Active Prelude to Calculus

Subsection 5.5.1 When a rational function has a “hole”

Example 5.5.1.

Consider the rational function \(r(x) = \frac{x^2 - 1}{x^2 - 3x - 4}\) from Preview Activity 5.5.1, whose numerator is \(p(x) = x^2 - 1\) and whose denominator is \(q(x) = x^2 - 3x - 4\text{.}\) Explain why the graph of \(r\) generated by Desmos or another computational device is incorrect, and also identify the locations of any zeros and vertical asymptotes of \(r\text{.}\)

Solution.

It is helpful with any rational function to factor the numerator and denominator. We note that \(p(x) = x^2 - 1 = (x-1)(x+1)\) and \(q(x) = x^2 - 3x - 4 = (x+1)(x-4)\text{.}\) The domain of \(r\) is thus the set of all real numbers except \(x = -1\) and \(x = 4\text{,}\) the set of all points where \(q(x) \ne 0\text{.}\)

Knowing that \(r\) is not defined at \(x = -1\text{,}\) it is natural to study the graph of \(r\) near that value. Plotting the function in Desmos, we get a result similar to the one shown in Figure 5.5.2, which appears to show no unusual behavior at \(x = -1\text{,}\) and even that \(r(-1)\) is defined. If we zoom in on that point, as shown in Figure 5.5.3, the technology still fails to visually demonstrate the fact that \(r(-1)\) is not defined. This is because graphing utilities sample functions at a finite number of points and then connect the resulting dots to generate the curve we see.

We know from our algebraic work with the denominator, \(q(x) = (x+1)(x-4)\text{,}\) that \(r\) is not defined at \(x = -1\text{.}\) While the denominator \(q\) gets closer and closer to \(0\) as \(x\) approaches \(-1\text{,}\) so does the numerator, since \(p(x) = (x-1)(x+1)\text{.}\) If we consider values close but not equal to \(x = -1\text{,}\) we see results in Table 5.5.5.

\(x\)	\(-1.1\)	\(-1.01\)	\(-1.001\)
\(r(x)\)	\(\frac{0.21}{0.51} \approx 0.4118 \)	\(\frac{0.0201}{0.0501} \approx 0.4012\)	\(\frac{0.002001}{0.005001} \approx 0.4001\)
\(x\)	\(-0.9\)	\(-0.99\)	\(-0.999\)
\(r(x)\)	\(\frac{-0.19}{-0.49} \approx 0.3878\)	\(\frac{-0.0199}{-0.0499} \approx 0.3989\)	\(\frac{-0.001999}{-0.004999} \approx 0.3999\)

In the table, we see that both the numerator and denominator get closer and closer to \(0\) as \(x\) gets closer and closer to \(-1\text{,}\) but that their quotient appears to be getting closer and closer to \(y = 0.4\text{.}\) Indeed, we see this behavior in the graph of \(r\text{,}\) though the graphing utility misses the fact that \(r(-1)\) is actually not defined. A precise graph of \(r\) near \(x = -1\) should look like the one presented in Figure 5.5.4, where we see an open circle at the point \((-1, 0.4)\) that demonstrates that \(r(-1)\) is not defined, and that \(r\) does not have a vertical asymptote or zero at \(x = -1\text{.}\)

Finally, we also note that \(p(1) = 0\) and \(q(1) = -6\text{,}\) so at \(x = 1\text{,}\) \(r(x)\) has a zero (its numerator is zero and its denominator is not). In addition, \(q(4) = 0\) and \(p(4) = 15\) (its denominator is zero and its numerator is not), so \(r(x)\) has a vertical asymptote at \(x = 4\text{.}\) These features are accurately represented by the original Desmos graph shown in Figure 5.5.2.

Subsection 5.5.2 Sign charts and finding formulas for rational functions

Example 5.5.7.

Construct a sign chart for the function \(q(x) = \frac{(x-2)(x^2-9)}{(x-3)(x-1)^2}\text{.}\) Then, graph the function \(q\) and compare the graph and sign chart.

Solution.

