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U1A L6 Linear, Quadratic & Polynomial Inequalities
UNIT 1A LESSON 6 Linear, Quadratic and Polynomial Inequalities
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INTERVALS & INEQUALITIES
U1A L6 Linear, Quadratic & Polynomial Inequalities INTERVALS & INEQUALITIES Interval Notation Inequality Notation Graph
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U1A L6 Linear, Quadratic & Polynomial Inequalities
Express the following intervals in terms of inequalities and graph the intervals
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U1A L6 Linear, Quadratic & Polynomial Inequalities
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.
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U1A L6 Linear, Quadratic & Polynomial Inequalities
Linear Inequalities Solve the inequalities. State the answer in inequality form, interval form and graph. EXAMPLE 2 5x + 7 > – 8 5x > – 15 x > – 3 inequality form [– 3, ∞) interval form (−𝟑, −𝟖) 𝒀 𝟐 =−𝟖 𝒀 𝟐 =𝟓𝒙 +𝟕 -3 graph
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U1A L6 Linear, Quadratic & Polynomial Inequalities
Linear Inequalities Solve the inequalities. State the answer in inequality form, interval form and graph. EXAMPLE 3 𝒀 𝟏 =𝟑𝒙 +𝟏 3x + 1 < 7x – 7 – 4x < – 8 x > 2 inequality form (𝟐, 𝟕) (2, ∞) interval form 2 graph 𝒀 𝟐 =𝟕𝒙−𝟕
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U1A L6 Linear, Quadratic & Polynomial Inequalities
Linear Inequalities Solve the inequalities and graph. State the answer in inequality form and interval form. 𝒀 𝟏 =𝟖−𝒙 𝒀 𝟐 =𝟓𝒙+𝟐 EXAMPLE 4 8 – x > 5x + 2 (𝟏, 𝟕) – 6x > – 6 −𝟔𝒙 −𝟔 < −𝟔 −𝟔 x < 1 inequality form (–∞, 1) interval form 1 graph
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SOLVING POYNOMIAL INEQUALITIES
U1A L6 Linear, Quadratic & Polynomial Inequalities SOLVING POYNOMIAL INEQUALITIES In order to solve any polynomial equation or inequality you must FACTOR first!!! EXAMPLE 5 : x2 + 2x – 8 = 0 Let’s use our heads (x + 4)(x – 2) = 0 x = –4 or x = 2 If x < – 4 If x is between –4 and 2 If x > 2 ( )( – 2) is ( )( – 2) is ( )( – 2) is −𝟏𝟎 −𝟓 −𝟏𝟎 𝟎 𝟎 𝟔 𝟑 𝟑 𝟔 −𝟓 negative positive positive + − + 2 –4
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SOLVING POYNOMIAL INEQUALITIES
U1A L6 Linear, Quadratic & Polynomial Inequalities SOLVING POYNOMIAL INEQUALITIES x2 + 2x – 8 = 0 x2 + 2x – 8 > 0 x2 + 2x – 8 < 0 (x + 4)(x – 2) = 0 (x + 4)(x – 2) > 0 (x + 4)(x – 2) < 0 x = – 4 or x = 2 x < – 4 or x > 2 – 4 < x < 2 −∞,−4 ∪ 2, ∞ (−𝟒, 𝟐) + + − 2 – 4
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𝒀 𝟏 =𝒙𝟐 + 𝟐𝒙 – 𝟖 𝒙<−𝟒 𝒙>𝟐 𝒙=−𝟒 𝒙=𝟐 −𝟒<𝒙<𝟐
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U1A L6 Linear, Quadratic & Polynomial Inequalities
EXAMPLE 6 Factor x3 + 2x2 – 5x – 6 Test potential zeros ±1, ±2, ±3, ±6. 23 + 2(2)2 – 5(2) – 6 = – 10 – 6 = 0 (x – 2) is a factor Subtraction Method Addition Method x3 + 2x2 – 5x – 6 = (x – 2)(1x2 + 4x + 3) = (x – 2)(x + 3)(x + 1)
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U1A L6 Linear, Quadratic & Polynomial Inequalities
EXAMPLE 6 continued x3 + 2x2 – 5x – 6 x3 + 2x2 – 5x – 6 = (x – 2)(x + 3)(x + 1) − + + − – 3 – 1 2 If x < – 3 ( – 2)( )( ) is −𝟒 neg −𝟒 −𝟒 If – 3 < x < – 1 ( – 2)( )( ) is pos −𝟐 −𝟐 −𝟐 If – 1 < x < 2 ( – 2)( )( ) is neg 𝟎 𝟎 𝟎 If x > 2 ( – 2)( )( ) is 𝟓 𝟓 pos 𝟓
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U1A L6 Linear, Quadratic & Polynomial Inequalities
EXAMPLE x3 + 2x2 – 5x – 6 EXAMPLE 6 x3 + 2x2 – 5x – 6 x3 + 2x2 – 5x – 6 = (x – 2)(x + 3)(x + 1) neg pos pos neg –3 –1 2 x3 + 2x2 – 5x – 6 < 0 x3 + 2x2 – 5x – 6 > 0 x < –3 or –1 < x < 2 –3 < x < –1 or x > 2 −∞, −𝟑 ∪ −𝟏, 𝟐 −𝟑, 𝟏 ∪[𝟐, ∞)
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x > 2 –3 < x < –1 –1 < x < 2 x < –3
𝒙 𝟑 +𝟐 𝒙 𝟐 −𝟓𝒙−𝟔≥𝟎 x > 2 –3 < x < –1 𝒙=−𝟑 𝒙=−𝟏 𝒙=𝟐 –1 < x < 2 x < –3 𝒙 𝟑 +𝟐 𝒙 𝟐 −𝟓𝒙−𝟔≤𝟎
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