smc

Consider the quadratic in completed square form: it must be $y=a(x-4{)}^2+2$. Now substitute $x=2$ and $y=0$ to give $a=-\frac{1}{2}$. Now expanding gives $y=-\frac{1}{2}{x}^2+4x-6$, so the product is $-\frac{1}{2}\cdot 4\cdot -6=3\cdot 4=12$, which is $\boxed{12}$.