Question

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

Answer 1

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Explanation:

graph{-x^2-x-(3x+2)^2 [-1.5, 0.5, -0.5, 1.5]}

It looks like this with an `x`-intercept at around `x = -0.5`. This will be helpful in checking our answer.

`1.` First, we have to expand. So

`-x^2 - x - 9x^2 - 4 - 12x`

`= -10x^2 - 13x -4`

Luckily, if we plug in `x = 0.5` for `x`, we get a value of zero in accordance to our `y` intercept, so we are on the right track.

`2.` Then we complete the square so that the equation is in vertex form (to find the vertex). So:

`y = -10x^2 - 13x -4`

`-y/10 = x^2 + 1.3x + 0.4`

`-y/10 + 0.0225 = x^2 + 1.3y + 0.4225`

`-y/10 + 0.0225 = (x+0.65)^2`

`-y/10 =(x+0.65)^2 - 0.0225`

`y = -10(x+0.65)^2 + 0.225`

So the vertex would be `(-0.65, 0.225)`