Module 9 Lecture Notes - PCCspot.pcc.edu/.../lecturenotes/Math_111_Module_9_Lecture_Notes.pdf ·...

Module 9 Lecture Notes

Contents

9.1 Summary of the Properties of Rational Functions . . . . . . . . . . . . 2

9.2 Graphing Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . 3

9.2.1 Examples Without Holes . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

9.2.2 Examples With Holes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

9.2.3 A Calculator-Assisted Example . . . . . . . . . . . . . . . . . . . . . . . . 15

9.3 Determining the Formula of a Rational Function Based on Its Graph 17

9.4 Improper Rational Functions and Oblique Asymptotes . . . . . . . . . 25

For comical asymptotes, view the following:

http://changestartsintheheart.wordpress.com/tag/asymptote/

http://www.theoildrum.com/node/5110

1

http://changestartsintheheart.wordpress.com/tag/asymptote/

http://www.theoildrum.com/node/5110

Math 111 Module 9 Lecture Notes

9.1 Summary of the Properties of Rational Functions

A rational function is of the form R(x) =p(x)

q(x)where p and q are polynomial functions.

The zeros of a rational function are the values of x for which p(x) = 0, as the function’s value

is zero where the value of the numerator is zero. Most of the time, the zeros will occur at a

when the factor (x− a) is in the numerator of R.

A rational function is undefined where q(x) = 0, as this would cause division by zero.

A vertical asymptote occurs when the denominator of the simplified form of R is equal to

zero.The vertical asymptote x = b will occur when the factor (x− b) is in the denominator of

the simplified form of R.

A hole occurs when both the numerator and denominator equal zero for some value of x.

We will identify a zero at c when the linear factor (x − c) occurs in both the numerator and

denominator of a rational function. Note that during simplification this factor cancels and

results in a domain restriction for R.

The long run behavior and horizontal asymptote of R can be determined by the ratio of

leading terms of p and q. Alternately, the following can be used to determine if a horizontal

or oblique asymptote exists:

Let m be the degree of the function p in the numerator and let n be the degree of the

function q in the denominator.

• If m < n, then the horizontal asymptote is y = 0.

• If m = n, then the horizontal asymptote is y = c where c is a real number determined

by the ratio of leading coefficients.

• If m > n, then no horizontal asymptote exists.

If m = n + 1, then an oblique asymptote exists.

Instructor: A.E.Cary Page 2 of 26


9.2 Graphing Rational Functions

9.2.1 Examples Without Holes

Example 1: Graph the rational function R(x) =6− x

x− 2by completing the following: factor and

simplify R(x); state the domain; determine any holes and/or vertical asymptotes; determine any

horizontal asymptotes; determine any horizontal and vertical intercepts; determine the zeros and

their multiplicity.

• R(x) is fully simplified.

• The domain of R is {x |x 6= 2} .

• There are no holes. The vertical asymptote is x = 2.

• Since the degree of the numerator is the same as the degree of the denominator, the horizontal

asymptote is determined by the ratio of leading coefficients and is y = −1.

• As R(0) = −3, the vertical intercept is (0,−3). The horizontal intercepts (and zeros) of R

occur where R(x) = 0:

R(x) = 0

6− x = 0

x = 6

The horizontal intercept is (6, 0)

• The zero of R 6, as found in the previous step. It has a multiplicity of 1, so the graph crosses

the horizontal axis at 6.

Table 9.1: Behavior of R

Zeros/ VA

Interval (−∞, 2) (2, 6) (6,∞)

x -2 4 7

R(x) R(−2) = −2 R(4) = 1 R(7) = −15

+/− - (below) + (above) - (below)

Point (−2,−2) (4, 1)(7,−1

5

)



Figure 9.1

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

x=2

y= -1

x

yy = 2x−6

x+4



Example 2: Graph the rational function R(x) =2x− 6

x + 4by completing the following: factor and

simplify R(x); state the domain; determine any holes and/or vertical asymptotes; determine any

horizontal asymptotes; determine any horizontal and vertical intercepts; determine the zeros and

their multiplicity.

• Simplifying R(x):

R(x) =2x− 6

x + 4

R(x) =2(x− 3)

x + 4

• The domain of R is {x |x 6= −4} .

