{"id":4995,"date":"2020-11-13T21:10:11","date_gmt":"2020-11-13T21:10:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebra\/?post_type=chapter&#038;p=4995"},"modified":"2020-11-20T18:17:30","modified_gmt":"2020-11-20T18:17:30","slug":"rational-functions-custom-edit","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/chapter\/rational-functions-custom-edit\/","title":{"raw":"Rational Functions","rendered":"Rational Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul class=\"ul1\">\r\n \t<li class=\"li2\"><span class=\"s1\">Use arrow notation to describe end behavior of rational functions.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Solve applied problems involving rational functions.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Find the domains of rational functions.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Identify vertical and horizontal asymptotes of graphs of rational functions.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Graph rational functions.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\nSuppose we know that the cost of making a product is dependent on the number of items, [latex]x[\/latex], produced. This is given by the equation [latex]C\\left(x\\right)=15,000x - 0.1{x}^{2}+1000[\/latex]. If we want to know the average cost for producing [latex]x[\/latex]\u00a0items, we would divide the cost function by the number of items, [latex]x[\/latex].\r\n\r\nThe average cost function, which yields the average cost per item for [latex]x[\/latex]\u00a0items produced, is\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{15,000x - 0.1{x}^{2}+1000}{x}[\/latex]<\/p>\r\nMany other application problems require finding an average value in a similar way, giving us variables in the denominator. Written without a variable in the denominator, this function will contain a negative integer power.\r\n\r\nIn the last few sections, we have worked with polynomial functions, which are functions with non-negative integers for exponents. In this section, we explore rational functions, which have variables in the denominator.\r\n<h2>Characteristics of Rational Functions<\/h2>\r\nWe have seen the graphs of the basic <strong>reciprocal function<\/strong> and the squared reciprocal function from our study of toolkit functions. Examine these graphs\u00a0and notice some of their features.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213907\/CNX_Precalc_Figure_03_07_0012.jpg\" alt=\"Graphs of f(x)=1\/x and f(x)=1\/x^2\" width=\"731\" height=\"453\" \/>\r\n\r\nSeveral things are apparent if we examine the graph of [latex]f\\left(x\\right)=\\dfrac{1}{x}[\/latex].\r\n<ol>\r\n \t<li>On the left branch of the graph, the curve approaches the [latex]x[\/latex]-axis [latex]\\left(y=0\\right) \\text{ as } x\\to -\\infty [\/latex].<\/li>\r\n \t<li>As the graph approaches [latex]x=0[\/latex] from the left, the curve drops, but as we approach zero from the right, the curve rises.<\/li>\r\n \t<li>Finally, on the right branch of the graph, the curves approaches the [latex]x[\/latex]<em>-<\/em>axis [latex]\\left(y=0\\right) \\text{ as } x\\to \\infty [\/latex].<\/li>\r\n<\/ol>\r\nTo summarize, we use <strong>arrow notation<\/strong> to show that [latex]x[\/latex]\u00a0or [latex]f\\left(x\\right)[\/latex] is approaching a particular value.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center;\" colspan=\"2\">Arrow Notation<\/th>\r\n<\/tr>\r\n<tr>\r\n<th style=\"text-align: center;\">Symbol<\/th>\r\n<th style=\"text-align: center;\">Meaning<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]x\\to {a}^{-}[\/latex]<\/td>\r\n<td>[latex]x[\/latex] approaches [latex]a[\/latex]\u00a0from the left ([latex]x&lt;a[\/latex] but close to [latex]a[\/latex])<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x\\to {a}^{+}[\/latex]<\/td>\r\n<td>[latex]x[\/latex] approaches [latex]a[\/latex]\u00a0from the right ([latex]x&gt;a[\/latex]\u00a0but close to [latex]a[\/latex])<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x\\to \\infty[\/latex]<\/td>\r\n<td>[latex]x[\/latex] approaches infinity ([latex]x[\/latex]\u00a0increases without bound)<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x\\to -\\infty [\/latex]<\/td>\r\n<td>[latex]x[\/latex] approaches negative infinity ([latex]x[\/latex]\u00a0decreases without bound)<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]f\\left(x\\right)\\to \\infty [\/latex]<\/td>\r\n<td>the output approaches infinity (the output increases without bound)<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]f\\left(x\\right)\\to -\\infty [\/latex]<\/td>\r\n<td>the output approaches negative infinity (the output decreases without bound)<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]f\\left(x\\right)\\to a[\/latex]<\/td>\r\n<td>the output approaches [latex]a[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Local Behavior of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/h2>\r\nLet\u2019s begin by looking at the reciprocal function, [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]. We cannot divide by zero, which means the function is undefined at [latex]x=0[\/latex]; so zero is not in the domain<em>.<\/em> As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in the table below.\r\n<table id=\"Table_03_07_002\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td>\u20130.1<\/td>\r\n<td>\u20130.01<\/td>\r\n<td>\u20130.001<\/td>\r\n<td>\u20130.0001<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] <\/strong><\/td>\r\n<td>\u201310<\/td>\r\n<td>\u2013100<\/td>\r\n<td>\u20131000<\/td>\r\n<td>\u201310,000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe write in arrow notation\r\n<p style=\"text-align: center;\">[latex]\\text{as }x\\to {0}^{-},f\\left(x\\right)\\to -\\infty [\/latex]<\/p>\r\nAs the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in the table below.\r\n<table id=\"Table_03_07_003\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td>0.1<\/td>\r\n<td>0.01<\/td>\r\n<td>0.001<\/td>\r\n<td>0.0001<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] <\/strong><\/td>\r\n<td>10<\/td>\r\n<td>100<\/td>\r\n<td>1000<\/td>\r\n<td>10,000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe write in arrow notation\r\n<p style=\"text-align: center;\">[latex]\\text{As }x\\to {0}^{+}, f\\left(x\\right)\\to \\infty [\/latex].<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213909\/CNX_Precalc_Figure_03_07_0022.jpg\" alt=\"Graph of f(x)=1\/x which denotes the end behavior. As x goes to negative infinity, f(x) goes to 0, and as x goes to 0^-, f(x) goes to negative infinity. As x goes to positive infinity, f(x) goes to 0, and as x goes to 0^+, f(x) goes to positive infinity.\" width=\"731\" height=\"474\" \/>\r\n\r\nThis behavior creates a <strong>vertical asymptote<\/strong>, which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line [latex]x=0[\/latex]\u00a0as the input becomes close to zero.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213912\/CNX_Precalc_Figure_03_07_0032.jpg\" alt=\"Graph of f(x)=1\/x with its vertical asymptote at x=0.\" width=\"487\" height=\"364\" \/>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Vertical Asymptote<\/h3>\r\nA <strong>vertical asymptote<\/strong> of a graph is a vertical line [latex]x=a[\/latex] where the graph tends toward positive or negative infinity as the inputs approach [latex]a[\/latex]. We write\r\n\r\n[latex]\\text{As }x\\to a,f\\left(x\\right)\\to \\infty , \\text{or as }x\\to a,f\\left(x\\right)\\to -\\infty [\/latex].\r\n\r\n<\/div>\r\n<h2>End Behavior of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/h2>\r\nAs the values of [latex]x[\/latex]\u00a0approach infinity, the function values approach 0. As the values of [latex]x[\/latex]\u00a0approach negative infinity, the function values approach 0. Symbolically, using arrow notation\r\n<p style=\"text-align: center;\">[latex]\\text{As }x\\to \\infty ,f\\left(x\\right)\\to 0,\\text{and as }x\\to -\\infty ,f\\left(x\\right)\\to 0[\/latex].<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213913\/CNX_Precalc_Figure_03_07_0042.jpg\" alt=\"Graph of f(x)=1\/x which highlights the segments of the turning points to denote their end behavior.\" width=\"731\" height=\"475\" \/>\r\n\r\nBased on this overall behavior and the graph, we can see that the function approaches 0 but never actually reaches 0; it seems to level off as the inputs become large. This behavior creates a <strong>horizontal asymptote<\/strong>, a horizontal line that the graph approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line [latex]y=0[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213915\/CNX_Precalc_Figure_03_07_0052.jpg\" alt=\"Graph of f(x)=1\/x with its vertical asymptote at x=0 and its horizontal asymptote at y=0.\" width=\"487\" height=\"364\" \/>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Horizontal Asymptote<\/h3>\r\nA <strong>horizontal asymptote<\/strong> of a graph is a horizontal line [latex]y=b[\/latex] where the graph approaches the line as the inputs increase or decrease without bound. We write\r\n\r\n[latex]\\text{As }x\\to \\infty \\text{ or }x\\to -\\infty ,\\text{ }f\\left(x\\right)\\to b[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Arrow Notation<\/h3>\r\nUse arrow notation to describe the end behavior and local behavior of the function below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213918\/CNX_Precalc_Figure_03_07_0062.jpg\" alt=\"Graph of f(x)=1\/(x-2)+4 with its vertical asymptote at x=2 and its horizontal asymptote at y=4.\" width=\"487\" height=\"477\" \/>\r\n\r\n[reveal-answer q=\"444547\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"444547\"]\r\n\r\nNotice that the graph is showing a vertical asymptote at [latex]x=2[\/latex], which tells us that the function is undefined at [latex]x=2[\/latex].\r\n<p style=\"text-align: center;\">As [latex]x\\to {2}^{-},\\hspace{2mm}f\\left(x\\right)\\to -\\infty[\/latex], and as [latex]x\\to {2}^{+},\\text{ }f\\left(x\\right)\\to \\infty [\/latex]<\/p>\r\nAnd as the inputs decrease without bound, the graph appears to be leveling off at output values of 4, indicating a horizontal asymptote at [latex]y=4[\/latex]. As the inputs increase without bound, the graph levels off at 4.\r\n<p style=\"text-align: center;\">As [latex]x\\to \\infty ,\\text{ }f\\left(x\\right)\\to 4[\/latex], and as [latex]x\\to -\\infty ,\\text{ }f\\left(x\\right)\\to 4[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse arrow notation to describe the end behavior and local behavior for the reciprocal squared function[latex]f\\left(x\\right)=\\dfrac{1}{x^2}[\/latex].\r\n\r\n[reveal-answer q=\"785291\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"785291\"]\r\n\r\nEnd behavior: as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to 0[\/latex]; Local behavior: as [latex]x\\to 0, f\\left(x\\right)\\to \\infty [\/latex] (there are no [latex]x[\/latex]- or [latex]y[\/latex]-intercepts)\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom25\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129042&amp;theme=oea&amp;iframe_resize_id=mom25\" width=\"100%\" height=\"550\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Transformations to Graph a Rational Function<\/h3>\r\nSketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any.\r\n\r\n[reveal-answer q=\"670271\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"670271\"]\r\n\r\nShifting the graph left 2 and up 3 would result in the function\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{1}{x+2}+3[\/latex]<\/p>\r\nor equivalently, by giving the terms a common denominator,\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{3x+7}{x+2}[\/latex]<\/p>\r\nThe graph of the shifted function is displayed\u00a0below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213922\/CNX_Precalc_Figure_03_07_0072.jpg\" alt=\"Graph of f(x)=1\/(x+2)+3 with its vertical asymptote at x=-2 and its horizontal asymptote at y=3.\" width=\"731\" height=\"441\" \/>\r\n\r\nNotice that this function is undefined at [latex]x=-2[\/latex], and the graph also is showing a vertical asymptote at [latex]x=-2[\/latex].\r\n<p style=\"text-align: center;\">As [latex]x\\to -{2}^{-}, f\\left(x\\right)\\to -\\infty[\/latex] , and as\u00a0 [latex]x\\to -{2}^{+}, f\\left(x\\right)\\to \\infty [\/latex]<\/p>\r\nAs the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at [latex]y=3[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\text{As }x\\to \\pm \\infty , f\\left(x\\right)\\to 3[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nNotice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units.\r\n\r\n[reveal-answer q=\"600482\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"600482\"]\r\n\r\nThe function and the asymptotes are shifted 3 units right and 4 units down. As [latex]x\\to 3,f\\left(x\\right)\\to \\infty[\/latex], and as [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to -4[\/latex].\r\n<p id=\"fs-id1165137823960\">The function is [latex]f\\left(x\\right)=\\dfrac{1}{{\\left(x - 3\\right)}^{2}}-4[\/latex].<\/p>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/12\/02000148\/CNX_Precalc_Figure_03_07_0082.jpg\"><img class=\"aligncenter size-full wp-image-2938\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/12\/02000148\/CNX_Precalc_Figure_03_07_0082.jpg\" alt=\"Graph of f(x)=1\/(x-3)^2-4 with its vertical asymptote at x=3 and its horizontal asymptote at y=-4.\" width=\"487\" height=\"365\" \/><\/a>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>A Mixing Problem<\/h2>\r\nIn the previous example, we shifted a toolkit function in a way that resulted in the function [latex]f\\left(x\\right)=\\dfrac{3x+7}{x+2}[\/latex]. This is an example of a rational function. 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.