5.6 Rational functions  (Page 4/16)

Vertical asymptotes

The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.

Identifying vertical asymptotes

Find the vertical asymptotes of the graph of $k(x)=\frac{5+2x^2}{2-x-x^2}.$

First, factor the numerator and denominator.

$$\begin{array}{ccc}\hfill k(x)& =& \frac{5+2x^2}{2-x-x^2}\hfill \\ & =& \frac{5+2x^2}{(2+x)(1-x)}\hfill \end{array}$$

To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:

$$\begin{array}{ccc}\hfill (2+x)(1-x)& =& 0\hfill \\ \hfill x& =& \mathrm{-2},1\hfill \end{array}$$

Neither $x=\mathrm{–2}$ nor $x=1$ are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in [link] confirms the location of the two vertical asymptotes.

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Removable discontinuities

Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity    .

For example, the function $f(x)=\frac{x^2-1}{x^2-2x-3}$ may be re-written by factoring the numerator and the denominator.

Notice that $x+1$ is a common factor to the numerator and the denominator. The zero of this factor, $x=\mathrm{-1},$ is the location of the removable discontinuity. Notice also that $x–3$ is not a factor in both the numerator and denominator. The zero of this factor, $x=3,$ is the vertical asymptote. See [link] . [Note that removable discontinuities may not be visible when we use a graphing calculator, depending upon the window selected.]

Identifying vertical asymptotes and removable discontinuities for a graph

Find the vertical asymptotes and removable discontinuities of the graph of $k(x)=\frac{x-2}{x^2-4}.$

Factor the numerator and the denominator.

$$k(x)=\frac{x-2}{(x-2)(x+2)}$$

Notice that there is a common factor in the numerator and the denominator, $x–2.$ The zero for this factor is $x=2.$ This is the location of the removable discontinuity.

Notice that there is a factor in the denominator that is not in the numerator, $x+2.$ The zero for this factor is $x=\mathrm{-2.}$ The vertical asymptote is $x=\mathrm{-2.}$ See [link] .

The graph of this function will have the vertical asymptote at $x=\mathrm{-2},$ but at $x=2$ the graph will have a hole.

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