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Chapter 2- Linear Functions

Chapter 2- Linear Functions. Algebra II. Table of Contents. 2.1- Solving Linear Equations and Inequalities 2.2- Proportional Reasoning 2.3- Graphing Linear Functions 2.4- Writing Linear Functions. 2.1. Algebra II (Bell work). Define the following words Equation pg. 90

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Chapter 2- Linear Functions

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  1. Chapter 2- Linear Functions Algebra II

  2. Table of Contents • 2.1- Solving Linear Equations and Inequalities • 2.2- Proportional Reasoning • 2.3- Graphing Linear Functions • 2.4- Writing Linear Functions

  3. 2.1 Algebra II (Bell work) • Define the following words • Equation pg. 90 • Linear Equation in one variable pg. 90 • Grab notebook/homework sheet off of the podium • I will also remind you of some classroom procedures

  4. 2.1- Solving Linear Equations and Inequalities Algebra II

  5. 2-1 Example 1 The local phone company charges $12.95 a month for the first 200 of air time, plus $0.07 for each additional minute. If Nina’s bill for the month was $14.56, how many additional minutes did she use? Let m represent the number of additional minutes that Nina used. Model additional minute charge number of additional minutes monthly charge total charge plus times = = 14.56 12.95 m + 0.07 *

  6. 2-1 –12.95 –12.95 The local phone company charges $12.95 a month for the first 200 of air time, plus $0.07 for each additional minute. If Nina’s bill for the month was $14.56, how many additional minutes did she use? Let m represent the number of additional minutes that Nina used. Solve. 12.95 + 0.07m = 14.56 Subtract 12.95 from both sides. 0.07m = 1.61 Divide both sides by 0.07. 0.07 0.07 m = 23 Nina used 23 additional minutes.

  7. 2-1 Example 1 Stacked cups are to be placed in a pantry. One cup is 3.25 in. high and each additional cup raises the stack 0.25 in. How many cups fit between two shelves 14 in. apart? additional cup height number of additional cups total height plus times = one cup = 14.00 c 3.25 + 0.25 *

  8. 2-1 –3.25 –3.25 0.25 0.25 Stacked cups are to be placed in a pantry. One cup is 3.25 in. high and each additional cup raises the stack 0.25 in. How many cups fit between two shelves 14 in. apart? Solve. 3.25 + 0.25c = 14.00 0.25c = 10.75 c = 43 44 cups fit between the 14 in. shelves.

  9. 2-1 4 4 –12–12 Example 2 Solving Equations with the Distributive Property Solve 4(m + 12) = –36 Method 1 The quantity (m + 12) is multiplied by 4, so divide by 4 first. 4(m + 12) = –36 Divide both sides by 4. m + 12 = –9 Subtract 12 from both sides. m = –21

  10. 2-1 –48 –48 4m –84 = 4 4 Solve 4(m + 12) = –36 Method 2 Distribute before solving. 4m + 48 = –36 4m = –84 m = –21

  11. 2-1 –6 –6 3 3 –9 –9 –2–2 –9p 36 = –3 –3 Solve 3(2 –3p) = 42. Solve 3(2 – 3p) = 42 . Method 1 Method 2 3(2 – 3p) = 42 6 – 9p = 42 –9p = 36 2 – 3p = 14 –3p = 12 p = –4 p = –4

  12. 2-1 12r 6 1212 = r = Solve –3(5 – 4r) = –9. –15 + 12r = –9 +15 +15 12r = 6

  13. Just Read If there are variables on both sides of the equation, (1) simplify each side. (2) collect all variable terms on one side and all constants terms on the other side. (3) isolate the variables

  14. 2-1 +2w +2w +11k +11k +10 + 10 –12 –12 5 5 3 3 Example 3 Solving Equations with Variables on Both Sides Solve 3k– 14k + 25 = 2 – 6k – 12. Solve3(w + 7) – 5w = w + 12. –2w+ 21 = w + 12 –11k + 25 = –6k– 10 25 = 5k – 10 21 = 3w + 12 35 = 5k 9 = 3w 3 = w 7 = k

  15. 2.1 Algebra II (Bell work) • Define the following words • Identity – An equation that is true for all values of the variable, such as x = x • Answer is all real, or R, or ( - ∞ , ∞ ) • Contradiction – An equation that has no solution, such as 3 = 5.

