Question

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

Answer 1

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

Explanation:

We start with the given
`3y=-2(x+3)(x-1)`

divide both sides of the equation by -2

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

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

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

Expand the right sides of the equation by multiplication

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

complete the square

`-3/2y=x^2+2x+1-1-3`

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

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

transpose the -4 to the left side

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

factor out the -3/2 so that coefficient of y is 1 inside the grouping symbol

`-3/2(y-8/3)=(x--1)^2" " "`this is now the vertex form

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