Question

How do you find the VERTEX of a parabola `f(x)= -x^2-6x-5`?

Answer 1

Convert the given equation into vertex form [`y = m(x-a)^2+b`]
to get `(x, f(x)) = (-3, 4)`
or use the formula (see below) for the `f(x)` vertex coordinate and solve for `x`

Explanation:

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)`