4.6 RATIONAL FUNCTIONS - websites.rcc.eduwebsites.rcc.edu/magana/files/2015/06/Section-4.6.pdf4.6...

CONCEPTUAL OBJECTIVES

n Understand arrow notation.

n Interpret the behavior of the graph of a rational function

near an asymptote.

SKI LLS OBJECTIVES

n Find the domain of a rational function.

n Determine vertical, horizontal, and slant asymptotes of

rational functions.

n Graph rational functions.

Domain of Rational Functions

So far in this chapter we have discussed polynomial functions. We now turn our attention

to rational functions, which are ratios of polynomial functions. Ratios of integers are

called rational numbers. Similarly, ratios of polynomial functions are called rational

functions.

A function f (x) is a rational function if

where the numerator, n(x), and the denominator, d(x), are polynomial functions.

The domain of f (x) is the set of all real numbers x such that .

Note: If d(x) is a constant, then f (x) is a polynomial function.

d(x) Z 0

f (x) =n(x)

d(x) d(x) Z 0

Rational FunctionDE F I N IT ION

The domain of any polynomial function is the set of all real numbers. When we divide two

polynomial functions, the result is a rational function, and we must exclude any values of

x that make the denominator equal to zero.

RATIONAL FUNCTIONS

SECTION

4.6

445

EXAMPLE 1 Finding the Domain of a Rational Function

Find the domain of the rational function Express the domain in

interval notation.

Solution:

Set the denominator equal to zero. x2 x 6 5 0

Factor. (x! 2)(x 3) 5 0

Solve for x. x 5 2 or x 5 3

Eliminate these values from the domain. or

State the domain in interval notation.

n YOUR TURN Find the domain of the rational function

Express the domain in interval notation.

It is important to note that there are not always restrictions on the domain. For example, if

the denominator is never equal to zero, the domain is the set of all real numbers.

f (x) =x - 2

x2 - 3x - 4.

(-`, -2) (-2, 3) (3, `)

x Z 3x Z -2

f (x) =x + 1

x2 - x - 6.

n Answer: The domain is the set of

all real numbers such that

or Interval notation:

(-`, -1) (-1, 4) (4, `)

x Z 4.

x Z -1

446 CHAPTER 4 Polynomial and Rational Functions

EXAMPLE 2 When the Domain of a Rational Function Is the Set ofAll Real Numbers

Find the domain of the rational function Express the domain in interval notation.

Solution:

Set the denominator equal to zero. x2 ! 9 5 0

Subtract 9 from both sides. x25 9

Solve for x. x 5 3i or x 5 3i

There are no real solutions; therefore

the domain has no restrictions. R, the set of all real numbers

State the domain in interval notation.

n YOUR TURN Find the domain of the rational function Express the

domain in interval notation.

g(x) =5x

x2 + 4.

(-`, `)

g(x) =3x

x2 + 9.

n Answer: The domain is the set of

all real numbers. Interval

notation: ( `, `)

It is important to note that , where , and are not

the same function. Although f (x) can be written in the factored form

g(x) = x - 2x Z -2f (x) =x2 - 4

x + 2

, its domain is different. The domain of g(x) is the set of f (x) =(x - 2)(x + 2)

x + 2= x - 2

all real numbers, whereas the domain of f(x) is the set of all real numbers such that

If we were to plot f(x) and g(x), they would both look like the line y 5 x 2. However, f (x)

would have a hole, or discontinuity, at the point x 5 2.

x Z -2.

x

y

5

5–5

–5

hole

4.6 Rational Functions 447

x

y

(–1, –1)

(1, 1)

x1

f (x) =

x approaching 0 from

the left

x approaching

0 from the right

We cannot let x 5 0 because that point is not in the domain of the function. We should,

however, ask the question, “how does f(x) behave as x approaches zero?” Let us take values

that get closer and closer to x5 0, such as , , , . . . (See the table above.) We use an

arrow to represent the word approach, a positive superscript to represent from the right, and a

negative superscript to represent from the left. A plot of this function can be generated using

point-plotting techniques. The following are observations of the graph f (x) =1

x.

11000

1100

110

x 0

10 100 1000 undefined 1000 100 10f (x) 51

x

1

10

1

100

1

1000-

1

1000-

1

100-

1

10

x f (x) 5

10

1 1

1 1

10 110

110

1

x

Vertical, Horizontal, and Slant Asymptotes

If a function is not defined at a point, then it is still useful to know how the function behaves

near that point. Let’s start with a simple rational function, the reciprocal function

This function is defined everywhere except at x 5 0.f (x) =1

x.

