{"id":16027,"date":"2019-09-26T17:24:04","date_gmt":"2019-09-26T17:24:04","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-model-an-application-with-a-linear-function\/"},"modified":"2024-05-02T15:54:38","modified_gmt":"2024-05-02T15:54:38","slug":"read-model-an-application-with-a-linear-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-model-an-application-with-a-linear-function\/","title":{"raw":"Applications of Linear Functions","rendered":"Applications of Linear Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write the equation for a linear function given an application<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165137594074\">In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know or can figure out the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.<\/p>\r\nWe will begin by looking at applications that that involve calculating slope and then we will move on to writing a linear equation given an application.\r\n\r\nThe units for slope are always [latex]\\dfrac{\\text{units for the output}}{\\text{units for the input}}[\/latex]. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input.\u00a0\u00a0Recall that the slope measures steepness.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe population of a city increased from\u00a0[latex]23,400[\/latex] in\u00a0[latex]2008[\/latex] to\u00a0[latex]27,800[\/latex] in [latex]2012[\/latex]. Find the change of population per year if we assume the change was constant from\u00a0[latex]2008[\/latex] to\u00a0[latex]2012[\/latex].\r\n\r\n[reveal-answer q=\"246268\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"246268\"]\r\n\r\nThe rate of change relates the change in population to the change in time. The population increased by [latex]27,800-23,400=4400[\/latex] people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{4,400\\text{ people}}{4\\text{ years}}=1,100\\text{ }\\dfrac{\\text{people}}{\\text{year}}[\/latex]<\/p>\r\nSo the population increased by\u00a0[latex]1,100[\/latex] people per year.\r\n\r\nBecause we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next video, we show an example where we determine the increase in cost for producing solar panels given two data points.\r\n\r\nhttps:\/\/youtu.be\/4RbniDgEGE4\r\n<div class=\"textbox shaded\">\r\n<h3 id=\"fs-id1165137404879\">How To: Given a linear function [latex]f[\/latex] and the initial value and rate of change, evaluate [latex]f(c)[\/latex]<\/h3>\r\n<ol id=\"fs-id1165137660790\">\r\n \t<li>Determine the initial value and the rate of change (slope).<\/li>\r\n \t<li>Substitute the values into [latex]f\\left(x\\right)=mx+b[\/latex].<\/li>\r\n \t<li>Evaluate the function at [latex]x=c[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\nInitial value is a term that is typically used in applications of functions. \u00a0It can be represented as the starting point of the relationship we are describing with a function. In the case of linear functions, the initial value is typically the <em>y<\/em>-intercept. Here are some characteristics of the initial value:\r\n<ul>\r\n \t<li>The point [latex](0,y)[\/latex] is often the initial value of a linear function<\/li>\r\n \t<li>The <em>y<\/em>-value of the initial value comes from <em>b<\/em> in slope-intercept form of a linear function,\u00a0[latex]f\\left(x\\right)=mx+b[\/latex]<\/li>\r\n \t<li>The initial value can be found by solving for <em>b<\/em>\u00a0or substituting\u00a0[latex]0[\/latex] in for <em>x<\/em> in a linear function.<\/li>\r\n<\/ul>\r\nIn our next example, we are given a scenario where Marcus wants to increase the number of songs in his music collection by a fixed amount each month. This is a perfect candidate for a linear function because the increase in the number of songs stays the same each month. We will identify the initial value for the music collection and write an equation that represents the number of songs in the collection for any number of months\u00a0<em>t<\/em>.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nMarcus currently has\u00a0[latex]200[\/latex] songs in his music collection. Every month he adds\u00a0[latex]15[\/latex] new songs. Write a formula for the number of songs, <em>N<\/em>, in his collection as a function of time, <em>t<\/em>, the number of months. How many songs will he own in a year?\r\n[reveal-answer q=\"14550\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"14550\"]\r\n<p id=\"fs-id1165135411394\">The initial value for this function is\u00a0[latex]200[\/latex] because he currently owns\u00a0[latex]200[\/latex] songs, so <i>N<\/i>(0) =\u00a0[latex]200[\/latex]. This means that <em>b<\/em> =[latex]200[\/latex].