Question

How do you find the vertex and intercepts for `y = - 2x^2 + 8x + 4`?

Answer 1

The x-intercepts are simply the roots of the quadratic, `x=0.58, x=3.4`.
The y-intercept is simply the constant term, `y=4`.
The vertex is the point at which the slope (gradient) is zero, so differentiate and then solve to find the point `(2,12)`.

Explanation:

I'll use the method that requires calculus. Another answerer may be able to answer using only algebra.

First visualise the curve: the `x^2` term is negative, so this is an 'upside-down' parabola, with its vertex at the top and the opening toward the bottom.

The x-intercepts occur when `y=0`, so:

`-2x^2+8x+4=0`

Solve using the quadratic formula or otherwise:

`x=(-8+-sqrt(64-4*(-2)*4))/(2(-2))=(-8+-sqrt32)/-4`
`x=3.4 or 0.58`

To find the vertex, we know that it is a point where the gradient = `0`. The first derivative of the expression gives the slope:

`y=-2x^2+8x+4`

`(dy)/(dx)=-4x+8`

Set this equal to zero:

`0=-4x+8`

Solving, `x=2`. To find the `y` value of the vertex, substitute this in the original expression:

`y=-2(2)^2+8(2)+4=12`

Therefore the coordinates of the vertex are `(2,12)`.