{"id":5338,"date":"2021-10-13T18:24:23","date_gmt":"2021-10-13T18:24:23","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/chapter\/introduction-graphs-of-exponential-functions\/"},"modified":"2022-04-14T19:09:59","modified_gmt":"2022-04-14T19:09:59","slug":"introduction-graphs-of-exponential-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/chapter\/introduction-graphs-of-exponential-functions\/","title":{"raw":"Graphs of Exponential and Logistic Functions","rendered":"Graphs of Exponential and Logistic Functions"},"content":{"raw":"As we discussed in the previous page, exponential functions are used for many real-world applications such as finance, forensics, computer science, and most of the life sciences. Working with an equation that describes a real-world situation gives us a method for making predictions. Most of the time, however, the equation itself is not enough. We learn a lot about things by seeing their visual representations, and that is exactly why graphing exponential equations is a powerful tool. It gives us another layer of insight for predicting future events.\r\n<h2>Characteristics of Graphs of Exponential Functions<\/h2>\r\nBefore we begin graphing, it is helpful to review the behavior of exponential growth. Recall the table of values for a function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex] whose base is greater than one. We\u2019ll use the function [latex]f\\left(x\\right)={2}^{x}[\/latex]. Observe how the output values in the table below\u00a0change as the input increases by 1.\r\n<table id=\"Table_04_02_01\" summary=\"Two rows and eight columns. The first row is labeled,\"><colgroup> <\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>\u20133<\/td>\r\n<td>\u20132<\/td>\r\n<td>\u20131<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\r\n<td>[latex]\\frac{1}{8}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{4}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nEach output value is the product of the previous output and the base, 2. We call the base 2 the <em>constant ratio<\/em>. In fact, for any exponential function with the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex], <em>b<\/em>\u00a0is the constant ratio of the function. This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of <em>a<\/em>.\r\n\r\nNotice from the table that:\r\n<ul>\r\n \t<li>the output values are positive for all values of <em>x<\/em><\/li>\r\n \t<li>as <em>x<\/em>\u00a0increases, the output values increase without bound<\/li>\r\n \t<li>as <em>x<\/em>\u00a0decreases, the output values grow smaller, approaching zero<\/li>\r\n<\/ul>\r\nThe graph below\u00a0shows the exponential growth function [latex]f\\left(x\\right)={2}^{x}[\/latex].\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\/02231129\/CNX_Precalc_Figure_04_02_0012.jpg\" alt=\"Graph of the exponential function, 2^(x), with labeled points at (-3, 1\/8), (-2, \u00bc), (-1, \u00bd), (0, 1), (1, 2), (2, 4), and (3, 8). The graph notes that the x-axis is an asymptote.\" width=\"487\" height=\"520\" \/> Notice that the graph gets close to the x-axis but never touches it.[\/caption]\r\n\r\nThe domain of [latex]f\\left(x\\right)={2}^{x}[\/latex] is all real numbers, the range is [latex]\\left(0,\\infty \\right)[\/latex], and the horizontal asymptote is [latex]y=0[\/latex].\r\n\r\nTo get a sense of the behavior of <strong>exponential decay<\/strong>, we can create a table of values for a function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex] whose base is between zero and one. We\u2019ll use the function [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex]. Observe how the output values in the table below\u00a0change as the input increases by 1.\r\n<table summary=\"Two rows and eight columns. The first row is labeled,\">\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>\u20133<\/td>\r\n<td>\u20132<\/td>\r\n<td>\u20131<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]g\\left(x\\right)=\\left(\\frac{1}{2}\\right)^{x}[\/latex]<\/strong><\/td>\r\n<td>8<\/td>\r\n<td>4<\/td>\r\n<td>2<\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{4}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{8}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAgain, because the input is increasing by 1, each output value is the product of the previous output and the base or constant ratio [latex]\\frac{1}{2}[\/latex].\r\n\r\nNotice from the table that:\r\n<ul>\r\n \t<li>the output values are positive for all values of <em>x<\/em><\/li>\r\n \t<li>as <em>x<\/em>\u00a0increases, the output values grow smaller, approaching zero<\/li>\r\n \t<li>as <em>x<\/em>\u00a0decreases, the output values grow without bound<\/li>\r\n<\/ul>\r\nThe graph below shows the exponential decay function, [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex].