{"id":5386,"date":"2021-10-13T21:33:27","date_gmt":"2021-10-13T21:33:27","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/?post_type=chapter&#038;p=5386"},"modified":"2022-04-25T19:46:11","modified_gmt":"2022-04-25T19:46:11","slug":"page","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/chapter\/page\/","title":{"raw":"Choosing Appropriate Models","rendered":"Choosing Appropriate Models"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Choose an appropriate model for data.<\/li>\r\n \t<li>Use a graphing utility to create an exponential regression from a set of data.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h3>Choosing an Appropriate Model<\/h3>\r\nNow that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015.\r\n\r\nThree kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model although other models, such as quadratic models, may also be considered.\r\n\r\nIn choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave down away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up and a logarithmic curve always concave down.\r\n\r\nA logistic curve changes concavity. It starts out concave up and then changes to concave down beyond a certain point, called a point of inflection.\r\n\r\nAfter using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Choosing a Mathematical Model<\/h3>\r\nDoes a linear, exponential, logarithmic, or logistic model best fit the values listed below? Find the model, and use a graph to check your choice.\r\n<table summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<td>4<\/td>\r\n<td>5<\/td>\r\n<td>6<\/td>\r\n<td>7<\/td>\r\n<td>8<\/td>\r\n<td>9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em><strong>y<\/strong><\/em><\/td>\r\n<td>0<\/td>\r\n<td>1.386<\/td>\r\n<td>2.197<\/td>\r\n<td>2.773<\/td>\r\n<td>3.219<\/td>\r\n<td>3.584<\/td>\r\n<td>3.892<\/td>\r\n<td>4.159<\/td>\r\n<td>4.394<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"852220\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"852220\"]\r\n\r\nFirst, plot the data on a graph as in the graph below. For the purpose of graphing, round the data to two significant digits.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181351\/CNX_Precalc_Figure_04_07_0082.jpg\" alt=\"Graph of the previous table\u2019s values.\" width=\"487\" height=\"476\" \/>\r\n\r\nClearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try [latex]y=a\\mathrm{ln}\\left(bx\\right)[\/latex]. Plugging in the first point, [latex]\\left(\\text{1,0}\\right)[\/latex], gives [latex]0=a\\mathrm{ln}b[\/latex]. We reject the case that <em>a\u00a0<\/em>= 0 (if it were, all outputs would be 0), so we know\r\n<p style=\"text-align: center;\">[latex]\\mathrm{ln}\\left(b\\right)=0[\/latex]. Thus <em>b\u00a0<\/em>= 1 and [latex]y=a\\mathrm{ln}\\left(\\text{x}\\right)[\/latex]. Next we can use the point [latex]\\left(\\text{9,4}\\text{.394}\\right)[\/latex] to solve for <em>a<\/em>:\r\n[latex]\\begin{array}{l}y=a\\mathrm{ln}\\left(x\\right)\\hfill \\\\ 4.394=a\\mathrm{ln}\\left(9\\right)\\hfill \\\\ a=\\frac{4.394}{\\mathrm{ln}\\left(9\\right)}\\hfill \\end{array}[\/latex]<\/p>\r\nBecause [latex]a=\\frac{4.394}{\\mathrm{ln}\\left(9\\right)}\\approx 2[\/latex], an appropriate model for the data is [latex]y=2\\mathrm{ln}\\left(x\\right)[\/latex].\r\n\r\nTo check the accuracy of the model, we graph the function together with the given points.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181353\/CNX_Precalc_Figure_04_07_009a2.jpg\" alt=\"Graph of previous table\u2019s values showing that it fits the function y=2ln(x) with an asymptote at x=0.\" width=\"487\" height=\"476\" \/> The graph of [latex]y=2\\mathrm{ln}x[\/latex].[\/caption]We can conclude that the model is a good fit to the data. Compare the figure above\u00a0to the graph of [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex] shown below.[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181356\/CNX_Precalc_Figure_04_07_009b2.jpg\" alt=\"Graph of previous table\u2019s values showing that it fits the function y=2ln(x) with an asymptote at x=0.\" width=\"487\" height=\"476\" \/> The graph of [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex][\/caption]The graphs appear to be identical when <em>x\u00a0<\/em>&gt; 0. A quick check confirms this conclusion: [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)=2\\mathrm{ln}\\left(x\\right)[\/latex] for <em>x\u00a0<\/em>&gt; 0. However, if <em>x\u00a0<\/em>&lt; 0, the graph of [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex] includes an \"extra\" branch as shown below. This occurs because while [latex]y=2\\mathrm{ln}\\left(x\\right)[\/latex] cannot have negative values in the domain (as such values would force the argument to be negative), the function [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex] can have negative domain values.<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181358\/CNX_Precalc_Figure_04_07_0102.jpg\" alt=\"Graph of y=ln(x^2).