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1 Straight Lines and Linear Functions The Cartesian Coordinate System Straight Lines Linear Functions and Mathematical Models Intersection of Straight Lines The Method of Least Squares 1.1 The Cartesian Coordinate System y P(x, y) C(h, k) k r x h The Cartesian Coordinate System 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 Negative Direction Positive Direction –4 –3 –2 –1 0 1 2 3 4 2 3 p The Cartesian Coordinate System The Cartesian coordinate system extends this concept to a plane (two dimensional space) by adding a vertical axis. 4 3 2 1 –4 –3 –2 –1 1 2 3 4 –1 –2 –3 –4 The Cartesian Coordinate System The horizontal line is called the x-axis, and the vertical line is called the y-axis. y 4 3 2 1 x –4 –3 –2 –1 1 2 3 4 –1 –2 –3 –4 The Cartesian Coordinate System The point where these two lines intersect is called the origin. y 4 3 2 1 Origin x –4 –3 –2 –1 1 2 3 4 –1 –2 –3 –4 The Cartesian Coordinate System In the x-axis, positive numbers are to the right and negative numbers are to the left of the origin. y 4 3 2 Negative Direction 1 Positive Direction x –4 –3 –2 –1 1 2 3 4 –1 –2 –3 –4 The Cartesian Coordinate System In the y-axis, positive numbers are above and negative numbers are below the origin. y 4 Positive Direction 3 2 1 x –4 –3 –2 –1 1 2 3 4 –1 Negative Direction –2 –3 –4 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). y (– 2, 4) 4 3 (4, 3) 2 1 x –4 –3 –2 –1 1 2 3 4 –1 (3, –1) (–1, – 2) –2 –3 –4 The Cartesian Coordinate System The axes divide the plane into four quadrants as shown below. y 4 Quadrant II 3 Quadrant I (–, +) (+, +) 2 1 x –4 –3 –2 –1 1 2 3 4 –1 Quadrant III –2 Quadrant IV (–, –) –3 (+, –) –4 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 P1(x1, y1) and P2(x2, y2) in the plane is given by d x2 x1 y2 y1 2 2 Examples Find the distance between the points (– 4, 3) and (2, 6). Solution Let P1(– 4, 3) and P2(2, 6) be points in the plane. We have x1 = – 4 y1 = 3 x2 = 2 y2 = 6 Using the distance formula, we have d x2 x1 y2 y1 2 2 2 ( 4) 6 3 2 2 62 32 45 3 5 Example 1, page 4 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 y we get P(x, y) x h y k r 2 2 C(h, k) k r Squaring both sides gives x h y k r2 2 2 x h Example 3, page 4 Equation of a Circle An equation of a circle with center C(h, k) and radius r is given by x h y k r2 2 2 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: x h y k r2 2 2 y x ( 1) y 3 22 2 2 (–1, 3) 3 x 1 y 3 4 2 2 2 x –1 Example 4, page 5 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: y x h y k r2 2 2 x 0 y 0 32 2 2 3 x x y 9 2 2 Example 4, page 5 1.2 Straight Lines Slope of a Vertical Line Let L denote the unique straight line that passes through the two distinct points (x1, y1) and (x2, y2). If x1 = x2, then L is a vertical line, and the slope is undefined. y L (x1, y1) (x2, y2) x Slope of a Nonvertical Line If (x1, y1) and (x2, y2) are two distinct points on a nonvertical line L, then the slope m of L is given by y y2 y1 m x x2 x1 y L (x2, y2) y2 – y1 = y (x1, y1) x2 – x1 = x x Slope of a Nonvertical Line If m > 0, the line slants upward from left to right. y m=1 L y = 1 x = 1 x Slope of a Nonvertical Line If m > 0, the line slants upward from left to right. y L m=2 y = 2 x = 1 x Slope of a Nonvertical Line If m < 0, the line slants downward from left to right. y m = –1 x = 1 y = –1 x L Slope of a Nonvertical Line If m < 0, the line slants downward from left to right. y m = –2 x = 1 y = –2 x L Examples Sketch the straight line that passes through the point (2, 5) and has slope – 4/3. Solution 1. Plot the point (2, 5). y 6 2. A slope of – 4/3 means x = 3 5 that if x increases by 3, (2, 5) 4 y decreases by 4. y = – 4 3 3. Plot the resulting 2 point (5, 1). 