Math\t107-6381\t-\tQuiz\t#4\t-\tSchultz\t-\tDue\tFebruary\t21,\t2016\t-\tpage\t1\tof 3 Follow\tthese\tdirections\tcarefully. This\tquiz\tis\tdue\tby\t11:59\tEastern\ttime\ton\tFebruary\t21,\t2016. o This\tis\tan\timportant\tassignment,\tcounting\t12%\tof\tyour\tgrade. o Submit\tthis\tassignment\tin\tyour\tassignments\tfolder\tby\tthe\tdue\tdate. Answer\tall\tthe\tquestions. There\tare\t9\tproblems\ton\t3\tpages. o There\tare\t150\tpoints\tpossible. o Show\tall\twork\tin\torder\tto\treceive\tany\tcredit. There\tis\tpartial\tcredit\tfor\tthese problems. Submit\tyour\tassignment\tas\tan\tattachment. Under\tno\tcircumstances\tshould\tyou type\tyour\tanswers\tinto\ta\ttext\tbox. 1. Refer\tto\tthe\tpolynomial = 4 + 2 ' 1 * ' 5 . (12\tpts\ttotal) This\tfunction\thas\tfive\t(5)\tzeros. Complete\tthe\ttable. (10\tpts) a) b) c) d) e) 2. f) (2\tpts) Write\teach\texpression\tin\tsimplest + form. (11\tpts\ttotal) a. (1 3)(8 + 7) (3\tpts) (8\tpts) Solve\teach\tquadratic\tequation. (15\tpts\ttotal) a. 6 ' 13 + 6 = 0 b. 4 ' 4 + 17 = 0 (5\tpts) (10\tpts) b. 4. Multiplicity Determine\tthe\tdegree\tof\tthis\tpolynomial. 3. Zero '67 '8*7 Determine\tthe\tend\tbehavior\tfor\teach\tpolynomial. Refer\tto\tsection\t3.1,\tpage\t240,\tfor guidance. (10\tpts\ttotal) a. = 5 * 6 ' + 2 1 (5\tpts) b. = 6 < 3 ' + 1 (5\tpts) Math\t107-6381\t-\tQuiz\t#4\t-\tSchultz\t-\tDue\tFebruary\t21,\t2016\t-\tpage\t2\tof 3 5. Solve\teach\tinequality. a. b. 6. ' 5 14 < 0 >8? >6' 0 (14\tpts\ttotal) (6\tpts) (8\tpts) Consider\tthe\tquadratic\tfunction = ' + 4 + 5. (28\tpts\ttotal) a. Graph\tthis\tfunction. (5\tpts) Answer\teach\tpart.\tYou'll\tneed\tto\tdo\tsome\tof\tthese\tparts\tto\tmake\tthe\tgraph.\t(Parts\tb and\tf\tshould\tbe\tparticularly\thelpful. Does\tthis\tparabola\topen\tup\tor\tdown? This should\thelp\tyou\tanswer\tparts\tc\tand\te,\tand\tfind\tthe\trange.) 7. b. Find\tthe\tvertex. (5\tpts) c. State\twhether\tthere\tis\ta\tmaximum\tand\tminimum\tvalue\tand\tfind\tthat\tvalue. (2\tpts\teach) d. Find\tthe\tdomain\tand\trange. (2\tpts\teach) e. Find\tthe\tinterval(s)\ton\twhich\tthe\tfunction\tis\tincreasing,\tand\tfind\tthe\tinterval(s) on\twhich\tthe\tfunction\tis\tdecreasing. (2\tpts\teach) f. Find\tthe\t(, )\tcoordinates\tof\tthe\tx-\tand\ty-\tintercepts. (6\tpts;\t2\tpts\teach) Consider\tthe\trational\tfunction ' 2 7 + 6 = ' . + 5 14 (20\tpts\ttotal) a. Graph\tthis\tfunction. Draw\teach\tasymptote\twith\ta\tdotted\tline,\tand\tlabel\tit\twith\tits equation. (6\tpts) Answer\teach\tpart. Each\tpart\tbelow\tshould\thelp\tyou\tmake\tthe\tgraph. b. Find\tthe\tdomain. (Make\tsure\tyou\tfind\tthe\tdomain\tbefore\tyou\tmake\tany cancellations\tof\tfactors.) (5\tpts) c. Find\tthe\tequation\tof\tthe\tvertical\tasymptote(s),\tif\tany,\tof\tthe\tgraph. (3\tpts) d. Find\tthe\tequation\tof\tthe\thorizontal\tasymptote(s),\tif\tany,\tof\tthe\tgraph. (3\tpts) e. The\tgraph\thas\tone\thole. Find\tthe\t(, )\tcoordinates\tof\tthe\thole. (3\tpts) Math\t107-6381\t-\tQuiz\t#4\t-\tSchultz\t-\tDue\tFebruary\t21,\t2016\t-\tpage\t3\tof 3 8. Consider\tthe\ttable\tof\tvalues\tgiven\tbelow. Refer\tto\tsection\t2.5,\tstarting\ton\tpage\t225, and\trefer\tto\tExample\t2.5.1. (20\tpts\ttotal) x -1 -1 0 1 2 3 4 y 2 3 4 3 7 6 8 a. Find\tthe\tleast\tsquares\tregression\tline\tand\tcomment\ton\tthe\tgoodness\tof\tfit. (5\tpts) b. Interpret\tthe\tslope\tof\tthe\tline\tof\tbest\tfit. (5\tpts) c. Use\tthe\tregression\tline\tto\tpredict\ty\twhen\tx\tequals\t5. (5\tpts) d. Use\tthe\tregression\tline\tto\tpredict\ty\twhen\tx\tequals\t3. (5\tpts) 9. Consider\tthe\ttable\tof\tvalues\tbelow. Refer\tto\tsection\t2.5,\tstarting\ton\tpage\t228, beginning\twith\tthe\tdiscussion\tstarting\tthere\tand\treading\tExample\t2.5.2\ton\tpage\t229. (20\tpts\ttotal) Time Temperature in Fahrenheit 7am 35 9am 50 11am 56 1pm 59 2pm 61 4pm 62 8pm 59 11pm 44 a. Fit\ta\tquadratic\tmodel\tfor\tthe\ttemperature\tas\ta\tfunction\tof\ttime,\tand\tcomment\ton the\tgoodness\tof\tfit. (10\tpts) b. What\tis\tyour\tpredicted\ttemperature\tat\t3pm? (5\tpts) c. At\twhat\ttime\tof\tday\tdoes\tthe\tmaximum\ttemperature\toccur? (5\tpts) Section 2.5: Regression, from College Algebra: Corrected Edition by Carl Stitz, Ph.D. and Jeff Zeager, Ph.D. is available under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 license. 2013, Carl Stitz. 2.5 Regression 2.5 225 Regression We have seen examples already in the text where linear and quadratic functions are used to model a wide variety of real world phenomena ranging from production costs to the height of a projectile above the ground. In this section, we use some basic tools from statistical analysis to quantify linear and quadratic trends that we may see in real world data in order to generate linear and quadratic models. Our goal is to give the reader an understanding of the basic processes involved, but we are quick to refer the reader to a more advanced course1 for a complete exposition of this material. Suppose we collected three data points: {(1, 2), (3, 1), (4, 3)}. By plotting these points, we can clearly see that they do not lie along the same line. If we pick any two of the points, we can nd a line containing both which completely misses the third, but our aim is to nd a line which is in some sense 'close' to all the points, even though it may go through none of them. The way we measure 'closeness' in this case is to nd the total squared error between the data points and the line. Consider our three data points and the line y = 1 x + 1 . For each of our data points, 2 2 we nd the vertical distance between the point and the line. To accomplish this, we need to nd a point on the line directly above or below each data point - in other words, a point on the line with the same x-coordinate as our data point. For example, to nd the point on the line directly 1 below (1, 2), we plug x = 1 into y = 1 x + 2 and we get the point (1, 1). Similarly, we get (3, 1) to 2 5 correspond to (3, 2) and 4, 2 for (4, 3). 4 3 2 1 1 2 3 4 We nd the total squared error E by taking the sum of the squares of the dierences of the ycoordinates of each data point and its corresponding point on the line. For the data and line above 2 9 E = (2 1)2 + (1 2)2 + 3 5 = 4 . Using advanced mathematical machinery,2 it is possible to 2 nd the line which results in the lowest value of E. This line is called the least squares regression line, or sometimes the 'line of best t'. The formula for the line of best t requires notation we won't present until Chapter 9.1, so we will revisit it then. The graphing calculator can come to our assistance here, since it has a built-in feature to compute the regression line. We enter the data and perform the Linear Regression feature and we get 1 2 and authors with more expertise in this area, Like Calculus and Linear Algebra 226 Linear and Quadratic Functions The calculator tells us that the line of best t is y = ax + b where the slope is a 0.214 and the y-coordinate of the y-intercept is b 1.428. (We will stick to using three decimal places for our approximations.) Using this line, we compute the total squared error for our data to be E 1.786. The value r is the correlation coecient and is a measure of how close the data is to being on the same line. The closer |r| is to 1, the better the linear t. Since r 0.327, this tells us that the line of best t doesn't t all that well - in other words, our data points aren't close to being linear. The value r2 is called the coecient of determination and is also a measure of the goodness of t.3 Plotting the data with its regression line results in the picture below. Our rst example looks at energy consumption in the US over the past 50 years.4 Year Energy Usage, in Quads5 1950 34.6 1960 45.1 1970 67.8 1980 78.3 1990 84.6 2000 98.9 Example 2.5.1. Using the energy consumption data given above, 1. Plot the data using a graphing calculator. 