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Rational Function<\/h3>\r\nA <strong>rational function<\/strong> is a function that can be written as the quotient of two polynomial functions [latex]P\\left(x\\right) \\text{and} Q\\left(x\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{a}_{p}{x}^{p}+{a}_{p - 1}{x}^{p - 1}+...+{a}_{1}x+{a}_{0}}{{b}_{q}{x}^{q}+{b}_{q - 1}{x}^{q - 1}+...+{b}_{1}x+{b}_{0}},Q\\left(x\\right)\\ne 0[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Applied Problem Involving a Rational Function<\/h3>\r\nA large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning?\r\n\r\n[reveal-answer q=\"527196\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"527196\"]\r\n\r\nLet [latex]t[\/latex] be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 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 12 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 17 pounds of sugar to 220 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 12 minutes than at the beginning.\r\n<h4>Analysis of the Solution<\/h4>\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 0.1 in the long term.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThere are 1,200 freshmen and 1,500 sophomores at a prep rally at noon. After 12 p.m., 20 freshmen arrive at the rally every five minutes while 15 sophomores leave the rally. Find the ratio of freshmen to sophomores at 1 p.m.\r\n\r\n[reveal-answer q=\"526334\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"526334\"]\r\n\r\n[latex]\\dfrac{12}{11}[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129067&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"550\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Domain and Its Effect on Vertical Asymptotes<\/h2>\r\nA <strong>vertical asymptote<\/strong> represents a value at which a rational function is undefined, so that value is not in the domain of the function. A reciprocal function cannot have values in its domain that cause the denominator to equal zero. In general, to find the domain of a rational function, we need to determine which inputs would cause division by zero.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Domain of a Rational Function<\/h3>\r\nThe domain of a rational function includes all real numbers except those that cause the denominator to equal zero.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a rational function, find the domain.<\/h3>\r\n<ol>\r\n \t<li>Set the denominator equal to zero.<\/li>\r\n \t<li>Solve to find the [latex]x[\/latex]-values that cause the denominator to equal zero.<\/li>\r\n \t<li>The domain is all real numbers except those found in Step 2.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Domain of a Rational Function<\/h3>\r\nFind the domain of [latex]f\\left(x\\right)=\\dfrac{x+3}{{x}^{2}-9}[\/latex].\r\n\r\n[reveal-answer q=\"860212\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"860212\"]\r\n\r\nBegin by setting the denominator equal to zero and solving.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} {x}^{2}-9&amp;=0 \\\\ {x}^{2}&amp;=9 \\\\ x&amp;=\\pm 3 \\end{align}[\/latex]<\/p>\r\nThe denominator is equal to zero when [latex]x=\\pm 3[\/latex]. The domain of the function is all real numbers except [latex]x=\\pm 3[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nA graph of this function confirms that the function is not defined when [latex]x=\\pm 3[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213925\/CNX_Precalc_Figure_03_07_0092.jpg\" alt=\"Graph of f(x)=1\/(x-3) with its vertical asymptote at x=3 and its horizontal asymptote at y=0.\" width=\"487\" height=\"364\" \/>\r\n\r\nThere is a vertical asymptote at [latex]x=3[\/latex] and a hole in the graph at [latex]x=-3[\/latex]. We will discuss these types of holes in greater detail later in this section.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the domain of [latex]f\\left(x\\right)=\\dfrac{4x}{5\\left(x - 1\\right)\\left(x - 5\\right)}[\/latex].\r\n\r\n[reveal-answer q=\"553731\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"553731\"]\r\n\r\nThe domain is all real numbers except [latex]x=1[\/latex] and [latex]x=5[\/latex].\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129068&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"400\">\r\n<\/iframe>\r\n\r\n<\/div>\r\nBy looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location.\r\n\r\nWatch the following video to see more examples of finding the domain of a rational function.\r\n\r\nhttps:\/\/youtu.be\/v0IhvIzCc_I\r\n<h2>Vertical Asymptotes<\/h2>\r\nThe vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a rational function, identify any vertical asymptotes of its graph.<\/h3>\r\n<ol>\r\n \t<li>Factor the numerator and denominator.<\/li>\r\n \t<li>Note any restrictions in the domain of the function.<\/li>\r\n \t<li>Reduce the expression by canceling common factors in the numerator and the denominator.<\/li>\r\n \t<li>Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.<\/li>\r\n \t<li>Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Identifying Vertical Asymptotes<\/h3>\r\nFind the vertical asymptotes of the graph of [latex]k\\left(x\\right)=\\dfrac{5+2{x}^{2}}{2-x-{x}^{2}}[\/latex].\r\n\r\n[reveal-answer q=\"787718\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"787718\"]\r\n\r\nFirst, factor the numerator and denominator.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}k\\left(x\\right)&amp;=\\dfrac{5+2{x}^{2}}{2-x-{x}^{2}} \\\\[1mm] &amp;=\\dfrac{5+2{x}^{2}}{\\left(2+x\\right)\\left(1-x\\right)} \\end{align}[\/latex]<\/p>\r\nTo find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:\r\n<p style=\"text-align: center;\">[latex]\\left(2+x\\right)\\left(1-x\\right)=0[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x=-2,1[\/latex]<\/p>\r\nNeither [latex]x=-2[\/latex] nor [latex]x=1[\/latex] are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph below confirms the location of the two vertical asymptotes.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213927\/CNX_Precalc_Figure_03_07_0102.jpg\" alt=\"Graph of k(x)=(5+2x)^2\/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.\" width=\"487\" height=\"514\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n<iframe id=\"mom110\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=24104&amp;theme=oea&amp;iframe_resize_id=mom110\" width=\"100%\" height=\"650\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Removable Discontinuities<\/h2>\r\nOccasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a <strong>removable discontinuity<\/strong>.\r\n\r\nFor example the function [latex]f\\left(x\\right)=\\dfrac{{x}^{2}-1}{{x}^{2}-2x - 3}[\/latex] may be re-written by factoring the numerator and the denominator.\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{\\left(x+1\\right)\\left(x - 1\\right)}{\\left(x+1\\right)\\left(x - 3\\right)}[\/latex]<\/p>\r\nNotice that [latex]x+1[\/latex] is a common factor to the numerator and the denominator. The zero of this factor, [latex]x=-1[\/latex], is the location of the removable discontinuity. Notice also that [latex]x - 3[\/latex] is not a factor in both the numerator and denominator. The zero of this factor, [latex]x=3[\/latex], is the vertical asymptote.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213929\/CNX_Precalc_Figure_03_07_0112.jpg\" alt=\"Graph of f(x)=(x^2-1)\/(x^2-2x-3) with its vertical asymptote at x=3 and a removable discontinuity at x=-1.\" width=\"487\" height=\"326\" \/>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Removable Discontinuities of Rational Functions<\/h3>\r\nA <strong>removable discontinuity<\/strong> occurs in the graph of a rational function at [latex]x=a[\/latex] if <em>a<\/em>\u00a0is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Identifying Vertical Asymptotes and Removable Discontinuities for a Graph<\/h3>\r\nFind the vertical asymptotes and removable discontinuities of the graph of [latex]k\\left(x\\right)=\\dfrac{x - 2}{{x}^{2}-4}[\/latex].\r\n\r\n[reveal-answer q=\"519186\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"519186\"]\r\n\r\nFactor the numerator and the denominator.\r\n<p style=\"text-align: center;\">[latex]k\\left(x\\right)=\\dfrac{x - 2}{\\left(x - 2\\right)\\left(x+2\\right)}[\/latex]<\/p>\r\nNotice that there is a common factor in the numerator and the denominator, [latex]x - 2[\/latex]. The zero for this factor is [latex]x=2[\/latex]. This is the location of the removable discontinuity.\r\n\r\nNotice that there is a factor in the denominator that is not in the numerator, [latex]x+2[\/latex]. The zero for this factor is [latex]x=-2[\/latex]. The vertical asymptote is [latex]x=-2[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213931\/CNX_Precalc_Figure_03_07_0122.jpg\" alt=\"Graph of k(x)=(x-2)\/(x-2)(x+2) with its vertical asymptote at x=-2 and a removable discontinuity at x=2.\" width=\"487\" height=\"364\" \/>\r\n\r\nThe graph of this function will have the vertical asymptote at [latex]x=-2[\/latex], but at [latex]x=2[\/latex] the graph will have a hole.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the vertical asymptotes and removable discontinuities of the graph of [latex]f\\left(x\\right)=\\dfrac{{x}^{2}-25}{{x}^{3}-6{x}^{2}+5x}[\/latex].\r\n\r\n[reveal-answer q=\"575454\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"575454\"]\r\n\r\nRemovable discontinuity at [latex]x=5[\/latex]. Vertical asymptotes: [latex]x=0,\\text{ }x=1[\/latex].\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom111\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=74565&amp;theme=oea&amp;iframe_resize_id=mom111\" width=\"100%\" height=\"400\">\r\n<\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom111\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=74565&amp;theme=oea&amp;iframe_resize_id=mom111\" width=\"100%\" height=\"400\">\r\n<\/iframe>\r\n\r\n<\/div>\r\n<h2>Horizontal Asymptotes and Intercepts<\/h2>\r\nWhile vertical asymptotes describe the behavior of a graph as the <em>output<\/em> gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the <em>input<\/em> gets very large or very small. Recall that a polynomial\u2019s end behavior will mirror that of the leading term. Likewise, a rational function\u2019s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions.\r\n\r\nThere are three distinct outcomes when checking for horizontal asymptotes:\r\n\r\n<strong>Case 1:<\/strong> If the degree of the denominator &gt; degree of the numerator, there is a <strong>horizontal asymptote<\/strong> at [latex]y=0[\/latex].\r\n<p style=\"text-align: center;\">Example: [latex]f\\left(x\\right)=\\dfrac{4x+2}{{x}^{2}+4x - 5}[\/latex]<\/p>\r\nIn this case the end behavior is [latex]f\\left(x\\right)\\approx \\frac{4x}{{x}^{2}}=\\frac{4}{x}[\/latex]. This tells us that, as the inputs increase or decrease without bound, this function will behave similarly to the function [latex]g\\left(x\\right)=\\frac{4}{x}[\/latex], and the outputs will approach zero, resulting in a horizontal asymptote at [latex]y=0[\/latex]. Note that this graph crosses the horizontal asymptote.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"900\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213933\/CNX_Precalc_Figure_03_07_0132.jpg\" alt=\"Graph of f(x)=(4x+2)\/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=0.\" width=\"900\" height=\"302\" \/> Horizontal Asymptote [latex]y=0[\/latex] when [latex]f\\left(x\\right)=\\dfrac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne{0}\\text{ where degree of }p&lt;\\text{degree of q}[\/latex].[\/caption]<strong>Case 2:<\/strong> If the degree of the denominator &lt; degree of the numerator by one, we get a slant asymptote.\r\n<p style=\"text-align: center;\">Example: [latex]f\\left(x\\right)=\\dfrac{3{x}^{2}-2x+1}{x - 1}[\/latex]<\/p>\r\nIn this case the end behavior is [latex]f\\left(x\\right)\\approx \\frac{3{x}^{2}}{x}=3x[\/latex]. This tells us that as the inputs increase or decrease without bound, this function will behave similarly to the function [latex]g\\left(x\\right)=3x[\/latex]. As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of [latex]g\\left(x\\right)=3x[\/latex] looks like a diagonal line, and since [latex]f[\/latex]\u00a0will behave similarly to [latex]g[\/latex], it will approach a line close to [latex]y=3x[\/latex]. This line is a slant asymptote.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213937\/CNX_Precalc_Figure_03_07_0142.jpg\" alt=\"Graph of f(x)=(3x^2-2x+1)\/(x-1) with its vertical asymptote at x=1 and a slant asymptote aty=3x+1.\" width=\"487\" height=\"440\" \/> Slant Asymptote when [latex]f\\left(x\\right)=\\dfrac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne 0[\/latex] where degree of [latex]p&gt;\\text{ degree of }q\\text{ by }1[\/latex].[\/caption]To find the equation of the slant asymptote, divide [latex]\\dfrac{3{x}^{2}-2x+1}{x - 1}[\/latex]. The quotient is [latex]3x+1[\/latex], and the remainder is 2. The slant asymptote is the graph of the line [latex]g\\left(x\\right)=3x+1[\/latex].\r\n\r\n<strong>Case 3:<\/strong> If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at [latex]y=\\frac{{a}_{n}}{{b}_{n}}[\/latex], where [latex]{a}_{n}[\/latex] and [latex]{b}_{n}[\/latex] are the leading coefficients of [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex] for [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne 0[\/latex].\r\n<p style=\"text-align: center;\">Example: [latex]f\\left(x\\right)=\\dfrac{3{x}^{2}+2}{{x}^{2}+4x - 5}[\/latex]<\/p>\r\nIn this case the end behavior is [latex]f\\left(x\\right)\\approx \\frac{3{x}^{2}}{{x}^{2}}=3[\/latex]. This tells us that as the inputs grow large, this function will behave like the function [latex]g\\left(x\\right)=3[\/latex], which is a horizontal line. As [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to 3[\/latex], resulting in a horizontal asymptote at [latex]y=3[\/latex]. Note that this graph crosses the horizontal asymptote.