# What is the vertex form of y= 2x^2-16x+32 ?

Mar 2, 2018

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

#### Explanation:

To find the vertex form, you need to complete the square. So set the equation equal to zero, then separate the coefficient of x, which is 2:
$0 = {x}^{2} - 8 x + 16$
Move the ones (16) to the other side, then add "c" to complete the square.
$- 16 + c = {x}^{2} - 8 x + c$

To find c, you need to divide the middle number by 2, and then square that number. so because $- \frac{8}{2} = - 4$, when you square that you get that c is 16. So add 16 to both sides:
$0 = {x}^{2} - 8 x + 16$
Because ${x}^{2} - 8 x + 16$ is a perfect square, you can factor that into ${\left(x - 4\right)}^{2}$ .

Then you need to multiply the coefficient back into the equation:
$0 = 2 {\left(x - 4\right)}^{2}$ Normally you would move the ones back over, but in this case the vertex is (4,0), so you don't need to do that. Then set the equation to y and you're done.