1

Warm-up• Solve the following rational equation.

xx

x

xx

2

4

1

4

122

: 1

: 4

Solution x

Extr x

2

Set Equation to ZERO

xx

x

xx

2

4

1

4

122

02

4

1

4

122

xx

x

xx

Next Slide

12 1 20

( 4) 4

x

x x x x

12 ( 1) 2( 4)0

( 4)

x x x

x x

3

Problem Continued

0)4(

)4(2)1(12

xx

xxx

08212 2 xxx

0432 xx

0432 xx

0)1)(4( xx

14 xorx

x = -4 Extraneous

x = 1

MUST CHECK ANSWERS

x = -4 does not work

Rational Function Discontinuities

Section 2-6

5

Objectives

• I can identify Graph Discontinuities– Vertical Asymptotes – Horizontal Asymptotes – Slant Asymptotes– Holes

• I can find “x” and “y” intercepts

6

Rational Functions

• A rational function is any ratio of two polynomials, where denominator cannot be ZERO!

• Examples:

1)(

x

xxf

103

1)(

2

xx

xxf

7

Asymptotes• Asymptotes are the boundary lines that a

rational function approaches, but never crosses.

• We draw these as Dashed Lines on our graphs.

• There are three types of asymptotes: – Vertical– Horizontal (Graph can cross these)– Slant

8

Vertical Asymptotes

• Vertical Asymptotes exist where the denominator would be zero.

• They are graphed as Vertical Dashed Lines

• There can be more than one!

• To find them, set the denominator equal to zero and solve for “x”

9

Example #1

• Find the vertical asymptotes for the following function:

1)(

x

xxf

•Set the denominator equal to zero

•x – 1 = 0, so x = 1

•This graph has a vertical asymptote at x = 1

10

1 2 63 4 5 7 8 9 10

4

3

2

7

5

6

8

9

x-axis

y-axis

0

1-2-6 -3-4-5-7-8-910

-4

-3

-2

-1

-7

-5

-6

-8

-9

0

-1

Vertical Asymptote at

X = 1

11

Other Examples:

• Find the vertical asymptotes for the following functions:

3

3)(

xxg

)5)(2(

1)(

xx

xxg

3: xVA

5:

2:

xVA

xVA

To find Vertical Asymptote(s)1) Set reduced denominator = 0

2) Solve for x = #.

3) Your answer is written as a line.

To find Vertical Asymptote(s)1) Set reduced denominator = 0

2) Solve for x = #.

3) Your answer is written as a line.

Horizontal Asymptotes

• Horizontal Asymptotes are also Dashed Lines drawn horizontally to represent another boundary.

• To find the horizontal asymptote you compare the degree of the numerator with the degree of the denominator

• See next slide:

Horizontal Asymptote (HA)

Given Rational Function:

Compare DEGREE of Numerator to Denominator

If N < D , then y = 0 is the HA

If N > D, then the graph has NO HA

If N = D, then the HA is

Numerator( )

Denominatorf x

N

D

LCy

LC

15

Example #1

• Find the horizontal asymptote for the following function:

1)(

x

xxf

•Since the degree of numerator is equal to degree of denominator (m = n)

•Then HA: y = 1/1 = 1

•This graph has a horizontal asymptote at y = 1

1 2 63 4 5 7 8 9 10

4

3

2

7

5

6

8

9

x-axis

y-axis

0

1-2-6 -3-4-5-7-8-910

-4

-3

-2

-1

-7

-5

-6

-8

-9

0

-1

Horizontal Asymptote at

y = 1

Other Examples:

• Find the horizontal asymptote for the following functions:

3

3)(

xxg

13

13)(

2

2

xx

xxg

5

1)(

3

x

xxg

0: yHA

3: yHA

NoneHA :

To find Horizontal Asymptote(s)

1) Compare DEGREE of numerator and denominator

Num BIGGER then NO HA

Num SMALLER then y = 0

Degree is SAME then

To find Horizontal Asymptote(s)

1) Compare DEGREE of numerator and denominator

Num BIGGER then NO HA

Num SMALLER then y = 0

Degree is SAME then Num

Den

LCy

LC

19

Slant Asymptotes (SA)

• Slant asymptotes exist when the degree of the numerator is one larger than the denominator.

• Cannot have both a HA and SA• To find the SA, divide the Numerator by the

Denominator. • The results is a line y = mx + b that is the

SA.

Example of SA

20

22 4 8( )

2

x xf x

x

2 4 8 -2

2

4

8

16

8

2 8y x

Remainder does not matter

To find Slant Asymptote(s)1) DEGREE of Numerator must be

ONE bigger than Denominator

2) Divide with Synthetic or Long Division

3) Don’t use the RemainderGet y = mx + b

To find Slant Asymptote(s)1) DEGREE of Numerator must be

ONE bigger than Denominator

2) Divide with Synthetic or Long Division

3) Don’t use the RemainderGet y = mx + b

22

Holes

• A hole exists when the same factor exists in both the numerator and denominator of the rational expression and the factor is eliminated when you reduce!

Example of Hole Discontinuity

23

( 4)( 1)( )

( 2)( 4)

x xf x

x x

Cancel LIKE factors

( 1)( )

( 2)

xf x

x

____, ____4

( 4 1) 5 5

( 4 2) 6 6

5

6

( 3)( 3)

( 3)( 1)

x xf x

x x

HOLES

To Find Holes1) Factor.2) Reduce. 3) A hole is formed when a factor is

eliminated from the denominator.4) Set eliminated factor = 0 and solve for

x.5) Find the y-value of the hole by

substituting the x-value into reduced form and solve for y.

6) Your answer is written as a point. (x, y)

To Find Holes1) Factor.2) Reduce. 3) A hole is formed when a factor is

eliminated from the denominator.4) Set eliminated factor = 0 and solve for

x.5) Find the y-value of the hole by

substituting the x-value into reduced form and solve for y.

6) Your answer is written as a point. (x, y)

To find x- intercept(s)

1) Set reduced numerator = 0

2) Solve for x.

3) Answer is written as a point.

(#, 0)

To find x- intercept(s)

1) Set reduced numerator = 0

2) Solve for x.

3) Answer is written as a point.

(#, 0)

To find y- intercept

1) Substitute 0 in for all x’s in reduced form.

2) Solve for y.3) Answer is a point. (0, #)

To find y- intercept

1) Substitute 0 in for all x’s in reduced form.

2) Solve for y.3) Answer is a point. (0, #)

27

Homework

• WS 5-3