Question

How do you graph `f(x) = (2/3)(x-2)^2+0.5`?

Answer 1

See explanation. Determining the key points.

Explanation:

This is the vertex form of a quadratic. standard form `y=a(x+b/(2a))^2+c+k` where `k` is a correction due to the error produced by `a(b/(2a))^2` so we have `a(b/(2a))^2+k=0`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given: `y=2/3(xcolor(red)(-2))^2color(green)(+0.5)` where `0.5=k+c`

`color(blue)("General shape")`

Note that `2/3xx x^2>0` so the graph is of form `uu`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the vertex")`

`x_("vertex")=(-1)xxcolor(red)(-2)= +2`
`y_("vertex")=color(green)(+5)`

`=>"Verex"->(x,y)=(2,5)`

Note that as the graph is of form `uu` and the vertex is above the x-axis then the graph does NOT cross the x-axis.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine "y" intercept")`

`y_("intercept")=2/3(color(red)(-2))^2color(green)(+0.5) = 19/6`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Method demonstration: determine "x" intercpts")`

Set `y=0=2/3(x-2)^2+0.5`

`(x-2)^2=3/2(-1/2) = -3/4`

`sqrt((x-2)^2)=sqrt(-3/4)`

`x=2+-sqrt(-3/4)`

You can go further with this using complex numbers. However, you do not need to as you are only plotting/sketching the graph.