# What is the vertex of y= -3x^2 + 12x +6 ?

Oct 16, 2016

The vertex is $\left(2 , 18\right)$

#### Explanation:

$y = - 3 {x}^{2} + 12 x + 6$
Differentiating with respect to x
$\frac{\mathrm{dy}}{\mathrm{dx}} = - 6 x + 12$
Any max or min is when $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$- 6 x + 12 = 0 \implies 6 x = 12 \implies x = 2$
To determine if it's a max or min , we must differentiate once more
So $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - 6$ (<0)
so it's a maximum
The vertex is when $x = 2$ and $y = - 3 \cdot 2 \cdot 2 + 12 \cdot 2 + 6 = - 12 + 24 + 6 = 18$
Here is a graph of the function

graph{-3x^2+12x+6 [-40, 40, -20, 20]}