3 We refer the interested reader to a course in Statistics to explore the signicance of r and r2 . See this Department of Energy activity 5 The unit 1 Quad is 1 Quadrillion = 1015 BTUs, which is enough heat to raise Lake Erie roughly 1 F 4 2.5 Regression 227 2. Find the least squares regression line and comment on the goodness of t. 3. Interpret the slope of the line of best t. 4. Use the regression line to predict the annual US energy consumption in the year 2013. 5. Use the regression line to predict when the annual consumption will reach 120 Quads. Solution. 1. Entering the data into the calculator gives The data certainly appears to be linear in nature. 2. Performing a linear regression produces We can tell both from the correlation coecient as well as the graph that the regression line is a good t to the data. 3. The slope of the regression line is a 1.287. To interpret this, recall that the slope is the rate of change of the y-coordinates with respect to the x-coordinates. Since the y-coordinates represent the energy usage in Quads, and the x-coordinates represent years, a slope of positive 1.287 indicates an increase in annual energy usage at the rate of 1.287 Quads per year. 4. To predict the energy needs in 2013, we substitute x = 2013 into the equation of the line of best t to get y = 1.287(2013) 2473.890 116.841. The predicted annual energy usage of the US in 2013 is approximately 116.841 Quads. 228 Linear and Quadratic Functions 5. To predict when the annual US energy usage will reach 120 Quads, we substitute y = 120 into the equation of the line of best t to get 120 = 1.287x 2473.908. Solving for x yields x 2015.454. Since the regression line is increasing, we interpret this result as saying the annual usage in 2015 won't yet be 120 Quads, but that in 2016, the demand will be more than 120 Quads. Our next example gives us an opportunity to nd a nonlinear model to t the data. According to the National Weather Service, the predicted hourly temperatures for Painesville on March 3, 2009 were given as summarized below. Time Temperature, F 10AM 17 11AM 19 12PM 21 1PM 23 2PM 24 3PM 24 4PM 23 To enter this data into the calculator, we need to adjust the x values, since just entering the numbers could cause confusion. (Do you see why?) We have a few options available to us. Perhaps the easiest is to convert the times into the 24 hour clock time so that 1 PM is 13, 2 PM is 14, etc.. If we enter these data into the graphing calculator and plot the points we get While the beginning of the data looks linear, the temperature begins to fall in the afternoon hours. This sort of behavior reminds us of parabolas, and, sure enough, it is possible to nd a parabola of best t in the same way we found a line of best t. The process is called quadratic regression and its goal is to minimize the least square error of the data with their corresponding points on the parabola. The calculator has a built in feature for this as well which yields 2.5 Regression 229 The coecient of determination R2 seems reasonably close to 1, and the graph visually seems to be a decent t. We use this model in our next example. Example 2.5.2. Using the quadratic model for the temperature data above, predict the warmest temperature of the day. When will this occur? Solution. The maximum temperature will occur at the vertex of the parabola. Recalling the b 9.464 Vertex Formula, Equation 2.4, x = 2a 2(0.321) 14.741. This corresponds to roughly 2 : 45 PM. To nd the temperature, we substitute x = 14.741 into y = 0.321x2 + 9.464x 45.857 to get y 23.899, or 23.899 F. The results of the last example should remind you that regression models are just that, models. Our predicted warmest temperature was found to be 23.899 F, but our data says it will warm to 24 F. It's all well and good to observe trends and guess at a model, but a more thorough investigation into why certain data should be linear or quadratic in nature is usually in order - and that, most often, is the business of scientists. 230 2.5.1 Linear and Quadratic Functions Exercises 1. According to this website6 , the census data for Lake County, Ohio is: Year Population 1970 197200 1980 212801 1990 215499 2000 227511 (a) Find the least squares regression line for these data and comment on the goodness of t.7 Interpret the slope of the line of best t. (b) Use the regression line to predict the population of Lake County in 2010. (The recorded gure from the 2010 census is 230,041) (c) Use the regression line to predict when the population of Lake County will reach 250,000. 2. According to this website8 , the census data for Lorain County, Ohio is: Year Population 1970 256843 1980 274909 1990 271126 2000 284664 (a) Find the least squares regression line for these data and comment on the goodness of t. Interpret the slope of the line of best t. (b) Use the regression line to predict the population of Lorain County in 2010. (The recorded gure from the 2010 census is 301,356) (c) Use the regression line to predict when the population of Lake County will reach 325,000. 3. Using the energy production data given below Year Production (in Quads) 1950 1960 1970 1980 1990 2000 35.6 42.8 63.5 67.2 70.7 71.2 (a) Plot the data using a graphing calculator and explain why it does not appear to be linear. (b) Discuss with your classmates why ignoring the rst two data points may be justied from a historical perspective. (c) Find the least squares regression line for the last four data points and comment on the goodness of t. Interpret the slope of the line of best t. (d) Use the regression line to predict the annual US energy production in the year 2010. (e) Use the regression line to predict when the annual US energy production will reach 100 Quads. 6 http://www.ohiobiz.com/census/Lake.pdf We'll develop more sophisticated models for the growth of populations in Chapter 6. For the moment, we use a theorem from Calculus to approximate those functions with lines. 8 http://www.ohiobiz.com/census/Lorain.pdf 7 2.5 Regression 231 4. The chart below contains a portion of the fuel consumption information for a 2002 Toyota Echo that I (Je) used to own. The rst row is the cumulative number of gallons of gasoline that I had used and the second row is the odometer reading when I relled the gas tank. So, for example, the fourth entry is the point (28.25, 1051) which says that I had used a total of 28.25 gallons of gasoline when the odometer read 1051 miles. Gasoline Used (Gallons) Odometer (Miles) 0 9.26 19.03 28.25 36.45 44.64 53.57 62.62 71.93 81.69 90.43 41 356 731 1051 1347 1631 1966 2310 2670 3030 3371 Find the least squares line for this data. Is it a good t? What does the slope of the line represent? Do you and your classmates believe this model would have held for ten years had I not crashed the car on the Turnpike a few years ago? (I'm keeping a fuel log for my 2006 Scion xA for future College Algebra books so I hope not to crash it, too.) 5. On New Year's Day, I (Je, again) started weighing myself every morning in order to have an interesting data set for this section of the book. (Discuss with your classmates if that makes me a nerd or a geek. Also, the professionals in the eld of weight management strongly discourage weighing yourself every day. When you focus on the number and not your overall health, you tend to lose sight of your objectives. I was making a noble sacrice for science, but you should not try this at home.) The whole chart would be too big to put into the book neatly, so I've decided to give only a small portion of the data to you. This then becomes a Civics lesson in honesty, as you shall soon see. There are two charts given below. One has my weight for the rst eight Thursdays of the year (January 1, 2009 was a Thursday and we'll count it as Day 1.) and the other has my weight for the rst 10 Saturdays of the year. Day # (Thursday) My weight in pounds Day # (Saturday) My weight in pounds 1 8 15 22 29 36 43 50 238.2 237.0 235.6 234.4 233.0 233.8 232.8 232.0 3 10 17 24 31 38 45 52 59 66 238.4 235.8 235.0 234.2 236.2 236.2 235.2 233.2 236.8 238.2 (a) Find the least squares line for the Thursday data and comment on its goodness of t. (b) Find the least squares line for the Saturday data and comment on its goodness of t. (c) Use Quadratic Regression to nd a parabola which models the Saturday data and comment on its goodness of t. (d) Compare and contrast the predictions the three models make for my weight on January 1, 2010 (Day #366). Can any of these models be used to make a prediction of my weight 20 years from now? Explain your answer. 