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

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Answer 1 · 2015-05-20T22:08:54

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 $x = (-3+1)/2 = -1$

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

$y = -2(x+3)(x-1) = -2(-1+3)(-1-1) = -2xx2xx-2 = 8$

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

Another way of calculating this is as follows:

$y = -2(x+3)(x-1) = -2(x^2+2x-3)$

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

Differentiate this by $x$ ...

$d/(dx)(-2x^2-4x+6) = -4x-4$

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

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