Question

How do you find the y intercept, axis of symmetry and the vertex to graph the function `f(x)=x^2-2/3x-8/9`?

Answer 1

`y` intercept: at `y=-8/9`
Axis of symmetry: `x=1/3`
Vertex: `(1/3,-1)`

Explanation:

y-intercept
The y-intercept is the value of `y` when `x=0`.
Substituting `0` for `x` in `y=x^2-2/3x-8/9`
gives `y=-8/9`

Axis of symmetry
For a parabola in the form `y=x^2+px+q`
the axis of symmetry is `x=p/2`
For the given example `p=-2/3`
so `p/2 =-1/3`
(see bottom for alternate derivation)

Vertex
Given an axis of symmetry `x=t`
The vertex is at `(a,f(t))`
In this case `f(-1/3) = (-1/3)^2-(-1/3) * 2/3 -8/9 = 1/9-2/9-8/9=-1`

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Alternate Deviation of Axis of Symmetry and Vertex
A parabola in the form `y=(x-a)^2+b`
has a vertex at `(a,b)`
and an axis of symmetry `x=a`

`y=2^2-2/3x-8/9` can be converted into this form by "completing the square".

`y=x^2-2/3xcolor(red)(+(-1/3)^2)-8/9color(red)(-(1/3)^2)`

`color(white)("X")=(x-1/3)^2-1`
giving:
`color(white)("XXX")`vertex at `(1/3,-1)` and
`color(white)("XXX")`axis of symmetry `x=1/3`

graph{x^2-2/3x-8/9 [-2.947, 3.215, -2.12, 0.958]}