• There are no holes. The vertical asymptote is x = −4.


asymptote is determined by the ratio of leading coefficients and is y = 2.

• As R(0) = −32, the vertical intercept is

(0,−3

2

). The horizontal intercepts (and zeros) of R


R(x) = 0

2(x− 3) = 0

x− 3 = 0

x = 3

The horizontal intercept is (3, 0)

• The zero of R 3, as found in the previous step. It has a multiplicity of 1, so the graph crosses

the horizontal axis at 3.




Zeros/ VA

Interval (−∞,−4) (−4, 3) (3,∞)

x -6 -2 4

R(x) R(−6) = 9 R(−2) = −5 R(4) = 14

+/− + (above) - (below) + (above)

Point (−6, 9) (−2,−5)(4, 1

4

)

Figure 9.2

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

x= -4

y=1 x

yy = 2x−6

x+4



Example 3: Graph the rational function R(x) =x2 − 5x− 6

x2 + x− 12by completing the following: factor

and simplify R(x); state the domain; determine any holes and/or vertical asymptotes; determine

the long-run behavior and any horizontal asymptotes; determine any horizontal and vertical inter-

cepts; determine the zeros and their multiplicity.


R(x) =x2 − 5x− 6

x2 + x− 12

R(x) =(x− 6)(x + 1)

(x + 4)(x− 3)

• The domain of R is {x |x 6= −4, 3} .

• There are no holes. The vertical asymptotes are x = −4 and x = 3.



• As R(0) = 12, the vertical intercept is

(0, 1

2

). The horizontal intercepts (and zeros) of R occur

where R(x) = 0:

R(x) = 0

(x− 6)(x + 1) = 0

x− 6 = 0 or x + 1 = 0

x = 6 or x = −1

The horizontal intercepts are (6, 0) and (−1, 0).

• The zeros of R are -1 and 6, as found in the previous step. Each has a multiplicity of 1, so

the graph crosses the horizontal axis at -1 and 6.




Zeros/ VA

Interval (−∞,−4) (−4,−1) (−1, 3) (3, 6) (6,∞)

x -6 -2 1 4 8

R(x) R(−6) = 103

R(−2) = −45

R(1) = 1 R(4) = −54

R(8) = 310

+/− + (above) - (below) + (above) - (below) + (above)

Point(−6, 10

3

) (−2,−4

5

)(0, 1)

(4,−5

4

) (8, 3

10

)

Figure 9.3

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

x= -4 x=3

y=1 x

yy = x2−5x−6

x2+x−12



Example 4: Graph the rational function R(x) =8

x2 − 4by completing the following:

factor and simplify R(x); state the domain; determine any holes and/or vertical asymptotes;

determine the long-run behavior and any horizontal asymptotes; determine any horizontal and

vertical intercepts; determine the zeros and their multiplicity.


R(x) =8

x2 − 4

R(x) =8

(x− 2)(x + 2)

The function is fully simplified.



• Since the degree of the numerator is less than the degree of the denominator, the horizontal

asymptote is y = 0.

• As R(0) = −2, the vertical intercept is (0,−2). The horizontal intercepts (and zeros) of R


R(x) = 0

8 6= 0

There are no horizontal intercepts.

• There are no zeros of R.




Zeros/ VA

Interval (−∞,−2) (−2, 2) (2,∞)

x -4 0 4

R(x) R(−4) = 23

R(0) = −2 R(4) = 23

+/− + (above) - (below) + (above)

Point(−4, 2

3

)(0,−2)

(4, 2

3

)

Figure 9.4

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

x= -2 x=2

y=0

x

yy = 8

x2−4



9.2.2 Examples With Holes

Example 5: Graph the rational function R(x) =x2 − 4x + 3

x− 3by completing the following: factor





R(x) =x2 − 4x + 3

x− 3

R(x) =(x− 3)(x− 1)

x− 3

R(x) = x− 1, x 6= 3

Thus R(x) is a linear function with a restricted domain.

• The domain of R is {x |x 6= 3} or (−∞, 3)⋃

(3,∞).

• There is a hole at 3. To determine the output of this hole, we use the simplified form of R:

R(3) ≈ 3− 1

= 2

Thus the point (3, 2) will be graphed as an open circle. There are no vertical asymptotes.

• No horizontal asymptote exists. In the long run, R(x) “looks like” the function defined by

f(x) = x.