\" width=\"487\" height=\"216\" \/>[\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nDoes a linear, exponential, or logarithmic model best fit the data in the table below? Find the model.\r\n<table summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<td>4<\/td>\r\n<td>5<\/td>\r\n<td>6<\/td>\r\n<td>7<\/td>\r\n<td>8<\/td>\r\n<td>9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em><strong>y<\/strong><\/em><\/td>\r\n<td>3.297<\/td>\r\n<td>5.437<\/td>\r\n<td>8.963<\/td>\r\n<td>14.778<\/td>\r\n<td>24.365<\/td>\r\n<td>40.172<\/td>\r\n<td>66.231<\/td>\r\n<td>109.196<\/td>\r\n<td>180.034<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"842897\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"842897\"]\r\n\r\nExponential. [latex]y=2{e}^{0.5x}[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Expressing an Exponential Model in Base <em>e<\/em><\/h3>\r\nWhile powers and logarithms of any base can be used in modeling, the two most common bases are [latex]10[\/latex] and [latex]e[\/latex]. In science and mathematics, the base <em>e<\/em>\u00a0is often preferred. We can use properties of exponents and properties of logarithms to change any base to base <em>e<\/em>.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a model with the form [latex]y=a{b}^{x}[\/latex], change it to the form [latex]y={A}_{0}{e}^{kx}[\/latex]<\/h3>\r\n<ol>\r\n \t<li>Rewrite [latex]y=a{b}^{x}[\/latex] as [latex]y=a{e}^{\\mathrm{ln}\\left({b}^{x}\\right)}[\/latex].<\/li>\r\n \t<li>Use the power rule of logarithms to rewrite as [latex]y=a{e}^{x\\mathrm{ln}\\left(b\\right)}=a{e}^{\\mathrm{ln}\\left(b\\right)x}[\/latex].<\/li>\r\n \t<li>Note that [latex]a={A}_{0}[\/latex] and [latex]k=\\mathrm{ln}\\left(b\\right)[\/latex] in the equation [latex]y={A}_{0}{e}^{kx}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Changing to base [latex]e[\/latex]<\/h3>\r\nChange the function [latex]y=2.5{\\left(3.1\\right)}^{x}[\/latex] so that this same function is written in the form [latex]y={A}_{0}{e}^{kx}[\/latex].\r\n\r\n[reveal-answer q=\"726035\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"726035\"]\r\n\r\nThe formula is derived as follows\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y &amp; =2.5{\\left(3.1\\right)}^{x}\\hfill \\\\ \\hfill &amp; =2.5{e}^{\\mathrm{ln}\\left({3.1}^{x}\\right)}\\hfill &amp; \\text{Insert exponential and its inverse}\\text{.}\\hfill \\\\ \\hfill &amp; =2.5{e}^{x\\mathrm{ln}3.1}\\hfill &amp; \\text{Properties of logs}\\text{.}\\hfill \\\\ \\hfill &amp; =2.5{e}^{\\left(\\mathrm{ln}3.1\\right)}{}^{x}\\hfill &amp; \\text{Commutative law of multiplication}\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nChange the function [latex]y=3{\\left(0.5\\right)}^{x}[\/latex] to one having <i>e<\/i>\u00a0as the base.\r\n\r\n[reveal-answer q=\"741534\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"741534\"][latex]y=3{e}^{\\left(\\mathrm{ln}0.5\\right)x}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Exponential Regression<\/h2>\r\nAs we have learned, there are a multitude of situations that can be modeled by exponential functions, such as investment growth, radioactive decay, atmospheric pressure changes, and temperatures of a cooling object. What do these phenomena have in common? For one thing, all the models either increase or decrease as time moves forward. But that\u2019s not the whole story. It\u2019s the <em>way<\/em> data increase or decrease that helps us determine whether it is best modeled by an exponential function. Knowing the behavior of exponential functions in general allows us to recognize when to use exponential regression, so let\u2019s review exponential growth and decay.\r\n\r\nRecall that exponential functions have the form [latex]y=a{b}^{x}[\/latex] or [latex]y={A}_{0}{e}^{kx}[\/latex]. When performing regression analysis, we use the form most commonly used on graphing utilities, [latex]y=a{b}^{x}[\/latex]. Take a moment to reflect on the characteristics we\u2019ve already learned about the exponential function [latex]y=a{b}^{x}[\/latex] (assume <em>a<\/em> &gt; 0):\r\n<ul>\r\n \t<li><em>b<\/em>\u00a0must be greater than zero and not equal to one.<\/li>\r\n \t<li>The initial value of the model is\u00a0<em>a<\/em>.\r\n<ul>\r\n \t<li>If <em>b\u00a0<\/em>&gt; 1, the function models exponential growth. As <em>x<\/em>\u00a0increases, the outputs of the model increase slowly at first, but then increase more and more rapidly, without bound.<\/li>\r\n \t<li>If 0 &lt; <em>b\u00a0<\/em>&lt; 1, the function models <strong>exponential decay<\/strong>. As <em>x<\/em>\u00a0increases, the outputs for the model decrease rapidly at first and then level off to become asymptotic to the <em>x<\/em>-axis. In other words, the outputs never become equal to or less than zero.