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213939\/CNX_Precalc_Figure_03_07_0152.jpg\" alt=\"Graph of f(x)=(3x^2+2)\/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=3.\" width=\"731\" height=\"364\" \/> Horizontal Asymptote when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne 0\\text{ where degree of }p=\\text{degree of }q[\/latex].[\/caption]Notice that, while the graph of a rational function will never cross a <strong>vertical asymptote<\/strong>, the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote.\r\n\r\nIt should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the <strong>end behavior<\/strong> of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{3{x}^{5}-{x}^{2}}{x+3}[\/latex]<\/p>\r\nwith end behavior\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)\\approx \\dfrac{3{x}^{5}}{x}=3{x}^{4}[\/latex],<\/p>\r\nthe end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.\r\n<p style=\"text-align: center;\">As [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to \\infty [\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Horizontal Asymptotes of Rational Functions<\/h3>\r\nThe <strong>horizontal asymptote<\/strong> of a rational function can be determined by looking at the degrees of the numerator and denominator.\r\n<ul id=\"fs-id1165137722720\">\r\n \t<li>Degree of numerator <em>is less than<\/em> degree of denominator: horizontal asymptote at [latex]y=0[\/latex].<\/li>\r\n \t<li>Degree of numerator <em>is greater than degree of denominator by one<\/em>: no horizontal asymptote; slant asymptote.<\/li>\r\n \t<li>Degree of numerator <em>is equal to<\/em> degree of denominator: horizontal asymptote at ratio of leading coefficients.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Identifying Horizontal and Slant Asymptotes<\/h3>\r\nFor the functions below, identify the horizontal or slant asymptote.\r\n<ol>\r\n \t<li>[latex]g\\left(x\\right)=\\dfrac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}[\/latex]<\/li>\r\n \t<li>[latex]h\\left(x\\right)=\\dfrac{{x}^{2}-4x+1}{x+2}[\/latex]<\/li>\r\n \t<li>[latex]k\\left(x\\right)=\\dfrac{{x}^{2}+4x}{{x}^{3}-8}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"968793\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"968793\"]\r\n\r\nFor these solutions, we will use [latex]f\\left(x\\right)=\\dfrac{p\\left(x\\right)}{q\\left(x\\right)}, q\\left(x\\right)\\ne 0[\/latex].\r\n<ol>\r\n \t<li>[latex]g\\left(x\\right)=\\dfrac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}[\/latex]: The degree of [latex]p[\/latex] and the degree of [latex]q[\/latex] are both equal to 3, so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at [latex]y=\\frac{6}{2}[\/latex] or [latex]y=3[\/latex].<\/li>\r\n \t<li>[latex]h\\left(x\\right)=\\dfrac{{x}^{2}-4x+1}{x+2}[\/latex]: The degree of [latex]p=2[\/latex] and degree of [latex]q=1[\/latex]. Since [latex]p&gt;q[\/latex] by 1, there is a slant asymptote found at [latex]\\dfrac{{x}^{2}-4x+1}{x+2}[\/latex].<img class=\"aligncenter wp-image-4515 size-medium\" style=\"text-align: center;\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1916\/2016\/11\/21031600\/CNX_CAT_Figure_05_01_011-300x71.jpg\" alt=\"Synthetic division of x^2-4x+1 by x+2, resulting in x-6 with a remainder of 13\" width=\"300\" height=\"71\" \/>The quotient is [latex]x - 6[\/latex] and the remainder is 13. There is a slant asymptote at [latex]y=x - 6[\/latex].<\/li>\r\n \t<li>[latex]k\\left(x\\right)=\\dfrac{{x}^{2}+4x}{{x}^{3}-8}[\/latex]: The degree of [latex]p=2\\text{ }&lt;[\/latex] degree of [latex]q=3[\/latex], so there is a horizontal asymptote [latex]y=0[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15836&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"400\">\r\n<\/iframe>\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=105058&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"400\">\r\n<\/iframe>\r\n\r\n<\/div>\r\nWatch this video to see more worked examples of determining which kind of horizontal asymptote a rational function will have.\r\n\r\nhttps:\/\/youtu.be\/A1tApZSE8nI\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Identifying Horizontal Asymptotes<\/h3>\r\nIn the sugar concentration problem earlier, we created the equation [latex]C\\left(t\\right)=\\dfrac{5+t}{100+10t}[\/latex].\r\n\r\nFind the horizontal asymptote and interpret it in context of the problem.\r\n\r\n[reveal-answer q=\"905213\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"905213\"]\r\n\r\nBoth the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is [latex]t[\/latex], with coefficient 1. In the denominator, the leading term is [latex]10t[\/latex], with coefficient 10. The horizontal asymptote will be at the ratio of these values:\r\n<p style=\"text-align: center;\">[latex]t\\to \\infty , C\\left(t\\right)\\to \\frac{1}{10}[\/latex]<\/p>\r\nThis function will have a horizontal asymptote at [latex]y=\\frac{1}{10}[\/latex].\r\n\r\nThis tells us that as the values of [latex]t[\/latex]\u00a0increase, the values of [latex]C[\/latex]\u00a0will approach [latex]\\frac{1}{10}[\/latex]. In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or [latex]\\frac{1}{10}[\/latex] pounds per gallon.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Identifying Horizontal and Vertical Asymptotes<\/h3>\r\nFind the horizontal and vertical asymptotes of the function\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{\\left(x - 2\\right)\\left(x+3\\right)}{\\left(x - 1\\right)\\left(x+2\\right)\\left(x - 5\\right)}[\/latex]<\/p>\r\n[reveal-answer q=\"228875\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"228875\"]\r\n\r\nFirst, note that this function has no common factors, so there are no potential removable discontinuities.\r\n\r\nThe function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at [latex]x=1,-2,\\text{and }5[\/latex], indicating vertical asymptotes at these values.\r\n\r\nThe numerator has degree 2, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to 0[\/latex]. This function will have a horizontal asymptote at [latex]y=0[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213942\/CNX_Precalc_Figure_03_07_0162.jpg\" alt=\"Graph of f(x)=(x-2)(x+3)\/(x-1)(x+2)(x-5) with its vertical asymptotes at x=-2, x=1, and x=5 and its horizontal asymptote at y=0.\" width=\"731\" height=\"514\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the vertical and horizontal asymptotes of the function:\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{\\left(2x - 1\\right)\\left(2x+1\\right)}{\\left(x - 2\\right)\\left(x+3\\right)}[\/latex]<\/p>\r\n[reveal-answer q=\"5590\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"5590\"]\r\n\r\nVertical asymptotes at [latex]x=2[\/latex] and [latex]x=-3[\/latex]; horizontal asymptote at [latex]y=4[\/latex].\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom50\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129072&amp;theme=oea&amp;iframe_resize_id=mom50\" width=\"100%\" height=\"400\">\r\n<\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Intercepts of Rational Functions<\/h3>\r\nA <strong>rational function<\/strong> will have a <em>y<\/em>-intercept when the input is zero, if the function is defined at zero. A rational function will not have a [latex]y[\/latex]-intercept if the function is not defined at zero.\r\n\r\nLikewise, a rational function will have [latex]x[\/latex]-intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, [latex]x[\/latex]-intercepts can only occur when the numerator of the rational function is equal to zero.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Intercepts of a Rational Function<\/h3>\r\nFind the intercepts of [latex]f\\left(x\\right)=\\dfrac{\\left(x - 2\\right)\\left(x+3\\right)}{\\left(x - 1\\right)\\left(x+2\\right)\\left(x - 5\\right)}[\/latex].\r\n\r\n[reveal-answer q=\"418596\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"418596\"]\r\n\r\nWe can find the [latex]y[\/latex]-intercept by evaluating the function at zero\r\n<p style=\"text-align: center;\">[latex]\\begin{align}f\\left(0\\right)&amp;=\\dfrac{\\left(0 - 2\\right)\\left(0+3\\right)}{\\left(0 - 1\\right)\\left(0+2\\right)\\left(0 - 5\\right)} \\\\[1mm] &amp;=\\frac{-6}{10} \\\\[1mm] &amp;=-\\frac{3}{5} \\\\[1mm] &amp;=-0.6 \\end{align}[\/latex]<\/p>\r\nThe [latex]x[\/latex]-intercepts will occur when the function is equal to zero. A rational function is equal to 0 when the numerator is 0, as long as the denominator is not also 0.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 0&amp;=\\frac{\\left(x - 2\\right)\\left(x+3\\right)}{\\left(x - 1\\right)\\left(x+2\\right)\\left(x - 5\\right)} \\\\[1mm] 0&amp;=\\left(x - 2\\right)\\left(x+3\\right)\\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x=2, -3[\/latex]<\/p>\r\nThe [latex]y[\/latex]-intercept is [latex]\\left(0,-0.6\\right)[\/latex], the [latex]x[\/latex]-intercepts are [latex]\\left(2,0\\right)[\/latex] and [latex]\\left(-3,0\\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213944\/CNX_Precalc_Figure_03_07_0172.jpg\" alt=\"Graph of f(x)=(x-2)(x+3)\/(x-1)(x+2)(x-5) with its vertical asymptotes at x=-2, x=1, and x=5, its horizontal asymptote at y=0, and its intercepts at (-3, 0), (0, -0.6), and (2, 0).\" width=\"731\" height=\"514\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGiven the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the [latex]x[\/latex]- and [latex]y[\/latex]-intercepts and the horizontal and vertical asymptotes.\r\n\r\n[reveal-answer q=\"937415\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"937415\"]\r\n\r\nFor the transformed reciprocal squared function, we find the rational form.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}f\\left(x\\right)&amp;=\\dfrac{1}{{\\left(x - 3\\right)}^{2}}-4 \\\\[1mm] &amp;=\\dfrac{1 - 4{\\left(x - 3\\right)}^{2}}{{\\left(x - 3\\right)}^{2}} \\\\[1mm] &amp;=\\dfrac{1 - 4\\left({x}^{2}-6x+9\\right)}{\\left(x - 3\\right)\\left(x - 3\\right)} \\\\[1mm] &amp;=\\frac{-4{x}^{2}+24x - 35}{{x}^{2}-6x+9}\\end{align}[\/latex]<\/p>\r\nBecause the numerator is the same degree as the denominator we know that as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to -4; \\text{so } y=-4[\/latex] is the horizontal asymptote. Next, we set the denominator equal to zero, and find that the vertical asymptote is [latex]x=3[\/latex], because as [latex]x\\to 3,f\\left(x\\right)\\to \\infty[\/latex]. We then set the numerator equal to 0 and find the [latex]x[\/latex]-intercepts are at [latex]\\left(2.5,0\\right)[\/latex] and [latex]\\left(3.5,0\\right)[\/latex]. Finally, we evaluate the function at 0 and find the [latex]y[\/latex]-intercept to be at [latex]\\left(0,\\frac{-35}{9}\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see more worked examples of finding asymptotes, intercepts and holes of rational functions.\r\n\r\nhttps:\/\/youtu.be\/UnVZs2EaEjI\r\n<h2>Graph Rational Functions<\/h2>\r\nPreviously we saw that the numerator of a rational function reveals the [latex]x[\/latex]-intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials.\r\n\r\nThe vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213946\/CNX_Precalc_Figure_03_07_0192.jpg\" alt=\"Graph of y=1\/x with its vertical asymptote at x=0.\" width=\"487\" height=\"364\" \/>\r\n\r\nWhen the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213948\/CNX_Precalc_Figure_03_07_0182.jpg\" alt=\"Graph of y=1\/x^2 with its vertical asymptote at x=0.\" width=\"487\" height=\"365\" \/>\r\n\r\nFor example the graph of [latex]f\\left(x\\right)=\\dfrac{{\\left(x+1\\right)}^{2}\\left(x - 3\\right)}{{\\left(x+3\\right)}^{2}\\left(x - 2\\right)}[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213950\/CNX_Precalc_Figure_03_07_0202.jpg\" alt=\"Graph of f(x)=(x+1)^2(x-3)\/(x+3)^2(x-2) with its vertical asymptotes at x=-3 and x=2, its horizontal asymptote at y=1, and its intercepts at (-1, 0), (0, 1\/6), and (3, 0).\" width=\"731\" height=\"626\" \/>\r\n<ul>\r\n \t<li>At the [latex]x[\/latex]-intercept [latex]x=-1[\/latex] corresponding to the [latex]{\\left(x+1\\right)}^{2}[\/latex] factor of the numerator, the graph bounces, consistent with the quadratic nature of the factor.<\/li>\r\n \t<li>At the [latex]x[\/latex]-intercept [latex]x=3[\/latex] corresponding to the [latex]\\left(x - 3\\right)[\/latex] factor of the numerator, the graph passes through the axis as we would expect from a linear factor.<\/li>\r\n \t<li>At the vertical asymptote [latex]x=-3[\/latex] corresponding to the [latex]{\\left(x+3\\right)}^{2}[\/latex] factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex].<\/li>\r\n \t<li>At the vertical asymptote [latex]x=2[\/latex], corresponding to the [latex]\\left(x - 2\\right)[\/latex] factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side, consistent with the behavior of the function [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex].<\/li>\r\n<\/ul>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a rational function, sketch a graph.<\/h3>\r\n<ol>\r\n \t<li>Evaluate the function at 0 to find the [latex]y[\/latex]-intercept.<\/li>\r\n \t<li>Factor the numerator and denominator.<\/li>\r\n \t<li>For factors in the numerator not common to the denominator, determine where each factor of the numerator is zero to find the [latex]x[\/latex]-intercepts.<\/li>\r\n \t<li>Find the multiplicities of the [latex]x[\/latex]-intercepts to determine the behavior of the graph at those points.<\/li>\r\n \t<li>For factors in the denominator, note the multiplicities of the zeros to determine the local behavior. For those factors not common to the numerator, find the vertical asymptotes by setting those factors equal to zero and then solve.<\/li>\r\n \t<li>For factors in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to 0 and then solve.