\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\/02231133\/CNX_Precalc_Figure_04_02_0022.jpg\" alt=\"Graph of decreasing exponential function, (1\/2)^x, with labeled points at (-3, 8), (-2, 4), (-1, 2), (0, 1), (1, 1\/2), (2, 1\/4), and (3, 1\/8). The graph notes that the x-axis is an asymptote.\" width=\"487\" height=\"520\" \/> The domain of [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex] is all real numbers, the range is [latex]\\left(0,\\infty \\right)[\/latex], and the horizontal asymptote is [latex]y=0[\/latex].[\/caption]\r\n<div class=\"textbox\">\r\n<h3>A General Note: Characteristics of the Graph of the Function [latex]f\\left(x\\right)={b}^{x}[\/latex]<\/h3>\r\nAn exponential function with the form [latex]f\\left(x\\right)={b}^{x}[\/latex], [latex]b&gt;0[\/latex], [latex]b\\ne 1[\/latex], has these characteristics:\r\n<ul>\r\n \t<li><strong>one-to-one<\/strong> function<\/li>\r\n \t<li>horizontal asymptote: [latex]y=0[\/latex]<\/li>\r\n \t<li>domain: [latex]\\left(-\\infty , \\infty \\right)[\/latex]<\/li>\r\n \t<li>range: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\r\n \t<li><em>x-<\/em>intercept: none<\/li>\r\n \t<li><em>y-<\/em>intercept: [latex]\\left(0,1\\right)[\/latex]<\/li>\r\n \t<li>increasing if [latex]b&gt;1[\/latex]<\/li>\r\n \t<li>decreasing if [latex]b&lt;1[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an exponential function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex], graph the function<\/h3>\r\n<ol>\r\n \t<li>Create a table of points.<\/li>\r\n \t<li>Plot at least 3\u00a0point from the table including the <em>y<\/em>-intercept [latex]\\left(0,1\\right)[\/latex].<\/li>\r\n \t<li>Draw a smooth curve through the points.<\/li>\r\n \t<li>State the domain, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], the range, [latex]\\left(0,\\infty \\right)[\/latex], and the horizontal asymptote, [latex]y=0[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Sketching the Graph of an Exponential Function of the Form [latex]f\\left(x\\right)={b}^{x}[\/latex]<\/h3>\r\nSketch a graph of [latex]f\\left(x\\right)={0.25}^{x}[\/latex]. State the domain, range, and asymptote.\r\n\r\n[reveal-answer q=\"410947\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"410947\"]\r\n\r\nBefore graphing, identify the behavior and create a table of points for the graph.\r\n<ul>\r\n \t<li>Since <em>b\u00a0<\/em>= 0.25 is between zero and one, we know the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote <em>y\u00a0<\/em>= 0.<\/li>\r\n \t<li>Create a table of points.\r\n<table summary=\"Two rows and eight columns. The first row is labeled,\">\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>\u20133<\/td>\r\n<td>\u20132<\/td>\r\n<td>\u20131<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)={0.25}^{x}[\/latex]<\/strong><\/td>\r\n<td>64<\/td>\r\n<td>16<\/td>\r\n<td>4<\/td>\r\n<td>1<\/td>\r\n<td>0.25<\/td>\r\n<td>0.0625<\/td>\r\n<td>0.015625<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t<li>Plot the <em>y<\/em>-intercept, [latex]\\left(0,1\\right)[\/latex], along with two other points. We can use [latex]\\left(-1,4\\right)[\/latex] and [latex]\\left(1,0.25\\right)[\/latex].<\/li>\r\n<\/ul>\r\nDraw a smooth curve connecting the 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\/02231140\/CNX_Precalc_Figure_04_02_0042.jpg\" alt=\"Graph of the decaying exponential function f(x) = 0.25^x with labeled points at (-1, 4), (0, 1), and (1, 0.25).\" width=\"487\" height=\"332\" \/> The domain is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], the range is [latex]\\left(0,\\infty \\right)[\/latex], and the horizontal asymptote is [latex]y=0[\/latex].[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSketch the graph of [latex]f\\left(x\\right)={4}^{x}[\/latex]. State the domain, range, and asymptote.\r\n\r\n[reveal-answer q=\"192861\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"192861\"]\r\n\r\nThe domain is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], the range is [latex]\\left(0,\\infty \\right)[\/latex], and the horizontal asymptote is [latex]y=0[\/latex].<span id=\"fs-id1165137437648\">\r\n<\/span>\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/08000344\/CNX_Precalc_Figure_04_02_0052.jpg\"><img class=\"aligncenter wp-image-3353 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/08000344\/CNX_Precalc_Figure_04_02_0052.jpg\" width=\"487\" height=\"332\" \/><\/a>\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3607&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\">\r\n<\/iframe>\r\n\r\n<\/div>\r\n<h2>Bounded Growth<\/h2>\r\nExponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, most models like this begin to approach some limiting value and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an <strong>exponential growth<\/strong> model although the exponential growth model is still useful over a short term before approaching the limiting value.