1. Slide 6-1 Copyright © 2005 Pearson Education, Inc. SEVENTH EDITION and EXPANDED SEVENTH EDITION

2. Copyright © 2005 Pearson Education, Inc. Chapter 6 Algebra, Graphs and Functions

3. Copyright © 2005 Pearson Education, Inc. 6.1 Order of Operations

4. Slide 6-4 Copyright © 2005 Pearson Education, Inc. Definitions Algebra: a generalized form of arithmetic. Variables: used to represent numbers Algebraic expression: a collection of variables, numbers, parentheses, and operation symbols.  Examples:

5. Slide 6-5 Copyright © 2005 Pearson Education, Inc. Order of Operations 1. First, perform all operations within parentheses or other grouping symbols (according to the following order). 2.Next, perform all exponential operations (that is, raising to powers or finding roots). 3.Next, perform all multiplication and divisions from left to right. 4.Finally, perform all additions and subtractions from left to right.

6. Slide 6-6 Copyright © 2005 Pearson Education, Inc. Example: Evaluating an Expression Evaluate the expression x 2 + 4x + 5 for x = 3. Solution: x 2 + 4x + 5 = 3 2 + 4(3) + 5 = 9 + 12 + 5 = 26

7. Slide 6-7 Copyright © 2005 Pearson Education, Inc. Example: Substituting for Two Variables Evaluate when x = 3 and y = 4. Solution:

8. Copyright © 2005 Pearson Education, Inc. 6.2 Linear Equations in One Variable

9. Slide 6-9 Copyright © 2005 Pearson Education, Inc. Definitions Like terms are terms that have the same variables with the same exponents on the variables. Unlike terms have different variables or different exponents on the variables.

10. Slide 6-10 Copyright © 2005 Pearson Education, Inc. Properties of the Real Numbers Associative property of multiplication (ab)c = a(bc) Associative property of addition (a + b) + c = a + (b + c) Commutative property of multiplication ab = ba Commutative property of addition a + b = b + a Distributive propertya(b + c) = ab + ac

11. Slide 6-11 Copyright © 2005 Pearson Education, Inc. Example: Combine Like Terms 8x + 4x = (8 + 4)x = 12x 5y  6y = (5  6)y =  y x + 15  5x + 9 = (1  5)x + (15+9) =  4x + 24 3x + 2 + 6y  4 + 7x = (3 + 7)x + 6y + (2  4) = 10x + 6y  2

12. Slide 6-12 Copyright © 2005 Pearson Education, Inc. Solving Equations Addition Property of Equality If a = b, then a + c = b + c for all real numbers a, b, and c. Find the solution to the equation x  9 = 24. x  9 + 9 = 24 + 9 x = 33 Check: x  9 = 24 33  9 = 24 ? 24 = 24 true

13. Slide 6-13 Copyright © 2005 Pearson Education, Inc. Solving Equations continued Subtraction Property of Equality If a = b, then a  c = b  c for all real numbers a, b, and c. Find the solution to the equation x + 12 = 31. x + 12  12 = 31  12 x = 19 Check: x + 12 = 31 19 + 12 = 31 ? 31 = 31 true

14. Slide 6-14 Copyright © 2005 Pearson Education, Inc. Solving Equations continued Multiplication Property of Equality If a = b, then a c = b c for all real numbers a, b, and c, where c  0. Find the solution to the equation

15. Slide 6-15 Copyright © 2005 Pearson Education, Inc. Solving Equations continued Division Property of Equality If a = b, then for all real numbers a, b, and c, c  0. Find the solution to the equation 4x = 48.

16. Slide 6-16 Copyright © 2005 Pearson Education, Inc. General Procedure for Solving Linear Equations If the equation contains fractions, multiply both sides of the equation by the lowest common denominator (or least common multiple). This step will eliminate all fractions from the equation. Use the distributive property to remove parentheses when necessary. Combine like terms on the same side of the equal sign when possible.

17. Slide 6-17 Copyright © 2005 Pearson Education, Inc. General Procedure for Solving Linear Equations continued Use the addition or subtraction property to collect all terms with a variable on one side of the equal sign and all constants on the other side of the equal sign. It may be necessary to use the addition or subtraction property more than once. This process will eventually result in an equation of the form ax = b, where a and b are real numbers.

