Answer
The vertical asymptote will be\[x=-5\], there will be a hole at\[x=5\].
Work Step by Step
$\begin{align}
& g\left( x \right)=\frac{x-5}{{{x}^{2}}-25} \\
& =\frac{p\left( x \right)}{q\left( x \right)}
\end{align}$
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 $x=5$.