As x→∞, R(x)→∞As x→ −∞, R(x)→ −∞

• As R(0) = −1, the vertical intercept is (0,−1).

The horizontal intercepts (and zeros) of R occur where R(x) = 0:

R(x) = 0

x− 1 = 0

x = 1



The horizontal intercept is (1, 0).

• The zero of R is 1 with multiplicity 1, so the graph crosses the horizontal axis at 1.

• When graphing this function, all that needs to be graphed is the line y = x− 1 with an open

circle at (3, 2).

Figure 9.5

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

x

yy = x2−4x+3

x−3



Example 6: Graph the rational function R(x) =3x− 6

x2 + x− 6by completing the following: factor





R(x) =3x− 6

x2 + x− 6

R(x) =3(x− 2)

(x + 3)(x− 2)

R(x) =3

x + 3, x 6= 2

• The domain of R is {x |x 6= −3, 2}.

• There is a hole at 2. To determine the output of this hole, we use the simplified form of R:

R(2) ≈ 3

2 + 3

=3

5

Thus the point(2, 3

5

)will be graphed as an open circle.

The vertical asymptote is x = −3.

• Since the degree of the numerator is less than the degree of the denominator, the horizontal

asymptote is y = 0.

• As R(0) = 1, the vertical intercept is (0, 1).

The horizontal intercepts (and zeros) of R occur where R(x) = 0:

R(x) = 0

3 6= 0

There is no horizontal intercept.

• There are no zeros of R.




Zeros/ VA

Interval (−∞,−3) (−3,∞)

x -4 -2

R(x) R(−4) = −3 R(−2) = 3

+/− - (below) + (above)

Point (−4,−3) (−2, 3)

Figure 9.6

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

x= -3

y=0 x

yy = 3x−6

x2+x−6



9.2.3 A Calculator-Assisted Example

Example 7: Graph the rational function R(x) =x2 − 2x− 8

x2 − 9by completing the following: factor





R(x) =x2 − 2x− 8

x2 − 9

=(x− 4)(x + 2)

(x− 3)(x + 3)





• As R(0) = 89, the vertical intercept is

(0, 8

9

). The horizontal intercepts (and zeros) of R occur

where R(x) = 0:

R(x) = 0

(x− 4)(x + 2) = 0

x− 4 = 0 or x + 2 = 0

x = 4 or x = −2

The horizontal intercepts are (4, 0) and (−2, 0).

• The zeros of R are -2 and 4, as found in the previous step. Each has a multiplicity of 1, so

the graph crosses the horizontal axis at -2 and 4.




Zeros/ VA

Interval (−∞,−3) (−3,−2) (−2, 3) (3, 4) (4,∞)

x -4 -2.5 0.5 3.5 6

R(x) R(−4) ≈ 2.3R(−2.5) ≈−1.2

R(0.5) = 1R(3.5) ≈−0.85

R(6) ≈ 0.6

+/− + (above) - (below) + (above) - (below) + (above)

Point (−4, 2.3) (−2.5,−1.2) (0.5, 1) (3.5,−0.85) (6, 0.6)

**Function values approximated with a calculator.**

Figure 9.7

−10 −8 −6 −4 −2 2 4 6 8 10

−6

−4

−2

2

4

6

x= -3 x=3

y=1 x

yy = x2−2x−8

x2−9



9.3 Determining the Formula of a Rational Function Based

on Its Graph

How to find a possible formula for a rational function:

• State any zeros. Use these to determine factors and the multiplicity of each factor that

appears in the numerator.

• State any vertical asymptotes. Use these to determine factors and the multiplicity of

each factor that appears in the denominator.

• If a “hole” appears at x = a, then put the factor (x − a) in both the numerator and

denominator.

• Use one other point to determine if there is a constant factor other than 1.

Example 8: Find a possible formula for the rational function graphed in Figure 9.8.

Figure 9.8

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

y=3

x=4

(3, 6)

x

y




Zeros/ VA Multiplicity Factor Numerator/denominator

zero: 5 odd (1) (x− 5) numerator

VA: x = 4 odd (1) (x− 4) denominator

Thus a possible formula for R(x) is of the form

R(x) =k(x− 5)

(x− 4)

where k is a constant factor.

