1. 1  The Cartesian Coordinate System  Straight Lines  Linear Functions and Mathematical Models  Intersection of Straight Lines  The Method of Least Squares Straight Lines and Linear Functions

2. 1.1 The Cartesian Coordinate System

3.  We can represent real numbers geometrically by points on a real number, or coordinate, line:  This line includes all real numbers.  Exactly one point on the line is associated with each real number, and vice-versa (one dimensional space). Origin Positive Direction Negative Direction  The Cartesian Coordinate System

4.  The Cartesian coordinate system extends this concept to a plane (two dimensional space) by adding a vertical axis. 4321 – 1 – 2 – 3 – 4

5. The Cartesian Coordinate System   The horizontal line is called the x-axis, and the vertical line is called the y-axis. 4 3 2 1 – 1 – 2 – 3 – 4 x y

6. The Cartesian Coordinate System  The point where these two lines intersect is called the origin. 4 3 2 1 – 1 – 2 – 3 – 4 x y Origin

7. The Cartesian Coordinate System  In the x-axis, positive numbers are to the right and negative numbers are to the left of the origin. 4 3 2 1 – 1 – 2 – 3 – 4 x y Positive Direction Negative Direction

8. The Cartesian Coordinate System  In the y-axis, positive numbers are above and negative numbers are below the origin. 4 3 2 1 – 1 – 2 – 3 – 4 x y Positive Direction Negative Direction

9. (– 2, 4) (–1, – 2) (4, 3) The Cartesian Coordinate System  A point in the plane can now be represented uniquely in this coordinate system by an ordered pair of numbers (x, y). 4 3 2 1 – 1 – 2 – 3 – 4 x y (3, –1)

10. The Cartesian Coordinate System  The axes divide the plane into four quadrants as shown below. 4 3 2 1 – 1 – 2 – 3 – 4 x y Quadrant I (+, +) Quadrant II (–, +) Quadrant IV (+, –) Quadrant III (–, –)

11. The Distance Formula  The distance between any two points in the plane can be expressed in terms of the coordinates of the points. Distance formula  The distance d between two points P 1 (x 1, y 1 ) and P 2 (x 2, y 2 ) in the plane is given by

12. Examples  Find the distance between the points (– 4, 3) and (2, 6). Solution  Let (– 4, 3 and (2, 6 be points in the plane.  Let P 1 (– 4, 3) and P 2 (2, 6) be points in the plane.  We have x1 = – 4 y1 = 3 x2 = 2 y2 = 6x1 = – 4 y1 = 3 x2 = 2 y2 = 6x1 = – 4 y1 = 3 x2 = 2 y2 = 6x1 = – 4 y1 = 3 x2 = 2 y2 = 6  Using the distance formula, we have Example 1, page 4

13. Examples  Let P(x, y) denote a point lying on the circle with radius r and center C(h, k). Find a relationship between x and y. Solution  By the definition of a circle, the distance between P(x, y) and C(h, k) is r.  With the distance formula we get  Squaring both sides gives C(h, k) h k r P(x, y) y x Example 3, page 4

14. Equation of a Circle  An equation of a circle with center C(h, k) and radius r is given by

15. Examples  Find an equation of the circle with radius 2 and center (–1, 3). Solution  We use the circle formula with r = 2, h = –1, and k = 3: (–1, 3) –1 3 2 y x Example 4, page 5

16. Examples  Find an equation of the circle with radius 3 and center located at the origin. Solution  We use the circle formula with r = 3, h = 0, and k = 0: 3 y x Example 4, page 5

17. 1.2 Straight Lines

18. Slope of a Vertical Line  Let L denote the unique straight line that passes through the two distinct points (x 1, y 1 ) and (x 2, y 2 ).  If x 1 = x 2, then L is a vertical line, and the slope is undefined. (x 1, y 1 ) (x 2, y 2 ) y x L

19. Slope of a Nonvertical Line  If (x 1, y 1 ) and (x 2, y 2 ) are two distinct points on a nonvertical line L, then the slope m of L is given by (x 1, y 1 ) (x 2, y 2 ) y x L y 2 – y 1 =  y x 2 – x 1 =  x