WORDS MATH

As x approaches zero from the right,

the function f(x) increases without bound.

As x approaches zero from the left, the

function f (x) decreases without bound.

As x approaches infinity (increases without bound),

the function f(x) approaches zero from above.

As x approaches negative infinity (decreases without

bound), the function f (x) approaches zero from below.

The symbol ` does not represent an actual real number. This symbol represents growing

without bound.

1. Notice that the function is not defined at x5 0. The y-axis, or the vertical line x5 0,

represents the vertical asymptote.

2. Notice that the value of the function is never equal to zero. The x-axis is never

touched by the function. The x-axis, or y 5 0, is a horizontal asymptote.

1

xS 0!

xS !ˆ

1

xS 0"

xSˆ

1

xS !ˆ

xS 0!

1

xSˆ

xS 0"

x

y

(–1, –1)

(1, 1)

x1

f (x) =


The line x5 a is a vertical asymptote for the graph of a function if f(x) either increases

or decreases without bound as x approaches a from either the left or the right.

Vertical AsymptotesDE F I N IT ION

x

y

x

y

x

y

x

y

x = a

f (x)

x = af (x)

x = a

f (x)

x = af (x)

Asymptotes are lines that the graph of a function approaches. Suppose a football

team’s defense is its own 8 yard line and the team gets an “offsides” penalty that results in

loss of “half the distance to the goal.” Then the offense would get the ball on the 4 yard

line. Suppose the defense gets another penalty on the next play that results in “half the

distance to the goal.” The offense would then get the ball on the 2 yard line. If the defense

received 10 more penalties all resulting in “half the distance to the goal,” would the

referees give the offense a touchdown? No, because although the offense may appear to

be snapping the ball from the goal line, technically it has not actually reached the goal

line. Asymptotes utilize the same concept.

We will start with vertical asymptotes.Although the function had one vertical

asymptote, in general, rational functions can have none, one, or several vertical asymptotes. We

will first formally define what a vertical asymptote is and then discuss how to find it.

f (x) =1

x

Vertical asymptotes assist us in graphing rational functions since they essentially “steer”

the function in the vertical direction. How do we locate the vertical asymptotes of a

rational function? Set the denominator equal to zero. If the numerator and denominator

have no common factors, then any numbers that are excluded from the domain of a rational

function locate vertical asymptotes.

A rational function is said to be in lowest terms if the numerator n(x) and

denominator d(x) have no common factors. Let be a rational function in lowest

terms; then any zeros of the numerator n(x) correspond to x-intercepts of the graph of f, and

any zeros of the denominator d(x) correspond to vertical asymptotes of the graph of f. If a

rational function does have a common factor (is not in lowest terms), then the common

factor(s) should be canceled, resulting in an equivalent rational function R(x) in lowest

f (x) =n(x)

d(x)

f (x) =n(x)

d(x)


Let be a rational function in lowest terms (that is, assume n(x) and d(x) are

polynomials with no common factors); then the graph of f has a vertical asymptote

at any real zero of the denominator d(x). That is, if d(a) 5 0, then x 5 a corresponds

to a vertical asymptote on the graph of f.

Note: If f is a rational function that is not in lowest terms, then divide out the

common factors, resulting in a rational function R that is in lowest terms. Any

common factor x a of the function f corresponds to a hole in the graph of f at

x 5 a provided the multiplicity of a in the numerator is greater than or equal to the

multiplicity of a in the denominator.

f (x) =n(x)

d(x)

LOCATING VERTICAL ASYMPTOTES Study Tip

The vertical asymptotes of a rational

function in lowest terms occur at

x-values that make the denominator

equal to zero.

EXAMPLE 3 Determining Vertical Asymptotes

Locate any vertical asymptotes of the rational function

Solution:

Factor the denominator.

The numerator and denominator have no common factors.

