# What is the vertex form of y=2x^2 +2x-8 ?

Dec 17, 2017

$2 {\left(x + \frac{1}{2}\right)}^{2} - \frac{17}{2}$

#### Explanation:

The vertex form of a quadratic equation looks like this:
$y = a {\left(x - h\right)}^{2} + k$

To get our equation into this form, we need to complete the square, but first I want to make the ${x}^{2}$ term have a coefficient of $1$ (you'll notice that the $x$ inside the vertex form has this):
$2 {x}^{2} + 2 x - 8 = 2 \left({x}^{2} + x - 4\right)$

To complete the square, we can use the following formula:
${x}^{2} + p x + q = {\left(x + \frac{p}{2}\right)}^{2} - {\left(\frac{p}{2}\right)}^{2} + q$

Applying this to ${x}^{2} + x - 4$, we get:
${x}^{2} + x - 4 = {\left(x + \frac{1}{2}\right)}^{2} - {\left(\frac{1}{2}\right)}^{2} - 4 = {\left(x + \frac{1}{2}\right)}^{2} - \frac{17}{4}$

Now we put this back into our original expression:
$2 \left({\left(x + \frac{1}{2}\right)}^{2} - \frac{17}{4}\right) = 2 {\left(x + \frac{1}{2}\right)}^{2} - \frac{17}{2}$

And this is in vertex form, so it is our answer.

Dec 17, 2017

$y = 2 {\left(x + \frac{1}{2}\right)}^{2} - \frac{17}{2}$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$\text{to express in this form use "color(blue)"completing the square}$

• " ensure the coefficient of the "x^2" term is 1"

$\Rightarrow y = 2 \left({x}^{2} + x - 4\right)$

• " add/subtract "(1/2"coefficient of x-term")^2" to"
${x}^{2} + x$

$y = 2 \left({x}^{2} + 2 \left(\frac{1}{2}\right) x \textcolor{red}{+ \frac{1}{4}} \textcolor{red}{- \frac{1}{4}} - 4\right)$

$\textcolor{w h i t e}{y} = 2 {\left(x + \frac{1}{2}\right)}^{2} + 2 \times - \frac{17}{4}$

$\Rightarrow y = 2 {\left(x + \frac{1}{2}\right)}^{2} - \frac{17}{2} \leftarrow \textcolor{red}{\text{in vertex form}}$