232 Linear and Quadratic Functions (e) Why is this a Civics lesson in honesty? Well, compare the two linear models you obtained above. One was a good t and the other was not, yet both came from careful selections of real data. In presenting the tables to you, I have not lied about my weight, nor have you used any bad math to falsify the predictions. The word we're looking for here is 'disingenuous'. Look it up and then discuss the implications this type of data manipulation could have in a larger, more complex, politically motivated setting. (Even Obi-Wan presented the truth to Luke only \"from a certain point of view.\") 6. (Data that is neither linear nor quadratic.) We'll close this exercise set with two data sets that, for reasons presented later in the book, cannot be modeled correctly by lines or parabolas. It is a good exercise, though, to see what happens when you attempt to use a linear or quadratic model when it's not appropriate. (a) This rst data set came from a Summer 2003 publication of the Portage County Animal Protective League called \"Tattle Tails\". They make the following statement and then have a chart of data that supports it. \"It doesn't take long for two cats to turn into 80 million. If two cats and their surviving ospring reproduced for ten years, you'd end up with 80,399,780 cats.\" We assume N (0) = 2. Year x Number of Cats N (x) 1 2 3 4 5 6 7 8 9 10 12 66 382 2201 12680 73041 420715 2423316 13968290 80399780 Use Quadratic Regression to nd a parabola which models this data and comment on its goodness of t. (Spoiler Alert: Does anyone know what type of function we need here?) (b) This next data set comes from the U.S. Naval Observatory. That site has loads of awesome stu on it, but for this exercise I used the sunrise/sunset times in Fairbanks, Alaska for 2009 to give you a chart of the number of hours of daylight they get on the 21st of each month. We'll let x = 1 represent January 21, 2009, x = 2 represent February 21, 2009, and so on. Month Number Hours of Daylight 1 2 3 4 5 6 7 8 9 10 11 12 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 Use Quadratic Regression to nd a parabola which models this data and comment on its goodness of t. (Spoiler Alert: Does anyone know what type of function we need here?) 2.5 Regression 2.5.2 233 Answers 1. (a) y = 936.31x 1645322.6 with r = 0.9696 which indicates a good t. The slope 936.31 indicates Lake County's population is increasing at a rate of (approximately) 936 people per year. (b) According to the model, the population in 2010 will be 236,660. (c) According to the model, the population of Lake County will reach 250,000 sometime between 2024 and 2025. 2. (a) y = 796.8x 1309762.5 with r = 0.8916 which indicates a reasonable t. The slope 796.8 indicates Lorain County's population is increasing at a rate of (approximately) 797 people per year. (b) According to the model, the population in 2010 will be 291,805. (c) According to the model, the population of Lake County will reach 325,000 sometime between 2051 and 2052. 3. (c) y = 0.266x459.86 with r = 0.9607 which indicates a good t. The slope 0.266 indicates the country's energy production is increasing at a rate of 0.266 Quad per year. (d) According to the model, the production in 2010 will be 74.8 Quad. (e) According to the model, the production will reach 100 Quad in the year 2105. 4. The line is y = 36.8x + 16.39. We have r = .99987 and r2 = .9997 so this is an excellent t to the data. The slope 36.8 represents miles per gallon. 5. (a) The line for the Thursday data is y = .12x + 237.69. We have r = .9568 and r2 = .9155 so this is a really good t. (b) The line for the Saturday data is y = 0.000693x + 235.94. We have r = 0.008986 and r2 = 0.0000807 which is horrible. This data is not even close to linear. (c) The parabola for the Saturday data is y = 0.003x2 0.21x+238.30. We have R2 = .47497 which isn't good. Thus the data isn't modeled well by a quadratic function, either. (d) The Thursday linear model had my weight on January 1, 2010 at 193.77 pounds. The Saturday models give 235.69 and 563.31 pounds, respectively. The Thursday line has my weight going below 0 pounds in about ve and a half years, so that's no good. The quadratic has a positive leading coecient which would mean unbounded weight gain for the rest of my life. The Saturday line, which mathematically does not t the data at all, yields a plausible weight prediction in the end. I think this is why grown-ups talk about \"Lies, Damned Lies and Statistics.\" 6. (a) The quadratic model for the cats in Portage county is y = 1917803.54x2 16036408.29x+ 24094857.7. Although R2 = .70888 this is not a good model because it's so far o for small values of x. Case in point, the model gives us 24,094,858 cats when x = 0 but we know N (0) = 2. 234 Linear and Quadratic Functions (b) The quadratic model for the hours of daylight in Fairbanks, Alaska is y = .51x2 +6.23x .36. Even with R2 = .92295 we should be wary of making predictions beyond the data. Case in point, the model gives 4.84 hours of daylight when x = 13. So January 21, 2010 will be \"extra dark\"? Obviously a parabola pointing down isn't telling us the whole story. Section 3.1: Graphs of Polynomials, from College Algebra: Corrected Edition by Carl Stitz, Ph.D. and Jeff Zeager, Ph.D. is available under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 license. 2013, Carl Stitz. Chapter 3 Polynomial Functions 3.1 Graphs of Polynomials Three of the families of functions studied thus far - constant, linear and quadratic - belong to a much larger group of functions called polynomials. We begin our formal study of general polynomials with a denition and some examples. Denition 3.1. A polynomial function is a function of the form f (x) = an xn + an1 xn1 + . . . + a2 x2 + a1 x + a0 , where a0 , a1 , . . . , an are real numbers and n 1 is a natural number. The domain of a polynomial function is (, ). There are several things about Denition 3.1 that may be o-putting or downright frightening. The best thing to do is look at an example. Consider f (x) = 4x5 3x2 + 2x 5. Is this a polynomial function? We can re-write the formula for f as f (x) = 4x5 + 0x4 + 0x3 + (3)x2 + 2x + (5). Comparing this with Denition 3.1, we identify n = 5, a5 = 4, a4 = 0, a3 = 0, a2 = 3, a1 = 2 and a0 = 5. In other words, a5 is the coecient of x5 , a4 is the coecient of x4 , and so forth; the subscript on the a's merely indicates to which power of x the coecient belongs. The business of restricting n to be a natural number lets us focus on well-behaved algebraic animals.1 Example 3.1.1. Determine if the following functions are polynomials. Explain your reasoning. 