Question

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

Answer 1

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).