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*}
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Guidance
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}{x3}\\
&(x3)(y)=(x3)\left(\frac{x}{x3}\right)\\
& xy3y=\cancel{(x3)}\left(\frac{x}{\cancel{x3}}\right)\\
& xy3y=x\\
& xyx=3y\\
& x(y1)=3y\\
&\boxed{x=\frac{3y}{y1}}\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*} and this function has a horizontal asymptote at \begin{align*}y = 1\end{align*}.
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 xintercepts (zeros) of the function. The zeros of the function will be the values for \begin{align*}x\end{align*} that cause the numerator, but not also the denominator, to be equal to zero.
Example A
Use technology to sketch the graph of the rational function. Find all zeros and asymptotes and label those on your sketch.
\begin{align*}y=\frac{1}{x^29}=\frac{1}{(x+3)(x3)}\end{align*}
Solution: 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 xintercepts 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*} and the other 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*} is to look at the degrees of the numerators and denominators. The degree is the highest exponent. The degree of the numerator is 0 and the degree of the denominator is 2. In general, if the degree of the numerator is less than the degree of the denominator, there will be a horizontal asymptote at \begin{align*}y=0\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.
Example B
Use technology to sketch the graph of the rational function. Find all zeros and asymptotes and label those on your sketch.
\begin{align*}y=\frac{1}{x}\end{align*}
Solution: 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 xintercepts 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*}. You can use algebra to solve the equation for \begin{align*}x\end{align*} and look for the values of \begin{align*}y\end{align*} that will cause the denominator to be equal to zero:

\begin{align*}& y=\frac{1}{x}\\
& xy=1\\
&\frac{xy}{y}=\frac{1}{y}\\
&\boxed{x=\frac{1}{y}}\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.
Example C
Use technology to sketch the graph of the rational function. Find all zeros and asymptotes and label those on your sketch.
\begin{align*}y=\frac{x+2}{x3}\end{align*}
Solution: 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 (xintercept) 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*}. You can use algebra to solve the equation for \begin{align*}x\end{align*} and look for the values of \begin{align*}y\end{align*} that will cause the denominator to be equal to zero:
\begin{align*}& y=\frac{x+2}{x3}\\ &(x3)(y)=\left(\frac{x+2}{x3}\right)(x3)\\ & xy3y=x+2\\ & xyx=2+3y\\ & x(y1)=2+3y\\ &\boxed{x=\frac{3y+2}{y1}}\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.
Concept Problem Revisited
\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*}.
Guided Practice
1. Sketch the graph of the rational function: \begin{align*}y=\frac{x+1}{x}\end{align*}
2. Sketch the graph of the rational function: \begin{align*}y=\frac{x^2}{x3}\end{align*}
3. Find the asymptotes of the function: \begin{align*}y=\frac{1}{x^216}\end{align*}
Answers:
1. The zero of the denominator is 0. This means that there will be a vertical asymptote at \begin{align*}x=0\end{align*} (the yaxis). 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}{y1}\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 xintercept at 1.
2. 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 xintercept at 0.
3. 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:
Explore More
Sketch the graph of each of the following rational functions.
 \begin{align*}y=\frac{2}{x+3}\end{align*}
 \begin{align*}y=\frac{x}{x1}\end{align*}
 \begin{align*}y=\frac{1}{x^24}\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^2x12}\end{align*}
 \begin{align*}y=\frac{x1}{x+3}\end{align*}
 \begin{align*}y=\frac{x1}{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^2x2}\end{align*}
 \begin{align*}y=\frac{2}{x4}\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{x1}{x+3}\end{align*}