Answer
Vertical asymptotes: none
Hole at $x=5$
Work Step by Step
Locating Vertical Asymptotes, (page 395) tells us that for $f(x)=\displaystyle \frac{p(x)}{q(x)}$,
if p(x) and q(x) have NO common factors,
and a is a zero of the denominator $q(x)$ ,
then the line $x=a$ is a vertical asymptote.
BUT, if they DO have common factors,
after REDUCING the form of the function's equation,
the number a may not cause the denominator to be zero any more.
in which case there will be a hole at x=a.
The point is that both p(x) and q(x) have to be factored,
and if we can, then we reduce the expression for f(x).
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$f(x)=\displaystyle \frac{x^{2}-25}{x-5}$
$p(x)=x^{2}-25=(x+5)(x-5)\qquad $
(difference of squares)
$q(x)=x-5\qquad $... fully factored
We can reduce the expression for f(x):
$f(x)=\displaystyle \frac{x^{2}-25}{x-5}=\frac{(x+5)(x-5)}{ (x-5)}=(x+5),\quad x\neq 5$
$f(x)$ has no vertical asymptotes,
$f(x)$ is not defined for $x=5$
Vertical asymptotes: none
Hole at $x=5$