<\/p>\r\n<img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201031\/CNX_Precalc_Figure_02_01_0102.jpg\" alt=\"The equation f of x equals mx plus b. And arrow labeled 15 points to m, and an arrow labeled 200 points to b. The equation becomes N of t equals 15 t plus 200.\" width=\"487\" height=\"131\" \/>\r\n<p id=\"fs-id1165137738190\">The number of songs increases by\u00a0[latex]15[\/latex] songs per month, so the rate of change is\u00a0[latex]15[\/latex] songs per month. Therefore, we know that [latex]m=15[\/latex]. We can substitute the initial value and the rate of change into slope-intercept form.<span id=\"fs-id1165137417445\">\r\n<\/span><\/p>\r\n<p id=\"fs-id1165137810258\">We can write the formula [latex]N\\left(t\\right)=15t+200[\/latex].<\/p>\r\n<p id=\"fs-id1165137454711\">With this formula, we can then predict how many songs Marcus will have in\u00a01 year (12 months). In other words, we can evaluate the function at [latex]t = 12 [\/latex] .<\/p>\r\n\r\n<div id=\"fs-id1165137462736\" class=\"equation unnumbered\">[latex]\\begin{array}{l}N\\left(12\\right) &amp; =15\\left(12\\right)+200\\hfill \\\\ &amp; =180+200\\hfill \\\\ &amp; =380\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137694205\">Marcus will have\u00a0[latex]380[\/latex] songs in\u00a0[latex]12[\/latex] months.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_02_01_10\" class=\"example\">\r\n<div id=\"fs-id1165137836747\" class=\"exercise\">\r\n<p id=\"fs-id1165134065131\">In the example we just completed, notice that <em>N<\/em> is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.<\/p>\r\nThe following video provides an example of how to write a function that will give the cost in dollars for a given number of credit hours taken, x.\u00a0 Notice how the function consists of an initial value (the cost of the registration fee) plus an increase in cost for every credit hour taken.\r\n\r\nhttps:\/\/youtu.be\/X3Sx2TxH-J0\r\n\r\nIn our next example, we will show that you can write the equation for a linear function given two data points. In this case, Ilya's weekly income depends on the number of insurance policies he sells. We are given his income for two different weeks and the number of policies sold. We first find the rate of change and then solve for the initial value.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWorking as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya\u2019s weekly income, <i>I<\/i>, depends on the number of new policies, <em>n<\/em>, he sells during the week. Last week, he sold\u00a0[latex]3[\/latex] new policies and earned\u00a0[latex]$760[\/latex] for the week. The week before, he sold\u00a0[latex]5[\/latex] new policies and earned\u00a0[latex]$920[\/latex]. Find an equation for <em>I<\/em>(<em>n<\/em>) and interpret the meaning of the components of the equation.\r\n[reveal-answer q=\"249315\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"249315\"]\r\n<p id=\"fs-id1165135169134\">The given information gives us two input-output pairs:\u00a0[latex](3, 760)[\/latex] and [latex](5, 920)[\/latex]. We start by finding the rate of change.<\/p>\r\n\r\n<div id=\"fs-id1165135195046\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}m &amp; =\\dfrac{920 - 760}{5 - 3}\\hfill \\\\ &amp; =\\dfrac{$160}{\\text{2 policies}}\\hfill \\\\ &amp; =$80\\text{ per policy}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137855034\">Keeping track of units can help us interpret this quantity. Income increased by\u00a0[latex]$160[\/latex] when the number of policies increased by\u00a0[latex]2[\/latex], so the rate of change is\u00a0[latex]$80[\/latex] per policy. Therefore, Ilya earns a commission of\u00a0[latex]$80[\/latex] for each policy sold during the week.<\/p>\r\n<p id=\"fs-id1165137855040\">We can then solve for the initial value.<\/p>\r\n\r\n<div id=\"fs-id1165135484088\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}\\text{ }I\\left(n\\right)=80n+b\\hfill &amp; \\hfill \\\\ \\text{ }760=80\\left(3\\right)+b\\hfill &amp; \\text{when }n=3, I\\left(3\\right)=760\\hfill \\\\ 760 - 80\\left(3\\right)=b\\hfill &amp; \\hfill \\\\ \\text{ }520=b\\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137400716\">The value of <em>b<\/em>\u00a0is the starting value for the function and represents Ilya\u2019s income when\u00a0\u00a0[latex]n = 0[\/latex], or when no new policies are sold. We can interpret this as Ilya\u2019s base salary for the week which does not depend upon the number of policies sold.<\/p>\r\n<p id=\"fs-id1165135203653\">We can now write the final equation.