{"id":1405,"date":"2019-03-07T15:34:21","date_gmt":"2019-03-07T15:34:21","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/rational-functions\/"},"modified":"2025-03-31T19:12:59","modified_gmt":"2025-03-31T19:12:59","slug":"rational-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/rational-functions\/","title":{"raw":"4.3 Rational Functions","rendered":"4.3 Rational Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nIn this section, you will:\r\n<ul>\r\n \t<li>Use arrow notation.<\/li>\r\n \t<li>Find the domain of rational functions.<\/li>\r\n \t<li>Identify horizontal and vertical asymptotes and removable discontinuities.<\/li>\r\n \t<li>Graph rational functions.<\/li>\r\n \t<li>Determine the formula for the graph of a rational function.<\/li>\r\n \t<li>Solve applied problems involving rational functions.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165137740975\">Suppose we know that the cost of making a product is dependent on the number of items, [latex]x,[\/latex] produced. and that it 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] items, we would divide the cost function by the number of items, [latex]x.[\/latex]<\/p>\r\n<p id=\"fs-id1165137768452\">The average cost function, which yields the average cost per item for [latex]x[\/latex] items produced, is<\/p>\r\n\r\n<div id=\"eip-634\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{15,000x-0.1{x}^{2}+1000}{x}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137863708\">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>\r\n<p id=\"fs-id1165133305345\">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 can be thought of as the ratio of two polynomial functions.<\/p>\r\n\r\n<div id=\"fs-id1165135397235\" class=\"bc-section section\">\r\n<h3>Using Arrow Notation<\/h3>\r\n<p id=\"fs-id1165137659469\">We have seen the graphs of the\u00a0<span class=\"no-emphasis\">reciprocal function<\/span> and the squared reciprocal function from our study of toolkit functions. Examine these graphs, as shown in <a class=\"autogenerated-content\" href=\"#Figure_03_07_001\">Figure 1<\/a>, and notice some of their features.<\/p>\r\n\r\n<div id=\"Figure_03_07_001\" class=\"medium\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"565\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153218\/CNX_Precalc_Figure_03_07_001.jpg\" alt=\"Graphs of f(x)=1\/x and f(x)=1\/x^2\" width=\"565\" height=\"350\" \/> Figure 1[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137935728\">Several things are apparent if we examine the graph of [latex]f\\left(x\\right)=\\frac{1}{x}.[\/latex]<\/p>\r\n\r\n<ol id=\"fs-id1165135438444\" type=\"1\">\r\n \t<li>On the left branch of the graph, the curve approaches the <em>x<\/em>-axis [latex]\\left(y=0\\right) \\textrm{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 <em>x-<\/em>axis [latex]\\left(y=0\\right) \\textrm{as } x\\to \\infty .[\/latex]<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1165137461553\">To summarize, we use arrow notation to show that [latex]x[\/latex] or [latex]f\\left(x\\right)[\/latex] is approaching a particular value. See <a class=\"autogenerated-content\" href=\"#Table_03_07_001\">Table 1<\/a>.<\/p>\r\n\r\n<table id=\"Table_03_07_001\" summary=\"..\"><caption>Table 1:\u00a0 Arrow Notation<\/caption>\r\n<thead>\r\n<tr>\r\n<th class=\"border\">Symbol<\/th>\r\n<th class=\"border\">Meaning<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\">[latex]x\\to {a}^{-}[\/latex]<\/td>\r\n<td class=\"border\">[latex]x[\/latex] approaches [latex]a[\/latex] from the left ([latex]x \\lt a[\/latex] but is increasing and getting closer and closer to\u00a0 [latex]a[\/latex])<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">[latex]x\\to {a}^{+}[\/latex]<\/td>\r\n<td class=\"border\">[latex]x[\/latex] approaches [latex]a[\/latex] from the right ([latex]x&gt;a[\/latex] but is decreasing and getting closer and closer to [latex]a[\/latex])<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">[latex]x\\to a[\/latex]<\/td>\r\n<td class=\"border\">the input approaches [latex]a[\/latex] from both sides<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">[latex]x\\to \\infty [\/latex]<\/td>\r\n<td class=\"border\">[latex]x[\/latex] goes toward infinity ([latex]x[\/latex] increases without bound)<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">[latex]x\\to -\\infty [\/latex]<\/td>\r\n<td class=\"border\">[latex]x[\/latex] goes toward negative infinity ([latex]x[\/latex] decreases without bound)<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">[latex]f\\left(x\\right)\\to \\infty [\/latex]<\/td>\r\n<td class=\"border\">the output goes toward infinity (the output increases without bound)<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">[latex]f\\left(x\\right)\\to -\\infty [\/latex]<\/td>\r\n<td class=\"border\">the output goes toward negative infinity (the output decreases without bound)<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">[latex]f\\left(x\\right)\\to a[\/latex]<\/td>\r\n<td class=\"border\">the output approaches [latex]a[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-id1165137759950\" class=\"bc-section section\">\r\n<h4>Local Behavior of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/h4>\r\n<p id=\"fs-id1165137755329\">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 (negative values very close to zero), the function values decrease without bound. We can see this behavior in <a class=\"autogenerated-content\" href=\"#Table_03_07_002\">Table 2<\/a>.<\/p>\r\n\r\n<table id=\"Table_03_07_002\" summary=\"..\"><caption>Table 2<\/caption>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\"><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td class=\"border\" style=\"text-align: center;\">\u20130.1<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">\u20130.01<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">\u20130.001<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">\u20130.0001<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\"><strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/strong><\/td>\r\n<td class=\"border\" style=\"text-align: center;\">\u201310<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">\u2013100<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">\u20131000<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">\u201310,000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165137542511\">We write in arrow notation, as [latex]x\\to {0}^{-},f\\left(x\\right)\\to -\\infty [\/latex].<\/p>\r\n<p id=\"fs-id1165137506235\">As the input values approach zero from the right side (positive values very close to zero), the function values increase without bound. We can see this behavior in <a class=\"autogenerated-content\" href=\"#Table_03_07_003\">Table 3<\/a>.<\/p>\r\n\r\n<table id=\"Table_03_07_003\" summary=\"..