1. Algebra 2 Two-Variable Inequalities Lesson 2-8

2. Goals Goal To graph Two-Variable Inequalities. Rubric Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems.

3. Vocabulary Linear Inequality Boundary Half-Plane Test Point

4. Essential Question Big Idea: Function Is graphing an inequality in two variables similar to graphing?

5. Definitions Linear functions form the basis of linear inequalities. A linear inequality in two variables relates two variables using an inequality symbol, such as y > 2x – 4. Its graph is a region of the coordinate plane bounded by a line. The line is a boundary line, which divides the coordinate plane into two regions. A solid boundary line indicates that points on the line are solutions of the inequality ( ≥, ≤ ). A dashed line means the points are not solutions of the inequality (>, <). The boundary line separates the coordinate plane into two half-planes, one of which consists of solutions of the inequality.

6. The boundary line y = 2x – 4, shown at right, divides the coordinate plane into half-planes: one where y > 2x – 4 and one where y 2x – 4 is above the boundary line where y = 2x – 4. Lower points have smaller y values, so the region where y < 2x – 4 is below the boundary line. Example: half-planes

7. A graph of a linear inequality in two variables x and y consists of all points (x, y) whose coordinates satisfy the inequality. A dashed boundary line indicates that the line is NOT part of the solution. y x 4 4 y < – 3x + 9 A solid boundary line indicates that the line IS part of the solution. y x 4 4 y  – 3x + 9 Example:

8. Test Point To determine which half-plane is the solution and to shade, pick a test point that is not on the boundary line. Check whether that point satisfies the inequality. If it does, shade the half-plane that includes the test point. If not, shade the other half-plane. The origin (0, 0) is usually an easy point to test, as long as it is not on the boundary line.

9. Inequality 2x + 5y  10. Graph the line 2x + 5y = 10. The solid boundary line indicates that the line is part of the solution. Look for a test point. Is (0, 0) a solution? 2x + 5y  10 2(0) + 5(0)  10 0  9 False (0, 0) is not included in the solution. y x 1 4 1 4 Example:

10. Inequality 2x – 3y < 0 Graph the line 2x – 3y = 0 The dashed boundary line indicates that the line is NOT part of the solution. Look for a test point. Is (1, 2) a solution? 2x – 3y < 0 2(1) – 3(2) < 0 – 4 < 0 True (1, 2) is included in the solution. x y 11 22 33 44 123 4 22 33 44 1 2 3 4 2 – 6 < 0 Example:

11. Graphing Linear Inequalities in Two Variables Step 1: Replace the inequality symbol with an equal sign and graph the resulting equation. If the inequality is ( ), use dashes to graph the line; if the inequality is (  or  ), use a solid line. The graph separates the xy-plane into two half- planes. Step 2: Select a test point P that is not on the line (that is, select a test point in one of the half-planes). (a) If the coordinates of P satisfy the inequality, then shade the half-plane containing P. (b) If the coordinates of P do not satisfy the inequality, then shade the half-plane that does not contain P. Procedure:

12. Graph 3x + 4y ≤ 12 using intercepts. Step 1 Find the intercepts. Substitute x = 0 and y = 0 into 3x + 4y = 12 to find the intercepts of the boundary line. y-interceptx-intercept 3x + 4y = 12 3(0) + 4y = 12 3x + 4(0) = 12 4y = 12 3x + 4y = 12 y = 3 3x = 12 x = 4 Example:

13. Step 2 Draw the boundary line. The line goes through (0, 3) and (4, 0). Draw a solid line for the boundary line because it is part of the graph. Step 3 Find the correct region to shade. Substitute (0, 0) into the inequality. Because 0 + 0 ≤ 12 is true, shade the region that contains (0, 0). (0, 3) (4, 0) Example: continued

14. Example: Graph 7x + y > –14 x y Pick a point not on the graph: (0,0) Graph 7x + y = –14 as a dashed line. Test it in the original inequality. 7(0) + 0 > – 14, 0 > – 14 True, so shade the side containing (0,0). (0, 0)

