Question

What is h(x) = –3x2 – 6x + 5 written in vertex form? h(x) = –3(x + 1)2 + 2 h(x) = –3(x + 1)2 + 8 h(x) = –3(x – 3)2 – 4 h(x) = –3(x – 3)2 + 32

Answer 1

ANSWER

The correct answer is



`h(x) = - 3(x + 1)^(2) +8`



Step-by-step explanation


The function we want to write in vertex form is


`h(x) = - 3 {x}^(2) - 6x + 5`

We factor -3 out of the first two terms to obtain,


`h(x) = - 3(x ^(2) + 2x) + 5`


We add and subtract half the coefficient of

`x`
multiplied by a factor of

`- 3`
to get;




`h(x) = - 3(x ^(2) + 2x) + - 3( {1})^(2) - - 3( {1})^(2) + 5`


The expression becomes


`h(x) = - 3(x ^(2) + 2x) - 3( {1})^(2) + 3( {1})^(2) + 5`



We factor -3 again out of the first two terms to get,




`h(x) = - 3(x ^(2) + 2x + ( {1})^(2) ) + 3 * 1 + 5`




`h(x) = - 3(x ^(2) + 2x + ( {1})^(2) ) + 3 + 5`





The expression in the parenthesis now becomes a perfect square.


This implies that;


`h(x) = - 3(x + 1)^(2) +8`