1 (5, 1) 4. Draw a line through x 1 2 3 4 5 6 the two points. L Examples Find the slope m of the line that goes through the points (–1, 1) and (5, 3). Solution Choose (x1, y1) to be (–1, 1) and (x2, y2) to be (5, 3). With x1 = –1, y1 = 1, x2 = 5, y2 = 3, we find y2 y1 3 1 2 1 m x2 x1 5 ( 1) 6 3 Example 2, page 11 Examples Find the slope m of the line that goes through the points (–2, 5) and (3, 5). Solution Choose (x1, y1) to be (–2, 5) and (x2, y2) to be (3, 5). With x1 = –2, y1 = 5, x2 = 3, y2 = 5, we find y2 y1 55 0 m 0 x2 x1 3 ( 2) 5 Example 3, page 11 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 6 (–2, 5) (3, 5) L 4 3 m=0 2 1 x –2 –1 1 2 3 4 Example 3, page 11 Parallel Lines Two distinct lines are parallel if and only if their slopes are equal or their slopes are undefined. Example Let L1 be a line that passes through the points (–2, 9) and (1, 3), and let L2 be the line that passes through the points (– 4, 10) and (3, – 4). Determine whether L1 and L2 are parallel. Solution The slope m1 of L1 is given by 39 m1 2 1 ( 2) The slope m2 of L2 is given by 4 10 m2 2 3 ( 4) Since m1 = m2, the lines L1 and L2 are in fact parallel. Example 4, page 12 Equations of Lines Let L be a straight line parallel to the y-axis. y Then L crosses the x-axis at L some point (a, 0) , with the (a, y ) x-coordinate given by x = a, where a is a real number. Any other point on L has the form (a, y), where y is an appropriate number. (a, 0) x The vertical line L can therefore be described as x=a Equations of Lines Let L be a nonvertical line with a slope m. Let (x1, y1) be a fixed point lying on L, and let (x, y) be a variable point on L distinct from (x1, y1). Using the slope formula by letting (x, y) = (x2, y2), we get y y1 m x x1 Multiplying both sides by x – x1 we get y y1 m( x x1 ) Point-Slope Form An equation of the line that has slope m and passes through point (x1, y1) is given by y y1 m( x x1 ) Examples Find an equation of the line that passes through the point (1, 3) and has slope 2. Solution Use the point-slope form y y1 m( x x1 ) Substituting for point (1, 3) and slope m = 2, we obtain y 3 2( x 1) Simplifying we get 2x y 1 0 Example 5, page 13 Examples Find an equation of the line that passes through the points (–3, 2) and (4, –1). Solution The slope is given by y y 1 2 3 m 2 1 x2 x1 4 (3) 7 Substituting in the point-slope form for point (4, –1) and slope m = – 3/7, we obtain 3 y 1 ( x 4) 7 7 y 7 3x 12 3x 7 y 5 0 Example 6, page 14 Perpendicular Lines If L1 and L2 are two distinct nonvertical lines that have slopes m1 and m2, respectively, then L1 is perpendicular to L2 (written L1 ┴ L2) if and only if 1 m1 m2 Example Find the equation of the line L1 that passes through the point (3, 1) and is perpendicular to the line L2 described by y 3 2( x 1) Solution L2 is described in point-slope form, so its slope is m2 = 2. Since the lines are perpendicular, the slope of L1 must be m1 = –1/2 Using the point-slope form of the equation for L1 we obtain 1 y 1 ( x 3) 2 2 y 2 x 3 x 2y 5 0 Example 7, page 14 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 y-intercept (0, b) x-intercept x (a, 0) L 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 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 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. 3x 4 y 8 4 y 8 3 x 3 y x2 4 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 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 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. General Form of a Linear Equation An equation of a straight line is a linear equation; conversely, every linear equation represents a straight line. 