Question

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

Answer 1

`y=2(x-4)^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^2-8x+16`
Move the ones (16) to the other side, then add "c" to complete the square.
`-16+c=x^2-8x+c`

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

Then you need to multiply the coefficient back into the equation:
`0=2(x-4)^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.