Set the denominator equal to zero. and

Solve for x. and

The vertical asymptotes are and .

n YOUR TURN Locate any vertical asymptotes of the following rational function:

f (x) =3x - 1

2x2 - x - 15

x =23x = -

12

x =2

3x = -

1

2

3x - 2 = 02x + 1 = 0

f (x) =5x + 2

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

f (x) =5x + 2

6x2 - x - 2.

n Answer: and x 5 3x = -52

terms. If (x a)p is a factor of the numerator and (x a)q is a factor of the denominator,

then there is a hole in the graph at x 5 a provided p " q and x 5 a is a vertical asymptote

if p # q.


EXAMPLE 4 Determining Vertical Asymptotes When the RationalFunction Is Not in Lowest Terms

Locate any vertical asymptotes of the rational function

Solution:

Factor the denominator. x3 3x2 10x 5 x(x2 3x 10)

f (x) =x + 2

x3 - 3x2 - 10x.

We now turn our attention to horizontal asymptotes. As we have seen, rational functions

can have several vertical asymptotes. However, rational functions can have at most one

horizontal asymptote. Horizontal asymptotes imply that a function approaches a constant

value as x becomes large in the positive or negative direction. Another difference between

vertical and horizontal asymptotes is that the graph of a function never touches a vertical

asymptote but, as you will see in the next box, the graph of a function may cross a horizontal

asymptote, just not at the “ends” (x S ;`).

The line y5 b is a horizontal asymptote of the graph of a function if f(x) approaches

b as x increases or decreases without bound. The following are three examples:

As x S `, f (x)S b

Horizontal AsymptoteDE F I N IT ION

x

y

x

y

y = b

f (x)

y = b

f (x)

x

y

y = b

f (x)

n Answer: x 5 3

Note: A horizontal asymptote steers a function as x gets large. Therefore, when x is

not large, the function may cross the asymptote.

5 x(x 5)(x! 2)

Write the rational function in factored form.

Cancel (divide out) the common factor (x! 2).

Find the values when the denominator of R is

equal to zero. x 5 0 and x 5 5

The vertical asymptotes are and .

Note: x5 2 is not in the domain of f(x), even though there is no vertical asymptote there.

There is a “hole” in the graph at x5 2. Graphing calculators do not always show such “holes.”

n YOUR TURN Locate any vertical asymptotes of the following rational function:

f (x) =x2 - 4x

x2 - 7x + 12

x = 5x = 0

x Z -2R(x) =1

x(x - 5)

f (x) =(x + 2)

x(x - 5)(x + 2)

How do we determine whether a horizontal asymptote exists? And, if it does, how do we

locate it? We investigate the value of the rational function as or as . One of

two things will happen: either the rational function will increase or decrease without bound

or the rational function will approach a constant value.

We say that a rational function is proper if the degree of the numerator is less than the

degree of the denominator. Proper rational functions, like , approach zero as x gets

large. Therefore, all proper rational functions have the specific horizontal asymptote,

y5 0 (see Example 5a).

We say that a rational function is improper if the degree of the numerator is greater than

or equal to the degree of the denominator. In this case, we can divide the numerator by the

denominator and determine how the quotient behaves as x increases without bound.

n If the quotient is a constant (resulting when the degrees of the numerator and

denominator are equal), then as or as , the rational function

approaches the constant quotient (see Example 5b).

n If the quotient is a polynomial function of degree 1 or higher, then the quotient

depends on x and does not approach a constant value as x increases (see Example 5c).

In this case, we say that there is no horizontal asymptote.

We find horizontal asymptotes by comparing the degree of the numerator and the degree

of the denominator. There are three cases to consider:

1. The degree of the numerator is less than the degree of the denominator.

2. The degree of the numerator is equal to the degree of the denominator.

3. The degree of the numerator is greater than the degree of the denominator.

xS -`xS `

f (x) =1

x

xS -`xS `

Let f be a rational function given by

where n(x) and d(x) are polynomials.

1. When n # m, the x-axis (y5 0) is the horizontal asymptote.

2. When n5m, the line (ratio of leading coefficients) is the horizontal asymptote.

3. When n $ m, there is no horizontal asymptote.

y =an

bm

f (x) =n(x)

d(x)=

anxn+ an-1x

n-1+ Á + a1x + a0

bmxm+ bm-1x

m-1+ Á + b1x + b0

LOCATING HORIZONTAL ASYMPTOTES

In other words,

1. When the degree of the numerator is less than the degree of the denominator, then

y 5 0 is the horizontal asymptote.

2. When the degree of the numerator is the same as the degree of the denominator,

then the horizontal asymptote is the ratio of the leading coefficients.

