= m + b slope -intercept This is called the slope-intercept form. 3 = m + b This is called the slope-intercept form. = 5 + 10 = 10 + 5 P = -0.2Q + 100 4 Page 2
= m + b -intercept b -intercept = point where line crosses the -ais Set = 0 and solve for = m 0 + b = b 5 = m + b b -b/m -intercept = point where line crosses the -ais Set = 0 and solve for 0 = m + b m + b = 0 m = -b = -b/m 6 Page 3
= m + b slope ( 2, 2 ) ( 1, 1 ) rise 1 2 2 1 Slope = = = = run 1 2 2 1 7 Lines Can Have Negative or Positive Slope. Negative Slope... Positive Slope rise 1 2 2 1 Slope = = = = run 1 2 2 1 8 Page 4
Etreme Slopes Zero and Infinit / / = /0 = Infinit / / = 0/ = Zero 9 Eercises (4 questions) 1. Label the lines in Figure 1 with a slope of 1/2, 1, and 2. Label the lines in Figure 2 with a slope of 1/2, -1, and 2. 2. Find the slope, -intercept, and -intercept of the following lines. Draw each line. (a) = 10 (b) = -4 + 64 (c) 2 + = 50 (d) = -0.25 + 16 FIGURE 1 FIGURE 2 10 Page 5
Eercises (4 questions) 3. A line representing the relationship between quantit demanded and price is given b: Q = 200 10P. Describe the slope of this line using the variables Q and P (instead of and ). What is the numerical value of the slope? 4. Refer to the line drawn in Figure 3. What is the equation of this line? What is the numerical value of the slope? Describe the slope of this line using the variables P and Q. P 20 FIGURE 3 200 Q 11 Eercises: **Answers** 1. The lines are labeled below. If ou think of the slope in absolute value terms (i.e., -2 becomes 2), then the largest slope (2) is associated with the steepest line, whether it is a positivel sloped or negativel sloped line. 2 1 ½ FIGURE 1 FIGURE 2 -½ -2-1 12 Page 6
Eercises: **Answers** 2. The lines are drawn on the net slide. (a) = 10: slope = 10; -intercept = 0; -intercept = 0. (b) = -4 + 64: slope = -4, -intercept = 64; -intercept = 16. (c) 2 + = 50: First rewrite the equation so that it conforms with = m + b. Isolating on the left-hand side and dividing through b 2 ields: = -0.5 + 25. Therefore, the slope = -0.5; -intercept = 25; -intercept = 50. (d) = -0.25 + 16: First rewrite the equation as we did above. The equation becomes = -4 + 64. This is the identical equation to part (b), with the identical slope and intercepts. The onl difference is that the equation began with on the left-hand side. 13 Eercises: **Answers** 2. The lines are sketched below (not to scale). (a) = 10 (b) & (d) = -4 + 64 64 0 25 16 (c) 2 + = 50 ( = -0.5 + 25) 50 14 Page 7
Eercises: **Answers** 3. The slope of a line is rise/run where the rise variable is on the left-hand side of the equation and the run variable is on the right. For the equation Q = 200 10P, we describe the slope in general terms as Q/ P. The numerical value of the slope is 10. Recall our discussion of price elasticit in the Basics module. The formula we derived was: Price elasticit = % Q % P = [ Q/ P ] * [ P 1 / Q 1 ] Calculate the price elasticit at P = 15 using the demand equation Q = 200 10P. If P = 15, then Q = 200 10(15) = 50. Now appl the formula: Price elasticit = -10(15/50) = -3 15 Eercises: **Answers** 4. Figure 3 shows that the vertical intercept (in this case, the P-intercept, since P is on the -ais) is 20. To calculate the slope, divide the rise b the run: 20/200 or 0.1. The equation of the line is therefore P = 20 0.1Q. In general notation, we describe the slope as P/ Q. P 20 FIGURE 3 200 Q 16 Page 8
Eercises: **Answers** 4. Figure 3 shows that the vertical intercept (in this case, the P-intercept, since P is on the -ais) is 20. To calculate the slope, divide the rise b the run: 20/200 or 0.1. The equation of the line is therefore P = 20 0.1Q. In general notation, we describe the slope as P/ Q. The onl difference between Eercise 3 (Q = 200 10P) and Eercise 4 (P = 20 0.1Q) is that the equation is inverted in one case Q is on the left and in the other case P is on the left, but it is the identical equation. The numerical value of the slope is 10 in one case and 0.1 in the other; one slope is the reciprocal of the other. The general notation for the slope is also the reciprocal: Q/ P when Q is on the left and P/ Q when P is on the left. 17 Rotations and Shifts A line can be described b its slope and -intercept. If onl the slope changes, then the line rotates. If onl the -intercept changes, the line shifts (a parallel shift). = 10 + 3000 = 2 + 5000 = 2 + 3000 5000 = 2 + 3000 3000 3000 18 Page 9
Rotations and Shifts Of course, both the slope and -intercept can change simultaneousl. = 10 + 5000 5000 = 2 + 3000 3000 19 Rotations and Shifts Eercises In each case below describe the difference (shift, rotation, or both) between the original line and the new line. It is not necessar to draw the lines. (a) = 10 and = 10-5 (b) = -4 + 64 and = -2 + 64 (c) = 0.1 + 10 and = 0.2 + 15 20 Page 10
