<s>[INST] Find the vertical asymptote of a function For an assignment, I was asked to find the vertical asymptote of the function $$g(x)= \frac{\frac{1}{2}x^3-4x^2+6x}{7x^2-56x+84}.$$ 
According to my text, a reliable method of finding the asymptote is to factor the numerator and denominator, and what left in the denominator that was not cancelled out is the asymptote.
I factored the numerator to $\frac{1}{2}(x^2-6x)(x-2)$, and the denominator factored to $7(x-2)(x-6)$, therefore $(x-2)$ cancelled out, leaving $(x-6)$ in the denominator.
However, 6 was not accepted as the answer, and I would like to know why.
 [/INST] You did not finish factoring the numerator. That's where the problem is.
You started out all right, but then you had to keep factoring that $x^2-6x$: it is not degree $1$, and it is not irreducible quadratic, so it can still be factored. In fact, it has a factor of $x$. So you really have:
$$\frac{1}{2}x^3 - 4x^2+6x = \frac{1}{2}\left(x^3 - 8x^2 + 12x\right) = \frac{1}{2}x\left(x^2-8x+12\right) = \frac{1}{2}x(x-6)(x-2).$$
In fact, both $x-2$ and $x-6$ cancel out in $g(x)$. 
As a function, $g(x)$ is equal to 
$$g(x) = \left\{\begin{array}{ll}
\frac{1}{14}x &\mbox{if $x\neq 2$ and $x\neq 6$,}\\
\mbox{undefined} &\mbox{if $x=2$ or $x=6$.}
\end{array}\right.$$
So it does not have any vertical asymptotes. 
</s>