Question

How do you write `-2x^2+2x-5` in vertex form and identify the vertex, y intercept and x intercept?

Answer 1

The vertex form of a parabola is
`y = m(x-a)^2 + b` where the vertex is at `(a,b)`

Given `-2x^2+2x-5` (note that I added a "squared" to the first term that was not there in the original question statement)

Extract the `m`
`y= (-2)(x^2-x)-5`

Complete the square
`y = (-2)(x^2-x+(1/2)^2) -5 - (-2)(1/2)^2`

`y = (-2)(x-1/2)^2+ (-9/2)`
which is in vertex form
with the vertex at `(1/2,-9/2)`

The y-intercept is the value of `y` when `x=0`
`y= -2x^2+2x-5` with `x=0` becomes
`y = -5`

The x-intercepts are the values of `x` when `y=0`
`y= -2x^2+2x-5`
or, more easily (using the vertex form),
`y = (-2)(x-1/2)^2+ (-9/2)` with `y=0` becomes

`(x-1/2)^2 = 9/2*(-1/2)`

`(x-1/2) = +-3/2i` (Note there are no Real solutions for the x-intercept)

`x = (1 +-3i)/2`