**Vertex form**

An equation in the form:

#color(white)("XXXX")##f(x)=m(x-a)^2+b#

has its vertex at #(a,b)#

Given: #f(x) = -x^2-6x-5#

Extract #m# for the vertex form:

#color(white)("XXXX")##f(x) = (-1)(x^2+6x) -5#

Complete the square

#color(white)("XXXX")##f(x) = (-1)(x^2+6x+9) -5 - (-1)(9)#

Simplifying

#color(white)("XXXX")##f(x) = (-1)(x+3)^2 +4#

#color(white)("XXXX")##color(white)("XXXX")#which is in vertex form with the vertex at #(-3,4)#

**y vertex coordinate formula**

An equation in the form:

#color(white)("XXXX")##f(x)=ax^2+bx+c#

has the #f(x)# coordinate at

#color(white)("XXXX")##c-b^2/(4a)#

#color(white)("XXXX")##color(white)("XXXX")#[this is a result of the significance of the discriminant]

For #f(x)=-x^2-6x-5#

#color(white)("XXXX")##a=-1##color(white)("XXXX")##b=-6##color(white)("XXXX")##c=-5#

and the #f(x)# coordinate of the vertex is at

#color(white)("XXXX")##-5 - ((-6)^2/(4(-1))) = 4#

Solving for the corresponding #x# coordinate:

#color(white)("XXXX")##-x^2-6x-5 = 4#

#color(white)("XXXX")##-x^2-6x-9 = 0#

#color(white)("XXXX")##(x^2+6x+9) = 0#

#color(white)("XXXX")##(x+3)^2 = 0#

#color(white)("XXXX")##rarr##color(white)("XXXX")##x=-3#

and again, we have the vertex is at #(-3,4)#