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

1 Answer

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

Explanation:

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.