Answer
Vertical asymptote at $x=0$ or $x=3$
Work Step by Step
$f(x)=\dfrac{4x^2+4x-24}{x^4-2x^3-9x^2+18x}=\dfrac{4(x+3)(x-2)}{x(x-2)(x+3)(x-3)}$
$=\dfrac{4}{x(x-3)}.$
Vertical asymptotes occur when the denominator alone is $0\to$
$x(x-3)=0\to x=0$ or $x=3.$
