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?    

The values of $x$ that make the denominator zero (in this case $3,-3,4,-4$) are the candidates for vertical asymptotes.
You also need to see what values of $x$ make the numerator zero.  If both are zero then it is possible you have a removable singularity, not an asymptote.  For example, consider $$\frac{x-2}{x-2}$$