Rotations and Shifts Eercises: **Answers** (a) = 10 and = 10 5 A parallel shift downward (b 5 units). (b) = -4 + 64 and = -2 + 64 A rotation: the new line still goes through the point (0, 64) but is flatter. (c) = 0.1 + 10 and = 0.2 + 15 A shift upward (b 5 units), as well as a rotation: the new line has a larger -intercept and is steeper. 21 Deriving a Line from Data Given All The Information Use the information below to describe the relationship between time and the value of the computer sstem as a linear equation. Write the equation of the line and provide a sketch. Value of Computer Computer Sstem Initial Cost $12,000 12,000 Useful Life 8 ears Depreciation Rate $1,500 per ear = -1,500 + 12,000 8 Age of Computer 22 Page 11
Deriving a Line from Data Given Two Points Use the data in the table to find the equation of the line. Draw the line and label the - intercept and the -intercept. 50 50 55 70 Four steps to finding the equation of a line given two points: 1. Slope = rise/run = (70 50)/(55 50) = 20/5 = 4 2. Write what ou have found so far: = 4 + b 3. Substitute in either point to find the -intercept 50 = 4(50) + b b = -150 or, 70 = 4(55) + b b = -150 4. Write the finished equation: = 4-150 23 Deriving a Line from Data Given Two Points Use the data in the table to find the equation of the line. Draw the line and label the - intercept and the -intercept. Solve for the intercepts and draw the line: 50 50 55 70 = 4-150 = 0 = 150 = 0 0 = 4 150 = 37.5-150 37.5 24 Page 12
Deriving a Line from Data Eercises 1. Write the equation of a line with a slope of 2 passing through the point (0,10). 2. Write the equation of a line with a slope of 2 passing through the point (5,40). 3. Write the equation of a line passing through (25,44) and (75,42). 4. The table below shows 2002 data for some U.S. Treasur securities. You would like to know the ield on a 1-ear Treasur bill, but the data are missing. Assume that there is a linear relationship between the time to maturit and the ield. What is our prediction of the one-ear ield? Maturit Yield (% per ear) 6 months 1.89 2 ears 3.01 25 Deriving a Line from Data Eercises: **Answers** 1. You are given the slope of 2. Also, knowing that the point (0,10) is on the line tells ou that the -intercept is 10. The line is therefore = 2 + 10. 2. Find the equation of a line with a slope of 2 passing through the point (5,40). Write the line given what ou know: = 2 + b Now substitute in the point ou are given and find the -intercept: 40 = 2(5) + b b = 30 Therefore, = 2 + 30 26 Page 13
Deriving a Line from Data Eercises: **Answers** 3. Given two points: (25,44) and (75,42). There are four steps to finding the equation of a line given two points: a) Slope = rise/run = (44 42)/(25 75) = 2/-50 = -0.04 b) Write what ou have found so far: = -0.04 + b c) Substitute in either point to find the -intercept 44 = -0.04(25) + b b = 45 d) Write the finished equation: = 45 0.04 27 Deriving a Line from Data Eercises: **Answers** 4. What is our prediction of the one-ear ield? In order to fit a line through these points ou first have to change the maturit variable into common units. Maturit Yield (% per ear) 6 months 1.89 2 ears 3.01 28 Page 14
Deriving a Line from Data Eercises: **Answers** 4. What is our prediction of the one-ear ield? In order to fit a line through these points ou first have to change the maturit variable into common units. If, for eample, ou want to define maturit in ears, then rewrite the table as: Maturit (ears) 0.5 1.89 2 3.01 Yield (% per ear) 29 Deriving a Line from Data Eercises: **Answers** 4. Maturit (ears) 0.5 1.89 2 3.01 Yield (% per ear) Now we can proceed. Let = maturit and = ield. The slope is rise/run = (3.01 1.89)/(2 0.5) = 0.75 (rounding). Use either point to find the -intercept, e.g., 3.01 = 0.75(2) + b b = 1.51. The equation is: Yield = 0.75*Maturit + 1.51 The ield for a securit with a 1-ear maturit is: Yield = 0.75*1 + 1.51 = 2.26 (the actual 1-ear ield was 2.51) 30 Page 15
Application Forecasting Trends in Sales In the last eercise we used a linear equation to find missing data. Here we will use a linear equation to project into the future. The sales of a local software compan increased from $715,000 in 1999 to $1,210,000 in 2001. The store manager believes that sales have been increasing linearl with time. (What does that mean in English?) We want to epress sales, S, as a linear function of time, t. 31 First, find m: Application Forecasting Trends in Sales S = mt + b Setup: Let 1999 = 0, 2000 = 1, 2001 = 2, and so on; t 0 = 1999, t 1 = 2000, and t 2 = 2001; S 0 = 715,000, S 2 = 1,210,000 m = = S t 1,210,000 715,000 = 2 0 = 247,500 32 Page 16
Application Forecasting Trends in Sales So far, we have S = 247,500t + b Now find b: 715,000 = 247,500(0) + b, or b = 715,000 Therefore, S(t) = 247,500t + 715,000 33 Application Forecasting Trends in Sales Use the line describing historical software sales to predict what sales will be in 2010. S ($1000s) S(t) = 247,500t + 715,000 3,437.5. 2010 is t = 11 715 S(11) = 247,500(11)+715,000 = $3,437,500 11 t 34 Page 17
Application Forecasting Trends in Sales Is our forecast reliable? Sales ($1000s) Sales ($1000s) Time Time Etrapolation (beond the sample) is much harder to do than interpolation (between points in the sample). Have to adjust for repeated seasonal variation, long-term business ccles, and so on. 35 Page 18