2 = x21x - 32f1x2 = x2

+ 2x - 8

f1x2 = x3- 4x2

- x + 4

3x27 2x + 5

ƒ2x - 5 ƒ 7 3

x3+ 2x2

- 5x - 6 = 0

9 +

3x

=

2

x2

3x2- 5x + 1 = 0

a. According to the graph, what proportion of responses arecorrect after 5 learning trials?

b. According to the graph, how many learning trials arenecessary for 0.95 of the responses to be correct?

c. Use the function’s equation to write the equation of thehorizontal asymptote.What does this mean in terms of thevariables modeled by the learning curve?

Solve each inequality in Exercises 28–29 and graph the solutionset on a real number line. Express each solution set in intervalnotation.

28. 29.

30. The intensity of light received at a source varies inversely asthe square of the distance from the source. A particular lighthas an intensity of 20 foot-candles at 15 feet. What is thelight’s intensity at 10 feet?

2x + 1x - 3

… 3x26 x + 12

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polynomial and rational functions

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