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13. Write

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

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

13. Write y = 2x^2 -12x+16 in vertex form. Please show work if possible so I know-example-1

1 Answer

5 votes

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 * {(x - 3)}^(2) ]+[ 2 * ( - 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.

User Tsurahman
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