1. 
2. 
3. 
4. 
Solve by CTS: 4x2 + 16x + 8 = -8
Solve by Quadratic formula: 3x2 + 5x = 5x2 + 6x – 8.
Use the discriminant to determine the number
of real solutions to 4x2 + 3x = 8x. Explain
what that means. Use your calculator to find the vertex and
intercepts of -10.2x2 + 2.8x + 13.75. Algebra II
1
Algebra II
2
1. 
2. 
3. 
4. 
5. 
6. 
4(x – 3)2 + 3 = 12
x2 + 6x + 9 = 10
x2 + 3x + 8 = 4
x2 – 10 = x2 + 12
5x2 + 7x – 12 = 3x2 + 2x – 8 3x2 + 5x – 2 = 0
Algebra II
3
Graphing and Solving
Quadratic Inequalities
Algebra II
¡  A
quadratic function with an
inequality symbol
Graphing a quadratic inequality:
1. 
Graph the parabola as the boundary line.
<, > à dotted boundary
≤, ≥ à solid boundary
2. 
Choose a test point and shade
appropriately.
Algebra II
5
1. y < x2 – 4x + 1
Vertex:
x = 4 / (2Ÿ1) = 2
y
= (2)2 – 4(2) + 1 = -3
(2, -3)
x
y
Table à
3
-2
Reflect
4
1
Test Point: (2, 0)
0 < 22 – 4(2) + 1
0 < –3 FALSE! Don't shade
this point! Algebra II
6
2. y ≤ -2x2 + 8x – 2
Vertex:
x = -8 / (2Ÿ-2) = 2
y = -2(2)2 + 8(2) – 2 = 6
(2, 6)
x
y
Table à
3
4
Reflect
4
-2
Test Point: (2, 0)
0 ≤ –2(2)2 + 8(2) – 2
0 ≤ 6
TRUE!
Shade this point!
Algebra II
7
3. y ≤ 2x2 – 3 Vertex:
x = 0/ (2Ÿ2) = 0
y = 2(0)2 – 3 = -3
(0, -3)
X
Y
Table à
1
-1
Reflect
2
5
Test Point: (0, 0)
0 ≤ 2(0)2 – 3
0 ≤ -3 FALSE! Don't shade
this point!
Algebra II
8
4. y > -(x – 3)2 + 1
Vertex: (3, 1)
Table à
Reflect
x
y
4
0
5
-3
Test Point: (0,0)
0 > -(0-3)2 + 1
0>-8
TRUE!
Shade this point!
Algebra II
9
5. y > (x – 4)(x + 2)
x-int: (4, 0), (-2, 0)
Vertex:
x = (4 + -2)/2 = 1
y = (1 – 4)(1 + 2) = -9
(1, -9)
Test Point: (0,0)
0 > (0 – 4)(0 + 2)
0 > -8 TRUE!
Shade this point!
Algebra II
10
¡  The
solution to a quadratic inequality is
always an interval of numbers that will
work
¡  Always
write as “ax2 + bx + c symbol 0”
¡  There
are two methods to solve
quadratic inequalities:
§  Algebraically
§  Graphically
Algebra II
11
1. 
Solve the quadratic like its an
equation. 2. 
The x-values are critical points
that split the number line into
three parts. Test an x-value in each section.
4.  Write the intervals in which the
test values work. 3. 
