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UNIT 1A LESSON 6 1 Linear, Quadratic and Polynomial Inequalities.

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Presentation on theme: "UNIT 1A LESSON 6 1 Linear, Quadratic and Polynomial Inequalities."— Presentation transcript:

1 UNIT 1A LESSON 6 1 Linear, Quadratic and Polynomial Inequalities

2 2 INTERVALS & INEQUALITIES Interval Notation Inequality Notation Graph

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

4 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 5 Linear Inequalities Solve the inequalities. State the answer in inequality form, interval form and graph. 5x + 7 > – 8 5x > – 15 x > – 3 inequality form [– 3, ∞) interval form -3 EXAMPLE 2 graph

6 6 Linear Inequalities Solve 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 2 EXAMPLE 3 graph

7 7 Linear Inequalities Solve 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 1 EXAMPLE 4 graph

8 8 SOLVING POYNOMIAL INEQUALITIES In order to solve any polynomial equation or inequality you must FACTOR first!!! EXAMPLE 5 : x 2 + 2x – 8 = 0 (x + 4)(x – 2) = 0 –4–4 2 x = –4 or x = 2 If x < – 4 ( + 4)( – 2) is positive If x is between –4 and 2 ( + 4)( – 2) is negative If x > 2 positive ( + 4)( – 2) is Let’s use our heads

9 9 SOLVING POYNOMIAL INEQUALITIES x 2 + 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 x 2 + 2x – 8 > 0 x 2 + 2x – 8 < 0 x 2 – 4 < x < 2

10 10

11 11 EXAMPLE 6 Factor x 3 + 2x 2 – 5x – 6 -2 1 2 -5 -6 -2 -8 -6 1 4 3 0 -2 1 2 -5 -6 -2 -8 -6 1 4 3 0 2 3 + 2(2) 2 – 5(2) – 6 = 8 + 8 – 10 – 6 = 0 (x – 2) is a factor x 3 + 2x 2 – 5x – 6 = (x – 2)(1x 2 + 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 2 1 2 -5 -6 2 8 6 1 4 3 0 Subtraction Method Addition Method

12 12 EXAMPLE 6 continued x 3 + 2x 2 – 5x – 6 x 3 + 2x 2 – 5x – 6 = (x – 2)(x + 3)(x + 1) – 3 – 12 ( – 2)( + 3)( + 1) is If x < – 3 pos neg If – 3 < x < – 1 ( – 2)( + 3)( + 1) is If – 1 < x < 2 If x > 2 neg pos

13 13 x 3 + 2x 2 – 5x – 6 = (x – 2)(x + 3)(x + 1) –3–3 –1–12 neg posneg pos x 3 + 2x 2 – 5x – 6 < 0 x 3 + 2x 2 – 5x – 6 > 0 x < –3 or –1 < x < 2 –3 2 EXAMPLE 6 x 3 + 2x 2 – 5x – 6 EXAMPLE x 3 + 2x 2 – 5x – 6

14 14 x < –3 –3 < x < –1 –1 < x < 2 x > 2


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