\r\n\r\nThe <strong>logistic growth model<\/strong> is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model\u2019s upper bound called the <strong>carrying capacity<\/strong>. For constants <em>a<\/em>, <em>b<\/em>, and <em>c<\/em>, the logistic growth of a population over time <em>x<\/em>\u00a0is represented by the model\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\r\nThe graph below shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate at which the rate of increase decreases.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181346\/CNX_Precalc_Figure_04_07_0062.jpg\" alt=\"Graph of f(x)=c\/(1+ae^(-tx)). The carrying capacity is the asymptote at y=c. The initial value of population is (0, c\/(1+a)). The point of maximum growth is (ln(a)\/b, c\/2).\" width=\"487\" height=\"367\" \/>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Logistic Growth<\/h3>\r\nThe logistic growth model is\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\r\nwhere\r\n<ul>\r\n \t<li>[latex]\\frac{c}{1+a}[\/latex] is the initial value<\/li>\r\n \t<li><em>c<\/em>\u00a0is the <em>carrying capacity\u00a0<\/em>or <em>limiting value<\/em><\/li>\r\n \t<li><em>b<\/em>\u00a0is a constant determined by the rate of growth.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Logistic-Growth Model<\/h3>\r\nAn influenza epidemic spreads through a population rapidly at a rate that depends on two factors. The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model good for studying the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.\r\n\r\nFor example, at time <em>t\u00a0<\/em>= 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is <em>b\u00a0<\/em>= 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.\r\n\r\n[reveal-answer q=\"346660\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"346660\"]\r\n\r\nWe substitute the given data into the logistic growth model\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\r\nBecause at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is <em>c\u00a0<\/em>= 1000. To find <em>a<\/em>, we use the formula that the number of cases at time <em>t\u00a0<\/em>= 0 is [latex]\\frac{c}{1+a}=1[\/latex], from which it follows that <em>a\u00a0<\/em>= 999. This model predicts that, after ten days, the number of people who have had the flu is [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}\\approx 293.8[\/latex]. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, <em>c\u00a0<\/em>= 1000.\r\n<h4>Analysis of the Solution<\/h4>\r\nRemember that because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values.\r\n\r\nThe graph below gives a good picture of how this model fits the data.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181349\/CNX_Precalc_Figure_04_07_0072.jpg\" alt=\"Graph of f(x)=1000\/(1+999e^(-0.5030x)) with the y-axis labeled as\" width=\"731\" height=\"492\" \/> The graph of [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}[\/latex].[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUsing the model in the previous example, estimate the number of cases of flu on day 15.\r\n\r\n[reveal-answer q=\"421768\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"421768\"]\r\n\r\n895 cases on day 15[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2><\/h2>\r\n<h2>Key Equations<\/h2>\r\n<table id=\"fs-id2306479\" summary=\"...\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 85px;\">definition of the exponential function<\/td>\r\n<td style=\"width: 739px;\">[latex]f\\left(x\\right)={b}^{x}\\text{, where }b&gt;0, b\\ne 1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 85px;\"><span style=\"color: #000000;\">definition of exponential growth<\/span><\/td>\r\n<td style=\"width: 739px;\"><span style=\"color: #000000;\">[latex]f\\left(x\\right)=a{b}^{x},\\text{ where }a&gt;0,b&gt;0,b\\ne 1[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 85px;\"><span style=\"color: #000000;\">continuous growth formula<\/span><\/td>\r\n<td style=\"width: 739px;\"><span style=\"color: #000000;\">[latex]A\\left(t\\right)=a{e}^{rt},\\text{ where }[\/latex]<em>t<\/em>\u00a0is the number of time periods of growth [latex]\\\\[\/latex]<em>a<\/em>\u00a0is the starting amount (in the continuous compounding formula a is replaced with P, the principal)[latex]\\\\[\/latex]<em>e<\/em>\u00a0is the mathematical constant,\u00a0[latex] e\\approx 2.718282[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ul>\r\n \t<li>An exponential function is defined as a function with a positive constant other than 1 raised to a variable exponent.