  16. 2-1 + v + v –2x –2x The equation has no solution. The solution set is the emptyset, which is represented by the symbol . Example 4 Identifying Identities and Contradictions Solve 3v – 9 – 4v = –(5 + v). Solve 2(x – 6) = –5x – 12 + 7x. 2x – 12 = 2x – 12 3v – 9 – 4v = –(5 + v) –9 – v = –5 – v –12 = –12  –9 ≠ –5 x The solutions set is all real number, or.

  17. 2.1 + 9x +9x –5x –5x The equation has no solution. The solution set is the empty set, which is represented by the symbol . Solve 3(2 –3x) = –7x – 2(x –3). Solve 5(x – 6) = 3x – 18 + 2x. 5(x – 6) = 3x – 18 + 2x 3(2 –3x) = –7x – 2(x –3) 5x – 30 = 5x – 18 6 – 9x = –9x + 6 –30 ≠ –18 x 6 = 6  The solutions set is all real numbers, or.

  18. Just Read If you multiply or divide by a negative number, flip the inequality sign These properties also apply to inequalities expressed with >, ≥, and ≤.

  19. 2-1 Example 5 Solving Inequalities Solve and graph 8a –2 ≥ 13a + 8. Solve and graph x + 8 ≥ 4x + 17. 8a – 2 ≥ 13a + 8 x + 8 ≥ 4x + 17 –13a –13a –x –x –5a – 2 ≥ 8 8 ≥ 3x +17 +2 +2 –17 –17 –5a ≥ 10 –9 ≥ 3x –5a ≤ 10 –9 ≥ 3x –5–5 33 –3 ≥ x or x ≤ –3 a ≤ –2

  20. HW pg.94 • 2.1- Due Monday Sep. 29th • Day 1: 2, 3-11 (Odd), 21- 25 (Odd), 41, 42, 45, 47 • Day 2: 13, 15, 33-35, 37-39, 49, 52, 53, 54, 56, 62-65 • Follow All HW Guidelines or ½ off

  21. 2-2 Copy boxed parts below

  22. 2.2- Proportional Reasoning Algebra II

  23. 2-2 16 24 = 1624 206.4 24p 14c p 12.9 = = = = 2424 88132 p12.9 88c 1848 Example 1 Solve each proportion. 14 c = A. B. 88132 206.4 = 24p 88c =1848 88 88 8.6 = p c = 21

  24. 2-2 Optional y 77 y77 152.5 = = = = = x7 1284 1284 924 84y 2.5x 105 = 2.5 2.5 8484 Solve each proportion. 15 2.5 A. B. x7 924 = 84y 2.5x =105 11 = y x = 42

  25. 2-2 Because percents can be expressed as ratios, you can use the proportion to solve percent problems. Remember! Percent is a ratio that means per hundred. For example: 30% = 0.30 = 30 100

  26. 2-2 Example 2 Solving Percent Problems A poll taken one day before an election showed that 22.5% of voters planned to vote for a certain candidate. If 1800 voters participated in the poll, how many indicated that they planned to vote for that candidate? Method 1 Use a proportion. Method 2 Use a percent equation. Percent (as decimal) whole = part 0.2251800 = x 405 = x x = 405 So 405 voters are planning to vote for that candidate.

  27. 2-2 At Clay High School, 434 students, or 35% of the students, play a sport. How many students does Clay High School have? Method 1 Use a proportion. Method 2 Use a percent equation. 35% = 0.35 Percent (as decimal)whole = part 0.35x = 434 Cross multiply. 100(434) = 35x x = 1240 Solve for x. x = 1240 Clay High School has 1240 students.

  28. 2-2 1.24 m 1 stride length 39.37 in. 1 m 49 in. 1 stride length  ≈ 600 m 482 strides x m 1 stride = Example 3 Fitness Application (Dimensional Analysis) Ryan ran 600 meters and counted 482 strides. How long is Ryan’s stride in inches? (Hint: 1 m ≈ 39.37 in.) Use a proportion to find the length of his stride in meters. 600 = 482x x ≈ 1.24 m Ryan’s stride length is approximately 49 inches.

  29. 2-2 Optional 1.35 m 1 stride length 39.37 in. 1 m 53 in. 1 stride length  ≈ 400 m 297 strides x m 1 stride = Luis ran 400 meters in 297 strides. Find his stride length in inches. Use a proportion to find the length of his stride in meters. 400 = 297x x ≈1.35 m Luis’s stride length is approximately 53 inches.