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\nAs part of the results, your calculator will display a number known as the <em>correlation coefficient<\/em>, labeled by the variable <em>r<\/em>\u00a0or [latex]{r}^{2}[\/latex]. (You may have to change the calculator\u2019s settings for these to be shown.) The values are an indication of the \"goodness of fit\" of the regression equation to the data. We more commonly use the value of [latex]{r}^{2}[\/latex] instead of <em>r<\/em>, but the closer either value is to 1, the better the regression equation approximates the data.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Exponential Regression<\/h3>\r\n<em>Exponential regression<\/em> is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. We use the command \"ExpReg\" on a graphing utility to fit an exponential function to a set of data points. This returns an equation of the form [latex]y=a{b}^{x}[\/latex].\r\n\r\nNote that:\r\n<ul>\r\n \t<li><em>b<\/em>\u00a0must be non-negative.<\/li>\r\n \t<li>When <em>b\u00a0<\/em>&gt; 1, we have an exponential growth model.<\/li>\r\n \t<li>When 0 &lt; <em>b\u00a0<\/em>&lt; 1, we have an exponential decay model.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Exponential Regression to Fit a Model to Data<\/h3>\r\nIn 2007, a university study was published investigating the crash risk of alcohol impaired driving. Data from 2,871 crashes were used to measure the association of a person\u2019s blood alcohol level (BAC) with the risk of being in an accident. The table below\u00a0shows results from the study.[footnote]Source: <em>Indiana University Center for Studies of Law in Action, 2007<\/em>[\/footnote] The <em>relative risk<\/em> is a measure of how many times more likely a person is to crash. So, for example, a person with a BAC of 0.09 is 3.54 times as likely to crash as a person who has not been drinking alcohol.\r\n<table summary=\"Two rows and thirteen columns. The first row is labeled,\">\r\n<tbody>\r\n<tr>\r\n<td><strong>BAC<\/strong><\/td>\r\n<td>0<\/td>\r\n<td>0.01<\/td>\r\n<td>0.03<\/td>\r\n<td>0.05<\/td>\r\n<td>0.07<\/td>\r\n<td>0.09<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Relative Risk of Crashing<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>1.03<\/td>\r\n<td>1.06<\/td>\r\n<td>1.38<\/td>\r\n<td>2.09<\/td>\r\n<td>3.54<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>BAC<\/strong><\/td>\r\n<td>0.11<\/td>\r\n<td>0.13<\/td>\r\n<td>0.15<\/td>\r\n<td>0.17<\/td>\r\n<td>0.19<\/td>\r\n<td>0.21<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Relative Risk of Crashing<\/strong><\/td>\r\n<td>6.41<\/td>\r\n<td>12.6<\/td>\r\n<td>22.1<\/td>\r\n<td>39.05<\/td>\r\n<td>65.32<\/td>\r\n<td>99.78<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ol>\r\n \t<li>Let <em>x<\/em>\u00a0represent the BAC level and let <em>y\u00a0<\/em>represent the corresponding relative risk. Use exponential regression to fit a model to these data.<\/li>\r\n \t<li>After 6 drinks, a person weighing 160 pounds will have a BAC of about 0.16. How many times more likely is a person with this weight to crash if they drive after having a 6-pack of beer? Round to the nearest hundredth.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"134040\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"134040\"]\r\n\r\n1. Using the STAT then EDIT menu on a graphing utility, list the BAC values in L1 and the relative risk values in L2. Then use the STATPLOT feature to verify that the scatterplot follows the exponential pattern shown below:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03175715\/CNX_Precalc_Figure_04_08_0022.jpg\" alt=\"Graph of a scattered plot.\" width=\"487\" height=\"475\" \/>\r\n\r\nUse the \u201cExpReg\u201d command from the STAT then CALC menu to obtain the exponential model,\r\n[latex]y=0.58304829{\\left(\\text{2.207202130E10}\\right)}^{x}[\/latex]\r\n\r\nConverting from scientific notation, we have:[latex]y=0.58304829{\\left(\\text{22,072,021,300}\\right)}^{x}[\/latex]\r\n\r\nNotice that [latex]r2 \\approx 0.97 [\/latex] which indicates the model is a good fit to the data. To see this, graph the model in the same window as the scatterplot to verify it is a good fit as shown in below:\r\n\r\n<img class=\"alignnone size-medium wp-image-4779\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1916\/2016\/11\/25183922\/CNX_Precalc_Figure_04_08_001-300x293.jpg\" alt=\"Graph of a scatter plot of Relative Risk of Crashing vs BAC\" width=\"300\" height=\"293\" \/>\r\n\r\n2. Use the model to estimate the risk associated with a BAC of 0.16. Substitute 0.16 for <em>x<\/em>\u00a0in the model and solve for <em>y<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y\\hfill &amp; =0.58304829{\\left(\\text{22,072,021,300}\\right)}^{x}\\hfill &amp; \\text{Use the regression model found in part (a)}\\text{.}\\hfill \\\\ \\hfill &amp; =0.58304829{\\left(\\text{22,072,021,300}\\right)}^{0.16}\\hfill &amp; \\text{Substitute 0}\\text{.16 for }x\\text{.