Paul was given the inequality y>or= -ax-b. He determined that all points in quadrant 2 will be solutions.

Question

Part A: write one possible inequality Paul could have graphed

Part B: Write all possible values of a.

Part C: Identify all possible values of b

Answer 1

If a ≤ 0 and b ≥ 0, then y = -ax - b has nonnegative slope and a nonpositive y-intercept.

So, all points in the second quadrant are points in the solution set of y ≥ -ax - b. The solution set contains other points as well.

Examples: a = -3 and b = 2 So, y ≥ 3x - 2

a = 0 and b = 1 So, y ≥ -1

a = -5 and b = 0 So, y ≥ 5x

Answer 2

The only way for all points in quadrant 2 to be a solution is for a and b to both be 0. Then it's the x-axis, so everything in the upper left (Quadrant 2) is a solution.

y >= 0

0 is the only possible value for a and b.

To see it, graph y >= -ax - b on Desmos. Make a and b sliders.