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 an

Answer 1

I don't see the function — the question looks cut off. Could you paste the full expression?

Meanwhile, here’s how to decide which option is correct:

- Factor the denominator. Real zeros of the denominator are where vertical asymptotes or holes can occur.
- If a factor of the denominator is canceled by the numerator, that x-value is a hole (no vertical asymptote there).
- If a factor of the denominator remains (not canceled), that x-value is a vertical asymptote.

Examples matching the answer choices:
- No vertical asymptote: numerator cancels all real zeros of the denominator (e.g. (x^2−64)/(x^2−64) = 1) or the denominator has no real zeros (e.g. 1/(x^2+1)).
- Vertical asymptote at x = 8 only: e.g. (x+8)/(x^2−64) = 1/(x−8) (hole at x=−8, asymptote at x=8).
- Vertical asymptote at x = −8 only: e.g. (x−8)/(x^2−64) = 1/(x+8).
- Vertical asymptotes at both x = 8 and x = −8: e.g. 1/(x^2−64).

Send the function and I’ll tell you which option applies.