html5-img
1 / 58

5-7

Solving Quadratic Inequalities. 5-7. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 2. Holt Algebra 2. Warm Up 1. Graph the inequality y < 2 x + 1. Solve using any method. 2. x 2 – 16 x + 63 = 0. 7, 9. 3. 3 x 2 + 8 x = 3. Objectives.

floyd
Download Presentation

5-7

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Solving Quadratic Inequalities 5-7 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

  2. Warm Up 1. Graph the inequalityy < 2x + 1. Solve using any method. 2. x2 – 16x + 63 = 0 7, 9 3. 3x2 + 8x = 3

  3. Objectives Solve quadratic inequalities by using tables and graphs. Solve quadratic inequalities by using algebra.

  4. Vocabulary quadratic inequality in two variables

  5. Many business profits can be modeled by quadratic functions. To ensure that the profit is above a certain level, financial planners may need to graph and solve quadratic inequalities. A quadratic inequality in two variables can be written in one of the following forms, where a, b, and c are real numbers and a ≠ 0. Its solution set is a set of ordered pairs (x, y).

  6. y < ax2 + bx + c y > ax2 + bx + c y ≤ ax2 + bx + c y ≥ ax2 + bx + c In Lesson 2-5, you solved linear inequalities in two variables by graphing. You can use a similar procedure to graph quadratic inequalities.

  7. Example 1: Graphing Quadratic Inequalities in Two Variables Graph y ≥ x2 – 7x + 10. Step 1 Graph the boundary of the related parabola y = x2 – 7x + 10 with a solid curve. Its y-intercept is 10, its vertex is (3.5, –2.25), and its x-intercepts are 2 and 5.

  8. Example 1 Continued Step 2 Shade above the parabola because the solution consists of y-values greater than those on the parabola for corresponding x-values.

  9. Example 1 Continued Check Use a test point to verify the solution region. y ≥ x2 – 7x + 10 0 ≥ (4)2–7(4) + 10 Try (4, 0). 0 ≥ 16 – 28 + 10 0 ≥ –2 

  10. Check It Out! Example 1a Graph the inequality. y ≥ 2x2 – 5x – 2 Step 1 Graph the boundary of the related parabola y = 2x2 – 5x – 2 with a solid curve. Its y-intercept is –2, its vertex is (1.3, –5.1), and its x-intercepts are –0.4 and 2.9.

  11. Check It Out! Example 1a Continued Step 2 Shade above the parabola because the solution consists of y-values greater than those on the parabola for corresponding x-values.

  12. Check It Out! Example 1a Continued Check Use a test point to verify the solution region. y < 2x2 – 5x – 2 0 ≥ 2(2)2– 5(2) – 2 Try (2, 0). 0 ≥ 8 – 10 – 2 0 ≥ –4 

  13. Check It Out! Example 1b Graph each inequality. y < –3x2 – 6x – 7 Step 1 Graph the boundary of the related parabola y = –3x2 – 6x – 7 with a dashed curve. Its y-intercept is –7.

  14. Check It Out! Example 1b Continued Step 2 Shade below the parabola because the solution consists of y-values less than those on the parabola for corresponding x-values.

  15. Check It Out! Example 1b Continued Check Use a test point to verify the solution region. y < –3x2 – 6x –7 –10 < –3(–2)2– 6(–2) – 7 Try (–2, –10). –10 < –12 + 12 – 7 –10 < –7 

  16. Reading Math For and statements, both of the conditions must be true. For or statements, at leastone of the conditions must be true. Quadratic inequalities in one variable, such as ax2 + bx + c > 0 (a ≠ 0), have solutions in one variable that are graphed on a number line.

  17. Example 2A: Solving Quadratic Inequalities by Using Tables and Graphs Solve the inequality by using tables or graphs. x2 + 8x + 20 ≥ 5 Use a graphing calculator to graph each side of the inequality. Set Y1 equal to x2 + 8x + 20 and Y2 equal to 5. Identify the values of x for which Y1≥ Y2.

  18. –6 –4 –2 0 2 4 6 Example 2A Continued The parabola is at or above the line when x is less than or equal to –5 or greater than or equal to –3. So, the solution set is x ≤ –5 or x ≥ –3 or (–∞, –5] U [–3, ∞). The table supports your answer. The number line shows the solution set.