4 + x3 x 4. f (x) = 3 x 1. g(x) = 1 2. p(x) = 4x + x3 x 5. h(x) = |x| 3. q(x) = 4x + x3 x2 + 4 6. z(x) = 0 Enjoy this while it lasts. Before we're through with the book, you'll have been exposed to the most terrible of algebraic beasts. We will tame them all, in time. 236 Polynomial Functions Solution. 3 +4 1. We note directly that the domain of g(x) = x x is x = 0. By denition, a polynomial has all real numbers as its domain. Hence, g can't be a polynomial. 3 2. Even though p(x) = x +4x simplies to p(x) = x2 + 4, which certainly looks like the form x given in Denition 3.1, the domain of p, which, as you may recall, we determine before we simplify, excludes 0. Alas, p is not a polynomial function for the same reason g isn't. 3. After what happened with p in the previous part, you may be a little shy about simplifying 3 q(x) = x 2+4x to q(x) = x, which certainly ts Denition 3.1. If we look at the domain of x +4 q before we simplied, we see that it is, indeed, all real numbers. A function which can be written in the form of Denition 3.1 whose domain is all real numbers is, in fact, a polynomial. 1 1 4. We can rewrite f (x) = 3 x as f (x) = x 3 . Since 3 is not a natural number, f is not a polynomial. 5. The function h(x) = |x| isn't a polynomial, since it can't be written as a combination of powers of x even though it can be written as a piecewise function involving polynomials. As we shall see in this section, graphs of polynomials possess a quality2 that the graph of h does not. 6. There's nothing in Denition 3.1 which prevents all the coecients an , etc., from being 0. Hence, z(x) = 0, is an honest-to-goodness polynomial. Denition 3.2. Suppose f is a polynomial function. Given f (x) = an xn + an1 xn1 + . . . + a2 x2 + a1 x + a0 with an = 0, we say - The natural number n is called the degree of the polynomial f . - The term an xn is called the leading term of the polynomial f . - The real number an is called the leading coecient of the polynomial f . - The real number a0 is called the constant term of the polynomial f . If f (x) = a0 , and a0 = 0, we say f has degree 0. If f (x) = 0, we say f has no degree.a a Some authors say f (x) = 0 has degree for reasons not even we will go into. The reader may well wonder why we have chosen to separate o constant functions from the other polynomials in Denition 3.2. Why not just lump them all together and, instead of forcing n to be a natural number, n = 1, 2, . . ., allow n to be a whole number, n = 0, 1, 2, . . .. We could unify all 2 One which really relies on Calculus to verify. 3.1 Graphs of Polynomials 237 of the cases, since, after all, isn't a0 x0 = a0 ? The answer is 'yes, as long as x = 0.' The function f (x) = 3 and g(x) = 3x0 are dierent, because their domains are dierent. The number f (0) = 3 is dened, whereas g(0) = 3(0)0 is not.3 Indeed, much of the theory we will develop in this chapter doesn't include the constant functions, so we might as well treat them as outsiders from the start. One good thing that comes from Denition 3.2 is that we can now think of linear functions as degree 1 (or 'rst degree') polynomial functions and quadratic functions as degree 2 (or 'second degree') polynomial functions. Example 3.1.2. Find the degree, leading term, leading coecient and constant term of the following polynomial functions. 1. f (x) = 4x5 3x2 + 2x 5 3. h(x) = 4x 5 2. g(x) = 12x + x3 4. p(x) = (2x 1)3 (x 2)(3x + 2) Solution. 1. There are no surprises with f (x) = 4x5 3x2 + 2x 5. It is written in the form of Denition 3.2, and we see that the degree is 5, the leading term is 4x5 , the leading coecient is 4 and the constant term is 5. 2. The form given in Denition 3.2 has the highest power of x rst. To that end, we re-write g(x) = 12x + x3 = x3 + 12x, and see that the degree of g is 3, the leading term is x3 , the leading coecient is 1 and the constant term is 0. 3. We need to rewrite the formula for h so that it resembles the form given in Denition 3.2: 4 1 h(x) = 4x = 5 x = 1 x + 4 . The degree of h is 1, the leading term is 5 x, the leading 5 5 5 5 1 4 coecient is 5 and the constant term is 5 . 4. It may seem that we have some work ahead of us to get p in the form of Denition 3.2. However, it is possible to glean the information requested about p without multiplying out the entire expression (2x 1)3 (x 2)(3x + 2). The leading term of p will be the term which has the highest power of x. The way to get this term is to multiply the terms with the highest power of x from each factor together - in other words, the leading term of p(x) is the product of the leading terms of the factors of p(x). Hence, the leading term of p is (2x)3 (x)(3x) = 24x5 . This means that the degree of p is 5 and the leading coecient is 24. As for the constant term, we can perform a similar trick. The constant term is obtained by multiplying the constant terms from each of the factors (1)3 (2)(2) = 4. Our next example shows how polynomials of higher degree arise 'naturally'4 in even the most basic geometric applications. 3 Technically, 00 is an indeterminant form, which is a special case of being undened. The authors realize this is beyond pedantry, but we wouldn't mention it if we didn't feel it was neccessary. 4 this is a dangerous word... 238 Polynomial Functions Example 3.1.3. A box with no top is to be fashioned from a 10 inch 12 inch piece of cardboard by cutting out congruent squares from each corner of the cardboard and then folding the resulting tabs. Let x denote the length of the side of the square which is removed from each corner. x x x x 12 in height x x x depth width x 10 in 1. Find the volume V of the box as a function of x. Include an appropriate applied domain. 2. Use a graphing calculator to graph y = V (x) on the domain you found in part 1 and approximate the dimensions of the box with maximum volume to two decimal places. What is the maximum volume? Solution. 1. From Geometry, we know that Volume = width height depth. The key is to nd each of these quantities in terms of x. From the gure, we see that the height of the box is x itself. The cardboard piece is initially 10 inches wide. Removing squares with a side length of x inches from each corner leaves 10 2x inches for the width.5 As for the depth, the cardboard is initially 12 inches long, so after cutting out x inches from each side, we would have 12 2x inches remaining. As a function6 of x, the volume is V (x) = x(10 2x)(12 2x) = 4x3 44x2 + 120x To nd a suitable applied domain, we note that to make a box at all we need x > 0. Also the shorter of the two dimensions of the cardboard is 10 inches, and since we are removing 2x inches from this dimension, we also require 10 2x > 0 or x < 5. Hence, our applied domain is 0 < x < 5. 2. Using a graphing calculator, we see that the graph of y = V (x) has a relative maximum. For 0 < x < 5, this is also the absolute maximum. Using the 'Maximum' feature of the calculator, we get x 1.81, y 96.77. This yields a height of x 1.81 inches, a width of 10 2x 6.38 inches, and a depth of 12 2x 8.38 inches. The y-coordinate is the maximum volume, which is approximately 96.77 cubic inches (also written in3 ). 5 There's no harm in taking an extra step here and making sure this makes sense. If we chopped out a 1 inch square from each side, then the width would be 8 inches, so chopping out x inches would leave 10 2x inches. 