First, we fully factor \(q\) and identify the \(x\)-values that are not in its domain. Since \(x^2-9 = (x-3)(x+3)\text{,}\) we see that

\begin{equation*} q(x) = \frac{(x-2)(x-3)(x+3)}{(x-3)(x-1)^2}\text{.} \end{equation*}

From the denominator, we observe that \(q\) is not defined at \(x = 3\) and \(x = 1\) since those values make the factors \(x - 3 = 0\) or \((x-1)^2 = 0\text{.}\) Thus, the domain of \(q\) is the set of all real numbers except \(x = 1\) and \(x = 3\text{.}\) From the numerator, we see that both \(x = 2\) and \(x = -3\) are zeros of \(q\) since these values make the numerator zero while the denominator is nonzero. We expect that \(q\) will have a hole at \(x = 3\) since this \(x\)-value is not in the domain and it makes both the numerator and denominator \(0\text{.}\) Indeed, computing values of \(q\) for \(x\) near \(x = 3\) suggests that

\begin{equation*} \lim_{x \to 3} q(x) = 1.5\text{,} \end{equation*}

and thus \(q\) does not change sign at \(x = 3\text{.}\)

Thus, we have three different \(x\)-values to place on the sign chart: \(x = -3\text{,}\) \(x = 1\text{,}\) and \(x = 2\text{.}\) We now analyze the sign of each of the factors in \(q(x) = \frac{(x-2)(x-3)(x+3)}{(x-3)(x-1)^2}\) on the various intervals. For \(x \lt -3\text{,}\) \((x-2) \lt 0\text{,}\) \((x-3) \lt 0\text{,}\) \((x+3) \lt 0\text{,}\) and \((x-1)^2 \gt 0\text{.}\) Thus, for \(x \lt -3\text{,}\) the sign of \(q\) is

\begin{equation*} \frac{- - -}{- +} = + \end{equation*}

since there are an even number of negative terms in the quotient.

On the interval \(-3 \lt x \lt 1\text{,}\) \((x-2) \lt 0\text{,}\) \((x-3) \lt 0\text{,}\) \((x+3) \gt 0\text{,}\) and \((x-1)^2 \gt 0\text{.}\) Thus, for these \(x\)-values, the sign of \(q\) is

\begin{equation*} \frac{- - +}{- +} = -\text{.} \end{equation*}

Using similar reasoning, we can complete the sign chart shown in Figure 5.5.8. A plot of the function \(q\text{,}\) as seen in Figure 5.5.9, shows behavior that matches the sign function, as well as the need to manually identify the hole at \((3, 1.5)\text{,}\) which is missed by the graphing software.

In both the sign chart and the figure, we see that \(q\) changes sign at each of its zeros, \(x = -3\) and \(x = 2\text{,}\) and that it does not change as it passes by its vertical asymptote at \(x = 1\text{.}\) The reason \(q\) doesn’t change sign at the asympotote is because of the repeated factor of \((x-1)^2\) which is always positive.

Subsection 5.5.3 Summary

• If a rational function \(r(x) = \frac{p(x)}{q(x)}\) has the properties that \(p(a) = 0\) and \(q(a) = 0\) and

  \begin{equation*} \lim_{x \to a} r(x) = L\text{,} \end{equation*}

  then \(r\) has a hole at the point \((a,L)\text{.}\) This behavior can occur when there is a matching factor of \((x-a)\) in both \(p\) and \(q\text{.}\)
• For a rational function \(r(x) = \frac{p(x)}{q(x)}\text{,}\) we determine where the function has zeros and where it has vertical asymptotes by considering where the numerator and denominator are \(0\text{.}\) In particular, if \(p(a) = 0\) and \(q(a) \ne 0\text{,}\) then \(r(a) = 0\text{,}\) so \(r\) has a zero at \(x = a\text{.}\) And if \(q(a) = 0\) and \(p(a) \ne 0\text{,}\) then \(r(a)\) is undefined and \(r\) has a vertical asymptote at \(x = a\text{.}\)
• By writing a rational function’s numerator in factored form, we can generate a sign chart for the function that takes into account all of the zeros and vertical asymptotes of the function, which are the only points where the function can possibly change sign. By testing \(x\)-values in various intervals between zeros and/or vertical asymptotes, we can determine where the rational function is positive and where the function is negative.