3. If the degree of the numerator is greater than the degree of the denominator, then

there is no horizontal asymptote.


Thus far we have discussed linear asymptotes: vertical and horizontal. There are

three types of lines: horizontal (slope is zero), vertical (slope is undefined), and slant

(nonzero slope). Similarly, there are three types of linear asymptotes: horizontal, vertical,

and slant.


n Answer: is the horizontal

asymptote.

y = -74

Study Tip

There are three types of linear

asymptotes: horizontal, vertical,

and slant.

b. Graph .g(x) =8x2 + 3

4x2 + 1

c. Graph .h(x) =8x3 + 3

4x2 + 1

EXAMPLE 5 Finding Horizontal Asymptotes

Determine whether a horizontal asymptote exists for the graph of each of the given rational

functions. If it does, locate the horizontal asymptote.

a. b. c.

Solution (a):

The degree of the numerator 8x ! 3 is 1. n 5 1

The degree of the denominator 4x2 ! 1 is 2. m 5 2

The degree of the numerator is less than the

degree of the denominator. n # m

The x-axis is the horizontal asymptote for the graph of f (x). y 5 0

The line is the horizontal asymptote for the graph of f (x).

Solution (b):

The degree of the numerator 8x2 ! 3 is 2. n 5 2


The degree of the numerator is equal to the

degree of the denominator. n 5 m

The ratio of the leading coefficients is the

horizontal asymptote for the graph of g(x).

The line is the horizontal asymptote for the graph of g(x).

If we divide the numerator by the denominator,

the resulting quotient is the constant 2.

Solution (c):

The degree of the numerator 8x3 ! 3 is 3. n 5 3


The degree of the numerator is greater than

the degree of the denominator. n $ m

The graph of the rational function h(x) has .

If we divide the numerator by the denominator,

the resulting quotient is a linear function.

n YOUR TURN Find the horizontal asymptote (if one exists) for the graph of the

rational function f (x) =7x3 + x - 2

-4x3 + 1.

h(x) =8x3 + 3

4x2 + 1= 2x +

-2x + 3

4x2 + 1

no horizontal asymptote

g(x) = 8x2 + 3

4x2 + 1= 2 +

1

4x2 + 1

y = 2

y =8

4= 2

y = 0

h(x) =8x3 + 3

4x2 + 1g(x) =

8x2 + 3

4x2 + 1f (x) =

8x + 3

4x2 + 1

Technology Tip

The following graphs correspond

to the rational functions given in

Example 5. The horizontal

asymptotes are apparent, but

are not drawn in the graph.

a. Graph .f(x) =8x + 3

4x2 + 1

EXAMPLE 6 Finding Slant Asymptotes

Determine the slant asymptote of the rational function .

Solution:

Divide the numerator by the

denominator with long division.

f (x) =4x3 + x2 + 3

x2 - x + 1

n Answer: y5 x ! 5

Let f be a rational function given by , where n(x) and d(x) are polynomials

and the degree of n(x) is one more than the degree of d(x). On dividing n(x) by d(x),

the rational function can be expressed as

where the degree of the remainder r (x) is less than the degree of d(x) and

the line y 5 mx ! b is a slant asymptote for the graph of f.

Note that as or , .f (x)S mx + bxS `xS -`

f (x) = mx + b +r(x)

d(x)

f (x) =n(x)

d(x)

SLANT ASYMPTOTES

4x + 5

x2 - x + 1 4x3 + x2 + 0x + 3

-(4x3 - 4x2 + 4x)

5x2 - 4x + 3

- (5x2 - 5x + 5)

x - 2

Note that as the rational

expression approaches 0.

The quotient is the slant asymptote.

n YOUR TURN Find the slant asymptote of the rational function .f (x) =x2 + 3x + 2

x - 2

y = 4x + 5

x S ;`

S 0 as

f (x) = 4x 5 +x - 2

x2 - x + 1xS ;`

Recall that when dividing polynomials the degree of the quotient is always the difference

between the degree of the numerator and the degree of the denominator. For example, a

cubic (third-degree) polynomial divided by a quadratic (second-degree) polynomial results

in a linear (first-degree) polynomial. A fifth-degree polynomial divided by a fourth-degree

polynomial results in a first-degree (linear) polynomial. When the degree of the numerator

is exactly one more than the degree of the denominator, the quotient is linear and represents

a slant asymptote.


