
3.5: Rational Functions


1. Rational Functions (Definition)

Definition: Rational Function

A rational function is a quotient of polynomials $$\dfrac{P(x)}{Q(x)}$$.

Example 1

$\dfrac{(x^2 + x - 1)}{(3x^3+ 1)},$

$\dfrac{(x - 1)}{(x^2 +1)}, \text{ and}$

$\dfrac{x^2}{(x + 1)}$

are all Rational Functions

Example 2

Find the domain of

$\dfrac{(x^2 + 1)}{(x^2 -1)}.$

Solution

The domain of this rational function is the set of all real numbers that do not make the denominator zero. We find

$x^2 -1 = 0$

solving

$x = 1, \;\;\; \text{or} \;\;\; x = -1.$

So that the domain is

$\{x | x \text{ is not }1 \text{ or } -1\}.$

2. Vertical Asymptotes

Definition: Vertical Asymptote

A Vertical Asymptote of a rational function occurs where the denominator is 0.

Example 3

Graph the vertical asymptotes of

$\dfrac{(x^2 + 1)}{(x^2 -1)}$

Solution

From the last example, we see that there are vertical asymptotes at 1 and -1.

Since $$f(x)$$ is positive a little to the left of -1, we say that as

$x \rightarrow -1^{-} \text{ ("x goes to -1 from the left")},$

$f(x) \rightarrow \infty.$

Similarly since $$f(x)$$ is negative a little to the right of -1, we say that as

$x\rightarrow -1^{+} \text{( "x goes to -1 from the right")},$

$f(x) \rightarrow -\infty.$

Since $$f(x)$$ is negative a little to the left of 1, as

$x \rightarrow1^{-},$

$f(x) \rightarrow -\infty.$

Similarly since $$f(x)$$ is positive a little to the right of 1, as

$x\rightarrow1^{+},$

$f(x) \rightarrow\infty.$

Four Types of Vertical Asymptotes

Below are the four types of vertical asymptotes:

3. Horizontal Asymptotes

Example 4

Consider the rational function

$f(x) = \dfrac{(3x^2 + x - 1)}{(x^2 - x - 2)}.$

For the numerator, the term $$3x^2$$ dominates when $$x$$ is large, while for the denominator, the term $$x^2$$ dominates when $$x$$ is large. Hence as

$x \rightarrow \infty,$

$f(x)\rightarrow\dfrac{3x^2}{x^2}=3.$

3 is called the horizontal asymptote and we have the the left and right behavior of the graph is a horizontal line $$y = 3$$.

4. Oblique Asymptotes

Consider the function

$f(x) = \dfrac{(x^2 - 3x - 4)}{(x + 3)}$

$$f(x)$$ does not have a horizontal asymptote, since

$\dfrac{x^2}{x}= x$

is not a constant, but we see (on the calculator) that the left and right behavior of the curve is like a line. Our goal is to find the equation of this line.

We use synthetic division to see that

$\dfrac{(x^2 - 3x - 4)}{(x + 3)} = x - 6 + \dfrac{14}{(x+3)}.$

For very large $$x$$,

$\dfrac{14}{x} + 3$

is very small, hence $$f(x)$$ is approximately equal to

$x - 6$

on the far left and far right of the graph. We call this line an Oblique Asymptote.

To graph, we see that there is a vertical asymptote at

$x = -3$

with behavior:

left down and right up

The graph has x-intercepts at 4 and -1, and a y intercept at $$-\frac{4}{3}$$.

Exercise

Graph

$\dfrac{(x^3 + 8)}{(x^2 - 3x - 4)}$

5. Rational Functions With Common Factors

Consider the graph of

$y = \dfrac{x-1}{x-1}$

What is wrong with the picture? When

$f(x) = \dfrac{g(x)(x - r)}{h(x)(x - r)}$

with neither $$g(r)$$ nor $$h(r)$$ zero, the graph will have a hole at $$x = r$$. We call this hole a removable discontinuity.

Example

Graph

\begin{align} f(x) &= \dfrac{(x^2 - 2)}{(x^2 - x - 2)} = \dfrac{(x - 2)(x + 2)}{(x - 2)(x + 1)}.\end{align}

This graph will have a vertical asymptote at $$x =-1$$ and a hole at $$(2,2)$$.

We end our discussion with a list of steps for graphing rational functions.

Steps in graphing rational functions:

• Step 1 Plug in $$x = 0$$ to find the y-intercept
• Step 2 Factor the numerator and denominator. Cancel any common factors remember to put in the appropriate holes if necessary.
• Step 3 Set the numerator = 0 to find the x-intercepts
• Step 4 Set the denominator = 0 to find the vertical asymptotes. Then plug in nearby values to fine the left and right behavior of the vertical asymptotes.
• Step 5 If the degree of the numerator = degree of the denominator then the graph has a horizontal asymptote. To determine the value of the horizontal asymptote, divide the term highest power of the numerator by the term of highest power of the denominator.
• If the degree of the numerator = degree of the denominator + 1, then use polynomial or synthetic division to determine the equation of the oblique asymptote.
• Step 6 Graph it!

Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.