Respuesta :
Answer:
[tex]Y=-\frac{1}{3}(x+6)^2+7[/tex] [Vertex form]
Step-by-step explanation:
Given function:
[tex]Y=-\frac{1}{3}x^2-4x-5[/tex]
We need to find the vertex form which is.,
[tex]y=a(x-h)^2+k[/tex]
where [tex](h,k)[/tex] represents the co-ordinates of vertex.
We apply completing square method to do so.
We have
[tex]Y=-\frac{1}{3}x^2-4x-5[/tex]
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.
[tex]-3\times Y=-3\times\frac{1}{3}x^2-(-3)\times4x-(-3)\times 5[/tex]
[tex]-3Y=x^2+12x+15[/tex]
Isolating [tex]x^2[/tex] and [tex]x[/tex] terms on one side.
Subtracting both sides by 15.
[tex]-3Y-15=x^2+12x-15-15[/tex]
[tex]-3Y-15=x^2+12x[/tex]
In order to make the right side a perfect square trinomial, we will take half of the co-efficient of [tex]x[/tex] term, square it and add it both sides side.
square of half of the co-efficient of [tex]x[/tex] term = [tex](\frac{1}{2}\times 12)^2=(6)^2=36[/tex]
Adding 36 to both sides.
[tex]-3Y-15+36=x^2+12x+36[/tex]
[tex]-3Y+21=x^2+12x+36[/tex]
Since [tex]x^2+12x+36[/tex] is a perfect square of [tex](x+6)[/tex], so, we can write as:
[tex]-3Y+21=(x+6)^2[/tex]
Subtracting 21 to both sides:
[tex]-3Y+21-21=(x+6)^2-21[/tex]
[tex]-3Y=(x+6)^2-21[/tex]
Dividing both sides by -3.
[tex]\frac{-3Y}{-3}=\frac{(x+6)^2}{-3}-\frac{21}{-3}[/tex]
[tex]Y=-\frac{1}{3}(x+6)^2+7[/tex] [Vertex form]
