Question

What is the vertex form of `y=-1/3(x-2)(2x+5)`?

Answer 1

The Vertex Form is `(x--1/4)^2=-3/2*(y-27/8)`

Explanation:

We start from the given
`y=-1/3(x-2)(2x+5)`
Expand first
`y=-1/3(2x^2-4x+5x-10)`
simplify
`y=-1/3(2x^2+x-10)`
insert a `1=2/2` to make factoring of 2 clear

`y=-1/3(2x^2+2/2x-10)`

now, factor out the 2

`y=-2/3(x^2+x/2-5)`
complete the square now by adding `1/16` and subtracting `1/16` inside the grouping symbol

`y=-2/3(x^2+x/2+1/16-1/16-5)`

the first 3 terms inside the grouping symbol is now a Perfect Square Trinomial so that the equation becomes

`y=-2/3((x+1/4)^2-81/16)`
Distribute the `-2/3` inside the grouping symbol
`y=-2/3(x+1/4)^2-2/3(-81/16)`

`y=-2/3(x--1/4)^2+27/8`
let us simplify now to the Vertex Form

`y-27/8=-2/3(x--1/4)^2`

Finally

`(x--1/4)^2=-3/2(y-27/8)`

graph{(x--1/4)^2=-3/2(y-27/8)[-20,20,-10,10]}

God bless...I hope the explanation is useful..