\"><caption>Table 3<\/caption>\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td class=\"border\" style=\"height: 14px; width: 232px;\"><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td class=\"border\" style=\"height: 14px; width: 18px; text-align: center;\">0.1<\/td>\r\n<td class=\"border\" style=\"height: 14px; width: 25px; text-align: center;\">0.01<\/td>\r\n<td class=\"border\" style=\"height: 14px; width: 32px; text-align: center;\">0.001<\/td>\r\n<td class=\"border\" style=\"height: 14px; width: 39px; text-align: center;\">0.0001<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td class=\"border\" style=\"height: 14px; width: 232px;\"><strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/strong><\/td>\r\n<td class=\"border\" style=\"height: 14px; width: 18px; text-align: center;\">10<\/td>\r\n<td class=\"border\" style=\"height: 14px; width: 25px; text-align: center;\">100<\/td>\r\n<td class=\"border\" style=\"height: 14px; width: 32px; text-align: center;\">1000<\/td>\r\n<td class=\"border\" style=\"height: 14px; width: 39px; text-align: center;\">10,000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165134338836\">We write in arrow notation as [latex]x\\to {0}^{+}, f\\left(x\\right)\\to \\infty .[\/latex]<\/p>\r\n\r\n<div><\/div>\r\n<p id=\"fs-id1165135397958\">See <a class=\"autogenerated-content\" href=\"#Figure_03_07_002\">Figure 2<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_002\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"463\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153221\/CNX_Precalc_Figure_03_07_002.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=\"463\" height=\"300\" \/> Figure 2[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137648090\">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] as the input becomes close to zero. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_003\">Figure 3<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_003\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"399\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153224\/CNX_Precalc_Figure_03_07_003.jpg\" alt=\"Graph of f(x)=1\/x with its vertical asymptote at x=0.\" width=\"399\" height=\"298\" \/> Figure 3[\/caption]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137732344\">\r\n<div class=\"textbox definitions\">\r\n<h3>Definition<\/h3>\r\n<p id=\"fs-id1165137561740\">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>\r\nfrom the left side,\r\n<p style=\"text-align: center;\">as [latex]x\\to a^{-},f\\left(x\\right)\\to \\infty ,[\/latex] or as\u00a0<span style=\"background-color: initial;\">[latex]x\\to a^{-},f\\left(x\\right)\\to -\\infty,[\/latex] and<\/span><\/p>\r\nfrom the right side,\r\n<p style=\"text-align: center;\">[latex]x\\to a^{+},f\\left(x\\right)\\to \\infty,[\/latex] \u00a0or as [latex]x\\to a^{+},f\\left(x\\right)\\to -\\infty .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137502160\" class=\"bc-section section\">\r\n<h4>End Behavior of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/h4>\r\n<p id=\"fs-id1165137408019\">As the values of [latex]x[\/latex] increase without bound, the function values approach 0. As the values of [latex]x[\/latex] decrease without bound, the function values approach 0. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_004\">Figure 4<\/a>. Symbolically, using arrow notation<\/p>\r\n<p id=\"fs-id1165137565255\" style=\"text-align: center;\">[latex]\\textrm{as }x\\to \\infty ,f\\left(x\\right)\\to 0,\\textrm{and as }x\\to -\\infty ,f\\left(x\\right)\\to 0.[\/latex]<\/p>\r\n\r\n<div id=\"Figure_03_07_004\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"465\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153227\/CNX_Precalc_Figure_03_07_004.jpg\" alt=\"Graph of f(x)=1\/x which highlights the segments of the turning points to denote their end behavior.\" width=\"465\" height=\"302\" \/> Figure 4[\/caption]\r\n\r\n<\/div>\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] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_005\">Figure 5<\/a><strong>.<\/strong>\r\n<div id=\"Figure_03_07_005\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"401\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153230\/CNX_Precalc_Figure_03_07_005.jpg\" alt=\"Graph of f(x)=1\/x with its vertical asymptote at x=0 and its horizontal asymptote at y=0.\" width=\"401\" height=\"300\" \/> Figure 5[\/caption]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137901226\">\r\n<div class=\"textbox definitions\">\r\n<h3>Definition<\/h3>\r\n<p id=\"fs-id1165137782455\">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 [latex]\\textrm{as }x\\to \\infty \\textrm{ or }x\\to -\\infty ,\\text{ }f\\left(x\\right)\\to b.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_03_07_01\" class=\"textbox examples\">\r\n<div id=\"fs-id1165133213902\">\r\n<div id=\"fs-id1165137657454\">\r\n<h3>Example 1:\u00a0 Using Arrow Notation<\/h3>\r\n<p id=\"fs-id1165137437543\">Use arrow notation to describe the end behavior and local behavior of the function graphed in <a class=\"autogenerated-content\" href=\"#Figure_03_07_006\">Figure 6<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_006\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"307\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153233\/CNX_Precalc_Figure_03_07_006.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=\"307\" height=\"301\" \/> Figure 6[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137870943\">[reveal-answer q=\"fs-id1165137870943\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137870943\"]\r\n<p id=\"fs-id1165137851860\">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>\r\n\r\n<div id=\"eip-id1165134070637\" class=\"unnumbered\" style=\"text-align: center;\">As [latex]x\\to {2}^{-},f\\left(x\\right)\\to -\\infty ,[\/latex] and as [latex]x\\to {2}^{+},\\text{ }f\\left(x\\right)\\to \\infty .[\/latex]<\/div>\r\n<p id=\"fs-id1165137696383\">Also, as the inputs decrease without bound, the graph appears to be leveling off with output values closer and closer to 4, indicating a horizontal asymptote at [latex]y=4.[\/latex] As the inputs increase without bound, the graph levels off approaching 4.<\/p>\r\n\r\n<div id=\"eip-id1165132961960\" class=\"unnumbered\" 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][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137597363\" class=\"precalculus tryit\">\r\n<h3>Try it #1<\/h3>\r\n<div id=\"ti_03_07_01\">\r\n<div id=\"fs-id1165137431673\">\r\n\r\nUse arrow notation to describe the end behavior and local behavior for the reciprocal squared function.\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135541751\">[reveal-answer q=\"fs-id1165135541751\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135541751\"]\r\n<p id=\"fs-id1165137828072\">End behavior: as [latex]x\\to \u00b1\\infty , f\\left(x\\right)\\to 0;[\/latex] There is a horizontal asymptote at\u00a0 [latex]y=0.[\/latex]<\/p>\r\nLocal behavior: as [latex]x\\to 0, f\\left(x\\right)\\to \\infty [\/latex] There is a vertical asymptote at [latex]x=0.[\/latex] There are no <em>x<\/em>- or <em>y<\/em>-intercepts.