What is the vertex of y=-2(x+3)^2+12x+4 ?

Apr 5, 2017

vertex $\left(0 , - 14\right)$

Explanation:

Given -

$y = - 2 {\left(x + 3\right)}^{2} + 12 x + 4$

$y = - 2 \left({x}^{2} + 6 x + 9\right) + 12 x + 4$

$y = - 2 {x}^{2} - 12 x - 18 + 12 x + 4$

$y = - 2 {x}^{2} - 14$

$x$ term is missing in the expression $- 2 {x}^{2} - 14$
Let us supply it.

$y = - 2 {x}^{2} + 0 x - 14$

$x = \frac{- b}{2 \times a} = \frac{0}{2 \times \left(- 2\right)} = 0$

At $x = 0$

$y = - 2 {\left(0\right)}^{2} - 14 = - 14$

vertex $\left(0 , - 14\right)$