# What is the vertex form of y=-8x^2 +8x+32 ?

Jun 3, 2016

$y = - 8 \left[{\left(x + \frac{1 x}{2}\right)}^{2} + 3 \frac{1}{2}\right]$

This gives the vertex as $\left(- \frac{1}{2} , 3 \frac{1}{2}\right)$

#### Explanation:

Vertex form is $y = a {\left(x b\right)}^{2} + c$ This is obtained by the process of completing the square.

Step 1. Divide the coefficient of ${x}^{2}$ out as a common factor.

$y = - 8 \left[{x}^{2} + x + 4\right]$

Step 2: Add in the missing square number to create the square of a binomial. Subtract it as well to keep the value of the right hand side the same.

$y = - 8 \left[{x}^{2} + x + {\textcolor{red}{\left(\frac{1}{2}\right)}}^{2} + 4 - {\textcolor{red}{\left(\frac{1}{2}\right)}}^{2}\right]$

Step 3: Write the first 3 terms in the bracket as ${\left(\text{binomial}\right)}^{2}$

$y = - 8 \left[{\left(x + \frac{1 x}{2}\right)}^{2} + 3 \frac{1}{2}\right]$

This gives the vertex as $\left(- \frac{1}{2} , 3 \frac{1}{2}\right)$