20. Slope of a Nonvertical Line  If m > 0, the line slants upward from left to right. x L  y = 1  x = 1 m = 1 y

21. Slope of a Nonvertical Line  If m > 0, the line slants upward from left to right. y x L  y = 2  x = 1 m = 2

22. m = –1 Slope of a Nonvertical Line  If m < 0, the line slants downward from left to right. x L  y = –1  x = 1 y

23. m = –2 Slope of a Nonvertical Line  If m < 0, the line slants downward from left to right. y x L  y = –2  x = 1

24. 123456123456123456123456 (2, 5) Examples  Sketch the straight line that passes through the point (2, 5) and has slope – 4/3. Solution 1.Plot the point (2, 5). 2.A slope of – 4/3 means that if x increases by 3, y decreases by 4. 3.Plot the resulting point (5, 1). 4.Draw a line through the two points. y x L  y = – 4  x = 3 654321 (5, 1)

25. Examples  Find the slope m of the line that goes through the points (–1, 1) and (5, 3). Solution  Choose (x 1, y 1 ) to be (–1, 1) and (x 2, y 2 ) to be (5, 3).  With x 1 = –1, y 1 = 1, x 2 = 5, y 2 = 3, we find Example 2, page 11

26. Examples  Find the slope m of the line that goes through the points (–2, 5) and (3, 5). Solution  Choose (x 1, y 1 ) to be (–2, 5) and (x 2, y 2 ) to be (3, 5).  With x 1 = –2, y 1 = 5, x 2 = 3, y 2 = 5, we find Example 3, page 11

27. –2 –1 1234 Examples  Find the slope m of the line that goes through the points (–2, 5) and (3, 5). Solution  The slope of a horizontal line is zero: y x L 64321 (–2, 5) (3, 5) m = 0 Example 3, page 11

28. Parallel Lines  Two distinct lines are parallel if and only if their slopes are equal or their slopes are undefined.

29. Example  Let L 1 be a line that passes through the points (–2, 9) and (1, 3), and let L 2 be the line that passes through the points (– 4, 10) and (3, – 4).  Determine whether L 1 and L 2 are parallel. Solution  The slope m 1 of L 1 is given by  The slope m 2 of L 2 is given by  Since m 1 = m 2, the lines L 1 and L 2 are in fact parallel. Example 4, page 12

30. Equations of Lines  Let L be a straight line parallel to the y-axis.  Then L crosses the x-axis at some point (a, 0), with the x-coordinate given by x = a, where a is a real number.  Any other point on L has the form (a, ), where is an appropriate number.  The vertical line L can therefore be described as x = a (a, ) y x L (a, 0)

31. Equations of Lines  Let L be a nonvertical line with a slope m.  Let (x 1, y 1 ) be a fixed point lying on L, and let (x, y) be a variable point on L distinct from (x 1, y 1 ).  Using the slope formula by letting (x, y) = (x 2, y 2 ), we get  Multiplying both sides by x – x 1 we get

32. Point-Slope Form  An equation of the line that has slope m and passes through point (x 1, y 1 ) is given by

33. Examples  Find an equation of the line that passes through the point (1, 3) and has slope 2. Solution  Use the point-slope form  Substituting for point (1, 3) and slope m = 2, we obtain  Simplifying we get Example 5, page 13

34. Examples  Find an equation of the line that passes through the points (–3, 2) and (4, –1). Solution  The slope is given by  Substituting in the point-slope form for point (4, –1) and slope m = – 3/7, we obtain Example 6, page 14

35. Perpendicular Lines  If L 1 and L 2 are two distinct nonvertical lines that have slopes m 1 and m 2, respectively, then L 1 is perpendicular to L 2 (written L 1 ┴ L 2 ) if and only if

36. Example  Find the equation of the line L 1 that passes through the point (3, 1) and is perpendicular to the line L 2 described by Solution  L 2 is described in point-slope form, so its slope is m 2 = 2.  Since the lines are perpendicular, the slope of L 1 must be m 1 = –1/2  Using the point-slope form of the equation for L 1 we obtain Example 7, page 14