15. Graph 3x – 4y > 12 using intercepts. Step 1 Find the intercepts. Substitute x = 0 and y = 0 into 3x – 4y = 12 to find the intercepts of the boundary line. y-interceptx-intercept 3x – 4y = 12 3(0) – 4y = 123x – 4(0) = 12 – 4y = 12 3x – 4y = 12 y = – 3 3x = 12 x = 4 Your Turn:

16. Step 2 Draw the boundary line. The line goes through (0, –3) and (4, 0). Draw the boundary line dashed because it is not part of the solution. Step 3 Find the correct region to shade. Substitute (0, 0) into the inequality. Because 0 + 0 >12 is false, shade the region that does not contain (0, 0). (4, 0) (0, –3) Your Turn: continued

17. Your Turn: Graph 3x + 5y  –2 x y Pick a point not on the graph: (0,0), but just barely Graph 3x + 5y = –2 as a solid line. Test it in the original inequality. 3(0) + 5(0) > – 2, 0 > – 2 False, so shade the side that does not contain (0,0). (0, 0)

18. Your Turn: Graph 3x < 15 x y Pick a point not on the graph: (0,0) Graph 3x = 15 as a dashed line. Test it in the original inequality. 3(0) < 15, 0 < 15 True, so shade the side containing (0,0). (0, 0)

19. 1. Change the inequality into slope-intercept form, y = mx + b. Graph the related equation. 2. If > or < then the line should be dashed. If > or < then the line should be solid. 3. If y > mx+b or y > mx+b, shade the half plane above the line. If y < mx+b or y < mx+b, shade the half plane below the line. Alternate Procedure

20. If the boundary line is y = 2x – 4 then; For y > 2x – 4, shade the region above the line. For y < 2x – 4, shade the region below the line. Shading

21. Example: STEP 1 STEP 3 STEP 2 6 4 2 5

22. Example: 6 4 2 5 STEP 3 STEP 1 STEP 2

23. Example: STEP 1 STEP 2 STEP 3

24. Graph the inequality. The boundary line is which has a y-intercept of 2 and a slope of. Draw the boundary line dashed because it is not part of the solution. Then shade the region above the boundary line to show. Your Turn:

25. Graph the inequality y ≤ –1. Recall that y= –1 is a horizontal line. Step 1 Draw a solid line for y=–1 because the boundary line is part of the graph. Step 2 Shade the region below the boundary line to show where y < –1.. Your Turn:

26. The boundary line is y = 3x – 2 which has a y–intercept of –2 and a slope of 3. Draw a solid line because it is part of the solution. Then shade the region above the boundary line to show y > 3x – 2. Graph the inequality y ≥ 3x –2. Your Turn:

27. Graph the inequality y < –3. Recall that y = –3 is a horizontal line. Step 1 Draw the boundary line dashed because it is not part of the solution. Step 2 Shade the region below the boundary line to show where y < –3.. Your Turn:

28. Lee Cooper writes and edits short articles for a local newspaper. It generally takes her an hour to write an article and about a half-hour to edit an article. If Lee works up to 8 hours a day, how many articles can she write and edit in one day? Step 1 Let x equal the number of articles Lee can write. Let y equal the number of articles that Lee can edit. Write an open sentence representing the situation. Number of articles she can writeplustimes number of articles she can editis up to8 hours. hour x+8y < Example: Application

29. Example 6-3a Step 2 Solve for y in terms of x. Original inequality Subtract x from each side. Simplify. Multiply each side by 2. Simplify.

30. Example 6-3a Step 3 Since the open sentence includes the equation, graph as a solid line. Test a point in one of the half-planes, for example, (0, 0). Shade the half-plane containing (0, 0) since is true. Answer:

31. Example 6-3a Step 4 Examine the situation.  Lee cannot work a negative number of hours. Therefore, the domain and range contain only nonnegative numbers.  Lee only wants to count articles that are completely written or completely edited. Thus, only points in the half-plane whose x - and y - coordinates are whole numbers are possible solutions.  One solution is (2, 3). This represents 2 written articles and 3 edited articles.

