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

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

#### Explanation:

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

Simplify

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

using the formula for vertex $\left(h , k\right)$

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

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

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

The vertex form is as follows

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

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

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