ASYMPTOTES TUTORIAL

HorizontalVertical

Slantand Holes

Do Now: graph the following and name the asymptotes:

y =1

x−4

Definition of an asymptote

An asymptote is a straight line which acts as a boundary for the graph of a function.

When a function has an asymptote (and not all functions have them) the function gets closer and closer to the asymptote as the input value to the function approaches either a specific value a or positive or negative infinity.

The functions most likely to have asymptotes are rational functions

Vertical Asymptotes

Vertical asymptotes occur when thefollowing condition is met: The denominator of the simplified rational function is equal to 0. Remember, the simplified rational function has cancelled any factors common to both the numerator and denominator.

Finding Vertical AsymptotesExample 1

Given the function

The first step is to cancel any factors common to both numerator and denominator. In this case there are none.

The second step is to see where the denominator of the simplified function equals 0.

x

xxf

22

52

1

01

012

022

x

x

x

x

Finding Vertical Asymptotes Example 1 Con’t.

The vertical line x = -1 is the only vertical asymptote for the function. As the input value x to this function gets closer and closer to -1 the function itself looks and acts more and more like the vertical line

x = -1.

Graph of Example 1

The vertical dotted line at x = –1 is the vertical asymptote.

Rational Functions and Asymptotes

Arrow notation:

€

x → a+

x → a−

x →+∞

x → −∞

Meaning:

x approaches a from the right

x approaches a from the left

x is approaching infinity, increasing forever

x is approaching infinity, decreasing forever

Vertical Asymptotes

The line x = a is a vertical asymptote of a function if f(x) increases or decreases without bound as x approaches a.

If

Then:

€

x → a+ or x → a−

€

f x( ) →+∞ or f x( ) → −∞

Graph of Example 1

The vertical dotted line at x = –1 is the vertical asymptote.

limx→−1+

=∞

limx→−1−

=−∞

Finding Vertical AsymptotesExample 2

If

First simplify the

function. Factor

both numerator

and denominator

and cancel any

common factors.

9

121022

2

x

xxxf

2x2 +10x+12x2 −9

=x+ 3( ) 2x+ 4( )x+ 3( ) x−3( )

=2x+ 4x−3

*There is no asymptote at x = -3!

Finding Vertical Asymptotes Example 2 Con’t.

The asymptote(s) occur where the simplified denominator equals 0.

The vertical line x=3 is the only vertical asymptote for this function.

As the input value x to this function gets closer and closer to 3 the function itself looks more and more like the vertical line x=3.

3

03

x

x

Graph of Example 2

The vertical dotted line at x = 3 is the vertical asymptote

limx→ 3+

=∞

limx→ 3−

=−∞

Finding Vertical AsymptotesExample 3

If

Factor both the numerator and denominator and cancel any common factors.

6

52

xx

xxg

Finding Vertical AsymptotesExample 3

If

Factor both the numerator and denominator and cancel any common factors.In this case there are no common factors to cancel.

6

52

xx

xxg

32

5

6

52

xx

x

xx

x

Finding Vertical AsymptotesExample 3 Con’t.The denominator equals zero whenever

either

or

This function has two vertical asymptotes, one at x = -2 and the other at x = 3

2

02

x

x

3

03

x

x

Graph of Example 3

The two vertical dotted lines at x = -2 and x = 3 are the vertical asymptotes

Write 2 limit notations for each one.

Horizontal asymptotes occur when either one of the following conditions is met (you should notice that both conditions cannot be true for the same function).Horizontal asymptotes guide the end behavior of graph.

If the degree of the numerator < degree of the denominator then the horizontal asymptote will be y= 0.

If the degree of the numerator = degree of denominator then the horizontal asymptote will be

If the degree of the numerator > degree of the denominator then there is no horizontal asymptote.

Find the horizontal asymptote:

1. f x( ) =

2x−1x+ 2

x. f x

x

3

2

12

H.A. : y 2

H.A. : none

H.A. : y 0

Exponents are the same; divide the coefficients

Bigger on Top; None

Bigger on Bottom; y=0

Finding Horizontal AsymptotesExample 4

If

then there is a horizontal asymptote at the line y=0 because the degree of the numerator (2) is less than the degree of the denominator (3). This means that as x gets larger and larger in both the positive and negative directions (x → ∞ and x → -∞)

the function itself looks more and more like the horizontal line y = 0

27

533

2

x

xxxf

Graph of Example 4

The horizontal line y = 0 is the horizontal asymptote.

limx→∞

=0 and limx→−∞

=0

Finding Horizontal Asymptotes Example 5

If

then because the degree of the numerator (2) is equal to the degree of the denominator (2)

there is a horizontal asymptote at the line y=6/5.

