Question

find the vertical asymptotes f(x) =x(x-1)/ x^3+16x

Answer 1

so you have
\[f(x)=\frac{x(x-1)}{x^3}+16x?\]

Answer 2

or did you mean to say f(x)=x(x-1)/(x^3+16x)?

Answer 3

\[f(x)=\frac{x(x-1)}{x^3+16x}\]?

Answer 4

16x goes with x^3 on the bottom

Answer 5

so that last one I wrote?

Answer 6

yes

Answer 7

ok since the vertical asymptote is somewhere the function is not defined then let's look for when the bottom is 0 (we need to be careful with this because we don't want to include a hole as a vertical asymptote)

Anyway
first step is to factor bottom

Answer 8

x(x^2+16)