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_03_07_02\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137694119\">\r\n<div id=\"fs-id1165137694121\">\r\n<h3>Example 2:\u00a0 Using Transformations to Graph a Rational Function<\/h3>\r\n<p id=\"fs-id1165137640093\">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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137911752\">[reveal-answer q=\"fs-id1165137911752\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137911752\"]\r\n<p id=\"fs-id1165137911755\">Shifting the graph left 2 and up 3 would result in the function<\/p>\r\n\r\n<div id=\"eip-id1165135547466\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{1}{x+2}+3,[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137640711\">or equivalently, by adding the terms after finding a common denominator,<\/p>\r\n\r\n<div id=\"eip-id1165131954096\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{3x+7}{x+2}.[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137446966\">The graph of the shifted function is displayed in <a class=\"autogenerated-content\" href=\"#Figure_03_07_007\">Figure 7<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_007\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"499\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153236\/CNX_Precalc_Figure_03_07_007.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=\"499\" height=\"301\" \/> Figure 7[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137891390\">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>\r\n\r\n<div id=\"eip-id1165134252833\" class=\"unnumbered\" style=\"text-align: center;\">As [latex]x\\to -{2}^{-}, f\\left(x\\right)\\to -\\infty ,[\/latex] and as [latex]x\\to -{2}^{+}, f\\left(x\\right)\\to \\infty .[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137736971\">As the inputs increase and decrease without bound, the graph appears to be leveling off with output values getting closer and closer to 3, indicating a horizontal asymptote of [latex]y=3.[\/latex]<\/p>\r\n\r\n<div id=\"eip-id1165134250660\" class=\"unnumbered\" style=\"text-align: center;\">As [latex]x\\to \u00b1\\infty , f\\left(x\\right)\\to 3.[\/latex]<\/div>\r\n<div id=\"fs-id1165137889779\">\r\n<h3>Analysis<\/h3>\r\n<p id=\"fs-id1165137401110\">Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137434563\" class=\"precalculus tryit\">\r\n<h3>Try it #2<\/h3>\r\n<div id=\"ti_03_07_02\">\r\n<div id=\"fs-id1165137824780\">\r\n<p id=\"fs-id1165137824781\">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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137812954\">[reveal-answer q=\"fs-id1165137812954\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137812954\"]\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"404\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153239\/CNX_Precalc_Figure_03_07_008.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=\"404\" height=\"303\" \/> Figure 8[\/caption]\r\n<p id=\"fs-id1165137416225\">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 \u00b1\\infty ,f\\left(x\\right)\\to -4.[\/latex] Therefore, the vertical asymptote is\u00a0 [latex]x=3,[\/latex] and the horizontal asymptote is\u00a0 [latex]y=-4.[\/latex]<\/p>\r\n<p id=\"fs-id1165137823960\">The function is [latex]f\\left(x\\right)=\\frac{1}{{\\left(x-3\\right)}^{2}}-4.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137534932\" class=\"bc-section section\">\r\n<h3>Solving Applied Problems Involving Rational Functions<\/h3>\r\n<p id=\"fs-id1165137639549\">In <a class=\"autogenerated-content\" href=\"#Example_03_07_02\">Example 2<\/a>, we shifted a toolkit function in a way that resulted in the function [latex]f\\left(x\\right)=\\frac{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>\r\n\r\n<div id=\"fs-id1165137386884\">\r\n<div class=\"textbox definitions\">\r\n<h3>Definition<\/h3>\r\n<p id=\"fs-id1165137754871\">A <strong>rational function<\/strong> is a function that can be written as the quotient of two polynomial functions [latex]P\\left(x\\right) \\textrm{and } Q\\left(x\\right).[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165137411279\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\displaystyle{\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{{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]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_03_07_03\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137536326\">\r\n<div id=\"fs-id1165137536328\">\r\n<h3>Example 3:\u00a0 Solving an Applied Problem Involving a Rational Function<\/h3>\r\n<p id=\"fs-id1165137694204\">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>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137526522\">[reveal-answer q=\"fs-id1165137526522\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137526522\"]\r\n<p id=\"fs-id1165137733797\">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>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{water: }W\\left(t\\right)=100+10t\\text{ in gallons}\\\\ \\text{sugar: }S\\left(t\\right)=5+1t\\text{ in pounds}\\end{array}[\/latex]<\/p>\r\n[latex]\\\\[\/latex]\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">The concentration, [latex]C,[\/latex] will be the ratio of pounds of sugar to gallons of water.<\/span>\r\n<p style=\"text-align: center;\"><span style=\"text-align: center;\">[latex]C\\left(t\\right)=\\frac{5+t}{100+10t}[\/latex]<\/span><\/p>\r\n[latex]\\\\[\/latex]\r\n\r\n<span style=\"text-align: left;\">The concentration after 12 minutes is given by evaluating [latex]C\\left(t\\right)[\/latex] at [latex]t=\\text{ }12.[\/latex]<\/span>\r\n<p style=\"text-align: center;\"><span style=\"text-align: center; font-size: 0.9em;\">[latex]C\\left(12\\right)=\\frac{5+12}{100+10\\left(12\\right)} \\text{ }=\\frac{17}{220}[\/latex]<\/span><\/p>\r\n[latex]\\\\[\/latex]\r\n\r\nThis means the concentration is 17 pounds of sugar to 220 gallons of water.\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">At the beginning, the concentration is<\/span>\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: center;\">[latex]C\\left(0\\right)=\\frac{5+0}{100+10\\left(0\\right)}\\text{ }=\\frac{1}{20}[\/latex]<\/span><\/p>\r\n[latex]\\\\[\/latex]\r\n\r\n<span style=\"font-size: 1rem;\">Since [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.<\/span>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135194535\" class=\"precalculus tryit\">\r\n<h3>Try it #3<\/h3>\r\n<div id=\"ti_03_07_03\">\r\n<div id=\"fs-id1165137640599\">\r\n<p id=\"fs-id1165137436098\">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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137644596\">[reveal-answer q=\"fs-id1165137644596\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137644596\"]\r\n<p id=\"fs-id1165137644598\">[latex]\\frac{12}{11}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bc-section section\">\r\n<h3>Finding the Domains of Rational Functions<\/h3>\r\n<p id=\"fs-id1165137530059\">A <span class=\"no-emphasis\">vertical asymptote<\/span> represents a value at which a rational function is undefined, so that value is not in the domain of the function. A rational 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.