Question

What is the axis of symmetry and vertex for the graph `y= -x^2-4x+5`?

Answer 1

I'll walk you through it!

Axis of symmetry is `x=-2`

Vertex is at `(-2,9)`

Explanation:

There are different ways of going about this:

`1.` Solve y=0 to find the `x`-intercepts
`y=-x^2-4x+5`
`0=-x^2-4x+5`
`0=(x+5)(-x+1)`

Therefore `x = -5 and x = 1` are the `x`-intercepts.

The axis of symmetry has to be between these two points:

-5, -4, -3, -2, -1, 0, 1 `" "` Or `x = (-5+1)/2 = -4/2 = -2`

Hence, our axis of symmetry is `x=-2`.

Because we know the graph will be at its lowest or highest (i.e. its vertex), on the line `x=-2`, we can plug `-2` into the original equation and find out the vertex point:

`y =-(-2)^2 -4(-2)+5`

`y=9`

Therefore the vertex is at `(-2,9)`

`2`. Complete the square
This is an alternative method, which is very commonly used in middle and high school. I will not get into how to complete the square, and just show you what I did:

`y=-x^2-4x+5`
`y=-(x^2+4x-5)`
`y =-(x^2 +4x +4-4-5)`
`y =-[(x+2)^2-9)]`

`y=-(x+ **2** )^2 **+9**`

Therefore the vertex is at (`-2,9)`

Answer 2

Vertex (-2,9) Axis of symmetry x=-2

Explanation:

The problem requires the given quadratic function, to be put in vertex form as demonstrated below. The quadratic function represents a vertical parabola.

`y= -(x^2 +4x)+5`

=`-(x^2 +4x +4 -4)+5`

=`-(x^2 +4x+4) +4+5`

=`-(x+2)^2 +9`

The given quadratic function is now in vertex form. This represents a vertical parabola, opening down wards, with its vertex at (-2,9) and the axis of symmetry being x=-2