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.

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