37. (a, 0) (0, b) Crossing the Axis  A straight line L that is neither horizontal nor vertical cuts the x-axis and the y-axis at, say, points (a, 0) and (0, b), respectively.  The numbers a and b are called the x-intercept and y-intercept, respectively, of L. y x L y-intercept x-intercept

38. Slope-Intercept Form  An equation of the line that has slope m and intersects the y-axis at the point (0, b) is given by y = mx + b

39. Examples  Find the equation of the line that has slope 3 and y-intercept of – 4. Solution  We substitute m = 3 and b = – 4 into y = mx + b and get y = 3x – 4 Example 8, page 15

40. Examples  Determine the slope and y-intercept of the line whose equation is 3x – 4y = 8. Solution  Rewrite the given equation in the slope-intercept form.  Comparing to y = mx + b, we find that m = ¾ and b = – 2.  So, the slope is ¾ and the y-intercept is – 2. Example 9, page 15

41. Applied Example  Suppose an art object purchased for $50,000 is expected to appreciate in value at a constant rate of $5000 per year for the next 5 years.  Write an equation predicting the value of the art object for any given year.  What will be its value 3 years after the purchase? Solution  Let x = time (in years) since the object was purchased y = value of object (in dollars)  Then, y = 50,000 when x = 0, so the y-intercept is b = 50,000.  Every year the value rises by 5000, so the slope is m = 5000.  Thus, the equation must be y = 5000x + 50,000.  After 3 years the value of the object will be $65,000: y = 5000(3) + 50,000 = 65,000 Applied Example 11, page 16

42. General Form of a Linear Equation  The equation Ax + By + C = 0 where A, B, and C are constants and A and B are not both zero, is called the general form of a linear equation in the variables x and y.

43. General Form of a Linear Equation  An equation of a straight line is a linear equation; conversely, every linear equation represents a straight line.

44. Example  Sketch the straight line represented by the equation 3x – 4y – 12 = 0 Solution  Since every straight line is uniquely determined by two distinct points, we need find only two such points through which the line passes in order to sketch it.  For convenience, let’s compute the x- and y-intercepts: ✦ Setting y = 0, we find x = 4; so the x-intercept is 4. ✦ Setting x = 0, we find y = –3; so the y-intercept is –3.  Thus, the line goes through the points (4, 0) and (0, –3). Example 12, page 17

45. Example  Sketch the straight line represented by the equation 3x – 4y – 12 = 0 Solution  Graph the line going through the points (4, 0) and (0, –3). 123456123456123456123456 (0, – 3) y x L 1 11– 1– 1– 2– 2– 3– 3– 4– 411– 1– 1– 2– 2– 3– 3– 4– 4 (4, 0) Example 12, page 17

46. Equations of Straight Lines Vertical line:x = a Horizontal line:y = b Point-slope form: y – y 1 = m(x – x 1 ) Slope-intercept form: y = mx + b General Form:Ax + By + C = 0

47. 1.3 Linear Functions and Mathematical Models Real-world problem Mathematical model Solution of real- world Problem Solution of mathematical model Formulate Interpret Solve Test

48. Mathematical Modeling  Mathematics can be used to solve real-world problems.  Regardless of the field from which the real-world problem is drawn, the problem is analyzed using a process called mathematical modeling.  The four steps in this process are: Real-world problem Mathematical model Solution of real- world Problem Solution of mathematical model Formulate Interpret Solve Test

49. Functions  A function f is a rule that assigns to each value of x one and only one value of y.  The value y is normally denoted by f(x), emphasizing the dependency of y on x.

50. Example  Let x and y denote the radius and area of a circle, respectively.  From elementary geometry we have y = p x 2  This equation defines y as a function of x, since for each admissible value of x there corresponds precisely one number y = p x 2 giving the area of the circle.  The area function may be written as f(x) = p x 2  To compute the area of a circle with a radius of 5 inches, we simply replace x in the equation by the number 5: f(5) = p( 5 2 )= 25 p

51. Domain and Range  Suppose we are given the function y = f(x).  The variable x is referred to as the independent variable, and the variable y is called the dependent variable.  The set of all the possible values of x is called the domain of the function f.  The set of all the values of f(x) resulting from all the possible values of x in its domain is called the range of f.  The output f(x) associated with an input x is unique: ✦ Each x must correspond to one and only one value of f(x).