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 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). y L 1 (4, 0) x 1 2 3 4 5 6 –1 –2 –3 (0, – 3) –4 Example 12, page 17 Equations of Straight Lines Vertical line: x=a Horizontal line: y=b Point-slope form: y – y1 = m(x – x1) Slope-intercept form: y = mx + b General Form: Ax + By + C = 0 1.3 Linear Functions and Mathematical Models Real - world Formulate Mathematical problem model Test Solve - Solution of real Solution of world Problem Interpret mathematical model 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 Formulate Mathematical model problem Test Solve Solution of real- Solution of world Problem Interpret mathematical model 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. Example Let x and y denote the radius and area of a circle, respectively. From elementary geometry we have y = px2 This equation defines y as a function of x, since for each admissible value of x there corresponds precisely one number y = px2 giving the area of the circle. The area function may be written as f(x) = px2 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) = p52)= 25p 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). Linear Function The function f defined by f ( x) mx b where m and b are constants, is called a linear function. 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: Year 2005 2006 2007 2008 2009 2010 Expenditure 2.00 2.17 2.34 2.50 2.69 2.90 A mathematical model giving the approximate U.S. health- care expenditures over the period in question is given by S (t ) 0.178t 1.989 where t is measured in years, with t = 0 corresponding to 2005. Applied Example 1, page 29 Applied Example: U.S. Health-Care Expenditures We have S (t ) 0.178t 1.989 Year 2005 2006 2007 2008 2009 2010 Expenditure 2.00 2.17 2.34 2.50 2.69 2.90 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? Applied Example 1, page 29 Applied Example: U.S. Health-Care Expenditures We have S (t ) 0.178t 1.989 Year 2005 2006 2007 2008 2009 2010 Expenditure 2.00 2.17 2.34 2.50 2.69 2.90 Solution a. The graph of the given data and of the function S is: S(t) 3.0 S (t ) 0.178t 1.989 2.8 2.6 2.4 2.2 2.0 t 1 2 3 4 5 Applied Example 1, page 29 Applied Example: U.S. Health-Care Expenditures We have S (t ) 0.178t 1.989 Year 2005 2006 2007 2008 2009 2010 Expenditure 2.00 2.17 2.34 2.50 2.69 2.90 Solution b. The projected U.S. health-care expenditures in 2011 is S (6) 0.178(6) 1.989 3.057 or approximately $3.06 trillion. Applied Example 1, page 29 Applied Example: U.S. Health-Care Expenditures We have S (t ) 0.178t 1.989 Year 2005 2006 2007 2008 2009 2010 Expenditure 2.00 2.17 2.34 2.50 2.69 2.90 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. Applied Example 1, page 29 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 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, C ( x) 20 x 20,000 R( x) 30 x P( x ) R( x ) C ( x ) 30 x (20 x 20,000) 10 x 20,000 Applied Example 2, page 31 1.4 Intersections of Straight Lines Finding the Point of Intersection Suppose we are given two straight lines L1 and L2 with equations y = m1x + b1 and y = m2x + b2 (where m1, b1, m2, and b2 are constants) that intersect at the point P(x0, y0). The point P(x0, y0) lies on the line L1 and so satisfies the equation y = m1x + b1. The point P(x0, y0) also lies on the line L2 and so satisfies y = m2x + b2 as well. Therefore, to find the point of intersection P(x0, y0) of the lines L1 and L2, we solve for x and y the system composed of the two equations y = m1x + b1 and y = m2x + b2 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 x 1 2 x 4 3x 3 x 1 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 Example 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 5 L1 4 3 2 (1, 2) 1 