What if you had a function like
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CK12 Foundation: 1202S RGraphs of Rational Functions
Guidance
Graphs of rational functions are very distinctive, because they get closer and closer to certain values but never reach those values. This behavior is called asymptotic behavior, and we will see that rational functions can have horizontal asymptotes, vertical asymptotes or oblique (or slant) asymptotes.
Now we’ll extend the domain and range of rational equations to include negative values of
As we graph rational functions, we need to always pay attention to values of
Example A
Graph the function
Solution
Before we make a table of values, we should notice that the function is not defined for
Let’s make two tables: one for







0.1 

4 

0.2 

3 

0.3 

2 

0.4 

1 

0.5 

0.5 

1 

0.4 

2 

0.3 

3 

0.2 

4 

0.1 

5 

We can see that as we pick positive values of
Notice on the graph that for values of
We also notice that as the absolute values of
Asymptotes are usually denoted as dashed lines on a graph. They are not part of the function; instead, they show values that the function approaches, but never gets to. A horizontal asymptote shows the value of \begin{align*}y\end{align*}
Now we’ll show the graph of a rational function that has a vertical asymptote at a nonzero value of \begin{align*}x\end{align*}
Example B
Graph the function \begin{align*}y = \frac{1}{(x  2)^2}\end{align*}
Solution
We can see that the function is not defined for \begin{align*}x = 2\end{align*}
Now let’s make a table of values.
\begin{align*}x\end{align*} 
\begin{align*}y = \frac {1}{(x2)^2}\end{align*} 

0 
\begin{align*}y= \frac {1}{(02)^2} = \frac {1}{4}\end{align*} 
1 
\begin{align*}y= \frac {1}{(12)^2} = 1\end{align*} 
1.5 
\begin{align*}y= \frac {1}{(1.52)^2} = 4\end{align*} 
2  \begin{align*}\text{undefined}\end{align*} 
2.5  \begin{align*}y = \frac {1}{(2.52)^2} = 4\end{align*} 
3  \begin{align*}y= \frac {1}{(32)^2} = 1\end{align*} 
4  \begin{align*}y= \frac {1}{(42)^2} = \frac {1}{4}\end{align*} 
Here’s the resulting graph:
Notice that we didn’t pick as many values for our table this time, because by now we have a pretty good idea what happens near the vertical asymptote.
We also know that for large values of \begin{align*}x\end{align*}, the value of \begin{align*}y\end{align*} could approach a constant value. In this case that value is \begin{align*}y = 0\end{align*}: this is the horizontal asymptote.
A rational function doesn’t have to have a vertical or horizontal asymptote. The next example shows a rational function with no vertical asymptotes.
Example C
Graph the function \begin{align*}y= \frac{x^2}{x^2 + 1}\end{align*}.
Solution
We can see that this function will have no vertical asymptotes because the denominator of the fraction will never be zero. Let’s make a table of values to see if the value of \begin{align*}y\end{align*} approaches a particular value for large values of \begin{align*}x\end{align*}, both positive and negative.
\begin{align*}x\end{align*}  \begin{align*}y= \frac {x^2}{x^2+1}\end{align*} 

\begin{align*}3\end{align*}  \begin{align*}y= \frac {(3)^2}{(3)^2+1} = \frac {9}{10} = 0.9\end{align*} 
2  \begin{align*}y= \frac {(2)^2}{(2)^2+1} = \frac {4}{5} = 0.8\end{align*} 
1  \begin{align*}y= \frac {(1)^2}{(1)^2+1} = \frac {1}{2} = 0.5\end{align*} 
0  \begin{align*}y= \frac {(0)^2}{(0)^2+1} = \frac {0}{1} = 0\end{align*} 
1  \begin{align*}y= \frac {(1)^2}{(1)^2+1} = \frac {1}{2} = 0.5\end{align*} 
2  \begin{align*}y= \frac {(2)^2}{(2)^2+1} = \frac {4}{5} = 0.8\end{align*} 
3  \begin{align*}y= \frac {(3)^2}{(3)^2+1} = \frac {9}{10} = 0.9\end{align*} 
Below is the graph of this function.
The function has no vertical asymptote. However, we can see that as the values of \begin{align*}x\end{align*} get larger, the value of \begin{align*}y\end{align*} gets closer and closer to 1, so the function has a horizontal asymptote at \begin{align*}y = 1\end{align*}.
Watch this video for help with the Examples above.
CK12 Foundation: Graphs of Rational Functions
Vocabulary
 Graphs of rational functions are very distinctive, because they get closer and closer to certain values but never reach those values. This behavior is called asymptotic behavior, and we will see that rational functions can have horizontal asymptotes, vertical asymptotes or oblique (or slant) asymptotes.
Guided Practice
Graph the function \begin{align*}y=\frac{3}{x1}\end{align*}.
Solution
Start by making a table of values:
\begin{align*}x\end{align*}  \begin{align*}y= \frac {3}{x1}\end{align*} 

\begin{align*}2\end{align*}  \begin{align*}y= \frac {3}{21} = \frac {3}{3} = 1\end{align*} 
1  \begin{align*}y= \frac {3}{11} = \frac {3}{2} = 1.5\end{align*} 
0  \begin{align*}y= \frac {3}{01} = \frac {3}{1} = 3\end{align*} 
1  undefined 
2  \begin{align*}y= \frac {3}{21} = \frac {3}{1} = 3\end{align*} 
3  \begin{align*}y= \frac {3}{31}= \frac {3}{2} = 1.5\end{align*} 
4  \begin{align*}y= \frac {3}{41}= \frac {3}{3} = 1\end{align*} 
Next, graph the points. Recall that the function \begin{align*}y=\frac{1}{x}\end{align*} has two curves, that are on either side of the vertical asymptote, which is where the function is undefined. The same is true for this function.
Practice
Graph the following rational functions. Draw dashed vertical and horizontal lines on the graph to denote asymptotes.
 \begin{align*}y=\frac {2}{x3}\end{align*}
 \begin{align*}y=\frac {3}{x^2}\end{align*}
 \begin{align*}y=\frac {x}{x1}\end{align*}
 \begin{align*}y=\frac {2x}{x+1}\end{align*}
 \begin{align*}y=\frac {1}{x^2+2}\end{align*}
 \begin{align*}y=\frac {x}{x^2+9}\end{align*}
 \begin{align*}y=\frac {x^2}{x^2+1}\end{align*}
 \begin{align*}y=\frac {1}{x^21}\end{align*}
 \begin{align*}y=\frac {2x}{x^29}\end{align*}
 \begin{align*}y=\frac {x^2}{x^216}\end{align*}
 \begin{align*}y=\frac {3}{x^24x+4}\end{align*}
 \begin{align*}y=\frac {x}{x^2x6}\end{align*}