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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) =
448 CHAPTER 4 Polynomial and Rational Functions
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)
4.6 Rational Functions 449
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.
450 CHAPTER 4 Polynomial and Rational Functions
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.
4.6 Rational Functions 451
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.
452 CHAPTER 4 Polynomial and Rational Functions
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 denominator 4x2 ! 1 is 2. m 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 denominator 4x2 ! 1 is 2. m 5 2
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.
4.6 Rational Functions 453
Technology Tip
The graph of
has a slant asymptote of
y = 4x + 5
f (x) =4x3
+ x2+ 3
x2- x + 1
454 CHAPTER 4 Polynomial and Rational Functions
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.
4.6 Rational Functions 455
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:
STEP 1 Find the domain.
Set the denominator equal to zero. x2 1 5 0
Solve for x. x 5 ;1
State the domain.
STEP 2 Find the intercepts.
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)
456 CHAPTER 4 Polynomial and Rational Functions
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 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:
STEP 1 Find the domain.
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
The behavior of each function as x
approaches or can be shown
using tables of values.
Graph f (x) =4x3 + 10x2 - 6x
8 - x3.
-``
4.6 Rational Functions 457
The graph of f (x) shows that the
vertical asymptote is at
and the horizontal asymptote is
at .y = -4
x = 2
STEP 2 Find the intercepts.
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).
STEP 4 Find the asymptotes.
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.
STEP 5 Find additional
points on the graph.
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
458 CHAPTER 4 Polynomial and Rational Functions
EXAMPLE 10 Graphing a Rational Function with a Slant Asymptote
Graph the rational function
Solution:
STEP 1 Find the domain.
Set the denominator equal to zero. x ! 2 5 0
Solve for x. x 5 2
State the domain.
STEP 2 Find the intercepts.
y-intercept: f (0) = -4
2= -2
(-`,-2)! (-2, `)
f (x) =x2 - 3x - 4
x + 2.Technology Tip
The behavior of each function as x
approaches or can be shown
using tables of values.
Graph f (x) =x2 - 3x - 4
x + 2.
-``
The graph of f (x) shows that the
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).
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 the asymptotes.
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 5 Find additional
points on the graph.
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)
4.6 Rational Functions 459
EXAMPLE 11 Graphing a Rational Function with a Hole in the Graph
Graph the rational function .
Solution:
STEP 1 Find the domain.
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.
STEP 2 Find the intercepts.
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.
STEP 3 Find any holes.
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.
STEP 4 Find the asymptotes.
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
The behavior of each function as x
approaches or can be shown
using tables of values.
Graph f (x) =x2
+ x - 6
x2- x - 2
.
-``
The graph of f (x) shows that the
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
460 CHAPTER 4 Polynomial and Rational Functions
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
4.6 Rational Functions 461
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