6 When we write V (x), it is in the context of function notation, not the volume V times the quantity x. 3.1 Graphs of Polynomials 239 In order to solve Example 3.1.3, we made good use of the graph of the polynomial y = V (x), so we ought to turn our attention to graphs of polynomials in general. Below are the graphs of y = x2 , y = x4 and y = x6 , side-by-side. We have omitted the axes to allow you to see that as the exponent increases, the 'bottom' becomes 'atter' and the 'sides' become 'steeper.' If you take the the time to graph these functions by hand,7 you will see why. y = x2 y = x4 y = x6 All of these functions are even, (Do you remember how to show this?) and it is exactly because the exponent is even.8 This symmetry is important, but we want to explore a dierent yet equally important feature of these functions which we can be seen graphically - their end behavior. The end behavior of a function is a way to describe what is happening to the function values (the y-values) as the x-values approach the 'ends' of the x-axis.9 That is, what happens to y as x becomes small without bound10 (written x ) and, on the ip side, as x becomes large without bound11 (written x ). For example, given f (x) = x2 , as x , we imagine substituting x = 100, x = 1000, etc., into f to get f (100) = 10000, f (1000) = 1000000, and so on. Thus the function values are becoming larger and larger positive numbers (without bound). To describe this behavior, we write: as x , f (x) . If we study the behavior of f as x , we see that in this case, too, f (x) . (We told you that the symmetry was important!) The same can be said for any function of the form f (x) = xn where n is an even natural number. If we generalize just a bit to include vertical scalings and reections across the x-axis,12 we have 7 Make sure you choose some x-values between 1 and 1. Herein lies one of the possible origins of the term 'even' when applied to functions. 9 Of course, there are no ends to the x-axis. 10 We think of x as becoming a very large (in the sense of its absolute value) negative number far to the left of zero. 11 We think of x as moving far to the right of zero and becoming a very large positive number. 12 See Theorems 1.4 and 1.5 in Section 1.7. 8 240 Polynomial Functions End Behavior of functions f (x) = axn , n even. Suppose f (x) = axn where a = 0 is a real number and n is an even natural number. The end behavior of the graph of y = f (x) matches one of the following: for a > 0, as x , f (x) and as x , f (x) for a < 0, as x , f (x) and as x , f (x) Graphically: a>0 a<0 we now turn our attention to functions of the form f (x) where n 3 is an odd natural number. (we ignore case when 1, since graph x a line and doesn't t general pattern higher-degree polynomials.) below have graphed y =x3 , x5 x7 . 'attening' 'steepening' that saw with even powers presents itself here as well, and, it should come no surprise all these are odd.13 end behavior same, x3 degreed studied earlier, can generalize their behavior. axn odd. suppose 0 real number matches one following: for> 0, as x , f (x) and as x , f (x) for a < 0, as x , f (x) and as x , f (x) Graphically: a>0 13 a<0 And are, perhaps, the inspiration for the moniker 'odd function'. 3.1 Graphs of Polynomials 241 Despite having dierent end behavior, all functions of the form f (x) = axn for natural numbers n share two properties which help distinguish them from other animals in the algebra zoo: they are continuous and smooth. While these concepts are formally dened using Calculus,14 informally, graphs of continuous functions have no 'breaks' or 'holes' in them, and the graphs of smooth functions have no 'sharp turns'. It turns out that these traits are preserved when functions are added together, so general polynomial functions inherit these qualities. Below we nd the graph of a function which is neither smooth nor continuous, and to its right we have a graph of a polynomial, for comparison. The function whose graph appears on the left fails to be continuous where it has a 'break' or 'hole' in the graph; everywhere else, the function is continuous. The function is continuous at the 'corner' and the 'cusp', but we consider these 'sharp turns', so these are places where the function fails to be smooth. Apart from these four places, the function is smooth and continuous. Polynomial functions are smooth and continuous everywhere, as exhibited in the graph on the right. 'hole' 'corner' 'cusp' 'break' Pathologies not found on graphs of polynomials The graph of a polynomial The notion of smoothness is what tells us graphically that, for example, f (x) = |x|, whose graph is the characteristic '' shape, cannot be a polynomial. The notion of continuity is what allowed us to construct the sign diagram for quadratic inequalities as we did in Section 2.4. This last result is formalized in the following theorem. Theorem 3.1. The Intermediate Value Theorem (Zero Version): Suppose f is a continuous function on an interval containing x = a and x = b with a < b. If f (a) and f (b) have dierent signs, then f has at least one zero between x = a and x = b; that is, for at least one real number c such that a < c < b, we have f (c) = 0. The Intermediate Value Theorem is extremely profound; it gets to the heart of what it means to be a real number, and is one of the most often used and under appreciated theorems in Mathematics. With that being said, most students see the result as common sense since it says, geometrically, that the graph of a polynomial function cannot be above the x-axis at one point and below the x-axis at another point without crossing the x-axis somewhere in between. The following example uses the Intermediate Value Theorem to establish a fact that that most students take for granted. Many students, and sadly some instructors, will nd it silly. 14 In fact, if you take Calculus, you'll nd that smooth functions are automatically continuous, so that saying 'polynomials are continuous and smooth' is redundant. 242 Polynomial Functions Example 3.1.4. Use the Intermediate Value Theorem to establish that 2 is a real number. x2 Solution. Consider the polynomial function f (x) = 2. Then f (1) = 1 and f (3) = 7. Since f (1) and f (3) have dierent signs, the Intermediate Value Theorem guarantees us a real number c between 1 and 3 with f (c) = 0. If c2 2 = 0 then c = 2. Since c is between 1 and 3, c is positive, so c = 2. Our primary use of the Intermediate Value Theorem is in the construction of sign diagrams, as in Section 2.4, since it guarantees us that polynomial functions are always positive (+) or always negative () on intervals which do not contain any of its zeros. The general algorithm for polynomials is given below. Steps for Constructing a Sign Diagram for a Polynomial Function Suppose f is a polynomial function. 1. Find the zeros of f and place them on the number line with the number 0 above them. 2. Choose a real number, called a test value, in each of the intervals determined in step 1. 3. Determine the sign of f (x) for each test value in step 2, and write that sign above the corresponding interval. Example 3.1.5. Construct a sign diagram for f (x) = x3 (x 3)2 (x + 2) x2 + 1 . Use it to give a rough sketch of the graph of y = f (x). Solution. First, we nd the zeros of f by solving x3 (x 3)2 (x + 2) x2 + 1 = 0. We get x = 0, x = 3 and x = 2. (The equation x2 + 1 = 0 produces no real solutions.) These three points divide the real number line into four intervals: (, 2), (2, 0), (0, 3) and (3, ). We select the test values x = 3, x = 1, x = 1 and x = 4. We nd f (3) is (+), f (1) is () and f (1) is (+) as is f (4). Wherever f is (+), its graph is above the x-axis; wherever f is (), its graph is below the x-axis. The x-intercepts of the graph of f are (2, 0), (0, 0) and (3, 0). Knowing f is smooth and continuous allows us to sketch its graph. y (+) 0 () 0 (+) 0 (+) 2 0 3 3 1 1 x 4 A sketch of y = f (x) A couple of notes about the Example 3.1.5 are in order. First, note that we purposefully did not label the y-axis in the sketch of the graph of y = f (x). This is because the sign diagram gives us the zeros and the relative position of the graph - it doesn't give us any information as to how high or low the graph strays from the x-axis. Furthermore, as we have mentioned earlier in the text, without Calculus, the values of the relative maximum and minimum can only be found approximately using a calculator. If we took the time to nd the leading term of f , we would nd it to be x8 . Looking 3.1 Graphs of Polynomials 243 at the end behavior of f , we notice that it matches the end behavior of y = x8 . This is no accident, as we nd out in the next theorem. Theorem 3.2. End Behavior for Polynomial Functions: The end behavior of a polynomial f (x) = an xn +an1 xn1 +. . .