Question

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

Answer 1

Answer:

$Y=-\frac{1}{3}(x+6)^2+7$   [Vertex form]

Step-by-step explanation:

Given function:

$Y=-\frac{1}{3}x^2-4x-5$

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  

$Y=-\frac{1}{3}x^2-4x-5$

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.

$-3\times Y=-3\times\frac{1}{3}x^2-(-3)\times4x-(-3)\times 5$

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

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 = $(\frac{1}{2}\times 12)^2=(6)^2=36$

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.

$\frac{-3Y}{-3}=\frac{(x+6)^2}{-3}-\frac{21}{-3}$

$Y=-\frac{1}{3}(x+6)^2+7$   [Vertex form]