<\/p>\r\n\r\n<div id=\"fs-id1165137506449\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]I\\left(n\\right)=80n+520[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137655487\">Our final interpretation is that Ilya\u2019s base salary is\u00a0[latex]$520[\/latex] per week and he earns an additional\u00a0[latex]$80[\/latex] commission for each policy sold.<\/p>\r\n&nbsp;\r\n\r\nNotice that we used units to help us verify that we were calculating the rate correctly. It makes sense to speak in terms of the price per policy. To calculate the initial value, we solved for <em>b<\/em> by substituting values from one of the points we were given for <em>n<\/em> and <em>I<\/em>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video example, we show how to identify the initial value, slope, and equation for a linear function.\r\n\r\nhttps:\/\/youtu.be\/JMQSdRFJ1S4\r\n\r\nWe will show one more example of how to write a linear function that represents the monthly cost to run a company given monthly fixed costs and production costs per item.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose Ben starts a company in which he incurs a fixed cost of\u00a0[latex]$1,250[\/latex] per month for the overhead which includes his office rent. His production costs are\u00a0[latex]$37.50[\/latex] per item. Write a linear function <em>C\u00a0<\/em>where <i>C<\/i>(<em>x<\/em>)\u00a0is the cost for <em>x<\/em>\u00a0items produced in a given month.\r\n\r\n[reveal-answer q=\"625547\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"625547\"]\r\n\r\nThe fixed cost is present every month,\u00a0[latex]$1,250[\/latex]. The costs that can vary include the cost to produce each item, which is\u00a0[latex]$37.50[\/latex] for Ben. The variable cost, called the marginal cost, is represented by\u00a0[latex]37.5[\/latex]. The cost Ben incurs is the sum of these two costs represented by [latex]C\\left(x\\right)=1250+37.5x[\/latex].\r\n\r\n&nbsp;\r\n<p id=\"fs-id1165135511326\">It is important to note that we are writing a function based on monthly costs, so the initial cost will be\u00a0[latex]$1,250[\/latex], because Ben has to pay that amount monthly for rent. If Ben produces\u00a0[latex]100[\/latex] items in a month, his monthly cost is represented by<\/p>\r\n\r\n<div id=\"fs-id1165137417815\" class=\"equation unnumbered\">[latex]\\begin{array}{l}C\\left(100\\right)=1250+37.5\\left(100\\right)=5000\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137862645\">So his monthly cost would be\u00a0[latex]$5,000[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video example shows how to write a linear function that represents how many miles you can travel in a rental car given a fixed amount of money.\r\n<div id=\"fs-id1165137767515\" class=\"commentary\">\r\n\r\nhttps:\/\/youtu.be\/H8KR3w2nXqs\r\n\r\n<\/div>\r\nIn the next example, we will take data that is in tabular (table) form to write an equation that describes the rate of change of a rat population.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<p id=\"fs-id1165135161325\">The table below relates the number of rats in a population to time (in weeks). Use the table to write a linear equation.<\/p>\r\n\r\n<table id=\"Table_02_01_02\" summary=\"Two rows and five columns. The first row is labeled, 'w, the numers of weeks'. The second row is labeled is labeled, 'P(w), number of rats'. Reading the remaining rows as ordered pairs (i.e., (w, P(w)), we have the following values: (0, 1000), (2, 1080), (4, 1160), and (6, 1240).\">\r\n<tbody>\r\n<tr>\r\n<td><strong><em>w<\/em>, number of weeks<\/strong><\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]6[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong><em>P(w)<\/em>, number of rats<\/strong><\/td>\r\n<td>[latex]1000[\/latex]<\/td>\r\n<td>[latex]1080[\/latex]<\/td>\r\n<td>[latex]1160[\/latex]<\/td>\r\n<td>[latex]1240[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"780371\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"780371\"]\r\n<p id=\"fs-id1165137530990\">We can see from the table that the initial value for the number of rats is\u00a0[latex]1000[\/latex], so <em>b<\/em> =[latex]1000[\/latex].<\/p>\r\n<p id=\"fs-id1165137935601\">Rather than solving for <em>m<\/em>, we can tell from looking at the table that the population increases by\u00a0[latex]80[\/latex] for every\u00a0[latex]2[\/latex] weeks that pass. This means that the rate of change is\u00a0[latex]80[\/latex] rats per\u00a0[latex]2[\/latex] weeks, which can be simplified to\u00a0[latex]40[\/latex] rats per week.