How do you graph y>= sqrt(3x+12)

Show how to get certain points.

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  1. 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.

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