Question

How do you graph and label the vertex and axis of symmetry `y=-4/3x^2+8/3x+8/3`?

Answer 1

The vertex form is `y=-4/3(x-1)^2+4`

Explanation:

We use

`a^2-2ab+b^2=(a-b)^2`

We complete the square and factorise in order to find the vertex form

`y=-4/3x^2+8/3x+8/3`

`y=-4/3(x^2-2x)+8/3`

`y=-4/3(x^2-2x+1)+8/3+4/3`

`y=-4/3(x-1)^2+12/3`

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

This is a parabola with vertex at `V=(1,4)`

The axis of symmetry is `x=1`

The intercept with the y-axis is `=(0,8/3)`

The intercepts with the x-axis is when `y=0`

`0=-4/3(x-1)^2+4`

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

`(x-1)^2=3`

`x-1=+-sqrt3`

`x=1+-sqrt3`

The intercepts are `(1+sqrt3,0)` and `(1-sqrt3,0)`

Now, we can draw the graph

graph{(y+4/3(x-1)^2-4)((x-1)^2+(y-4)^2-0.02)(y-1000(x-1))=0 [-11.25, 11.25, -5.625, 5.625]}