<p>Is x=-2 a vertical asymptote and a removable discontinuity(hole) for </p>
<p>y = (x+2)/(x^2-4)</p>
<p>or would it just be a vertical asymptote?</p>
<p>According to your equation…</p>
<p>Do a bit of algebra, and you would be left with 1/(x-2). There would be a vertical asymptote on x=2 (and a discontinuity at that point)</p>
<p>^Just to clarify, there would still be a hole at x=-2.</p>
<p>^ Oh yeah, that is true. Forgot about that. xD</p>
<p>It is not a vertical asymptote. It is solely a hole.</p>
<p>^x=2 is an asymptote while x=-2 is a hole.</p>
<p>Yes, I know. I was merely answering the OP’s question without extraneous information about x=2.</p>
<p>Basically, you can’t have a vertical asymptote and a hole at the same x = k location, where k is a constant. Generally speaking, if evaluation at x = k yields 0 / 0, there’s a hole. If evaluation at x = k yields C / 0, where C is a non-zero constant, there’s a vertical asymptote.</p>
<p>The only type of function that I can see that has an exception to this rule is if x = k is a zero of the denominator with higher multiplicity than that of x = k in the numerator. For instance, the function f(x) = (x + 2)(x + 4) / (x + 4)^2 has a vertical asymptote at x = -4. When you cancel the (x + 4) in the numerator with one of the (x + 4)'s in the denominator, the function that remains still has asymptotic behavior at x = -4.</p>