1. Quadratic Function Models Consider the Following Table of the Number of Americans That Are Over 100 in Various Years

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CHAPTER 5

Nonlinear Models

1. Quadratic Function Models Consider the following table of the number of Americans that are over 100 in various years. Year Number (in 1000’s) 1994 50 1996 56 1998 65 2000 75 2002 94 2004 110 How many would you predict would be over 100 in 2008?

Let’s plot the data in our calculator, using L1 for Year and L2 for Number. Set a WINDOW of [1992, 2008] [0, 180]. ⇥

The data is more curved than linear, but not so much that linear regression is not an option. The line of best fit for linear regression is y = 6.057x 12, 033.229 or f(x) = 6.057x 12, 033.229. 78 1. QUADRATIC FUNCTION MODELS 79 Add this to the plot as Y1 and look at the graph. Evaluating f(x) at x = 2008 (2nd/CALC/1:value/X=2008/ENTER) gives f(2008) = 130. Visually, this appears to be an underestimation. Recall that regression equations work best with values of x within the boundaries of the data (1994 x 2004). Working outside these boundaries to make predictions is called extrap olation, and is often not as accurate as working inside the boundaries. Note also that r2 = .96 here, so there is a linear relationship, but how strong this holds beyond the boundaries of the data is questionable. What about finding a curve to fit the data? We will try the parabola which best fits the data. A parabola is the graph of a quadratic function y = ax2 + bx + c. We obtain the quadratic equation of best fit by using quadratic regression (STAT/CALC/5:QuadReg/VARS/Y-VARS/1:Function/1;Y1/ENTER). Note that r2 = .997 here, so we expect a very good fit: y = .402x2 1600.282x + 1593498.2.

This appears to be a good ﬁt and we expect a good approximation for 2008. If y = f(x), f(2008) = 157. 80 5. NONLINEAR MODELS Parabolas – graphs of y = ax2 + bx + c a > 0 concave up a < 0 concave down a = magnitude of a parabola. The greater the magnitude, the steeper the parab| | ola. On the TI: (1) Compare x2 4x + 2 and 4x2 4x + 2 on [ 3, 6] [ 3, 20]. ⇥ (2) Compare x2 3x + 5 and 3x2 3x + 5 on [ 6, 4] [ 10, 10]. ⇥ Notice that c (or more precisely, (0, c)), is the y-intercept, while b a↵ects the horizontal and vertical placement of the parabola. Since the roots of a parabola are the solutions of ax2 + bx + c = 0, they are

b pb2 4ac b pb2 4ac x = ± = 2a 2a ± 2a

b by the quadratic formula, and the x-coordinate of the vertex is x = , 2a b which is halfway between the roots. The vertical line x = is called the 2a axis of symmetry. 1. QUADRATIC FUNCTION MODELS 81 The vertex is the point on the graph where the function changes from decreasing to increasing or vice-versa. Example. Find the concavity, intercepts, and vertex of f(x) = 2x2 2.4x + 14.96 Solution f is concave down with a magnitude of 2. The axis of symmetry is b 2.4 x = = = 0.6. 2a 2( 2) Since f( 0.6) = 15.68, the vertex is ( .6, 15.68). The y-intercept is c = 14.96 = f(0). The x-intercepts are b pb2 4ac b pb2 4ac x = ± = = .6 ( 2.8) 2a 2a ± 2a ± or x = 3.4 and x = 2.2. 82 5. NONLINEAR MODELS Problem (Page 299 #8). We are given that the graph is a parabola, so it is the graph of f(x) = ax2 + bx + c. Three noncollinear points are sucient to provide 3 linear equations in a, b, and c that has a unique solution. The graph here appears to pass through ( 1, 1), (0, 1), and (1, 3). From f(0) = c = 1, the y-intercept, we have f(x) = ax2 + bx + 1. From f( 1) = a b + 1 = 1, we have a b = 0. From f(1) = a + b + 1 = 3, we have a + b = 2. Adding the last two equations on the right, we have 2a = 2 a = 1 b = 1. ) ) Thus f(x) = x2 + x + 1. Problem (Page 300 #14). We ﬁnd the unique quadratic function that ﬁts the three points (0, 6), (20, 68), and (40, 438), the middle point along with the two extremes. From f(0) = c = 6, the y-intercept, we have f(x) = ax2 + bx + 6. From f(20) = 400a + 20b + 6 = 68, we have 400a + 20b = 62. From f(40) = 1600a + 40b + 6 = 438, we have 1600a + 40b = 432. From the last two equations on the right, we have

