How do you write the vertex form equation of the parabola #y=x^2-16x+71#?

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Answer 1 · 2017-12-29T11:52:55

#y=(x-8)^2+7#

The #x#-coordinate of the vertex, which is often called #h#, can be found by computing #h=-b/(2a)#.

For this problem: #h = -(-16)/(2(1)) = 8#.

To find the #y#-coordinate of the vertex, often called #k#, we substitute #h# in for all the #x#s we see in the original:

#k = (8)^2-16(8)+71 = 64 - 128 +71 =7#.

The vertex is #(h,k)#, so for this problem #(8,7)#.

Vertex form of the equation is: #y=a*(x-h)^2+k#. We've found #h# and #k#. The value of #a# is the coefficient of #x^2# in the original, so #a=1#.

The equation we're looking for is: #y=(x-8)^2+7#.