Question

How do you graph the inequality `y>=x^2+3x-18`?

Answer 1

It is the region comprising the parabola `x+3/2)^2=y+81/4` and its interior. See the illustrative Socratic graph.

Explanation:

The boundary for the region that comprises point ( x, y ) such that

`y>= x^2+3x-18` is the parabola `y = x^2+3x-18`

This has the form `(x+3/2)^2=y+81/4` disclosing that the vertex of

the parabola is `(-3/2, -81/4)` and its axis is `x = -3/2 uarr`.

If (x, y) is on the parabola, for any interior point (x, Y), `Y >=y`.

graph{y-x^2-3x+9>=0 [-40, 40, -20, 20]}