MA001: College Algebra

In Example 9, we see that the numerator of a rational function reveals the $x$-intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials.

The vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity. See Figure 17.

Figure 17

When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides. See Figure 18.

Figure 18

For example, the graph of $f(x)=\frac{(x+1)^{2}(x-3)}{(x+3)^{2}(x-2)}$ is shown in Figure 19.

Figure 19

HOW TO

Given a rational function, sketch a graph.

EXAMPLE 11

Graphing a Rational Function

Sketch a graph of $f(x)=\frac{(x+2)(x-3)}{(x+1)^{2}(x-2)}$.

Solution

We can start by noting that the function is already factored, saving us a step.

Next, we will find the intercepts. Evaluating the function at zero gives the $y$-intercept:

$\begin{aligned} f(0) &=\frac{(0+2)(0-3)}{(0+1)^{2}(0-2)} \\ &=3 \end{aligned}$

To find the $x$-intercepts, we determine when the numerator of the function is zero. Setting each factor equal to zero, we find $x$-intercepts at $x=–2$ and $x=3$. At each, the behavior will be linear (multiplicity 1), with the graph passing through the intercept.

We have a $y$-intercept at $(0,3)$ and $x$-intercepts at $(–2,0)$ and $(3,0)$.

To find the vertical asymptotes, we determine when the denominator is equal to zero. This occurs when $x+1=0$ and when $x–2=0$, giving us vertical asymptotes at $x=–1$ and $x=2$.

There are no common factors in the numerator and denominator. This means there are no removable discontinuities.

Finally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal asymptote at $y=0$.

To sketch the graph, we might start by plotting the three intercepts. Since the graph has no $x$-intercepts between the vertical asymptotes, and the $y$-intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph as shown in Figure 20.

Figure 20

The factor associated with the vertical asymptote at $x=−1$ was squared, so we know the behavior will be the same on both sides of the asymptote. The graph heads toward positive infinity as the inputs approach the asymptote on the right, so the graph will head toward positive infinity on the left as well.

For the vertical asymptote at $x=2$, the factor was not squared, so the graph will have opposite behavior on either side of the asymptote. See Figure 21. After passing through the $x$-intercepts, the graph will then level off toward an output of zero, as indicated by the horizontal asymptote.

Figure 21

TRY IT #8

Given the function $f(x)=\frac{(x+2)^{2}(x-2)}{2(x-1)^{2}(x-3)}$, use the characteristics of polynomials and rational functions to describe its behavior and sketch the function.

Source: Rice University, https://openstax.org/books/college-algebra/pages/5-6-rational-functions
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