Question

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

Answer 1

I would graph the equation `y = -2x^2-x+3`. This divides the plane into two regions. Test points in each region to see if the region contains solutions to the inequality.

Explanation:

Graph `y = -2x^2-x+3`.

The graph is a parabola.
We could graph this by finding the vertex and so on, but I find it simpler to just find the `x` intercepts

`-2x^2-x+3 - -(2x^2+x-3) = -(2xcolor(white)"XXX")(xcolor(white)"XXX")`

`= -(2x+3)(x-1)`

So the `x` intercepts are `-3/2` and `1`.

The `y` intercept is `3`, so the graph of the equation is

graph{ -2x^2-x+3 [-8.62, 7.186, -3.96, 3.94]}

Testing `(0,0)` we see that `0 >= -2(0)^2-(0)+3` is false so there are no solutions in the region containing the origin.

Testing `(0,5)` (remember the `y` intercept is `3`) or `(1,3)` or `(5,0)` or some other point outside the region containing `(0,0)`, we learn the the region not containing `(0,0)` contains the solutions

OR We can reason the the values of `y` greater than -2x^2-x+3 are above the curve.

graph{y >= -2x^2-x+3 [-8.62, 7.186, -3.96, 3.94]} .