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).
