unit 1a lesson 6 1 linear, quadratic and polynomial inequalities
TRANSCRIPT
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UNIT 1ALESSON 6
Linear, Quadratic and Polynomial Inequalities
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INTERVALS & INEQUALITIESInterval Notation
InequalityNotation
Graph
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Express the following intervals in terms of inequalities and graph the intervals
1,4
2
14
2x
[ 2,1) 2 1x
( 4, ) 4x
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REMEMBER:Linear inequalities are solved the same as equations
EXCEPT when the final step involves dividing by a NEGATIVE.
You must change the direction of the sign.
10 4
0 4
2 2
1
5 2
20 5
20 5
5 5
4 1
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Linear InequalitiesSolve the inequalities.State the answer in inequality form, interval form and graph.
5x + 7 > – 8
5x > – 15
x > – 3 inequality form
[– 3, ∞) interval form
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EXAMPLE 2
graph
(−𝟑 ,−𝟖)
𝒀 𝟐=𝟓 𝒙+𝟕
𝒀 𝟐=−𝟖
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Linear InequalitiesSolve the inequalities.State the answer in inequality form, interval form and graph.
3x + 1 < 7x – 7
– 4x < – 8
x > 2 inequality form
(2, ∞) interval form
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EXAMPLE 3
(𝟐 ,𝟕)
𝒀 𝟏=𝟑 𝒙+𝟏
𝒀 𝟐=𝟕 𝒙−𝟕graph
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Linear InequalitiesSolve the inequalities and graph. State the answer in inequality form and interval form.
8 – x > 5x + 2
– 6x > – 6
x < 1 inequality form
(–∞, 1) interval form
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EXAMPLE 4
−𝟔 𝒙−𝟔
<−𝟔−𝟔
(𝟏 ,𝟕)
𝒀 𝟏=𝟖−𝒙 𝒀 𝟐=𝟓 𝒙+𝟐
graph
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SOLVING POYNOMIAL INEQUALITIES In order to solve any polynomial equation or inequality you must FACTOR first!!!
EXAMPLE 5 : x2 + 2x – 8 = 0
(x + 4)(x – 2) = 0
–4 2
x = –4 or x = 2
If x < – 4
( + 4)( – 2) is
positive
If x is between –4 and 2
( + 4)( – 2) isnegative
If x > 2
positive
( + 4)( – 2) is
+¿ +¿−
Let’s use our heads
−𝟏𝟎−𝟏𝟎−𝟓 −𝟓 𝟑𝟑 𝟔𝟔𝟎𝟎
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SOLVING POYNOMIAL INEQUALITIES
x2 + 2x – 8 = 0
(x + 4)(x – 2) = 0
– 4 2
x = – 4 or x = 2
(x + 4)(x – 2) > 0 (x + 4)(x – 2) < 0
x2 + 2x – 8 > 0 x2 + 2x – 8 < 0
x < – 4 or x > 2 – 4 < x < 2
(−𝟒 ,𝟐)(−∞ ,−4 )∪ (2 ,∞ )
+¿ +¿−
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𝒀 𝟏=𝒙𝟐+𝟐𝒙 –𝟖
𝒙>𝟐𝒙<−𝟒
−𝟒<𝒙<𝟐
𝒙=𝟐𝒙=−𝟒
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EXAMPLE 6 Factor x3 + 2x2 – 5x – 6
-2 1 2 -5 -6 -2 -8 -6 1 4 3 0
23 + 2(2)2 – 5(2) – 6 = 8 + 8 – 10 – 6 = 0
(x – 2) is a factor
x3 + 2x2 – 5x – 6 = (x – 2)(1x2 + 4x + 3)
= (x – 2)(x + 3)(x + 1)
Test potential zeros ±1, ±2, ±3, ±6.
2 1 2 -5 -6 2 8 6 1 4 3 0
Subtraction Method Addition Method
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EXAMPLE 6 continued x3 + 2x2 – 5x – 6
x3 + 2x2 – 5x – 6 = (x – 2)(x + 3)(x + 1)
– 3 – 1 2
( – 2)( + 3)( + 1) is
− +¿
If x < – 3
pos
neg
If – 3 < x < – 1 ( – 2)( + 3)( + 1) is
( – 2)( + 3)( + 1) is
( – 2)( + 3)( + 1) is
− +¿
If – 1 < x < 2
If x > 2
neg
pos
−𝟐
−𝟒−𝟒 −𝟒
−𝟐−𝟐
𝟎 𝟎 𝟎
𝟓 𝟓 𝟓
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x3 + 2x2 – 5x – 6 = (x – 2)(x + 3)(x + 1)
–3 –1 2
neg pos neg pos
x3 + 2x2 – 5x – 6 < 0 x3 + 2x2 – 5x – 6 > 0
x < –3 or –1 < x < 2
–3 < x < –1 or x > 2
EXAMPLE 6 x3 + 2x2 – 5x – 6 EXAMPLE x3 + 2x2 – 5x – 6
(−∞ ,−𝟑 )∪ [−𝟏 ,𝟐 ] [−𝟑 ,𝟏 ]∪¿
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𝒙=−𝟑 𝒙=−𝟏 𝒙=𝟐
x < –3
–3 < x < –1
–1 < x < 2
x > 2 𝒙𝟑+𝟐𝒙𝟐−𝟓𝒙 −𝟔≥𝟎
𝒙𝟑+𝟐𝒙𝟐−𝟓𝒙 −𝟔≤𝟎