{"id":1017,"date":"2025-07-21T19:13:47","date_gmt":"2025-07-21T19:13:47","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1017"},"modified":"2026-01-14T19:24:27","modified_gmt":"2026-01-14T19:24:27","slug":"applications-of-rational-functions-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/applications-of-rational-functions-learn-it-2\/","title":{"raw":"Applications of Rational Functions: Learn It 2","rendered":"Applications of Rational Functions: Learn It 2"},"content":{"raw":"<h2>Solve Applied Problems Involving Rational Functions<\/h2>\r\nA <strong>rational function<\/strong> is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Horizontal Asymptote<\/strong>\r\n<ul>\r\n \t<li>A horizontal line of the form [latex]y=c[\/latex]<\/li>\r\n \t<li>The constant [latex]c[\/latex] represents a\u00a0number that the function value (output) approaches in the long run, either as the input grows very small or very large.<\/li>\r\n \t<li>Horizontal asymptotes represent the long-run behavior (the end behavior) of the graph of the funtion.<\/li>\r\n \t<li>A function's graph may cross a horizontal asymptote briefly, even more than once, but will eventually settle down near it, as the value of the function approaches the constant [latex]c[latex].<\/li>\r\n<\/ul>\r\n<strong>Vertical Asymptote<\/strong>\r\n<ul>\r\n \t<li>A vertical line of the form [latex]x=a[\/latex]<\/li>\r\n \t<li>The constant [latex]a[\/latex] represents an input for which the function value (output) is undefined.<\/li>\r\n \t<li>Substituting the value of [latex]a[\/latex] into the function will result in a zero in the function's denominator.<\/li>\r\n \t<li>The graph of the function \"bends around\", either increasing or decreasing without bound as the input nears [latex]a[\/latex]<\/li>\r\n \t<li>A function's graph will never cross a vertical asymptote.<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">A large mixing tank currently contains [latex]100[\/latex] gallons of water into which [latex]5[\/latex] pounds of sugar have been mixed. A tap will open pouring [latex]10[\/latex] gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of [latex]1[\/latex] pound per minute.Find the concentration (pounds per gallon) of sugar in the tank after [latex]12[\/latex] minutes. Is that a greater concentration than at the beginning?[reveal-answer q=\"527196\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"527196\"]Let [latex]t[\/latex] be the number of minutes since the tap opened. Since the water increases at [latex]10[\/latex] gallons per minute, and the sugar increases at [latex]1[\/latex] pound per minute, these are constant rates of change. This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each:\r\n<p style=\"text-align: center;\">[latex]\\text{water: }W\\left(t\\right)=100+10t\\text{ in gallons}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{sugar: }S\\left(t\\right)=5+1t\\text{ in pounds}[\/latex]<\/p>\r\nThe concentration, [latex]C[\/latex], will be the ratio of pounds of sugar to gallons of water\r\n<p style=\"text-align: center;\">[latex]C\\left(t\\right)=\\dfrac{5+t}{100+10t}[\/latex]<\/p>\r\nThe concentration after [latex]12[\/latex] minutes is given by evaluating [latex]C\\left(t\\right)[\/latex] at [latex]t=12[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}C\\left(12\\right)&amp;=\\dfrac{5+12}{100+10\\left(12\\right)}\\\\&amp;=\\dfrac{17}{220}\\end{align}[\/latex]<\/p>\r\nThis means the concentration is [latex]17[\/latex] pounds of sugar to [latex]220[\/latex] gallons of water.\r\n\r\nAt the beginning the concentration is\r\n<p style=\"text-align: center;\">[latex]\\begin{align}C\\left(0\\right)&amp;=\\dfrac{5+0}{100+10\\left(0\\right)} \\\\ &amp;=\\dfrac{1}{20}\\hfill \\end{align}[\/latex]<\/p>\r\nSince [latex]\\frac{17}{220}\\approx 0.08&gt;\\frac{1}{20}=0.05[\/latex], the concentration is greater after [latex]12[\/latex] minutes than at the beginning.\r\n\r\n<strong>Analysis of the Solution<\/strong>\r\n\r\nTo find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the denominator:\r\n<p style=\"text-align: center;\">[latex]\\dfrac{1}{10}=0.1[\/latex]<\/p>\r\nNotice the horizontal asymptote is [latex]y=0.1[\/latex]. This means the concentration, [latex]C[\/latex], the ratio of pounds of sugar to gallons of water, will approach [latex]0.1[\/latex] in the long term.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24666[\/ohm2_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]318955[\/ohm_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]318956[\/ohm_question]<\/section>","rendered":"<h2>Solve Applied Problems Involving Rational Functions<\/h2>\n<p>A <strong>rational function<\/strong> is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Horizontal Asymptote<\/strong><\/p>\n<ul>\n<li>A horizontal line of the form [latex]y=c[\/latex]<\/li>\n<li>The constant [latex]c[\/latex] represents a\u00a0number that the function value (output) approaches in the long run, either as the input grows very small or very large.