# What is the vertex form of y=-(-2x-13)(x+5) ?

Feb 24, 2016

$\textcolor{b l u e}{\text{vertex form "->" } y = 2 {\left(x + \frac{23}{4}\right)}^{2} + \frac{9}{8}}$

#### Explanation:

$\textcolor{b l u e}{\text{Determine the structure of the vertex form}}$

Multiply out the brackets giving:

$y = 2 {x}^{2} + 10 x + 13 x + 65$

$y = 2 {x}^{2} + 23 x + 65 \text{ }$...................................(1)

write as:
$y = 2 \left({x}^{2} + \frac{23}{2} x\right) + 65$

What we are about to do will introduce an error for the constant. We get round this by introducing a correction.

Let the correction be k then we have

$\textcolor{b r o w n}{y = 2 {\left(x + \frac{23}{4}\right)}^{2} + k + 65 \text{ }}$..................................(2)

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To get to this point I moved the square from ${x}^{2}$ to outside the brackets. I also multiplied the coefficient of $\frac{23}{2} x$ by $\frac{1}{2}$ giving the $\frac{23}{4}$ inside the brackets.
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$\textcolor{b l u e}{\text{Determine the value of the correction}}$

We need the values of a point for substitution so that k can be calculated.

Using equation (1) set $x = 0$ giving

$y = 2 {\left(0\right)}^{2} + 23 \left(0\right) + 65 \implies y = 65$

So we have our ordered pair of $\left(x , y\right) \to \left(0 , 65\right)$

Substitute this into equation (2) giving:

$\cancel{65} = 2 {\left(0 + \frac{23}{4}\right)}^{2} + k + \cancel{65} \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({2}_{a}\right)$

$k = - \frac{529}{8}$

$y = 2 {\left(x + \frac{23}{4}\right)}^{2} - \frac{529}{8} + 65 \text{ }$..................................(3)

But$\text{ } 65 - \frac{529}{8} = \frac{9}{8}$

Substitute into equation (3) gives:

$\textcolor{b l u e}{\text{vertex form "->" } y = 2 {\left(x + \frac{23}{4}\right)}^{2} + \frac{9}{8}}$

$\textcolor{b r o w n}{\text{Note that "(-1)xx23/4 = -5 3/4 ->" axis if symmetry}}$ 