Question

How do you graph and identify the vertex and axis of symmetry for `y=1/3x^2-2x-3`?

Answer 1

The Axis of symmetry is the vertical line `x=3`

The coordinates of the vertex are `(3,-6)`

Explanation:

This is a quadratic function in the form `ax^2+bx+c=y`. With `a=(1/3), b=-2, and c=-3`
When in this form the axis of symmetry is the vertical line `x=-b/(2a)`
Here this becomes `x=-(-2)/(2(1/3))=3`.
This is the line of symmetry and the x value of the vertex.

The `y` value of the vertex is then found by plugging the `x` value of the line of symmetry into the original equation and solving for `y`
`y=(1/3)x^2-2x-3=(1/3)(3)^2-2(3)-3=-6`

So the coordinates of the vertex are `(3,-6)`