What is the vertex form of #y=4x^2-32x+63#?

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Answer 1 · 2016-01-15T04:44:02

#y=4(x-4)^2-1#

If the standard form of a quadratic equation is -

#y=ax^2+bx+c#
Then -

Its vertex form is -

#y=a(x-h)^2+k#
Where -

#a = #co-efficient of #x#
#h=(-b)/(2a)#
#k=ah^2+bh+c#

Use the formula to change it to vertex form -

#y=4x^2-32x+63#
#a=4#
#h=(-(-32))/(2 xx 4)=32/8=4#
#k=4(4)^2-32(4)+63#
#k=64-128+63#
#k=127-128=-1#

Substitute #a=4; h=4 : k=-1# in

#y=a(x-h)^2+k#
#y=4(x-4)^2-1#