# What is the vertex form of y=(6x-2)(2x+11) ?

Mar 27, 2018

$y = 6 {\left(x + \frac{31}{12}\right)}^{2} - \frac{1225}{24}$

#### Explanation:

$y = \left(3 x - 1\right) \left(2 x + 11\right)$

Multiply the brackets

$y = 6 {x}^{2} + 33 x - 2 x - 11$

$y = 6 {x}^{2} + 31 x - 11 \leftarrow \text{ Starting point}$
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$\textcolor{b l u e}{\text{Discussing what is happening}}$

Note that for standardised form $y = a {x}^{2} + b x + c$ we intend to make this $y = a {\left(x + \frac{b}{2 a}\right)}^{2} + k + c \textcolor{w h i t e}{.} \leftarrow \text{ completed square format}$

If you multiply out the whole thing we get:

$y = a {x}^{2} + b x \textcolor{red}{+ a {\left(\frac{b}{2 a}\right)}^{2}} + k + c$

The $\textcolor{red}{+ a {\left(\frac{b}{2 a}\right)}^{2}} + k$ is not in the original equation.

To 'force' this back to the original equation we

set $\textcolor{red}{+ a {\left(\frac{b}{2 a}\right)}^{2}} + k = 0$
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$\textcolor{b l u e}{\text{Returning to the solution}}$

$y = 6 {x}^{2} + 31 x - 11 \textcolor{w h i t e}{\text{d")->color(white)("d}} y = 6 {\left(x + \frac{31}{6 \times 2}\right)}^{2} + k - 11$

However:
$\textcolor{red}{+ a {\left(\frac{b}{2 a}\right)}^{2}} + k = 0 \textcolor{w h i t e}{\text{d")->color(white)("dddd}} \textcolor{red}{6 {\left(\frac{31}{2 \times 6}\right)}^{2}} + k = 0$

$\textcolor{w h i t e}{\text{dddddddddddddddd")->color(white)("dddd}} {31}^{2} / \left(4 \times 6\right) + k = 0$

$\textcolor{w h i t e}{\text{dddddddddddddddd")->color(white)("dddd}} k = - \frac{961}{24}$

So we now have:

$y = 6 {x}^{2} + 31 x - 11 \textcolor{w h i t e}{\text{d")->color(white)("ddd}} y = 6 {\left(x + \frac{31}{6 \times 2}\right)}^{2} - \frac{1225}{24}$

$\textcolor{w h i t e}{\text{dddddddddddddddd")->color(white)("dddd}} y = 6 {\left(x + \frac{31}{12}\right)}^{2} - \frac{1225}{24}$