Graphing Inequalities Lesson #29 – Graphing Inequalities

Recall that an inequality contains the symbols _________ <,>,≤, and ≥ so too can y < ax +b and y ≥ a(x-h)2 +k! If the line y=ax+b and the paraboloa y=a(x-h)2+k can be graphed, The solution set is a region of space where the inequality relation holds true. Eg. 1 Sketch y > 2x +1 Solution Set: Set of all possible solutions Dashed line states that y cannot equal 2x+1, only values greater (>) than that.

are the x and y intercepts. Eg. 2 Sketch Recall that (5,0) and (0,2) are the x and y intercepts. CHECK shading sub in (0,0) Solution Set Since 0 IS less than 1. (0,0) is in the solution set and the shading is below the line.

Math Gods Eg. 3 Sketch 3x + 6y + 12 ≥ 0, Rearrange first - 12 -3x - 12 -6y ≥ -3x - 12 -6 -6 Solution Set y ≤ 1x + 2 2 Check with (0,0) Math Gods 3x - 6y + 12 ≥ 0 “When you divide by a negative switch the direction of the inequality symbol.” 3(0) – 6(0) + 12 ≥ 0 12 ≥ 0 True! so we shaded properly

Eg. 4 Sketch the parabola y < 0.5(x-2)(x+4) Find the vertex Xsymmetry = -1 2+(-4) = 2 y < 0.5(x-2)(x+4) y < 0.5((-1)-2)((-1)+4) y < 0.5(-3)(3) y < -4.5 Check shading with (0,0) y < 0.5(x-2)(x+4) y < 0.5(0-2)(0+4) False! 0 is not < -4 so (0,0) is not in the set and we shaded properly. 0 < 0.5(-2)(4) 0 < -4

Homework Pg. 159 #1-2 Pg. 169 #1 Pg. 171 #3