Note, 6 is the leading coefficient of the numerator and 5 is the leading coefficient of the denominator. As x→∞ and as x→-∞ g(x) looks

more and more like the line y=6/5

975

5362

2

xx

xxxg

Graph of Example 5

The horizontal dotted line at y = 6/5 is the

horizontal asymptote.

limx→∞

=65and lim

x→−∞=65

Finding End behavior Asymptotes Do Now: graph:

If

There are no horizontal asymptotes because the degree of the numerator is greater than the degree of the denominator.

1

9522

3

x

xxxf

Graph of previous function

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 using synthetic division, (if possible) or long division – dividing the numerator by the denominator.

Finding slant asymptotes with synthetic division:

f (x) =2x2 −3x+5

x+2

This one has a linear denominator so we can use synthetic division

Note: slants and horizontals don’t happen together!

Finding slant asymptotes with synthetic division is only possible when the denominator is linear!

f (x) =2x2 −3x+5

x+2

y =2x−7(Remainder was 19)

End behavior Asymptotes

Graph the following function:

The end behavior asymptote will be a quadraticfunction because the numerator is 2 degreeslarger in the numerator

Use synthetic division to find it.

f (x) =x3 −2xx+1

End behavior Asymptotes

Graph the following function:

Note that the remainders don’t matter here!

f (x) =x3 −2xx+1

y =x2 −x−1

Using long division

Find the quotient and remainder:

1

9522

3

x

xxxf

The slant asymptote

We divide: note that here we will put the first answer over the 5x.

1

9522

3

x

xxxf

The slant asymptote

We divide and find it….

1

9522

3

x

xxxf

x2 +1 −2x3 +0x2 +5x−9

(−2x3

−2x

−2x )

(7x −9)

The slant asymptote

We divide and find it….Y = -2x is the slant asymptote!

1

9522

3

x

xxxf

x2 +1 −2x3 +0x2 +5x−9

(−2x3

−2x

−2x )

(7x −9)

7x-9 is just a remainder

Graph of Example 6

Y=-2x

Finding a Slant Asymptote Example 7

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 AsymptoteExample 7 Con’t.

xxx 23

943 2 xx

333 2 xx

127 x

Finding a Slant AsymptoteExample 7 Con’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

3x

3xy

Graph of Example 7

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

Finding slant asymptotes with synthetic division:

f (x) =2x2 −3x+5

x+2

divide using synthetic division

Finding slant asymptotes with synthetic division:

f (x) =2x2 −3x+5

x+2

divide using long division

Finding slant asymptotes with synthetic division is only possible when the denominator is linear!

f (x) =2x2 −3x+5

x+2

y =2x−7(Remainder was 19)

Finding Intercepts of Rational Functions

x-intercept: solve f (x) = 0

y-intercept: evaluate f (0)

Do now: find the x and y intercepts:y=2x+8

Intercepts of Rational Functions

Example

x-intercept: solve f (x) = 0

.352

1)(

2

xxx

xf

.1 isintercept - The 1

01

0352

12

xx

xxx

x

Setting the function = 0 is essentially setting the numerator= to 0!

Intercepts of Rational Functions

y-intercept: evaluate f (0)

31

isintercept - The 31

3)0(5)0(210

)0( 2

yf

Find all asymptotes and intercepts

Example Graph

Vertical Asymptote:

Horizontal Asymptote:

x-intercept:

y-intercept:

.312

)(

xx

xf

Find the asymptotes and intercepts

Example

Solution Vertical Asymptote:

Horizontal Asymptote:

x-intercept:

y-intercept:

.312

)(

xx

xf

303 xx

212 yy

21or 01203

12 xxx

x

31

301)0(2

)0( f

Graph all the asymptotes and interceptsx=3, y=2 (-.5,0) (0, -1/3)

.312

)(

xx

xf

Graph all the asymptotes and interceptsx=3, y=2 (-.5,0) (0, -1/3)

.312

)(

xx

xf

But find a point for themissing part-try f(6)

Graph all the asymptotes and interceptsx=3, y=2 (-.5,0) (0, -1/3)