\u00a0The <strong>domain of a rational function<\/strong> includes all real numbers except those that cause the denominator to equal zero.<\/p>\r\n\r\n<div id=\"fs-id1165135530461\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137695173\"><strong>Given a rational function, find the domain.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137724163\" type=\"1\">\r\n \t<li>Set the denominator equal to zero.<\/li>\r\n \t<li>Solve to find the <em>x<\/em>-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.\u00a0 Express your answer using interval notation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_07_04\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135587815\">\r\n<div id=\"fs-id1165137647179\">\r\n<h3>Example 4:\u00a0 Finding the Domain of a Rational Function<\/h3>\r\n<p id=\"fs-id1165137501029\">Find the domain of [latex]f\\left(x\\right)=\\frac{x+3}{{x}^{2}-9}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137656082\">[reveal-answer q=\"fs-id1165137656082\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137656082\"]\r\n<p id=\"fs-id1165135445735\">Begin by setting the denominator equal to zero and solving.<\/p>\r\n\r\n<div id=\"eip-id1165134254379\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*} {x}^{2}-9&amp;=0\\\\ {x}^{2}&amp;=9 \\\\ x&amp;=\u00b13\\end{align*}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<div class=\"unnumbered\"><\/div>\r\n<p id=\"fs-id1165137642958\">The denominator is equal to zero when [latex]x=\u00b13.[\/latex] The domain of the function is all real numbers except [latex]x=\u00b13.[\/latex] In interval notation we write,\u00a0 [latex]\\left(-\\infty, -3\\right)\\cup\\left(-3, 3\\right)\\cup\\left(3,\\infty\\right).[\/latex]<\/p>\r\n\r\n<h3>Analysis<\/h3>\r\n<p id=\"fs-id1165133276227\">A graph of this function, as shown in <a class=\"autogenerated-content\" href=\"#Figure_03_07_009\">Figure 9<\/a>, confirms that the function is not defined when [latex]x=\u00b13.[\/latex]<\/p>\r\n\r\n<div id=\"Figure_03_07_009\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"408\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153241\/CNX_Precalc_Figure_03_07_009.jpg\" alt=\"Graph of f(x)=1\/(x-3) with its vertical asymptote at x=3 and its horizontal asymptote at y=0.\" width=\"408\" height=\"305\" \/> Figure 9[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137455163\">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>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134192966\" class=\"precalculus tryit\">\r\n<h3>Try it #4<\/h3>\r\n<div id=\"ti_03_07_04\">\r\n<div id=\"fs-id1165137414038\">\r\n\r\nFind the domain of [latex]f\\left(x\\right)=\\frac{4x}{5\\left(x-1\\right)\\left(x-5\\right)}.[\/latex]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137474960\">[reveal-answer q=\"fs-id1165137474960\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137474960\"]\r\n<p id=\"fs-id1165137474961\">The domain is all real numbers except [latex]x=1[\/latex] and [latex]x=5[\/latex] or [latex]\\left(-\\infty, 1\\right)\\cup\\left(1, 5\\right)\\cup\\left(5,\\infty\\right).[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137836721\" class=\"bc-section section\">\r\n<h3>Identifying Vertical Asymptotes and Removable Discontinuities of Rational Functions<\/h3>\r\n<p id=\"fs-id1165135439868\">By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are vertical 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 vertical asymptotes, and calculate their location.<\/p>\r\n\r\n<div id=\"fs-id1165135194731\" class=\"bc-section section\">\r\n<div class=\"textbox\">\r\n\r\nThe <strong>vertical asymptotes<\/strong> of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator.\u00a0 Vertical asymptotes occur at the zeros of such factors.\r\n\r\n<\/div>\r\n<div class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137451766\"><strong>Given a rational function, identify any vertical asymptotes and removable discontinuties of its graph.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165134079627\" type=\"1\">\r\n \t<li>Factor the numerator and denominator.<\/li>\r\n \t<li>List values not 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>Any values that cause the denominator to be zero in this simplified version are where the vertical asymptotes occur.<\/li>\r\n \t<li>Any remaining values not in the domain are removable discontinuities also known as holes.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_07_05\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137937688\">\r\n<div id=\"fs-id1165137645463\">\r\n<h3>Example 5:\u00a0 Identifying Vertical Asymptotes<\/h3>\r\n<p id=\"fs-id1165137627104\">Find the vertical asymptotes of the graph of [latex]k\\left(x\\right)=\\frac{5+2{x}^{2}}{2-x-{x}^{2}}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137389408\">[reveal-answer q=\"fs-id1165137389408\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137389408\"]\r\n<p id=\"fs-id1165137601646\">First, factor the numerator and denominator.<\/p>\r\n\r\n<div id=\"eip-id1165133027629\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}k\\left(x\\right)&amp;=\\frac{5+2{x}^{2}}{2-x-{x}^{2}} \\\\ \\text{ }&amp;=\\frac{5+2{x}^{2}}{\\left(2+x\\right)\\left(1-x\\right)}\\end{align*}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137745213\">To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero and solving:<\/p>\r\n\r\n<div id=\"eip-id1165132921420\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(2+x\\right)\\left(1-x\\right)=0\\hfill \\\\ \\text{ }x=-2,1\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165135241250\">Neither [latex]x=\u20132[\/latex] nor [latex]x=1[\/latex] are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in <a class=\"autogenerated-content\" href=\"#Figure_03_07_010\">Figure 10<\/a> confirms the location of the two vertical asymptotes.<\/p>\r\n\r\n<div id=\"Figure_03_07_010\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"310\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153245\/CNX_Precalc_Figure_03_07_010.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=\"310\" height=\"327\" \/> Figure 10[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137640086\" class=\"bc-section section\">\r\n\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 removable discontinuity.\u00a0 These often cannot be seen when technology is used to create a graph.\r\n<p id=\"fs-id1165137470947\">For example, the function [latex]f\\left(x\\right)=\\frac{{x}^{2}-1}{{x}^{2}-2x-3}[\/latex] may be re-written by factoring the numerator and the denominator.<\/p>\r\n\r\n<div id=\"eip-589\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{\\left(x+1\\right)\\left(x-1\\right)}{\\left(x+1\\right)\\left(x-3\\right)}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137470356\">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 NOT in the domain and 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. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_011\">Figure 11<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_011\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"412\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153247\/CNX_Precalc_Figure_03_07_011.