Download 1 or x - TeacherWeb

UNIT 1A
LESSON 6B
Quadratic and Polynomial
Inequalities
1
INTERVALS & INEQUALITIES
Interval
Notation
Inequality
Notation
Graph
2
Solving Quadratic Inequalities by Graphing
𝑦 = 𝑥 2 − 6𝑥 + 5
𝑥 2 − 6𝑥 + 5 > 𝟎
𝒙|𝒙 < 𝟏 𝒐𝒓 𝒙 > 𝟓
𝑥 2 − 6𝑥 + 5 ≥ 𝟎
𝒙|𝒙 ≤ 𝟏 𝒐𝒓 𝒙 ≥ 𝟓, 𝒙 ∈ 𝑹
𝑥 2 − 6𝑥 + 5 < 𝟎
𝒙|𝟏 < 𝒙 < 𝟓, 𝒙 ∈ 𝑹
𝑥 2 − 6𝑥 + 5 ≤ 𝟎
𝒙|𝟏 ≤ 𝒙 ≤ 𝟓, 𝒙 ∈ 𝑹
3
SOLVING POLYNOMIAL INEQUALITIES
In order to solve any polynomial equation or inequality you must FACTOR first!!!
x2 – 6x + 5 = 0
Let’s use
our heads
(x – 1 )(x – 5) = 0
x = 1or x = 5
If x < 1
If x > 5
If x is between 1 and 5
( 𝟒 – 1 )( 𝟒 – 5) is ( 𝟔 – 1 )( 𝟔 – 5) is
( 𝟎 – 1 )( 𝟎 – 5) is
positive
negative
+
−
1
positive
+
5
4
SOLVING POLYNOMIAL INEQUALITIES
In order to solve any polynomial equation or inequality you must FACTOR first!!!
+
+
1
5
x2 – 6x +5 > 0
x2 – 6x +5 > 0
(x – 1 )(x – 5) > 0
(x – 1 )(x – 5) > 0
x < 1 or x > 5
x < 1 or x > 5
(−∞, 𝟏) ∪ (𝟓, ∞)
−∞, 𝟏 ∪ [𝟓, ∞)
5
SOLVING POLYNOMIAL INEQUALITIES
In order to solve any polynomial equation or inequality you must FACTOR first!!!
−
1
5
x2 – 6x +5 < 0
x2 – 6x +5 < 0
(x – 1 )(x – 5) < 0
(x – 1 )(x – 5) < 0
1<x<5
1<x<5
(𝟏, 𝟓)
[𝟏, 𝟓]
6
Do Questions 1- 4 on pages 2 and 3
on your own
7
Special Cases
𝒚 = 𝒙𝟐 + 𝟔𝒙 + 𝟗
𝒙𝟐 + 𝟔𝒙 + 𝟗 > 𝟎
𝒙|𝒙 ≠ −𝟑, 𝒙 ∈ 𝑹
𝒙𝟐 + 𝟔𝒙 + 𝟗 ≥ 𝟎
𝒙|𝒙 ∈ 𝑹
𝒙𝟐 + 𝟔𝒙 + 𝟗 < 𝟎
𝒏𝒐 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
𝒙𝟐 + 𝟔𝒙 + 𝟗 ≤ 𝟎
𝒙|𝒙 = −𝟑
8
𝐒𝐨𝐥𝐯𝐞 𝐛𝐲 𝐟𝐚𝐜𝐭𝐨𝐫𝐢𝐧𝐠
𝒙𝟐 + 𝟔𝒙 + 𝟗 = 𝟎
𝒙+𝟑
𝟐
−𝟑
=𝟎
𝒙 = −𝟑
If 𝑥 < −3
−𝟒 + 𝟑
If 𝑥 > −3
𝟐
is
𝟎+𝟑
positive
𝒙𝟐 + 𝟔𝒙 + 𝟗 > 𝟎
𝒙 ≠ −𝟑
𝒙𝟐 + 𝟔𝒙 + 𝟗 < 𝟎
𝒏𝒐 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
