{"id":538,"date":"2024-05-23T19:46:16","date_gmt":"2024-05-23T19:46:16","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/?post_type=chapter&#038;p=538"},"modified":"2026-02-05T09:56:57","modified_gmt":"2026-02-05T09:56:57","slug":"1-4-modeling-with-linear-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/1-4-modeling-with-linear-functions\/","title":{"raw":"1.4 Modeling with Linear Functions","rendered":"1.4 Modeling with Linear Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Construct and solve applications using linear functions.<\/li>\r\n<\/ul>\r\n<\/div>\r\nTo use a linear function to describe a real-world application, we must first determine the known quantities and define the unknown quantities as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. For example, think of the rental price for a car. Say the company charges [latex]$0.10[\/latex] per mile in addition to a flat rate. In this case, to find the total cost to rent the car the known cost of [latex]$0.10[\/latex] per mile will be multiplied by an unknown quantity; the number of miles driven. If we use [latex]x[\/latex] to represent the unknown number of miles driven, then [latex]0.10x[\/latex] represents the variable cost because it changes according to the number of miles driven.\r\n\r\nIf a quantity is independent of a variable we usually just add or subtract it according to the problem. As these amounts do not change, we call them fixed costs. If the car rental company charges [latex]$0.10[\/latex] per mile plus a daily fee of [latex]$50[\/latex], then the equation that describes the daily car rental cost, [latex]C[\/latex], is [latex]C=0.10x+50[\/latex].\r\n<h2>Creating a Model<\/h2>\r\nAt the beginning of section 1.1, we discussed that climate change is one of the most serious threats to our planet today and how the increasing levels of CO<sub>2<\/sub> in the earth's atmosphere parallels the increase in global temperatures. In 2016, the global temperature was [latex]1.1[\/latex] degrees C ([latex]2[\/latex] degrees F) warmer than pre-industrial levels. The Paris Agreement was made and aims to keep the global temperature increase well below [latex]2[\/latex] degrees C, and hopefully limit that increase to [latex]1.5[\/latex] degrees C. (https:\/\/e360.yale.edu\/features\/how-the-world-passed-a-carbon-threshold-400ppm-and-why-it-matters) Although 196 countries signed this agreement, global warming is stilling increasing. From 2015 - 2019 the global temperature has risen [latex]0.2[\/latex] degrees C. If this trend keeps rising like it has, our planet will be in serious danger.\r\n\r\nWe can use data from the follow graph to make a linear equation that models the increase of CO<sub>2<\/sub> levels and use that model to predict the CO<sub>2<\/sub> levels in the future, if the CO<sub>2<\/sub> levels keep increasing at a similar rate. Data on CO<sub>2<\/sub> levels have\u00a0been collected since 1958 and is presented graphically below (red graph). We will be focusing on data starting with January 2005 up to present time. When we look at that portion of the graph the trend appears to follow an overall increasing linear model.\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/1-1-graphs\/screenshot-2024-08-03-at-1-07-01%e2%80%afpm\/\" rel=\"attachment wp-att-1720\"><img class=\"alignnone wp-image-1720 size-large\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-03-at-1.07.01\u202fPM-1024x768.png\" alt=\"A graph showing changes in atmospheric CO\u2082, ocean CO\u2082, and ocean pH in the North Pacific over time. The red data shows atmospheric CO\u2082 at Mauna Loa, fluctuating, but rising steadily from about 315 ppm in 1958 to over 420 ppm in 2024. The green data shows ocean CO\u2082 (pCO\u2082) at Station ALOHA, increasing from about 315 \u00b5atm in 1988 to over 400 \u00b5atm in 2024. The blue data shows ocean pH at Station ALOHA, which decreases from about 8.2 to just below 8.05 over the same time.\" width=\"1024\" height=\"768\" \/><\/a>\r\n\r\nThis link <a href=\"https:\/\/gml.noaa.gov\/ccgg\/trends\/graph.html\" target=\"_blank\" rel=\"noopener\">https:\/\/gml.noaa.gov\/ccgg\/trends\/graph.html<\/a>\u00a0will take you to an interactive graph that you can use to fill out the following table.\r\n<table style=\"border-collapse: collapse; width: 50%; border: 1px solid black; height: 300px; font-size: 110%;\">\r\n<thead>\r\n<tr style=\"height: 12px;\">\r\n<th style=\"width: 30%; border: 1px solid #999999; height: 12px;\">Date of Monthly Averages<\/th>\r\n<th style=\"width: 30%; border: 1px solid #999999; height: 12px;\">CO<sub>2<\/sub> levels (in ppm)<\/th>\r\n<th style=\"width: 70%; border: 1px solid #999999; height: 12px;\"><strong>Ordered Pair <\/strong>[latex](x,y)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2005<\/td>\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">378.63<\/td>\r\n<td style=\"width: 70%; border: 1px solid #999999;\">(2005, 378.63)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2008<\/td>\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\"><\/td>\r\n<td