18. Slide 6-18 Copyright © 2005 Pearson Education, Inc. General Procedure for Solving Linear Equations continued Solve for the variable using the division or multiplication property. This will result in an answer in the form x = c, where c is a real number.

19. Slide 6-19 Copyright © 2005 Pearson Education, Inc. Example: Solving Equations Solve 3x  4 = 17.

20. Slide 6-20 Copyright © 2005 Pearson Education, Inc. Solve 21 = 6 + 3(x + 2)

21. Slide 6-21 Copyright © 2005 Pearson Education, Inc. Solve 8x + 3 = 6x + 21

22. Slide 6-22 Copyright © 2005 Pearson Education, Inc. Solve 6(x  2) + 2x + 3 = 4(2x  3) + 2 False, the equation has no solution. The equation is inconsistent.

23. Slide 6-23 Copyright © 2005 Pearson Education, Inc. Solve 4(x + 1)  6(x + 2) =  2(x + 4) True, 0 = 0 the solution is all real numbers.

24. Slide 6-24 Copyright © 2005 Pearson Education, Inc. Proportions A proportion is a statement of equality between two ratios. Cross Multiplication  If then ad = bc, b  0, d  0.

25. Slide 6-25 Copyright © 2005 Pearson Education, Inc. To Solve Application Problems Using Proportions Represent the unknown quantity by a variable. Set up the proportion by listing the given ratio on the left-hand side of the equal sign and the unknown and other given quantity on the right-hand side of the equal sign. When setting up the right-hand side of the proportion, the same respective quantities should occupy the same respective positions on the left and right.

26. Slide 6-26 Copyright © 2005 Pearson Education, Inc. To Solve Application Problems Using Proportions continued For example, an acceptable proportion might be Once the proportion is properly written, drop the units and use cross multiplication to solve the equation. Answer the question or questions asked.

27. Slide 6-27 Copyright © 2005 Pearson Education, Inc. Example A 50 pound bag of fertilizer will cover an area of 15,000 ft 2. How many pounds are needed to cover an area of 226,000 ft 2 ? 754 pounds of fertilizer would be needed.

28. Copyright © 2005 Pearson Education, Inc. 6.3 Formulas

29. Slide 6-29 Copyright © 2005 Pearson Education, Inc. Definitions A formula is an equation that typically has a real-life application.

30. Slide 6-30 Copyright © 2005 Pearson Education, Inc. Perimeter The formula for the perimeter of a rectangle is Perimeter = 2 length + 2 width or P = 2l + 2w. Use the formula to find the perimeter of a yard when l = 150 feet and w = 100 feet.  P = 2l + 2w P = 2(150) + 2(100) P = 300 + 200 P = 500 feet

31. Slide 6-31 Copyright © 2005 Pearson Education, Inc. Example The formula for the volume of a cylinder is V =  r 2 h. Use the formula to find the height of a cylinder with a radius of 6 inches and a volume of 565.49 in 3.   The height of the cylinder is 5 inches.

32. Slide 6-32 Copyright © 2005 Pearson Education, Inc. Exponential Equations: Carbon Dating Carbon dating is used by scientists to find the age of fossils, bones, and other items. The formula used in carbon dating is If 15 mg of C 14 is present in an animal bone recently excavated, how many milligrams will be present in 4000 years?

33. Slide 6-33 Copyright © 2005 Pearson Education, Inc. Exponential Equations: Carbon Dating continued In 4000 years, approximately 9.2 mg of the original 15 mg of C 14 will remain.

34. Slide 6-34 Copyright © 2005 Pearson Education, Inc. Solving for a Variable in a Formula or Equation Solve the equation 3x + 8y  9 = 0 for y.

35. Slide 6-35 Copyright © 2005 Pearson Education, Inc. Solve for b 2.

36. Copyright © 2005 Pearson Education, Inc. 6.4 Applications of Linear Equations in One Variable

37. Slide 6-37 Copyright © 2005 Pearson Education, Inc. Translating Words to Expressions 3x  9 The sum of three times a number decreased by 9. 2x + 8Eight more than twice a number 2x2xTwice a number x  5 Five less than a number x + 10Ten more than a number Mathematical ExpressionPhrase

38. Slide 6-38 Copyright © 2005 Pearson Education, Inc. To Solve a Word Problem Read the problem carefully at least twice to be sure that you understand it. If possible, draw a sketch to help visualize the problem. Determine which quantity you are being asked to find. Choose a letter to represent this unknown quantity. Write down exactly what this letter represents. Write the word problem as an equation. Solve the equation for the unknown quantity. Answer the question or questions asked. Check the solution.

