## Calculus 10th Edition

Vertical asymptotes at $x=3$ and $x=-3.$
A function has vertical asymptotes at values that make the denominator only $0$. $f(x)=\dfrac{x^3}{x^2-9}=\dfrac{x^3}{(x-3)(x+3)}\to(x-3)(x+3)=0\to x=3$ or $x=-3.$