# What is the vertex of the parabola y= -2(x+3)(x-1)?

Apr 27, 2018

$\text{vertex } = \left(- 1 , 8\right)$

#### Explanation:

$\text{the vertex lies on the axis of symmetry which is situated}$
$\text{at the midpoint of the zeros}$

$\text{ to find zeros let y = 0}$

$\Rightarrow - 2 \left(x + 3\right) \left(x - 1\right) = 0$

$\text{equate each factor to zero and solve for x}$

$x - 1 = 0 \Rightarrow x = 1$

$x + 3 = 0 \Rightarrow x = - 3$

$\text{axis of symmetry is } x = \frac{1 - 3}{2} = - 1$

$\text{x-coordinate of vertex } = - 1$

$\text{substitute "x=-1" into the equation for y-coordinate}$

$\Rightarrow y = - 2 \left(2\right) \left(- 2\right) = 8$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- 1 , 8\right)$
graph{(y+2x^2+4x-6)((x+1)^2+(y-8)^2-0.04)=0 [-20, 20, -10, 10]}