What is the vertex form of # y= (13x-4)(2x-12)+2x^2+2x#?

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Answer 1 · 2017-12-17T03:16:23

The Vertex form is
#y--5217/28=28(x-81/28)^2# where #(h, k)=(81/28, -5217/28)# the vertex

From the given #y=(13x-4)(2x-12)+2x^2+2x#

Simplify

#y=(13x-4)(2x-12)+2x^2+2x#
#y=26x^2-8x-156x+48+2x^2+2x#
#y=28x^2-162x+48#

using the formula for vertex #(h, k)#

with #a=28# and #b=-162# and #c=48#

#h=-b/(2a)=(-(-162))/(2*28)=81/28#

#k=c-(b^2)/(4a)=48-(-162)^2/(4*28)=-5217/28#

The vertex form is as follows

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

#y--5217/28=28(x-81/28)^2#

God bless ..... I hope the explanation is useful.