<\/p>\r\n\r\n<div id=\"fs-id1165137737900\" class=\"equation unnumbered\">[latex]P\\left(w\\right)=40w+1000[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137465125\">If we did not notice the rate of change from the table, we could still solve for the slope using any two points from the table. For example, using\u00a0[latex](2, 1080)[\/latex] and\u00a0[latex](6, 1240)[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1165137627069\" class=\"equation unnumbered\">[latex]\\begin{array}{l}m &amp; =\\dfrac{1240 - 1080}{6 - 2}\\hfill \\\\ &amp; =\\dfrac{160}{4}\\hfill \\\\ &amp; =40\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nIs the initial value always provided in a table of values like the table in\u00a0the previous example? Write your ideas in the textbox below before you look at the answer.\r\n\r\nIf your answer is no, give a description of how you would find the initial value.\r\n\r\n[practice-area rows=\"2\"][\/practice-area]\r\n[reveal-answer q=\"298919\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"298919\"]\r\n\r\nNo. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of\u00a0[latex]0[\/latex], then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to\u00a0\u00a0[latex]0[\/latex], find the slope, substitute one coordinate pair and the slope into [latex]f\\left(x\\right)=mx+b[\/latex], and solve for b.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Summary<\/h2>\r\n<ul>\r\n \t<li>Sometimes we are given an initial value and sometimes we have to solve for it.<\/li>\r\n \t<li>Using units can help you verify that you have calculated slope correctly.<\/li>\r\n \t<li>We can write the equation for a line given a slope and a data point or from a table of data.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write the equation for a linear function given an application<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137594074\">In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know or can figure out the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.<\/p>\n<p>We will begin by looking at applications that that involve calculating slope and then we will move on to writing a linear equation given an application.<\/p>\n<p>The units for slope are always [latex]\\dfrac{\\text{units for the output}}{\\text{units for the input}}[\/latex]. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input.\u00a0\u00a0Recall that the slope measures steepness.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The population of a city increased from\u00a0[latex]23,400[\/latex] in\u00a0[latex]2008[\/latex] to\u00a0[latex]27,800[\/latex] in [latex]2012[\/latex]. Find the change of population per year if we assume the change was constant from\u00a0[latex]2008[\/latex] to\u00a0[latex]2012[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q246268\">Show Solution<\/span><\/p>\n<div id=\"q246268\" class=\"hidden-answer\" style=\"display: none\">\n<p>The rate of change relates the change in population to the change in time. The population increased by [latex]27,800-23,400=4400[\/latex] people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{4,400\\text{ people}}{4\\text{ years}}=1,100\\text{ }\\dfrac{\\text{people}}{\\text{year}}[\/latex]<\/p>\n<p>So the population increased by\u00a0[latex]1,100[\/latex] people per year.<\/p>\n<p>Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next video, we show an example where we determine the increase in cost for producing solar panels given two data points.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  Slope Application Involving Production Costs\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/4RbniDgEGE4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox shaded\">\n<h3 id=\"fs-id1165137404879\">How To: Given a linear function [latex]f[\/latex] and the initial value and rate of change, evaluate [latex]f(c)[\/latex]<\/h3>\n<ol id=\"fs-id1165137660790\">\n<li>Determine the initial value and the rate of change (slope).<\/li>\n<li>Substitute the values into [latex]f\\left(x\\right)=mx+b[\/latex].<\/li>\n<li>Evaluate the function at [latex]x=c[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p>Initial value is a term that is typically used in applications of functions. \u00a0It can be represented as the starting point of the relationship we are describing with a function. In the case of linear functions, the initial value is typically the <em>y<\/em>-intercept. Here are some characteristics of the initial value:<\/p>\n<ul>\n<li>The point [latex](0,y)[\/latex] is often the initial value of a linear function<\/li>\n<li>The <em>y<\/em>-value of the initial value comes from <em>b<\/em> in slope-intercept form of a linear function,\u00a0[latex]f\\left(x\\right)=mx+b[\/latex]<\/li>\n<li>The initial value can be found by solving for <em>b<\/em>\u00a0or substituting\u00a0[latex]0[\/latex] in for <em>x<\/em> in a linear function.