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Find the vertical asymptote(s) and hole(s) of the rational function \(\displaystyle f(x)=\frac{ x^{2} + 5 x - 24 }{ x^{2} - x - 6 }\)

Factoring gives \(\displaystyle f(x)=\frac{ \left(x - 3\right) \left(x + 8\right) }{ \left(x - 3\right) \left(x + 2\right) }\). The factor \(\displaystyle x - 3\) reduces out. The reduced function is \(\displaystyle f(x) = \frac{x + 8}{x + 2}\). This gives a hole at \(\displaystyle \left(3,\frac{11}{5}\right) \). The limit from the left is \(\displaystyle \lim_{x \to -2^-}\left(\frac{x + 8}{x + 2}\right) = -\infty\). The limit from the right is \(\displaystyle \lim_{x \to -2^+}\left(\frac{x + 8}{x + 2}\right) = \infty\). This gives the vertical asymptote as \(\displaystyle x = -2\).

Download \(\LaTeX\) 

\begin{question}Find the vertical asymptote(s) and hole(s) of the rational function $f(x)=\frac{ x^{2} + 5 x - 24 }{ x^{2} - x - 6 }$
    \soln{6cm}{Factoring gives $f(x)=\frac{ \left(x - 3\right) \left(x + 8\right) }{ \left(x - 3\right) \left(x + 2\right) }$. The factor $x - 3$ reduces out. The reduced function is $f(x) = \frac{x + 8}{x + 2}$. This gives a hole at $\left(3,\frac{11}{5}\right) $. The limit from the left is $\lim_{x \to -2^-}\left(\frac{x + 8}{x + 2}\right) = -\infty$. The limit from the right is $\lim_{x \to -2^+}\left(\frac{x + 8}{x + 2}\right) = \infty$. This gives the vertical asymptote as $x = -2$.}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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\begin{document}\begin{question}(10pts) The question goes here!
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HTML for Canvas 
<p> <p>Find the vertical asymptote(s) and hole(s) of the rational function  <img class="equation_image" title=" \displaystyle f(x)=\frac{ x^{2} + 5 x - 24 }{ x^{2} - x - 6 } " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%3D%5Cfrac%7B%20x%5E%7B2%7D%20%2B%205%20x%20-%2024%20%7D%7B%20x%5E%7B2%7D%20-%20x%20-%206%20%7D%20" alt="LaTeX:  \displaystyle f(x)=\frac{ x^{2} + 5 x - 24 }{ x^{2} - x - 6 } " data-equation-content=" \displaystyle f(x)=\frac{ x^{2} + 5 x - 24 }{ x^{2} - x - 6 } " /> </p> </p>
HTML for Canvas
<p> <p>Factoring gives  <img class="equation_image" title=" \displaystyle f(x)=\frac{ \left(x - 3\right) \left(x + 8\right) }{ \left(x - 3\right) \left(x + 2\right) } " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%3D%5Cfrac%7B%20%5Cleft%28x%20-%203%5Cright%29%20%5Cleft%28x%20%2B%208%5Cright%29%20%7D%7B%20%5Cleft%28x%20-%203%5Cright%29%20%5Cleft%28x%20%2B%202%5Cright%29%20%7D%20" alt="LaTeX:  \displaystyle f(x)=\frac{ \left(x - 3\right) \left(x + 8\right) }{ \left(x - 3\right) \left(x + 2\right) } " data-equation-content=" \displaystyle f(x)=\frac{ \left(x - 3\right) \left(x + 8\right) }{ \left(x - 3\right) \left(x + 2\right) } " /> . The factor  <img class="equation_image" title=" \displaystyle x - 3 " src="/equation_images/%20%5Cdisplaystyle%20x%20-%203%20" alt="LaTeX:  \displaystyle x - 3 " data-equation-content=" \displaystyle x - 3 " />  reduces out. The reduced function is  <img class="equation_image" title=" \displaystyle f(x) = \frac{x + 8}{x + 2} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cfrac%7Bx%20%2B%208%7D%7Bx%20%2B%202%7D%20" alt="LaTeX:  \displaystyle f(x) = \frac{x + 8}{x + 2} " data-equation-content=" \displaystyle f(x) = \frac{x + 8}{x + 2} " /> . This gives a hole at  <img class="equation_image" title=" \displaystyle \left(3,\frac{11}{5}\right)  " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%283%2C%5Cfrac%7B11%7D%7B5%7D%5Cright%29%20%20" alt="LaTeX:  \displaystyle \left(3,\frac{11}{5}\right)  " data-equation-content=" \displaystyle \left(3,\frac{11}{5}\right)  " /> . The limit from the left is  <img class="equation_image" title=" \displaystyle \lim_{x \to -2^-}\left(\frac{x + 8}{x + 2}\right) = -\infty " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20-2%5E-%7D%5Cleft%28%5Cfrac%7Bx%20%2B%208%7D%7Bx%20%2B%202%7D%5Cright%29%20%3D%20-%5Cinfty%20" alt="LaTeX:  \displaystyle \lim_{x \to -2^-}\left(\frac{x + 8}{x + 2}\right) = -\infty " data-equation-content=" \displaystyle \lim_{x \to -2^-}\left(\frac{x + 8}{x + 2}\right) = -\infty " /> . The limit from the right is  <img class="equation_image" title=" \displaystyle \lim_{x \to -2^+}\left(\frac{x + 8}{x + 2}\right) = \infty " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20-2%5E%2B%7D%5Cleft%28%5Cfrac%7Bx%20%2B%208%7D%7Bx%20%2B%202%7D%5Cright%29%20%3D%20%5Cinfty%20" alt="LaTeX:  \displaystyle \lim_{x \to -2^+}\left(\frac{x + 8}{x + 2}\right) = \infty " data-equation-content=" \displaystyle \lim_{x \to -2^+}\left(\frac{x + 8}{x + 2}\right) = \infty " /> . This gives the vertical asymptote as  <img class="equation_image" title=" \displaystyle x = -2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-2%20" alt="LaTeX:  \displaystyle x = -2 " data-equation-content=" \displaystyle x = -2 " /> .</p> </p>