Finding asymptotes for $f(x)=\frac{x^2+3x-10}{3x^2+13x-10}$ $$f(x)=\frac{x^2+3x-10}{3x^2+13x-10}$$
I know that the horizontal asymptote is $1/3$. To find the vertical asymptotes, I set the denominator equal to zero and used the quadratic formula, and I got $-5$ and $2/3$, and this is wrong. 
How do you find the vertical asymptotes for this problem? 
thanks. 

Note that
$$f(x)=\frac{(x+5)(x-2)}{(x+5)(3x-2)}.$$
Therefore
$$\lim_{x\to-5} f(x) = \lim_{x\to-5}\frac{(x+5)(x-2)}{(x+5)(3x-2)} =\lim_{x\to-5}\frac{x-2}{3x-2} = \frac{7}{17}.$$
Since this limit is actually not infinite, there is no vertical asymptote when $x=-5$.
In general, vertical asymptotes of a rational function occur at $x=a$ provided that the largest power of $(x-a)$ dividing the denominator is greater than that of the numerator. Here, $(x+5)$ divided the numerator and the denominator exactly once, so there was no asymptote when $x=-5$.
