section 2.6 rational functions hand out rational functions sheet!

Section 2.6 Rational Functions

Hand out Rational Functions Sheet!

Section 2.6 Rational Functions

Objective: To find asymptotes and domain

of rational functions.

The ratio of two polynomial functions is called a rational function.

f (x) x

x 1

Examples

1

( )3 8

xf x

x x

Denominator cannot be zero!

1x 3, 8 x

What is an asymptote?

An asymptote is a line that a graph approaches as it moves away from the origin.

•Asymptotes can be vertical or horizontal.

•Vertical asymptotes cannot ever be crossed, but horizontal ones can.

•Vertical asymptotes are determined by the zeros of the denominator.

•Horizontal asymptotes are determined by comparing the degrees of the numerator vs. denominator.

ASYMPTOTE RULES Vertical Asymptotes: located at zeros of q(x). 

Horizontal Asymptotes: (at most one) a. If degree of the numerator < degree of the denominator: y = 0 b. If degree of the numerator = degree of the denominator y = ratio of lead coefficients

c. If degree of the numerator > degree of the denominator no horizontal asymptote  Slant Asymptote: only if the degree of the numerator is exactly one more than the degree of denominator. Divide numerator by denominator. y = ax+b is the slant asymptote. **In order to get a good sketch of the graph, you must plot some points between and beyond each x-intercept and vertical asymptote.

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Example 4For each rational function, determine any

horizontal or vertical asymptotes.

a)

b)

c)

f (x)

6x 1

3x 3

g(x)

x 1

x2 4

h(x)

x2 1

x 1

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ExampleSolution

a)

HA: Degree of numerator and denominator are both 1. Since the ratio of the leading coefficients is 6/3, the horizontal asymptote is y = 2.

VA: When x = –1, the denominator, 3x + 3, equals 0 and the numerator, 6x – 1 does not equal 0, so the vertical asymptote is x = 1.

f (x)

6x 1

3x 3

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ExampleSolution continued

a)

Here’s a graph of f(x).

f (x)

6x 1

3x 3

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ExampleSolution continued

b)

HA: Degree of numerator is one less than the degree of the denominator so the x-axis, or y = 0, is a horizontal asymptote.

VA: When x = ±2, the denominator, x2 – 4, equals 0 and the numerator, x + 1 does not equal 0, so the vertical asymptotes are x = 2 and x = 2.

g(x)

x 1

x2 4

Copyright © 2010 Pearson Education, Inc.

ExampleSolution continued

b)

Here’s a graph of g(x).

g(x)

x 1

x2 4

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ExampleSolution

c)

HA: Degree of numerator is greater than the degree of the denominator so there are no horizontal asymptotes.

VA: When x = –1, both the numerator and denominator equal 0 so the expression is not in lowest terms: g(x) = x – 1, x ≠ –1. There are no vertical asymptotes.

h(x)

x2 1

x 1

Copyright © 2010 Pearson Education, Inc.

ExampleSolution

c)

Here’s the graph of h(x).A straight line with thepoint (–1, –2) missing.

Why isn’t there a vertical asymptote at x=-1?

h(x)

x2 1

x 1

Graph f (x) 2

x 4

x

y

Vertical asymptote at x = 4

Horizontal asymptote at y = 0

x intercept?

y intercept?

x f(x)

5

2

0 -1/2

2

-1

None1

0,2

x

y

Graph 2

6( )

6f x

x x

Vertical asymptote ?

Horizontal asymptote ?

X intercepts?

Y intercepts?

Domain?

x = 2 and x = -3

y = 0

(0, -1)

None

2, 3x

Evaluate at convenient values of x.

x

y

Graph f (x) x2 25

x 5

5x

Factor:

5 5

( )5

x xf x

x

Graph looks like ( ) 5

with a hole at 5.

f x x

x

5xDomain:

Analyze the function. Find the following:

Vertical asymptotes:

horizontal asymptotes:

x intercepts:

y intercepts:

Domain:

2 1

3

xy

x

x

y

Analyze the function. Find the following:

Vertical asymptotes:

horizontal asymptotes:

x intercepts (reduce 1st):

y intercepts:

Domain:

2

2 8

9 20

xy

x x

2( 4)

4 5

x

x x

2

5x

Analyze the graph of:

2

2

2 3( )

1

xf x

x

X intercepts?

Vertical asymptote ?

Horizontal asymptote ?

Domain?

Range?

Y intercepts?

Homework

• P.148

• 7-19 all, 31-34 all, 35, 39

• (table of values not required)