Vertical asymptote of $\frac {3x^2 - 18x - 81}{6x^2 - 54}$ Vertical asymptote of  $f(x) = \frac {3x^2 - 18x - 81}{6x^2 - 54}$ is 3, but why not -3?
The original function is already expanded, however, factoring it out a little bit, we get:
$ \frac {3(x-9) (x+3)}{6(x-3)(x+3)}$
The way Salman Khan explained it here didn't really answer that question.
To find vertical asymptote of a rational function we see where it's undefined. And it's undefined when denominator is equal to $0$. As you can see from the factored version of the function, x = -3 does make the denominator equal to $0$!
 A: The function $f$ is just a slightly disguised form of
$$
g(x) = \frac{x-9}{2(x-3)}
$$
So, we have
$$
\lim_{x \to -3}f(x) = g(-3) = \frac{-12}{-12} = 1
$$
So, there is no vertical asymptote at $x=-3$. Actually, the original function $f$ is undefined at $x=-3$ because its defining formula gives $0/0$. So there is a "hole" in its graph, if you like. But that's not the same thing as the graph shooting off to infinity (which is what you get at an asymptote).
Here's what the graph of $f$ looks like

The little hollow circle at $(-3,1)$ is meant to indicate that the function $f$ is undefined when $x=-3$.
A: A vertical asymptote of a function is a line $x=k$ being approximated by the function as $x\to k^{\pm}$, meaning that it has to go to $\pm\infty$ in that point.  Thus, $x=-3$ is not a vertical asymptote of your $f$ because $$\require\cancel \lim\limits_{x\to -3}  \frac {3(x-9) \cancel{(x+3)}}{6(x-3)\cancel{(x+3)}}=\frac{-36}{-36}=1.$$
In other words, a function has a vertical asymptote in a point of discontinuity of second kind, whereas $x=-3$ is a point of removable discontinuity (third kind) of $f(x)$.
A: A line with equation $x=k$ is an asymptote for a function $f$ when 
$$
\lim_{x\to k^{\pm}} f(x) =\pm\infty,
$$
meaning that at least one of the four combinations occurs.
For rational functions of the form $f(x)=p(x)/q(x)$ where $p$ and $q$ are polynomials, the condition $q(k)=0$ is a necessary condition for $f$ having $x=k$ as asymptote. Indeed, if $q(k)\ne0$, the function $f$ is continuous at $k$, which implies that $\lim_{x\to k}f(x)=f(k)$ exists and is finite.
However, the condition is not sufficient: the simple example of $f(x)=2x/x$ shows this. It's true that the denominator vanishes at $0$, but
$$
\lim_{x\to0}f(x)=2
$$
so none of the four combinations above occurs.
Your case is very similar:
$$
f(x) = \frac {3x^2 - 18x - 81}{6x^2 - 54}=\frac{(x-9)(x+3)}{2(x-3)(x+3)}
$$
so
$$
\lim_{x\to -3}f(x)=\lim_{x\to-3}\frac{x-9}{2(x-3)}=\frac{-12}{-12}=1
$$
and none of the four combinations mentioned at the start occurs.
You can observe that $\lim_{x\to-3}(3x^2 - 18x - 81)=0$, so $\lim_{x\to-3}f(x)$ is in the indeterminate form $0/0$, meaning that you can't draw conclusions about the limit without further work (in this case, factoring out the common factor $x+3$).
It is possible that both numerator and denominator vanish at a point $k$ and still the line $x=k$ be an asymptote for the function. Consider
$$
f(x)=\frac{x^2-3x+2}{x^2-2x+1}=\frac{(x-1)(x-2)}{(x-1)^2}
$$
Then both numerator and denominator vanish at $1$, but
$$
\lim_{x\to1^-}f(x)=\lim_{x\to1^-}\frac{x-2}{x-1}=\infty,
\qquad
\lim_{x\to1^+}f(x)=\lim_{x\to1^+}\frac{x-2}{x-1}=-\infty,
$$
so $x=1$ is an asymptote for $f$.
