# How do you find the vertex in y= -2(x+3)(x-1)?

May 20, 2015

The vertex will have an $x$ coordinate that is the average of the two zeros, which are at $x = - 3$ and $x = 1$.

So the $x$ coordinate is at $x = \frac{- 3 + 1}{2} = - 1$

Substitute this value of $x$ back into the equation to get

$y = - 2 \left(x + 3\right) \left(x - 1\right) = - 2 \left(- 1 + 3\right) \left(- 1 - 1\right) = - 2 \times 2 \times - 2 = 8$

So the vertex is at (-1, 8).

Another way of calculating this is as follows:

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

$= - 2 {x}^{2} - 4 x + 6$

Differentiate this by $x$...

$\frac{d}{\mathrm{dx}} \left(- 2 {x}^{2} - 4 x + 6\right) = - 4 x - 4$

The derivative, which represents the slope of the curve at any point will be zero at the vertex, when $- 4 x - 4 = 0$, that is when $x = - 1$.

Substitute this value of $x$ back into the equation as before to get $y = 8$, giving the vertex as (-1, 8).