Can you use your graphing calculator to help you sketch the graph of the rational function? Does this function have any zeros or asymptotes?

\begin{align*}y=\frac{4}{x^2+1}\end{align*}

### Graphing Rational Functions

A rational function is the quotient of two polynomial functions. In general,

\begin{align*}f(x)=\frac{A(x)}{B(x)}\end{align*}

where \begin{align*}A\end{align*}

Notice that the function has two pieces. In between those two pieces are the asymptotes.

- A vertical asymptote occurs at the \begin{align*}x\end{align*}
x value(s) that cause the denominator of the function to be equal to zero (which is undefined). This function has a vertical asymptote at \begin{align*}x=3\end{align*}x=3 . - A horizontal asymptote occurs at the \begin{align*}y\end{align*}
y value(s) that cause the denominator of the function to be zero if the function is rewritten and solved for \begin{align*}x\end{align*}x instead of \begin{align*}y\end{align*}y . This is what it looks like to solve for \begin{align*}x\end{align*}x :

- \begin{align*}& y=\frac{x}{x-3}\\
&(x-3)(y)=(x-3)\left(\frac{x}{x-3}\right)\\
& xy-3y=\cancel{(x-3)}\left(\frac{x}{\cancel{x-3}}\right)\\
& xy-3y=x\\
& xy-x=3y\\
& x(y-1)=3y\\
&\boxed{x=\frac{3y}{y-1}}\end{align*}
y=xx−3(x−3)(y)=(x−3)(xx−3)xy−3y=(x−3)(xx−3)xy−3y=xxy−x=3yx(y−1)=3yx=3yy−1

- Thus, \begin{align*}y \ne 1\end{align*}
y≠1 and this function has a horizontal asymptote at \begin{align*}y = 1\end{align*}y=1 .

The image below shows the graph with the asymptotes drawn in and labelled. For rational functions, the asymptotes represent the lines that the function will approach but never touch.

It is also important to note the x-intercepts (zeros) of the function. The zeros of the function will be the values for \begin{align*}x\end{align*}

#### Let's graph the following rational functions:

- \begin{align*}y=\frac{1}{x^2-9}=\frac{1}{(x+3)(x-3)}\end{align*}
y=1x2−9=1(x+3)(x−3)

Here is the sketch from the calculator:

It can sometimes be hard to interpret what you see on the graphing calculator screen. Use algebra to find the asymptotes and zeros and sketch those first.

There are no zeros for the numerator. Therefore, there are no x-intercepts for the function.

The zeros of the denominator are 3 and –3. This means that there are two vertical asymptotes. One vertical asymptote is the line \begin{align*}x = 3\end{align*}

Another way to determine a horizontal asymptote besides solving the equation for \begin{align*}x\end{align*}

Once you have sketched the asymptotes, use the table and/or graph from the graphing calculator to decide what the rest of the graph looks like. Here is the graph with the asymptotes labelled.

- \begin{align*}y=-\frac{1}{x}\end{align*}
y=−1x

Here is the sketch from the calculator:

Use algebra to find the asymptotes and zeros and sketch those first.

There are no zeros for the numerator. Therefore, there are no x-intercepts for the function.

The zero of the denominator is 0. This means that there is one vertical asymptote, the line \begin{align*}x = 0\end{align*}

The horizontal asymptote is \begin{align*}y = 0\end{align*}

- \begin{align*}& y=-\frac{1}{x}\\
& xy=-1\\
&\frac{xy}{y}=\frac{-1}{y}\\
&\boxed{x=-\frac{1}{y}}\end{align*}
y=−1xxy=−1xyy=−1yx=−1y

Once you have sketched the asymptotes, use the table and/or graph from the graphing calculator to decide what the rest of the graph looks like. Here is the graph with the asymptotes labelled.

- \begin{align*}y=\frac{x+2}{x-3}\end{align*}
y=x+2x−3

Here is the sketch from the calculator:

Use algebra to find the asymptotes and zeros and sketch those first.