<\/li>\r\n \t<li>Scientific and graphing calculators have the key [latex]\\left[{e}^{x}\\<span style=\"color: #000000;\">right][\/latex] or [latex]\\left[\\mathrm{exp}\\left(x\\right)\\right][\/latex] for calculating powers of <em>e<\/em>.<\/span><\/li>\r\n \t<li><span style=\"color: #000000;\">Continuous growth or decay models are exponential models that use <em>e<\/em>\u00a0as the base. Continuous growth and decay models can be found when the initial value and growth or decay rate are known.<\/span><\/li>\r\n \t<li>The graph of the function [latex]f\\left(x\\right)={b}^{x}[\/latex] has a <em>y-<\/em>intercept at [latex]\\left(0, 1\\right)[\/latex], domain of [latex]\\left(-\\infty , \\infty \\right)[\/latex], range of [latex]\\left(0, \\infty \\right)[\/latex], and horizontal asymptote of [latex]y=0[\/latex].<\/li>\r\n \t<li>If [latex]b&gt;1[\/latex], the function is increasing. The left tail of the graph will approach the asymptote [latex]y=0[\/latex], and the right tail will increase without bound.<\/li>\r\n \t<li>If 0 &lt;\u00a0<em>b<\/em> &lt; 1, the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote [latex]y=0[\/latex].<\/li>\r\n \t<li>We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/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>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>\r\n<section id=\"fs-id1165137846440\" class=\"key-concepts\">\r\n<div>\r\n<dl id=\"fs-id1165135397912\" class=\"definition\">\r\n \t<dt><strong>annual percentage rate (APR)<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135397918\">the yearly interest rate earned by an investment account, also called <em>nominal rate<\/em><\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>exponential growth<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">a model that grows by a rate proportional to the amount present<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137838644\" class=\"definition\">\r\n \t<dt><strong>nominal rate<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838650\">the yearly interest rate earned by an investment account, also called <em>annual percentage rate<\/em><\/dd>\r\n<\/dl>\r\n<\/div>\r\n<\/section>","rendered":"<p>As we discussed in the previous page, exponential functions are used for many real-world applications such as finance, forensics, computer science, and most of the life sciences. Working with an equation that describes a real-world situation gives us a method for making predictions. Most of the time, however, the equation itself is not enough. We learn a lot about things by seeing their visual representations, and that is exactly why graphing exponential equations is a powerful tool. It gives us another layer of insight for predicting future events.<\/p>\n<h2>Characteristics of Graphs of Exponential Functions<\/h2>\n<p>Before we begin graphing, it is helpful to review the behavior of exponential growth. Recall the table of values for a function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex] whose base is greater than one. We\u2019ll use the function [latex]f\\left(x\\right)={2}^{x}[\/latex]. Observe how the output values in the table below\u00a0change as the input increases by 1.<\/p>\n<table id=\"Table_04_02_01\" summary=\"Two rows and eight columns. The first row is labeled,\">\n<colgroup> <\/colgroup>\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>\u20133<\/td>\n<td>\u20132<\/td>\n<td>\u20131<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\n<td>[latex]\\frac{1}{8}[\/latex]<\/td>\n<td>[latex]\\frac{1}{4}[\/latex]<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>8<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Each output value is the product of the previous output and the base, 2. We call the base 2 the <em>constant ratio<\/em>. In fact, for any exponential function with the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex], <em>b<\/em>\u00a0is the constant ratio of the function. This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of <em>a<\/em>.<\/p>\n<p>Notice from the table that:<\/p>\n<ul>\n<li>the output values are positive for all values of <em>x<\/em><\/li>\n<li>as <em>x<\/em>\u00a0increases, the output values increase without bound<\/li>\n<li>as <em>x<\/em>\u00a0decreases, the output values grow smaller, approaching zero<\/li>\n<\/ul>\n<p>The graph below\u00a0shows the exponential growth function [latex]f\\left(x\\right)={2}^{x}[\/latex].