x –1 1 2 3 4 5 L2 Example 1, page 40 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 R( x) 10 x and C( x) 4 x 12,000. Setting R(x) = C(x), we obtain 10 x 4 x 12,000 6 x 12,000 x 2000 Applied Example 2, page 41 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 R(2000) 10(2000) 20,000 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 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 5x 3 p 30 0 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 52 x 30 p 45 0 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 Applied Example: Market Equilibrium Solution We need to solve the system of equations 5x 3 p 30 0 52 x 30 p 45 0 for x and p. Let’s solve the first equation for p in terms of x: 5x 3 p 30 0 3 p 5x 30 5 p x 10 3 Applied Example 6, page 44 Applied Example: Market Equilibrium Solution We need to solve the system of equations 5x 3 p 30 0 52 x 30 p 45 0 for x and p. Now we substitute the value of p into the second equation: 5 52 x 30 x 10 45 0 3 52 x 50x 300 45 0 102x 255 0 102 x 255 255 5 x Applied Example 6, page 44 102 2 Applied Example: Market Equilibrium Solution We need to solve the system of equations 5x 3 p 30 0 52 x 30 p 45 0 for x and p. Finally, we substitute the value x = 5/2 into the first equation that we already solved: 5 5 p 10 3 2 25 10 6 35 5.83 Applied Example 6, page 44 6 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 1.5 The Method of Least Squares The Method of Least Squares 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 Suppose we are given five data points P1(x1, y1), P2(x2, y2), P3(x3, y3), P4(x4, y4), and P5(x5, y5) describing the relationship between two variables x and y. By plotting these data points, we obtain a scatter diagram: y P5 10 P3 P2 P4 5 P1 x 5 10 The Method of Least Squares Suppose we try to fit a straight line L to the data points P1, P2, P3, P4, and P5. The line will miss these points by the amounts d1, d2, d3, d4, and d5 respectively. y d5 L 10 d3 d4 5 d2 d1 x 5 10 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 d1, d2, d3, d4, and d5, that is d12 d 2 d 32 d 4 d 52 2 2 be made as small as possible. y d5 L 10 d3 d4 5 d2 d1 x 5 10 The Method of Least Squares Suppose we are given n data points: P1(x1, y1), P2(x2, y2), P3(x3, y3), . . . , Pn(xn, yn) 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 ( x1 x2 x3 xn )m nb y1 y2 y3 yn and ( x12 x2 x3 xn )m ( x1 x2 x3 xn )b 2 2 2 y1 x1 y2 x2 y3 x3 yn xn simultaneously. These last two equations are called normal equations. Example Find the equation of the least-squares line for the data P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6) Solution Here, we have n = 5 and x1 = 1 x2 = 2 x3 = 3 x4 = 4 x5 = 5 y1 = 1 y2 = 3 y3 = 4 y4 = 3 y5 = 6 Before using the x y x2 xy equations it is convenient 1 1 1 1 to summarize these data 2 3 4 6 in the form of a table: 3 4 9 12 4 3 16 12 5 6 25 30 15 17 55 61 Example 1, page 53 Example Find the equation of the least-squares line for the data P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6) Solution Here, we have n = 5 and x1 = 1 x2 = 2 x3 = 3 x4 = 4 x5 = 5 y1 = 1 y2 = 3 y3 = 4 y4 = 3 y5 = 6 Using the table to substitute in the second equation we get ( x12 x2 x3 ... xn )m ( x1 x2 x3 ... xn )b 2 2 2 y1 x1 y2 x2 y3 x3 ... yn xn 55m 15b 61 Example 1, page 53 Example Find the equation of the least-squares line for the data P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6) Solution Here, we have n = 5 and x1 = 1 x2 = 2 x3 = 3 x4 = 4 x5 = 5 y1 = 1 y2 = 3 y3 = 4 y4 = 3 y5 = 6 Using the table to substitute in the first equation we get ( x1 x2 x3 ... xn )m nb y1 y2 y3 ... yn 15m 5b 17 Example 1, page 53 Example Find the equation of the least-squares line for the data