52. Linear Function  The function f defined by where m and b are constants, is called a linear function.

53. Applied Example: U.S. Health-Care Expenditures  Because the over-65 population will be growing more rapidly in the next few decades, health-care spending is expected to increase significantly in the coming decades.  The following table gives the projected U.S. health-care expenditures (in trillions of dollars) from 2005 through 2010:  A mathematical model giving the approximate U.S. health- care expenditures over the period in question is given by where t is measured in years, with t = 0 corresponding to 2005. Year Year200520062007200820092010 Expenditure Expenditure2.002.172.342.502.692.90 Applied Example 1, page 29

54. Applied Example: U.S. Health-Care Expenditures  We have a.Sketch the graph of the function S and the given data on the same set of axes. b.Assuming that the trend continues, how much will U.S. health-care expenditures be in 2011? c.What is the projected rate of increase of U.S. health- care expenditures over the period in question? Year Year200520062007200820092010 Expenditure Expenditure2.002.172.342.502.692.90 Applied Example 1, page 29

55. Applied Example: U.S. Health-Care Expenditures  We have Solution a.The graph of the given data and of the function S is: Year Year200520062007200820092010 Expenditure Expenditure2.002.172.342.502.692.90 S(t) t 3.02.82.62.42.22.0 12345123451234512345 Applied Example 1, page 29

56. Applied Example: U.S. Health-Care Expenditures  We have Solution b.The projected U.S. health-care expenditures in 2011 is or approximately $3.06 trillion. Year Year200520062007200820092010 Expenditure Expenditure2.002.172.342.502.692.90 Applied Example 1, page 29

57. Applied Example: U.S. Health-Care Expenditures  We have Solution c.The function S is linear, so the rate of increase of the U.S. health-care expenditures is given by the slope of the straight line represented by S, which is approximately $0.18 trillion per year. Year Year200520062007200820092010 Expenditure Expenditure2.002.172.342.502.692.90 Applied Example 1, page 29

58. Cost, Revenue, and Profit Functions  Let x denote the number of units of a product manufactured or sold.  Then, the total cost function is C(x)= Total cost of manufacturing x units of the product  The revenue function is R(x)= Total revenue realized from the sale of x units of the product  The profit function is P(x)= Total profit realized from manufacturing and selling x units of the product

59. Applied Example: Profit Function  Puritron, a manufacturer of water filters, has a monthly fixed cost of $20,000, a production cost of $20 per unit, and a selling price of $30 per unit.  Find the cost function, the revenue function, and the profit function for Puritron. Solution  Let x denote the number of units produced and sold.  Then, Applied Example 2, page 31

60. 1.4 Intersections of Straight Lines

61. Finding the Point of Intersection  Suppose we are given two straight lines L 1 and L 2 with equations y = m 1 x + b 1 and y = m 2 x + b 2 (where m 1, b 1, m 2, and b 2 are constants) that intersect at the point P(x 0, y 0 ).  The point P(x 0, y 0 ) lies on the line L 1 and so satisfies the equation y = m 1 x + b 1.  The point P(x 0, y 0 ) also lies on the line L 2 and so satisfies y = m 2 x + b 2 as well.  Therefore, to find the point of intersection P(x 0, y 0 ) of the lines L 1 and L 2, we solve for x and y the system composed of the two equations y = m 1 x + b 1 and y = m 2 x + b 2

62. Example  Find the point of intersection of the straight lines that have equations y = x + 1 and y = – 2x + 4 Solution  Substituting the value y as given in the first equation into the second equation, we obtain  Substituting this value of x into either one of the given equations yields y = 2.  Therefore, the required point of intersection is (1, 2). Example 1, page 40

63. –11 2 3 45 –11 2 3 45 54321Example  Find the point of intersection of the straight lines that have equations y = x + 1 and y = – 2x + 4 Solution  The graph shows the point of intersection (1, 2) of the two lines: y x (1, 2) L1L1L1L1 L2L2L2L2 Example 1, page 40

64. Applied Example: Break-Even Level  Prescott manufactures its products at a cost of $4 per unit and sells them for $10 per unit.  If the firm’s fixed cost is $12,000 per month, determine the firm’s break-even point. Solution  The revenue function R and the cost function C are given respectively by  Setting R(x) = C(x), we obtain Applied Example 2, page 41