The constant factor of k can be determined to be 3 based upon the horizontal asymptote y = 3.

Therefore possible formula for R(x) is

R(x) =3(x− 5)

(x− 4)

Alternately, we could determine k by using one point on the graph that is not a horizontal intercept.

Here, the point (3, 6) is given:

R(3) = 6

k(3− 5)

(3− 4)= 6

2k = 6

k = 3




Figure 9.9

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

y= -2

x= -1

x

y




VA: x = −1 odd (1) (x + 1) denominator


R(x) =k(x− 3)

(x + 1)

where k is a constant factor.

The constant factor of k can be determined to be −2 based upon the horizontal asymptote y = −2.

Therefore a possible formula for R(x) is

R(x) =−2(x− 3)

(x + 1)


Here, the point (0, 6) is given:

R(0) = 6

k(0− 3)

(0 + 1)= 6

k = −2



Example 10: Find a possible formula for the rational function graphed in Figure 9.10 below.

Figure 9.10

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

y=2

x=1 x=3

(0, 8

3

)x

y



zero: 2 even (2) (x− 2)2 numerator




R(x) =k(x− 2)2

(x− 1)(x− 3)

where k is a constant factor. The constant factor of k can be determined to be 2 based upon the

horizontal asymptote y = 2. Therefore a possible formula for R(x) is

R(x) =2(x− 2)2

(x− 1)(x− 3)



Alternately, we could determine k by using the point(0, 8

3

):

R(0) =8

3k(0− 2)2

(0− 1)(0− 3)=

8

34

3k =

8

3

k = 2



Example 11: Find a possible formula for the rational function graphed in Figure 9.11 below.

Figure 9.11

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

y=0

x=2x= -4

(0, 1)x

y



VA: x = −4 odd (1) (x + 4) denominator

VA: x = 2 even (2) (x− 2)2 denominator


R(x) =k

(x + 4)(x− 2)2

where k is a constant factor. To determine k, we must use one point on the graph that is not a

horizontal intercept. Here, the point (0, 1) is given:

R(0) = 1

k

(0 + 4)(0− 2)2= 1

k

16= 1

k = 16


R(x) =16

(x + 4)(x− 2)2




Figure 9.12

−12 −8 −4 4 8 12

−6

−4

−2

2

4

6

y=2

x=4x= -5

x

y




zero: -2 even (2) (x + 2)2 numerator

VA: x = −5 even (2) (x + 5)2 denominator



R(x) =k(x− 3)(x + 2)2

(x + 5)2(x− 4)

where k is a constant factor. The constant factor of k can be determined to be 2 based upon the

horizontal asymptote y = 2.


R(x) =2(x− 3)(x + 2)2

(x + 5)2(x− 4)




Here, the point (−4, 7) is given:

R(−4) = 7

k(−4− 3)(−4 + 2)2

(−4 + 5)2(−4− 4)= 7

k(−28)

−8= 7

k(7)

2= 7

k = 7 · 2

7

k = 2



9.4 Improper Rational Functions and Oblique Asymptotes

Example 13: The graph of R(x) = x2−4x−54x−8

“looks like” the function defined by f(x) = 14x in the

long run. We know that this function has an oblique asymptote as the degree of the numerator is

greater than the degree of the denominator. To determine the equation of the oblique asymptote,

either polynomial long division or a graphing calculator are needed. Using the “expand” key on a

graphing calculator to perform polynomial long division, we find:

We use this to write:

R(x) =x2 − 4x− 5

4x− 8

=−9

4x− 8+

1

4x− 1

2

The first term in the expanded R(x) is the remainder of this long division. The expression 14x− 1

2

is used to determine the equation of the oblique asymptote, which is y = 14x − 1

2. The function

and its oblique asymptote are graphed in Figure 9.13 below.

Figure 9.13

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

x=2

x

yy = 1

4x − 1

2

y = x2−4x−54x−8



Example 14: The graph of R(x) =3− x7

4x2 − 1“looks like” the function defined by f(x) = −1

4x5 in

the long run. Graphs of y = R(x) and y = f(x) are shown in Figure 9.14 below.

Figure 9.14

−3 −2 −1 1 2 3

−30

−20

−10

10

20

30

x=0.5x= -0.5

x

y

y = − 14x5

y = 3−x7

4x2−1


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