How do you graph y>= sqrt(3x+12)
Show how to get certain points.
The only points you should plot are on the y = sqrt (3x + 12) line. Let's assume that we only take the positive square root; otherwise there will be two lines.
If x = 0, y = sqrt 12 = 3.464
If x = 2, y = sqry 21 = 4.583
If x = 8, y = sqrt 36 = 6
Since you have a >= sign and not an = sign in your case, all of the territory above the line that you plot will be a solution.