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UNIT 1A LESSON 6

UNIT 1A LESSON 6. Linear, Quadratic and Polynomial Inequalities. INTERVALS & INEQUALITIES. Interval Notation. Inequality Notation. Graph. Express the following intervals in terms of inequalities and graph the intervals. REMEMBER:

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UNIT 1A LESSON 6

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  1. UNIT 1ALESSON 6 Linear, Quadratic and Polynomial Inequalities

  2. INTERVALS & INEQUALITIES Interval Notation Inequality Notation Graph

  3. Express the following intervals in terms of inequalities and graph the intervals

  4. 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.

  5. 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 graph

  6. Linear Inequalities Solve the inequalities. State the answer in inequality form, interval form and graph. EXAMPLE 3 3x + 1 < 7x – 7 – 4x< – 8 x > 2 inequality form (2, ∞) interval form 2 graph

  7. Linear Inequalities Solve the inequalities and graph. State the answer in inequality form and interval form. EXAMPLE 4 8 – x > 5x + 2 – 6x> – 6 x < 1inequality form (–∞, 1) interval form 1 graph

  8. 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 > 2 If x is between –4 and 2 If x < – 4 ( + 4)( – 2)is ( + 4)( – 2)is ( + 4)( – 2)is negative positive positive 2 –4

  9. 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 2 –4

  10. -2 1 2 -5 -6 -2 -8 -6 1 4 3 0 2 1 2 -5 -6 2 8 6 1 4 3 0 EXAMPLE 6 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 x3 + 2x2 – 5x – 6 = (x – 2)(1x2 + 4x + 3) = (x – 2)(x + 3)(x + 1)

  11. EXAMPLE 6 continued x3+ 2x2 – 5x – 6 x3 + 2x2 – 5x – 6 = (x – 2)(x + 3)(x + 1) – 3 – 1 2 ( – 2)(+ 3)( + 1) is neg If x< – 3 pos If – 3 < x < – 1 ( – 2)( + 3)( + 1) is neg If – 1 < x < 2 ( – 2)( + 3)( + 1) is ( – 2)( + 3)( + 1) is pos If x > 2

  12. EXAMPLE x3 + 2x2 – 5x – 6 EXAMPLE 6 x3 + 2x2 – 5x – 6 x3 + 2x2 – 5x – 6 = (x – 2)(x + 3)(x + 1) pos neg pos neg –3 –1 2 x3 + 2x2 – 5x – 6 < 0 x3 + 2x2 – 5x – 6 > 0 –3 <x<–1 or x> 2 x< –3 or –1 <x< 2

  13. x> 2 –3 <x<–1 –1 <x< 2 x< –3

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