The vertex form of a parabola is in the form #y=a(x-h)^2+k#, where the vertex is at the point #(h,k)#.
In order to find the vertex, we must complete the square. When we have #y=x^2-16x+72#, we should think about it as #y=color(red)(x^2-16x+?)+72#, so that #color(red)(x^2-16x+?)# is a perfect square.
Perfect squares appear in the form #(x+a)^2=x^2+2ax+a^2#. We already have an #x^2# in both, and we know that #-16x=2ax#, that is, #2# times #x# times some other number. If we divide #-16x# by #2x#, we see that #a=-8#. Therefore, the completed square is #x^2-16x+64#, which is equivalent to #(x-8)^2#.
However, we're not done. If we plug #64# into our equation, we must counteract that somewhere else to keep both sides equal. So, we can say that #y=color(red)(x^2-16x+64)+72-64#. This way, we have added and subtracted #64# to the same side, so the equation has not actually been changed because #64-64=0#.
We can rewrite #y=color(red)(x^2-16x+64)+72-64# to resemble the form #y=a(x-h)^2+k#.
#y=color(red)(x^2-16x+64)+72-64#
#y=color(red)((x-8)^2)+72-64#
#color(blue)(y=(x-8)^2+8#
With this equation, we can determine that the vertex #(h,k)# is at the point #(8,8)#.