<\/li>\n<\/ul>\n<p>In our next example, we are given a scenario where Marcus wants to increase the number of songs in his music collection by a fixed amount each month. This is a perfect candidate for a linear function because the increase in the number of songs stays the same each month. We will identify the initial value for the music collection and write an equation that represents the number of songs in the collection for any number of months\u00a0<em>t<\/em>.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Marcus currently has\u00a0[latex]200[\/latex] songs in his music collection. Every month he adds\u00a0[latex]15[\/latex] new songs. Write a formula for the number of songs, <em>N<\/em>, in his collection as a function of time, <em>t<\/em>, the number of months. How many songs will he own in a year?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q14550\">Show Solution<\/span><\/p>\n<div id=\"q14550\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135411394\">The initial value for this function is\u00a0[latex]200[\/latex] because he currently owns\u00a0[latex]200[\/latex] songs, so <i>N<\/i>(0) =\u00a0[latex]200[\/latex]. This means that <em>b<\/em> =[latex]200[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201031\/CNX_Precalc_Figure_02_01_0102.jpg\" alt=\"The equation f of x equals mx plus b. And arrow labeled 15 points to m, and an arrow labeled 200 points to b. The equation becomes N of t equals 15 t plus 200.\" width=\"487\" height=\"131\" \/><\/p>\n<p id=\"fs-id1165137738190\">The number of songs increases by\u00a0[latex]15[\/latex] songs per month, so the rate of change is\u00a0[latex]15[\/latex] songs per month. Therefore, we know that [latex]m=15[\/latex]. We can substitute the initial value and the rate of change into slope-intercept form.<span id=\"fs-id1165137417445\"><br \/>\n<\/span><\/p>\n<p id=\"fs-id1165137810258\">We can write the formula [latex]N\\left(t\\right)=15t+200[\/latex].<\/p>\n<p id=\"fs-id1165137454711\">With this formula, we can then predict how many songs Marcus will have in\u00a01 year (12 months). In other words, we can evaluate the function at [latex]t = 12[\/latex] .<\/p>\n<div id=\"fs-id1165137462736\" class=\"equation unnumbered\">[latex]\\begin{array}{l}N\\left(12\\right) & =15\\left(12\\right)+200\\hfill \\\\ & =180+200\\hfill \\\\ & =380\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137694205\">Marcus will have\u00a0[latex]380[\/latex] songs in\u00a0[latex]12[\/latex] months.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_02_01_10\" class=\"example\">\n<div id=\"fs-id1165137836747\" class=\"exercise\">\n<p id=\"fs-id1165134065131\">In the example we just completed, notice that <em>N<\/em> is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.<\/p>\n<p>The following video provides an example of how to write a function that will give the cost in dollars for a given number of credit hours taken, x.\u00a0 Notice how the function consists of an initial value (the cost of the registration fee) plus an increase in cost for every credit hour taken.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Write and Graph a Linear Function by Making a Table of Values (Intro)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/X3Sx2TxH-J0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In our next example, we will show that you can write the equation for a linear function given two data points. In this case, Ilya&#8217;s weekly income depends on the number of insurance policies he sells. We are given his income for two different weeks and the number of policies sold. We first find the rate of change and then solve for the initial value.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya\u2019s weekly income, <i>I<\/i>, depends on the number of new policies, <em>n<\/em>, he sells during the week. Last week, he sold\u00a0[latex]3[\/latex] new policies and earned\u00a0[latex]$760[\/latex] for the week. The week before, he sold\u00a0[latex]5[\/latex] new policies and earned\u00a0[latex]$920[\/latex]. Find an equation for <em>I<\/em>(<em>n<\/em>) and interpret the meaning of the components of the equation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q249315\">Show Solution<\/span><\/p>\n<div id=\"q249315\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135169134\">The given information gives us two input-output pairs:\u00a0[latex](3, 760)[\/latex] and [latex](5, 920)[\/latex]. We start by finding the rate of change.