Domain and vertical asymptotes

Factor: numerator $x^2-4 = (x-2)(x+2)$, denominator $x^3-x = x(x-1)(x+1)$. So $f(x) = \dfrac{(x-2)(x+2)}{x(x-1)(x+1)}$. There are no common factors between numerator and denominator, so no holes occur.

The denominator is zero at $x=0,1,-1$, and the numerator is nonzero at each of these points, so all three are vertical asymptotes. The domain is all real numbers except $x=0,1,-1$, i.e. $(-\infty,-1)\cup(-1,0)\cup(0,1)\cup(1,\infty)$.

Set the numerator equal to zero: $x^2 - 4 = 0$ gives $x = 2$ and $x = -2$. Both values are in the domain (they don't make the denominator zero), so the $x$-intercepts are $(2,0)$ and $(-2,0)$.

For the $y$-intercept, we need $f(0)$, but $x=0$ is not in the domain since the denominator vanishes there. Therefore there is no $y$-intercept.

Divide numerator and denominator by $x^3$: $f(x) = \dfrac{\frac{1}{x} - \frac{4}{x^3}}{1 - \frac{1}{x^2}}$. As $x\to\infty$, every term with $x$ in the denominator goes to 0, leaving $\dfrac{0}{1} = 0$.

Since the degree of the numerator (2) is less than the degree of the denominator (3), the limit as $x\to \pm\infty$ is 0 in both directions, so $y=0$ is a horizontal asymptote.