<\/li>\n<li>Horizontal asymptotes represent the long-run behavior (the end behavior) of the graph of the funtion.<\/li>\n<li>A function&#8217;s graph may cross a horizontal asymptote briefly, even more than once, but will eventually settle down near it, as the value of the function approaches the constant [latex]c[latex].<\/li>\n<\/ul>\n<p>  <strong>Vertical Asymptote<\/strong>  <\/p>\n<ul>\n<li>A vertical line of the form [latex]x=a[\/latex]<\/li>\n<li>The constant [latex]a[\/latex] represents an input for which the function value (output) is undefined.<\/li>\n<li>Substituting the value of [latex]a[\/latex] into the function will result in a zero in the function's denominator.<\/li>\n<li>The graph of the function \"bends around\", either increasing or decreasing without bound as the input nears [latex]a[\/latex]<\/li>\n<li>A function's graph will never cross a vertical asymptote.<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">A large mixing tank currently contains [latex]100[\/latex] gallons of water into which [latex]5[\/latex] pounds of sugar have been mixed. A tap will open pouring [latex]10[\/latex] gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of [latex]1[\/latex] pound per minute.Find the concentration (pounds per gallon) of sugar in the tank after [latex]12[\/latex] minutes. Is that a greater concentration than at the beginning?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q527196\">Show Solution<\/button><\/p>\n<div id=\"q527196\" class=\"hidden-answer\" style=\"display: none\">Let [latex]t[\/latex] be the number of minutes since the tap opened. Since the water increases at [latex]10[\/latex] gallons per minute, and the sugar increases at [latex]1[\/latex] pound per minute, these are constant rates of change. This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{water: }W\\left(t\\right)=100+10t\\text{ in gallons}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\text{sugar: }S\\left(t\\right)=5+1t\\text{ in pounds}[\/latex]<\/p>\n<p>The concentration, [latex]C[\/latex], will be the ratio of pounds of sugar to gallons of water<\/p>\n<p style=\"text-align: center;\">[latex]C\\left(t\\right)=\\dfrac{5+t}{100+10t}[\/latex]<\/p>\n<p>The concentration after [latex]12[\/latex] minutes is given by evaluating [latex]C\\left(t\\right)[\/latex] at [latex]t=12[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}C\\left(12\\right)&=\\dfrac{5+12}{100+10\\left(12\\right)}\\\\&=\\dfrac{17}{220}\\end{align}[\/latex]<\/p>\n<p>This means the concentration is [latex]17[\/latex] pounds of sugar to [latex]220[\/latex] gallons of water.<\/p>\n<p>At the beginning the concentration is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}C\\left(0\\right)&=\\dfrac{5+0}{100+10\\left(0\\right)} \\\\ &=\\dfrac{1}{20}\\hfill \\end{align}[\/latex]<\/p>\n<p>Since [latex]\\frac{17}{220}\\approx 0.08>\\frac{1}{20}=0.05[\/latex], the concentration is greater after [latex]12[\/latex] minutes than at the beginning.<\/p>\n<p><strong>Analysis of the Solution<\/strong><\/p>\n<p>To find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the denominator:<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{1}{10}=0.1[\/latex]<\/p>\n<p>Notice the horizontal asymptote is [latex]y=0.1[\/latex]. This means the concentration, [latex]C[\/latex], the ratio of pounds of sugar to gallons of water, will approach [latex]0.1[\/latex] in the long term.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24666\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24666&theme=lumen&iframe_resize_id=ohm24666&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm318955\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318955&theme=lumen&iframe_resize_id=ohm318955&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm318956\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318956&theme=lumen&iframe_resize_id=ohm318956&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":13,"menu_order":16,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":508,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1017"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1017\/revisions"}],"predecessor-version":[{"id":5359,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1017\/revisions\/5359"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/508"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1017\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=1017"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1017"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=1017"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=1017"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}