Question

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

Answer 1

The vertex is at `(-3/4,-7/4)`

Explanation:

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

Expand the polynomial:
`y=-3x^2-2x-(x^2+4x+4)`

Combine like terms:
`y=-4x^2-6x-4`

Factor out `-4`:
`y=-4[x^2+3/2x+1]`

Complete the square:
`y=-4[(x+3/4)^2-(3/4)^2+1]`

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

`y=-4(x+3/4)^2-7/4`

From vertex form, the vertex is at `(-3/4,-7/4)`

Answer 2

Vertex: `(-3/4, -55/16)~~(-0.75, -3.4375)`

Explanation:

1) Rewrite this equation in standard form
`y=-3x^2-2x-(x+2)^2`
`y=-3x^2-2x-(x^2+4x+4)`
`y=-4x^2-6x-4`

2) Rewrite this equation in vertex form by completing the square
`y=(-4x^2-6x)-4`
`y=-4(x^2+3/2x)-4`
`y=-4(x^2+3/2x+(3/4)^2)-4+(3/4)^2`
`y=-4(x+3/4)^2-55/16`

The vertex form is `y=a(x-h)^2+k` reveals the vertex at `(h,k)`

Vertex: `(-3/4, -55/16)~~(-0.75, -3.4375)`

You can see this if you graph the equation
graph{y=-4x^2-6x-4 [-3, 2, -7, 0.1]}