Skip to main content
\(\require{cancel}\newcommand\degree[0]{^{\circ}} \newcommand\Ccancel[2][black]{\renewcommand\CancelColor{\color{#1}}\cancel{#2}} \newcommand{\alert}[1]{\boldsymbol{\color{magenta}{#1}}} \newcommand{\blert}[1]{\boldsymbol{\color{blue}{#1}}} \newcommand{\bluetext}[1]{\color{blue}{#1}} \delimitershortfall-1sp \newcommand\abs[1]{\left|#1\right|} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \)

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
\(4\) \(2\)
\(4\) \(\dfrac{1}{2}\)
\(6\) \(\dfrac{2}{9}\)
graph showing 6 points about a vertical asymptote
Figure335

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.

table of values and graph of rational function
Figure336

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.

4 cases of behavior near vertical asymptote
Figure338
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.

table and graph of rational funciton
Figure340
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. \(x\) \(-3\) \(-1\) \(1\) \(3\)
    \(y\) \(\frac{1}{5} \) \(\frac{-1}{3} \) \(\frac{-1}{3} \) \(\frac{1}{5} \)
    rational function

SubsectionExercises