Question

How do you find the axis of symmetry and vertex point of the function: `y = -x^2 - x + 9`?

Answer 1

Vertex: `(-1/2,9 1/4)`
Line of symmetry: `x=-1/2`

Explanation:

use `-b/(2a)` to find the vertex:

`-(-1)/(2xx(-1)) = -1/2`

plug `-1/2` in for `x`

`= 9 1/4`

vertex: `(-1/2,9 1/4)`

on parabolic functions the axis of symmetry is the vertical line where the vertex is, so the equation is:

`x=-1/2`

Answer 2

An alternate way, using the vertex form:

Explanation:

We can change the standard form of the equation to the vertex form, which has the general form:

`y=a(x-h)^2+k`

We get there by completing the square (we do that by taking the constant of the `x` term, halving it, squaring the half, then adding and subtracting it into the equation:

`y=-x^2-x+9`

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

`y=-(x^2+x+1/4-1/4)+9`

`y=-((x+1/2)^2-1/4)+9`

`y=-(x+1/2)^2+1/4+9`

`y=-(x+1/2)^2+9 1/4`

The vertex is given by `(h,k)`, which in this case is `(-1/2, 9 1/4)`.

The axis of symmetry runs through the vertex vertically and so takes the form of `x= "the " x " value of the vertex"`, which in this case is

`x=-1/2`

This can all be seen in the graph:

graph{-x^2-x+9[-3,3,8,10]}

note that I've adjusted the graph unequally vertically and horizontally to better show the vertex