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=\"412\" height=\"276\" \/> Figure 11[\/caption]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137891255\">\r\n<p id=\"fs-id1165137558555\">A <strong>removable discontinuity<\/strong> occurs in the graph of a rational function at [latex]x=a[\/latex] if [latex]a[\/latex] is a zero for a factor that is both in the denominator and 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 may be the location of a removable discontinuity. This is a removable discontinuity if the multiplicity of this factor in the numerator 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 a vertical asymptote at that value.<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_03_07_06\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135168126\">\r\n<div id=\"fs-id1165137807554\">\r\n<h3>Example 6:\u00a0 Identifying Vertical Asymptotes and Removable Discontinuities for a Graph<\/h3>\r\n<p id=\"fs-id1165137727527\">Find the vertical asymptotes and removable discontinuities of the graph of [latex]k\\left(x\\right)=\\frac{x-2}{{x}^{2}-4}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137425735\">[reveal-answer q=\"fs-id1165137425735\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137425735\"]\r\n<p id=\"fs-id1165137425737\">Factor the numerator and the denominator.<\/p>\r\n\r\n<div id=\"eip-id1165134315756\" class=\"unnumbered\" style=\"text-align: center;\">[latex]k\\left(x\\right)=\\frac{x-2}{\\left(x-2\\right)\\left(x+2\\right)}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165135309767\">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>\r\n<p id=\"fs-id1165137550074\">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] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_012\">Figure 12<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_012\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"401\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153250\/CNX_Precalc_Figure_03_07_012.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=\"401\" height=\"300\" \/> Figure 12[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165135634122\">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>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"precalculus tryit\">\r\n<h3>Try it #5<\/h3>\r\n<div id=\"ti_03_07_05\">\r\n<div id=\"fs-id1165135686722\">\r\n<p id=\"fs-id1165135686724\">Find the vertical asymptotes and removable discontinuities of the graph of<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{{x}^{2}-25}{{x}^{3}-6{x}^{2}+5x}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135203443\">[reveal-answer q=\"fs-id1165135203443\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135203443\"]\r\n<p id=\"fs-id1165135203445\">Removable discontinuity at [latex]x=5.[\/latex] Vertical asymptotes: [latex]x=0,\\text{ }x=1.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134222349\" class=\"bc-section section\">\r\n<h3>Identifying Horizontal Asymptotes of Rational Functions<\/h3>\r\n<p id=\"fs-id1165135501072\">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>\r\n<p id=\"fs-id1165137503143\">There are three distinct outcomes when checking for horizontal asymptotes:<\/p>\r\n<p id=\"fs-id1165137503146\"><strong>Case 1:<\/strong> If the degree of the denominator is greater than the degree of the numerator, there is a <span class=\"no-emphasis\">horizontal asymptote<\/span> at [latex]y=0.[\/latex]<\/p>\r\n\r\n<div id=\"eip-83\" class=\"unnumbered\" style=\"text-align: center;\">Example:\u00a0 [latex]f\\left(x\\right)=\\frac{4x+2}{{x}^{2}+4x-5}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165135530372\">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] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_013\">Figure 13<\/a>. Note that this graph crosses the horizontal asymptote.<\/p>\r\n\r\n<div id=\"Figure_03_07_013\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"927\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153254\/CNX_Precalc_Figure_03_07_013.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=\"927\" height=\"311\" \/> Figure 13 Horizontal Asymptote [latex]y=0[\/latex] when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},\\text{ }q\\left(x\\right)\\ne 0[\/latex] where degree of [latex]p&lt;[\/latex] degree of [latex]q.[\/latex][\/caption]<\/div>\r\n<div><\/div>\r\n<div class=\"wp-caption-text\">\r\n\r\n<strong style=\"font-size: 1rem; text-align: initial;\">Case 2:<\/strong> If the degree of the denominator is less than the degree of the numerator by one, we get a slant asymptote.\r\n\r\n<\/div>\r\n<div id=\"eip-417\" class=\"unnumbered\" style=\"text-align: center;\">Example:\u00a0 [latex]f\\left(x\\right)=\\frac{3{x}^{2}-2x+1}{x-1}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137646911\">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] will behave similarly to [latex]g,[\/latex] it will approach a line close to [latex]y=3x.[\/latex] This line is a slant asymptote.<\/p>\r\n<p id=\"fs-id1165137419715\">To find the equation of the slant asymptote, divide [latex]\\frac{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] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_014\">Figure 14<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_014\" class=\"small\">[caption id=\"\" align=\"aligncenter\" width=\"336\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153257\/CNX_Precalc_Figure_03_07_014.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=\"336\" height=\"304\" \/> Figure 14\u00a0Slant Asymptote when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},\\text{ }q\\left(x\\right)\\ne 0[\/latex] where degree of [latex]p&gt;[\/latex] degree of [latex]q[\/latex] by 1.[\/caption]<\/div>\r\n<div class=\"wp-caption-text\"><\/div>\r\n<strong>Case 3:<\/strong> If the degree of the denominator equals the 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<div id=\"eip-773\" class=\"unnumbered\" style=\"text-align: center;\">Example:\u00a0 [latex]f\\left(x\\right)=\\frac{3{x}^{2}+2}{{x}^{2}+4x-5}[\/latex]<\/div>\r\n<div><strong>[latex]\\\\[\/latex]<\/strong><\/div>\r\n<p id=\"fs-id1165135593620\">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 \u00b1\\infty ,f\\left(x\\right)\\to 3,[\/latex] resulting in a horizontal asymptote of [latex]y=3.[\/latex] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_015\">Figure 15<\/a>. Note that this graph crosses the horizontal asymptote.<\/p>\r\n\r\n<div id=\"Figure_03_07_015\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"558\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153301\/CNX_Precalc_Figure_03_07_015.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=\"558\" height=\"278\" \/> Figure 15\u00a0Horizontal Asymptote when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},\\text{ }q\\left(x\\right)\\ne 0[\/latex] where degree of [latex]p=[\/latex] degree of [latex]q.