<\/li>\r\n \t<li>Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes.<\/li>\r\n \t<li>Sketch the graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing a Rational Function<\/h3>\r\nSketch a graph of [latex]f\\left(x\\right)=\\dfrac{\\left(x+2\\right)\\left(x - 3\\right)}{{\\left(x+1\\right)}^{2}\\left(x - 2\\right)}[\/latex].\r\n\r\n[reveal-answer q=\"829751\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"829751\"]\r\n\r\nWe can start by noting that the function is already factored, saving us a step.\r\n\r\nNext, we will find the intercepts. Evaluating the function at zero gives the <em>y<\/em>-intercept:\r\n<p style=\"text-align: center;\">[latex]f\\left(0\\right)=\\frac{\\left(0+2\\right)\\left(0 - 3\\right)}{{\\left(0+1\\right)}^{2}\\left(0 - 2\\right)}=3[\/latex]<\/p>\r\nTo find the [latex]x[\/latex]-intercepts, we determine when the numerator of the function is zero. Setting each factor equal to zero, we find [latex]x[\/latex]-intercepts at [latex]x=-2[\/latex] and [latex]x=3[\/latex]. At each, the behavior will be linear (multiplicity 1), with the graph passing through the intercept.\r\n\r\nWe have a [latex]y[\/latex]-intercept at [latex]\\left(0,3\\right)[\/latex] and <em>x<\/em>-intercepts at [latex]\\left(-2,0\\right)[\/latex] and [latex]\\left(3,0\\right)[\/latex].\r\n\r\nTo find the vertical asymptotes, we determine when the denominator is equal to zero. This occurs when [latex]x+1=0[\/latex] and when [latex]x - 2=0[\/latex], giving us vertical asymptotes at [latex]x=-1[\/latex] and [latex]x=2[\/latex].\r\n\r\nThere are no common factors in the numerator and denominator. This means there are no removable discontinuities.\r\n\r\nFinally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal asymptote at [latex]y=0[\/latex].\r\n\r\nTo sketch the graph, we might start by plotting the three intercepts. Since the graph has no [latex]x[\/latex]-intercepts between the vertical asymptotes, and the [latex]y[\/latex]-intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213952\/CNX_Precalc_Figure_03_07_0212.jpg\" alt=\"Graph of only the middle portion of f(x)=(x+2)(x-3)\/(x+1)^2(x-2) with its intercepts at (-2, 0), (0, 3), and (3, 0).\" width=\"487\" height=\"440\" \/>\r\n\r\nThe factor associated with the vertical asymptote at [latex]x=-1[\/latex] was squared, so we know the behavior will be the same on both sides of the asymptote. The graph heads toward positive infinity as the inputs approach the asymptote on the right, so the graph will head toward positive infinity on the left as well.\r\n\r\nFor the vertical asymptote at [latex]x=2[\/latex], the factor was not squared, so the graph will have opposite behavior on either side of the asymptote. After passing through the [latex]x[\/latex]-intercepts, the graph will then level off toward an output of zero, as indicated by the horizontal asymptote.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213954\/CNX_Precalc_Figure_03_07_022.jpg\" alt=\"Graph of f(x)=(x+2)(x-3)\/(x+1)^2(x-2) with its vertical asymptotes at x=-1 and x=2, its horizontal asymptote at y=0, and its intercepts at (-2, 0), (0, 3), and (3, 0).\" width=\"487\" height=\"439\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGiven the function [latex]f\\left(x\\right)=\\dfrac{{\\left(x+2\\right)}^{2}\\left(x - 2\\right)}{2{\\left(x - 1\\right)}^{2}\\left(x - 3\\right)}[\/latex], use the characteristics of polynomials and rational functions to describe its behavior and sketch the function.\r\n\r\n[reveal-answer q=\"451047\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"451047\"]\r\n\r\nHorizontal asymptote at [latex]y=\\frac{1}{2}[\/latex]. Vertical asymptotes at [latex]x=1[\/latex] and [latex]x=3[\/latex]. [latex]y[\/latex]-intercept at [latex]\\left(0,\\frac{4}{3}.\\right)[\/latex]\r\n<p id=\"fs-id1165135168380\"><em>x<\/em>-intercepts at [latex]\\left(2,0\\right) \\text{ and }\\left(-2,0\\right)[\/latex]. [latex]\\left(-2,0\\right)[\/latex] is a zero with multiplicity 2, and the graph bounces off the [latex]x[\/latex]-axis at this point. [latex]\\left(2,0\\right)[\/latex] is a single zero and the graph crosses the axis at this point.<span id=\"fs-id1165137745200\">\r\n<\/span><\/p>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/12\/02213511\/CNX_Precalc_Figure_03_07_023.jpg\"><img class=\"aligncenter size-full wp-image-2947\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/12\/02213511\/CNX_Precalc_Figure_03_07_023.jpg\" alt=\"cnx_precalc_figure_03_07_023\" width=\"731\" height=\"477\" \/><\/a>\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom20\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129075&amp;theme=oea&amp;iframe_resize_id=mom20\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\nWatch the following video to see another worked example of how to match different kinds of rational functions with their graphs.\r\n\r\nhttps:\/\/youtu.be\/vMVYaFptvkk\r\n<h2>Writing Rational Functions<\/h2>\r\nNow that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. A rational function written in factored form will have an [latex]x[\/latex]-intercept where each factor of the numerator is equal to zero. (An exception occurs in the case of a removable discontinuity.) As a result, we can form a numerator of a function whose graph will pass through a set of [latex]x[\/latex]-intercepts by introducing a corresponding set of factors. Likewise, because the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Writing Rational Functions from Intercepts and Asymptotes<\/h3>\r\nIf a <strong>rational function<\/strong> has [latex]x[\/latex]-intercepts at [latex]x={x}_{1}, {x}_{2}, ..., {x}_{n}[\/latex], vertical asymptotes at [latex]x={v}_{1},{v}_{2},\\dots ,{v}_{m}[\/latex], and no [latex]{x}_{i}=\\text{any }{v}_{j}[\/latex], then the function can be written in the form:\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=a\\frac{{\\left(x-{x}_{1}\\right)}^{{p}_{1}}{\\left(x-{x}_{2}\\right)}^{{p}_{2}}\\cdots {\\left(x-{x}_{n}\\right)}^{{p}_{n}}}{{\\left(x-{v}_{1}\\right)}^{{q}_{1}}{\\left(x-{v}_{2}\\right)}^{{q}_{2}}\\cdots {\\left(x-{v}_{m}\\right)}^{{q}_{n}}}[\/latex]<\/p>\r\nwhere the powers [latex]{p}_{i}[\/latex] or [latex]{q}_{i}[\/latex] on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor [latex]a[\/latex]<em>\u00a0<\/em>can be determined given a value of the function other than the [latex]x[\/latex]-intercept or by the horizontal asymptote if it is nonzero.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a graph of a rational function, write the function.<\/h3>\r\n<ol>\r\n \t<li>Determine the factors of the numerator. Examine the behavior of the graph at the <em>x<\/em>-intercepts to determine the zeroes and their multiplicities. (This is easy to do when finding the \"simplest\" function with small multiplicities\u2014such as 1 or 3\u2014but may be difficult for larger multiplicities\u2014such as 5 or 7, for example.)<\/li>\r\n \t<li>Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote to determine the factors and their powers.<\/li>\r\n \t<li>Use any clear point on the graph to find the stretch factor.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing a Rational Function from Intercepts and Asymptotes<\/h3>\r\nWrite an equation for the rational function\u00a0below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213956\/CNX_Precalc_Figure_03_07_024.jpg\" alt=\"Graph of a rational function.\" width=\"487\" height=\"475\" \/>\r\n[reveal-answer q=\"666867\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"666867\"]\r\n\r\nThe graph appears to have [latex]x[\/latex]-intercepts at [latex]x=-2[\/latex] and [latex]x=3[\/latex]. At both, the graph passes through the intercept, suggesting linear factors. The graph has two vertical asymptotes. The one at [latex]x=-1[\/latex] seems to exhibit the basic behavior similar to [latex]\\frac{1}{x}[\/latex], with the graph heading toward positive infinity on one side and heading toward negative infinity on the other. The asymptote at [latex]x=2[\/latex] is exhibiting a behavior similar to [latex]\\frac{1}{{x}^{2}}[\/latex], with the graph heading toward negative infinity on both sides of the asymptote.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213958\/CNX_Precalc_Figure_03_07_025.jpg\" alt=\"Graph of a rational function denoting its vertical asymptotes and x-intercepts.\" width=\"731\" height=\"475\" \/>\r\n\r\nWe can use this information to write a function of the form\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=a\\dfrac{\\left(x+2\\right)\\left(x - 3\\right)}{\\left(x+1\\right){\\left(x - 2\\right)}^{2}}[\/latex].<\/p>\r\nTo find the stretch factor, we can use another clear point on the graph, such as the [latex]y[\/latex]-intercept [latex]\\left(0,-2\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-2&amp;=a\\dfrac{\\left(0+2\\right)\\left(0 - 3\\right)}{\\left(0+1\\right){\\left(0 - 2\\right)}^{2}} \\\\[1mm] -2&amp;=a\\frac{-6}{4} \\\\[1mm] a=\\frac{-8}{-6}=\\frac{4}{3} \\end{align}[\/latex]<\/p>\r\nThis gives us a final function of [latex]f\\left(x\\right)=\\dfrac{4\\left(x+2\\right)\\left(x - 3\\right)}{3\\left(x+1\\right){\\left(x - 2\\right)}^{2}}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n<iframe id=\"mom25\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1802&amp;theme=oea&amp;iframe_resize_id=mom25\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<section id=\"fs-id1165137659195\" class=\"key-equations\">\r\n<h2>Key Equations<\/h2>\r\n<table id=\"eip-id1362369\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>Rational Function<\/td>\r\n<td>[latex]f\\left(x\\right)=\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{a}_{p}{x}^{p}+{a}_{p - 1}{x}^{p - 1}+...+{a}_{1}x+{a}_{0}}{{b}_{q}{x}^{q}+{b}_{q - 1}{x}^{q - 1}+...+{b}_{1}x+{b}_{0}}, Q\\left(x\\right)\\ne 0[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section><section id=\"fs-id1165137793507\" class=\"key-concepts\">\r\n<h2>Key Concepts<\/h2>\r\n<ul id=\"fs-id1165137603314\">\r\n \t<li>We can use arrow notation to describe local behavior and end behavior of the toolkit functions [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex].<\/li>\r\n \t<li>A function that levels off at a horizontal value has a horizontal asymptote. A function can have more than one vertical asymptote.<\/li>\r\n \t<li>Application problems involving rates and concentrations often involve rational functions.<\/li>\r\n \t<li>The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.<\/li>\r\n \t<li>The vertical asymptotes of a rational function will occur where the denominator of the function is equal to zero and the numerator is not zero.<\/li>\r\n \t<li>A removable discontinuity might occur in the graph of a rational function if an input causes both numerator and denominator to be zero.<\/li>\r\n \t<li>A rational function\u2019s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions.<\/li>\r\n \t<li>Graph rational functions by finding the intercepts, behavior at the intercepts and asymptotes, and end behavior.<\/li>\r\n \t<li>If a rational function has <em>x<\/em>-intercepts at [latex]x={x}_{1},{x}_{2},\\dots ,{x}_{n}[\/latex], vertical asymptotes at [latex]x={v}_{1},{v}_{2},\\dots ,{v}_{m}[\/latex], and no [latex]{x}_{i}=\\text{any }{v}_{j}[\/latex], then the function can be written in the form\u00a0[latex]f\\left(x\\right)=a\\dfrac{{\\left(x-{x}_{1}\\right)}^{{p}_{1}}{\\left(x-{x}_{2}\\right)}^{{p}_{2}}\\cdots {\\left(x-{x}_{n}\\right)}^{{p}_{n}}}{{\\left(x-{v}_{1}\\right)}^{{q}_{1}}{\\left(x-{v}_{2}\\right)}^{{q}_{2}}\\cdots {\\left(x-{v}_{m}\\right)}^{{q}_{n}}}[\/latex]<\/li>\r\n<\/ul>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137758530\" class=\"definition\">\r\n \t<dt><strong>arrow notation<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135154402\">a way to symbolically represent the local and end behavior of a function by using arrows to indicate that an input or output approaches a value<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135154407\" class=\"definition\">\r\n \t<dt><strong>horizontal asymptote<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135154413\">a horizontal line [latex]y=b[\/latex] where the graph approaches the line as the inputs increase or decrease without bound.<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135192626\" class=\"definition\">\r\n \t<dt><strong>rational function<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165134401081\">a function that can be written as the ratio of two polynomials<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134401085\" class=\"definition\">\r\n \t<dt><strong>removable discontinuity<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165134401090\">a single point at which a function is undefined that, if filled in, would make the function continuous; it appears as a hole on the graph of a function<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137426312\" class=\"definition\">\r\n \t<dt><strong>vertical asymptote<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137426317\">a vertical line [latex]x=a[\/latex] where the graph tends toward positive or negative infinity as the inputs approach [latex]a[\/latex]<\/dd>\r\n<\/dl>\r\n<\/div>\r\n<\/section>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul class=\"ul1\">\n<li class=\"li2\"><span class=\"s1\">Use arrow notation to describe end behavior of rational functions.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Solve applied problems involving rational functions.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Find the domains of rational functions.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Identify vertical and horizontal asymptotes of graphs of rational functions.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Graph rational functions.