  19. Example 2B: Solving Quadratics Inequalities by Using Tables and Graphs Solve the inequality by using tables and graph. x2 + 8x + 20 < 5 Use a graphing calculator to graph each side of the inequality. Set Y1 equal to x2 + 8x + 20 and Y2 equal to 5. Identify the values of which Y1< Y2.

  20. –6 –4 –2 0 2 4 6 Example 2B Continued The parabola is below the line when x is greater than –5 and less than –3. So, the solution set is–5 < x < –3 or (–5, –3). The table supports your answer. The number line shows the solution set.

  21. Check It Out! Example 2a Solve the inequality by using tables and graph. x2 – x + 5 < 7 Use a graphing calculator to graph each side of the inequality. Set Y1 equal to x2 – x + 5 and Y2 equal to 7. Identify the values of which Y1< Y2.

  22. –6 –4 –2 0 2 4 6 Check It Out! Example 2a Continued The parabola is below the line when x is greater than –1 and less than 2. So, the solution set is–1 < x < 2 or (–1, 2). The table supports your answer. The number line shows the solution set.

  23. Check It Out! Example 2b Solve the inequality by using tables and graph. 2x2 – 5x + 1 ≥ 1 Use a graphing calculator to graph each side of the inequality. Set Y1 equal to 2x2 – 5x + 1 and Y2 equal to 1. Identify the values of which Y1≥ Y2.

  24. –6 –4 –2 0 2 4 6 Check It Out! Example 2b Continued The parabola is at or above the line when x is less than or equal to 0 or greater than or greater than or equal to 2.5. So, the solution set is(–∞, 0] U [2.5, ∞) The number line shows the solution set.

  25. The number lines showing the solution sets in Example 2 are divided into three distinct regions by the points –5 and –3. These points are called critical values. By finding the critical values, you can solve quadratic inequalities algebraically.

  26. Example 3: Solving Quadratic Equations by Using Algebra Solve the inequality x2 – 10x + 18 ≤ –3 by using algebra. Step 1 Write the related equation. x2 – 10x + 18 = –3

  27. Example 3 Continued Step 2 Solve the equation for x to find the critical values. x2 –10x+ 21 = 0 Write in standard form. (x– 3)(x – 7) = 0 Factor. Zero Product Property. x– 3 = 0 or x – 7 = 0 Solve for x. x=3 or x = 7 The critical values are 3 and 7. The critical values divide the number line into three intervals: x ≤ 3, 3 ≤ x ≤ 7, x ≥ 7.

  28. Critical values –3 –2 –1 0 1 2 3 4 5 6 7 8 9 Test points Example 3 Continued Step 3 Test an x-value in each interval. x2 – 10x + 18 ≤ –3 Try x = 2. (2)2 – 10(2) + 18 ≤ –3 x (4)2 – 10(4) + 18 ≤ –3 Try x = 4.  Try x = 8. x (8)2 – 10(8) + 18 ≤ –3

  29. –3 –2 –1 0 1 2 3 4 5 6 7 8 9 Example 3 Continued Shade the solution regions on the number line. Use solid circles for the critical values because the inequality contains them. The solution is 3 ≤ x ≤ 7 or [3, 7].

  30. Check It Out! Example 3a Solve the inequality by using algebra. x2 – 6x + 10 ≥ 2 Step 1 Write the related equation. x2 – 6x + 10 = 2

  31. Check It Out! Example 3a Continued Step 2 Solve the equation for x to find the critical values. x2 – 6x+ 8 = 0 Write in standard form. Factor. (x– 2)(x – 4) = 0 Zero Product Property. x– 2 = 0 or x – 4 = 0 Solve for x. x=2 or x = 4 The critical values are 2 and 4. The critical values divide the number line into three intervals: x ≤ 2, 2 ≤ x ≤ 4, x ≥ 4.

  32. Critical values Test points –3 –2 –1 0 1 2 3 4 5 6 7 8 9 Check It Out! Example 3a Continued Step 3 Test an x-value in each interval. x2 – 6x + 10 ≥ 2 Try x = 1. (1)2 – 6(1) + 10 ≥ 2  (3)2 – 6(3) + 10 ≥ 2 x Try x = 3. (5)2 – 6(5) + 10 ≥ 2 Try x = 5. 

  33. –3 –2 –1 0 1 2 3 4 5 6 7 8 9 Check It Out! Example 3a Continued Shade the solution regions on the number line. Use solid circles for the critical values because the inequality contains them. The solution is x ≤ 2 or x ≥ 4.