+a2 x2 +a1 x+a0 with an = 0 matches the end behavior of y = an xn . To see why Theorem 3.2 is true, let's rst look at a specic example. Consider f (x) = 4x3 x + 5. If we wish to examine end behavior, we look to see the behavior of f as x . Since we're concerned with x's far down the x-axis, we are far away from x = 0 so can rewrite f (x) for these values of x as 1 5 f (x) = 4x3 1 2 + 3 4x 4x As x becomes unbounded (in either direction), the terms 0, as the table below indicates. 1 4x2 x 1000 100 10 10 100 1000 1 4x2 and 5 4x3 become closer and closer to 5 4x3 0.00000025 0.00000000125 0.000025 0.00000125 0.0025 0.00125 0.0025 0.00125 0.000025 0.00000125 0.00000025 0.00000000125 In other words, as x , f (x) 4x3 (1 0 + 0) = 4x3 , which is the leading term of f . The formal proof of Theorem 3.2 works in much the same way. Factoring out the leading term leaves f (x) = an xn 1 + an1 a2 a1 a0 + ... + + + an x an xn2 an xn1 an xn As x , any term with an x in the denominator becomes closer and closer to 0, and we have f (x) an xn . Geometrically, Theorem 3.2 says that if we graph y = f (x) using a graphing calculator, and continue to 'zoom out', the graph of it and its leading term become indistinguishable. Below are the graphs of y = 4x3 x + 5 (the thicker line) and y = 4x3 (the thinner line) in two dierent windows. A view 'close' to the origin. A 'zoomed out' view. 244 Polynomial Functions Let's return to the function in Example 3.1.5, f (x) = x3 (x3)2 (x+2) x2 + 1 , whose sign diagram and graph are reproduced below for reference. Theorem 3.2 tells us that the end behavior is the same as that of its leading term x8 . This tells us that the graph of y = f (x) starts and ends above the x-axis. In other words, f (x) is (+) as x , and as a result, we no longer need to evaluate f at the test values x = 3 and x = 4. Is there a way to eliminate the need to evaluate f at the other test values? What we would really need to know is how the function behaves near its zeros does it cross through the x-axis at these points, as it does at x = 2 and x = 0, or does it simply touch and rebound like it does at x = 3. From the sign diagram, the graph of f will cross the x-axis whenever the signs on either side of the zero switch (like they do at x = 2 and x = 0); it will touch when the signs are the same on either side of the zero (as is the case with x = 3). What we need to determine is the reason behind whether or not the sign change occurs. y (+) 0 () 0 (+) 0 (+) 2 0 3 3 1 1 x 4 A sketch of y = f (x) Fortunately, f was given to us in factored form: f (x) = x3 (x 3)2 (x + 2). When we attempt to determine the sign of f (4), we are attempting to nd the sign of the number (4)3 (7)2 (2), which works out to be ()(+)() which is (+). If we move to the other side of x = 2, and nd the sign of f (1), we are determining the sign of (1)3 (4)2 (+1), which is ()(+)(+) which gives us the (). Notice that signs of the rst two factors in both expressions are the same in f (4) and f (1). The only factor which switches sign is the third factor, (x + 2), precisely the factor which gave us the zero x = 2. If we move to the other side of 0 and look closely at f (1), we get the sign pattern (+1)3 (2)2 (+3) or (+)(+)(+) and we note that, once again, going from f (1) to f (1), the only factor which changed sign was the rst factor, x3 , which corresponds to the zero x = 0. Finally, to nd f (4), we substitute to get (+4)3 (+2)2 (+5) which is (+)(+)(+) or (+). The sign didn't change for the middle factor (x 3)2 . Even though this is the factor which corresponds to the zero x = 3, the fact that the quantity is squared kept the sign of the middle factor the same on either side of 3. If we look back at the exponents on the factors (x + 2) and x3 , we see that they are both odd, so as we substitute values to the left and right of the corresponding zeros, the signs of the corresponding factors change which results in the sign of the function value changing. This is the key to the behavior of the function near the zeros. We need a denition and then a theorem. Denition 3.3. Suppose f is a polynomial function and m is a natural number. If (x c)m is a factor of f (x) but (x c)m+1 is not, then we say x = c is a zero of multiplicity m. Hence, rewriting f (x) = x3 (x 3)2 (x + 2) as f (x) = (x 0)3 (x 3)2 (x (2))1 , we see that x = 0 is a zero of multiplicity 3, x = 3 is a zero of multiplicity 2 and x = 2 is a zero of multiplicity 1. 3.1 Graphs of Polynomials 245 Theorem 3.3. The Role of Multiplicity: Suppose f is a polynomial function and x = c is a zero of multiplicity m. If m is even, the graph of y = f (x) touches and rebounds from the x-axis at (c, 0). If m is odd, the graph of y = f (x) crosses through the x-axis at (c, 0). Our last example shows how end behavior and multiplicity allow us to sketch a decent graph without appealing to a sign diagram. Example 3.1.6. Sketch the graph of f (x) = 3(2x 1)(x + 1)2 using end behavior and the multiplicity of its zeros. Solution. The end behavior of the graph of f will match that of its leading term. To nd the leading term, we multiply by the leading terms of each factor to get (3)(2x)(x)2 = 6x3 . This tells us that the graph will start above the x-axis, in Quadrant II, and nish below the x-axis, in Quadrant IV. Next, we nd the zeros of f . Fortunately for us, f is factored.15 Setting each factor 1 1 equal to zero gives is x = 2 and x = 1 as zeros. To nd the multiplicity of x = 2 we note that it corresponds to the factor (2x 1). This isn't strictly in the form required in Denition 3.3. If we factor out the 2, however, we get (2x 1) = 2 x 1 , and we see that the multiplicity of x = 1 2 2 is 1. Since 1 is an odd number, we know from Theorem 3.3 that the graph of f will cross through the x-axis at 1 , 0 . Since the zero x = 1 corresponds to the factor (x + 1)2 = (x (1))2 , 2 we nd its multiplicity to be 2 which is an even number. As such, the graph of f will touch and rebound from the x-axis at (1, 0). Though we're not asked to, we can nd the y-intercept by nding f (0) = 3(2(0) 1)(0 + 1)2 = 3. Thus (0, 3) is an additional point on the graph. Putting this together gives us the graph below. y x 15 Obtaining the factored form of a polynomial is the main focus of the next few sections. 246 3.1.1 Polynomial Functions Exercises In Exercises 1 - 10, nd the degree, the leading term, the leading coecient, the constant term and the end behavior of the given polynomial. 1. f (x) = 4 x 3x2 2. g(x) = 3x5 2x2 + x + 1 3. q(r) = 1 16r4 5. f (x) = 3x17 + 22.5x10 x7 + 4. Z(b) = 42b b3 1 3 7. P (x) = (x 1)(x 2)(x 3)(x 4) 9. f (x) = 2x3 (x + 1)(x + 2)2 6. s(t) = 4.9t2 + v0 t + s0 8. p(t) = t2 (3 5t)(t2 + t + 4) 10. G(t) = 4(t 2)2 t + 1 2 In Exercises 11 - 20, nd the real zeros of the given polynomial and their corresponding multiplicities. Use this information along with a sign chart to provide a rough sketch of the graph of the polynomial. Compare your answer with the result from a graphing utility. 