Technology Tip

The graph of

has a slant asymptote of

y = 4x + 5

f (x) =4x3

+ x2+ 3

x2- x + 1


EXAMPLE 7 Graphing a Rational Function

Graph the rational function .

Solution:

STEP 1 Find the domain.

Set the denominator equal to zero. x2 4 5 0

Solve for x. x 5 ;2

State the domain.

STEP 2 Find the intercepts.

y-intercept: y 5 0

x-intercepts: x 5 0

The only intercept is at the point .(0, 0)

f (x) =x

x2- 4

= 0

f (0) =0

-4= 0

(-`, -2)! (-2, 2)! (2, `)

f (x) =x

x2- 4

Study Tip

Any real number excluded from

the domain of a rational function

corresponds to either a vertical

asymptote or a hole on its graph.

It is important to note that any real number eliminated from the domain of a rational

function corresponds to either a vertical asymptote or a hole on its graph.

Let f be a rational function given by .

Step 1: Find the domain of the rational function f.

Step 2: Find the intercept(s).

n y-intercept: evaluate f(0).

n x-intercept: solve the equation n(x) 5 0 for x in the domain of f.

Step 3: Find any holes.

n Factor the numerator and denominator.

n Divide out common factors.

n A common factor x a corresponds to a hole in the graph of f at x 5 a

if the multiplicity of a in the numerator is greater than or equal to the

multiplicity of a in the denominator.

n The result is an equivalent rational function in lowest terms.

Step 4: Find any asymptotes.

n Vertical asymptotes: solve q(x) 5 0.

n Compare the degree of the numerator and the degree of the denominator to

determine whether either a horizontal or slant asymptote exists. If one exists,

find it.

Step 5: Find additional points on the graph of f—particularly near asymptotes.

Step 6: Sketch the graph; draw the asymptotes, label the intercept(s) and additional

points, and complete the graph with a smooth curve between and beyond

the vertical asymptotes.

R(x) =p(x)

q(x)

f (x) =n(x)

d(x)

GRAPHING RATIONAL FUNCTIONS

Study Tip

Common factors need to be divided

out first; then the remaining x-values

corresponding to a denominator

value of 0 are vertical asymptotes.

Graphing Rational Functions

We can now graph rational functions using asymptotes as graphing aids. The following box

summarizes the five-step procedure for graphing rational functions.


x 3 1 1 3

f (x)35-

13

13-

35

x

y

–5 5

5

–5

n Answer:

EXAMPLE 8 Graphing a Rational Function with No Horizontalor Slant Asymptotes

State the asymptotes (if there are any) and graph the rational function .

Solution:


Set the denominator equal to zero. x2 1 5 0

Solve for x. x 5 ;1

State the domain.


y-intercept:

x-intercepts: n(x) 5 x4 x3

6x25 0

Factor. x2(x 3)(x! 2) 5 0

Solve. x 5 0, x 5 3, and x 5 2

The intercepts are the points (0, 0), (3, 0), and ("2, 0).

f (0) =0

-1= 0

(-`, -1)! (-1, 1)! (1, `)

f (x) =x4

- x3- 6x2

x2- 1

Technology Tip

The behavior of each function as x

approaches or can be shown

using tables of values.

Graph f (x) =x4

- x3- 6x2

x2- 1

.

-` ̀

x

y

5

5

–5

–5

STEP 3 Find any holes.

There are no common factors, so f is in lowest terms.

Since there are no common factors, there are no holes on the graph of f.

STEP 4 Find any asymptotes.

Vertical asymptotes: d(x) 5 (x ! 2)(x 2) 5 0

and

Horizontal asymptote:

Degree of numerator " Degree of denominator y 5 0

STEP 5 Find additional points on the graph.

STEP 6 Sketch the graph; label the intercepts,

asymptotes, and additional points

and complete with a smooth curve

approaching the asymptotes.

n YOUR TURN Graph the rational function .f (x) =x

x2- 1

Degree of numerator = 1

Degree of denominator = 2

x = 2x = -2

f (x) =x

(x + 2)(x - 2)





STEP 4 Find the asymptotes.

Vertical asymptote: d(x) 5 x2 1 5 0

Factor. (x ! 1)( x 1) 5 0

Solve. x 5 1 and x 5 1

No horizontal asymptote: degree of n(x) # degree of d(x) [4 # 2]

No slant asymptote: degree of n(x) degree of d(x) # 1 [4 2 5 2 # 1]

The asymptotes are x5"1 and x 5 1.