<\/p>\n<div id=\"fs-id1165135195046\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}m & =\\dfrac{920 - 760}{5 - 3}\\hfill \\\\ & =\\dfrac{$160}{\\text{2 policies}}\\hfill \\\\ & =$80\\text{ per policy}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137855034\">Keeping track of units can help us interpret this quantity. Income increased by\u00a0[latex]$160[\/latex] when the number of policies increased by\u00a0[latex]2[\/latex], so the rate of change is\u00a0[latex]$80[\/latex] per policy. Therefore, Ilya earns a commission of\u00a0[latex]$80[\/latex] for each policy sold during the week.<\/p>\n<p id=\"fs-id1165137855040\">We can then solve for the initial value.<\/p>\n<div id=\"fs-id1165135484088\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}\\text{ }I\\left(n\\right)=80n+b\\hfill & \\hfill \\\\ \\text{ }760=80\\left(3\\right)+b\\hfill & \\text{when }n=3, I\\left(3\\right)=760\\hfill \\\\ 760 - 80\\left(3\\right)=b\\hfill & \\hfill \\\\ \\text{ }520=b\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137400716\">The value of <em>b<\/em>\u00a0is the starting value for the function and represents Ilya\u2019s income when\u00a0\u00a0[latex]n = 0[\/latex], or when no new policies are sold. We can interpret this as Ilya\u2019s base salary for the week which does not depend upon the number of policies sold.<\/p>\n<p id=\"fs-id1165135203653\">We can now write the final equation.<\/p>\n<div id=\"fs-id1165137506449\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]I\\left(n\\right)=80n+520[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137655487\">Our final interpretation is that Ilya\u2019s base salary is\u00a0[latex]$520[\/latex] per week and he earns an additional\u00a0[latex]$80[\/latex] commission for each policy sold.<\/p>\n<p>&nbsp;<\/p>\n<p>Notice that we used units to help us verify that we were calculating the rate correctly. It makes sense to speak in terms of the price per policy. To calculate the initial value, we solved for <em>b<\/em> by substituting values from one of the points we were given for <em>n<\/em> and <em>I<\/em>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video example, we show how to identify the initial value, slope, and equation for a linear function.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex: Savings Linear Function Application (Slope, Intercept Meaning)\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/JMQSdRFJ1S4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We will show one more example of how to write a linear function that represents the monthly cost to run a company given monthly fixed costs and production costs per item.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose Ben starts a company in which he incurs a fixed cost of\u00a0[latex]$1,250[\/latex] per month for the overhead which includes his office rent. His production costs are\u00a0[latex]$37.50[\/latex] per item. Write a linear function <em>C\u00a0<\/em>where <i>C<\/i>(<em>x<\/em>)\u00a0is the cost for <em>x<\/em>\u00a0items produced in a given month.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q625547\">Show Solution<\/span><\/p>\n<div id=\"q625547\" class=\"hidden-answer\" style=\"display: none\">\n<p>The fixed cost is present every month,\u00a0[latex]$1,250[\/latex]. The costs that can vary include the cost to produce each item, which is\u00a0[latex]$37.50[\/latex] for Ben. The variable cost, called the marginal cost, is represented by\u00a0[latex]37.5[\/latex]. The cost Ben incurs is the sum of these two costs represented by [latex]C\\left(x\\right)=1250+37.5x[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165135511326\">It is important to note that we are writing a function based on monthly costs, so the initial cost will be\u00a0[latex]$1,250[\/latex], because Ben has to pay that amount monthly for rent. If Ben produces\u00a0[latex]100[\/latex] items in a month, his monthly cost is represented by<\/p>\n<div id=\"fs-id1165137417815\" class=\"equation unnumbered\">[latex]\\begin{array}{l}C\\left(100\\right)=1250+37.5\\left(100\\right)=5000\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137862645\">So his monthly cost would be\u00a0[latex]$5,000[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video example shows how to write a linear function that represents how many miles you can travel in a rental car given a fixed amount of money.<\/p>\n<div id=\"fs-id1165137767515\" class=\"commentary\">\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex:  Linear Equation Application (Cost of a Rental Car)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/H8KR3w2nXqs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>In the next example, we will take data that is in tabular (table) form to write an equation that describes the rate of change of a rat population.