Question

What is the vertex of `y= -3x^2-5x-(3x-2)^2`?

Answer 1

The vertex is `(7/(24), -143/48)`.

Explanation:

First expand `(3x-2)^2=9x^2-12x+4`.

Substituting that in we have:

`y=-3x^2-5x-(9x^2-12x+4)`

Distribute the negative:

`y=-3x^2-5x-9x^2+12x-4`

Collect like terms:

`y=-12x^2+7x-4`

The vertex is `(h,k)` where `h=-b/(2a)` and `k` is the value of `y` when `h` is substituted.

`h=-(7)/(2(-12))=7/(24)`.

`k=-12(7/(24))^2+7(7/(24))-4=-143/48` (I used a calculator...)

The vertex is `(7/(24), -143/48)`.

Answer 2

`(7/24,-143/48)`

Explanation:

`"we require to express in standard form"`

`rArry=-3x^2-5x-(9x^2-12x+4)`

`color(white)(rArry)=-3x^2-5x-9x^2+12x-4`

`color(white)(rArry)=-12x^2+7x-4larrcolor(blue)"in standard form"`

`"given the equation of a parabola in standard form then"`
`"the x-coordinate of the vertex is"`

`x_(color(red)"vertex")=-b/(2a)`

`"here "a=-12,b=7,c=-4`

`rArrx_(color(red)"vertex")=-7/(-24)=7/24`

`"substitute this value into the equation for y"`

`y=-12(7/24)^2+7(7/24)-4=-143/48`

`rArrcolor(magenta)"vertex "=(7/24,-143/48)`