Consider the function f(x)=(x-9)/(x^3-81x) find the vertical asymptote(s) of f(x)

Well, here's a harder one
Basically, to find a vertical asymptote, we set the denominator to 0. Like this:
x^3 - 81x = 0
Then we solve for x.
x(x^2 - 81) = 0
x(x+9)(x-9) = 0
x = 0, 9, -9
There are vertical asymptotes at x = 0, x = 9, and x = -9.

The correct answer is -20y

Ithink ghe answer to this question is b just trying to others