# What is the vertex of y=2(x +3)^2 -8x ?

The vertex is $\left(- 1 , 16\right)$.
$y = 2 {x}^{2} + 12 x + 18 - 8 x = 2 {x}^{2} + 4 x + 18$. The coefficient of ${x}^{2}$ is positive so we know the vertex is a minimum. This vertex will be the zero of the derivative of this trinomial. So we need its derivative.
$f \left(x\right) = 2 {x}^{2} + 4 x + 18$ so $f ' \left(x\right) = 4 x + 4 = 4 \left(x + 1\right)$. This derivative is zero for $x = - 1$ so the vertex is at the point $\left(- 1 , f \left(- 1\right)\right) = \left(- 1 , 16\right)$