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p id=\"fs-id1165135161325\">The table below relates the number of rats in a population to time (in weeks). Use the table to write a linear equation.<\/p>\n<table id=\"Table_02_01_02\" summary=\"Two rows and five columns. The first row is labeled, 'w, the numers of weeks'. The second row is labeled is labeled, 'P(w), number of rats'. Reading the remaining rows as ordered pairs (i.e., (w, P(w)), we have the following values: (0, 1000), (2, 1080), (4, 1160), and (6, 1240).\">\n<tbody>\n<tr>\n<td><strong><em>w<\/em>, number of weeks<\/strong><\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]6[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong><em>P(w)<\/em>, number of rats<\/strong><\/td>\n<td>[latex]1000[\/latex]<\/td>\n<td>[latex]1080[\/latex]<\/td>\n<td>[latex]1160[\/latex]<\/td>\n<td>[latex]1240[\/latex]<\/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=\"q780371\">Show Solution<\/span><\/p>\n<div id=\"q780371\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137530990\">We can see from the table that the initial value for the number of rats is\u00a0[latex]1000[\/latex], so <em>b<\/em> =[latex]1000[\/latex].<\/p>\n<p id=\"fs-id1165137935601\">Rather than solving for <em>m<\/em>, we can tell from looking at the table that the population increases by\u00a0[latex]80[\/latex] for every\u00a0[latex]2[\/latex] weeks that pass. This means that the rate of change is\u00a0[latex]80[\/latex] rats per\u00a0[latex]2[\/latex] weeks, which can be simplified to\u00a0[latex]40[\/latex] rats per week.<\/p>\n<div id=\"fs-id1165137737900\" class=\"equation unnumbered\">[latex]P\\left(w\\right)=40w+1000[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137465125\">If we did not notice the rate of change from the table, we could still solve for the slope using any two points from the table. For example, using\u00a0[latex](2, 1080)[\/latex] and\u00a0[latex](6, 1240)[\/latex],<\/p>\n<div id=\"fs-id1165137627069\" class=\"equation unnumbered\">[latex]\\begin{array}{l}m & =\\dfrac{1240 - 1080}{6 - 2}\\hfill \\\\ & =\\dfrac{160}{4}\\hfill \\\\ & =40\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Is the initial value always provided in a table of values like the table in\u00a0the previous example? Write your ideas in the textbox below before you look at the answer.<\/p>\n<p>If your answer is no, give a description of how you would find the initial value.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"2\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q298919\">Show Solution<\/span><\/p>\n<div id=\"q298919\" class=\"hidden-answer\" style=\"display: none\">\n<p>No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of\u00a0[latex]0[\/latex], then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to\u00a0\u00a0[latex]0[\/latex], find the slope, substitute one coordinate pair and the slope into [latex]f\\left(x\\right)=mx+b[\/latex], and solve for b.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Summary<\/h2>\n<ul>\n<li>Sometimes we are given an initial value and sometimes we have to solve for it.<\/li>\n<li>Using units can help you verify that you have calculated slope correctly.<\/li>\n<li>We can write the equation for a line given a slope and a data point or from a table of data.<\/li>\n<\/ul>\n<\/div>\n<\/div>\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-16027\">\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>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>Ex: Linear Equation Application (Cost of a Rental Car). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/H8KR3w2nXqs\">https:\/\/youtu.be\/H8KR3w2nXqs<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Savings Linear Function Application (Slope, Intercept Meaning). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/JMQSdRFJ1S4\">https:\/\/youtu.be\/JMQSdRFJ1S4<\/a>. <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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":169554,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"\",\"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 Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Linear Equation Application (Cost of a Rental Car)\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/H8KR3w2nXqs\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Savings Linear Function Application (Slope, Intercept 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