800a 40b = 124 800a = 308 a = .385 (1600a + 40b = 432 ) ) ) ) 154 + 20b = 62 20b = 92 b = 4.6. ) ) Thus f(x) = .385x2 4.6x + 6. Note that f(10) = 1.5 = 16 and f(30) = 214.5 = 233, so the 2nd and 4th points are missed. 6 6 1. QUADRATIC FUNCTION MODELS 83

The graph on the left shows f(x) and the 5 points. Since health care costs are predicted to rise into the future, this model looks reasonably solid for predicting future costs despite the discrepancy at x = 10. The model on the right comes from using the ﬁrst three points, f(x) = .21x2 1.1x + 6, and clearly begins to underestimate costs after year 20. This is nota good model. 84 5. NONLINEAR MODELS Problem (Page 306 #36). The data is decreasing and concave up, so a quadratic model seems feasible. From quadratic regression, we get f(x) = 2.562x2 58.9x + 1321. Note that r2 = .995, so this is a good ﬁt.

Graphically, we also see a good ﬁt during a time when the active army was downsizing. However, the downsizing model will only last a couple more years in the future. 2. HIGHER POLYNOMIAL FUNCTION MODELS 85 Problem (Page 304 #28). The data does not appear to be quadratic. The quadratic equation of best ﬁt, f(x) = .024x2 + 5.46x + 71.84, only has r2 = .922, quite di↵erent from our usual .99’s. Also, while a parabola is concave up or down, the concavity appears to change here.

2. Higher Polynomial Function Models What kind of model can we use for the previous data? Since our quadratic model for this data is practically linear, let’s also rule out a linear model for now. Definition. An inflection point is a point where a function changes concavity. Note. (1) A line has no concavity. (2) A parabola has no inflection point. 86 5. NONLINEAR MODELS Definition. A cubic function is a polynomial of degree 3 and has the form f(x) = ax3 + bx2 + cx + d. Note. (1) A cubic function has exactly one inflection point. (2) If a > 0, we get one of these two shapes (concave down, followed by concave up).

(3) If a < 0, we get one of these two shapes (concave up, followed by concave down).

But the data of #28 has two regions of concave up and one of concave down for 2 inﬂection points. Definition. A quartic function is a polynomial of degree 4 and has the form f(x) = ax4 + bx3 + cx2 + dx + e. 2. HIGHER POLYNOMIAL FUNCTION MODELS 87 Note. (1) A quartic function may have 0 or 2 inﬂection points. (2) If a > 0, we get one of the three basic shapes below.

(3) If a < 0, we get one of the three basic shapes below.

Based on the above, a quartic model with a > 0 is suggested for our data. We use 7:QuartReg to do quartic regression. With r2 = 1, we have an excellent ﬁt. We have f(x) = .014513x4 .56287x3 + 6.59517x2 16.6583x + 72.3. The graph is on the next page. 88 5. NONLINEAR MODELS

Although the data does an adequate job of modeling the data for the years 1980– 2000, we need to be careful about extrapolating this data past 2000, especially given the steep rise of the curve after x = 20 (2000). So perhaps, in the end, a linear model might be the best predictor of the future. 2. HIGHER POLYNOMIAL FUNCTION MODELS 89 Problem (Page 329 #18). A plot of the data points shows one change of concavity, so we use cubic regression. We have f(x) = .4514x3 + 6.1247x2 6.1706x + 1.9944 with r2 = .998, suggesting a good ﬁt.

We see a good fit, but the projection for 2006, f(11) = 210, is troublesome, since it indicates a decline in internet usage at a time when it would be expected to grow. Perhaps linear regression would be better for predicting users for the near future. 90 5. NONLINEAR MODELS Problem (Page 324 #4). The data appears to be linear, but we are asked to use a quadratic, cubic, or quartic model, the simplest to fit the data. Since the data appears slightly concave down, we choose the quadratic model. We get f(x) = 13.9512x2 + 2777.65x + 29155.78 with r2 = .998, suggesting a good fit. Linear regression gives r2 = .997.