𝟐
is
positive
𝒙𝟐 + 𝟔𝒙 + 𝟗 ≥ 𝟎
𝒙∈𝑹
𝒙𝟐 + 𝟔𝒙 + 𝟗 ≤ 𝟎
𝒙 = −𝟑
9
Do Questions 1- 4 on page 5
on your own
10
Higher Order Polynomials
𝒚 = 𝒙𝟑 + 𝟐𝒙𝟐 − 𝟓𝒙 − 𝟔
𝒙𝟑 + 𝟐𝒙𝟐 − 𝟓𝒙 − 𝟔 > 𝟎
𝒙| − 𝟑 < 𝒙 < −𝟏 𝒐𝒓 𝒙 > 𝟐, 𝒙 ∈ 𝑹
𝒙𝟑 + 𝟐𝒙𝟐 − 𝟓𝒙 − 𝟔 ≥ 𝟎
𝒙| − 𝟑 ≤ 𝒙 ≤ −𝟏 𝒐𝒓 𝒙 ≥ 𝟐, 𝒙 ∈ 𝑹
𝒙𝟑 + 𝟐𝒙𝟐 − 𝟓𝒙 − 𝟔 < 𝟎
𝒙|𝒙 < −𝟑 𝒐𝒓 − 𝟏 < 𝒙 < 𝟐, 𝒙 ∈ 𝑹
𝒙𝟑 + 𝟐𝒙𝟐 − 𝟓𝒙 − 𝟔 ≤ 𝟎
𝒙|𝒙 ≤ −𝟑 𝒐𝒓 − 𝟏 ≤ 𝒙 ≤ 𝟐, 𝒙 ∈ 𝑹
11
Factor x3 + 2x2 – 5x – 6
Test potential zeros ±1, ±2, ±3, ±6.
23 + 2(2)2 – 5(2) – 6 = 8 + 8 – 10 – 6 = 0
(x – 2) is a factor
Addition Method
Subtraction Method
-2 1
2
-2
1 4
-5
-8
3
-6
-6
0
2 1
1
2
2
4
-5
8
3
-6
6
0
x3 + 2x2 – 5x – 6 = (x – 2)(1x2 + 4x + 3)
= (x – 2)(x + 3)(x + 1)
12
EXAMPLE continued
x3 + 2x2 – 5x – 6
x3 + 2x2 – 5x – 6 = (x – 2)(x + 3)(x + 1)
−
+
–3
+
−
–1
2
If x < – 3
( −𝟒 – 2)( −𝟒 + 3)( −𝟒 + 1) is
neg
If – 3 < x < – 1
( −𝟐 – 2)( −𝟐 + 3)( −𝟐 + 1) is
pos
If – 1 < x < 2
( 𝟎 – 2)( 𝟎
+ 3)( 𝟎
+ 1) is
neg
If x > 2
( 𝟓 – 2)( 𝟓
+ 3)( 𝟓 + 1) is
pos
13
x3 6+ 2x2 – 5xx–3 6+ 2x2 – 5x – 6
EXAMPLE
x3 + 2x2 – 5x – 6 = (x – 2)(x + 3)(x + 1)
neg
pos
–3
pos
neg
–1
2
x3 + 2x2 – 5x – 6 > 0
x3 + 2x2 – 5x – 6 < 0
–3 < x < –1 or x > 2
x < –3 or –1 < x < 2
x3 + 2x2 – 5x – 6 > 0
x3 + 2x2 – 5x – 6 < 0
–3 < x < –1 or x > 2
x < –3 or –1 < x < 2
14
Do Questions 1- 4 on page 7
on your own
15
Special Cases of Cubics
16
Higher Order Polynomials
𝒚 = 𝒙𝟑 + 𝒙𝟐 − 𝟓𝒙 + 𝟑
𝒙𝟑 + 𝒙𝟐 − 𝟓𝒙 + 𝟑 > 𝟎
𝒙|𝒙 > −𝟑, 𝒙 ≠ 𝟏𝒙 ∈ 𝑹
𝒙𝟑 + 𝒙𝟐 − 𝟓𝒙 + 𝟑 ≥ 𝟎
𝒙|𝒙 ≥ −𝟑, 𝒙 ∈ 𝑹
𝒙𝟑 + 𝒙𝟐 − 𝟓𝒙 + 𝟑 < 𝟎
𝒙|𝒙 < −𝟑 𝒙 ∈ 𝑹
𝒙𝟑 + 𝒙𝟐 − 𝟓𝒙 + 𝟑 ≤ 𝟎
𝒙|𝒙 ≤ −𝟑 , 𝒙 = 𝟏, 𝒙 ∈ 𝑹
17
Factor x3 + x2 – 5x +3
Test potential zeros ±1, ±2, ±3.
