1. Splash Screen

2. Five-Minute Check (over Lesson 2-5) Then/Now New Vocabulary
Example 1: Solve a Polynomial Inequality
Example 2: Solve a Polynomial Inequality Using End Behavior
Example 3: Polynomial Inequalities with Unusual Solution Sets
Example 4: Solve a Rational Inequality
Example 5: Real-World Example: Solve a Rational Inequality
Lesson Menu

3. A. D ; horizontal asymptote at y = –2
Find the domain and the equations of any vertical or horizontal asymptotes of
A. D ; horizontal asymptote at y = –2
B. D ; vertical asymptote at x = –2
C. D ; vertical asymptote at x = –2
D. D ; horizontal asymptote at y = –3
5–Minute Check 1

4. A. D ; horizontal asymptote at y = –2
Find the domain and the equations of any vertical or horizontal asymptotes of
A. D ; horizontal asymptote at y = –2
B. D ; vertical asymptote at x = –2
C. D ; vertical asymptote at x = –2
D. D ; horizontal asymptote at y = –3
5–Minute Check 1

5. A. Determine any asymptotes, holes, and intercepts of
A. 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);
5–Minute Check 2

6. A. Determine any asymptotes, holes, and intercepts of
A. 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);
5–Minute Check 2

7. B. Graph and state its domain.C. D.
5–Minute Check 2

8. B. Graph and state its domain.C. D.
5–Minute Check 2

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

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

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

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

13. You solved polynomial and rational equations. (Lessons 2-3 and 2-4)
Solve polynomial inequalities.
Solve rational inequalities.
Then/Now

14. polynomial inequality sign chart rational inequality
Vocabulary

15. Solve a Polynomial Inequality
Subtracting 1 from each side, you get x 2 – 8x + 15 ≤ 0. Let f (x) = x2 – 8x 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 1

16. Think: (x – 5) and (x – 3) are
Solve a Polynomial Inequality
f (x) = (x – 5)(x – 3)
Think: (x – 5) and (x – 3) are
both negative when x = 2.
f (x) = (x – 5)(x – 3)
Example 1

17. Solve a Polynomial Inequality
Because 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.
Answer:
Example 1

18. Solve a Polynomial Inequality
Because 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.
Answer: [3, 5]
Example 1

19. Solve x 2 – 9x + 10 < 46.
A. (–3, 12) B. C.
D. (–12, 3)
Example 1

20. Solve x 2 – 9x + 10 < 46.
A. (–3, 12) B. C.
D. (–12, 3)
Example 1

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

22. Solve a Polynomial Inequality Using End Behavior
Step 3 Determine the end behavior of f (x). Because the degree of f is odd and its leading coefficient is positive, you know This means that the function starts off negative at the left and ends positive at the right.
Example 2

23. Solve a Polynomial Inequality Using End Behavior
Step 4 Because 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 2

24. Solve a Polynomial Inequality Using End Behavior
Answer:
Example 2

25. Solve a Polynomial Inequality Using End Behavior
Answer: (–4, 1) 
CHECK The graph of f (x) = x 3 – 3x 2 – 22x + 24 is above the x-axis on (–4, 1)  (6, ∞).
Example 2

26. Solve 2x3 + 9x 2 ≥ – 3x + 4. A. (–∞, –4]  B. (–∞, –4) 
C. [–4, –1] or
D. (–4, –1) or
Example 2

27. Solve 2x3 + 9x 2 ≥ – 3x + 4. A. (–∞, –4]  B. (–∞, –4) 
C. [–4, –1] or
D. (–4, –1) or
Example 2

28. Polynomial Inequalities with Unusual Solution Sets
A. 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 3

29. Polynomial Inequalities with Unusual Solution Sets
A. 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 3

30. Polynomial Inequalities with Unusual Solution Sets
B. 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 (–∞, ∞).
Answer:
Example 3

31. Polynomial Inequalities with Unusual Solution Sets
B. 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 (–∞, ∞).
Answer:
Example 3

