Question

How do you find the vertex of the parabola `y=1/2(x+1)(x-5)`?

Answer 1

vertex: `(2,-4 1/2)`

Explanation:

Our objective will be to transform the given equation into vertex form:
`color(white)("XXX")y=m(x-color(red)(a))^2+color(blue)(b)` with vertex at `(color(red)(a),color(blue)(b))`
If `y=1/2(x+1)(x-5)`
then
`color(white)("XXX")2y=x^2-4x-5`

`color(white)("XXX")2y= (x^2-4xcolor(green)(+4)) -5 color(green)(-4)` (completing the square)

`color(white)("XXX")2y=(x-2)^2-9`

`color(white)("XXX")y=1/2(x-color(red)(2))^2+color(blue)("("-9/2")")`

which is the vertex form with vertex at `(color(red)(2),color(blue)("("-9/2")"))`

Here's the graph to show that this result is reasonable.
graph{1/2*(x^2-4x-5) [-5.086, 7.4, -5.96, 0.28]}