Holt McDougal Algebra 2

Rational FunctionsRational Functions

Holt Algebra 2

Holes & Slant AsymptotesHoles & Slant Asymptotes

Holt McDougal Algebra 2

Holt McDougal Algebra 2

Rational Functions

Now we find out why we have to start out factoring….to find holes in the graph.

Holt McDougal Algebra 2

Rational Functions

In some cases, both the numerator and the denominator of a rational function will equal 0 for a particular value of x. As a result, the function will be undefined at this x-value. If this is the case, the graph of the function may have a hole. A hole is an omitted point in a graph.

Holt McDougal Algebra 2

Rational Functions

Example : Graphing Rational Functions with Holes

(x – 3)(x + 3)x – 3

f(x) =

Identify holes in the graph of f(x) = . Then graph.

x2 – 9 x – 3

Factor the numerator.

The expression x – 3 is a factor of both the numerator and the denominator. Set it = to 0 to find the x- coordinate of the hole.

There is a hole in the graph at x = 3.

Divide out common factors.

(x – 3)(x + 3)(x – 3)

For x ≠ 3,

f(x) = = x + 3

Holt McDougal Algebra 2

Rational Functions

Example Continued

The graph of f is the same as the graph of y = x + 3, except for the hole at x = 3. On the graph, indicate the hole with an open circle. The domain of f is all real #’s except 3

Hole at x = (3, 6)

To find the y-coordinate of the hole plug 3 into the reduced equation.

Holt McDougal Algebra 2

Rational Functions

Check It Out! Example 5

(x – 2)(x + 3)x – 2

f(x) =

Identify holes in the graph of f(x) = . Then graph.

x2 + x – 6 x – 2

Factor the numerator.

The expression x – 2 is a factor of both the numerator and the denominator. Set it = to 0 to find the x- coordinate of the hole.

There is a hole in the graph at x = 2.

Divide out common factors.

For x ≠ 2,

f(x) = = x + 3(x – 2)(x + 3)(x – 2)

Holt McDougal Algebra 2

Rational Functions

Check It Out! Example 5 Continued

The graph of f is the same as the graph of y = x + 3, except for the hole at x = 2. On the graph, indicate the hole with an open circle. The domain of f is {x|x ≠ 2}.

Hole at x = 2

x 2

f xx 2 x 2

Domain: , 2 , 2, 2 , 2,

Vertical Asymptotes: x 2

Horizontal Asymptotes: y 0

Holes: 1

2,4

Intercepts:

1

x 2

10,

2

10,

2

2

2

x 5x 6f x

x 2x 3

Domain: , 1 , 1, 3 , 3,

Vertical Asymptotes: x 1

Horizontal Asymptotes: y 1

Holes: 1

3,4

Intercepts: 0, 2

2, 0

x 2 x 3

x 1 x 3

Domain: , 3 , 3, 3 , 3,

Vertical Asymptotes: x 3

Horizontal Asymptotes: y 0

Holes: 1

3,2

Intercepts: 0, 1

2

3x 9f x

x 9

3 x 3

x 3 x 3

3

x 3

Holt McDougal Algebra 2

Rational Functions

Let’s go back and look at the worksheet and find the problems with holes.

Slant Asymptotes

Slant asymptotes occur when the degree of the numerator is exactly one bigger than the degree of the denominator. In this case a slanted line (not horizontal and not vertical) is the function’s asymptote.

To find the equation of the asymptote we need to use long division – dividing the numerator by the denominator.

Holt McDougal Algebra 2

Rational Functions

When dividing to find slant asymptotes:

• Do synthetic division (if possible); if not, do long division!

• The resulting polynomial (ignoring the remainder) is the equation of the slant asymptote.

EXAMPLE: Finding the Slant Asymptoteof a Rational Function

Find the slant asymptotes of f (x) 2 4 5 .

