Question

How do you graph `y = -x^2 + 3`?

Answer 1

graph{-x^2+3 [-10, 10, -5, 5]}

Explanation:

The equation is of the kind

`y=ax^2+bx+c` `a=-1, b=0, c=3` which means it is a parabola whose axis is parallel to the y axis. Notice that `a<0` so it is a downward facing parabola.

Now you have to calculate the vertex:

use the formula
`2ax_v+b=0 => x_v=-b/2a => x_v=0`
`y_v=ax_v^2+bx_v+c=0+0+3=3`

So the vertex is in `(0,3)`, the axis is parallel to the y axis and intersect the curve in `(0,3) =>` it's the y axis `=>` the parabola is symmetric on the y axis.

Then you calculate the intersection with the x axis. Trivially, they are `(sqrt(3),0)` and `(-sqrt(3),0)`

Now use what you know and draw the curve, as in the graph.