<p>If f(x) = (x+2)/ ((x-2)(x^2-4)), its graph will have </p>
<p>A. 1 Horiz. and 2 vertical asymptotes
B. 1 Horiz. and 1 vertical asymptote</p>
<p>You're probably wondering why I didn't give more answer choices. The answer is that they were ridiculous choices. Anyway, I put A, because I thought that y ≠ 0 and x ≠2,-2, and my calculator seemed to agree that this was the case. However, Barrons put B, and said that y≠0 and x≠2.</p>
<p>Thoughts?</p>
<p>If you factor the (x^2-4) out is becomes (x-2)(x+2) and the (x+2) cancels out so x=-2 is a hole, not an asymptote.</p>
<p>Wow, Barrons is SO ANNOYING. But knowing this is good. Thanks.</p>
<p>Vertical asymptotes occur when you have some number k divided by zero, where k is nonzero. For example, f(x) = 1/x has a vertical asymptote at x = 0. f(x) = (x+2)/((x-2)(x^2 - 4)) does not have a vertical asymptote at x = -2 because as x → -2, f(x) → 0/0, which is indeterminate (not the same as undefined).</p>
<p>Or you could’ve graphed it on your calculator if that mathematical operation didn’t come to you.</p>