Question

How do you graph the quadratic function and identify the vertex and axis of symmetry for `y=y=-3x^2+5`?

Answer 1

Vertex `= (0, +5)`
Axis of symmetry `x=0`

Explanation:

For a quadratic in standard form: `ax^2+bx+c`
The axis of symmetry and vertex will occur where `x=(-b)/(2a)`

In our example: `-3x^2+5 -> a=-3, b=0, c=+5`

`(-b)/(2a) = 0/-6 =0`

Hence the axis of symmetry is `x=0` [i.e. the `y-`axis]

And the y-coordinate of the vertex of the parabola is:
`0+5=5`

`:.` Vertex `=(0, +5)`

Also, since the coefficient of `x^2 <0 -> y(0)=5` is the absolute maximum of `y`.

The vertex and axis of symmetry can be seen on the graph of `y` below.

graph{-x^2+5 [-11.25, 11.25, -5.63, 5.62]}