1. Graphing Inequalities
Lesson #29 –
Graphing Inequalities

2. 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.

3. 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.

4. 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

5. 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

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