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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

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ANSWER

The correct answer is


[tex]h(x) = - 3(x + 1)^{2} +8[/tex]



EXPLANATION


The function we want to write in vertex form is

[tex]h(x) = - 3 {x}^{2} - 6x + 5[/tex]

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

[tex]h(x) = - 3(x ^{2} + 2x) + 5[/tex]


We add and subtract half the coefficient of
[tex]x[/tex]
multiplied by a factor of
[tex] - 3[/tex]
to get;



[tex]h(x) = - 3(x ^{2} + 2x) + - 3( {1})^{2} - - 3( {1})^{2} + 5[/tex]


The expression becomes

[tex]h(x) = - 3(x ^{2} + 2x) - 3( {1})^{2} + 3( {1})^{2} + 5[/tex]



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



[tex]h(x) = - 3(x ^{2} + 2x + ( {1})^{2} ) + 3 \times 1 + 5[/tex]



[tex]h(x) = - 3(x ^{2} + 2x + ( {1})^{2} ) + 3 + 5[/tex]





The expression in the parenthesis now becomes a perfect square.


This implies that;

[tex]h(x) = - 3(x + 1)^{2} +8[/tex]




abidemiokin

The equation in vertex form is expressed as -3(x+1)^2 + 4

How to write equation in vertex form

Given the equation –3x2 – 6x + 5

Factoring out 3 from the first two terms will give:

= –3x^2 – 6x + 5

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

= -3(x^2 + 2x + 1) + 5 - 1

= -3(x+1)^2 + 4

Hence the equation in vertex form is expressed as -3(x+1)^2 + 4

Learn more on vertex form here; https://brainly.com/question/525947

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