jmc

We will complete the square on the given quadratic expression to find the vertex. Dividing by 3 and factoring $2$ from the first two terms, we have $$y=\frac{2}{3}(x^2-8x)+6$$ In order to make the expression inside the parentheses a perfect square, we need to add and subtract $(8/2{)}^2=16$ inside the parentheses. Doing this, we get $$y=\frac{2}{3}(x^2-8x+16-16)+6$$ so $$y=\frac{2}{3}(x-4{)}^2-\frac{32}{3}+6=\frac{2}{3}(x-4{)}^2-\frac{14}{3}$$ The graph of an equation of the form $y=a(x-h{)}^2+k$ is a parabola with vertex at $(h,k)$, so the vertex of our parabola is at $\left(4,-\frac{14}{3}\right)$. Thus, $m+n=4-\frac{14}{3}=\boxed{-\frac{2}{3}}$.