.312

)(

xx

xf

But find a point for themissing part-try f(6)

f (6) =2(6)+16−3

=133=4

13

End behavior:

Use long division and divide to find the slant asymptote:

Do Now:

Use long division and divide:

x3 −x2 2 x4 0x3 −2x2 3x6−(x4 − x3 +2x)

x3 −2x2 + x+6

x +1

−(x3 − x2 2)

−x2 + x + 4 remainder

Slant asymptote:Y = x + 1

Quadratic and cubic end behavior asymptotes….Use long division to find the quotient:

f (x) x4 2x3 −2x1x2 −3x 4

Quadratic and cubic end behavior asymptotes….Use long division to find the quotient:

Quadratic end behavior asymptote:

(Note: the remainder was 11x-43)

f (x) x4 2x3 −2x1x2 −3x 4

y =x2 +5x+11

do now:find the asymptotes

Holes occur in the graph of a rational function whenever the numerator and denominator have common factors. The holes occur at the x value(s) that make the common factors equal to 0.

The hole is known as a removable discontinuity.When you graph the function on your calculator you

won’t be able to see the hole but the function is still discontinuous.

f x( ) =2x2 +10x+12

x2 −9

Finding a Hole

We were able to cancel the (x + 3) in the numerator and denominator before finding the vertical asymptote.

Because (x + 3) is a common factor there will be a hole at the point where

Finding the coordinates of the hole: substitute -3 into the remaining factors:

f x( ) =2x2 +10x+12

x2 −9=

x+ 3( ) 2x+ 4( )x+ 3( ) x−3( )

3

03

x

x

2x + 4x−3

=2(−3)+ 4−3−3

=13

Finding a Hole

There is a hole at

Where are the intercepts and asymptotes?

f x( ) =2x2 +10x+12

x2 −9=

x+ 3( ) 2x+ 4( )x+ 3( ) x−3( )

(−3,13)

Finding a Hole

is a hole

(-2,0),and are intercepts

x=3, y=2 are the asymptotes

Sketch the function

f x( ) =2x2 +10x+12

x2 −9=

x+ 3( ) 2x+ 4( )x+ 3( ) x−3( )

(0,−43)

(−3,13)

Graph of Example

Notice there is a hole in the graph at the point where x = -3. You would not be able to see this hole if you graphed the curve on your calculator (but it’s there just the same.)

limx→ 3+

=∞

limx→ 3_

=−∞

Finding a Hole

Use the rational root theorem to factor the numerator

Factor both numerator and denominator to see if there are any common factors.

Find all the intercepts, holes and asymptotes toSketch the function

4

82

3

x

xxf

Finding a Hole

Factor both numerator and denominator to see if there are any common factors.

there will be a hole at (2,3) asymptotes are x=-2And a slant at y = x (long division) and an

intercept at (0,2). No x intercepts.

22

422

4

8 2

2

3

xx

xxx

x

xxf

4

82

3

x

xxf

Graph of last example

There is a hole in the curve at the point where x = 2. This curve also has a vertical asymptote at x = -2 and a slant asymptote y = x.

ProblemsFind the vertical asymptotes, horizontal asymptotes, slant asymptotes and holes and intercepts for each of the following functions. Not all will be found for each. Graph g(x)

2

2

2 15

7 10

x xf x

x x

Vertical: Horizontal : Slant: Hole:

22 5 7

3

x xg x

x

Vertical: Horizontal : Slant: Hole:

Problems

2

2

2 15

7 10

x xf x

x x

Vertical: x = -2Horizontal : y = 1Slant: noneHole: at x = - 5x int -3y int -.5

22 5 7

3

x xg x

x

Vertical: x = 3Horizontal : noneSlant: y = 2x +11Hole: nonex int: -3.5,1y int: 7/3

Graph of g(x) 22 5 7

3

x xg x

x

Vertical: x = 3Horizontal : noneSlant: y = 2x +11Hole: none int:

-3.5,1y int: 7/3

Graphing a rational function.

1. Find the vertical asymptotes if they exist.2. Find the horizontal asymptote if it exists.3. Find the x intercept(s) by finding f(x) = 0.4. Find the y intercept by finding f(0).5. Find any crossing points on the horizontal

asymptote by setting f(x) = the horizontal asymptote.

6. Graph the asymptotes, intercepts, and crossing points.

7. Find additional points in each section of the graph as needed.

8. Make a smooth curve in each section through known points and following asymptotes and end behavior.