The graph of the rational function $\frac{1}{q(x)}$ is shown below. If $q(x)$ is a quadratic and $q(2) = 6$, find $q(x).$

[asy]
size(8cm);
import graph;

Label f; 
f.p=fontsize(6); 

real f(real x) {return 1/(2*(x+1)*(x-1));}

int gridsize = 5;
draw((-gridsize,0)--(gridsize,0), black+1bp, Arrows(8));
draw((0,-gridsize)--(0, gridsize), black+1bp, Arrows(8));
label("$x$", (gridsize, 0), E);
label("$y$", (0, gridsize), N);
label("$0$", (0,0),SE, p=fontsize(8pt));
for (int i=-gridsize+1; i<0; ++i){
    label("$"+string(i)+"$",(i,0),S, p=fontsize(8pt));
    label("$"+string(i)+"$",(0,i),E, p=fontsize(8pt));}
for (int i=1; i<=gridsize-1; ++i){
    label("$"+string(i)+"$",(i,0),S, p=fontsize(8pt));
    label("$"+string(i)+"$",(0,i),E, p=fontsize(8pt));}



draw(graph(f,-5,-1.05));
draw(graph(f,-.95,.95));
draw(graph(f,1.05,5));
draw((-1,-5)--(-1,5), dashed);
draw((1,-5)--(1,5), dashed);
[/asy]
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 $q(x) = 2(x - 1)(x + 1) = \boxed{2x^2 - 2}.$