11. a(x) = x(x + 2)2 12. g(x) = x(x + 2)3 13. f (x) = 2(x 2)2 (x + 1) 14. g(x) = (2x + 1)2 (x 3) 15. F (x) = x3 (x + 2)2 16. P (x) = (x 1)(x 2)(x 3)(x 4) 17. Q(x) = (x + 5)2 (x 3)4 18. h(x) = x2 (x 2)2 (x + 2)2 19. H(t) = (3 t)(t2 + 1) 20. Z(b) = b(42 b2 ) In Exercises 21 - 26, given the pair of functions f and g, sketch the graph of y = g(x) by starting with the graph of y = f (x) and using transformations. Track at least three points of your choice through the transformations. State the domain and range of g. 21. f (x) = x3 , g(x) = (x + 2)3 + 1 22. f (x) = x4 , g(x) = (x + 2)4 + 1 23. f (x) = x4 , g(x) = 2 3(x 1)4 24. f (x) = x5 , g(x) = x5 3 25. f (x) = x5 , g(x) = (x + 1)5 + 10 26. f (x) = x6 , g(x) = 8 x6 27. Use the Intermediate Value Theorem to prove that f (x) = x3 9x + 5 has a real zero in each of the following intervals: [4, 3], [0, 1] and [2, 3]. 28. Rework Example 3.1.3 assuming the box is to be made from an 8.5 inch by 11 inch sheet of paper. Using scissors and tape, construct the box. Are you surprised?16 16 Consider decorating the box and presenting it to your instructor. If done well enough, maybe your instructor will issue you some bonus points. Or maybe not. 3.1 Graphs of Polynomials 247 In Exercises 29 - 31, suppose the revenue R, in thousands of dollars, from producing and selling x hundred LCD TVs is given by R(x) = 5x3 + 35x2 + 155x for 0 x 10.07. 29. Use a graphing utility to graph y = R(x) and determine the number of TVs which should be sold to maximize revenue. What is the maximum revenue? 30. Assume that the cost, in thousands of dollars, to produce x hundred LCD TVs is given by C(x) = 200x + 25 for x 0. Find and simplify an expression for the prot function P (x). (Remember: Prot = Revenue - Cost.) 31. Use a graphing utility to graph y = P (x) and determine the number of TVs which should be sold to maximize prot. What is the maximum prot? 32. While developing their newest game, Sasquatch Attack!, the makers of the PortaBoy (from Example 2.1.5) revised their cost function and now use C(x) = .03x3 4.5x2 + 225x + 250, for x 0. As before, C(x) is the cost to make x PortaBoy Game Systems. Market research indicates that the demand function p(x) = 1.5x + 250 remains unchanged. Use a graphing utility to nd the production level x that maximizes the prot made by producing and selling x PortaBoy game systems. 33. According to US Postal regulations, a rectangular shipping box must satisfy the inequality \"Length + Girth 130 inches\" for Parcel Post and \"Length + Girth 108 inches\" for other services. Let's assume we have a closed rectangular box with a square face of side length x as drawn below. The length is the longest side and is clearly labeled. The girth is the distance around the box in the other two dimensions so in our case it is the sum of the four sides of the square, 4x. (a) Assuming that we'll be mailing a box via Parcel Post where Length + Girth = 130 inches, express the length of the box in terms of x and then express the volume V of the box in terms of x. (b) Find the dimensions of the box of maximum volume that can be shipped via Parcel Post. (c) Repeat parts 33a and 33b if the box is shipped using \"other services\". x length x 248 Polynomial Functions 34. We now revisit the data set from Exercise 6b in Section 2.5. In that exercise, you were given a chart of the number of hours of daylight they get on the 21st of each month in Fairbanks, Alaska based on the 2009 sunrise and sunset data found on the U.S. Naval Observatory website. We let x = 1 represent January 21, 2009, x = 2 represent February 21, 2009, and so on. The chart is given again for reference. Month Number Hours of Daylight 1 2 3 4 5 6 7 8 9 10 11 12 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 Find cubic (third degree) and quartic (fourth degree) polynomials which model this data and comment on the goodness of t for each. What can we say about using either model to make predictions about the year 2020? (Hint: Think about the end behavior of polynomials.) Use the models to see how many hours of daylight they got on your birthday and then check the website to see how accurate the models are. Knowing that Sasquatch are largely nocturnal, what days of the year according to your models are going to allow for at least 14 hours of darkness for eld research on the elusive creatures? 35. An electric circuit is built with a variable resistor installed. For each of the following resistance values (measured in kilo-ohms, k), the corresponding power to the load (measured in milliwatts, mW ) is given in the table below. 17 Resistance: (k) Power: (mW ) 1.012 1.063 2.199 1.496 3.275 1.610 4.676 1.613 6.805 1.505 9.975 1.314 (a) Make a scatter diagram of the data using the Resistance as the independent variable and Power as the dependent variable. (b) Use your calculator to nd quadratic (2nd degree), cubic (3rd degree) and quartic (4th degree) regression models for the data and judge the reasonableness of each. (c) For each of the models found above, nd the predicted maximum power that can be delivered to the load. What is the corresponding resistance value? (d) Discuss with your classmates the limitations of these models - in particular, discuss the end behavior of each. 36. Show that the end behavior of a linear function f (x) = mx + b is as it should be according to the results we've established in the section for polynomials of odd degree.18 (That is, show that the graph of a linear function is \"up on one side and down on the other\" just like the graph of y = an xn for odd numbers n.) 17 The authors wish to thank Don Anthan and Ken White of Lakeland Community College for devising this problem and generating the accompanying data set. 18 Remember, to be a linear function, m = 0. 3.1 Graphs of Polynomials 249 37. There is one subtlety about the role of multiplicity that we need to discuss further; specically we need to see 'how' the graph crosses the x-axis at a zero of odd multiplicity. In the section, we deliberately excluded the function f (x) = x from the discussion of the end behavior of f (x) = xn for odd numbers n and we said at the time that it was due to the fact that f (x) = x didn't t the pattern we were trying to establish. You just showed in the previous exercise that the end behavior of a linear function behaves like every other polynomial of odd degree, so what doesn't f (x) = x do that g(x) = x3 does? It's the 'attening' for values of x near zero. It is this local behavior that will distinguish between a zero of multiplicity 1 and one of higher odd multiplicity. Look again closely at the graphs of a(x) = x(x + 2)2 and F (x) = x3 (x + 2)2 from Exercise 3.1.1. Discuss with your classmates how the graphs are fundamentally dierent at the origin. It might help to use a graphing calculator to zoom in on the origin to see the dierent crossing behavior. Also compare the behavior of a(x) = x(x + 2)2 to that of g(x) = x(x + 2)3 near the point (2, 0). What do you predict will happen at the zeros of f (x) = (x 1)(x 2)2 (x 3)3 (x 4)4 (x 5)5 ? 38. Here are a few other questions for you to discuss with your classmates. (a) How many local extrema could a polynomial of degree n have? How few local extrema can it have? (b) Could a polynomial have two local maxima but no local minima? (c) If a polynomial has two local maxima and two local minima, can it be of odd degree? Can it be of even degree? (d) Can a polynomial have local extrema without having any real zeros? (e) Why must every polynomial of odd degree have at least one real zero? (f) Can a polynomial have two distinct real zeros and no local extrema? (g) Can an x-intercept yield a local extrema? Can it yield an absolute extrema? (h) If the y-intercept yields an absolute minimum, what can we say about the degree of the polynomial and the sign of the leading coecient? 