jmc

Suppose we consider the equation $y=(6x+12)(x-8)$, which is equivalent to $y=6x^2-36x-96$. Then the graph of this equation is a parabola opening upward, with a minimum at the vertex. The vertex of a quadratic equation is located at the point where $x=-b/(2a)$. (This is the first part of the quadratic formula.)

In this case, we have $x=-(-36)/(2\times 6)=36/12=3$. The $y$-value at this point is $y=(6\times 3+12)(3-8)=(30)(-5)=\boxed{-150}$, which is also our minimum value of $k$.