Question

What is the vertex of `y= -x^2-4x-3-2(x-3)^2`?

Answer 1

The vertex is `(4/3,-47/3)`

Explanation:

`y=-x^2-4x-3-2(x-3)^2`

This is not in vertex form yet, so we need to expand and organize the quadratic, complete the square, then determine the vertex.

Expand:
`y=-x^2-4x-3-2(x^2-6x+9)`

`y=-x^2-4x-3-2x^2+12x-18`

Organize:
`y=-3x^2+8x-21`

Complete the square:
`y=-3[x^2-(8x)/3+7]`

`y=-3[(x-4/3)^2-16/9+7]`

`y=-3[(x-4/3)^2+47/9]`

`y=-3(x-4/3)^2-3(47/9)`

`y=-3(x-4/3)^2-47/3`

Determine vertex:
Vertex form is `y=a(x-color(red)(h))^2+color(blue)(k)` where `(color(red)(h),color(blue)(k))` is the vertex of the parabola.

The vertex is therefore at `(color(red)(4/3),color(blue)(-47/3))`.

Double check with graph:
graph{y=-x^2-4x-3-2(x-3)^2 [-30, 30, -30, 5]}