<\/span><\/li>\n<\/ul>\n<\/div>\n<p>Suppose we know that the cost of making a product is dependent on the number of items, [latex]x[\/latex], produced. This is given by the equation [latex]C\\left(x\\right)=15,000x - 0.1{x}^{2}+1000[\/latex]. If we want to know the average cost for producing [latex]x[\/latex]\u00a0items, we would divide the cost function by the number of items, [latex]x[\/latex].<\/p>\n<p>The average cost function, which yields the average cost per item for [latex]x[\/latex]\u00a0items produced, is<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{15,000x - 0.1{x}^{2}+1000}{x}[\/latex]<\/p>\n<p>Many other application problems require finding an average value in a similar way, giving us variables in the denominator. Written without a variable in the denominator, this function will contain a negative integer power.<\/p>\n<p>In the last few sections, we have worked with polynomial functions, which are functions with non-negative integers for exponents. In this section, we explore rational functions, which have variables in the denominator.<\/p>\n<h2>Characteristics of Rational Functions<\/h2>\n<p>We have seen the graphs of the basic <strong>reciprocal function<\/strong> and the squared reciprocal function from our study of toolkit functions. Examine these graphs\u00a0and notice some of their features.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213907\/CNX_Precalc_Figure_03_07_0012.jpg\" alt=\"Graphs of f(x)=1\/x and f(x)=1\/x^2\" width=\"731\" height=\"453\" \/><\/p>\n<p>Several things are apparent if we examine the graph of [latex]f\\left(x\\right)=\\dfrac{1}{x}[\/latex].<\/p>\n<ol>\n<li>On the left branch of the graph, the curve approaches the [latex]x[\/latex]-axis [latex]\\left(y=0\\right) \\text{ as } x\\to -\\infty[\/latex].<\/li>\n<li>As the graph approaches [latex]x=0[\/latex] from the left, the curve drops, but as we approach zero from the right, the curve rises.<\/li>\n<li>Finally, on the right branch of the graph, the curves approaches the [latex]x[\/latex]<em>&#8211;<\/em>axis [latex]\\left(y=0\\right) \\text{ as } x\\to \\infty[\/latex].<\/li>\n<\/ol>\n<p>To summarize, we use <strong>arrow notation<\/strong> to show that [latex]x[\/latex]\u00a0or [latex]f\\left(x\\right)[\/latex] is approaching a particular value.<\/p>\n<table>\n<thead>\n<tr>\n<th style=\"text-align: center;\" colspan=\"2\">Arrow Notation<\/th>\n<\/tr>\n<tr>\n<th style=\"text-align: center;\">Symbol<\/th>\n<th style=\"text-align: center;\">Meaning<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]x\\to {a}^{-}[\/latex]<\/td>\n<td>[latex]x[\/latex] approaches [latex]a[\/latex]\u00a0from the left ([latex]x<a[\/latex] but close to [latex]a[\/latex])<\/td>\n<\/tr>\n<tr>\n<td>[latex]x\\to {a}^{+}[\/latex]<\/td>\n<td>[latex]x[\/latex] approaches [latex]a[\/latex]\u00a0from the right ([latex]x>a[\/latex]\u00a0but close to [latex]a[\/latex])<\/td>\n<\/tr>\n<tr>\n<td>[latex]x\\to \\infty[\/latex]<\/td>\n<td>[latex]x[\/latex] approaches infinity ([latex]x[\/latex]\u00a0increases without bound)<\/td>\n<\/tr>\n<tr>\n<td>[latex]x\\to -\\infty[\/latex]<\/td>\n<td>[latex]x[\/latex] approaches negative infinity ([latex]x[\/latex]\u00a0decreases without bound)<\/td>\n<\/tr>\n<tr>\n<td>[latex]f\\left(x\\right)\\to \\infty[\/latex]<\/td>\n<td>the output approaches infinity (the output increases without bound)<\/td>\n<\/tr>\n<tr>\n<td>[latex]f\\left(x\\right)\\to -\\infty[\/latex]<\/td>\n<td>the output approaches negative infinity (the output decreases without bound)<\/td>\n<\/tr>\n<tr>\n<td>[latex]f\\left(x\\right)\\to a[\/latex]<\/td>\n<td>the output approaches [latex]a[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Local Behavior of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/h2>\n<p>Let\u2019s begin by looking at the reciprocal function, [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]. We cannot divide by zero, which means the function is undefined at [latex]x=0[\/latex]; so zero is not in the domain<em>.<\/em> As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in the table below.<\/p>\n<table id=\"Table_03_07_002\" summary=\"..\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>\u20130.1<\/td>\n<td>\u20130.01<\/td>\n<td>\u20130.001<\/td>\n<td>\u20130.0001<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] <\/strong><\/td>\n<td>\u201310<\/td>\n<td>\u2013100<\/td>\n<td>\u20131000<\/td>\n<td>\u201310,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We write in arrow notation<\/p>\n<p style=\"text-align: center;\">[latex]\\text{as }x\\to {0}^{-},f\\left(x\\right)\\to -\\infty[\/latex]<\/p>\n<p>As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in the table below.<\/p>\n<table id=\"Table_03_07_003\" summary=\"..\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>0.1<\/td>\n<td>0.01<\/td>\n<td>0.001<\/td>\n<td>0.0001<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] <\/strong><\/td>\n<td>10<\/td>\n<td>100<\/td>\n<td>1000<\/td>\n<td>10,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We write in arrow notation<\/p>\n<p style=\"text-align: center;\">[latex]\\text{As }x\\to {0}^{+}, f\\left(x\\right)\\to \\infty[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213909\/CNX_Precalc_Figure_03_07_0022.jpg\" alt=\"Graph of f(x)=1\/x which denotes the end behavior. As x goes to negative infinity, f(x) goes to 0, and as x goes to 0^-, f(x) goes to negative infinity. As x goes to positive infinity, f(x) goes to 0, and as x goes to 0^+, f(x) goes to positive infinity.\" width=\"731\" height=\"474\" \/><\/p>\n<p>This behavior creates a <strong>vertical asymptote<\/strong>, which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line [latex]x=0[\/latex]\u00a0as the input becomes close to zero.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213912\/CNX_Precalc_Figure_03_07_0032.jpg\" alt=\"Graph of f(x)=1\/x with its vertical asymptote at x=0.\" width=\"487\" height=\"364\" \/><\/p>\n<div class=\"textbox\">\n<h3>A General Note: Vertical Asymptote<\/h3>\n<p>A <strong>vertical asymptote<\/strong> of a graph is a vertical line [latex]x=a[\/latex] where the graph tends toward positive or negative infinity as the inputs approach [latex]a[\/latex]. We write<\/p>\n<p>[latex]\\text{As }x\\to a,f\\left(x\\right)\\to \\infty , \\text{or as }x\\to a,f\\left(x\\right)\\to -\\infty[\/latex].<\/p>\n<\/div>\n<h2>End Behavior of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/h2>\n<p>As the values of [latex]x[\/latex]\u00a0approach infinity, the function values approach 0. As the values of [latex]x[\/latex]\u00a0approach negative infinity, the function values approach 0. Symbolically, using arrow notation<\/p>\n<p style=\"text-align: center;\">[latex]\\text{As }x\\to \\infty ,f\\left(x\\right)\\to 0,\\text{and as }x\\to -\\infty ,f\\left(x\\right)\\to 0[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213913\/CNX_Precalc_Figure_03_07_0042.jpg\" alt=\"Graph of f(x)=1\/x which highlights the segments of the turning points to denote their end behavior.\" width=\"731\" height=\"475\" \/><\/p>\n<p>Based on this overall behavior and the graph, we can see that the function approaches 0 but never actually reaches 0; it seems to level off as the inputs become large. This behavior creates a <strong>horizontal asymptote<\/strong>, a horizontal line that the graph approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line [latex]y=0[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213915\/CNX_Precalc_Figure_03_07_0052.jpg\" alt=\"Graph of f(x)=1\/x with its vertical asymptote at x=0 and its horizontal asymptote at y=0.\" width=\"487\" height=\"364\" \/><\/p>\n<div class=\"textbox\">\n<h3>A General Note: Horizontal Asymptote<\/h3>\n<p>A <strong>horizontal asymptote<\/strong> of a graph is a horizontal line [latex]y=b[\/latex] where the graph approaches the line as the inputs increase or decrease without bound. We write<\/p>\n<p>[latex]\\text{As }x\\to \\infty \\text{ or }x\\to -\\infty ,\\text{ }f\\left(x\\right)\\to b[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using Arrow Notation<\/h3>\n<p>Use arrow notation to describe the end behavior and local behavior of the function below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213918\/CNX_Precalc_Figure_03_07_0062.jpg\" alt=\"Graph of f(x)=1\/(x-2)+4 with its vertical asymptote at x=2 and its horizontal asymptote at y=4.\" width=\"487\" height=\"477\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q444547\">Show Solution<\/span><\/p>\n<div id=\"q444547\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that the graph is showing a vertical asymptote at [latex]x=2[\/latex], which tells us that the function is undefined at [latex]x=2[\/latex].<\/p>\n<p style=\"text-align: center;\">As [latex]x\\to {2}^{-},\\hspace{2mm}f\\left(x\\right)\\to -\\infty[\/latex], and as [latex]x\\to {2}^{+},\\text{ }f\\left(x\\right)\\to \\infty[\/latex]<\/p>\n<p>And as the inputs decrease without bound, the graph appears to be leveling off at output values of 4, indicating a horizontal asymptote at [latex]y=4[\/latex]. As the inputs increase without bound, the graph levels off at 4.<\/p>\n<p style=\"text-align: center;\">As [latex]x\\to \\infty ,\\text{ }f\\left(x\\right)\\to 4[\/latex], and as [latex]x\\to -\\infty ,\\text{ }f\\left(x\\right)\\to 4[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function[latex]f\\left(x\\right)=\\dfrac{1}{x^2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q785291\">Show Solution<\/span><\/p>\n<div id=\"q785291\" class=\"hidden-answer\" style=\"display: none\">\n<p>End behavior: as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to 0[\/latex]; Local behavior: as [latex]x\\to 0, f\\left(x\\right)\\to \\infty[\/latex] (there are no [latex]x[\/latex]&#8211; or [latex]y[\/latex]-intercepts)<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom25\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129042&amp;theme=oea&amp;iframe_resize_id=mom25\" width=\"100%\" height=\"550\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using Transformations to Graph a Rational Function<\/h3>\n<p>Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q670271\">Show Solution<\/span><\/p>\n<div id=\"q670271\" class=\"hidden-answer\" style=\"display: none\">\n<p>Shifting the graph left 2 and up 3 would result in the function<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{1}{x+2}+3[\/latex]<\/p>\n<p>or equivalently, by giving the terms a common denominator,<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{3x+7}{x+2}[\/latex]<\/p>\n<p>The graph of the shifted function is displayed\u00a0below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213922\/CNX_Precalc_Figure_03_07_0072.jpg\" alt=\"Graph of f(x)=1\/(x+2)+3 with its vertical asymptote at x=-2 and its horizontal asymptote at y=3.\" width=\"731\" height=\"441\" \/><\/p>\n<p>Notice that this function is undefined at [latex]x=-2[\/latex], and the graph also is showing a vertical asymptote at [latex]x=-2[\/latex].<\/p>\n<p style=\"text-align: center;\">As [latex]x\\to -{2}^{-}, f\\left(x\\right)\\to -\\infty[\/latex] , and as\u00a0 [latex]x\\to -{2}^{+}, f\\left(x\\right)\\to \\infty[\/latex]<\/p>\n<p>As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at [latex]y=3[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\text{As }x\\to \\pm \\infty , f\\left(x\\right)\\to 3[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q600482\">Show Solution<\/span><\/p>\n<div id=\"q600482\" class=\"hidden-answer\" style=\"display: none\">\n<p>The function and the asymptotes are shifted 3 units right and 4 units down. As [latex]x\\to 3,f\\left(x\\right)\\to \\infty[\/latex], and as [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to -4[\/latex].<\/p>\n<p id=\"fs-id1165137823960\">The function is [latex]f\\left(x\\right)=\\dfrac{1}{{\\left(x - 3\\right)}^{2}}-4[\/latex].<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/12\/02000148\/CNX_Precalc_Figure_03_07_0082.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-2938\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/12\/02000148\/CNX_Precalc_Figure_03_07_0082.jpg\" alt=\"Graph of f(x)=1\/(x-3)^2-4 with its vertical asymptote at x=3 and its horizontal asymptote at y=-4.\" width=\"487\" height=\"365\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>A Mixing Problem<\/h2>\n<p>In the previous example, we shifted a toolkit function in a way that resulted in the function [latex]f\\left(x\\right)=\\dfrac{3x+7}{x+2}[\/latex]. This is an example of a rational function. 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<div class=\"textbox\">\n<h3>A General Note: Rational Function<\/h3>\n<p>A <strong>rational function<\/strong> is a function that can be written as the quotient of two polynomial functions [latex]P\\left(x\\right) \\text{and} Q\\left(x\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{a}_{p}{x}^{p}+{a}_{p - 1}{x}^{p - 1}+...+{a}_{1}x+{a}_{0}}{{b}_{q}{x}^{q}+{b}_{q - 1}{x}^{q - 1}+...+{b}_{1}x+{b}_{0}},Q\\left(x\\right)\\ne 0[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Applied Problem Involving a Rational Function<\/h3>\n<p>A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q527196\">Show Solution<\/span><\/p>\n<div id=\"q527196\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]t[\/latex] be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 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 12 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 17 pounds of sugar to 220 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 12 minutes than at the beginning.<\/p>\n<h4>Analysis of the Solution<\/h4>\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 0.1 in the long term.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>There are 1,200 freshmen and 1,500 sophomores at a prep rally at noon. After 12 p.m., 20 freshmen arrive at the rally every five minutes while 15 sophomores leave the rally. Find the ratio of freshmen to sophomores at 1 p.m.