[\/latex][\/caption]<\/div>\r\n<div class=\"wp-caption-text\"><\/div>\r\n<p id=\"fs-id1165137726840\">Notice that, while the graph of a rational function will never cross a <span class=\"no-emphasis\">vertical asymptote<\/span>, 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>\r\n<p id=\"fs-id1165137557874\">It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the <span class=\"no-emphasis\">end behavior<\/span> of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function<\/p>\r\n\r\n<div class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{3{x}^{5}-{x}^{2}}{x+3}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137723405\">with end behavior [latex]f\\left(x\\right)\\approx \\frac{3{x}^{5}}{x}=3{x}^{4},[\/latex] the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.\u00a0 We write as [latex]x\\to \u00b1\\infty , f\\left(x\\right)\\to \\infty.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165137659475\">\r\n<div class=\"textbox\">\r\n<h3>Horizontal Asymptotes of Rational Functions<\/h3>\r\n<p id=\"fs-id1165137528688\">The <span class=\"no-emphasis\">horizontal asymptote<\/span> of a rational function can be determined by looking at the degrees of the numerator and denominator.<\/p>\r\n\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>\r\n<div id=\"Example_03_07_07\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137812574\">\r\n<h3>Example 7:\u00a0 Identifying Horizontal Asymptotes<\/h3>\r\n<p id=\"fs-id1165134148527\">For the functions below, identify the horizontal or slant asymptote.<\/p>\r\n\r\n<ol type=\"a\">\r\n \t<li>[latex]g\\left(x\\right)=\\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}[\/latex]<\/li>\r\n \t<li>[latex]k\\left(x\\right)=\\frac{{x}^{2}+4x}{{x}^{3}-8}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1165137431476\">[reveal-answer q=\"fs-id1165137431476\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137431476\"]\r\n<p id=\"fs-id1165137542371\">For these solutions, we will use [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)}, q\\left(x\\right)\\ne 0.[\/latex]<\/p>\r\n\r\n<ol id=\"fs-id1165137610755\" type=\"a\">\r\n \t<li>[latex]g\\left(x\\right)=\\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}:[\/latex] The degree of [latex]p[\/latex] is [latex]3[\/latex] and the degree of [latex]q[\/latex] is [latex]3,[\/latex] so we can find the horizontal asymptote by taking the ratio of the leading terms, [latex]\\frac{6x^3}{2x^3}.[\/latex] There is a horizontal asymptote at [latex]y=\\frac{6}{2}[\/latex] or [latex]y=3.[\/latex]<\/li>\r\n \t<li>[latex]k\\left(x\\right)=\\frac{{x}^{2}+4x}{{x}^{3}-8}:[\/latex] The degree of [latex]p[\/latex] is [latex]2[\/latex] which is less than the degree of [latex]q[\/latex] which is [latex]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>\r\n<div id=\"Example_03_07_08\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137836670\">\r\n<div id=\"fs-id1165137836672\">\r\n<h3>Example 8:\u00a0 Identifying Horizontal Asymptotes<\/h3>\r\n<p id=\"fs-id1165137892264\">In the sugar concentration problem earlier, we created the equation [latex]C\\left(t\\right)=\\frac{5+t}{100+10t}.[\/latex]<\/p>\r\n<p id=\"fs-id1165135208611\">Find the horizontal asymptote and interpret it in context of the problem.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137559522\">[reveal-answer q=\"fs-id1165137559522\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137559522\"]\r\n<p id=\"fs-id1165137559524\">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>\r\n\r\n<div id=\"eip-id1165134167259\" class=\"unnumbered\" style=\"text-align: center;\">[latex]t\\to \\infty , C\\left(t\\right)\\to \\frac{1}{10}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137806518\">This function will have a horizontal asymptote at [latex]y=\\frac{1}{10}.[\/latex]<\/p>\r\n<p id=\"fs-id1165135450368\">This tells us that as the values of <em>t<\/em> increase, the values of [latex]C[\/latex] will 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.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_03_07_09\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137843849\">\r\n<div id=\"fs-id1165137843851\">\r\n<h3>Example 9:\u00a0 Identifying Horizontal and Vertical Asymptotes<\/h3>\r\n<p id=\"fs-id1165137419765\">Find the horizontal and vertical asymptotes of the function [latex]f\\left(x\\right)=\\frac{\\left(x-2\\right)\\left(x+3\\right)}{\\left(x-1\\right)\\left(x+2\\right)\\left(x-5\\right)}[\/latex]<\/p>\r\n\r\n<div><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137731893\">[reveal-answer q=\"fs-id1165137731893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137731893\"]\r\n<p id=\"fs-id1165137731895\">First, note that this function has no common factors, so there are no potential removable discontinuities.<\/p>\r\n<p id=\"fs-id1165137416868\">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,\u20132,\\text{and }5,[\/latex] indicating vertical asymptotes at these values.<\/p>\r\n<p id=\"fs-id1165137535648\">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 \u00b1\\infty , f\\left(x\\right)\\to 0.[\/latex] This function will have a horizontal asymptote at [latex]y=0.[\/latex] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_016\">Figure 16<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_016\" class=\"medium\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"425\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153303\/CNX_Precalc_Figure_03_07_016.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=\"425\" height=\"299\" \/> Figure 16[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137782334\" class=\"precalculus tryit\">\r\n<h3>Try it #6<\/h3>\r\n<div id=\"ti_03_07_06\">\r\n<div id=\"fs-id1165137761664\">\r\n<p id=\"fs-id1165137761666\">Find the vertical and horizontal asymptotes of the function [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\r\n<\/div>\r\n<div id=\"fs-id1165137731188\">[reveal-answer q=\"fs-id1165137731188\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137731188\"]\r\n<p id=\"fs-id1165137749296\">Vertical asymptotes at [latex]x=2[\/latex] and [latex]x=-3;[\/latex] horizontal asymptote at [latex]y=4.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134259298\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 10:\u00a0 Numerical Support for Asymptotes and Discontinuities<\/h3>\r\nFind the horizontal and vertical asymptotes and removal discontinuities of [latex]f\\left(x\\right)=\\frac{\\left(x+2\\right)\\left(x-1\\right)}{\\left(x+3\\right)\\left(x-1\\right)}[\/latex].\u00a0 Give numerical support for your conclusions.\r\n\r\n[reveal-answer q=\"269452\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"269452\"]\r\n\r\nSetting the denominator [latex]\\left(x+3\\right)\\left(x-1\\right)[\/latex] equal to zero and solving, we find that [latex]x=-3[\/latex] and [latex]x=1[\/latex] are not in the domain of the function. Since [latex]\\left(x-1\\right)[\/latex] is in the numerator and denominator an equal number of times, we conclude that there is a removable discontinuity or hole at [latex]x=1.[\/latex]\u00a0 Since [latex]\\left(x+3\\right)[\/latex] does not have a common factor in the numerator, we conclude that there is a vertical asymptote of [latex]x=-3.