style=\"width: 70%; border: 1px solid #999999;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2011<\/td>\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\"><\/td>\r\n<td style=\"width: 70%; border: 1px solid #999999;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2014<\/td>\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\"><\/td>\r\n<td style=\"width: 70%; border: 1px solid #999999;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2017<\/td>\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\"><\/td>\r\n<td style=\"width: 70%; border: 1px solid #999999;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2020<\/td>\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\"><\/td>\r\n<td style=\"width: 70%; border: 1px solid #999999;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2023<\/td>\r\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\"><\/td>\r\n<td style=\"width: 70%; border: 1px solid #999999;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nLet's find the average rate of change from Jan 2005 to Jan 2014, using the slope formula.\u00a0Recall that the slope [latex]m[\/latex], or rate of change of a linear function between two points [latex]\\left({x}_{2,\\text{ }}{y}_{2}\\right)[\/latex] and [latex]\\left({x}_{1},\\text{ }{y}_{1}\\right)[\/latex] is: [latex]\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex].\r\n\r\n[latex]\\dfrac{Change\\ in\\ CO_{2}\\ levels}{Change\\ in\\ years} = \\dfrac{398.04-378.63}{2014-2005} = \\dfrac{19.41}{9} = 2.16[\/latex]\r\n\r\nThe average CO<sub>2<\/sub> increase per year is [latex]2.16[\/latex] ppm.\r\n\r\nWe can use this information to write a linear equation that models the increase of atmospheric\u00a0CO<sub>2<\/sub> since Jan 2005. Let\u00a0[latex]x[\/latex] represent the number of years since 2005 ([latex]x=0[\/latex] for Jan 2005) and let [latex]y[\/latex] represent the amount of atmospheric CO<sub>2<\/sub>\u00a0in that given year, measured in parts per million (ppm).\r\n\r\nWe will write this equation in [latex]y=mx+b[\/latex] form. As discussed earlier in this module, [latex]m[\/latex] is the slope and [latex]b[\/latex] is the [latex]y[\/latex]-intercept. When we are writing a linear model, [latex]m[\/latex] is the average rate of change and [latex]b[\/latex] is the start amount (or initial value). Above, we found the average rate of change of CO<sub>2<\/sub>\u00a0to be [latex]2.16[\/latex] ppm. The start amount is the amount of atmospheric CO<sub>2<\/sub> in Jan 2005, which is [latex]378.63[\/latex] ppm.\r\n\r\nNow substituting these values in place of [latex]m[\/latex] and [latex]b[\/latex] we get [latex]y=2.16x+378.63[\/latex].\r\n\r\nWe can now use this linear model to make some predictions (assuming that the average rate of change stays the same).\r\n\r\nPredict the amount of atmospheric CO<sub>2<\/sub> (in ppm) in January 2030 and January 2045\r\n<ul>\r\n \t<li><span style=\"text-decoration: underline;\">In Jan 2030:<\/span> In order to use the above linear model we need to determine [latex]x[\/latex]. Remember that [latex]x=0[\/latex] for Jan 2005. To find the [latex]x[\/latex] we will find the difference in the years, [latex]x=2030-2005=25[\/latex]. Now substitute [latex]\\color{Green}{25}[\/latex] in place of [latex]\\color{Green}{x}[\/latex] and solve for [latex]y[\/latex].<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} y &amp;= 2.16\\color{Green}{x}\\color{black}{+378.63}\\\\ y &amp;= 2.16(\\color{Green}{25}\\color{black}{)+378.63}\\\\ y &amp;= 54+378.63\\\\ y &amp;= 432.63 \\end{align}[\/latex]<\/p>\r\nSo in 2030, the CO<sub>2<\/sub> levels will have risen to approximately [latex]432.63[\/latex] ppm.\r\n<ul>\r\n \t<li><span style=\"text-decoration: underline;\">In Jan 2045:<\/span>\u00a0To find the [latex]x[\/latex] we will find the difference of the years, [latex]x=2045-2005=\\color{Green}{40}[\/latex].<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} y &amp;= 2.16x+378.63\\\\ y &amp;= 2.16(\\color{Green}{40}\\color{black}{)+378.63}\\\\ y &amp;= 86.4+378.63\\\\ y &amp;= 465.03 \\end{align}[\/latex]<\/p>\r\nSo in 2045, the CO<sub>2<\/sub> levels will have risen to approximately [latex]465.03[\/latex] ppm.\r\n\r\nWhen will the atmospheric CO<sub>2<\/sub> levels reach [latex]500[\/latex] ppm, according to the model? (Round to nearest year)\r\n<ul>\r\n \t<li>Replace [latex]\\color{Green}{y}[\/latex] with [latex]\\color{Green}{500}[\/latex] and solve for x.<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} y &amp;= 2.16x+378.63\\\\ \\color{Green}{500} &amp;= 2.16x+378.63\\\\ 500-378.63 &amp;= 2.16x &amp;&amp; \\color{blue}{\\textsf{subtract}}\\\\ 121.37 &amp;= 2.16x &amp;&amp; \\color{blue}{\\textsf{divide}}\\\\ 56.189... &amp;= x \\end{align}[\/latex]<\/p>\r\nSo, the CO<sub>2<\/sub> levels will reach [latex]500[\/latex] ppm approximately 56 years after 2005. The CO<sub>2<\/sub> levels will be over [latex]500[\/latex] ppm in the year 2061.