250 3.1.2 Polynomial Functions Answers 1. f (x) = 4 x 3x2 Degree 2 Leading term 3x2 Leading coecient 3 Constant term 4 As x , f (x) As x , f (x) 2. g(x) = 3x5 2x2 + x + 1 Degree 5 Leading term 3x5 Leading coecient 3 Constant term 1 As x , g(x) As x , g(x) 3. q(r) = 1 16r4 Degree 4 Leading term 16r4 Leading coecient 16 Constant term 1 As r , q(r) As r , q(r) 4. Z(b) = 42b b3 Degree 3 Leading term b3 Leading coecient 1 Constant term 0 As b , Z(b) As b , Z(b) 5. f (x) = 3x17 + 22.5x10 x7 + Degree 17 Leading term 3x17 Leading coecient 3 Constant term 1 3 As x , f (x) As x , f (x) 1 3 7. P (x) = (x 1)(x 2)(x 3)(x 4) Degree 4 Leading term x4 Leading coecient 1 Constant term 24 As x , P (x) As x , P (x) 6. s(t) = 4.9t2 + v0 t + s0 Degree 2 Leading term 4.9t2 Leading coecient 4.9 Constant term s0 As t , s(t) As t , s(t) 8. p(t) = t2 (3 5t)(t2 + t + 4) Degree 5 Leading term 5t5 Leading coecient 5 Constant term 0 As t , p(t) As t , p(t) 3.1 Graphs of Polynomials 9. f (x) = 2x3 (x + 1)(x + 2)2 Degree 6 Leading term 2x6 Leading coecient 2 Constant term 0 As x , f (x) As x , f (x) 11. a(x) = x(x + 2)2 x = 0 multiplicity 1 x = 2 multiplicity 2 251 1 10. G(t) = 4(t 2)2 t + 2 Degree 3 Leading term 4t3 Leading coecient 4 Constant term 8 As t , G(t) As t , G(t) 12. g(x) = x(x + 2)3 x = 0 multiplicity 1 x = 2 multiplicity 3 y 2 y x 1 2 13. f (x) = 2(x 2)2 (x + 1) x = 2 multiplicity 2 x = 1 multiplicity 1 1 x 14. g(x) = (2x + 1)2 (x 3) 1 x = 2 multiplicity 2 x = 3 multiplicity 1 y 2 1 y 1 2 x 1 1 2 3 x 252 Polynomial Functions 15. F (x) = x3 (x + 2)2 x = 0 multiplicity 3 x = 2 multiplicity 2 16. P (x) = (x 1)(x 2)(x 3)(x 4) x = 1 multiplicity 1 x = 2 multiplicity 1 x = 3 multiplicity 1 x = 4 multiplicity 1 y y 2 x 1 1 17. Q(x) = (x + 5)2 (x 3)4 x = 5 multiplicity 2 x = 3 multiplicity 4 2 3 x 4 18. f (x) = x2 (x 2)2 (x + 2)2 x = 2 multiplicity 2 x = 0 multiplicity 2 x = 2 multiplicity 2 y y 54321 1 2 3 4 5 x 2 19. H(t) = (3 t) t2 + 1 x = 3 multiplicity 1 y 1 1 x 2 20. Z(b) = b(42 b2 ) b = 42 multiplicity 1 b = multiplicity 1 0 b = 42 multiplicity 1 y 1 2 3 t 654321 1 2 3 4 5 6 b 3.1 Graphs of Polynomials 253 22. g(x) = (x + 2)4 + 1 domain: (, ) range: [1, ) 21. g(x) = (x + 2)3 + 1 domain: (, ) range: (, ) y y 12 11 10 9 8 7 6 5 4 3 2 1 4 3 2 1 1 2 3 4 5 6 7 8 9 10 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 x 4 23. g(x) = 2 3(x 1)4 domain: (, ) range: (, 2] 3 1 24. g(x) = x5 3 domain: (, ) range: (, ) y y 2 1 1 2 3 4 5 6 7 8 9 10 11 12 13 2 1 2 x 10 9 8 7 6 5 4 3 2 1 1 1 2 3 4 5 6 7 8 9 10 1 x x 254 Polynomial Functions 26. g(x) = 8 x6 domain: (, ) range: (, 8] 25. g(x) = (x + 1)5 + 10 domain: (, ) range: (, ) y 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 4 3 2 1 y 10 9 8 7 6 5 4 3 2 1 x 1 1 2 3 4 5 6 7 8 9 10 1 x 27. We have f (4) = 23, f (3) = 5, f (0) = 5, f (1) = 3, f (2) = 5 and f (3) = 5 so the Intermediate Value Theorem tells us that f (x) = x3 9x + 5 has real zeros in the intervals [4, 3], [0, 1] and [2, 3]. 28. V (x) = x(8.5 2x)(11 2x) = 4x3 39x2 + 93.5x, 0 < x < 4.25. Volume is maximized when x 1.58, so the dimensions of the box with maximum volume are: height 1.58 inches, width 5.34 inches, and depth 7.84 inches. The maximum volume is 66.15 cubic inches. 29. The calculator gives the location of the absolute maximum (rounded to three decimal places) as x 6.305 and y 1115.417. Since x represents the number of TVs sold in hundreds, x = 6.305 corresponds to 630.5 TVs. Since we can't sell half of a TV, we compare R(6.30) 1115.415 and R(6.31) 1115.416, so selling 631 TVs results in a (slightly) higher revenue. Since y represents the revenue in thousands of dollars, the maximum revenue is $1,115,416. 30. P (x) = R(x) C(x) = 5x3 + 35x2 45x 25, 0 x 10.07. 31. The calculator gives the location of the absolute maximum (rounded to three decimal places) as x 3.897 and y 35.255. Since x represents the number of TVs sold in hundreds, x = 3.897 corresponds to 389.7 TVs. Since we can't sell 0.7 of a TV, we compare P (3.89) 35.254 and P (3.90) 35.255, so selling 390 TVs results in a (slightly) higher revenue. Since y represents the revenue in thousands of dollars, the maximum revenue is $35,255. 32. Making and selling 71 PortaBoys yields a maximized prot of $5910.67. 3.1 Graphs of Polynomials 255 33. (a) Our ultimate goal is to maximize the volume, so we'll start with the maximum Length + Girth of 130. This means the length is 130 4x. The volume of a rectangular box is always length width height so we get V (x) = x2 (130 4x) = 4x3 + 130x2 . (b) Graphing y = V (x) on [0, 33] [0, 21000] shows a maximum at (21.67, 20342.59) so the dimensions of the box with maximum volume are 21.67in. 21.67in. 43.32in. for a volume of 20342.59in.3 . (c) If we start with Length + Girth = 108 then the length is 108 4x and the volume is V (x) = 4x3 + 108x2 . Graphing y = V (x) on [0, 27] [0, 11700] shows a maximum at (18.00, 11664.00) so the dimensions of the box with maximum volume are 18.00in. 18.00in. 36in. for a volume of 11664.00in.3 . (Calculus will conrm that the measurements which maximize the volume are exactly 18in. by 18in. by 36in., however, as I'm sure you are aware by now, we treat all calculator results as approximations and list them as such.) 34. The cubic regression model is p3 (x) = 0.0226x3 0.9508x2 + 8.615x 3.446. It has R2 = 0.93765 which isn't bad. The graph of y = p3 (x) in the viewing window [1, 13] [0, 24] along with the scatter plot is shown below on the left. Notice that p3 hits the x-axis at about x = 12.45 making this a bad model for future predictions. To use the model to approximate the number of hours of sunlight on your birthday, you'll have to gure out what decimal value of x is close enough to your birthday and then plug it into the model. My (Je's) birthday is July 31 which is 10 days after July 21 (x = 7). Assuming 30 days in a month, I think x = 7.33 should work for my birthday and p3 (7.33) 17.5. The website says there will be about 18.25 hours of daylight that day. To have 14 hours of darkness we need 10 hours of daylight. We see that p3 (1.96) 10 and p3 (10.05) 10 so it seems reasonable to say that we'll have at least 14 hours of darkness from December 21, 2008 (x = 0) to February 21, 2009 (x = 2) and then again from October 21,2009 (x = 10) to December 21, 2009 (x = 12). The quartic regression model is p4 (x) = 0.0144x4 0.3507x3 + 2.259x2 1.571x+ 5.513. It has R2 = 0.98594 which is good. The graph of y = p4 (x) in the viewing window [1, 15] [0, 35] along with the scatter plot is shown below on the right. Notice that p4 (15) is above 24 making this a bad model as well for future predictions. However, p4 (7.33) 18.71 making it much better at predicting the hours of daylight on July 31 (my birthday). This model says we'll have at least 14 hours of darkness from December 21, 2008 (x = 0) to about March 1, 2009 (x = 2.30) and then again from October 10, 2009 (x = 9.667) to December 21, 2009 (x = 12). y = p3 (x) y = p4 (x) 256 Polynomial Functions 35. (a) The scatter plot is shown below with each of the three regression models. (b) The quadratic model is P2 (x) = 0.02x2 + 0.241x + 0.956 with R2 = 0.77708. The cubic model is P3 (x) = 0.005x3 0.103x2 + 0.602x + 0.573 with R2 = 0.98153. The quartic model is P4 (x) = 0.000969x4 + 0.0253x3 0.240x2 + 0.944x + 0.330 with R2 = 0.99929. (c) The maximums predicted by the three models are P2 (5.737) 1.648, P3 (4.232) 1.657 and P4 (3.784) 1.630, respectively. y = P2 (x) y = P3 (x) y = P4 (x)