  34. Check It Out! Example 3b Solve the inequality by using algebra. –2x2 + 3x + 7 < 2 Step 1 Write the related equation. –2x2 + 3x + 7 = 2

  35. Check It Out! Example 3b Continued Step 2 Solve the equation for x to find the critical values. –2x2 + 3x+ 5 = 0 Write in standard form. (–2x+ 5)(x + 1) = 0 Factor. Zero Product Property. –2x+ 5 = 0 or x + 1 = 0 Solve for x. x=2.5 or x = –1 The critical values are 2.5 and –1. The critical values divide the number line into three intervals: x < –1, –1 < x < 2.5, x > 2.5.

  36. Critical values –3 –2 –1 0 1 2 3 4 5 6 7 8 9 Test points Check It Out! Example 3b Continued Step 3 Test an x-value in each interval. –2x2 + 3x + 7 < 2 –2(–2)2 + 3(–2) + 7 < 2 Try x = –2.  –2(1)2 + 3(1) + 7 < 2 x Try x = 1. –2(3)2 + 3(3) + 7 < 2  Try x = 3.

  37. –3 –2 –1 0 1 2 3 4 5 6 7 8 9 Check It Out! Example 3 Shade the solution regions on the number line. Use open circles for the critical values because the inequality does not contain or equal to. The solution is x < –1 or x > 2.5.

  38. Remember! A compound inequality such as 12 ≤ x ≤ 28 can be written as {x|x ≥12 U x ≤ 28}, or x ≥ 12 and x ≤ 28. (see Lesson 2-8).

  39. Example 4: Problem-Solving Application The monthly profit P of a small business that sells bicycle helmets can be modeled by the function P(x) = –8x2 + 600x – 4200, where x is the average selling price of a helmet. What range of selling prices will generate a monthly profit of at least $6000?

  40. 1 Understand the Problem Example 4 Continued The answer will be the average price of a helmet required for a profit that is greater than or equal to $6000. • List the important information: • The profit must be at least $6000. • The function for the business’s profit is P(x) = –8x2 + 600x– 4200.

  41. Make a Plan 2 Example 4 Continued Write an inequality showing profit greater than or equal to $6000. Then solve the inequality by using algebra.

  42. 3 Solve Example 4 Continued Write the inequality. –8x2 + 600x – 4200 ≥ 6000 Find the critical values by solving the related equation. –8x2 + 600x – 4200 = 6000 Write as an equation. –8x2 + 600x – 10,200 = 0 Write in standard form. Factor out –8 to simplify. –8(x2 – 75x + 1275) = 0

  43. 3 Solve Example 4 Continued Use the Quadratic Formula. Simplify. x ≈ 26.04 or x ≈ 48.96

  44. 3 Solve 10 20 30 40 50 60 70 Example 4 Continued Test an x-value in each of the three regions formed by the critical x-values. Critical values Test points

  45. 3 Solve Example 4 Continued –8(25)2 + 600(25) – 4200 ≥ 6000 Try x = 25. x 5800 ≥ 6000 –8(45)2 + 600(45) – 4200 ≥ 6000 Try x = 45.  6600 ≥ 6000 –8(50)2 + 600(50) – 4200 ≥ 6000 Try x = 50. x 5800 ≥ 6000 Write the solution as an inequality. The solution is approximately 26.04 ≤ x ≤ 48.96.

  46. 3 Solve Example 4 Continued For a profit of $6000, the average price of a helmet needs to be between $26.04 and $48.96, inclusive.

  47. Look Back 4 Example 4 Continued Enter y = –8x2 + 600x – 4200 into a graphing calculator, and create a table of values. The table shows that integer values of x between 26.04 and 48.96 inclusive result in y-values greater than or equal to 6000.

  48. Check It Out! Example 4 A business offers educational tours to Patagonia, a region of South America that includes parts of Chile and Argentina . The profit P for x number of persons is P(x) = –25x2 + 1250x – 5000. The trip will be rescheduled if the profit is less $7500. How many people must have signed up if the trip is rescheduled?

  49. 1 Understand the Problem Check It Out! Example 4 Continued The answer will be the number of people signed up for the trip if the profit is less than $7500. • List the important information: • The profit will be less than $7500. • The function for the profit is P(x) = –25x2 + 1250x– 5000.

  50. Make a Plan 2 Check It Out! Example 4 Continued Write an inequality showing profit less than $7500. Then solve the inequality by using algebra.

More Related