\r\n<h2>Initial Value and Rate of Change<\/h2>\r\n<p id=\"fs-id1165137594074\">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\n\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 start amount (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 sometimes called the start amount. \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 [latex]y[\/latex]-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 [latex]y[\/latex]-value of the initial value comes from [latex]b[\/latex] 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 [latex]b[\/latex] or substituting [latex]0[\/latex] in for [latex]x[\/latex] in a linear function.<\/li>\r\n<\/ul>\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, [latex]N[\/latex], in his collection as a function of time, [latex]t[\/latex], the number of months.\r\n\r\n1) How many songs will he own in a year?\r\n\r\n2) How many months will it take to grow his collection to 500 songs?\r\n\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 [latex]N(0)=200[\/latex]. This means that [latex]b=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=\"Equation: f of x equals m x plus b. Two arrows point from the equation from m to 15 and b to 200. Below these numbers is another linear equation, 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\">1) To predict how many songs Marcus will have in\u00a01 year (12 months), 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\n2) To determine how many months it will take to grow his collection to [latex]500[\/latex] songs, replace [latex]N(t)[\/latex] with [latex]500[\/latex] and then solve for [latex]t[\/latex].\r\n\r\n[latex]\\begin{align} N(t) &amp;= 15t+200\\\\ 500 &amp;= 15t+200\\\\ 500-200 &amp;= 15t &amp;&amp; \\color{blue}{\\textsf{subtract}}\\\\ 300 &amp;= 15t\\\\ 20 &amp;= t &amp;&amp; \\color{blue}{\\textsf{divide}} \\end{align}[\/latex]\r\n\r\nIn 20 months, Marcus will have 500 songs in his collection.\r\n\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 above, notice that [latex]N[\/latex] is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<h2>Comparisons with Linear Functions<\/h2>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThere are two cell phone companies that offer different packages. Company A charges a monthly service fee of [latex]$34[\/latex] plus [latex]$0.05[\/latex] per minute of talk-time. Company B charges a monthly service fee of [latex]$40[\/latex] plus [latex]$0.04[\/latex] per minute of talk-time.\r\n<ol>\r\n \t<li>Write a linear equation that models the packages offered by both companies.<\/li>\r\n \t<li>If the average number of minutes used each month is [latex]1,160[\/latex], which company offers the better plan?<\/li>\r\n \t<li>If the average number of minutes used each month is [latex]420[\/latex], which company offers the better plan?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"14868\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"14868\"]\r\n<ol>\r\n \t<li>The model for Company [latex]A[\/latex] can be written as [latex]A=0.05x+34[\/latex]. This includes the variable cost of [latex]0.05x[\/latex] plus the monthly service charge of $34. Company [latex]B[\/latex]\u2019s package charges a higher monthly fee of [latex]$40[\/latex], but a lower variable cost of [latex]0.04x[\/latex]. Company [latex]B[\/latex]\u2019s model can be written as [latex]B=0.04x+\\$40[\/latex].<\/li>\r\n \t<li>If the average number of minutes used each month is [latex]1,160[\/latex], we have the following:\r\n<div>[latex]\\begin{array}{l}\\text{Company }A\\hfill&amp;=0.05\\left(1,160\\right)+34\\hfill \\\\ \\hfill&amp;=58+34\\hfill \\\\ \\hfill&amp;=92\\hfill \\\\ \\hfill \\\\ \\text{Company }B\\hfill&amp;=0.04\\left(1,160\\right)+40\\hfill \\\\ \\hfill&amp;=46.4+40\\hfill \\\\ \\hfill&amp;=86.4\\hfill \\end{array}[\/latex]<\/div>\r\nSo, Company [latex]B[\/latex] offers the lower monthly cost of [latex]$86.40[\/latex] as compared with the [latex]$92[\/latex] monthly cost offered by Company [latex]A[\/latex] when the average number of minutes used each month is [latex]1,160[\/latex].<\/li>\r\n \t<li>If the average number of minutes used each month is [latex]420[\/latex], we have the following:\r\n<div>[latex]\\begin{array}{l}\\text{Company }A\\hfill&amp;=0.05\\left(420\\right)+34\\hfill \\\\ \\hfill&amp;=21+34\\hfill \\\\ \\hfill&amp;=55\\hfill \\\\ \\hfill \\\\ \\text{Company }B\\hfill&amp;=0.04\\left(420\\right)+40\\hfill \\\\ \\hfill&amp;=16.8+40\\hfill \\\\ \\hfill&amp;=56.8\\hfill \\end{array}[\/latex]<\/div>\r\nIf the average number of minutes used each month is [latex]420[\/latex], then Company [latex]A[\/latex] offers a lower monthly cost of $55 compared to Company [latex]B[\/latex]\u2019s monthly cost of \u00a0[latex]$56.80[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows another example of writing two equations that will allow you to compare two different cell phone plans.