P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6) Solution Now we need to solve the simultaneous equations 15m 5b 17 55m 15b 61 Solving the first equation for b gives 15m 5b 17 5b 15m 17 17 b 3m 5 Example 1, page 53 Example Find the equation of the least-squares line for the data P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6) Solution Now we need to solve the simultaneous equations 15m 5b 17 55m 15b 61 Substituting b into the second equation gives 55m 15b 61 17 55m 15 3m 61 5 55m 45m 51 61 10m 10 Example 1, page 53 m 1 Example Find the equation of the least-squares line for the data P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6) Solution Now we need to solve the simultaneous equations 15m 5b 17 55m 15b 61 Finally, substituting the value m = 1 into the first equation that we already solved gives 17 b 3m 5 17 3(1) 5 Example 1, page 53 0.4 Example Find the equation of the least-squares line for the data P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6) Solution Now we need to solve the simultaneous equations 15m 5b 17 55m 15b 61 Thus, we find that m = 1 and b = 0.4. Therefore, the required least-squares line is y mx b y x 0.4 Example 1, page 53 Example Find the equation of the least-squares line for the data P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6) Solution Below is the graph of the required least-squares line y = x + 0.4 y 6 L 5 4 3 2 1 x 1 2 3 4 5 Example 1, page 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 U.S. health expenditures (in trillions of dollars) from 2005 through 2010: Year, t 0 1 2 3 4 5 Expenditure, y 2.00 2.17 2.34 2.50 2.69 2.90 Find a function giving the U.S. health-care spending between 2005 and 2010, using the least-squares technique. Applied Example 3, page 55 Applied Example: U.S. Health-Care Expenditures Solution The calculations required for obtaining the normal equations are summarized in the following table: t y t2 ty 0 2.00 0 0 1 2.17 1 2.17 2 2.34 4 4.68 3 2.50 9 7.50 4 2.69 16 10.76 5 2.90 25 14.50 15 14.60 55 39.61 Use the table to obtain the second normal equation: (t12 t2 t3 ... tn )m (t1 t2 t3 ... tn )b 2 2 2 y1t1 y2t2 y3t3 ... yntn Applied Example 3, page 55 55m 15b 39.61 Applied Example: U.S. Health-Care Expenditures Solution The calculations required for obtaining the normal equations are summarized in the following table: t y t2 ty 0 2.00 0 0 1 2.17 1 2.17 2 2.34 4 4.68 3 2.50 9 7.50 4 2.69 16 10.76 5 2.90 25 14.50 15 14.60 55 39.61 Use the table to obtain the first normal equation: (t1 t2 t3 ... tn )m nb y1 y2 y3 ... yn 15m 6b 14.60 Applied Example 3, page 55 Applied Example: U.S. Health-Care Expenditures Solution Now we solve the simultaneous equations 15m 6b 14.60 55m 15b 39.61 Solving the first equation for b gives 15m 6b 14.60 6b 15m 14.60 b 2.5m 2.4333 Applied Example 3, page 55 Applied Example: U.S. Health-Care Expenditures Solution Now we solve the simultaneous equations 15m 6b 14.60 55m 15b 39.61 Substituting b into the second equation gives 55m 15b 39.61 55m 15 2.5m 2.4333 39.61 55m 37.5m 36.4995 39.61 17.5m 3.1105 m 0.1777 Applied Example 3, page 55 Applied Example: U.S. Health-Care Expenditures Solution Now we solve the simultaneous equations 15m 6b 14.60 55m 15b 39.61 Finally, substituting the value m ≈ 0.1777 into the first equation that we already solved gives b 2.5m 2.4333 2.5(0.1777) 2.4333 1.9891 Applied Example 3, page 55 Applied Example: U.S. Health-Care Expenditures Solution Now we solve the simultaneous equations 15m 6b 14.60 55m 15b 39.61 Thus, we find that m ≈ 0.1777 and b ≈ 1.9891. Therefore, the required least-squares function is S (t ) y 1.178t 1.989 Applied Example 3, page 55 End of Chapter

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