This is a model where extrapolation seems plausible over the next period of years. It appears salaries will reach $100,000 between years 2010–2011 (x = 30 or x = 31). f(30) = 99, 929 f(31) = 101, 856 3. EXPONENTIAL FUNCTION MODELS 91 3. Exponential Function Models Consider the following data points. x 0 5 10 15 20 25 30 f(x) 10 76 577 4379 33,253 252,512 1,917,511 We consider the polynomial models. 92 5. NONLINEAR MODELS

All of these models have regions of negative values and miss points by 50,000 or more, so they are unsatisfactory. Let’s look at something new – exponential regression. 3. EXPONENTIAL FUNCTION MODELS 93

Clearly, exponential regression gives an almost perfect ﬁt. Looking back to the data, x 0 5 10 15 20 25 30 f(x) 10 76 577 4379 33,253 252,512 1,917,511 we see equal gaps of 5 for x, but not equal gaps for f(x) (as in the linear case). However, consider successive ratios: 76 577 4379 33, 253 252, 512 = 7.6, = 7.59, = 7.59, = 7.59, = 7.59, 10 76 577 4379 33, 253 1, 917, 511 = 7.59 252, 512 We notice a property of exponential functions – for equal increments of x, the ratio of function values is constant (ignoring some rounding o↵ error). 94 5. NONLINEAR MODELS Definition. If a and b are real numbers with a = 0, b > 0, and b = 1, then the function 6 6 y = f(x) = abx is called an exponential function. The value b is the base of the exponential function. Note. (1) Exponential functions are often used to model growth and decay of a quantity. In such cases, x (or t, which is more commonly used), represents time, and since f(0) = ab0 = a 1 = a, · a is called the initial value of the function. (2) If a = 0, b = 0, or b = 1, we would have constant functions. (3) If b < 0, any noninteger powers would be complex or imaginary numbers. (4) a is the y-intercept of the graph of f(x). (5) For b > 1, b is called the growth factor. Example. f(x) = 10(1.5)x. As x increases by 1, f(x) increases by a factor of 1.5. f(0) = 10, f(1) = 15 = 1.5(10), f(2) = 22.5 = 1.5(15) As x increases, the graph moves away from the x-axis.

(6) For 0 < b < 1, b is called a decay factor. 1 x Example. f(x) = 10(2) . 1 As x increases by 1, f(x) decreases by a factor of 2. 1 1 f(0) = 10, f(1) = 5 = (10), f(2) = 2.5 = (5) 2 2 As x increases, the graph moves towards the x-axis. 3. EXPONENTIAL FUNCTION MODELS 95

Problem (Page 344 #6). y = 0.1(0.2)x. (a) a < 0 and 0 < b < 1 = increasing. ) (b) Concave down. (c) y-intercept is (0, 0.1). 96 5. NONLINEAR MODELS Problem (Page 344 #18). Two points with distinct y-coordinates determine a unique exponential function, y = abx. To find the two numbers a and b, we need two equations, i.e., two points. You may use any two points, but write the point with the larger x first. (5, 32) = 32 = ab5 ) (2, 4) = 4 = ab2 ) (divide) 8 = b3 = 81/3 = b = 2 = b. ) ) Thus y = a 2x. · (2, 4) = 4 = a 22 = 4 = 4a = a = 1. ) · ) ) Then y = 1 2x. · Problem (Page 344 #20). y = abx. (0, 10000) is the y-intercept, so 10000 = ab0 = a 1 = a. Thus · y = 10000bx. Now take any other point. 24.3 (5, 24.3) = 24.3 = 10000 b5 = b5 = = .00243 = ) · ) 10000 ) b = .002431/5 = .3. Then y = 10000(.3)x. 3. EXPONENTIAL FUNCTION MODELS 97 Problem (Page 345 #22). Since t increases by 5’s (evenly), we check successive ratios of I: 100.2 114.4 133.6 151.6 1.12, 1.14, 1.17, 1.13. 89.5 ⇡ 100.2 ⇡ 114.4 ⇡ 133.6 ⇡ Since these ratios are approximately the same, exponential regression makes sense. We get I(t) = 88.50876368(1.027197187)t. For 2005 (t = 25), we have I(25) = 173.1. This means that at the end of 2005, the price of wine consumed at home is expected to be 173.1% of its 1984 price. For example, a bottle of wine costing $20 in 1984 is expected to cost $34.62 in 2005. Problem (Page 303 #24). Since the points appear to be concave down, the quadratic model gives us f(x) = .052x2 + 3.177x + 34.644 with r2 = .994. 98 5. NONLINEAR MODELS Even though it fits the data well, it forecasts that the percentage of new homes with central air will go down, which doesn’t seem likely, so this function is not a good model. The quartic model also has no inflection points for some cases, so we try that and get f(x) = .0000365x4 .000139x3 .101x2 + 3.891x + 34 with r2 = 1.

