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

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

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Answer 1 · 2017-12-17T03:16:23

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

Explanation:

From the given $y = (13 x - 4) (2 x - 12) + 2 x^{2} + 2 x$ $y=(13x-4)(2x-12)+2x^2+2x$

Simplify

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

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

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

$h = - \frac{b}{2 a} = \frac{- (- 162)}{2 \cdot 28} = \frac{81}{28}$ $h=-b/(2a)=(-(-162))/(2*28)=81/28$

$k = c - \frac{b^{2}}{4 a} = 48 - \frac{{(- 162)}^{2}}{4 \cdot 28} = - \frac{5217}{28}$ $k=c-(b^2)/(4a)=48-(-162)^2/(4*28)=-5217/28$

The vertex form is as follows

$y - k = a {(x - h)}^{2}$ $y-k=a(x-h)^2$

$y - - \frac{5217}{28} = 28 {(x - \frac{81}{28})}^{2}$ $y--5217/28=28(x-81/28)^2$

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

1 Answer

Explanation:

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What is the vertex form of y=(13x−4)(2x−12)+2x2+2x y= (13x-4)(2x-12)+2x^2+2x?

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Explanation:

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