}\\hfill \\\\ \\hfill &amp; \\approx \\text{26}\\text{.35}\\hfill &amp; \\text{Round to the nearest hundredth}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\r\nIf a 160-pound person drives after having 6 drinks, he or she is about 26.35 times more likely to crash than if driving while sober.\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe table below shows a recent graduate\u2019s credit card balance each month after graduation.\r\n<table summary=\"Two rows and ten columns. The first row is labeled,\">\r\n<tbody>\r\n<tr>\r\n<td><strong>Month<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<td>4<\/td>\r\n<td>5<\/td>\r\n<td>6<\/td>\r\n<td>7<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Debt ($)<\/strong><\/td>\r\n<td>620.00<\/td>\r\n<td>761.88<\/td>\r\n<td>899.80<\/td>\r\n<td>1039.93<\/td>\r\n<td>1270.63<\/td>\r\n<td>1589.04<\/td>\r\n<td>1851.31<\/td>\r\n<td>2154.92<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Use exponential regression to fit a model to these data.<\/li>\r\n \t<li>If spending continues at this rate, what will the graduate\u2019s credit card debt be one year after graduating?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"99197\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"99197\"]\r\n\r\na.\r\n\r\nThe exponential regression model that fits these data is [latex]y=522.88585984{\\left(1.19645256\\right)}^{x}[\/latex].\r\n\r\nb.\r\n\r\nIf spending continues at this rate, the graduate\u2019s credit card debt will be $4,499.38 after one year.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Is it reasonable to assume that an exponential regression model will represent a situation indefinitely?<\/strong>\r\n\r\n<em>No. Remember that models are formed by real-world data gathered for regression. It is usually reasonable to make estimates within the interval of original observation (interpolation). However, when a model is used to make predictions, it is important to use reasoning skills to determine whether the model makes sense for inputs far beyond the original observation interval (extrapolation).<\/em>\r\n\r\n<\/div>\r\n<h2><span style=\"font-size: 1.15em;\">\r\nKey Concepts<\/span><\/h2>\r\n<ul>\r\n \t<li>We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data.<\/li>\r\n \t<li style=\"list-style-type: none;\"><\/li>\r\n \t<li>Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero.<\/li>\r\n \t<li style=\"list-style-type: none;\"><\/li>\r\n \t<li>We use the command \"ExpReg\" on a graphing utility to fit function of the form [latex]y=a{b}^{x}[\/latex] to a set of data points.<\/li>\r\n \t<li style=\"list-style-type: none;\"><\/li>\r\n \t<li>Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound or where decay begins rapidly and then slows down to get closer and closer to zero.<\/li>\r\n \t<li style=\"list-style-type: none;\"><\/li>\r\n \t<li>Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time.<\/li>\r\n \t<li style=\"list-style-type: none;\"><\/li>\r\n \t<li>We use the command \"LnReg\" on a graphing utility to fit a function of the form [latex]y=a+b\\mathrm{ln}\\left(x\\right)[\/latex] to a set of data points.<\/li>\r\n \t<li style=\"list-style-type: none;\"><\/li>\r\n \t<li>Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows as the function approaches an upper limit.<\/li>\r\n \t<li style=\"list-style-type: none;\"><\/li>\r\n \t<li>We use the command \"Logistic\" on a graphing utility to fit a function of the form [latex]y=\\frac{c}{1+a{e}^{-bx}}[\/latex] to a set of data points.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>carrying capacity<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">in a logistic model, the limiting value of the output<\/dd>\r\n<\/dl>\r\n<dl class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>doubling time<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">the time it takes for a quantity to double<\/dd>\r\n<\/dl>\r\n<\/dt>\r\n<\/dl>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>half-life<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">the length of time it takes for a substance to exponentially decay to half of its original quantity<\/dd>\r\n<\/dl>\r\n<dl class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>logistic growth model<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">a function of the form [latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex] where [latex]\\frac{c}{1+a}[\/latex] is the initial value, <em>c<\/em>\u00a0is the carrying capacity, or limiting value, and <em>b<\/em>\u00a0is a constant determined by the rate of growth<\/dd>\r\n<\/dl>\r\n<\/dt>\r\n<\/dl>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Choose an appropriate model for data.<\/li>\n<li>Use a graphing utility to create an exponential regression from a set of data.<\/li>\n<\/ul>\n<\/div>\n<h3>Choosing an Appropriate Model<\/h3>\n<p>Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015.