<\/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\/02231129\/CNX_Precalc_Figure_04_02_0012.jpg\" alt=\"Graph of the exponential function, 2^(x), with labeled points at (-3, 1\/8), (-2, \u00bc), (-1, \u00bd), (0, 1), (1, 2), (2, 4), and (3, 8). The graph notes that the x-axis is an asymptote.\" width=\"487\" height=\"520\" \/><\/p>\n<p class=\"wp-caption-text\">Notice that the graph gets close to the x-axis but never touches it.<\/p>\n<\/div>\n<p>The domain of [latex]f\\left(x\\right)={2}^{x}[\/latex] is all real numbers, the range is [latex]\\left(0,\\infty \\right)[\/latex], and the horizontal asymptote is [latex]y=0[\/latex].<\/p>\n<p>To get a sense of the behavior of <strong>exponential decay<\/strong>, we can create a table of values for a function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex] whose base is between zero and one. We\u2019ll use the function [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex]. Observe how the output values in the table below\u00a0change as the input increases by 1.<\/p>\n<table summary=\"Two rows and eight columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>\u20133<\/td>\n<td>\u20132<\/td>\n<td>\u20131<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]g\\left(x\\right)=\\left(\\frac{1}{2}\\right)^{x}[\/latex]<\/strong><\/td>\n<td>8<\/td>\n<td>4<\/td>\n<td>2<\/td>\n<td>1<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]\\frac{1}{4}[\/latex]<\/td>\n<td>[latex]\\frac{1}{8}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Again, because the input is increasing by 1, each output value is the product of the previous output and the base or constant ratio [latex]\\frac{1}{2}[\/latex].<\/p>\n<p>Notice from the table that:<\/p>\n<ul>\n<li>the output values are positive for all values of <em>x<\/em><\/li>\n<li>as <em>x<\/em>\u00a0increases, the output values grow smaller, approaching zero<\/li>\n<li>as <em>x<\/em>\u00a0decreases, the output values grow without bound<\/li>\n<\/ul>\n<p>The graph below shows the exponential decay function, [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex].<\/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\/02231133\/CNX_Precalc_Figure_04_02_0022.jpg\" alt=\"Graph of decreasing exponential function, (1\/2)^x, with labeled points at (-3, 8), (-2, 4), (-1, 2), (0, 1), (1, 1\/2), (2, 1\/4), and (3, 1\/8). The graph notes that the x-axis is an asymptote.\" width=\"487\" height=\"520\" \/><\/p>\n<p class=\"wp-caption-text\">The domain of [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex] is all real numbers, the range is [latex]\\left(0,\\infty \\right)[\/latex], and the horizontal asymptote is [latex]y=0[\/latex].<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Characteristics of the Graph of the Function [latex]f\\left(x\\right)={b}^{x}[\/latex]<\/h3>\n<p>An exponential function with the form [latex]f\\left(x\\right)={b}^{x}[\/latex], [latex]b>0[\/latex], [latex]b\\ne 1[\/latex], has these characteristics:<\/p>\n<ul>\n<li><strong>one-to-one<\/strong> function<\/li>\n<li>horizontal asymptote: [latex]y=0[\/latex]<\/li>\n<li>domain: [latex]\\left(-\\infty , \\infty \\right)[\/latex]<\/li>\n<li>range: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\n<li><em>x-<\/em>intercept: none<\/li>\n<li><em>y-<\/em>intercept: [latex]\\left(0,1\\right)[\/latex]<\/li>\n<li>increasing if [latex]b>1[\/latex]<\/li>\n<li>decreasing if [latex]b<1[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an exponential function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex], graph the function<\/h3>\n<ol>\n<li>Create a table of points.<\/li>\n<li>Plot at least 3\u00a0point from the table including the <em>y<\/em>-intercept [latex]\\left(0,1\\right)[\/latex].<\/li>\n<li>Draw a smooth curve through the points.<\/li>\n<li>State the domain, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], the range, [latex]\\left(0,\\infty \\right)[\/latex], and the horizontal asymptote, [latex]y=0[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Sketching the Graph of an Exponential Function of the Form [latex]f\\left(x\\right)={b}^{x}[\/latex]<\/h3>\n<p>Sketch a graph of [latex]f\\left(x\\right)={0.25}^{x}[\/latex]. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q410947\">Show Solution<\/span><\/p>\n<div id=\"q410947\" class=\"hidden-answer\" style=\"display: none\">\n<p>Before graphing, identify the behavior and create a table of points for the graph.<\/p>\n<ul>\n<li>Since <em>b\u00a0<\/em>= 0.25 is between zero and one, we know the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote <em>y\u00a0<\/em>= 0.<\/li>\n<li>Create a table of points.