## Graphing Rational Functions

In our final section on rational functions, we will bring all the skills we have learned together to draw graphs of ration functions without the help of a calculator. We will use asymptotes, intercepts, and general characteristics of the function in question. You will also be able to determine the equation of a rational function given a graph.

### Graphing Rational Functions

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

• At the $x$-intercept $x=−1$ corresponding to the $(x+1)^{2}$ factor of the numerator, the graph "bounces", consistent with the quadratic nature of the factor.

• At the $x$-intercept $x=3$ corresponding to the $(x−3)$ factor of the numerator, the graph passes through the axis as we would expect from a linear factor.

• At the vertical asymptote $x=−3$ corresponding to the $(x+3)^{2}$ factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function $f(x)=\frac{1}{x^{2}}$.

• At the vertical asymptote $x=2$, corresponding to the $(x−2)$ factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side.

#### HOW TO

Given a rational function, sketch a graph.

1. Evaluate the function at $0$ to find the $y$-intercept.

2. Factor the numerator and denominator.

3. For factors in the numerator not common to the denominator, determine where each factor of the numerator is zero to find the $x$-intercepts.

4. Find the multiplicities of the $x$-intercepts to determine the behavior of the graph at those points.

5. For factors in the denominator, note the multiplicities of the zeros to determine the local behavior. For those factors not common to the numerator, find the vertical asymptotes by setting those factors equal to zero and then solve.

6. For factors in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to $0$ and then solve.

7. Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes.

8. Sketch the 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