39. Slide 6-39 Copyright © 2005 Pearson Education, Inc. Example The bill (parts and labor) for the repairs of a car was $496.50. The cost of the parts was $339. The cost of the labor was $45 per hour. How many hours were billed? Let h = the number of hours billed Cost of parts + labor = total amount 339 + 45h = 496.50

40. Slide 6-40 Copyright © 2005 Pearson Education, Inc. Example continued The car was worked on for 3.5 hours.

41. Slide 6-41 Copyright © 2005 Pearson Education, Inc. Example Sandra Cone wants to fence in a rectangular region in her backyard for her lambs. She only has 184 feet of fencing to use for the perimeter of the region. What should the dimensions of the region be if she wants the length to be 8 feet greater than the width?

42. Slide 6-42 Copyright © 2005 Pearson Education, Inc. continued, 184 feet of fencing, length 8 feet longer than width Let x = width of region Let x + 8 = length P = 2l + 2w x + 8 x The width of the region is 42 feet and the length is 50 feet.

43. Copyright © 2005 Pearson Education, Inc. 6.5 Variation

44. Slide 6-44 Copyright © 2005 Pearson Education, Inc. Direct Variation Variation is an equation that relates one variable to one or more other variables. In direct variation, the values of the two related variables increase or decrease together. If a variable y varies directly with a variable x, then y = kx where k is the constant of proportionality (or the variation constant).

45. Slide 6-45 Copyright © 2005 Pearson Education, Inc. Example The amount of interest earned on an investment, I, varies directly as the interest rate, r. If the interest earned is $50 when the interest rate is 5%, find the amount of interest earned when the interest rate is 7%. I = rx $50 = 0.05x 1000 = x

46. Slide 6-46 Copyright © 2005 Pearson Education, Inc. Example continued x = 1000, r = 7% I = rx I = 0.07(1000) I = $70 The amount of interest earned is $70.

47. Slide 6-47 Copyright © 2005 Pearson Education, Inc. Inverse Variation When two quantities vary inversely, as one quantity increases, the other quantity decreases, and vice versa. If a variable y varies inversely with a variable, x, then y = k/x where k is the constant of proportionality.

48. Slide 6-48 Copyright © 2005 Pearson Education, Inc. Example Suppose y varies inversely as x. If y = 12 when x = 18, find y when x = 21. Now substitute 216 for k in y = k/x and find y when x = 21.

49. Slide 6-49 Copyright © 2005 Pearson Education, Inc. Joint Variation One quantity may vary directly as the product of two or more other quantities. The general form of a joint variation, where y, varies directly as x and z, is y = kxz where k is the constant of proportionality.

50. Slide 6-50 Copyright © 2005 Pearson Education, Inc. Example The area, A, of a triangle varies jointly as its base, b, and height, h. If the area of a triangle is 48 in 2 when its base is 12 in. and its height is 8 in., find the area of a triangle whose base is 15 in. and whose height is 20 in. A = kbh

51. Slide 6-51 Copyright © 2005 Pearson Education, Inc. Example continued

52. Slide 6-52 Copyright © 2005 Pearson Education, Inc. Combined Variation A varies jointly as B and C and inversely as the square of D. If A = 1 when B = 9, C = 4, and D = 6, find A when B = 8, C = 12, and D = 5. Write the equation.

53. Slide 6-53 Copyright © 2005 Pearson Education, Inc. Combined Variation continued Find the constant of proportionality. Now find A.

54. Copyright © 2005 Pearson Education, Inc. 6.6 Linear Inequalities

55. Slide 6-55 Copyright © 2005 Pearson Education, Inc. Symbols of Inequality a < b means that a is less than b. a  b means that a is less than or equal to b. a > b means that a is greater than b. a  b means that a is greater than or equal to b.

