Answer
Vertical asymptote: $x=-3$
Hole at $x=3$
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).
------------------
$g(x)=\displaystyle \frac{x-3}{x^{2}-9}$
$p(x)=x-3\qquad $... fully factored
$q(x)=x^{2}-9=(x+3)(x-3)\qquad $
(difference of squares)
We can reduce the expression for $g(x):$
$g(x)=\displaystyle \frac{x-3}{x^{2}-9}=\frac{(x-3)}{ (x+3)(x-3)}=\frac{1}{x+3},\quad x\neq 3$
Vertical asymptote: $x=-3$
Hole at $x=3$
