calculus - Finding vertical asymptotes

So I am trying to find the behaviour of this function around an asymptote at $x=0$.

$a>0$

$$y=\frac{(x-a)(x^2+a)}{x^2}$$

I know that as $x \to 0^+$, $y \to -\infty $ and $x \to 0^-$, $y \to -\infty $.

I've tried dividing the top and bottom of $y$ by $x^2$ and ended with:

$$x-a+\frac{a}{x}-\frac{a^2}{x^2}$$

Which does not help me as I end with $\infty -\infty$. Any help would be greatly appreciated.

Answer

Expanding my comment above.

I know that as $x \to 0^+$, $y \to -\infty $ and $x \to 0^-$, $y \to
-\infty $.

Only the first assertion is correct.

Since $\left( x-a\right) (x^{2}+a)=x^{3}-ax^{2}+ax-a^{2}$, we have
$$\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\left( x-a\right) (x^{2}+a)}{x^{2}}
&=&\lim_{x\rightarrow 0}\left( x^{3}-ax^{2}+ax-a^{2}\right) \times

\lim_{x\rightarrow 0}\frac{1}{x^{2}} \\
&=&-a^{2}\times \lim_{x\rightarrow 0}\frac{1}{x^{2}}=-a^{2}\left( +\infty
\right) =-\infty, \qquad\text{ for }a\ne 0.
\end{eqnarray*}$$

So as $x$ approaches $0$, either from the left or from the right, $y\rightarrow -\infty $. The limit does not depend on the sign of $a$.

Plot of $y$ for $a=1$ (black) and $a=-1$ (green).