32. Polynomial Inequalities with Unusual Solution Sets
C. Solve x x + 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 x + 36 > 0 is (–∞, –6)  (–6, ∞). The graph of f (x) supports this conclusion.
Answer:
Example 3

33. Polynomial Inequalities with Unusual Solution Sets
C. Solve x x + 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 x + 36 > 0 is (–∞, –6)  (–6, ∞). The graph of f (x) supports this conclusion.
Answer: (–∞, –6)  (–6, ∞)
Example 3

34. Polynomial Inequalities with Unusual Solution Sets
D. Solve x x + 36 ≤ 0.
The related function f (x) = x x + 36 has a zero at x = –6. For all other values of x, the function is positive. Therefore, x x + 36 ≤ 0 has just one solution, x = –6.
Answer:
Example 3

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

36. Solve x 2 + 6x + 9 > 0. A. no solution B. (–∞, ∞) C. x = –3
D. (–∞, –3)  (–3, ∞)
Example 3

37. Solve x 2 + 6x + 9 > 0. A. no solution B. (–∞, ∞) C. x = –3
D. (–∞, –3)  (–3, ∞)
Example 3

38. Use the LCD, (x + 2) to rewrite each fraction. Then add.
Solve a Rational Inequality
Solve
Original inequality
Use the LCD, (x + 2) to rewrite each fraction. Then add.
Simplify.
Example 4

39. Solve a Rational Inequality
Let The zeros and undefined points of the inequality are the zeros of the numerator, none, and denominator, x = –2. Create a sign chart using this number. Then choose and test x-values in each interval to determine if f (x) is positive or negative.
Example 4

40. Solve a Rational Inequality
Example 4

41. Solve a Rational Inequality
The solution set of the original inequality is the interval for which f (x) is positive, (–∞, –2). The graph of shown here supports this conclusion.
Answer:
Example 4

42. Solve a Rational Inequality
The solution set of the original inequality is the interval for which f (x) is positive, (–∞, –2). The graph of shown here supports this conclusion.
Answer:
Example 4

43. Solve . A. (–∞, 3)  [11, ∞) B. [–∞, 3]  [11, ∞) C. (3, 11]
D. [3, 11]
Example 4

44. Solve . A. (–∞, 3)  [11, ∞) B. [–∞, 3]  [11, ∞) C. (3, 11]
D. [3, 11]
Example 4

45. Solve a Rational Inequality
CARPENTRY 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 of the tables.
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 5

46. l(10 – l) ≥ 24 Write the inequality.
Solve a Rational Inequality
l(10 – l) ≥ 24 Write the inequality.
10l – l 2 ≥ 24 Distributive Property
0 ≥ l 2 – 10l + 24 Subtract 10l – l 2 from each side.
0 ≥ (l – 6)(l – 4) Factor.
Example 5

47. Solve a Rational Inequality
The zeroes of the related function 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 5

48. So, the solution set of l(10 – l) ≥ 24 is [4, 6].
Solve a Rational Inequality
So, the solution set of l(10 – l) ≥ 24 is [4, 6].
The length must be between 4 feet and 6 feet, inclusive.
Answer:
Example 5

49. So, the solution set of l(10 – l) ≥ 24 is [4, 6].
Solve a Rational Inequality
So, the solution set of l(10 – l) ≥ 24 is [4, 6].
The length must be between 4 feet and 6 feet, inclusive.
Answer: l(10 – l) ≥ 24; 4 ft to 6 ft
Example 5

50. A. l(36 – l) ≥ 80; 0 ft < l ≤ 8 ft or l ≥ 10 ft
GARDENING 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 ft
D. l(18 – l) ≥ 80; 4 ft ≤ l ≤ 5 ft
Example 5

51. A. l(36 – l) ≥ 80; 0 ft < l ≤ 8 ft or l ≥ 10 ft
GARDENING 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 ft
D. l(18 – l) ≥ 80; 4 ft ≤ l ≤ 5 ft
Example 5

52. End of the Lesson