56. Slide 6-56 Copyright © 2005 Pearson Education, Inc. Example: Graphing Graph the solution set of x  4, where x is a real number, on the number line. The numbers less than or equal to 4 are all the points on the number line to the left of 4 and 4 itself. The closed circle at 4 shows that 4 is included in the solution set.

57. Slide 6-57 Copyright © 2005 Pearson Education, Inc. Example: Graphing Graph the solution set of x > 3, where x is a real number, on the number line. The numbers greater than 3 are all the points on the number line to the right of 3. The open circle at 3 is used to indicate that 3 is not included in the solution set.

58. Slide 6-58 Copyright © 2005 Pearson Education, Inc. Solve 3x  8 < 10 and graph the solution set. The solution set is all real numbers less than 6.

59. Slide 6-59 Copyright © 2005 Pearson Education, Inc. Compound Inequality Graph the solution set of the inequality  4 < x  3 a) where x is an integer. The solution set is the integers between  4 and 3, including 3.

60. Slide 6-60 Copyright © 2005 Pearson Education, Inc. Compound Inequality continued b) where x is a real number The solution set consists of all real numbers between  4 and 3, including the 3 but not the  4.

61. Slide 6-61 Copyright © 2005 Pearson Education, Inc. Example A student must have an average (the mean) on five tests that is greater than or equal to 85% but less than 92% to receive a final grade of B. Jamal’s grade on the first four tests were 98%, 89%, 88%, and 93%. What range of grades on the fifth test will give him a B in the course?

62. Slide 6-62 Copyright © 2005 Pearson Education, Inc. Example continued

63. Copyright © 2005 Pearson Education, Inc. 6.7 Graphing Linear Equations

64. Slide 6-64 Copyright © 2005 Pearson Education, Inc. Rectangular Coordinate System x-axis y-axis origin Quadrant I Quadrant II Quadrant III Quadrant IV The horizontal line is called the x-axis. The vertical line is called the y-axis. The point of intersection is the origin.

65. Slide 6-65 Copyright © 2005 Pearson Education, Inc. Plotting Points Each point in the xy-plane corresponds to a unique ordered pair (a, b). Plot the point (2, 4).  Move 2 units right  Move 4 units up 2 units 4 units

66. Slide 6-66 Copyright © 2005 Pearson Education, Inc. Graphing Linear Equations Graph the equation y = 5x + 2 33 11 0  2/5 20 yx

67. Slide 6-67 Copyright © 2005 Pearson Education, Inc. To Graph Equations by Plotting Points Solve the equation for y. Select at least three values for x and find their corresponding values of y. Plot the points. The points should be in a straight line. Draw a line through the set of points and place arrow tips at both ends of the line.

68. Slide 6-68 Copyright © 2005 Pearson Education, Inc. Graphing Using Intercepts The x-intercept is found by letting y = 0 and solving for x.  Example: y =  3x + 6 0 =  3x + 6  6 =  3x 2 = x The y-intercept is found by letting x = 0 and solving for y.  Example: y =  3x + 6 y =  3(0) + 6 y = 6

69. Slide 6-69 Copyright © 2005 Pearson Education, Inc. Graph 3x + 2y = 6 Find the x-intercept.  3x + 2y = 6  3x + 2(0) = 6  3x = 6  x = 2 Find the y-intercept.  3x + 2y = 6  3(0) + 2y = 6  2y = 6  y = 3

70. Slide 6-70 Copyright © 2005 Pearson Education, Inc. Slope The ratio of the vertical change to the horizontal change for any two points on the line.

71. Slide 6-71 Copyright © 2005 Pearson Education, Inc. Types of Slope The slope of a vertical line is undefined. The slope of a horizontal line is zero. zero negative undefined positive

72. Slide 6-72 Copyright © 2005 Pearson Education, Inc. Example: Finding Slope Find the slope of the line through the points (5,  3) and (  2,  3)

73. Slide 6-73 Copyright © 2005 Pearson Education, Inc. Graphing Equations by Using the Slope and y-Intercept Slope-Intercept Form of the Equation of the Line y = mx + b where m is the slope of the line and (0, b) is the y-intercept of the line.