This is an extremly good ﬁt, but would surpass 100% in just a few years, limiting its usefulness. What kind of curve does the shape of our points suggest? Our data would seem to approach 100 as a horizontal asymptote. The points suggest an exponential curve like the bottom curve on page 333. But all those curves have y = 0 as a horizontal asymptote, not y = 100. So we align the data.

We subtract 100 from each y-value in L2 and put the result in L3. You can do this easily by putting the cursor on L3 and typing 2nd 2 -100 ENTER. You then get the following graph using Plot2 with L1 and L3. 3. EXPONENTIAL FUNCTION MODELS 99

But ExpReg doesn’t work with negative y’s, so we multiply each y in L3 by 1 and put the result in L4. Put the cursor on L4 and type -1 2nd 3 ENTER. Plot3 with L1 and L4 gives a graph like the second one on page 333. Then ExpReg L1, L4, Y1 gives f(x) = 63.32(.954)x. Taking its negative and adding 100 (reversing what we did earlier) gives a ﬁnal model of f(x) = 63.32(.954)x + 100, whose graph follows. 100 5. NONLINEAR MODELS

Adjusting Y1 to the ﬁnal model also gives this graph as Plot1. Definition. Let y = abt model the amount of a quantity at time t years. The annual growth rate r of the quantity y is given by r = b 1. Similarly, the annual growth factor is b = r + 1. Note. (1) r is the decimal form of the percentage. (2) r > 0 is called appreciation. (3) r < 0 is called depreciation. 3. EXPONENTIAL FUNCTION MODELS 101 Problem (Page 345 #28). y = abt r = 16% = .16 = b = .16 + 1 = .84. ) y = a(.84)t From y(2002) = 599.99, we have 599.99 = a(.84)2002. But the calculator cannot handle .84 2002. ^ So let t = years after 2002. Then a = 599.99 = ) y = 599.99(.84)t. Since t = 5 for 2007, y(5) = 599.99(.84)5 = 250.92. Thus the TV would cost $250.92 in 2007. Problem (Page 346 #32). Solve (0.9)x = 0.5 graphically.

(1) Graph Y1 = 0.9 x. ^ (2) Graph Y2 = 0.5. (3) Set a WINDOW [0, 10] [0, 2]. ⇥ (4) Use 2nd/Calc/5:Intersect to get x = 6.5788. 102 5. NONLINEAR MODELS 4. Logarithmic Function Models How would we solve an equation like the one in the previous problem, (0.9)x = 0.5, algebraically, where the variable is an exponent? Let’s consider the function y = 3x and the following points on its graph: x 0 1 2 3 4 5 y 1 3 9 27 81 243

For what value of x is y = 100, or equivalently, solve 3x = 100. 4. LOGARITHMIC FUNCTION MODELS 103 x is the exponent we place on 3 to get 100. x = log base 3 of 100 |{z}Definition.| {zLet b and x}b|{z}e real| num{z bers} with b > 0, b = 1, and x > 0. The function 6 y = logb(x) [or y = logb x] is called a logarithmic function. The value b is called the base of the logarithmic function. We read the expression logb(x) as “log base b of x.” Note. A logarithm is an exponent, and every exponent needs a base.

We view the graph of x = g(y) = log3 y along with an illustration of log3 100, which is about 4.2. A partial list of points is y 1 3 9 27 81 243 x 0 1 2 3 4 5

Example. Find y = log4 16. y = 2 since 42 = 16.

Example. Find y = log5 125. y = 3 since 53 = 125. 104 5. NONLINEAR MODELS 1 Example. Find y = log . 3 81 ⇣ ⌘ 4 1 1 y = 4 since 3 = = . 34 81

The most common logarithms are log10, usually just written as log, and loge, usually written as ln for natural logarithm. These have corresponding keys on your calculator. The number e is generated by letting n in ! 1 1 n n + 1 n 1 + or . n n For instance, if n = 1010, w⇣e get ⌘ ⇣ ⌘ 2.7 1828 1828 to 9 decimal places for this irrational number. Theorem (Change of base form|ula){z}.|{z} Any logarithm can be written in terms of log and ln by log x ln x log x = = . b log b ln b 4. LOGARITHMIC FUNCTION MODELS 105 We view the graphs of y = ln x and y = log x together.

All logarithmic graphs pass through the point (1, 0) and have the y-axis as a vertical asymptote. For x > 1, the smaller the base, the larger the values of the logarithm. For x < 1, the smaller the base, the smaller (more negative) the values of the logarithm.

When b > 1, logb x increases at a decreasing rate (concave down).

