Question

Which statement describes the vertical asymptotes of the graph of ?

The graph has no vertical asymptote.
The graph has a vertical asymptote at x = 8 only.
The graph has a vertical asymptote at x = –8 only.
The graph has vertical asymptotes at both x = 8 and x = –8.

Answer 1

I’m missing the function — the problem text didn’t include it. Can you paste the expression whose graph you’re asking about?

How to decide once you give it:
- Factor the denominator and set it = 0. Any x that makes the denominator 0 is a candidate for a vertical asymptote.
- If that factor cancels with an identical factor in the numerator, the x-value is a removable hole (not a vertical asymptote). If it does not cancel, it’s a vertical asymptote.

Quick examples that match the answer choices:
- f(x) = 1/(x^2 − 64) → vertical asymptotes at x = ±8 (both).
- f(x) = (x + 8)/(x^2 − 64) → simplifies to 1/(x − 8) with a hole at x = −8, so vertical asymptote at x = 8 only.
- f(x) = (x − 8)/(x^2 − 64) → vertical asymptote at x = −8 only.
- f(x) = (x^2 − 64)/(x − 8) → simplifies to x + 8 with a hole at x = 8, so no vertical asymptote.

Send the specific function and I’ll pick the correct choice.