Finding vertical asymptotes of $\frac{3x^4 + 3x^3 - 36x^2}{x^4 - 25x^2 + 144}$ I'm trying to find the vertical asymptotes for $$f(x) = \frac{3x^4 + 3x^3 - 36x^2}{x^4 - 25x^2 + 144}$$
If I understand correctly, the vertical asymptote exists at $x=a$ when a value $a$ is found such that $f(a)$ increases to $∞$.
So, we must find a number that is infinitely small on the denominator, or in other words, $0$. 
Setting $$x^4 - 25^2 + 144 = 0$$
$$(x^2 - 9)(x^2 - 16) = 0$$
$$(x+3)(x-3)(x+4)(x-4) = 0$$
So based on my understanding, $[-4, -3, 3, 4]$ should be values of $a$ where $x = a$ is a vertical asymptote.
Graphing this function, I can clearly see that there are only vertical asymptotes at $x = 4$ and $x = -3$.
Where did I go wrong in my theory?    

Hint:
By factoring the numerator we have
\begin{align}
3x^4+3x^3-36x^2&=3x^2(x^2+x-12)\\
&=3x^2(x+4)(x-3)
\end{align}
Then 
$$\frac{3x^4+3x^3-36x^2}{x^4 - 25x^2 + 144}=\frac{3x^2(x+4)(x-3)}{(x+3)(x-3)(x+4)(x-4)}=\frac{3x^2}{(x+3)(x-4)}\qquad x\neq -4, x\neq 3$$
So, the limit of the function when $x\to -4$ and $x\to 3$ are finite, hence no asymptotes are at these points.