<\/p>\n<p>Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model although other models, such as quadratic models, may also be considered.<\/p>\n<p>In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave down away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up and a logarithmic curve always concave down.<\/p>\n<p>A logistic curve changes concavity. It starts out concave up and then changes to concave down beyond a certain point, called a point of inflection.<\/p>\n<p>After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Choosing a Mathematical Model<\/h3>\n<p>Does a linear, exponential, logarithmic, or logistic model best fit the values listed below? Find the model, and use a graph to check your choice.<\/p>\n<table summary=\"..\">\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<td>6<\/td>\n<td>7<\/td>\n<td>8<\/td>\n<td>9<\/td>\n<\/tr>\n<tr>\n<td><em><strong>y<\/strong><\/em><\/td>\n<td>0<\/td>\n<td>1.386<\/td>\n<td>2.197<\/td>\n<td>2.773<\/td>\n<td>3.219<\/td>\n<td>3.584<\/td>\n<td>3.892<\/td>\n<td>4.159<\/td>\n<td>4.394<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q852220\">Show Solution<\/span><\/p>\n<div id=\"q852220\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, plot the data on a graph as in the graph below. For the purpose of graphing, round the data to two significant digits.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181351\/CNX_Precalc_Figure_04_07_0082.jpg\" alt=\"Graph of the previous table\u2019s values.\" width=\"487\" height=\"476\" \/><\/p>\n<p>Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try [latex]y=a\\mathrm{ln}\\left(bx\\right)[\/latex]. Plugging in the first point, [latex]\\left(\\text{1,0}\\right)[\/latex], gives [latex]0=a\\mathrm{ln}b[\/latex]. We reject the case that <em>a\u00a0<\/em>= 0 (if it were, all outputs would be 0), so we know<\/p>\n<p style=\"text-align: center;\">[latex]\\mathrm{ln}\\left(b\\right)=0[\/latex]. Thus <em>b\u00a0<\/em>= 1 and [latex]y=a\\mathrm{ln}\\left(\\text{x}\\right)[\/latex]. Next we can use the point [latex]\\left(\\text{9,4}\\text{.394}\\right)[\/latex] to solve for <em>a<\/em>:<br \/>\n[latex]\\begin{array}{l}y=a\\mathrm{ln}\\left(x\\right)\\hfill \\\\ 4.394=a\\mathrm{ln}\\left(9\\right)\\hfill \\\\ a=\\frac{4.394}{\\mathrm{ln}\\left(9\\right)}\\hfill \\end{array}[\/latex]<\/p>\n<p>Because [latex]a=\\frac{4.394}{\\mathrm{ln}\\left(9\\right)}\\approx 2[\/latex], an appropriate model for the data is [latex]y=2\\mathrm{ln}\\left(x\\right)[\/latex].<\/p>\n<p>To check the accuracy of the model, we graph the function together with the given points.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181353\/CNX_Precalc_Figure_04_07_009a2.jpg\" alt=\"Graph of previous table\u2019s values showing that it fits the function y=2ln(x) with an asymptote at x=0.\" width=\"487\" height=\"476\" \/><\/p>\n<p class=\"wp-caption-text\">The graph of [latex]y=2\\mathrm{ln}x[\/latex].<\/p>\n<\/div>\n<p>We can conclude that the model is a good fit to the data. Compare the figure above\u00a0to the graph of [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex] shown below.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181356\/CNX_Precalc_Figure_04_07_009b2.jpg\" alt=\"Graph of previous table\u2019s values showing that it fits the function y=2ln(x) with an asymptote at x=0.\" width=\"487\" height=\"476\" \/><\/p>\n<p class=\"wp-caption-text\">The graph of [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex]<\/p>\n<\/div>\n<p>The graphs appear to be identical when <em>x\u00a0<\/em>&gt; 0. A quick check confirms this conclusion: [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)=2\\mathrm{ln}\\left(x\\right)[\/latex] for <em>x\u00a0<\/em>&gt; 0. However, if <em>x\u00a0<\/em>&lt; 0, the graph of [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex] includes an &#8220;extra&#8221; branch as shown below. This occurs because while [latex]y=2\\mathrm{ln}\\left(x\\right)[\/latex] cannot have negative values in the domain (as such values would force the argument to be negative), the function [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex] can have negative domain values.<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181358\/CNX_Precalc_Figure_04_07_0102.jpg\" alt=\"Graph of y=ln(x^2).\" width=\"487\" height=\"216\" \/><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Does a linear, exponential, or logarithmic model best fit the data in the table below? Find the model.<\/p>\n<table summary=\"..\">\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<td>6<\/td>\n<td>7<\/td>\n<td>8<\/td>\n<td>9<\/td>\n<\/tr>\n<tr>\n<td><em><strong>y<\/strong><\/em><\/td>\n<td>3.297<\/td>\n<td>5.437<\/td>\n<td>8.963<\/td>\n<td>14.778<\/td>\n<td>24.365<\/td>\n<td>40.172<\/td>\n<td>66.231<\/td>\n<td>109.196<\/td>\n<td>180.034<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q842897\">Show Solution<\/span><\/p>\n<div id=\"q842897\" class=\"hidden-answer\" style=\"display: none\">\n<p>Exponential. [latex]y=2{e}^{0.5x}[\/latex].