65. Applied Example: Break-Even Level  Prescott manufactures its products at a cost of $4 per unit and sells them for $10 per unit.  If the firm’s fixed cost is $12,000 per month, determine the firm’s break-even point. Solution  Substituting x = 2000 into R(x) = 10x gives  So, Prescott’s break-even point is 2000 units of the product, resulting in a break-even revenue of $20,000 per month. Applied Example 2, page 41

66. Applied Example: Market Equilibrium  The management of ThermoMaster, which manufactures an indoor-outdoor thermometer at its Mexico subsidiary, has determined that the demand equation for its product is where p is the price of a thermometer in dollars and x is the quantity demanded in units of a thousand.  The supply equation of these thermometers is where x (in thousands) is the quantity that ThermoMaster will make available in the market at p dollars each.  Find the equilibrium quantity and price. Applied Example 6, page 44

67. Applied Example: Market Equilibrium Solution  We need to solve the system of equations for x and p.  Let’s solve the first equation for p in terms of x: Applied Example 6, page 44

68. Applied Example: Market Equilibrium Solution  We need to solve the system of equations for x and p.  Now we substitute the value of p into the second equation: Applied Example 6, page 44

69. Applied Example: Market Equilibrium Solution  We need to solve the system of equations for x and p.  Finally, we substitute the value x = 5/2 into the first equation that we already solved: Applied Example 6, page 44

70. Applied Example: Market Equilibrium Solution  We conclude that the equilibrium quantity is 2500 units and the equilibrium price is $5.83 per thermometer. Applied Example 6, page 44

71. 1.5 The Method of Least Squares

72.  In this section, we describe a general method known as the method for least squares for determining a straight line that, in a sense, best fits a set of data points when the points are scattered about a straight line. The Method of Least Squares

73. 510 The Method of Least Squares  Suppose we are given five data points P 1 (x 1, y 1 ), P 2 (x 2, y 2 ), P 3 (x 3, y 3 ), P 4 (x 4, y 4 ), and P 5 (x 5, y 5 ) describing the relationship between two variables x and y.  By plotting these data points, we obtain a scatter diagram: x y P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P5P5P5P5 105

74. 105 510 The Method of Least Squares  Suppose we try to fit a straight line L to the data points P 1, P 2, P 3, P 4, and P 5.  The line will miss these points by the amounts d 1, d 2, d 3, d 4, and d 5 respectively. x y L

75. 510 The Method of Least Squares  The principle of least squares states that the straight line L that fits the data points best is the one chosen by requiring that the sum of the squares of d 1, d 2, d 3, d 4, and d 5, that is be made as small as possible. be made as small as possible. x y L 105

76. The Method of Least Squares  Suppose we are given n data points: P 1 (x 1, y 1 ), P 2 (x 2, y 2 ), P 3 (x 3, y 3 ),..., P n (x n, y n )  Then, the least-squares (regression) line for the data is given by the linear equation y = f(x) = mx + b where the constants m and b satisfy the equations andsimultaneously.  These last two equations are called normal equations.

77. Example  Find the equation of the least-squares line for the data P 1 (1, 1), P 2 (2, 3), P 3 (3, 4), P 4 (4, 3), and P 5 (5, 6) Solution  Here, we have n = 5 and x 1 = 1x 2 = 2x 3 = 3x 4 = 4x 5 = 5 y 1 = 1y 2 = 3y 3 = 4y 4 = 3y 5 = 6 xy x2x2x2x2xy 1111 2346 34912 431612 562530 15175561  Before using the equations it is convenient to summarize these data in the form of a table: Example 1, page 53

78. Example  Find the equation of the least-squares line for the data P 1 (1, 1), P 2 (2, 3), P 3 (3, 4), P 4 (4, 3), and P 5 (5, 6) Solution  Here, we have n = 5 and x 1 = 1x 2 = 2x 3 = 3x 4 = 4x 5 = 5 y 1 = 1y 2 = 3y 3 = 4y 4 = 3y 5 = 6  Using the table to substitute in the second equation we get Example 1, page 53