\r\n\r\nhttps:\/\/youtu.be\/Q5hlC_VPKGM\r\n<div id=\"Example_02_01_10\" class=\"example\">\r\n<div id=\"fs-id1165137836747\" class=\"exercise\">\r\n<h2>Writing a Linear Function Given Two Data Points<\/h2>\r\nThe next example demonstrating writing 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, [latex]I[\/latex], depends on the number of new policies, [latex]n[\/latex], 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].\r\n\r\n1) Find an equation for [latex]I(n)[\/latex] and interpret the meaning of the components of the equation.\r\n\r\n2) What is Ilya's weekly income if he sells [latex]12[\/latex] policies?\r\n\r\n3) How many policies does he need to sell to make a weekly income of $2040?\r\n\r\n[reveal-answer q=\"249315\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"249315\"]\r\n<p id=\"fs-id1165135169134\">1) The given information gives us two input-output pairs (number of policies sold, amount of money earned:\u00a0[latex](3, 760)[\/latex] and [latex](5, 920)[\/latex]. We start by finding the rate of change.<\/p>\r\nRecall that the slope [latex]m[\/latex], or rate of change of a linear function between two points [latex]\\left({x}_{1,\\text{ }}{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},\\text{ }{y}_{2}\\right)[\/latex] is: [latex]\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex].\r\n<div id=\"fs-id1165135195046\" class=\"equation unnumbered\">[latex]\\begin{align}m&amp;=\\dfrac{920 - 760}{5 - 3}\\\\&amp;=\\dfrac{$160}{\\text{2 policies}}\\\\&amp;=$80\\text{ per policy}\\end{align}[\/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\">[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 [latex]b[\/latex] is 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\">[latex]I\\left(n\\right)=80n+520[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137655487\">We can interpret the equation 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\n2) To find Ilya's weekly income if he sells [latex]12[\/latex] policies, let's use the function [latex]I(n)=80n+520[\/latex]. Start by replacing [latex]n[\/latex] with [latex]12[\/latex] in the function, then solve for [latex]I(n)[\/latex].\r\n<p style=\"text-align: center;\">$$\\begin{align}\r\nI(\\color{Green}{n}) &amp;= 80\\color{Green}{n} + 520 \\\\\r\nI(\\color{Green}{12}) &amp;= 80(\\color{Green}{12}) + 520 \\\\\r\n&amp;= 960 + 520 \\\\\r\n&amp;= 1480\r\n\\end{align}$$<\/p>\r\nIlya will make [latex]$1480[\/latex] if he sells [latex]12[\/latex] policies.\r\n\r\n3) If Ilya wants to make [latex]$2040[\/latex] a week, we need to find how many policies he would need to sell. Replace [latex]I(n)[\/latex] with [latex]2040[\/latex] and then solve for [latex]n[\/latex].\r\n<p style=\"text-align: center;\">$$\\begin{align}\r\n\\color{Green}{I(n)} &amp;= 80n + 520 \\\\\r\n\\color{Green}{2040} &amp;= 80n + 520 \\\\\r\n2040-520 &amp;= 80n \\\\\r\n1520 &amp;= 80n \\\\ 19 &amp;= n\r\n\\end{align}$$<\/p>\r\nIlya will need to sell [latex]19[\/latex] policies if he wants to make [latex]$2040[\/latex].\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<h2>Modeling Monthly Cost<\/h2>\r\nThis example of modeling real-world scenarios using linear functions involves the monthly cost to run a company given monthly fixed costs and production costs per item manufactured.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose Sofia starts a company in which she incurs a fixed cost of\u00a0[latex]$1,250[\/latex] per month for the overhead which includes office rent. Her production costs are\u00a0[latex]$37.50[\/latex] per item. Write a linear function [latex]C[\/latex] where [latex]C(x)[\/latex] is the cost for [latex]x[\/latex] items produced in a given month. What will be the monthly cost if Sofia produces [latex]100[\/latex] items in a 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 the same each month; [latex]$1,250[\/latex]. The costs that can vary include the cost to produce each item, which is\u00a0[latex]$37.50[\/latex] for Sofia's company. The variable cost (called the marginal cost) is represented by\u00a0[latex]37.5[\/latex]. The cost Sofia incurs is the sum of these two costs: [latex]C\\left(x\\right)=1250+37.5x[\/latex].\r\n\r\n&nbsp;\r\n<p id=\"fs-id1165135511326\">If Sofia produces\u00a0[latex]100[\/latex] items in a month, her monthly cost is [latex]C(100)=1250+37.5(100)=5000[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nA final 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\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Construct and solve applications using linear functions.<\/li>\n<\/ul>\n<\/div>\n<p>To use a linear function to describe a real-world application, we must first determine the known quantities and define the unknown quantities as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. For example, think of the rental price for a car. Say the company charges [latex]$0.10[\/latex] per mile in addition to a flat rate. In this case, to find the total cost to rent the car the known cost of [latex]$0.10[\/latex] per mile will be multiplied by an unknown quantity; the number of miles driven. If we use [latex]x[\/latex] to represent the unknown number of miles driven, then [latex]0.10x[\/latex] represents the variable cost because it changes according to the number of miles driven.