Answer
$f(x) = \frac{-(x-2)}{(x^2)(x-3)}$
Work Step by Step
$(x-3)$ is a term in the denominator since $x=3$ is a vertical asymptote.
$x^2$ is a term in the denominator since $x$ is a vertical asymptote and $\lim\limits_{x \to 0^-}f(x) = \lim\limits_{x \to 0^+}f(x) = -\infty$
$(x-2)$ is a term in the numerator since $f(2) = 0$
We can write the function $f$:
$f(x) = \frac{c~(x-2)}{(x^2)(x-3)}$
where $c$ is a constant
$\lim\limits_{x \to 3^-}f(x) = \infty$
Thus, we can let $c = -1$
We can write the function $f$:
$f(x) = \frac{-(x-2)}{(x^2)(x-3)}$
We can see that this function satisfies all the required conditions.
