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

Dec 26, 2016

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

$y \ge {x}^{2} + 3 x - 18$ is the parabola $y = {x}^{2} + 3 x - 18$

This has the form ${\left(x + \frac{3}{2}\right)}^{2} = y + \frac{81}{4}$ disclosing that the vertex of

the parabola is $\left(- \frac{3}{2} , - \frac{81}{4}\right)$ and its axis is $x = - \frac{3}{2} \uparrow$.

If (x, y) is on the parabola, for any interior point (x, Y), $Y \ge y$.

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