Find $a,b$ at $f(x)=\frac{x^2+x-12}{x^2-ax+b}$ An High school question:

Given :$$f(x)=\frac{x^2+x-12}{x^2-ax+b}$$ it's given that $x=3$ is a vertical asymptote find $a$ and $b$.

I tried: 
Since  $x=3$ is a vertical asymptote then $3^2-3a+b=0$, but now what

You can try also like that: Since $b= 3a-9$ we have $$f(x) = {(x+4)(x-3)\over x^2-ax+3a-9}= {(x+4)(x-3)\over (x-3)(x+3-a)} ={x+4\over x+3-a}$$
so $3-a=-3\implies a = 6$ and $b= 9$.
