## Precalculus (6th Edition) Blitzer

The vertical asymptote will be$x=-5$, there will be a hole at$x=5$.
\begin{align} & g\left( x \right)=\frac{x-5}{{{x}^{2}}-25} \\ & =\frac{p\left( x \right)}{q\left( x \right)} \end{align} So, $p\left( x \right)=x-5$ and $q\left( x \right)={{x}^{2}}-25$ For the vertical asymptotes, put $q\left( x \right)=0$: \begin{align} & q\left( x \right)={{x}^{2}}-25=0 \\ & {{x}^{2}}-25=0 \\ & \left( x-5 \right)\left( x+5 \right)=0 \end{align} Therefore, the vertical asymptotes are: $x=5,\text{ or }x=-5$. But, after rearranging, $x-5$ from the denominator will be eliminated. So, the vertical asymptote will be $x=-5$. To find the holes, observe the expression: \begin{align} & g\left( x \right)=\frac{x-5}{{{x}^{2}}-25} \\ & =\frac{x-5}{\left( x-5 \right)\left( x+5 \right)} \\ & =\frac{1}{x+5} \end{align} Here, $x-5$ will be eliminated; when x approaches 5, the polynomial approaches to $\frac{1}{10}$ , but at the point $x=5$ the polynomial will be discontinuous. So, there will be a hole at $x=5$ Thus, the vertical asymptote is $x=-5$; there will be a hole at $x=5$. Hence, the vertical asymptote will be $x=-5$; there will be a hole at $x=5$.