32. You offer to go to the local deli and pick up sandwiches for lunch. You have $30 to spend. Chicken sandwiches cost $3.00 each and tuna sandwiches are $1.50 each. How many sandwiches can you purchase for $30? Answer: Your Turn:

33. The open sentence that represents this situation is where x is the number of chicken sandwiches, and y is the number of tuna sandwiches. One solution is (4, 10). This means that you could purchase 4 chicken sandwiches and 10 tuna sandwiches. Your Turn: continued

34. Your Turn: A. One tutoring company advertises that it specializes in helping students who have a combined score on the SAT that is 900 or less. Write an inequality to describe the combined scores of students who are prospective tutoring clients. Let x represent the verbal score and y the math score. B. Does a student with verbal score of 480 and a math score of 410 fit the tutoring company’s guidelines?

35. Solution A. One tutoring company advertises that it specializes in helping students who have a combined score on the SAT that is 900 or less. Write an inequality to describe the combined scores of students who are prospective tutoring clients. Let x represent the verbal score and y the math score. Let x be the verbal part of the SAT and let y be the math part. Since the scores must be 900 or less, use the  symbol.

36. Solution Answer: x + y ≤ 900 Verbalmath part partand together are at most900. xy  900+

37. Solution B. Does a student with verbal score of 480 and a math score of 410 fit the tutoring company’s guidelines? The point (480, 410) is in the shaded region, so it satisfies the inequality. Answer: Yes, this student fits the tutoring company’s guidelines.

38. Absolute Value Inequalities You can graph two-variable absolute value inequalities in the same way that you graph linear inequalities. Graph the absolute value function then shade above OR below

39. Absolute Value Inequalities You can graph two-variable absolute value inequalities in the same way that you graph linear inequalities. Graph the absolute value function then shade above OR below Solid line…y Dashed line…y Shade above… y>, y> Shade below…y<, y<

40. Absolute Value Inequalities Example: Graph y < |x – 2| + 3

41. Absolute Value Inequalities Example: Graph y < |x – 2| + 3 DASHED line Shade BELOW slope = 1 Vertex = (2, 3)

42. Absolute Value Inequalities Example: Graph y < |x – 2| + 3 DASHED line Shade BELOW slope = 1Vertex = (2, 3)

43. Absolute Value Inequalities Example: Graph y < |x – 2| + 3 DASHED line Shade BELOW slope = 1 Vertex = (2, 3)

44. Absolute Value Inequalities Example: Graph y < |x – 2| + 3 DASHED line Shade BELOW slope = 1Vertex = (2, 3)

45. Absolute Value Inequalities Example: Graph y < |x – 2| + 3 DASHED line Shade BELOW slope = 1Vertex = (2, 3)

46. Absolute Value Inequalities Example: Graph y < |x – 2| + 3 DASHED line Shade BELOW slope = 1Vertex = (2, 3)

47. Absolute Value Inequalities Example: Graph –y + 1 < -2|x + 2|

48. Absolute Value Inequalities Example: Graph –y + 1 < -2|x + 2| -y < -2|x + 2| - 1 y > 2|x + 2| + 1 -y so CHANGE the direction of the inequality

49. Absolute Value Inequalities y > 2|x + 2| + 1

50. Absolute Value Inequalities y > 2|x + 2| + 1 Vertex = (-2, 1)Slope = 2 Solid line Shade above

51. Absolute Value Inequalities y > 2|x + 2| + 1

52. Absolute Value Inequalitie s y > 2|x + 2| + 1

53. Absolute Value Inequalities y > 2|x + 2| + 1

54. Absolute Value Inequalities y > 2|x + 2| + 1

55. Your Turn: Graph y ≤ |x| + 3 Solution:

56. Your Turn: Graph 2y ≥ 6|x| Solution:

57. Your Turn: Graph |x| < y + 2 Solution:

58. Essential Question Big Idea: Function Is graphing an inequality in two variables similar to graphing? Graphing an inequality is similar to graphing a line. The graph of an inequality contains all points on one side of the line. The graph may or may not include the points on the line.

59. Assignment Section 2-8, Pg 132 – 134; #1 – 6 all, 8 – 44 even.