Question

How do you graph and label the vertex and axis of symmetry of `y=2/3(x-2)^2-5/3`?

Answer 1

Vertex`->(x,y)=(2,-5/3)`

Axis of symmetry `->x_("vertex")->x=2`

`y_("intercept")->(x,y)=(0,1)`

`x_("intercept")=2+-sqrt(10)/3 larr" Exact answer"`
`x_("intercept")=3.58 and 0.42` to 2 decimal places (2dp)

Explanation:

This is the vertex form of a quadratic from which, with a small adjustment, you can directly read off the vertex.

Initial form: `y=ax^2+bx+c`

Vertex form `y=a(x+b/(2a))^2+k+c`

Where `axx[b/(2a)]^2+k=0`

In this case `k+c=-5/3`

`k` is needed as changing `y=ax^2+bx+c` into this form introduces an error. So the addition of `k` acts as a correction factor.

Given:`y=2/3(x-2)^2-5/3`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the vertex")`

`x_("vertex")=(-1)xxb/(2a)" "=" "(-1)xx(-2)=+2`

`y_("vertex")=k+c" "=" "=-5/3`

Vertex`->(x,y)=(2,-5/3)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine y-intercept")`

Set `x=0`

`y=2/3(0-2)^2-5/3`

`y_("intercept")=8/3-5/3=1`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine x-intercept")`

Set `y=0=2/3(x-2)^2-5/3`

Add `5/3` to both sides

`5/3=2/3(x-2)^2`

Multiply both sides by `3/2`

`15/6=(x-2)^2`

Square root both sides

`+-sqrt(15/6)=x-2`

`x_("intercept")=2+-sqrt(15/6) larr" Exact answer"`

Rounding decimals is never precise so the approximate answer is:

`x_("intercept")=3.58 and 0.42` to 2 decimal places (2dp)