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Lesson Menu 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

Over Lesson 2-5 5–Minute Check 1 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

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

Over Lesson 2-5 5–Minute Check 2 B. Graph and state its domain. A. B. C. D.

Over Lesson 2-5 5–Minute Check 3 A.2 B.4 C.2, 4 D.no solution

Over Lesson 2-5 5–Minute Check 4 A.2, –4 B.2 C.3 D.no solution

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

Vocabulary polynomial inequality sign chart rational inequality

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

Example 1 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 Answer: [3, 5] 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.

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

Example 2 Solve a Polynomial Inequality Using End Behavior Solve 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) = x 3 – 3x 2 – 22x 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 2 Solve a Polynomial Inequality Using End Behavior Step 3Determine 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 Solve a Polynomial Inequality Using End Behavior Step 4Because each zero listed is the location of a sign change, you can complete the sign chart. The solutions of x 3 – 3x 2 – 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 Solve a Polynomial Inequality Using End Behavior Answer: (–4, 1)  CHECKThe graph of f (x) = x 3 – 3x 2 – 22x + 24 is above the x-axis on (–4, 1)  (6, ∞).

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

Example 3 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 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 Polynomial Inequalities with Unusual Solution Sets C. Solve x x + 36 > 0. The related function f (x) = x x + 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 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 Solve x 2 + 6x + 9 > 0. A.no solution B.(–∞, ∞) C.x = –3 D.(–∞, –3)  (–3, ∞)

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

Example 4 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 Solve a Rational Inequality

Example 4 Solve a Rational Inequality Answer: 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.

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

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

Example 5 Solve a Rational Inequality The 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)

Example 5 Answer:l(10 – l) ≥ 24; 4 ft to 6 ft 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.

Example 5 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

End of the Lesson