jmc

The graph has vertical asymptotes at $x=-1$ and $x=1$. Since there is a vertical asymptote at $x=-1$, there must be a factor of $x+1$ in the denominator $q(x)$. Similarly, since there is a vertical asymptote at $x=1$, there must be a factor of $x-1$ in the denominator $q(x)$. Since $q(x)$ is quadratic, we have that $q(x)=a(x-1)(x+1)$, for some constant $a$. Since $q(2)=6$, we have $a(2-1)(2+1)=6$ and hence $a=2$.

So $q(x)=2(x-1)(x+1)=\boxed{2x^2-2}.$