[\/latex]\r\n\r\nTo support the hole at [latex]x=1,[\/latex] we choose input values just above and just below [latex]x=1[\/latex] that get closer and closer to [latex]x=1[\/latex] and then evaluate the function there.\u00a0 We will choose our [latex]x[\/latex] values equal to 1.1, 1.01, and 1.001 for the values slightly larger than [latex]x=1[\/latex] and 0.9, 0.99 and 0.999 for the values slightly smaller than [latex]x=1.[\/latex] See Table 4 and 5.\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\"><caption>Table 4<\/caption>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"width: 25%;\">[latex]x[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1.1<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1.01<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1.001<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"width: 25%;\">[latex]f\\left(x\\right)[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.7561<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.75062<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">.750062<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\"><caption>Table 5<\/caption>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"width: 25%;\">[latex]x[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.9<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.99<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.999<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"width: 25%;\">[latex]f\\left(x\\right)[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.74359<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.74937<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.749937<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNotice, in Table 4, that the output has more zeros following 0.75 as we move across the table.\u00a0 We therefore conclude that the output is going to 0.75 as [latex]x[\/latex] goes to 1 from above.\u00a0 We write, as [latex]x\\to 1^{+},[\/latex] [latex]f\\left(x\\right)\\to 0.75.[\/latex]\u00a0 In Table 5, we see that the output increases the number of nines that follow 0.74, indicating that the output goes to 0.75.\u00a0 We write, as [latex]x\\to1^{-},\\text{ }f\\left(x\\right)\\to 0.75.[\/latex]\u00a0 This means that there is a hole at [latex]\\left(1, 0.75\\right).[\/latex]\r\n\r\nTo support the vertical asymptote at [latex]x=-3,[\/latex] we choose values just above and just below [latex]x=-3[\/latex] that get closer and closer to [latex]x=-3[\/latex] and then evaluate the function there.\u00a0 We will choose our [latex]x[\/latex] values equal to -2.9, -2.99, and -2.999 for the values to the right of\u00a0 [latex]x=-3[\/latex] and -3.1, -3.01 and -3.001 for the values to the left of [latex]x=-3.[\/latex] See Table 6 and 7.\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\"><caption>Table 6<\/caption>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"width: 25%;\">[latex]x[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-2.9<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-2.99<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-2.999<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"width: 25%;\">[latex]f\\left(x\\right)[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-9<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-99<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-999<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\"><caption>Table 7<\/caption>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"width: 25%;\">[latex]x[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-3.1<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-3.01<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-3.001<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"width: 25%;\">[latex]f\\left(x\\right)[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">11<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">101<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1001<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nTable 6 shows that as [latex]x[\/latex] gets closer and closer to -3 from the right, the output is decreasing without bound or as [latex]x\\to-3^{+},\\text{ }f\\left(x\\right)\\to-\\infty[\/latex].\u00a0 Table 7 shows that [latex]x[\/latex] gets closer and closer to -3 from the left, the output is increasing without bound or\u00a0as [latex]x\\to-3^{-},\\text{ }f\\left(x\\right)\\to\\infty[\/latex].\u00a0 This supports that there is a vertical asymptote at [latex]x=-3.[\/latex]\r\n\r\nFinally, to determine the horizontal asymptote, we look at the ratio of the leading terms of the numerator and denominator:\r\n<p style=\"text-align: center;\">[latex]\\frac{x^2}{x^2}=1.[\/latex]<\/p>\r\nThere is a horizontal asymptote of [latex]y=1[\/latex].\u00a0 To support this numerically, we choose very large values for [latex]x[\/latex] such as 100, 1 000 and 10 000 and very negative numbers such as -100, -1 000, -10 000 and evaluate the function at these points.\u00a0 See Table 8 and 9.\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\"><caption>Table 8<\/caption>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"width: 25%;\">[latex]x[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">100<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1000<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">10000<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"width: 25%;\">[latex]f\\left(x\\right)[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.99029<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.99900299<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.99990002999<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\"><caption>Table 9<\/caption>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"width: 25%;\">[latex]x[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-100<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-1000<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-10000<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"width: 25%;\">[latex]f\\left(x\\right)[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1.0103<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1.001003<\/td>\r\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1.00010003<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nLook at the pattern in the output of each table.\u00a0 We see that as [latex]x[\/latex] increases or decreases without bound, the output becomes closer and closer to 1, supporting the asymptote we found algebraically.\u00a0 We write, as [latex]x\\to\\pm\\infty,\\text{ }f\\left(x\\right)\\to 1.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Intercepts of Rational Functions<\/h3>\r\n<p id=\"fs-id1165137452078\">A <span class=\"no-emphasis\">rational function<\/span> 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 <em>y<\/em>-intercept if the function is not defined at zero.<\/p>\r\n<p id=\"fs-id1165135192756\">Likewise, a rational function will have <em>x<\/em>-intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, <em>x<\/em>-intercepts can only occur when the numerator of the rational function is equal to zero.