<\/p>\n<p>If a quantity is independent of a variable we usually just add or subtract it according to the problem. As these amounts do not change, we call them fixed costs. If the car rental company charges [latex]$0.10[\/latex] per mile plus a daily fee of [latex]$50[\/latex], then the equation that describes the daily car rental cost, [latex]C[\/latex], is [latex]C=0.10x+50[\/latex].<\/p>\n<h2>Creating a Model<\/h2>\n<p>At the beginning of section 1.1, we discussed that climate change is one of the most serious threats to our planet today and how the increasing levels of CO<sub>2<\/sub> in the earth&#8217;s atmosphere parallels the increase in global temperatures. In 2016, the global temperature was [latex]1.1[\/latex] degrees C ([latex]2[\/latex] degrees F) warmer than pre-industrial levels. The Paris Agreement was made and aims to keep the global temperature increase well below [latex]2[\/latex] degrees C, and hopefully limit that increase to [latex]1.5[\/latex] degrees C. (https:\/\/e360.yale.edu\/features\/how-the-world-passed-a-carbon-threshold-400ppm-and-why-it-matters) Although 196 countries signed this agreement, global warming is stilling increasing. From 2015 &#8211; 2019 the global temperature has risen [latex]0.2[\/latex] degrees C. If this trend keeps rising like it has, our planet will be in serious danger.<\/p>\n<p>We can use data from the follow graph to make a linear equation that models the increase of CO<sub>2<\/sub> levels and use that model to predict the CO<sub>2<\/sub> levels in the future, if the CO<sub>2<\/sub> levels keep increasing at a similar rate. Data on CO<sub>2<\/sub> levels have\u00a0been collected since 1958 and is presented graphically below (red graph). We will be focusing on data starting with January 2005 up to present time. When we look at that portion of the graph the trend appears to follow an overall increasing linear model.<\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/1-1-graphs\/screenshot-2024-08-03-at-1-07-01%e2%80%afpm\/\" rel=\"attachment wp-att-1720\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1720 size-large\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-03-at-1.07.01\u202fPM-1024x768.png\" alt=\"A graph showing changes in atmospheric CO\u2082, ocean CO\u2082, and ocean pH in the North Pacific over time. The red data shows atmospheric CO\u2082 at Mauna Loa, fluctuating, but rising steadily from about 315 ppm in 1958 to over 420 ppm in 2024. The green data shows ocean CO\u2082 (pCO\u2082) at Station ALOHA, increasing from about 315 \u00b5atm in 1988 to over 400 \u00b5atm in 2024. The blue data shows ocean pH at Station ALOHA, which decreases from about 8.2 to just below 8.05 over the same time.\" width=\"1024\" height=\"768\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-03-at-1.07.01\u202fPM-1024x768.png 1024w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-03-at-1.07.01\u202fPM-300x225.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-03-at-1.07.01\u202fPM-768x576.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-03-at-1.07.01\u202fPM-65x49.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-03-at-1.07.01\u202fPM-225x169.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-03-at-1.07.01\u202fPM-350x263.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-03-at-1.07.01\u202fPM.png 1298w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/a><\/p>\n<p>This link <a href=\"https:\/\/gml.noaa.gov\/ccgg\/trends\/graph.html\" target=\"_blank\" rel=\"noopener\">https:\/\/gml.noaa.gov\/ccgg\/trends\/graph.html<\/a>\u00a0will take you to an interactive graph that you can use to fill out the following table.<\/p>\n<table style=\"border-collapse: collapse; width: 50%; border: 1px solid black; height: 300px; font-size: 110%;\">\n<thead>\n<tr style=\"height: 12px;\">\n<th style=\"width: 30%; border: 1px solid #999999; height: 12px;\">Date of Monthly Averages<\/th>\n<th style=\"width: 30%; border: 1px solid #999999; height: 12px;\">CO<sub>2<\/sub> levels (in ppm)<\/th>\n<th style=\"width: 70%; border: 1px solid #999999; height: 12px;\"><strong>Ordered Pair <\/strong>[latex](x,y)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2005<\/td>\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">378.63<\/td>\n<td style=\"width: 70%; border: 1px solid #999999;\">(2005, 378.63)<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2008<\/td>\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\"><\/td>\n<td style=\"width: 70%; border: 1px solid #999999;\"><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2011<\/td>\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\"><\/td>\n<td style=\"width: 70%; border: 1px solid #999999;\"><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2014<\/td>\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\"><\/td>\n<td style=\"width: 70%; border: 1px solid #999999;\"><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2017<\/td>\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\"><\/td>\n<td style=\"width: 70%; border: 1px solid #999999;\"><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2020<\/td>\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\"><\/td>\n<td style=\"width: 70%; border: 1px solid #999999;\"><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\">Jan 2023<\/td>\n<td style=\"height: 14px; width: 30%; border: 1px solid #999999;\"><\/td>\n<td style=\"width: 70%; border: 1px solid #999999;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Let&#8217;s find the average rate of change from Jan 2005 to Jan 2014, using the slope formula.