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q526334\">Show Solution<\/span><\/p>\n<div id=\"q526334\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\dfrac{12}{11}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129067&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"550\"><\/iframe><\/p>\n<\/div>\n<h2>Domain and Its Effect on Vertical Asymptotes<\/h2>\n<p>A <strong>vertical asymptote<\/strong> represents a value at which a rational function is undefined, so that value is not in the domain of the function. A reciprocal function cannot have values in its domain that cause the denominator to equal zero. In general, to find the domain of a rational function, we need to determine which inputs would cause division by zero.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Domain of a Rational Function<\/h3>\n<p>The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a rational function, find the domain.<\/h3>\n<ol>\n<li>Set the denominator equal to zero.<\/li>\n<li>Solve to find the [latex]x[\/latex]-values that cause the denominator to equal zero.<\/li>\n<li>The domain is all real numbers except those found in Step 2.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Domain of a Rational Function<\/h3>\n<p>Find the domain of [latex]f\\left(x\\right)=\\dfrac{x+3}{{x}^{2}-9}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q860212\">Show Solution<\/span><\/p>\n<div id=\"q860212\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by setting the denominator equal to zero and solving.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} {x}^{2}-9&=0 \\\\ {x}^{2}&=9 \\\\ x&=\\pm 3 \\end{align}[\/latex]<\/p>\n<p>The denominator is equal to zero when [latex]x=\\pm 3[\/latex]. The domain of the function is all real numbers except [latex]x=\\pm 3[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>A graph of this function confirms that the function is not defined when [latex]x=\\pm 3[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213925\/CNX_Precalc_Figure_03_07_0092.jpg\" alt=\"Graph of f(x)=1\/(x-3) with its vertical asymptote at x=3 and its horizontal asymptote at y=0.\" width=\"487\" height=\"364\" \/><\/p>\n<p>There is a vertical asymptote at [latex]x=3[\/latex] and a hole in the graph at [latex]x=-3[\/latex]. We will discuss these types of holes in greater detail later in this section.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the domain of [latex]f\\left(x\\right)=\\dfrac{4x}{5\\left(x - 1\\right)\\left(x - 5\\right)}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q553731\">Show Solution<\/span><\/p>\n<div id=\"q553731\" class=\"hidden-answer\" style=\"display: none\">\n<p>The domain is all real numbers except [latex]x=1[\/latex] and [latex]x=5[\/latex].<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129068&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"400\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<p>By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location.<\/p>\n<p>Watch the following video to see more examples of finding the domain of a rational function.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  The Domain of Rational Functions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/v0IhvIzCc_I?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Vertical Asymptotes<\/h2>\n<p>The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a rational function, identify any vertical asymptotes of its graph.<\/h3>\n<ol>\n<li>Factor the numerator and denominator.<\/li>\n<li>Note any restrictions in the domain of the function.<\/li>\n<li>Reduce the expression by canceling common factors in the numerator and the denominator.<\/li>\n<li>Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.<\/li>\n<li>Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Vertical Asymptotes<\/h3>\n<p>Find the vertical asymptotes of the graph of [latex]k\\left(x\\right)=\\dfrac{5+2{x}^{2}}{2-x-{x}^{2}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q787718\">Show Solution<\/span><\/p>\n<div id=\"q787718\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, factor the numerator and denominator.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}k\\left(x\\right)&=\\dfrac{5+2{x}^{2}}{2-x-{x}^{2}} \\\\[1mm] &=\\dfrac{5+2{x}^{2}}{\\left(2+x\\right)\\left(1-x\\right)} \\end{align}[\/latex]<\/p>\n<p>To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(2+x\\right)\\left(1-x\\right)=0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=-2,1[\/latex]<\/p>\n<p>Neither [latex]x=-2[\/latex] nor [latex]x=1[\/latex] are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph below confirms the location of the two vertical asymptotes.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213927\/CNX_Precalc_Figure_03_07_0102.jpg\" alt=\"Graph of k(x)=(5+2x)^2\/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.\" width=\"487\" height=\"514\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom110\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=24104&amp;theme=oea&amp;iframe_resize_id=mom110\" width=\"100%\" height=\"650\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Removable Discontinuities<\/h2>\n<p>Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a <strong>removable discontinuity<\/strong>.<\/p>\n<p>For example the function [latex]f\\left(x\\right)=\\dfrac{{x}^{2}-1}{{x}^{2}-2x - 3}[\/latex] may be re-written by factoring the numerator and the denominator.<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{\\left(x+1\\right)\\left(x - 1\\right)}{\\left(x+1\\right)\\left(x - 3\\right)}[\/latex]<\/p>\n<p>Notice that [latex]x+1[\/latex] is a common factor to the numerator and the denominator. The zero of this factor, [latex]x=-1[\/latex], is the location of the removable discontinuity. Notice also that [latex]x - 3[\/latex] is not a factor in both the numerator and denominator. The zero of this factor, [latex]x=3[\/latex], is the vertical asymptote.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213929\/CNX_Precalc_Figure_03_07_0112.jpg\" alt=\"Graph of f(x)=(x^2-1)\/(x^2-2x-3) with its vertical asymptote at x=3 and a removable discontinuity at x=-1.\" width=\"487\" height=\"326\" \/><\/p>\n<div class=\"textbox\">\n<h3>A General Note: Removable Discontinuities of Rational Functions<\/h3>\n<p>A <strong>removable discontinuity<\/strong> occurs in the graph of a rational function at [latex]x=a[\/latex] if <em>a<\/em>\u00a0is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Vertical Asymptotes and Removable Discontinuities for a Graph<\/h3>\n<p>Find the vertical asymptotes and removable discontinuities of the graph of [latex]k\\left(x\\right)=\\dfrac{x - 2}{{x}^{2}-4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q519186\">Show Solution<\/span><\/p>\n<div id=\"q519186\" class=\"hidden-answer\" style=\"display: none\">\n<p>Factor the numerator and the denominator.<\/p>\n<p style=\"text-align: center;\">[latex]k\\left(x\\right)=\\dfrac{x - 2}{\\left(x - 2\\right)\\left(x+2\\right)}[\/latex]<\/p>\n<p>Notice that there is a common factor in the numerator and the denominator, [latex]x - 2[\/latex]. The zero for this factor is [latex]x=2[\/latex]. This is the location of the removable discontinuity.<\/p>\n<p>Notice that there is a factor in the denominator that is not in the numerator, [latex]x+2[\/latex]. The zero for this factor is [latex]x=-2[\/latex]. The vertical asymptote is [latex]x=-2[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213931\/CNX_Precalc_Figure_03_07_0122.jpg\" alt=\"Graph of k(x)=(x-2)\/(x-2)(x+2) with its vertical asymptote at x=-2 and a removable discontinuity at x=2.\" width=\"487\" height=\"364\" \/><\/p>\n<p>The graph of this function will have the vertical asymptote at [latex]x=-2[\/latex], but at [latex]x=2[\/latex] the graph will have a hole.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the vertical asymptotes and removable discontinuities of the graph of [latex]f\\left(x\\right)=\\dfrac{{x}^{2}-25}{{x}^{3}-6{x}^{2}+5x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q575454\">Show Solution<\/span><\/p>\n<div id=\"q575454\" class=\"hidden-answer\" style=\"display: none\">\n<p>Removable discontinuity at [latex]x=5[\/latex]. Vertical asymptotes: [latex]x=0,\\text{ }x=1[\/latex].<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom111\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=74565&amp;theme=oea&amp;iframe_resize_id=mom111\" width=\"100%\" height=\"400\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom111\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=74565&amp;theme=oea&amp;iframe_resize_id=mom111\" width=\"100%\" height=\"400\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<h2>Horizontal Asymptotes and Intercepts<\/h2>\n<p>While vertical asymptotes describe the behavior of a graph as the <em>output<\/em> gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the <em>input<\/em> gets very large or very small. Recall that a polynomial\u2019s end behavior will mirror that of the leading term. Likewise, a rational function\u2019s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions.<\/p>\n<p>There are three distinct outcomes when checking for horizontal asymptotes:<\/p>\n<p><strong>Case 1:<\/strong> If the degree of the denominator &gt; degree of the numerator, there is a <strong>horizontal asymptote<\/strong> at [latex]y=0[\/latex].<\/p>\n<p style=\"text-align: center;\">Example: [latex]f\\left(x\\right)=\\dfrac{4x+2}{{x}^{2}+4x - 5}[\/latex]<\/p>\n<p>In this case the end behavior is [latex]f\\left(x\\right)\\approx \\frac{4x}{{x}^{2}}=\\frac{4}{x}[\/latex]. This tells us that, as the inputs increase or decrease without bound, this function will behave similarly to the function [latex]g\\left(x\\right)=\\frac{4}{x}[\/latex], and the outputs will approach zero, resulting in a horizontal asymptote at [latex]y=0[\/latex]. Note that this graph crosses the horizontal asymptote.<\/p>\n<div style=\"width: 910px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213933\/CNX_Precalc_Figure_03_07_0132.jpg\" alt=\"Graph of f(x)=(4x+2)\/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=0.\" width=\"900\" height=\"302\" \/><\/p>\n<p class=\"wp-caption-text\">Horizontal Asymptote [latex]y=0[\/latex] when [latex]f\\left(x\\right)=\\dfrac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne{0}\\text{ where degree of }p&lt;\\text{degree of q}[\/latex].<\/p>\n<\/div>\n<p><strong>Case 2:<\/strong> If the degree of the denominator &lt; degree of the numerator by one, we get a slant asymptote.<\/p>\n<p style=\"text-align: center;\">Example: [latex]f\\left(x\\right)=\\dfrac{3{x}^{2}-2x+1}{x - 1}[\/latex]<\/p>\n<p>In this case the end behavior is [latex]f\\left(x\\right)\\approx \\frac{3{x}^{2}}{x}=3x[\/latex]. This tells us that as the inputs increase or decrease without bound, this function will behave similarly to the function [latex]g\\left(x\\right)=3x[\/latex]. As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of [latex]g\\left(x\\right)=3x[\/latex] looks like a diagonal line, and since [latex]f[\/latex]\u00a0will behave similarly to [latex]g[\/latex], it will approach a line close to [latex]y=3x[\/latex]. This line is a slant asymptote.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213937\/CNX_Precalc_Figure_03_07_0142.jpg\" alt=\"Graph of f(x)=(3x^2-2x+1)\/(x-1) with its vertical asymptote at x=1 and a slant asymptote aty=3x+1.\" width=\"487\" height=\"440\" \/><\/p>\n<p class=\"wp-caption-text\">Slant Asymptote when [latex]f\\left(x\\right)=\\dfrac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne 0[\/latex] where degree of [latex]p&gt;\\text{ degree of }q\\text{ by }1[\/latex].<\/p>\n<\/div>\n<p>To find the equation of the slant asymptote, divide [latex]\\dfrac{3{x}^{2}-2x+1}{x - 1}[\/latex]. The quotient is [latex]3x+1[\/latex], and the remainder is 2. The slant asymptote is the graph of the line [latex]g\\left(x\\right)=3x+1[\/latex].<\/p>\n<p><strong>Case 3:<\/strong> If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at [latex]y=\\frac{{a}_{n}}{{b}_{n}}[\/latex], where [latex]{a}_{n}[\/latex] and [latex]{b}_{n}[\/latex] are the leading coefficients of [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex] for [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne 0[\/latex].<\/p>\n<p style=\"text-align: center;\">Example: [latex]f\\left(x\\right)=\\dfrac{3{x}^{2}+2}{{x}^{2}+4x - 5}[\/latex]<\/p>\n<p>In this case the end behavior is [latex]f\\left(x\\right)\\approx \\frac{3{x}^{2}}{{x}^{2}}=3[\/latex]. This tells us that as the inputs grow large, this function will behave like the function [latex]g\\left(x\\right)=3[\/latex], which is a horizontal line. As [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to 3[\/latex], resulting in a horizontal asymptote at [latex]y=3[\/latex]. Note that this graph crosses the horizontal asymptote.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213939\/CNX_Precalc_Figure_03_07_0152.jpg\" alt=\"Graph of f(x)=(3x^2+2)\/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=3.\" width=\"731\" height=\"364\" \/><\/p>\n<p class=\"wp-caption-text\">Horizontal Asymptote when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne 0\\text{ where degree of }p=\\text{degree of }q[\/latex].<\/p>\n<\/div>\n<p>Notice that, while the graph of a rational function will never cross a <strong>vertical asymptote<\/strong>, the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote.<\/p>\n<p>It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the <strong>end behavior<\/strong> of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{3{x}^{5}-{x}^{2}}{x+3}[\/latex]<\/p>\n<p>with end behavior<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)\\approx \\dfrac{3{x}^{5}}{x}=3{x}^{4}[\/latex],<\/p>\n<p>the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.