When 0 < b < 1, logb x decreases at an increasing rate (concave up), i.e., at a rate that is less and less negative. 106 5. NONLINEAR MODELS Theorem (Relationship between exponential and logarithmic functions). For b > 0, b = 1, and x > 0, the following statements are equivalent: 6 (1) y = logb x. (2) by = x.

Consider tables of values for f(x) = 10x and g(x) = log(x).

The output of one function is the input of the other. A point (x, y) is on one graph if and only if (y, x) is a point on the other graph. Functions with this property are called inverse functions, with each being the inverse of the other. This also means that the line y = x acts as a mirror between the two graphs. 4. LOGARITHMIC FUNCTION MODELS 107 Example. Solve 6x = 216. With calculator, using inverse relationship: log 216 x = log 216 = = 3. 6 log 6 Without calculator: 6x = 216 6x = 63 x = 3

x 1 Example. Solve 5 = 125. With calculator, using inverse relationship: 1 log 1 x = log = 125 = 3. 5 125 log 5 Without calculator: 1 5x = 125 1 5x = 53 x 3 5 = 5 x = 3 Example. Solve log3(x) = 5. With or without calculator:

log3(x) = 5 x = 35 x = 243 108 5. NONLINEAR MODELS Rules of Logarithms

Rule 7 is used extensively in solving exponential equations. Rules 1, 5, 6, 7 apply to any base. Rules 2,3,4 can be easily reformulated for any base. Example. Use rules of logarithms to write as a single logarithm: (1) log(3) log(27) = 3 1 2 log = log = log(3 ) = 27 9 2⇣log(3)⌘ ⇣ ⌘ (2) 4 log(x2) 5 log(x2) + log(2x3) = log(x2) + log(2x3) = log(2x3) log(x2) = 2x3 log = log(2x) x2 ⇣ ⌘ 4. LOGARITHMIC FUNCTION MODELS 109 (3) 2 3 log log(x3) = 5x ⇣2 ⌘3 8 log log(x3) = log log(x3) 5x 125x3 ⇣ 8⌘ ⇣ ⌘ 3 8 2 log 125x = log = 3 log x3 125x6 5x2 (4) ⇣ ⌘ ⇣ ⌘ ⇣ ⌘ 2 2 ln(3x) 3 ln(x ) 5 ln(x) = ln(9x2) + 6 ln(x) 5 ln(x) = ln(9x2) + ln(x) = ln(x) ln(9x2) = x 1 ln = ln = 9x2 9x ln(1)⇣ ⌘ln(9x) ⇣= ⌘ ln(9x) Problem (Page 358 #44). We notice that the function is increasing at a decreasing rate, so logarithmic regression is indicated. We get t = 148.3836356 + 32.91527261 ln i with t(125) = 10.54. this means the price of distilled spirits increased by 25% from 1984 to halfway through 1991. Graphs with two di↵erent ranges follow. 110 5. NONLINEAR MODELS 5. CHOOSING A MATHEMATICAL MODEL 111 5. Choosing a Mathematical Model Definition. Logistic Functions are functions of the form c y = f(x) = bx + k 1 + ae where a, b, c, and k are all constants.

The graph comes in two styles as an increasing or decreasing S-curve. The increasing curve increases first slowly, then fast, then slowly; it is concave up followed by concave down. The decreasing curve decreases first slowly, then fast, then slowly; it is concave down followed by concave up. Both versions are bounded by horizontal asymptotes y = k and y = c + k. For the increasing curve, c > 0; for the decreasing curve, c < 0. 112 5. NONLINEAR MODELS Selecting a Mathematical Model. (1) Draw a scatter plot. (2) Determine whether the scatter plot exhibits the behavior of the graph of one or more of the standard mathematical functions: linear, quadratic, cubic, quartic, exponential, logarithmic, or logistic. (3) Find a mathematical model for each function type selected in step 2. (4) Use all available information to anticipate the expected behavior of the thing being modeled outside of the data set. Eliminate models that don’t exhibit the expected behavior. (Sometimes it is convenient to switch the order of Steps 3 and 4.) (5) Choose the simplest model from among the models that meet your criteria. Key features of the various models are given below. 5. CHOOSING A MATHEMATICAL MODEL 113 Problem (Page 371 #6). With the exception of a couple of points, the data appears to be sort of linear, but a concave up region followed by a concave down region is evident. Since tuition is not likely to level o↵, ruling out the logistic model, we look to the cubic model. We get f(x) = .249x3 + 7.447x2 + 9.024x + 634.538. For 2005, the cubic model gives f(21) 1801, ⇡ which seems low. Thus we look to the linear model: f(x) = 69.612x + 534.076. For 2005, this gives f(21) 1995, ⇡ which seems to make more sense based on the data, even though the linear model may not fit the points as well. 114 5. NONLINEAR MODELS Problem (Page 370 #2). The data appears slightly concave up. Candidate models are then quadratic and exponential. We check ratios over the last 4 points where the x-gaps are equal: 965 946 932 = .978, = .980, = .985. 987 965 946 We view the graphs of the models for 30 years (put the quadratic model as Y1 and the exponential as Y2).

