Question

How do you graph `f(x)= 2/3(x-3)^2 - 6`?

Answer 1

Plot the vertex, the intercepts (and possibly a few other points) then connect the points with a smooth curve.

Explanation:

`f(x) = 2/3(x-3)^2 - 6` is in vertex form
so as one of our points we have the vertex `(3,-6)`

The `f(x)` (or `y`) intercept occurs where `x=0`
`color(white)("XXXX")` `f(0) = 2/3(-3)^2 -6 =0`
giving us another point: `(0,0)`

The x-intercepts occur where `f(x)=0`
`color(white)("XXXX")` `2/3(x-3)^2-6 = 0`

`color(white)("XXXX")` `(x-3)^2 = 6*(3/2)`

`color(white)("XXXX")` `x-3 = +-3`

`color(white)("XXXX")` `x = 0 or x=6`
Unfortunately this only gives us 1 more point (we already had `(0,0)`)
`color(white)("XXXX")` `(6,0)`

We could evaluate `f(x)` for a few more values of `x` (`x=+-6` and `x=+-9` would seem promising; but the 3 points we have will do for now:

graph{2/3(x-3)^2-6 [-4.39, 13.39, -6.404, 2.485]}