What are the vertical asymptotes of f(x) = (x+2)/(x²-16)?

A vertical asymptote occurs where the denominator is zero and the factor does not cancel with the numerator. Set $x^2-16=0$ to get $(x-4)(x+4)=0$, so $x=4$ or $x=-4$. The numerator, $x+2$, is not zero at either value, so nothing cancels and both are asymptotes, choice A. Choice B ($x=-2$) is where the numerator is zero, which gives an $x$-intercept, not an asymptote. Choice C ($x=16$) confuses $x^2=16$ with $x=16$. Choice D ($x=4$ only) finds just one of the two roots of $x^2=16$.