Rational Functions

Rational Functions

In this chapter, you'll learn what a rational function is, and you'll learn how to sketch the graph of a rational function.

Rational functions

A rational function is a fraction of polynomials. That is, if p(x) and q(x) are polynomials, then

p(x) () qx is a rational function. The numerator is p(x) and the denominator is q(x).

Examples.

?

3( 5) x

(x 1)

?1

x

?

23 x 1

=

23 x

The last example is both a polynomial and a rational function. In a similar way, any polynomial is a rational function.

In this class, from this point on, most of the rational functions that we'll see will have both their numerators and their denominators completely factored.

We will also only see examples where the numerator and the denominator have no common factors. (If they did have a common factor, we could just cancel them.)

*************

Implied domains

The implied domain of a rational function is the set of all real numbers for the roots of the denominator. That's because it doesn't make

except sense to divide by 0.

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x-intercepts

The -intercepts of a rational function p(x) (if there are any) are the numbers

x

2 R where

q(x)

()

p =0

()

q

If is such a number, then we can multiply by q() to find that

p() = 0 ? q() = 0

In other words, is a root of ( ). Thus, the roots of the numerator are

px

exactly the x-intercepts.

Example. 2 is the only -intercept of the rational function x

7( 2)( 2 + 1) xx

8(x 4)(x 6)

*************

In between x-intercepts and vertical asymptotes

When graphing a rational polynomial, first mark the vertical asymptotes

and the x-intercepts. Then choose a number c 2 R between any consecutive

pairs of these marked points on the -axis and see if the rational function is x

positive or negative when x = c. If it's positive, draw a dot above the x-axis

whose first coordinate is . If it's negative, draw a dot below the -axis whose

c

x

first coordinate is . c

Example. Let's look at the function

7( 2)( 2 + 1) ( )= x x r x 8( 4)( 6)

xx

again. The -intercept of its graph is at = 2 and it has vertical asymptotes

x

x

at = 4 and = 6. We need to decide whether ( ) is positive or negative

x

x

rx

between 2 and 4 on the x-axis, and between 4 and 6 on the x-axis.

Let's start by choosing a number between 2 and 4, say 3. Then

7(3 2)(32 + 1) r(3) = 8(3 4)(3 6)

199

Notice that 7, (3 4), and (3 6) are negative, while 8, (3 2), and (32 + 1) are positive.

If you are multiplying and dividing a collection of numbers that aren't equal to 0, just count how many negative numbers there are. If there is an even number of negatives, the result will be positive. If there is an odd number of negatives, the result will be negative. In the previous paragraph, there are three negative numbers -- 7, (3 4), and (3 6) -- so (3) 0.

r< The number 5 is a number that is in between 4 and 6, and

7(5 2)(32 + 1)

(5) =

0

r

8(5 4)(5 6) >

*************

Far right and far left

Let n be the leading term of ( ) and let m be the leading term of ( ).

ax

px

bx

qx

Recall that far to the right and left, ( ) looks like its leading term, n.

px

ax

And far to the right and left, ( ) looks like its leading term, m. It follows

qx

bx

that the far right and left portion of the graph of,

() px

() qx

looks like

n

ax

m

bx

and this is a function that we know how to graph.

Example. The leading term of 7( 2)( 2 + 1) is 7 3, and the leading

xx

x

term of 8( 4)( 6) is 8 2. Therefore, the graph of

xx

x

7( 2)( 2 + 1)

( )= rx

x 8(

x 4)(

6)

xx

looks like the graph of

73 x

=

7

82 8x

x

on the far left and far right part of its graph.

200

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