The zero of the numerator is –2 so the zero (x-intercept) of the function is (–2, 0).

The zero of the denominator is 3. This means that there is one vertical asymptote, the line \begin{align*}x = 3\end{align*}

The horizontal asymptote is \begin{align*}y = 1\end{align*}

\begin{align*}& y=\frac{x+2}{x-3}\\ &(x-3)(y)=\left(\frac{x+2}{x-3}\right)(x-3)\\ & xy-3y=x+2\\ & xy-x=2+3y\\ & x(y-1)=2+3y\\ &\boxed{x=\frac{3y+2}{y-1}}\end{align*}

Once you have sketched the asymptotes, use the table and/or graph from the graphing calculator to decide what the rest of the graph looks like. Here is the graph with the asymptotes labelled.

### Examples

#### Example 1

Earlier, you were asked if the following function has any zeros or asymptotes:

\begin{align*}y=\frac{4}{x^2+1}\end{align*}

Here is a sketch of the function:

- There are no zeros for this function since there are no zeros for the numerator. The graph does not cross the \begin{align*}x\end{align*}-axis.
- There are no zeros for the denominator. Therefore, there are no vertical asymptotes.
- Because the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at \begin{align*}y=0\end{align*}.

#### Example 2

Sketch the graph of the rational function: \begin{align*}y=\frac{x+1}{x}\end{align*}

The zero of the denominator is 0. This means that there will be a vertical asymptote at \begin{align*}x=0\end{align*} (the y-axis). Because the degree of the denominator and numerator are the same, you can solve the equation for \begin{align*}x\end{align*} and get \begin{align*}x=\frac{1}{y-1}\end{align*}. The zero of the denominator is now 1. This means there is a horizontal asymptote at \begin{align*}y=1\end{align*}. The numerator has a zero at -1, so there is an x-intercept at -1.

#### Example 3

Sketch the graph of the rational function: \begin{align*}y=\frac{x^2}{x-3}\end{align*}

The zero of the denominator is 3. This means that there will be a vertical asymptote at \begin{align*}x=3\end{align*}. Because the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes. The zero of the numerator is 0, so there is an x-intercept at 0.

#### Example 4

Find the asymptotes of the function: \begin{align*}y=\frac{1}{x^2-16}\end{align*}

The zeros of the denominator are 4 and –4. This means that there will be two vertical asymptotes. One vertical asymptote will be the line \begin{align*}x = 4\end{align*} and the other will be the line \begin{align*}x = -4\end{align*}. Because the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at \begin{align*}y=0\end{align*}. Though you did not need to sketch the graph, here is the graph of the function:

### Review

Sketch the graph of each of the following rational functions.

- \begin{align*}y=\frac{2}{x+3}\end{align*}
- \begin{align*}y=\frac{x}{x-1}\end{align*}
- \begin{align*}y=\frac{1}{x^2-4}\end{align*}
- \begin{align*}y=\frac{x+2}{x}\end{align*}
- \begin{align*}y=\frac{1}{x^2+2}\end{align*}
- \begin{align*}y=\frac{x}{x+2}\end{align*}
- \begin{align*}y=\frac{1}{x^2-x-12}\end{align*}
- \begin{align*}y=\frac{x-1}{x+3}\end{align*}
- \begin{align*}y=\frac{x-1}{x+4}\end{align*}
- \begin{align*}y=\frac{5}{x^2+1}\end{align*}

Without graphing the following rational functions, state what you know about their asymptotes and zeros.

- \begin{align*}y=\frac{1}{x^2-x-2}\end{align*}
- \begin{align*}y=-\frac{2}{x-4}\end{align*}
- \begin{align*}y=-\frac{2}{x^2+1}\end{align*}
- \begin{align*}y=\frac{6}{x^2+1}\end{align*}
- \begin{align*}y=\frac{x-1}{x+3}\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 8.4.