# splash screen. lesson menu five-minute check (over lesson 2-5) then/now new vocabulary example...

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• Splash Screen

• Lesson MenuFive-Minute Check (over Lesson 2-5)Then/NowNew VocabularyExample 1:Solve a Polynomial InequalityExample 2:Solve a Polynomial Inequality Using End BehaviorExample 3: Polynomial Inequalities with Unusual Solution SetsExample 4:Solve a Rational InequalityExample 5:Real-World Example: Solve a Rational Inequality

Over Lesson 2-5

5Minute Check 1

Over Lesson 2-5

5Minute Check 2A.asymptotes: x = 0, x = 1 and y = 1; intercept: (3, 0)B.asymptotes: x = 1 and y = 3; hole: x = 1; intercept: (3, 0); C.asymptotes: x = 1 and y = 0; hole: x = 0; intercept (0, 3); D.asymptotes: x = 0 and y = 1; hole: x = 1; intercept: (3, 0);

Over Lesson 2-5

5Minute Check 2

Over Lesson 2-5

5Minute Check 3A.2B.4C.2, 4D.no solution

Over Lesson 2-5

5Minute Check 4A.2, 4B.2C.3D.no solution

• Then/NowYou solved polynomial and rational equations. (Lessons 2-3 and 2-4)Solve polynomial inequalities.Solve rational inequalities.

• Vocabularypolynomial inequalitysign chartrational inequality

• Example 1Solve a Polynomial InequalitySubtracting 1 from each side, you get x 2 8x + 15 0. Let f (x) = x2 8x + 15. Factoring yields f (x) = (x 5)(x 3), so f (x) has real zeros at x = 5 and x = 3. Create a sign chart using these zeros. Then substitute an x-value in each test interval into the factored form of the polynomial to determine if f (x) is positive or negative at that point.

• Example 1Solve a Polynomial Inequalityf (x) = (x 5)(x 3)Think: (x 5) and (x 3) areboth negative when x = 2.f (x) = (x 5)(x 3)

• Example 1Answer: [3, 5]Solve a Polynomial InequalityBecause f (x) is negative in the middle interval, the solution of x 2 8x + 16 1 is [3, 5]. The graph of f (x) supports this conclusion, because f (x) is below the x-axis on this same interval.

• Example 1Solve x 2 9x + 10 < 46.

• Example 2Solve a Polynomial Inequality Using End BehaviorSolve x 3 22 x > 3x 2 24.Step 1Subtract 3x 2 24 from each side to get x 3 3x 2 22x + 24 > 0.Step 2Let f (x) = x3 3x2 22x + 24. Use the techniques from Lesson 2-4 to determine that f has real zeros with multiplicity 1 at x = 4, x = 1, and x = 6. Set up a sign chart.

• Example 2Solve a Polynomial Inequality Using End Behavior

• Example 2Solve a Polynomial Inequality Using End BehaviorStep 4Because each zero listed is the location of a sign change, you can complete the sign chart.The solutions of x3 3x2 22x + 24 > 0 are the x-values such that f (x) is positive. From the sign chart, you can see that the solution set is (4, 1) (6, ).

• Example 2Solve a Polynomial Inequality Using End BehaviorCHECKThe graph of f (x) = x 3 3x 2 22x + 24 is above the x-axis on (4, 1) (6, ).

• Example 2Solve 2x3 + 9x 2 3x + 4.

• Example 3Polynomial Inequalities with Unusual Solution SetsA. Solve x 2 + 2x + 3 < 0.The related function f (x) = x 2 + 2x + 3 has no real zeros, so there are no sign changes. This function is positive for all real values of x. Therefore, x 2 + 2x + 3 < 0 has no solution. The graph of f (x) supports this conclusion, because the graph is never below the x-axis. Answer:

• Example 3Polynomial Inequalities with Unusual Solution SetsB. Solve x 2 + 2x + 3 0.Because the related function f (x) = x 2 + 2x + 3 is positive for all real values of x, the solution set of x 2 + 2x + 3 0 is all real numbers or (, ).

• Example 3Polynomial Inequalities with Unusual Solution SetsC. Solve x 2 + 12x + 36 > 0.The related function f (x) = x2 + 12x + 36 has one real zero, x = 6, with multiplicity 2, so the value of f (x) does not change signs. This function is positive for all real values of x except x = 6. Therefore, the solution of x 2 + 12x + 36 > 0 is (, 6) (6, ). The graph of f (x) supports this conclusion. Answer: (, 6) (6, )

• Example 3Polynomial Inequalities with Unusual Solution SetsD. Solve x 2 + 12x + 36 0.The related function f (x) = x 2 + 12x + 36 has a zero at x = 6. For all other values of x, the function is positive. Therefore, x 2 + 12x + 36 0 has just one solution, x = 6.Answer: {6}

• Example 3Solve x 2 + 6x + 9 > 0.A.no solutionB.(, )C.x = 3D.(, 3) (3, )

• Example 4Solve a Rational InequalityOriginal inequalityUse the LCD, (x + 2) to rewrite each fraction. Then add.

• Example 4Solve a Rational Inequality

• Example 4Solve a Rational Inequality

• Example 4Solve a Rational Inequality

• Example 4A.(, 3) [11, )B.[, 3] [11, )C.(3, 11]D.[3, 11]

• Example 5Solve a Rational InequalityCARPENTRY A carpenter is making tables. The tables have rectangular tops with a perimeter of 20 feet and an area of at least 24 square feet. Write and solve an inequality that can be used to determine the possible lengths to which the tables can be made.Let l represent the length of the table top. Let w represent the width of the table top. Thus, 2w + 2l = 20. Then 10 l represents the width of the table top.

• Example 5Solve a Rational InequalityThen 10 l represents the width of the table top.l(10 l) 24Write the inequality.10l l 2 24Distributive Property0 l 2 10l + 24Subtract 10l l 2 from each side.0 (l 6)(l 4)Factor.

• Example 5Solve a Rational InequalityThe zeroes of this inequality are 6 and 4. Use these numbers to create and complete a sign chart for this function.f(x) = (l 6)(l 4)f(x) = (l 6)(l 4)

• Example 5Answer:l(10 l) 24; 4 ft to 6 ftSolve a Rational InequalitySo, the solution set of l(10 l) 24 is [4, 6].The length must be between 4 feet and 6 feet, inclusive.

• Example 5GARDENING A gardener is marking off rectangular garden plots. The perimeter of each plot is 36 feet and the area is at least 80 square feet. Write and solve an inequality that can be used to find the possible lengths of each plot.A.l(36 l) 80; 0 ft < l 8 ft or l 10 ft B.l(18 l) 80; 8 ft l 10 ft C.I 2 36l 80; 0 ft < l 8 ft or 10 ft l 36 ftD.l(18 l) 80; 4 ft l 5 ft

• End of the Lesson

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