Question

Write Y=3x^2+6x+8 In Vertex form

Answer 1

ANSWER

The vertex form is

$y = 3 {(x + 1)}^{2} + 5$

EXPLANATION

We want to write

$y = 3 {x}^{2} + 6x + 8$
in the vertex form which of the form


$y = a {(x - h)}^{2} + k$

We achieve this BG completing the square.

We proceed as follows,

$y = 3( {x}^{2} + 2x) + 8$


We add and subtract half the coefficient of x multiplied by a factor of 3 which is
$3( \frac{1}{2} \times 2)^{2} = 3( {1)}^{2}$


This gives us,


$y = 3( {x}^{2} + 2x) + 3 {(1)}^{2} -3 {(1)}^{2} + 8$

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

$y = 3( {x}^{2} + 2x + {1}^{2} ) -3 {(1)}^{2} + 8$

We now got a perfect square, which factors to,


$y = 3 {(x + 1)}^{2} + 5$
The equation is now in the vertex form.