Question 889080
First find when {{{y=0}}}
{{{x^2+x-12=0}}}
{{{(x+4)(x-3)=0}}}
Two solutions:
{{{x+4=0}}}
{{{x=-4}}}
and
{{{x-3=0}}}
{{{x=3}}}
Now you have the number line broken into three distinct regions.
{{{x<-4}}}
{{{-4<x<3}}}
{{{x>3}}}
Choose a value in each region and test the inequality.
{{{x=-5}}}
{{{x^2+x-12>=0}}}
{{{(-5)^2-5-12>=0}}}
{{{25-5-12>=0}}}
{{{8>=0}}}
True
.
.
{{{x=0}}}
{{{(0)^2+0-12>=0}}}
{{{-12>=0}}}
False
.
.
{{{x=4}}}
{{{(4)^2+4-12>=0}}}
{{{16+4-12>=0}}}
{{{8>=0}}}
True
So the first and third regions are the solution regions and you include the endpoints since it is greater than or equal to.
.
.

({{{-infinity}}},{{{-4}}}]U[{{{3}}},{{{infinity}}})
.
.
.
{{{graph(300,300,-6,6,-6,6,x^2+x-12)}}}