<\/p>\n<p style=\"text-align: center;\">As [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to \\infty[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Horizontal Asymptotes of Rational Functions<\/h3>\n<p>The <strong>horizontal asymptote<\/strong> of a rational function can be determined by looking at the degrees of the numerator and denominator.<\/p>\n<ul id=\"fs-id1165137722720\">\n<li>Degree of numerator <em>is less than<\/em> degree of denominator: horizontal asymptote at [latex]y=0[\/latex].<\/li>\n<li>Degree of numerator <em>is greater than degree of denominator by one<\/em>: no horizontal asymptote; slant asymptote.<\/li>\n<li>Degree of numerator <em>is equal to<\/em> degree of denominator: horizontal asymptote at ratio of leading coefficients.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Horizontal and Slant Asymptotes<\/h3>\n<p>For the functions below, identify the horizontal or slant asymptote.<\/p>\n<ol>\n<li>[latex]g\\left(x\\right)=\\dfrac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}[\/latex]<\/li>\n<li>[latex]h\\left(x\\right)=\\dfrac{{x}^{2}-4x+1}{x+2}[\/latex]<\/li>\n<li>[latex]k\\left(x\\right)=\\dfrac{{x}^{2}+4x}{{x}^{3}-8}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q968793\">Show Solution<\/span><\/p>\n<div id=\"q968793\" class=\"hidden-answer\" style=\"display: none\">\n<p>For these solutions, we will use [latex]f\\left(x\\right)=\\dfrac{p\\left(x\\right)}{q\\left(x\\right)}, q\\left(x\\right)\\ne 0[\/latex].<\/p>\n<ol>\n<li>[latex]g\\left(x\\right)=\\dfrac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}[\/latex]: The degree of [latex]p[\/latex] and the degree of [latex]q[\/latex] are both equal to 3, so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at [latex]y=\\frac{6}{2}[\/latex] or [latex]y=3[\/latex].<\/li>\n<li>[latex]h\\left(x\\right)=\\dfrac{{x}^{2}-4x+1}{x+2}[\/latex]: The degree of [latex]p=2[\/latex] and degree of [latex]q=1[\/latex]. Since [latex]p>q[\/latex] by 1, there is a slant asymptote found at [latex]\\dfrac{{x}^{2}-4x+1}{x+2}[\/latex].<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-4515 size-medium\" style=\"text-align: center;\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1916\/2016\/11\/21031600\/CNX_CAT_Figure_05_01_011-300x71.jpg\" alt=\"Synthetic division of x^2-4x+1 by x+2, resulting in x-6 with a remainder of 13\" width=\"300\" height=\"71\" \/>The quotient is [latex]x - 6[\/latex] and the remainder is 13. There is a slant asymptote at [latex]y=x - 6[\/latex].<\/li>\n<li>[latex]k\\left(x\\right)=\\dfrac{{x}^{2}+4x}{{x}^{3}-8}[\/latex]: The degree of [latex]p=2\\text{ }<[\/latex] degree of [latex]q=3[\/latex], so there is a horizontal asymptote [latex]y=0[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15836&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"400\"><br \/>\n<\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=105058&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"400\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<p>Watch this video to see more worked examples of determining which kind of horizontal asymptote a rational function will have.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Determine Horizontal Asymptotes of Rational Functions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/A1tApZSE8nI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Horizontal Asymptotes<\/h3>\n<p>In the sugar concentration problem earlier, we created the equation [latex]C\\left(t\\right)=\\dfrac{5+t}{100+10t}[\/latex].<\/p>\n<p>Find the horizontal asymptote and interpret it in context of the problem.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q905213\">Show Solution<\/span><\/p>\n<div id=\"q905213\" class=\"hidden-answer\" style=\"display: none\">\n<p>Both the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is [latex]t[\/latex], with coefficient 1. In the denominator, the leading term is [latex]10t[\/latex], with coefficient 10. The horizontal asymptote will be at the ratio of these values:<\/p>\n<p style=\"text-align: center;\">[latex]t\\to \\infty , C\\left(t\\right)\\to \\frac{1}{10}[\/latex]<\/p>\n<p>This function will have a horizontal asymptote at [latex]y=\\frac{1}{10}[\/latex].<\/p>\n<p>This tells us that as the values of [latex]t[\/latex]\u00a0increase, the values of [latex]C[\/latex]\u00a0will approach [latex]\\frac{1}{10}[\/latex]. In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or [latex]\\frac{1}{10}[\/latex] pounds per gallon.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Horizontal and Vertical Asymptotes<\/h3>\n<p>Find the horizontal and vertical asymptotes of the function<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{\\left(x - 2\\right)\\left(x+3\\right)}{\\left(x - 1\\right)\\left(x+2\\right)\\left(x - 5\\right)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q228875\">Show Solution<\/span><\/p>\n<div id=\"q228875\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, note that this function has no common factors, so there are no potential removable discontinuities.<\/p>\n<p>The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at [latex]x=1,-2,\\text{and }5[\/latex], indicating vertical asymptotes at these values.<\/p>\n<p>The numerator has degree 2, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to 0[\/latex]. This function will have a horizontal asymptote at [latex]y=0[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213942\/CNX_Precalc_Figure_03_07_0162.jpg\" alt=\"Graph of f(x)=(x-2)(x+3)\/(x-1)(x+2)(x-5) with its vertical asymptotes at x=-2, x=1, and x=5 and its horizontal asymptote at y=0.\" width=\"731\" height=\"514\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the vertical and horizontal asymptotes of the function:<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{\\left(2x - 1\\right)\\left(2x+1\\right)}{\\left(x - 2\\right)\\left(x+3\\right)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q5590\">Show Solution<\/span><\/p>\n<div id=\"q5590\" class=\"hidden-answer\" style=\"display: none\">\n<p>Vertical asymptotes at [latex]x=2[\/latex] and [latex]x=-3[\/latex]; horizontal asymptote at [latex]y=4[\/latex].<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom50\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129072&amp;theme=oea&amp;iframe_resize_id=mom50\" width=\"100%\" height=\"400\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Intercepts of Rational Functions<\/h3>\n<p>A <strong>rational function<\/strong> will have a <em>y<\/em>-intercept when the input is zero, if the function is defined at zero. A rational function will not have a [latex]y[\/latex]-intercept if the function is not defined at zero.<\/p>\n<p>Likewise, a rational function will have [latex]x[\/latex]-intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, [latex]x[\/latex]-intercepts can only occur when the numerator of the rational function is equal to zero.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Intercepts of a Rational Function<\/h3>\n<p>Find the intercepts of [latex]f\\left(x\\right)=\\dfrac{\\left(x - 2\\right)\\left(x+3\\right)}{\\left(x - 1\\right)\\left(x+2\\right)\\left(x - 5\\right)}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q418596\">Show Solution<\/span><\/p>\n<div id=\"q418596\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can find the [latex]y[\/latex]-intercept by evaluating the function at zero<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}f\\left(0\\right)&=\\dfrac{\\left(0 - 2\\right)\\left(0+3\\right)}{\\left(0 - 1\\right)\\left(0+2\\right)\\left(0 - 5\\right)} \\\\[1mm] &=\\frac{-6}{10} \\\\[1mm] &=-\\frac{3}{5} \\\\[1mm] &=-0.6 \\end{align}[\/latex]<\/p>\n<p>The [latex]x[\/latex]-intercepts will occur when the function is equal to zero. A rational function is equal to 0 when the numerator is 0, as long as the denominator is not also 0.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} 0&=\\frac{\\left(x - 2\\right)\\left(x+3\\right)}{\\left(x - 1\\right)\\left(x+2\\right)\\left(x - 5\\right)} \\\\[1mm] 0&=\\left(x - 2\\right)\\left(x+3\\right)\\end{align}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=2, -3[\/latex]<\/p>\n<p>The [latex]y[\/latex]-intercept is [latex]\\left(0,-0.6\\right)[\/latex], the [latex]x[\/latex]-intercepts are [latex]\\left(2,0\\right)[\/latex] and [latex]\\left(-3,0\\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213944\/CNX_Precalc_Figure_03_07_0172.jpg\" alt=\"Graph of f(x)=(x-2)(x+3)\/(x-1)(x+2)(x-5) with its vertical asymptotes at x=-2, x=1, and x=5, its horizontal asymptote at y=0, and its intercepts at (-3, 0), (0, -0.6), and (2, 0).\" width=\"731\" height=\"514\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-intercepts and the horizontal and vertical asymptotes.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q937415\">Show Solution<\/span><\/p>\n<div id=\"q937415\" class=\"hidden-answer\" style=\"display: none\">\n<p>For the transformed reciprocal squared function, we find the rational form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}f\\left(x\\right)&=\\dfrac{1}{{\\left(x - 3\\right)}^{2}}-4 \\\\[1mm] &=\\dfrac{1 - 4{\\left(x - 3\\right)}^{2}}{{\\left(x - 3\\right)}^{2}} \\\\[1mm] &=\\dfrac{1 - 4\\left({x}^{2}-6x+9\\right)}{\\left(x - 3\\right)\\left(x - 3\\right)} \\\\[1mm] &=\\frac{-4{x}^{2}+24x - 35}{{x}^{2}-6x+9}\\end{align}[\/latex]<\/p>\n<p>Because the numerator is the same degree as the denominator we know that as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to -4; \\text{so } y=-4[\/latex] is the horizontal asymptote. Next, we set the denominator equal to zero, and find that the vertical asymptote is [latex]x=3[\/latex], because as [latex]x\\to 3,f\\left(x\\right)\\to \\infty[\/latex]. We then set the numerator equal to 0 and find the [latex]x[\/latex]-intercepts are at [latex]\\left(2.5,0\\right)[\/latex] and [latex]\\left(3.5,0\\right)[\/latex]. Finally, we evaluate the function at 0 and find the [latex]y[\/latex]-intercept to be at [latex]\\left(0,\\frac{-35}{9}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see more worked examples of finding asymptotes, intercepts and holes of rational functions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex:  Find the Intercepts, Asymptotes, and Hole of a Rational Function\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/UnVZs2EaEjI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Graph Rational Functions<\/h2>\n<p>Previously we saw that the numerator of a rational function reveals the [latex]x[\/latex]-intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials.<\/p>\n<p>The vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213946\/CNX_Precalc_Figure_03_07_0192.jpg\" alt=\"Graph of y=1\/x with its vertical asymptote at x=0.\" width=\"487\" height=\"364\" \/><\/p>\n<p>When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213948\/CNX_Precalc_Figure_03_07_0182.jpg\" alt=\"Graph of y=1\/x^2 with its vertical asymptote at x=0.\" width=\"487\" height=\"365\" \/><\/p>\n<p>For example the graph of [latex]f\\left(x\\right)=\\dfrac{{\\left(x+1\\right)}^{2}\\left(x - 3\\right)}{{\\left(x+3\\right)}^{2}\\left(x - 2\\right)}[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213950\/CNX_Precalc_Figure_03_07_0202.jpg\" alt=\"Graph of f(x)=(x+1)^2(x-3)\/(x+3)^2(x-2) with its vertical asymptotes at x=-3 and x=2, its horizontal asymptote at y=1, and its intercepts at (-1, 0), (0, 1\/6), and (3, 0).\" width=\"731\" height=\"626\" \/><\/p>\n<ul>\n<li>At the [latex]x[\/latex]-intercept [latex]x=-1[\/latex] corresponding to the [latex]{\\left(x+1\\right)}^{2}[\/latex] factor of the numerator, the graph bounces, consistent with the quadratic nature of the factor.<\/li>\n<li>At the [latex]x[\/latex]-intercept [latex]x=3[\/latex] corresponding to the [latex]\\left(x - 3\\right)[\/latex] factor of the numerator, the graph passes through the axis as we would expect from a linear factor.<\/li>\n<li>At the vertical asymptote [latex]x=-3[\/latex] corresponding to the [latex]{\\left(x+3\\right)}^{2}[\/latex] factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex].<\/li>\n<li>At the vertical asymptote [latex]x=2[\/latex], corresponding to the [latex]\\left(x - 2\\right)[\/latex] factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side, consistent with the behavior of the function [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex].<\/li>\n<\/ul>\n<div class=\"textbox\">\n<h3>How To: Given a rational function, sketch a graph.<\/h3>\n<ol>\n<li>Evaluate the function at 0 to find the [latex]y[\/latex]-intercept.<\/li>\n<li>Factor the numerator and denominator.<\/li>\n<li>For factors in the numerator not common to the denominator, determine where each factor of the numerator is zero to find the [latex]x[\/latex]-intercepts.<\/li>\n<li>Find the multiplicities of the [latex]x[\/latex]-intercepts to determine the behavior of the graph at those points.<\/li>\n<li>For factors in the denominator, note the multiplicities of the zeros to determine the local behavior. For those factors not common to the numerator, find the vertical asymptotes by setting those factors equal to zero and then solve.<\/li>\n<li>For factors in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to 0 and then solve.<\/li>\n<li>Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes.<\/li>\n<li>Sketch the graph.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing a Rational Function<\/h3>\n<p>Sketch a graph of [latex]f\\left(x\\right)=\\dfrac{\\left(x+2\\right)\\left(x - 3\\right)}{{\\left(x+1\\right)}^{2}\\left(x - 2\\right)}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q829751\">Show Solution<\/span><\/p>\n<div id=\"q829751\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can start by noting that the function is already factored, saving us a step.