Question

Y=-1/3x^2-4x-5 in vertex form

Answer 1

`Y=-(1)/(3)(x+6)^2+7` [Vertex form]

Explanation:

Given function:

`Y=-(1)/(3)x^2-4x-5`

We need to find the vertex form which is.,

`y=a(x-h)^2+k`

where
`(h,k)` represents the co-ordinates of vertex.

We apply completing square method to do so.

We have

`Y=-(1)/(3)x^2-4x-5`

First of all we make sure that the leading co-efficient is =1.

In order to make the leading co-efficient is =1, we multiply each term with -3.

`-3* Y=-3*(1)/(3)x^2-(-3)*4x-(-3)* 5`

`-3Y=x^2+12x+15`

Isolating
`x^2` and
`x` terms on one side.

Subtracting both sides by 15.

`-3Y-15=x^2+12x-15-15`

`-3Y-15=x^2+12x`

In order to make the right side a perfect square trinomial, we will take half of the co-efficient of
`x` term, square it and add it both sides side.

square of half of the co-efficient of
`x` term =
`((1)/(2)* 12)^2=(6)^2=36`

Adding 36 to both sides.

`-3Y-15+36=x^2+12x+36`

`-3Y+21=x^2+12x+36`

Since
`x^2+12x+36` is a perfect square of
`(x+6)`, so, we can write as:

`-3Y+21=(x+6)^2`

Subtracting 21 to both sides:

`-3Y+21-21=(x+6)^2-21`

`-3Y=(x+6)^2-21`

Dividing both sides by -3.

`(-3Y)/(-3)=((x+6)^2)/(-3)-(21)/(-3)`

`Y=-(1)/(3)(x+6)^2+7` [Vertex form]