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.