<\/p><\/div>\n<\/div>\n<\/div>\n<h3>Expressing an Exponential Model in Base <em>e<\/em><\/h3>\n<p>While powers and logarithms of any base can be used in modeling, the two most common bases are [latex]10[\/latex] and [latex]e[\/latex]. In science and mathematics, the base <em>e<\/em>\u00a0is often preferred. We can use properties of exponents and properties of logarithms to change any base to base <em>e<\/em>.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a model with the form [latex]y=a{b}^{x}[\/latex], change it to the form [latex]y={A}_{0}{e}^{kx}[\/latex]<\/h3>\n<ol>\n<li>Rewrite [latex]y=a{b}^{x}[\/latex] as [latex]y=a{e}^{\\mathrm{ln}\\left({b}^{x}\\right)}[\/latex].<\/li>\n<li>Use the power rule of logarithms to rewrite as [latex]y=a{e}^{x\\mathrm{ln}\\left(b\\right)}=a{e}^{\\mathrm{ln}\\left(b\\right)x}[\/latex].<\/li>\n<li>Note that [latex]a={A}_{0}[\/latex] and [latex]k=\\mathrm{ln}\\left(b\\right)[\/latex] in the equation [latex]y={A}_{0}{e}^{kx}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Changing to base [latex]e[\/latex]<\/h3>\n<p>Change the function [latex]y=2.5{\\left(3.1\\right)}^{x}[\/latex] so that this same function is written in the form [latex]y={A}_{0}{e}^{kx}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q726035\">Show Solution<\/span><\/p>\n<div id=\"q726035\" class=\"hidden-answer\" style=\"display: none\">\n<p>The formula is derived as follows<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y & =2.5{\\left(3.1\\right)}^{x}\\hfill \\\\ \\hfill & =2.5{e}^{\\mathrm{ln}\\left({3.1}^{x}\\right)}\\hfill & \\text{Insert exponential and its inverse}\\text{.}\\hfill \\\\ \\hfill & =2.5{e}^{x\\mathrm{ln}3.1}\\hfill & \\text{Properties of logs}\\text{.}\\hfill \\\\ \\hfill & =2.5{e}^{\\left(\\mathrm{ln}3.1\\right)}{}^{x}\\hfill & \\text{Commutative law of multiplication}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Change the function [latex]y=3{\\left(0.5\\right)}^{x}[\/latex] to one having <i>e<\/i>\u00a0as the base.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q741534\">Show Solution<\/span><\/p>\n<div id=\"q741534\" class=\"hidden-answer\" style=\"display: none\">[latex]y=3{e}^{\\left(\\mathrm{ln}0.5\\right)x}[\/latex]<\/div>\n<\/div>\n<\/div>\n<h2>Exponential Regression<\/h2>\n<p>As we have learned, there are a multitude of situations that can be modeled by exponential functions, such as investment growth, radioactive decay, atmospheric pressure changes, and temperatures of a cooling object. What do these phenomena have in common? For one thing, all the models either increase or decrease as time moves forward. But that\u2019s not the whole story. It\u2019s the <em>way<\/em> data increase or decrease that helps us determine whether it is best modeled by an exponential function. Knowing the behavior of exponential functions in general allows us to recognize when to use exponential regression, so let\u2019s review exponential growth and decay.<\/p>\n<p>Recall that exponential functions have the form [latex]y=a{b}^{x}[\/latex] or [latex]y={A}_{0}{e}^{kx}[\/latex]. When performing regression analysis, we use the form most commonly used on graphing utilities, [latex]y=a{b}^{x}[\/latex]. Take a moment to reflect on the characteristics we\u2019ve already learned about the exponential function [latex]y=a{b}^{x}[\/latex] (assume <em>a<\/em> &gt; 0):<\/p>\n<ul>\n<li><em>b<\/em>\u00a0must be greater than zero and not equal to one.<\/li>\n<li>The initial value of the model is\u00a0<em>a<\/em>.\n<ul>\n<li>If <em>b\u00a0<\/em>&gt; 1, the function models exponential growth. As <em>x<\/em>\u00a0increases, the outputs of the model increase slowly at first, but then increase more and more rapidly, without bound.<\/li>\n<li>If 0 &lt; <em>b\u00a0<\/em>&lt; 1, the function models <strong>exponential decay<\/strong>. As <em>x<\/em>\u00a0increases, the outputs for the model decrease rapidly at first and then level off to become asymptotic to the <em>x<\/em>-axis. In other words, the outputs never become equal to or less than zero.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>As part of the results, your calculator will display a number known as the <em>correlation coefficient<\/em>, labeled by the variable <em>r<\/em>\u00a0or [latex]{r}^{2}[\/latex]. (You may have to change the calculator\u2019s settings for these to be shown.) The values are an indication of the &#8220;goodness of fit&#8221; of the regression equation to the data. We more commonly use the value of [latex]{r}^{2}[\/latex] instead of <em>r<\/em>, but the closer either value is to 1, the better the regression equation approximates the data.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Exponential Regression<\/h3>\n<p><em>Exponential regression<\/em> is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. We use the command &#8220;ExpReg&#8221; on a graphing utility to fit an exponential function to a set of data points. This returns an equation of the form [latex]y=a{b}^{x}[\/latex].