STEP 5 Find additional

points on the graph.

f (x) =x2(x - 3)(x + 2)

(x - 1)(x + 1)

The graph of f (x) shows that the

vertical asymptotes are at

and there is no horizontal asymptote

or slant asymptote.

x = ;1

x 3 0.5 0.5 2 4

f (x) 6.75 1.75 2.08 5.33 6.4

STEP 6 Sketch the graph; label the intercepts

and asymptotes, and complete with a

smooth curve between and beyond

the vertical asymptote.

–5 5

–10

10

x

y

(3, 0)(–2, 0)

x = 1

x = –1

n YOUR TURN State the asymptotes (if there are any) and graph the rational function

f (x) =x3 - 2x2 - 3x

x + 2.

–5 10

–25

50

x

y

(3, 0)

(–1, 0)

x = –2

n Answer: Vertical asymptote:

x5 2. No horizontal or slant

asymptotes.

EXAMPLE 9 Graphing a Rational Function with aHorizontal Asymptote

State the asymptotes (if there are any) and graph the rational function

Solution:


Set the denominator equal to zero. 8 x35 0

Solve for x. x 5 2

State the domain. (-`, 2)! (2, `)

f (x) =4x3 + 10x2 - 6x

8 - x3

Technology Tip




Graph f (x) =4x3 + 10x2 - 6x

8 - x3.

-``



vertical asymptote is at

and the horizontal asymptote is

at .y = -4

x = 2


y-intercept:

x-intercepts: n(x) 5 4x3 ! 10x2 6x 5 0

Factor. 2x (2x 1)(x! 3) 5 0

Solve. x 5 0, and x 5 3

The intercepts are the points (0, 0), , and ("3, 0).

STEP 3 Find the holes.

There are no common factors, so f is in lowest terms (no holes).


Vertical asymptote: d(x) 5 8 x3 5 0

Solve. x 5 2

Horizontal asymptote: degree of n(x) 5 degree of d(x)

Use leading coefficients.

The asymptotes are x5 2 and y 5"4.



y =4

-1= -4

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

(2 - x)(x2 + 2x + 4)

A12, 0 B

x =1

2,

f (0) =0

8= 0

STEP 6 Sketch the graph; label the intercepts

and asymptotes and complete with a

smooth curve.

n YOUR TURN Graph the rational function Give equations

of the vertical and horizontal asymptotes and state the intercepts.

f (x) =2x2 - 7x + 6

x2 - 3x - 4.

–10 10

–10

10

x

y

(–3, 0) x = 2

y = –4

( , 0)1

2

n Answer: Vertical asymptotes:

x 5 4, x 5 1

Horizontal asymptote: y 5 2

Intercepts: A0, - 32 B , A32, 0 B , (2, 0)

–4 6

x

y

–5

5

y = 2

x = –1 x = 4

x 4 1 1 3

f (x) 1 1.33 0.10 1.14 9.47

14


EXAMPLE 10 Graphing a Rational Function with a Slant Asymptote

Graph the rational function

Solution:


Set the denominator equal to zero. x ! 2 5 0

Solve for x. x 5 2

State the domain.


y-intercept: f (0) = -4

2= -2

(-`,-2)! (-2, `)

f (x) =x2 - 3x - 4

x + 2.Technology Tip




Graph f (x) =x2 - 3x - 4

x + 2.

-``


vertical asymptote is at and

the slant asymptote is at y = x - 5.

x = -2

–20 20

–20

20

x

y

(4, 0)(0, –2)

(–1, 0)

x = –2

y = x – 5

x 6 5 3 5 6

f (x) 12.5 12 14 0.86 1.75

n Answer:

Horizontal asymptote: x 5 3

Slant asymptote: y5 x ! 4

–10 10

–10

20

x

y

x = 3

y = x + 4

(–2, 0)

(1, 0)

(0, )2

3

x-intercepts: n(x) 5 x2 3x 4 5 0

Factor. (x ! 1)(x 4) 5 0

Solve. x 5 1 and x 5 4

The intercepts are the points (0, "2), ("1, 0), and (4, 0).