74. Slide 6-74 Copyright © 2005 Pearson Education, Inc. Steps Solve the equation for y to place the equation in slope-intercept form. Determine the slope and y-intercept from the equation. Plot the y-intercept. Obtain a second point using the slope. Draw a straight line through the points.

75. Slide 6-75 Copyright © 2005 Pearson Education, Inc. Example Graph 2x  3y = 9 Write in slope-intercept form. The y-intercept is (0,  3) and the slope is 2/3.

76. Slide 6-76 Copyright © 2005 Pearson Education, Inc. Example continued Plot a point at (0,  3) on the y-axis, then move up 2 units and to the right 3 units.

77. Slide 6-77 Copyright © 2005 Pearson Education, Inc. Horizontal Lines Graph y =  3 y is always equal to  3, the value of y can never be 0. The graph is parallel to the x-axis.

78. Slide 6-78 Copyright © 2005 Pearson Education, Inc. Vertical Lines Graph x =  3 x always equals  3, the value of x can never be 0. The graph is parallel to the y-axis.

79. Copyright © 2005 Pearson Education, Inc. 6.8 Linear Inequalities in Two Variables

80. Slide 6-80 Copyright © 2005 Pearson Education, Inc. To Graph Inequalities in Two Variables Mentally substitute the equal sign for the inequality sign and plot points as if you were graphing the equation. If the inequality is, draw a dashed line through the points. If the inequality is  or , draw a solid line through the points. Select a test point not on the line and substitute the x and y-coordinates into the inequality. If the substitution results in a true statement, shade the area on the same side of the line as the test point. If the test point results in a false statement, shade the area on the opposite side of the line as the test point.

81. Slide 6-81 Copyright © 2005 Pearson Education, Inc. Graph 3x + 4y > 12 Draw a dashed line. Select a test point.  Try (0, 0)  3x + 4y > 12 3(0) + 4(0) > 12 0 + 0 > 12 0 > 12 false Shade the opposite half plane.

82. Slide 6-82 Copyright © 2005 Pearson Education, Inc. Graph 3x + y   6 Draw a solid line. Select a test point.  Try (0, 0)  3x + y   6  3(0) + (0)   6  0 + 0   6  0   6 false  Shade the opposite half plane.

83. Slide 6-83 Copyright © 2005 Pearson Education, Inc. Graph y >  2 Draw a dashed line. Select a test point.  Try (0, 0)  0 >  2 true  Shade the half plane containing (0, 0).

84. Copyright © 2005 Pearson Education, Inc. 6.9 Solving Quadratic Equations by Using Factoring and by Using the Quadratic Formula

85. Slide 6-85 Copyright © 2005 Pearson Education, Inc. FOIL A binomial is an expression that contains two terms. To multiply two binomials, we use the FOIL method. F = First O = Outer I = Inner L = Last (a + b)(c + d) = ac + ad + bc + bd F I O L

86. Slide 6-86 Copyright © 2005 Pearson Education, Inc. Example Multiply: (2x + 4)(x + 6)

87. Slide 6-87 Copyright © 2005 Pearson Education, Inc. To Factor Trinomial Expressions of the Form x 2 + bx + c Find two numbers whose product is c and whose sum is b. Write the factors in the form  (x + ) (x + ) Check your answer by multiplying the factors using the FOIL method. One number from step 1 Other number from step 1

88. Slide 6-88 Copyright © 2005 Pearson Education, Inc. Factoring Example Factor x 2  7x + 12. We need to find two numbers whose product is 12 and whose sum is  7. (x  3)(x  4) 3 + 4 = 7 2 + 6 = 8 1 + 12 = 13 Sum of Factors (  3)(  4) (  2)(  6)  1(  12) Factors of 12  3 +  4 =  7 (3)(4)  2 +  6 =  8 (2)(6)  1 +  12 =  13 1(12) Sum of FactorsFactors of 12

89. Slide 6-89 Copyright © 2005 Pearson Education, Inc. Factoring Trinomials of the Form ax 2 + bc + c, a  1. Write all pairs of factors of the coefficient of the squared term, a. Write all pairs of the factors of the constant, c. Try various combinations of these factors until the sum of the products of the outer and inner terms is bx. Check your answer by multiplying the factors using the FOIL method.