Algebra II
12
1. 0 < x2 – 3x + 2
x2 – 3x + 2 > 0
x2 – 3x + 2 = 0
(x – 2)(x – 1) = 0
x = 2 x = 1
0
1
1.5
Check with calculator Solution: (-∞, 1) U (2, ∞)
2
3
0 < 02 – 3(0) + 2
0 < 32 – 3(3) + 2
0<2
0<2
0 < (1.5)2 – 3(1.5) + 2
True
True
0 < -.25
Algebra II
False
13
2. x2 – 3x – 10 ≤ 0
x2 – 3x – 10 = 0
(x – 5)(x + 2) = 0
x=5
x = -2
Check with calculator Solution: [-2, 5]
-3
0
6
-2
5
2
(-3) – 3(-3) – 10 ≤ 0
62 – 3(6) – 10 ≤ 0
8 ≤ 0
8≤0
2 – 3(0) – 10 ≤ 0
(0)
False
False
-10 ≤ 0
True
Algebra II
14
3. (x – 2)2 + 3 > 7
(x – 2)2 + 3 = 7
(x – 2)2 = 4
x–2=+2 x=0
-1
(-1-2)2 + 3 > 7
12 > 7
True
Algebra II
Check with calculator Solution: (-∞, 0) U (4, ∞)
x = 4
0
2
(2-2)2 + (3) > 7
3 > 7
False
4
5
(5 - 2)2 + 3 > 7
12 > 7 True
15
4. 2x2 + 5x ≤ 12
2x2 +5x – 12 ≤ 0
(2x – 3)(x + 4) = 0
x = 3/2 x = -4
Check with calculator Solution: [-4, 3/2]
2
0
-4
3/2
2(-5)2 + 5(-5) ≤ 12
2(2)2 + 5(2) ≤ 12
25 ≤ 12
18 ≤ 12 2(0)2 + 5(0) ≤ 12
False
False
0 ≤ 12
True
-5
Algebra II
16
5. 2x2 + 4x ≥ 13
2x2 + 4x – 13 ≥ 0
x = −4 ±
x=
16 − 4(2)(−13)
4
−4 ± 2 30 −2 ± 30
=
4
2
Check with calculator Solution: (-∞, -3.7]U [1.7, ∞)
2
0
-3.7
1.7
2(-4)2 + 4(-4) ≥ 13
2(2)2 + 4(2) ≤ 13
16 ≥ 12
16 ≥ 12 2(0)2 + 4(0) ≤ 13
False
False
0 ≥ 13
Algebra II
True
-4
17
6. -3x2 + x + 8 > 4
Check with calculator -3x2 + x + 4 = 0
-(3x2 – x – 4) = 0
(3x – 4)(x + 1) = 0
x = 4/3 x = -1
Solution: (-1 , 4/3)
-2
0
2
-1
4/3
-3(-2)2 + (-2) + 8 > 4
-3(2)2 + (2) + 8 > 4
-6 > 4
-2 > -4 2 + (0) + 8 > 4
-3(0)
False
False
8 > 4
Algebra II
True
18
1. 
Solve the quadratic like its an
equation. 2. 
Make a “rough” sketch of the
graph with the x-values.
Choose a test point and shade
appropriately.
4.  Write the interval in which the
test point work. 3. 
Algebra II
19
1. y ≤ -x2 + 2x + 8 -x2 + 2x + 8 = 0
-1(x2 – 2x – 8)= 0
Test Point: (0,0)
0 ≤ -(0)2 + 2(0) + 8
0 ≤ 8
TRUE!
Shade inside!
-1(x – 4)(x + 2) = 0
x = 4 x = -2 Solution: [-2,4]
Algebra II
20
2. y < 2x2 – 6x – 20
2x2 – 6x – 20 = 0 2x2 – 6x – 20 = 0
Test Point: (0,0)
0 < 2(0)2 – 6(0) – 20
0 < -20
False!
Shade outside!
2(x2 – 3x – 10) = 0
2(x – 5)(x + 2) = 0
x = 5 x = -2
Solution: (-∞,-2)U(5,∞)
Algebra II
21
3. y > 2x2 – 6x – 20
2x2 – 6x – 20 = 0
2(x2 – 3x – 10) = 0
Test Point: (0,0)
0 > 2(0)2 – 6(0) – 20
0 > -20
TRUE!
Shade inside!
2(x – 5)(x + 2) = 0
x = 5 x = -2
Solution: (-2,5)
Algebra II
22
4. y ≥ 3x2 – 4x – 4
3x2 – 4x – 4 = 0
3x2 – 4x – 4 = 0
Test Point: (0,0)
0 ≥ 3(0)2 – 4(0) – 4
0 ≥ -4
False!
Shade inside!
(3x – 2)(x – 2) = 0
x=⅔ x=2
Solution: [⅔, 2]
Algebra II
23
1. Graph: y < 2x2 – 4x + 1
2. Solve algebraically: y>x2 – 3x + 2
3. Solve graphically: y<2x2 – 3x – 2 Algebra II
24