Vertical asymptote: d(x) 5 x 2 5 0

Solve. x 5!2

Slant asymptote: degree of n(x) ! degree of d(x) 5 1

Divide n(x) by d(x).

Write the equation of the asymptote. y 5 x ! 5

The asymptotes are x 5 2 and y 5 x 5.



STEP 6 Sketch the graph; label the

intercepts and asymptotes,

and complete with a smooth

curve between and beyond

the vertical asymptote.

n YOUR TURN For the function , state the asymptotes (if any exist)

and graph the function.

f (x) =x2

+ x - 2

x - 3

f (x) =x2

- 3x - 4

x + 2= x - 5 +

6

x + 2

f (x) =(x - 4)(x + 1)

(x + 2)


EXAMPLE 11 Graphing a Rational Function with a Hole in the Graph

Graph the rational function .

Solution:


Set the denominator equal to zero. x2! x ! 2 5 0

Solve for x. (x ! 2)(x 1) 5 0

x 5!1 or x 5 2

State the domain.


y-intercept: y 5 3

x-intercepts: n(x) 5 x2 x ! 6 5 0

(x 3)(x! 2) 5 0

x 5!3 or x 5 2

The intercepts correspond to the points and . The point (2, 0)

appears to be an x-intercept; however, x5 2 is not in the domain of the function.


Since x! 2 is a common factor, there is a

hole in the graph of f at x 5 2.

Dividing out the common factor generates

an equivalent rational function in lowest terms.


Vertical asymptotes: x 1 5 0

Horizontal

asymptote:

Since the degree of the numerator

equals the degree of the denominator,

use the leading coefficients.

STEP 5 Find additional points on the graph.

STEP 6 Sketch the graph; label the intercepts,

asymptotes, and additional points and

complete with a smooth curve approaching

asymptotes. Recall the hole at .

Note that so the open “hole”

is located at the point (2, 5/3).

n YOUR TURN Graph the rational function .f (x) =x2

- x - 2

x2+ x - 6

R(2) =53

x = 2

y =1

1= 1

Degree of numerator of f5 Degree of denominator of f5 2

and

Degree of numerator of R 5 Degree of denominator of R5 1

x = -1

R(x) =(x + 3)

(x + 1)

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

(x - 2)(x + 1)

(-3, 0)(0, 3)

f (0) =-6

-2= 3

(-`, -1)! (-1, 2)! (2, `)

f (x) =x2

+ x - 6

x2- x - 2

Technology Tip




Graph f (x) =x2

+ x - 6

x2- x - 2

.

-``


vertical asymptote is at and

the horizontal asymptote is at .y = 1

x = -1

Notice that the hole at is not

apparent in the graph. A table of

values supports the graph.

x = 2

x !4 !2 1 3

f (x) or R(x) !1 5 232

13

-12

x

y

5

–5

–5

5

n Answer:

x

–7 3

y

5

–5


1. If degree of the numerator ! degree of the denominator

5 1, then there is a slant asymptote.

2. Divide the numerator by the denominator. The quotient

corresponds to the equation of the line (slant asymptote).

Procedure for Graphing Rational Functions

1. Find the domain of the function.

2. Find the intercept(s).n y-interceptn x-intercepts (if any)

3. Find any holes.n If x! a is a common factor of the numerator and

denominator, then x 5 a corresponds to a hole in the

graph of the rational function if the multiplicity of a in

the numerator is greater than or equal to the multiplicity

of a in the denominator. The result after the common

factor is canceled is an equivalent rational function in

lowest terms (no common factor).

4. Find any asymptotes.n Vertical asymptotesn Horizontal/slant asymptotes

5. Find additional points on the graph.

6. Sketch the graph: draw the asymptotes and label the intercepts

and points and connect with a smooth curve.

SUMMARY

In this section, rational functions were discussed.

n Domain:All real numbers except the x-values that make

the denominator equal to zero, d(x) 5 0.n Vertical Asymptotes: Vertical lines, x5 a, where d(a) 5 a,

after all common factors have been divided out. Vertical

asymptotes steer the graph and are never touched.n Horizontal Asymptotes: Horizontal lines, y5 b, that steer

the graph as .

1. If degree of the numerator " degree of the denominator,

then y5 0 is a horizontal asymptote.

2. If degree of the numerator 5 degree of the denominator,

then y 5 c is a horizontal asymptote where c is the

ratio of the leading coefficients of the numerator and

denominator, respectively.