<\/p>\n<p>Note that:<\/p>\n<ul>\n<li><em>b<\/em>\u00a0must be non-negative.<\/li>\n<li>When <em>b\u00a0<\/em>&gt; 1, we have an exponential growth model.<\/li>\n<li>When 0 &lt; <em>b\u00a0<\/em>&lt; 1, we have an exponential decay model.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using Exponential Regression to Fit a Model to Data<\/h3>\n<p>In 2007, a university study was published investigating the crash risk of alcohol impaired driving. Data from 2,871 crashes were used to measure the association of a person\u2019s blood alcohol level (BAC) with the risk of being in an accident. The table below\u00a0shows results from the study.<a class=\"footnote\" title=\"Source: Indiana University Center for Studies of Law in Action, 2007\" id=\"return-footnote-5386-1\" href=\"#footnote-5386-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> The <em>relative risk<\/em> is a measure of how many times more likely a person is to crash. So, for example, a person with a BAC of 0.09 is 3.54 times as likely to crash as a person who has not been drinking alcohol.<\/p>\n<table summary=\"Two rows and thirteen columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><strong>BAC<\/strong><\/td>\n<td>0<\/td>\n<td>0.01<\/td>\n<td>0.03<\/td>\n<td>0.05<\/td>\n<td>0.07<\/td>\n<td>0.09<\/td>\n<\/tr>\n<tr>\n<td><strong>Relative Risk of Crashing<\/strong><\/td>\n<td>1<\/td>\n<td>1.03<\/td>\n<td>1.06<\/td>\n<td>1.38<\/td>\n<td>2.09<\/td>\n<td>3.54<\/td>\n<\/tr>\n<tr>\n<td><strong>BAC<\/strong><\/td>\n<td>0.11<\/td>\n<td>0.13<\/td>\n<td>0.15<\/td>\n<td>0.17<\/td>\n<td>0.19<\/td>\n<td>0.21<\/td>\n<\/tr>\n<tr>\n<td><strong>Relative Risk of Crashing<\/strong><\/td>\n<td>6.41<\/td>\n<td>12.6<\/td>\n<td>22.1<\/td>\n<td>39.05<\/td>\n<td>65.32<\/td>\n<td>99.78<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol>\n<li>Let <em>x<\/em>\u00a0represent the BAC level and let <em>y\u00a0<\/em>represent the corresponding relative risk. Use exponential regression to fit a model to these data.<\/li>\n<li>After 6 drinks, a person weighing 160 pounds will have a BAC of about 0.16. How many times more likely is a person with this weight to crash if they drive after having a 6-pack of beer? Round to the nearest hundredth.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q134040\">Show Solution<\/span><\/p>\n<div id=\"q134040\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. Using the STAT then EDIT menu on a graphing utility, list the BAC values in L1 and the relative risk values in L2. Then use the STATPLOT feature to verify that the scatterplot follows the exponential pattern shown below:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03175715\/CNX_Precalc_Figure_04_08_0022.jpg\" alt=\"Graph of a scattered plot.\" width=\"487\" height=\"475\" \/><\/p>\n<p>Use the \u201cExpReg\u201d command from the STAT then CALC menu to obtain the exponential model,<br \/>\n[latex]y=0.58304829{\\left(\\text{2.207202130E10}\\right)}^{x}[\/latex]<\/p>\n<p>Converting from scientific notation, we have:[latex]y=0.58304829{\\left(\\text{22,072,021,300}\\right)}^{x}[\/latex]<\/p>\n<p>Notice that [latex]r2 \\approx 0.97[\/latex] which indicates the model is a good fit to the data. To see this, graph the model in the same window as the scatterplot to verify it is a good fit as shown in below:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-4779\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1916\/2016\/11\/25183922\/CNX_Precalc_Figure_04_08_001-300x293.jpg\" alt=\"Graph of a scatter plot of Relative Risk of Crashing vs BAC\" width=\"300\" height=\"293\" \/><\/p>\n<p>2. Use the model to estimate the risk associated with a BAC of 0.16. Substitute 0.16 for <em>x<\/em>\u00a0in the model and solve for <em>y<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y\\hfill & =0.58304829{\\left(\\text{22,072,021,300}\\right)}^{x}\\hfill & \\text{Use the regression model found in part (a)}\\text{.}\\hfill \\\\ \\hfill & =0.58304829{\\left(\\text{22,072,021,300}\\right)}^{0.16}\\hfill & \\text{Substitute 0}\\text{.16 for }x\\text{.}\\hfill \\\\ \\hfill & \\approx \\text{26}\\text{.35}\\hfill & \\text{Round to the nearest hundredth}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\n<p>If a 160-pound person drives after having 6 drinks, he or she is about 26.35 times more likely to crash than if driving while sober.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The table below shows a recent graduate\u2019s credit card balance each month after graduation.<\/p>\n<table summary=\"Two rows and ten columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><strong>Month<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<td>6<\/td>\n<td>7<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td><strong>Debt ($)<\/strong><\/td>\n<td>620.00<\/td>\n<td>761.88<\/td>\n<td>899.80<\/td>\n<td>1039.93<\/td>\n<td>1270.63<\/td>\n<td>1589.04<\/td>\n<td>1851.31<\/td>\n<td>2154.92<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Use exponential regression to fit a model to these data.