<br \/>\n<table summary=\"Two rows and eight columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>\u20133<\/td>\n<td>\u20132<\/td>\n<td>\u20131<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)={0.25}^{x}[\/latex]<\/strong><\/td>\n<td>64<\/td>\n<td>16<\/td>\n<td>4<\/td>\n<td>1<\/td>\n<td>0.25<\/td>\n<td>0.0625<\/td>\n<td>0.015625<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>Plot the <em>y<\/em>-intercept, [latex]\\left(0,1\\right)[\/latex], along with two other points. We can use [latex]\\left(-1,4\\right)[\/latex] and [latex]\\left(1,0.25\\right)[\/latex].<\/li>\n<\/ul>\n<p>Draw a smooth curve connecting the 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\/02231140\/CNX_Precalc_Figure_04_02_0042.jpg\" alt=\"Graph of the decaying exponential function f(x) = 0.25^x with labeled points at (-1, 4), (0, 1), and (1, 0.25).\" width=\"487\" height=\"332\" \/><\/p>\n<p class=\"wp-caption-text\">The domain is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], the range is [latex]\\left(0,\\infty \\right)[\/latex], and the horizontal asymptote is [latex]y=0[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Sketch the graph of [latex]f\\left(x\\right)={4}^{x}[\/latex]. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q192861\">Show Solution<\/span><\/p>\n<div id=\"q192861\" class=\"hidden-answer\" style=\"display: none\">\n<p>The domain is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], the range is [latex]\\left(0,\\infty \\right)[\/latex], and the horizontal asymptote is [latex]y=0[\/latex].<span id=\"fs-id1165137437648\"><br \/>\n<\/span><\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/08000344\/CNX_Precalc_Figure_04_02_0052.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3353 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/08000344\/CNX_Precalc_Figure_04_02_0052.jpg\" width=\"487\" height=\"332\" alt=\"image\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3607&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<h2>Bounded Growth<\/h2>\n<p>Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, most models like this begin to approach some limiting value and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an <strong>exponential growth<\/strong> model although the exponential growth model is still useful over a short term before approaching the limiting value.<\/p>\n<p>The <strong>logistic growth model<\/strong> is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model\u2019s upper bound called the <strong>carrying capacity<\/strong>. For constants <em>a<\/em>, <em>b<\/em>, and <em>c<\/em>, the logistic growth of a population over time <em>x<\/em>\u00a0is represented by the model<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\n<p>The graph below shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate at which the rate of increase decreases.<\/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\/03181346\/CNX_Precalc_Figure_04_07_0062.jpg\" alt=\"Graph of f(x)=c\/(1+ae^(-tx)). The carrying capacity is the asymptote at y=c. The initial value of population is (0, c\/(1+a)). The point of maximum growth is (ln(a)\/b, c\/2).\" width=\"487\" height=\"367\" \/><\/p>\n<div class=\"textbox\">\n<h3>A General Note: Logistic Growth<\/h3>\n<p>The logistic growth model is<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\n<p>where<\/p>\n<ul>\n<li>[latex]\\frac{c}{1+a}[\/latex] is the initial value<\/li>\n<li><em>c<\/em>\u00a0is the <em>carrying capacity\u00a0<\/em>or <em>limiting value<\/em><\/li>\n<li><em>b<\/em>\u00a0is a constant determined by the rate of growth.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Logistic-Growth Model<\/h3>\n<p>An influenza epidemic spreads through a population rapidly at a rate that depends on two factors. The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model good for studying the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.<\/p>\n<p>For example, at time <em>t\u00a0<\/em>= 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is <em>b\u00a0<\/em>= 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q346660\">Show Solution<\/span><\/p>\n<div id=\"q346660\" class=\"hidden-answer\" style=\"display: none\">\n<p>We substitute the given data into the logistic growth model<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\n<p>Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is <em>c\u00a0<\/em>= 1000. To find <em>a<\/em>, we use the formula that the number of cases at time <em>t\u00a0<\/em>= 0 is [latex]\\frac{c}{1+a}=1[\/latex], from which it follows that <em>a\u00a0<\/em>= 999. This model predicts that, after ten days, the number of people who have had the flu is [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}\\approx 293.8[\/latex]. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, <em>c\u00a0<\/em>= 1000.