\u00a0Recall that the slope [latex]m[\/latex], or rate of change of a linear function between two points [latex]\\left({x}_{2,\\text{ }}{y}_{2}\\right)[\/latex] and [latex]\\left({x}_{1},\\text{ }{y}_{1}\\right)[\/latex] is: [latex]\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex].<\/p>\n<p>[latex]\\dfrac{Change\\ in\\ CO_{2}\\ levels}{Change\\ in\\ years} = \\dfrac{398.04-378.63}{2014-2005} = \\dfrac{19.41}{9} = 2.16[\/latex]<\/p>\n<p>The average CO<sub>2<\/sub> increase per year is [latex]2.16[\/latex] ppm.<\/p>\n<p>We can use this information to write a linear equation that models the increase of atmospheric\u00a0CO<sub>2<\/sub> since Jan 2005. Let\u00a0[latex]x[\/latex] represent the number of years since 2005 ([latex]x=0[\/latex] for Jan 2005) and let [latex]y[\/latex] represent the amount of atmospheric CO<sub>2<\/sub>\u00a0in that given year, measured in parts per million (ppm).<\/p>\n<p>We will write this equation in [latex]y=mx+b[\/latex] form. As discussed earlier in this module, [latex]m[\/latex] is the slope and [latex]b[\/latex] is the [latex]y[\/latex]-intercept. When we are writing a linear model, [latex]m[\/latex] is the average rate of change and [latex]b[\/latex] is the start amount (or initial value). Above, we found the average rate of change of CO<sub>2<\/sub>\u00a0to be [latex]2.16[\/latex] ppm. The start amount is the amount of atmospheric CO<sub>2<\/sub> in Jan 2005, which is [latex]378.63[\/latex] ppm.<\/p>\n<p>Now substituting these values in place of [latex]m[\/latex] and [latex]b[\/latex] we get [latex]y=2.16x+378.63[\/latex].<\/p>\n<p>We can now use this linear model to make some predictions (assuming that the average rate of change stays the same).<\/p>\n<p>Predict the amount of atmospheric CO<sub>2<\/sub> (in ppm) in January 2030 and January 2045<\/p>\n<ul>\n<li><span style=\"text-decoration: underline;\">In Jan 2030:<\/span> In order to use the above linear model we need to determine [latex]x[\/latex]. Remember that [latex]x=0[\/latex] for Jan 2005. To find the [latex]x[\/latex] we will find the difference in the years, [latex]x=2030-2005=25[\/latex]. Now substitute [latex]\\color{Green}{25}[\/latex] in place of [latex]\\color{Green}{x}[\/latex] and solve for [latex]y[\/latex].<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[latex]\\begin{align} y &= 2.16\\color{Green}{x}\\color{black}{+378.63}\\\\ y &= 2.16(\\color{Green}{25}\\color{black}{)+378.63}\\\\ y &= 54+378.63\\\\ y &= 432.63 \\end{align}[\/latex]<\/p>\n<p>So in 2030, the CO<sub>2<\/sub> levels will have risen to approximately [latex]432.63[\/latex] ppm.<\/p>\n<ul>\n<li><span style=\"text-decoration: underline;\">In Jan 2045:<\/span>\u00a0To find the [latex]x[\/latex] we will find the difference of the years, [latex]x=2045-2005=\\color{Green}{40}[\/latex].<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[latex]\\begin{align} y &= 2.16x+378.63\\\\ y &= 2.16(\\color{Green}{40}\\color{black}{)+378.63}\\\\ y &= 86.4+378.63\\\\ y &= 465.03 \\end{align}[\/latex]<\/p>\n<p>So in 2045, the CO<sub>2<\/sub> levels will have risen to approximately [latex]465.03[\/latex] ppm.<\/p>\n<p>When will the atmospheric CO<sub>2<\/sub> levels reach [latex]500[\/latex] ppm, according to the model? (Round to nearest year)<\/p>\n<ul>\n<li>Replace [latex]\\color{Green}{y}[\/latex] with [latex]\\color{Green}{500}[\/latex] and solve for x.<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[latex]\\begin{align} y &= 2.16x+378.63\\\\ \\color{Green}{500} &= 2.16x+378.63\\\\ 500-378.63 &= 2.16x && \\color{blue}{\\textsf{subtract}}\\\\ 121.37 &= 2.16x && \\color{blue}{\\textsf{divide}}\\\\ 56.189... &= x \\end{align}[\/latex]<\/p>\n<p>So, the CO<sub>2<\/sub> levels will reach [latex]500[\/latex] ppm approximately 56 years after 2005. The CO<sub>2<\/sub> levels will be over [latex]500[\/latex] ppm in the year 2061.<\/p>\n<h2>Initial Value and Rate of Change<\/h2>\n<p id=\"fs-id1165137594074\">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<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 start amount (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 sometimes called the start amount. \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 [latex]y[\/latex]-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 [latex]y[\/latex]-value of the initial value comes from [latex]b[\/latex] 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 [latex]b[\/latex] or substituting [latex]0[\/latex] in for [latex]x[\/latex] in a linear function.