Rational FunctionsRational Functions

An introductionAn introduction

L. Waihman

A function is CONTINUOUS if you can draw the graph without lifting your pencil.

A POINT OF DISCONTINUITY occurs whenthere is a break in the graph.

Note the break in the graphwhen x=3. Why?

Look at the equation of the graph. Where is this equation undefined?

2 93

xx

We can factor the numerator and reducethe fraction to determine that the graphwill be a line; however, the undefined pointremains, so there is a point of discontinuitypoint of discontinuity here.

3 33

3

x xx

x

There are three basic kinds of discontinuity: point, jump, and infinitepoint, jump, and infinite.

The greatest integer function is an exampleof a jump discontinuityjump discontinuity.

Tangent, Cotangent, secant and cosecant functions are all examples of infinite infinite discontinuitiesdiscontinuities.

The previous function was an example of apoint of discontinuity.point of discontinuity.

A rational function rational function is the quotientof at least two polynomials.

The graphs of rational functions frequentlydisplay infinite and point discontinuities.

Rational functions have vertical asymptotesand may have horizontal asymptotes as well.

Let’s look at the parent function: 1xf x

If x = 0, then the entire function is undefined.Thus, there is a vertical asymptote at x=0.

Looking at the graph, you can see that the value of thefunction , as the values of x 0 from the positiveside; and the value of the function - , as the valuesof x 0 from the negative side. These are the limits ofthe function and are written as:

0lim ( )x

f x

0

lim ( )x

f x

DomainDomain

• The domain is then limited to:The domain is then limited to:

,0 0,

• To find the domain of a rational function,To find the domain of a rational function, set the denominator equal to zero.set the denominator equal to zero.

• The domain will always be all real The domain will always be all real numbers except those values found by numbers except those values found by solving this equation.solving this equation.

Determine the domain of these rational functions:

2

1. 2

xf x

x

2

2

2 92.

xf x

x

3

2

13.

4

xf x

x

: , 2 2,D

: ,0 0,D

: , 2 2,2 2,D

Recall that a vertical asymptote occurs whenthere is a value for which the function isundefined. This means, if there are nocommon factors, anywhere the denominator equals zero.

Remember that asymptotes are lines. When you label a vertical asymptote, you must write the equation of the vertical line.

Just make x equal everything it couldn’t in the domain.

State the vertical asymptotes:

2

1. 2

xf x

x

3

2

12.

4

xf x

x

V.A. : x 2

V.A. : x 2

Let’s say x is any positive number. As that value increases,the value of the entire function decreases; but, it willnever become zero or negative.

So this part of the graph will nevercross the x-axis. We express this using limit notation as:

What if x is a negative number? As that value decreases, the value of the entire function increases; but, it will never become zero or positive.So this part of the graph will never

cross the x-axis, either and:

lim 0x

f x

lim 0x

f x

Thus, the line is a horizontal asymptote.

As x ∞, f(x) 0, and as x ∞, f(x) 0.

0y

Given: is a polynomial of degree n , is a polynomial of degree m , and , 3 possibleconditions determinea horizontal asymptote:

f x g x f xg x

•If n<m, then is a horizontal asymptote.•If n>m, then there is NO horizontal asymptote.•If n=m, then is a horizontal asymptote, where cis the quotient of the leading coefficients.

0y

y c

Horizontal Asymptotes

BOBO BOTN EATS D/C

Bigger On Bottom y=0 Bigger On Top – None

Exponents Are The Same; Divide the Coefficients

Find the horizontal asymptote:

x

. f xx

2 11

2

x. f x

x

3

2

12

x

. f xx x2

23

20

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

Suppose that you were asked to graph:

3 51

xf xx

1st, determine where the graph is undefined. (Set the denominator to zero and solve for the variable.)

2nd , find the x-intercept by setting the numerator = to 0and solving for the variable.

1x

So, the graph crosses the x-axis at

5,0

3

1 0

1

x

x

There is a vertical asymptotevertical asymptote here.

3 5 0

3 5

5

3

x

x

x

Draw a dotted line at:

0, 5

3rd , find the y-intercept by letting x=0 and solving for y.

3(0) 5

0 15

1

y

y

4th , find the horizontal asymptote. (Note: the exponents are the same so divide the coefficients - EATS D/C) 3/1 = 3

3y

So, the graph crosses the y-axis at

The horizontal asymptote is:

Now, put all the information together and sketchthe graph:

Graph:

4 35

xf xx

1st, find the vertical asymptote.

2nd , find the x-intercept .

3rd , find the y-intercept.

5x

3,0

4

30,

5

4th , find the horizontal asymptote. 4y

5th , sketch the graph.

Graph: 2

2

2( )

6

x xf x

x x

1st, factor the entire equation:

1 2

3 2

x x

x x

2nd , find the x-intercepts:

3rd , find the y-intercept:

3

2

x

x

1,0

2,0

10,

3

4th , find the horizontal asymptote: 1y

5th , sketch the graph.

Then find the vertical asymptotes:

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 . Notice that as the values of x increase, the fractional part decreases (goes to 0), so thefunction approaches the line . Thus the lineis a slant asymptote.

512 5 xx

2 5x 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.