29Hole vs vertical asymptote

$\displaystyle f(x) = \dfrac{x + 2}{x^2 - x - 6}$. Find the vertical asymptote and the $\displaystyle x$-coordinate of the hole.

There's a vertical asymptote at x = 3 and a hole at x = -2. Factoring gives f(x) = (x+2)/[(x-3)(x+2)); the (x+2) cancels, causing a hole at x = -2, leaving 1/(x-3), which is undefined (asymptote) at x = 3.