Question 1015555
y=x^2−16x+63
The vertex has an x-value of -b/2a=-(-16)/2=8
f(8)= -1
Therefore, the vertex is at (8,-1)
The vertex form is (x+h)^2 +k, where you change the sign of the x value of the vertex (h) and keep the y component (k)
y=(x-8)^2 - 1
Expand that for x^2-16x+64-1
x^2-16x+63