# What is the vertex form of y=-3x^2-6x-3?

Jun 15, 2017

$y = - 3 {\left(x + 1\right)}^{2}$

#### Explanation:

Given -

$y = - 3 {x}^{2} - 6 x - 3$

Vertex -

x-coordinate of the vertex

$x = \frac{- b}{2 a} = \frac{- \left(- 6\right)}{2 \times \left(- 3\right)} = \frac{6}{- 6} = - 1$

Y-coordinate of the vertex

At $x = - 1$

$y = - 3 {\left(- 1\right)}^{2} - 6 \left(- 1\right) - 3 = - 3 + 6 - 3 = 0$

Vertex $\left(- 1 , 0\right)$

Vertex form of the equation is -

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

$a = - 3$ - coefficient of ${x}^{2}$
$h = - 1$
$k = 2$

$y = - 3 {\left(x + 1\right)}^{2} + 0$

$y = - 3 {\left(x + 1\right)}^{2}$

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