First, distribute the binomials terms.
#y=x^2+4x-12x-48#
#y=x^2-8x-48#
From here, complete the square with the first two terms of the quadratic equation.
Recall that vertex form is #y=a(x-h)^2+k# where the vertex of the parabola is at the point #(h,k)#.
#y=(x^2-8xcolor(red)(+16))-48color(red)(-16)#
Two things just happened:
The #16# was added inside the parentheses so that a perfect square term will be formed. This is because #(x^2-8x+16)=(x-4)^2#.
The #-16# was added outside the parentheses to keep the equation balanced. There is a net change of #0# now thanks to the addition of #16# and #-16#, but the face of the equation is changed.
Simplify:
#y=(x-4)^2-64#
This tells us that the parabola has a vertex at #(4,-64)#. graph{(x-12)(x+4) [-133.4, 133.5, -80, 40]}