79. Example  Find the equation of the least-squares line for the data P 1 (1, 1), P 2 (2, 3), P 3 (3, 4), P 4 (4, 3), and P 5 (5, 6) Solution  Here, we have n = 5 and x 1 = 1x 2 = 2x 3 = 3x 4 = 4x 5 = 5 y 1 = 1y 2 = 3y 3 = 4y 4 = 3y 5 = 6  Using the table to substitute in the first equation we get Example 1, page 53

80. Example  Find the equation of the least-squares line for the data P 1 (1, 1), P 2 (2, 3), P 3 (3, 4), P 4 (4, 3), and P 5 (5, 6) Solution  Now we need to solve the simultaneous equations  Solving the first equation for b gives Example 1, page 53

81. Example  Find the equation of the least-squares line for the data P 1 (1, 1), P 2 (2, 3), P 3 (3, 4), P 4 (4, 3), and P 5 (5, 6) Solution  Now we need to solve the simultaneous equations  Substituting b into the second equation gives Example 1, page 53

82. Example  Find the equation of the least-squares line for the data P 1 (1, 1), P 2 (2, 3), P 3 (3, 4), P 4 (4, 3), and P 5 (5, 6) Solution  Now we need to solve the simultaneous equations  Finally, substituting the value m = 1 into the first equation that we already solved gives Example 1, page 53

83. Example  Find the equation of the least-squares line for the data P 1 (1, 1), P 2 (2, 3), P 3 (3, 4), P 4 (4, 3), and P 5 (5, 6) Solution  Now we need to solve the simultaneous equations  Thus, we find that m = 1 and b = 0.4.  Therefore, the required least-squares line is Example 1, page 53

84. Example  Find the equation of the least-squares line for the data P 1 (1, 1), P 2 (2, 3), P 3 (3, 4), P 4 (4, 3), and P 5 (5, 6) Solution  Below is the graph of the required least-squares line y = x + 0.4 654321 12345123451234512345 x y L Example 1, page 53

85. Applied Example: U.S. Health-Care Expenditures  Because the over-65 population will be growing more rapidly in the next few decades, health-care spending is expected to increase significantly in the coming decades.  The following table gives the U.S. health expenditures (in trillions of dollars) from 2005 through 2010:  Find a function giving the U.S. health-care spending between 2005 and 2010, using the least-squares technique. Year, t Year, t012345 Expenditure, y Expenditure, y2.002.172.342.502.692.90 Applied Example 3, page 55

86. Applied Example: U.S. Health-Care Expenditures Solution  The calculations required for obtaining the normal equations are summarized in the following table:  Use the table to obtain the second normal equation: ty t2t2t2t2ty02.0000 12.171 2.17 2.17 22.344 4.68 4.68 32.509 7.50 7.50 42.6916 10.76 10.76 52.9025 14.50 14.50 15 14.60 14.6055 39.61 39.61 Applied Example 3, page 55

87. Applied Example: U.S. Health-Care Expenditures Solution  The calculations required for obtaining the normal equations are summarized in the following table:  Use the table to obtain the first normal equation: ty t2t2t2t2ty02.0000 12.171 2.17 2.17 22.344 4.68 4.68 32.509 7.50 7.50 42.6916 10.76 10.76 52.9025 14.50 14.50 15 14.60 14.6055 39.61 39.61 Applied Example 3, page 55

88. Applied Example: U.S. Health-Care Expenditures Solution  Now we solve the simultaneous equations  Solving the first equation for b gives Applied Example 3, page 55

89. Applied Example: U.S. Health-Care Expenditures Solution  Now we solve the simultaneous equations  Substituting b into the second equation gives Applied Example 3, page 55

90. Applied Example: U.S. Health-Care Expenditures Solution  Now we solve the simultaneous equations  Finally, substituting the value m ≈ 0.1777 into the first equation that we already solved gives Applied Example 3, page 55

91. Applied Example: U.S. Health-Care Expenditures Solution  Now we solve the simultaneous equations  Thus, we find that m ≈ 0.1777 and b ≈ 1.9891.  Therefore, the required least-squares function is Applied Example 3, page 55

92. End of Chapter