90. Slide 6-90 Copyright © 2005 Pearson Education, Inc. Example: Factoring Factor 3x 2 + 14x + 8. (3x + )(x + ) Thus, 3x 2 + 14x + 8 = (3x + 2)(x + 4). 14x Correct middle term (3x + 2)(x + 4) 10x(3x + 4)(x + 2) 11x(3x + 8)(x + 1) 25x(3x + 1)(x + 8) Sum of Outer and Inner Terms Possible Factors

91. Slide 6-91 Copyright © 2005 Pearson Education, Inc. Solving Quadratic Equations by Factoring Standard Form of a Quadratic Equation  ax 2 + bx + c = 0, a  0 Zero-Factor Property  If a b = 0, then a = 0 or b = 0.

92. Slide 6-92 Copyright © 2005 Pearson Education, Inc. To Solve a Quadratic Equation by Factoring Use the addition or subtraction property to make one side of the equation equal to 0. Factor the side of the equation not equal to 0. Use the zero-factor property to solve the equation.

93. Slide 6-93 Copyright © 2005 Pearson Education, Inc. Example: Solve by Factoring Solve 4x 2 + 17x  15 = 0. The solutions are  5 and ¾.

94. Slide 6-94 Copyright © 2005 Pearson Education, Inc. Quadratic Formula For a quadratic equation in standard form, ax 2 + bx + c = 0, a  0, the quadratic formula is

95. Slide 6-95 Copyright © 2005 Pearson Education, Inc. Example: Using the Quadratic Formula Solve the equation 3x 2 + 2x  7 = 0. a = 3, b = 2 and c =  7

96. Copyright © 2005 Pearson Education, Inc. 6.10 Functions and Their Graphs

97. Slide 6-97 Copyright © 2005 Pearson Education, Inc. Function A function is a special type of relation where each value of the independent variable corresponds to a unique value of the dependent variable. Domain the set of values used for the independent variable. Range The resulting set of values obtained for the dependent variable.

98. Slide 6-98 Copyright © 2005 Pearson Education, Inc. Vertical Line Test If a vertical line can be drawn so that it intersects the graph at more than one point, then each x does not have a unique y.

99. Slide 6-99 Copyright © 2005 Pearson Education, Inc. Types of Functions Linear: y = ax + b Quadratic: y = ax 2 + bx + c

100. Slide 6-100 Copyright © 2005 Pearson Education, Inc. Graphs of Quadratic Functions axis of symmetry

101. Slide 6-101 Copyright © 2005 Pearson Education, Inc. Graphs of Quadratic Functions continued Axis of Symmetry of a Parabola

102. Slide 6-102 Copyright © 2005 Pearson Education, Inc. General Procedure to Sketch the Graph of a Quadratic Equation Determine whether the parabola opens upward or downward. Determine the equation of the axis of symmetry. Determine the vertex of the parabola. Determine the y-intercept by substituting x = 0 into the equation. Determine the x-intercept (if they exist) by substituting y = 0 into the equation and solving for x. Draw the graph, making use of the information gained in steps 1 through 5. Remember the parabola will be symmetric with respect to the axis of symmetry.

103. Slide 6-103 Copyright © 2005 Pearson Education, Inc. Graph y = x 2 + 2x  3. Since a = 1, the parabola opens up Axis: y-coordinate of vertex (  1,  4) y-intercept: (0,  3)

104. Slide 6-104 Copyright © 2005 Pearson Education, Inc. Graph y = x 2 + 2x  3 continued x-intercepts: Plot the points and sketch.

105. Slide 6-105 Copyright © 2005 Pearson Education, Inc. Graphs of Exponential Functions Graph y = 3 x. Domain: all real numbers Range is y > 0 (  3, 1/27) 1/27 33 (  2, 1/9) 1/9 22 (  1, 1/3) 1/3 11 (3, 27)273 9 3 1 y = 3 x (2, 9)2 (1, 3)1 (0, 1)0 (x, y)x

106. Slide 6-106 Copyright © 2005 Pearson Education, Inc. Graph Domain: all real numbers Range is y > 0 (3, 1/27)1/273 (2, 1/9)1/92 (1, 1/3)1/31 (  3, 27) 27 33 9 3 1 (  2, 9) 22 (  1, 3) 11 (0, 1)0 (x, y)x