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.
 A: No factorization is needed. First, separate the rational function $\rm\:f/g\:$ into its integral and fractional parts by using the division algorithm to divide the numerator by the denominator, namely
$$\rm f\ =\ q\ g + r\ \ \Rightarrow\ \ \frac{f}g\ =\ q + \frac{r}g,\quad\quad deg\ r\ <\ deg\ g$$
Since the integral part $\rm\:q\:$ is a polynomial it has no vertical asymptotes. Rather, such asymptotes arise only from the "proper" fractional part $\rm\:r/g\:,\: $ namely at roots of the denominator $\rm\:g\:,\: $ which are not also roots of the numerator $\rm\:r\:.\: $ But in your example, performing the simple division shows that remainder $\rm\:r = 0\:,\: $ so there are no such asymptotes. Generally this will be a much simpler approach than immediate factorization since generally division (and gcd) algorithms are much simpler than factorization algorithms.
A: 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. 
A: Note that $\displaystyle f(x) = \frac{\frac{1}{2}x^3-4x^2+6x}{7x^2-56x+84} = \frac{1}{2}\frac{x^3-8x^2+12x}{7x^2-56x+84} = \frac{1}{2}\frac{1}{7}\frac{x^3-8x^2+12x}{x^2-8x+12} = \frac{x}{14}$.
So the function is "almost" a straight line passing through origin with a slope $\frac{1}{14}$ except at $x=2$ and $x=6$.
The function is not defined at $x=2$ and $x=6$. But $\displaystyle \lim_{x \rightarrow 2} = \frac{1}{7}$, $\displaystyle \lim_{x \rightarrow 6} = \frac{3}{7}$.
There are no asymptotes for the problem.