<\/li>\n<\/ul>\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, [latex]N[\/latex], in his collection as a function of time, [latex]t[\/latex], the number of months.<\/p>\n<p>1) How many songs will he own in a year?<\/p>\n<p>2) How many months will it take to grow his collection to 500 songs?<\/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 [latex]N(0)=200[\/latex]. This means that [latex]b=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=\"Equation: f of x equals m x plus b. Two arrows point from the equation from m to 15 and b to 200. Below these numbers is another linear equation, 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\">1) To predict how many songs Marcus will have in\u00a01 year (12 months), 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<p>2) To determine how many months it will take to grow his collection to [latex]500[\/latex] songs, replace [latex]N(t)[\/latex] with [latex]500[\/latex] and then solve for [latex]t[\/latex].<\/p>\n<p>[latex]\\begin{align} N(t) &= 15t+200\\\\ 500 &= 15t+200\\\\ 500-200 &= 15t && \\color{blue}{\\textsf{subtract}}\\\\ 300 &= 15t\\\\ 20 &= t && \\color{blue}{\\textsf{divide}} \\end{align}[\/latex]<\/p>\n<p>In 20 months, Marcus will have 500 songs in his collection.<\/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 above, notice that [latex]N[\/latex] is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.<\/p>\n<\/div>\n<\/div>\n<h2>Comparisons with Linear Functions<\/h2>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>There are two cell phone companies that offer different packages. Company A charges a monthly service fee of [latex]$34[\/latex] plus [latex]$0.05[\/latex] per minute of talk-time. Company B charges a monthly service fee of [latex]$40[\/latex] plus [latex]$0.04[\/latex] per minute of talk-time.<\/p>\n<ol>\n<li>Write a linear equation that models the packages offered by both companies.<\/li>\n<li>If the average number of minutes used each month is [latex]1,160[\/latex], which company offers the better plan?<\/li>\n<li>If the average number of minutes used each month is [latex]420[\/latex], which company offers the better plan?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q14868\">Show Solution<\/span><\/p>\n<div id=\"q14868\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>The model for Company [latex]A[\/latex] can be written as [latex]A=0.05x+34[\/latex]. This includes the variable cost of [latex]0.05x[\/latex] plus the monthly service charge of $34. Company [latex]B[\/latex]\u2019s package charges a higher monthly fee of [latex]$40[\/latex], but a lower variable cost of [latex]0.04x[\/latex]. Company [latex]B[\/latex]\u2019s model can be written as [latex]B=0.04x+\\$40[\/latex].<\/li>\n<li>If the average number of minutes used each month is [latex]1,160[\/latex], we have the following:\n<div>[latex]\\begin{array}{l}\\text{Company }A\\hfill&=0.05\\left(1,160\\right)+34\\hfill \\\\ \\hfill&=58+34\\hfill \\\\ \\hfill&=92\\hfill \\\\ \\hfill \\\\ \\text{Company }B\\hfill&=0.04\\left(1,160\\right)+40\\hfill \\\\ \\hfill&=46.4+40\\hfill \\\\ \\hfill&=86.4\\hfill \\end{array}[\/latex]<\/div>\n<p>So, Company [latex]B[\/latex] offers the lower monthly cost of [latex]$86.40[\/latex] as compared with the [latex]$92[\/latex] monthly cost offered by Company [latex]A[\/latex] when the average number of minutes used each month is [latex]1,160[\/latex].<\/li>\n<li>If the average number of minutes used each month is [latex]420[\/latex], we have the following:\n<div>[latex]\\begin{array}{l}\\text{Company }A\\hfill&=0.05\\left(420\\right)+34\\hfill \\\\ \\hfill&=21+34\\hfill \\\\ \\hfill&=55\\hfill \\\\ \\hfill \\\\ \\text{Company }B\\hfill&=0.04\\left(420\\right)+40\\hfill \\\\ \\hfill&=16.8+40\\hfill \\\\ \\hfill&=56.8\\hfill \\end{array}[\/latex]<\/div>\n<p>If the average number of minutes used each month is [latex]420[\/latex], then Company [latex]A[\/latex] offers a lower monthly cost of $55 compared to Company [latex]B[\/latex]\u2019s monthly cost of \u00a0[latex]$56.80[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video shows another example of writing two equations that will allow you to compare two different cell phone plans.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Write Linear Equations to Model and Compare Cell Phone Plans with Data Usage\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Q5hlC_VPKGM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div id=\"Example_02_01_10\" class=\"example\">\n<div id=\"fs-id1165137836747\" class=\"exercise\">\n<h2>Writing a Linear Function Given Two Data Points<\/h2>\n<p>The next example demonstrating writing 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, [latex]I[\/latex], depends on the number of new policies, [latex]n[\/latex], 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].<\/p>\n<p>1) Find an equation for [latex]I(n)[\/latex] and interpret the meaning of the components of the equation.