
## SectionShort-Run Behavior of Rational Functions

### SubsectionVertical Asymptotes and Holes

We have previously seen that a polynomial function is defined for all values of $x\text{,}$ and its graph is a smooth curve without any breaks or holes. The graph of a rational function, on the other hand, will have breaks or holes at those $x$-values where it is undefined.

###### Example334

Investigate the graph of $f(x) = \displaystyle{\frac{2}{(x - 3)^2}}$ near $x = 3\text{.}$

Solution

This function is undefined for $x = 3\text{,}$ so there is no point on the graph with $x$-coordinate $3\text{.}$ However, we can make a table of values for other values of $x\text{.}$ Plotting the ordered pairs in the table results in the points shown in Figure335.

 $x$ $y$ $0$ $\dfrac{2}{9}$ $1$ $\dfrac{1}{2}$ $2$ $2$ $3$ undefined $4$ $2$ $4$ $\dfrac{1}{2}$ $6$ $\dfrac{2}{9}$

Next, we make a table showing $x$-values close to $3\text{,}$ as in Figure336a. As we choose $x$-values closer and closer to $3\text{,}$ $(x - 3)^2$ gets closer to $0\text{,}$ so the fraction $\dfrac{2}{(x-3)^2}$ gets very large. This means that the graph approaches, but never touches, the vertical line $x=3\text{.}$ In other words, the graph has a vertical asymptote at $x=3\text{.}$ We indicate the vertical asymptote by a dashed line, as shown in Figure336b.

The vertical line $x=3$ in the example above is an example of a vertical asymptote. In general, we have the following result.

###### Vertical Asymptotes

If $Q(a) = 0$ but $P(a) \ne 0\text{,}$ then the graph of the rational function $f(x) = \displaystyle{\frac{P(x)}{Q(x)}}$ has a vertical asymptote at $x=a\text{.}$

However, if both the numerator and denominator of the rational function share a root, we no longer call this a vertical asymptote.

If $P(a)$ and $Q(a)$ are both zero in the rational function $f(x)=\frac{P(x)}{Q(x)}\text{,}$ but the reduced expression for $f(x)$ is defined at $x=a\text{,}$ then the graph of $f(x)$ has a hole at $x=a$ rather than an asymptote.

###### Example337

Find the vertical asymptotes of $G(x) = \displaystyle{\frac{4x^2}{x^2 - 4}}\text{.}$

Solution

To find the vertical asyptotes, we are looking for values of $a$ such that $4a^2 \neq 0$ and $a^2-4=0\text{.}$ The equation $a^2-4=0$ has solutions $a=2,-2\text{,}$ and both $4(\alert{2})^2\neq0$ and $4(\alert{-2})^2\neq0\text{.}$ This means that there are two vertical asymptotes, namely $x=2$ and $x=-2\text{.}$

Near a vertical asymptote, the graph of a rational function has one of the four characteristic shapes, illustrated in Figure338. Locating the vertical asymptotes can help us make a quick sketch of a rational function.

###### Example339

Locate the vertical asymptotes and sketch the graph of $g(x) = \displaystyle{\frac{x}{x + 1}}\text{.}$

Solution

The denominator, $x+1\text{,}$ equals zero when $x = -1\text{.}$ Because the numerator does not equal zero when $x = -1\text{,}$ there is a vertical asymptote at $x = -1\text{.}$ The asymptote separates the graph into two pieces.

We can use the Table feature of a calculator to evaluate $g(x)$ for several values of $x$ on either side of the asymptote, as shown in Figure340a. We plot the points found in this way; then connect the points on either side of the asymptote to obtain the graph shown in Figure340b.

###### Example341
1. Find the vertical asymptotes of $f(x) = \displaystyle{\frac{1}{x^2 - 4}}$ . Locate any $x$-intercepts.

2. Evaluate the function at $x = -3\text{,}$ $-1\text{,}$ $1\text{,}$ and $3\text{.}$ Sketch a graph of the function.

Solution
1. The denominator, $x^2-4\text{,}$ equals zero when $x=2$ and $x=-2\text{.}$ Because the numerator does not equal zero for either of these, there are two vertical asymptotes, at $x=2$ and $x=-2\text{.}$

To find the $x$-intercepts, we are looking for values of $x$ such that the function is equal to zero:

\begin{equation*} \dfrac{1}{x^2-4} = 0 \end{equation*}

Since $1$ divided by any real number is nonzero, there is no such $x\text{.}$ This means that there are no $x$-intercepts.

2.  $x$ $-3$ $-1$ $1$ $3$ $y$ $\frac{1}{5}$ $\frac{-1}{3}$ $\frac{-1}{3}$ $\frac{1}{5}$