For the quadratic model, r2 = .998, while for the exponential, r2 = .977, indicating a better ﬁt for the quadratic model, as is also clear from the graph. Further, the quadratic model matches the leveling o↵ of the loss of farms, while the exponential model (whose graph in this window is almost linear) will not level o↵ until much closer to the x-axis. So we choose the quadratic model f(x) = .1249001315x2 6.654950848x + 1013.888611. For the year 2000, this gives us f(22) 928. ⇡ This makes sense since we expect the number of farms to continue to decrease into the near future. 5. CHOOSING A MATHEMATICAL MODEL 115 116 5. NONLINEAR MODELS Problem (Page 373 #12). We ﬁrst create a data table. Let t = days since March 12. Let P = price in $1000’s. t P (in 1000’s) 0 198.9 4 205.9 18 209.9 32 212.9 The plot appears to be concave down and ever more slowly increasing, i.e., logarithmic. But a logarithmic curve cannot have 0 in its domain. So we align the data by letting t = days since March 11. Then we have t P (in 1000’s) 1 198.9 5 205.9 19 209.9 33 212.9 Then logarithmic regression gives P = 199.1067901 + 3.872217515 ln t for the new home style. Problem (Page 373 #14). We are given an annual growth rate of 4%, suggesting an exponential model y = abt. We have r = .04, so b = 1 .04 = .96. If t = years from the present, and M = # of members still alive, a = M(0) = 95, so the model is M(t) = 95(.96)t. 5. CHOOSING A MATHEMATICAL MODEL 117 Problem (Page 373 #17). We let x = years since 1990 and y = population. A data table is x y 0 2667 10 4316 11 4940 12 5555 Recall logistic regression c y = f(x) = bx + k 1 + ae has horizontal asymptotes y = k andy = c + k with c > 0 for the increasing curve and c < 0 for the decreasing curve. However, the TI-83 Plus and TI-84 Plus, but not the TI-89, assume that k = 0. For our data, k = 2600 seems reasonable, so we align the data by making the right hand column y 2600. x y 2600 0 67 10 1716 11 2340 12 2955 On the TI, put the cursor on L3, and punch 2nd 2-2600/ENTER to place the aligned data in L3, and then punch STAT/CALC/B:Logistic/2nd 1/,/2nd 3/,/Y1/ENTER. We get 6929.817052 y 2600 = .4019008339x 1 + 166.2278249e or 6929.817052 y = .4019008339x + 2600. 1 + 166.2278249e Since the denominator approaches 1 as x gets large, the model projects a max- imum population of 6930 + 2600 = 9530. 118 5. NONLINEAR MODELS

After adding 2600 to Y1, view the graph in a window [ 1, 20] [0, 10000]. Our model projects a 2003 population of ⇥ y(13) = 6258, quite a bit less than the department of Commerce projection of 7480. We look for a di↵erent model: (1) The quadratic and cubic models show population decreases between 1990 and 2000, so are rejected. Also, their predictions for 2003 are about the same as for the logistic model. (2) We don’t have enough points for the quartic model to work. (3) We go to the exponential model with the aligned data since we want y = 0 as a horizontal asymptote. We use

ExpReg L1, L3, Y1 to get y 2600 = 67.58587649(1.376627411)x or y = 67.58587649(1.376627411)x + 2600. Now y(13) = 6910, which is closer to the Department of Commerce projection. Problem (Page 373 #20). Let x = joint yearly salary and T = taxes owed. .1x if 0 x 14, 000   T (x) = .1(14, 000) + .15(x 14, 000) if 14, 000 < x 56, 800 . 8  <>.1(14, 000) + .15(42, 800) + .25(x 56, 800) if x > 56, 800 > .1x if 0 x 14, 000 :   Then T (x) = .15x 700 if 14, 000 < x 56, 800 . 8  <>.25x 6380 if 56, 800 < x 114, 650  :>