  30. 2-2 = = Z height of ∆XAB width of ∆XAB Y height of ∆XYZ width of ∆XYZ 3x A B 9 6 X Example 4 Scaling Geometric Figures in the Coordinate Plane 9x = 18, so x = 2

  31. Math Joke • Q: Why were the similar triangles weighing themselves • A: they were finding their Scale

  32. 2-2 = h ft 9 ft 6 ft 6 22 Shadow of tree Height of tree Shadow of house Height of house 22 ft = 9 h The tree in front of Luka’s house casts a 6-foot shadow at the same time as the house casts a 22-foot shadow. If the tree is 9 feet tall, how tall is the house? 6h = 198 h = 33 The house is 33 feet high.

  33. 2-2 • Optional = h ft 6 ft 20 ft 20 90 Shadow of climber Height of climber = 90 ft 6 h Shadow of tree Height of tree A 6-foot-tall climber casts a 20-foot long shadow at the same time that a tree casts a 90-foot long shadow. How tall is the tree? 20h = 540 h = 27 The tree is 27 feet high.

  34. HW pg. 100 • 2.2- Due Monday Sep. 29th • 7, 9- 11, 13, 18, 19, 21, 22, 25, 33, 36, 38, 68-71 • Ch: 35 • Follow All HW Guidelines or ½ off • Don’t forget to write assignments on hw sheet (or planner)

  35. 2-3 Algebra II (Bell work) Copy down the definition of a linear function Functions with a constant rate of change are called linear functions. A linear function can be written in the form f(x) = mx + b, where x is the independent variable and m and b are constants. (m = slope) (b = y-intercept) The graph of a linear function is a straight line made up of all points that satisfy y = f(x).

  36. 2.3- Grapping Linear Functions Algebra II

  37. 2-3 +2 +2 +2 –1 –1 –1 The rate of change, = , is constant So the data set is linear. Example 1 Recognizing Linear Functions Determine whether the data set could represent a linear function.

  38. 2-3 +1 +1 +1 +2 +4 +8 The rate of change, 2 ≠ 4 ≠ 8, is not constant. So the data set is not linear. Determine whether the data set could represent a linear function.

  39. 2-3 –2 –2 –2 –4 –4 –8 The rate of change, , is not constant. So the data set is not linear. Determine whether the data set could represent a linear function.

  40. 2-3 The constant rate of change for a linear function is its slope. The slope of a linear function is the ratio or m= =

  41. 2-3 Graph the line with slope m= that passes through (–1, –3). Example 2 Graphing Lines using Slope and a Point

  42. 2-3 Graph the line with slope m = that passes through (0, 2).

  43. 2-3 Copy the definitions below The y-intercept is the y-coordinate of a point where the line crosses the x-axis. The x-intercept is the x-coordinate of a point where the line crosses the y-axis.

  44. 2-3 x-intercept y-intercept Example 3 Graphing Lines Using the Intercepts Find the intercepts of 4x – 2y = 16, and graph the line. Find the x-intercept: 4x – 2y = 16 4x – 2(0) = 16 4x = 16 x = 4 Find the y-intercept: 4x – 2y = 16 4(0) – 2y = 16 –2y = 16 y = –8

  45. 2-3 y-intercept x-intercept Find the intercepts of 6x – 2y = –24, and graph the line. Find the x-intercept: 6x – 2y = –24 6x – 2(0) = –24 6x = –24 x = –4 Find the y-intercept: 6x – 2y = –24 6(0) – 2y = –24 –2y = –24 y = 12

  46. 2-3 Algebra II (Bell work) Linear functions can also be expressed as linear equations of the form y = mx+ b. When a linear function is written in the form y = mx + b The function is said to be in slope-intercept form because m is the slope of the graph and b is the y-intercept. Notice that slope-intercept form is the equation solved for y. So slope-intercept form is: y = mx + b or f(x) = mx + b, where m = slope, b = y-int

  47. Math Joke • Q: Why was the student afraid of the y-intercept • A: She thought she’d be stung by the b

  48. 2-3 +4x +4x The line has y-intercept –1 and slope 4, which is . Plot the point (0, –1). Then move up 4 and right 1 to find other points. Example 4 Graphing Functions in Slope-Intercept Form Write the function –4x + y = –1 in slope-intercept form. Solve for y first. –4x + y = –1 y = 4x – 1

  49. 2-3 The line has y-intercept 8 and slope . Plot the point (0, 8). Then move down 4 and right 3 to find other points. Write the function in slope-intercept form. Then graph the function. Solve for y first.

  50. 2-3 –30 –30 The line has y-intercept –2 and slope . Plot the point (0, –2). Then move up 1 and right 3 to find other points. Write the function 5x = 15y + 30 in slope-intercept form. Then graph the function. Solve for y first. 5x = 15y + 30 5x – 30 = 15y

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