13 + (1)2 – 5(1) + 3 = 1 + 1 – 5 + 3 = 0
(x – 1) is a factor
Addition Method
Subtraction Method
-1 1
1
-1
1 2
-5
-2
-3
3
3
0
1 1
1
1
1
2
-5 3
2 -3
-3
0
x3 + 2x2 – 5x – 6 = (x – 1)(1x2 + 2x – 3)
= (x – 1)(x + 3)(x – 1) = 𝑥 − 1 2 (x + 3)
18
EXAMPLE continued
x3 + x2 – 5x +3
= 𝑥 − 1 2 (x + 3)
x3 + x2 – 5x +3
+
−
–3
+
1
If x < – 3
( −𝟒 – 1)2( −𝟒 + 3) is
neg
If – 3 < x < 1
( −𝟐 – 1)2( −𝟐 + 3) is
pos
If x > 1
( 𝟓 – 1)2( 𝟓 + 3) is
pos
19
EXAMPLE continued
= 𝑥 − 1 2 (x + 3)
x3 + x2 – 5x +3
+
−
–3
x3 + x2 – 5x + 3 > 0
𝒙 > −𝟑, 𝒙 ≠ 𝟏
x3 + x2 – 5x + 3 > 0
𝒙 ≥ −𝟑
x3 + x2 – 5x +3
+
1
x3 + x2 – 5x + 3 < 0
x < –3
x3 + x2 – 5x + 3 < 0
𝒙 ≤– 𝟑, 𝒙 = 𝟏
20
Higher Order Polynomials
𝒚 = 𝒙𝟑 + 𝒙𝟐 + 𝟐𝒙 + 𝟐
𝒙𝟑 + 𝒙𝟐 + 𝟐𝒙 + 𝟐 > 𝟎
𝒙|𝒙 > −𝟏, 𝒙 ∈ 𝑹
𝒙𝟑 + 𝒙𝟐 + 𝟐𝒙 + 𝟐 ≥ 𝟎
𝒙|𝒙 ≥ −𝟏, 𝒙 ∈ 𝑹
𝒙𝟑 + 𝒙𝟐 + 𝟐𝒙 + 𝟐 < 𝟎
𝒙|𝒙 < −𝟏 𝒙 ∈ 𝑹
𝒙𝟑 + 𝒙𝟐 + 𝟐𝒙 + 𝟐 ≤ 𝟎
𝒙|𝒙 ≤ −𝟏, 𝒙 ∈ 𝑹
21
Factor x3 + x2 + 2x + 2
Test potential zeros ±1, ±2.
(–1)3 + (–1)2 + 2(–1) + 2 = – 1 + 1 – 2 + 2 = 0
(x + 1) is a factor
Addition Method
Subtraction Method
1 1
1
1 2
1 0
0 2
2
2
0
–11
1
–1
1 0
2 2
0 –2
2
0
x3 + x2 + 2x + 2 = (x + 1)(1x2 + 2)
= (x + 1)(x2 + 2)
22
x3 + x2 + 2x +2
EXAMPLE continued
x3 + x2 + 2x + 2
= (x + 1)(x2 + 2)
−
+
–1
If x < – 1
If x > – 1
( −𝟒 + 1)( −𝟒 𝟐 + 2) is
( 𝟒 + 1)( 𝟒
𝟐
+ 2) is
neg
pos
23
EXAMPLE continued
x3 + x2 + 2x + 2
x3 + x2 + 2x + 2
= (x + 1)(x2 + 2)
−
+
–1
x3 + x2 + 2x + 2 > 0
x3 + x2 + 2x + 2 < 0
𝒙 > −𝟏
𝒙 < −𝟏
x3
+
x2
+ 2x + 2 > 0
𝒙 ≥ −𝟏
x3 + x2 + 2x + 2 < 0
𝒙 ≤ −𝟏
24
Do Questions 1- 4 on page 10
on your own
25