What is the vertex form of y= 2x^2-16x+32 y=2x2−16x+32?

1 Answer
Bridget K.
Mar 2, 2018

y=2(x-4)^2y=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+160=x2−8x+16
Move the ones (16) to the other side, then add "c" to complete the square.
-16+c=x^2-8x+c−16+c=x2−8x+c

To find c, you need to divide the middle number by 2, and then square that number. so because -8/2=-4−82=−4, when you square that you get that c is 16. So add 16 to both sides:
0=x^2-8x+160=x2−8x+16
Because x^2-8x+16x2−8x+16 is a perfect square, you can factor that into (x-4)^2(x−4)2 .

Then you need to multiply the coefficient back into the equation:
0=2(x-4)^20=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.

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