3. If degree of the numerator # degree of the denominator,

then there is no horizontal asymptote.n Slant Asymptotes: Slant lines, y 5 mx b, that steer the

graph as .xS ;`

xS ;`

f (x) =n(x)

d(x)

SECTION

4.6

In Exercises 1–10, find the domain of each rational function.

1. 2. 3. 4.

5. 6. 7. 8.

9. 10.

In Exercises 11–20, find all vertical asymptotes and horizontal asymptotes (if there are any).

11. 12. 13. 14.

15. 16. 17. 18.

19. 20. f (x) =0.8x4

- 1

x2- 0.25

f (x) =(0.2x - 3.1)(1.2x + 4.5)

0.7(x - 0.5)(0.2x + 0.3)

f (x) =

110 Ax2

- 2x +3

10 B

2x - 1f (x) =

13x

2+

13x -

14

x2+

19

f (x) =6x2

+ 3x + 1

3x2- 5x - 2

f (x) =6x5

- 4x2+ 5

6x2+ 5x - 4

f (x) =2 - x3

2x - 7f (x) =

7x3+ 1

x + 5f (x) =

1

5 - xf (x) =

1

x + 2

f (x) =5(x2

- 2x - 3)

(x2- x - 6)

f (x) = -

3(x2+ x - 2)

2(x2- x - 6)

f (x) = -2x

x2+ 9

f (x) =7x

x2+ 16

f (x) =x - 1

x2+ 2x - 3

f (x) =x + 4

x2+ x - 12

f (x) =5 - 3x

(2 - 3x)(x - 7)f (x) =

2x + 1

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

3

4 - xf (x) =

1

x + 3

n SKILLS

EXERCISES

SECTION

4.6


In Exercises 21–26, find the slant asymptote corresponding to the graph of each rational function.

21. 22. 23.

24. 25. 26.

In Exercises 27–32, match the function to the graph.

27. 28. 29.

30. 31. 32.

a. b. c.

d. e. f.

In Exercises 33–58, use the graphing strategy outlined in this section to graph the rational functions.

33. 34. 35. 36. 37. 38.

39. 40. 41. 42.

43. 44. 45. 46.

47. 48. 49. 50.

51. 52. 53. 54.

55. 56. 57. 58. f (x) =-2x(x - 3)

x(x2+ 1)

f (x) =3x(x - 1)

x(x2- 4)

f (x) =(x - 1)(x2

- 9)

(x - 3)(x2+ 1)

f (x) =(x - 1)(x2

- 4)

(x - 2)(x2+ 1)

f (x) =(x + 1)2

(x2- 1)

f (x) =(x - 1)2

(x2- 1)

f (x) = x -4

xf (x) = 3x +

4

x

f (x) =25x2

- 1

(16x2- 1)2

f (x) =1 - 9x2

(1 - 4x2)3f (x) =

12x4

(3x + 1)4f (x) =

7x2

(2x + 1)2

f (x) =1 - x2

x2+ 1

f (x) =x2

+ 1

x2- 1

f (x) =3x3

+ 5x2- 2x

x2+ 4

f (x) =2x3

- x2- x

x2- 4

f (x) =x2

- 9

x + 2f (x) =

x2

x + 1f (x) =

3(x2- 1)

x2- 3x

f (x) =2(x2

- 2x - 3)

x2+ 2x

f (x) =2 + x

x - 1f (x) =

x - 1

xf (x) =

4x

x + 2f (x) =

2x

x - 1f (x) =

4

x - 2f (x) =

2

x + 1

–5 5

–5

5

x

y

–5 5

–5

5

x

y

–10 10

–10

10

x

y

–10 10

–150

150

x

y

–10 10

–10

10

x

y

–10 10

–10

10

x

y

f (x) =3x2

x + 4f (x) =

3x2

4 - x2f (x) = -

3x2

x2+ 4

f (x) =3x2

x2- 4

f (x) =3x

x - 4f (x) =

3

x - 4

f (x) =2x6

+ 1

x5- 1

f (x) =8x4

+ 7x3+ 2x - 5

2x3- x2

+ 3x - 1f (x) =

3x3+ 4x2

- 6x + 1

x2- x - 30

f (x) =2x2

+ 14x + 7

x - 5f (x) =

x2+ 9x + 20

x - 3f (x) =

x2+ 10x + 25

x + 4

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