Question

13. Write
y = 2x^2 -12x+16 in vertex form.

Please show work if possible so I know how to do it

Answer 1

First off, we factor out the expression:

$\displaystyle \large{y = 2 {x}^{2} - 12x + 16} \\ \displaystyle \large{y = 2 ( {x}^{2} - 6x + 8) }$

In the bracket, separate 8 out of the expression.

$\displaystyle \large{y = 2[ ( {x}^{2} - 6x + 8)] }\\ \displaystyle \large{y = 2[ ( {x}^{2} - 6x) + 8]}$

In x^2-6x, find the third term that can make up or convert it to a perfect square form. The third term is 9 because:

$\displaystyle \large{ {(x - 3)}^{2} = {x}^{2} - 6x + 9}$

So we add +9 in x^2-6x.

$\displaystyle \large{y = 2[ ( {x}^{2} - 6x + 9) + 8]}$

Convert the expression in the small bracket to perfect square.

$\displaystyle \large{y = 2[ {(x - 3)}^{2} + 8]}$

Since we add +9 in the small bracket, we have to subtract 8 with 9 as well.

$\displaystyle \large{y = 2[ {(x - 3)}^{2} + 8 - 9]} \\ \displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]}$

Then we distribute 2 in.

$\displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]}$

$\displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]} \\ \displaystyle \large{y = [2 \times {(x - 3)}^{2} ]+[ 2 \times ( - 1)] } \\ \displaystyle \large{y = 2 {(x - 3)}^{2} - 2 }$

Remember that negative multiply positive = negative.

Hence the vertex form is y = 2(x-3)^2-2 or first choice.