<\/li>\n<li>If spending continues at this rate, what will the graduate\u2019s credit card debt be one year after graduating?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q99197\">Show Solution<\/span><\/p>\n<div id=\"q99197\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.<\/p>\n<p>The exponential regression model that fits these data is [latex]y=522.88585984{\\left(1.19645256\\right)}^{x}[\/latex].<\/p>\n<p>b.<\/p>\n<p>If spending continues at this rate, the graduate\u2019s credit card debt will be $4,499.38 after one year.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Is it reasonable to assume that an exponential regression model will represent a situation indefinitely?<\/strong><\/p>\n<p><em>No. Remember that models are formed by real-world data gathered for regression. It is usually reasonable to make estimates within the interval of original observation (interpolation). However, when a model is used to make predictions, it is important to use reasoning skills to determine whether the model makes sense for inputs far beyond the original observation interval (extrapolation).<\/em><\/p>\n<\/div>\n<h2><span style=\"font-size: 1.15em;\"><br \/>\nKey Concepts<\/span><\/h2>\n<ul>\n<li>We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data.<\/li>\n<li style=\"list-style-type: none;\"><\/li>\n<li>Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero.<\/li>\n<li style=\"list-style-type: none;\"><\/li>\n<li>We use the command &#8220;ExpReg&#8221; on a graphing utility to fit function of the form [latex]y=a{b}^{x}[\/latex] to a set of data points.<\/li>\n<li style=\"list-style-type: none;\"><\/li>\n<li>Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound or where decay begins rapidly and then slows down to get closer and closer to zero.<\/li>\n<li style=\"list-style-type: none;\"><\/li>\n<li>Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time.<\/li>\n<li style=\"list-style-type: none;\"><\/li>\n<li>We use the command &#8220;LnReg&#8221; on a graphing utility to fit a function of the form [latex]y=a+b\\mathrm{ln}\\left(x\\right)[\/latex] to a set of data points.<\/li>\n<li style=\"list-style-type: none;\"><\/li>\n<li>Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows as the function approaches an upper limit.<\/li>\n<li style=\"list-style-type: none;\"><\/li>\n<li>We use the command &#8220;Logistic&#8221; on a graphing utility to fit a function of the form [latex]y=\\frac{c}{1+a{e}^{-bx}}[\/latex] to a set of data points.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137838635\" class=\"definition\">\n<dt><strong>carrying capacity<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">in a logistic model, the limiting value of the output<\/dd>\n<\/dl>\n<dl class=\"definition\">\n<dt>\n<\/dt>\n<dt><strong>doubling time<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">the time it takes for a quantity to double<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137838635\" class=\"definition\">\n<dt><strong>half-life<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">the length of time it takes for a substance to exponentially decay to half of its original quantity<\/dd>\n<\/dl>\n<dl class=\"definition\">\n<dt>\n<\/dt>\n<dt><strong>logistic growth model<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">a function of the form [latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex] where [latex]\\frac{c}{1+a}[\/latex] is the initial value, <em>c<\/em>\u00a0is the carrying capacity, or limiting value, and <em>b<\/em>\u00a0is a constant determined by the rate of growth<\/dd>\n<\/dl>\n<hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-5386-1\">Source: <em>Indiana University Center for Studies of Law in Action, 2007<\/em> <a href=\"#return-footnote-5386-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":167848,"menu_order":14,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5386","chapter","type-chapter","status-publish","hentry"],"part":5352,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5386","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/users\/167848"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5386\/revisions"}],"predecessor-version":[{"id":5775,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5386\/revisions\/5775"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/5352"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5386\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/media?parent=5386"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=5386"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/contributor?post=5386"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/license?post=5386"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}