Question

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

Answer 1

the vertex is at `(-0.875, 9.0625)`

Explanation:

`y = −3x^2 −x −3 −(x −3)^2`

Simplify the RHS
`y = -3x^2 -x -3 - x^2 - 6x +9`
`y = -4x^2 -7x +6`

The general quadratic form is `y = ax2 + bx + c`
The vertex can be found at `(h,k)`
where `h = -b/2a`

Substitute in what we know
`h = -(-7)/(2*-4) = -7/8 = -0.875`

Substititue the value of `h` for `x` in the original equation
`y = -4(-7/8)^2 -7(-7/8) +6 = 9.0625`

the vertex is at `(-0.875, 9.0625)`