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Remember that because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values.<\/p>\n<p>The graph below gives a good picture of how this model fits the data.<\/p>\n<div style=\"width: 741px\" 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\/03181349\/CNX_Precalc_Figure_04_07_0072.jpg\" alt=\"Graph of f(x)=1000\/(1+999e^(-0.5030x)) with the y-axis labeled as\" width=\"731\" height=\"492\" \/><\/p>\n<p class=\"wp-caption-text\">The graph of [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using the model in the previous example, estimate the number of cases of flu on day 15.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q421768\">Show Solution<\/span><\/p>\n<div id=\"q421768\" class=\"hidden-answer\" style=\"display: none\">\n<p>895 cases on day 15<\/p><\/div>\n<\/div>\n<\/div>\n<h2><\/h2>\n<h2>Key Equations<\/h2>\n<table id=\"fs-id2306479\" summary=\"...\">\n<tbody>\n<tr>\n<td style=\"width: 85px;\">definition of the exponential function<\/td>\n<td style=\"width: 739px;\">[latex]f\\left(x\\right)={b}^{x}\\text{, where }b>0, b\\ne 1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 85px;\"><span style=\"color: #000000;\">definition of exponential growth<\/span><\/td>\n<td style=\"width: 739px;\"><span style=\"color: #000000;\">[latex]f\\left(x\\right)=a{b}^{x},\\text{ where }a>0,b>0,b\\ne 1[\/latex]<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 85px;\"><span style=\"color: #000000;\">continuous growth formula<\/span><\/td>\n<td style=\"width: 739px;\"><span style=\"color: #000000;\">[latex]A\\left(t\\right)=a{e}^{rt},\\text{ where }[\/latex]<em>t<\/em>\u00a0is the number of time periods of growth [latex]\\\\[\/latex]<em>a<\/em>\u00a0is the starting amount (in the continuous compounding formula a is replaced with P, the principal)[latex]\\\\[\/latex]<em>e<\/em>\u00a0is the mathematical constant,\u00a0[latex]e\\approx 2.718282[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li>An exponential function is defined as a function with a positive constant other than 1 raised to a variable exponent.<\/li>\n<li>Scientific and graphing calculators have the key [latex]\\left[{e}^{x}\\<span style=\"color: #000000;\">right][\/latex] or [latex]\\left[\\mathrm{exp}\\left(x\\right)\\right][\/latex] for calculating powers of <em>e<\/em>.<\/span><\/li>\n<li><span style=\"color: #000000;\">Continuous growth or decay models are exponential models that use <em>e<\/em>\u00a0as the base. Continuous growth and decay models can be found when the initial value and growth or decay rate are known.<\/span><\/li>\n<li>The graph of the function [latex]f\\left(x\\right)={b}^{x}[\/latex] has a <em>y-<\/em>intercept at [latex]\\left(0, 1\\right)[\/latex], domain of [latex]\\left(-\\infty , \\infty \\right)[\/latex], range of [latex]\\left(0, \\infty \\right)[\/latex], and horizontal asymptote of [latex]y=0[\/latex].<\/li>\n<li>If [latex]b>1[\/latex], the function is increasing. The left tail of the graph will approach the asymptote [latex]y=0[\/latex], and the right tail will increase without bound.<\/li>\n<li>If 0 &lt;\u00a0<em>b<\/em> &lt; 1, the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote [latex]y=0[\/latex].<\/li>\n<li>We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/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>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<section id=\"fs-id1165137846440\" class=\"key-concepts\">\n<div>\n<dl id=\"fs-id1165135397912\" class=\"definition\">\n<dt><strong>annual percentage rate (APR)<\/strong><\/dt>\n<dd id=\"fs-id1165135397918\">the yearly interest rate earned by an investment account, also called <em>nominal rate<\/em><\/dd>\n<\/dl>\n<dl id=\"fs-id1165137838635\" class=\"definition\">\n<dt><strong>exponential growth<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">a model that grows by a rate proportional to the amount present<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137838644\" class=\"definition\">\n<dt><strong>nominal rate<\/strong><\/dt>\n<dd id=\"fs-id1165137838650\">the yearly interest rate earned by an investment account, also called <em>annual percentage rate<\/em><\/dd>\n<\/dl>\n<\/div>\n<\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-5338\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 63064. <strong>Authored by<\/strong>: Brin, Leon. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License, CC-BY + GPL<\/li><li>Question ID 129498. <strong>Authored by<\/strong>: Day, Alyson. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":167848,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen 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