<\/p>\n<p>Next, we will find the intercepts. Evaluating the function at zero gives the <em>y<\/em>-intercept:<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(0\\right)=\\frac{\\left(0+2\\right)\\left(0 - 3\\right)}{{\\left(0+1\\right)}^{2}\\left(0 - 2\\right)}=3[\/latex]<\/p>\n<p>To find the [latex]x[\/latex]-intercepts, we determine when the numerator of the function is zero. Setting each factor equal to zero, we find [latex]x[\/latex]-intercepts at [latex]x=-2[\/latex] and [latex]x=3[\/latex]. At each, the behavior will be linear (multiplicity 1), with the graph passing through the intercept.<\/p>\n<p>We have a [latex]y[\/latex]-intercept at [latex]\\left(0,3\\right)[\/latex] and <em>x<\/em>-intercepts at [latex]\\left(-2,0\\right)[\/latex] and [latex]\\left(3,0\\right)[\/latex].<\/p>\n<p>To find the vertical asymptotes, we determine when the denominator is equal to zero. This occurs when [latex]x+1=0[\/latex] and when [latex]x - 2=0[\/latex], giving us vertical asymptotes at [latex]x=-1[\/latex] and [latex]x=2[\/latex].<\/p>\n<p>There are no common factors in the numerator and denominator. This means there are no removable discontinuities.<\/p>\n<p>Finally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal asymptote at [latex]y=0[\/latex].<\/p>\n<p>To sketch the graph, we might start by plotting the three intercepts. Since the graph has no [latex]x[\/latex]-intercepts between the vertical asymptotes, and the [latex]y[\/latex]-intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213952\/CNX_Precalc_Figure_03_07_0212.jpg\" alt=\"Graph of only the middle portion of f(x)=(x+2)(x-3)\/(x+1)^2(x-2) with its intercepts at (-2, 0), (0, 3), and (3, 0).\" width=\"487\" height=\"440\" \/><\/p>\n<p>The factor associated with the vertical asymptote at [latex]x=-1[\/latex] was squared, so we know the behavior will be the same on both sides of the asymptote. The graph heads toward positive infinity as the inputs approach the asymptote on the right, so the graph will head toward positive infinity on the left as well.<\/p>\n<p>For the vertical asymptote at [latex]x=2[\/latex], the factor was not squared, so the graph will have opposite behavior on either side of the asymptote. After passing through the [latex]x[\/latex]-intercepts, the graph will then level off toward an output of zero, as indicated by the horizontal asymptote.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213954\/CNX_Precalc_Figure_03_07_022.jpg\" alt=\"Graph of f(x)=(x+2)(x-3)\/(x+1)^2(x-2) with its vertical asymptotes at x=-1 and x=2, its horizontal asymptote at y=0, and its intercepts at (-2, 0), (0, 3), and (3, 0).\" width=\"487\" height=\"439\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Given the function [latex]f\\left(x\\right)=\\dfrac{{\\left(x+2\\right)}^{2}\\left(x - 2\\right)}{2{\\left(x - 1\\right)}^{2}\\left(x - 3\\right)}[\/latex], use the characteristics of polynomials and rational functions to describe its behavior and sketch the function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q451047\">Show Solution<\/span><\/p>\n<div id=\"q451047\" class=\"hidden-answer\" style=\"display: none\">\n<p>Horizontal asymptote at [latex]y=\\frac{1}{2}[\/latex]. Vertical asymptotes at [latex]x=1[\/latex] and [latex]x=3[\/latex]. [latex]y[\/latex]-intercept at [latex]\\left(0,\\frac{4}{3}.\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135168380\"><em>x<\/em>-intercepts at [latex]\\left(2,0\\right) \\text{ and }\\left(-2,0\\right)[\/latex]. [latex]\\left(-2,0\\right)[\/latex] is a zero with multiplicity 2, and the graph bounces off the [latex]x[\/latex]-axis at this point. [latex]\\left(2,0\\right)[\/latex] is a single zero and the graph crosses the axis at this point.<span id=\"fs-id1165137745200\"><br \/>\n<\/span><\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/12\/02213511\/CNX_Precalc_Figure_03_07_023.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-2947\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/12\/02213511\/CNX_Precalc_Figure_03_07_023.jpg\" alt=\"cnx_precalc_figure_03_07_023\" width=\"731\" height=\"477\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom20\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129075&amp;theme=oea&amp;iframe_resize_id=mom20\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<p>Watch the following video to see another worked example of how to match different kinds of rational functions with their graphs.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex:  Match Equations of Rational Functions to Graphs\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/vMVYaFptvkk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Writing Rational Functions<\/h2>\n<p>Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. A rational function written in factored form will have an [latex]x[\/latex]-intercept where each factor of the numerator is equal to zero. (An exception occurs in the case of a removable discontinuity.) As a result, we can form a numerator of a function whose graph will pass through a set of [latex]x[\/latex]-intercepts by introducing a corresponding set of factors. Likewise, because the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Writing Rational Functions from Intercepts and Asymptotes<\/h3>\n<p>If a <strong>rational function<\/strong> has [latex]x[\/latex]-intercepts at [latex]x={x}_{1}, {x}_{2}, ..., {x}_{n}[\/latex], vertical asymptotes at [latex]x={v}_{1},{v}_{2},\\dots ,{v}_{m}[\/latex], and no [latex]{x}_{i}=\\text{any }{v}_{j}[\/latex], then the function can be written in the form:<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=a\\frac{{\\left(x-{x}_{1}\\right)}^{{p}_{1}}{\\left(x-{x}_{2}\\right)}^{{p}_{2}}\\cdots {\\left(x-{x}_{n}\\right)}^{{p}_{n}}}{{\\left(x-{v}_{1}\\right)}^{{q}_{1}}{\\left(x-{v}_{2}\\right)}^{{q}_{2}}\\cdots {\\left(x-{v}_{m}\\right)}^{{q}_{n}}}[\/latex]<\/p>\n<p>where the powers [latex]{p}_{i}[\/latex] or [latex]{q}_{i}[\/latex] on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor [latex]a[\/latex]<em>\u00a0<\/em>can be determined given a value of the function other than the [latex]x[\/latex]-intercept or by the horizontal asymptote if it is nonzero.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a graph of a rational function, write the function.<\/h3>\n<ol>\n<li>Determine the factors of the numerator. Examine the behavior of the graph at the <em>x<\/em>-intercepts to determine the zeroes and their multiplicities. (This is easy to do when finding the &#8220;simplest&#8221; function with small multiplicities\u2014such as 1 or 3\u2014but may be difficult for larger multiplicities\u2014such as 5 or 7, for example.)<\/li>\n<li>Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote to determine the factors and their powers.<\/li>\n<li>Use any clear point on the graph to find the stretch factor.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing a Rational Function from Intercepts and Asymptotes<\/h3>\n<p>Write an equation for the rational function\u00a0below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213956\/CNX_Precalc_Figure_03_07_024.jpg\" alt=\"Graph of a rational function.\" width=\"487\" height=\"475\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q666867\">Show Solution<\/span><\/p>\n<div id=\"q666867\" class=\"hidden-answer\" style=\"display: none\">\n<p>The graph appears to have [latex]x[\/latex]-intercepts at [latex]x=-2[\/latex] and [latex]x=3[\/latex]. At both, the graph passes through the intercept, suggesting linear factors. The graph has two vertical asymptotes. The one at [latex]x=-1[\/latex] seems to exhibit the basic behavior similar to [latex]\\frac{1}{x}[\/latex], with the graph heading toward positive infinity on one side and heading toward negative infinity on the other. The asymptote at [latex]x=2[\/latex] is exhibiting a behavior similar to [latex]\\frac{1}{{x}^{2}}[\/latex], with the graph heading toward negative infinity on both sides of the asymptote.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213958\/CNX_Precalc_Figure_03_07_025.jpg\" alt=\"Graph of a rational function denoting its vertical asymptotes and x-intercepts.\" width=\"731\" height=\"475\" \/><\/p>\n<p>We can use this information to write a function of the form<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=a\\dfrac{\\left(x+2\\right)\\left(x - 3\\right)}{\\left(x+1\\right){\\left(x - 2\\right)}^{2}}[\/latex].<\/p>\n<p>To find the stretch factor, we can use another clear point on the graph, such as the [latex]y[\/latex]-intercept [latex]\\left(0,-2\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-2&=a\\dfrac{\\left(0+2\\right)\\left(0 - 3\\right)}{\\left(0+1\\right){\\left(0 - 2\\right)}^{2}} \\\\[1mm] -2&=a\\frac{-6}{4} \\\\[1mm] a=\\frac{-8}{-6}=\\frac{4}{3} \\end{align}[\/latex]<\/p>\n<p>This gives us a final function of [latex]f\\left(x\\right)=\\dfrac{4\\left(x+2\\right)\\left(x - 3\\right)}{3\\left(x+1\\right){\\left(x - 2\\right)}^{2}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom25\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1802&amp;theme=oea&amp;iframe_resize_id=mom25\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<section id=\"fs-id1165137659195\" class=\"key-equations\">\n<h2>Key Equations<\/h2>\n<table id=\"eip-id1362369\" summary=\"..\">\n<tbody>\n<tr>\n<td>Rational Function<\/td>\n<td>[latex]f\\left(x\\right)=\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{a}_{p}{x}^{p}+{a}_{p - 1}{x}^{p - 1}+...+{a}_{1}x+{a}_{0}}{{b}_{q}{x}^{q}+{b}_{q - 1}{x}^{q - 1}+...+{b}_{1}x+{b}_{0}}, Q\\left(x\\right)\\ne 0[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<section id=\"fs-id1165137793507\" class=\"key-concepts\">\n<h2>Key Concepts<\/h2>\n<ul id=\"fs-id1165137603314\">\n<li>We can use arrow notation to describe local behavior and end behavior of the toolkit functions [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex].<\/li>\n<li>A function that levels off at a horizontal value has a horizontal asymptote. A function can have more than one vertical asymptote.<\/li>\n<li>Application problems involving rates and concentrations often involve rational functions.<\/li>\n<li>The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.<\/li>\n<li>The vertical asymptotes of a rational function will occur where the denominator of the function is equal to zero and the numerator is not zero.<\/li>\n<li>A removable discontinuity might occur in the graph of a rational function if an input causes both numerator and denominator to be zero.<\/li>\n<li>A rational function\u2019s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions.<\/li>\n<li>Graph rational functions by finding the intercepts, behavior at the intercepts and asymptotes, and end behavior.<\/li>\n<li>If a rational function has <em>x<\/em>-intercepts at [latex]x={x}_{1},{x}_{2},\\dots ,{x}_{n}[\/latex], vertical asymptotes at [latex]x={v}_{1},{v}_{2},\\dots ,{v}_{m}[\/latex], and no [latex]{x}_{i}=\\text{any }{v}_{j}[\/latex], then the function can be written in the form\u00a0[latex]f\\left(x\\right)=a\\dfrac{{\\left(x-{x}_{1}\\right)}^{{p}_{1}}{\\left(x-{x}_{2}\\right)}^{{p}_{2}}\\cdots {\\left(x-{x}_{n}\\right)}^{{p}_{n}}}{{\\left(x-{v}_{1}\\right)}^{{q}_{1}}{\\left(x-{v}_{2}\\right)}^{{q}_{2}}\\cdots {\\left(x-{v}_{m}\\right)}^{{q}_{n}}}[\/latex]<\/li>\n<\/ul>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137758530\" class=\"definition\">\n<dt><strong>arrow notation<\/strong><\/dt>\n<dd id=\"fs-id1165135154402\">a way to symbolically represent the local and end behavior of a function by using arrows to indicate that an input or output approaches a value<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135154407\" class=\"definition\">\n<dt><strong>horizontal asymptote<\/strong><\/dt>\n<dd id=\"fs-id1165135154413\">a horizontal line [latex]y=b[\/latex] where the graph approaches the line as the inputs increase or decrease without bound.<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135192626\" class=\"definition\">\n<dt><strong>rational function<\/strong><\/dt>\n<dd id=\"fs-id1165134401081\">a function that can be written as the ratio of two polynomials<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134401085\" class=\"definition\">\n<dt><strong>removable discontinuity<\/strong><\/dt>\n<dd id=\"fs-id1165134401090\">a single point at which a function is undefined that, if filled in, would make the function continuous; it appears as a hole on the graph of a function<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137426312\" class=\"definition\">\n<dt><strong>vertical asymptote<\/strong><\/dt>\n<dd id=\"fs-id1165137426317\">a vertical line [latex]x=a[\/latex] where the graph tends toward positive or negative infinity as the inputs approach [latex]a[\/latex]<\/dd>\n<\/dl>\n<\/div>\n<\/section>\n","protected":false},"author":167848,"menu_order":17,"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-4995","chapter","type-chapter","status-publish","hentry"],"part":764,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/chapters\/4995","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/wp\/v2\/users\/167848"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/chapters\/4995\/revisions"}],"predecessor-version":[{"id":5079,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/chapters\/4995\/revisions\/5079"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/parts\/764"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/chapters\/4995\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/wp\/v2\/media?parent=4995"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/chapter-type?post=4995"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/wp\/v2\/contributor?post=4995"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/wp\/v2\/license?post=4995"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}