<\/p>\n<p>2) What is Ilya&#8217;s weekly income if he sells [latex]12[\/latex] policies?<\/p>\n<p>3) How many policies does he need to sell to make a weekly income of $2040?<\/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\">1) The given information gives us two input-output pairs (number of policies sold, amount of money earned:\u00a0[latex](3, 760)[\/latex] and [latex](5, 920)[\/latex]. We start by finding the rate of change.<\/p>\n<p>Recall that the slope [latex]m[\/latex], or rate of change of a linear function between two points [latex]\\left({x}_{1,\\text{ }}{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},\\text{ }{y}_{2}\\right)[\/latex] is: [latex]\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex].<\/p>\n<div id=\"fs-id1165135195046\" class=\"equation unnumbered\">[latex]\\begin{align}m&=\\dfrac{920 - 760}{5 - 3}\\\\&=\\dfrac{$160}{\\text{2 policies}}\\\\&=$80\\text{ per policy}\\end{align}[\/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\">[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 [latex]b[\/latex] is 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\">[latex]I\\left(n\\right)=80n+520[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137655487\">We can interpret the equation 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>2) To find Ilya&#8217;s weekly income if he sells [latex]12[\/latex] policies, let&#8217;s use the function [latex]I(n)=80n+520[\/latex]. Start by replacing [latex]n[\/latex] with [latex]12[\/latex] in the function, then solve for [latex]I(n)[\/latex].<\/p>\n<p style=\"text-align: center;\">$$\\begin{align}<br \/>\nI(\\color{Green}{n}) &amp;= 80\\color{Green}{n} + 520 \\\\<br \/>\nI(\\color{Green}{12}) &amp;= 80(\\color{Green}{12}) + 520 \\\\<br \/>\n&amp;= 960 + 520 \\\\<br \/>\n&amp;= 1480<br \/>\n\\end{align}$$<\/p>\n<p>Ilya will make [latex]$1480[\/latex] if he sells [latex]12[\/latex] policies.<\/p>\n<p>3) If Ilya wants to make [latex]$2040[\/latex] a week, we need to find how many policies he would need to sell. Replace [latex]I(n)[\/latex] with [latex]2040[\/latex] and then solve for [latex]n[\/latex].<\/p>\n<p style=\"text-align: center;\">$$\\begin{align}<br \/>\n\\color{Green}{I(n)} &amp;= 80n + 520 \\\\<br \/>\n\\color{Green}{2040} &amp;= 80n + 520 \\\\<br \/>\n2040-520 &amp;= 80n \\\\<br \/>\n1520 &amp;= 80n \\\\ 19 &amp;= n<br \/>\n\\end{align}$$<\/p>\n<p>Ilya will need to sell [latex]19[\/latex] policies if he wants to make [latex]$2040[\/latex].<\/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-2\" 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<h2>Modeling Monthly Cost<\/h2>\n<p>This example of modeling real-world scenarios using linear functions involves the monthly cost to run a company given monthly fixed costs and production costs per item manufactured.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose Sofia starts a company in which she incurs a fixed cost of\u00a0[latex]$1,250[\/latex] per month for the overhead which includes office rent. Her production costs are\u00a0[latex]$37.50[\/latex] per item. Write a linear function [latex]C[\/latex] where [latex]C(x)[\/latex] is the cost for [latex]x[\/latex] items produced in a given month. What will be the monthly cost if Sofia produces [latex]100[\/latex] items in a 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 the same each month; [latex]$1,250[\/latex]. The costs that can vary include the cost to produce each item, which is\u00a0[latex]$37.50[\/latex] for Sofia&#8217;s company. The variable cost (called the marginal cost) is represented by\u00a0[latex]37.5[\/latex]. The cost Sofia incurs is the sum of these two costs: [latex]C\\left(x\\right)=1250+37.5x[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165135511326\">If Sofia produces\u00a0[latex]100[\/latex] items in a month, her monthly cost is [latex]C(100)=1250+37.5(100)=5000[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>A final 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-3\" 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<\/div>\n<\/div>\n","protected":false},"author":773621,"menu_order":4,"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-538","chapter","type-chapter","status-publish","hentry"],"part":24,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/538","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/users\/773621"}],"version-history":[{"count":41,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/538\/revisions"}],"predecessor-version":[{"id":2106,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/538\/revisions\/2106"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/parts\/24"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/538\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/media?parent=538"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=538"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/contributor?post=538"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/license?post=538"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}