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

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1. Quadratic Function Models Consider the Following Table of the Number of Americans That Are Over 100 in Various Years 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 equa- tions 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 sufficient 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 not−a 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 con- cavity. 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 re- gression. 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.
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