3x x

x

Solution Because the degree of the numerator, 2, is exactly one more than the degree of the denominator, 1, the graph of f has a slant asymptote. To find the equation of the slant asymptote, divide x 3 into x2 4x 5:

2 1 4 51 3 3 1 1 8

3

2

81 13

3 4 5

xx

x x x

Remainder

Rational Functions and Their Graphs

moremore

EXAMPLE: Finding the Slant Asymptoteof a Rational Function

Find the slant asymptotes of f (x) 2 4 5 .

3x x

x

Solution The equation of the slant asymptote is y x 1. Using our

strategy for graphing rational functions, the graph of f (x) is

shown.

2 4 53

x xx

-2 -1 4 5 6 7 8321

7

6

5

4

3

1

2

-1

-3

-2

Vertical asymptote: x = 3

Vertical asymptote: x = 3

Slant asymptote: y = x - 1

Slant asymptote: y = x - 1

3.6: Rational Functions and Their Graphs

Graph: 22 3

( )1

x xf x

x

Notice that in this function, the degree of the numeratoris larger than the denominator. Thus n>m and there is nohorizontal asymptote. However, if n is one more than m,the rational function will have a slant asymptote.

To find the slant asymptote, divide the numerator by the denominator:

2

2

2 5

1 2 3

2 2

5

5 5

5

x

x x x

x x

x

x

The result is . We ignore the remainder and the line is a slant asymptote.

512 5 xx

2 5y x

Graph:

22 3

1

x x

x

1st, find the vertical asymptote.

2nd , find the x-intercepts: and

3rd , find the y-intercept:

4th , find the horizontal asymptote. none

0,0

0,0

3,0

2

1x

2 3

1

x x

x

5th , find the slant asymptote: 2 5y x

6th , sketch the graph.

A Rational Function with a Slant Asymptote

Graph the rational function

• Factoring:

2 4 5( )

3

x xr x

x

( 1)( 5)

( 3)

x xy

x

A Rational Function with a Slant Asymptote

Finding the x-intercepts:

• –1 and 5 (from x + 1 = 0 and x – 5 = 0)

Finding the y-intercepts:

• 5/3 (because )20 4 0 5 5

(0)0 3 3

r

A Rational Function with a Slant Asymptote

Finding the horizontal asymptote:• None (because degree of numerator is greater

than degree of denominator)

Finding the vertical asymptote:• x = 3 (from the zero of the denominator)

A Rational Function with a Slant Asymptote

Finding the slant asymptote:

• Since the degree of the numerator is one more than the degree of the denominator, the function has a slant asymptote.

• Dividing, we obtain:

• Thus, y = x – 1 is the slant asymptote.

8( ) 1

3r x x

x

A Rational Function with a Slant Asymptote

Here are additional values and

the graph.

So far, we have considered only horizontal

and slant asymptotes as end behaviors for

rational functions.

• In the next example, we graph a function whose end behavior is like that of a parabola.

Slant Asymptotes and End Behavior

Finding a Slant Asymptote

If

There will be a slant asymptote because the degree of the numerator (3) is one bigger than the degree of the denominator (2).

Using long division, divide the numerator by the denominator.

1

9522

23

xx

xxxxf

Finding a Slant AsymptoteCon’t.

39521 232

xxxxxx

xxx 23

943 2 xx

333 2 xx

127 x

Finding a Slant AsymptoteCon’t.

We can ignore the remainder

The answer we are looking for is the quotient

and the equation of the slant asymptote is

127 x

3x3xy

Graph of Example 7

The slanted line y = x + 3 is the slant asymptote

Graph the rational function which has the following characteristics

Vert Asymp at x = 1, x = -3

Horz Asymp at y = 1

Intercepts (-2, 0), (3, 0), (0, 2)

Passes through (-5, 2)

Graph the rational function which has the following characteristics

Vert Asymp at x = 1, x = -1

Horz Asymp at y = 0

Intercepts (0, 0)

Passes through (-0.7, 1), (0.7, -1), (-2, -0.5), (2, 0.5)

Holt McDougal Algebra 2

Rational Functions

zero: (2,0); asymptotes: x = 0, y = 1; hole at (1, -1)

Identify the zeros, asymptotes, and holes in the graph of . Then graph. x2 – 3x + 2

x2 – x f(x) =