\u00a0 However, if the numerator is zero at a value not in the domain, then there is not an x-intercept at that point.<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_03_07_10\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135181577\">\r\n<div id=\"fs-id1165135181579\">\r\n<h3>Example 11:\u00a0 Finding the Intercepts of a Rational Function<\/h3>\r\n<p id=\"fs-id1165135638521\">Find the intercepts of [latex]f\\left(x\\right)=\\frac{\\left(x-2\\right)\\left(x+3\\right)}{\\left(x-1\\right)\\left(x+2\\right)\\left(x-5\\right)}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134037668\">[reveal-answer q=\"fs-id1165134037668\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134037668\"]\r\n<p id=\"fs-id1165134037670\">We can find the <em>y<\/em>-intercept by evaluating the function at zero.<\/p>\r\n\r\n<div id=\"eip-id1165134118154\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(0\\right)&amp;=\\frac{\\left(0-2\\right)\\left(0+3\\right)}{\\left(0-1\\right)\\left(0+2\\right)\\left(0-5\\right)} \\\\ \\text{ }&amp;=\\frac{-6}{10}\\hfill \\\\ \\text{ }&amp;=-\\frac{3}{5}\\hfill \\\\ \\text{ }\\text{ }&amp;=-0.6.\\hfill \\end{align*}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137466527\">The <em>x<\/em>-intercepts will occur when the function is equal to zero:<\/p>\r\n\r\n<div id=\"eip-id1165134388942\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}\\hfill &amp; \\hfill \\\\ 0=\\frac{\\left(x-2\\right)\\left(x+3\\right)}{\\left(x-1\\right)\\left(x+2\\right)\\left(x-5\\right)}\\hfill &amp; \\textrm{This is zero when the numerator is zero}.\\hfill \\\\ 0=\\left(x-2\\right)\\left(x+3\\right)\\hfill &amp; \\hfill \\\\ x=2, -3\\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137605789\">The <em>y<\/em>-intercept is [latex]\\left(0,-0.6\\right),[\/latex] and the <em>x<\/em>-intercepts are [latex]\\left(2,0\\right)[\/latex] and [latex]\\left(-3,0\\right).[\/latex] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_017\">Figure 17<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_017\" class=\"medium\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"461\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153307\/CNX_Precalc_Figure_03_07_017.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=\"461\" height=\"324\" \/> Figure 17[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137447991\" class=\"precalculus tryit\">\r\n<h3>Try it #7<\/h3>\r\n<div id=\"ti_03_07_07\">\r\n<div id=\"fs-id1165137460979\">\r\n<p id=\"fs-id1165137460980\">Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the <em>x<\/em>- and <em>y<\/em>-intercepts and the horizontal and vertical asymptotes.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135183976\">[reveal-answer q=\"fs-id1165135183976\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135183976\"]\r\n<div>For the transformed reciprocal squared function, we find the rational form.[latex]\\\\[\/latex]<\/div>\r\n<div style=\"text-align: center;\">\r\n\r\n[latex]\\begin{align*}f\\left(x\\right)&amp;=\\frac{1}{{\\left(x-3\\right)}^{2}}-4\\\\&amp;=\\frac{1-4{\\left(x-3\\right)}^{2}}{{\\left(x-3\\right)}^{2}}\\\\&amp;=\\frac{1-4\\left({x}^{2}-6x+9\\right)}{\\left(x-3\\right)\\left(x-3\\right)}\\\\&amp;=\\frac{-4{x}^{2}+24x-35}{{x}^{2}-6x+9}\\end{align*}[\/latex]\r\n<p id=\"fs-id1165137925364\">Because the numerator is the same degree as the denominator we know that as [latex]x\\to \u00b1\\infty , f\\left(x\\right)\\to -4; \\textrm{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 <em>x<\/em>-intercepts are at [latex]\\left(2.5,0\\right)[\/latex] and [latex]\\left(3.5,0\\right),[\/latex] since [latex]2.5[\/latex] and\u00a0[latex]3.5[\/latex] are in the domain.\u00a0 \u00a0Finally, we evaluate the function at 0 and find the <em>y<\/em>-intercept to be at [latex]\\left(0,\\frac{-35}{9}\\right).[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137461340\" class=\"bc-section section\">\r\n<h3>Graphing Rational Functions<\/h3>\r\n<p id=\"fs-id1165137461346\">In <a class=\"autogenerated-content\" href=\"#Example_03_07_09\">Example 11<\/a>, we see that the numerator of a rational function reveals the <em>x<\/em>-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>\r\n<p id=\"fs-id1165137695332\">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. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_019\">Figure 18<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_019\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"400\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153310\/CNX_Precalc_Figure_03_07_019.jpg\" alt=\"Graph of y=1\/x with its vertical asymptote at x=0.\" width=\"400\" height=\"299\" \/> Figure 18[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137661229\">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. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_018\">Figure 19<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_018\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"396\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153313\/CNX_Precalc_Figure_03_07_018.jpg\" alt=\"Graph of y=1\/x^2 with its vertical asymptote at x=0.\" width=\"396\" height=\"297\" \/> Figure 19[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137678097\">For example, the graph of [latex]f\\left(x\\right)=\\frac{{\\left(x+1\\right)}^{2}\\left(x-3\\right)}{{\\left(x+3\\right)}^{2}\\left(x-2\\right)}[\/latex] is shown in <a class=\"autogenerated-content\" href=\"#Figure_03_07_020\">Figure 20<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_020\" class=\"medium\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"468\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153316\/CNX_Precalc_Figure_03_07_020.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=\"468\" height=\"401\" \/> Figure 20[\/caption]\r\n\r\n<\/div>\r\n<ul id=\"fs-id1165137723939\">\r\n \t<li>At the <em>x<\/em>-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 <em>x<\/em>-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 id=\"fs-id1165135192767\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137640498\"><strong>Given a rational function, sketch a graph.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137640503\" type=\"1\">\r\n \t<li>Evaluate the function at 0 to find the <em>y<\/em>-intercept.<\/li>\r\n \t<li>Factor the numerator and denominator.<\/li>\r\n \t<li>For factors in the numerator that are not in the denominator, determine where each factor of the numerator is zero to find the <em>x<\/em>-intercepts.<\/li>\r\n \t<li>Find the multiplicities of the <em>x<\/em>-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 that are in both the denominator AND 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 id=\"Example_03_07_11\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137657378\">\r\n<div id=\"fs-id1165137657380\">\r\n<h3>Example 12:\u00a0 Graphing a Rational Function<\/h3>\r\n<p id=\"fs-id1165135701460\">Sketch a graph of [latex]f\\left(x\\right)=\\frac{\\left(x+2\\right)\\left(x-3\\right)}{{\\left(x+1\\right)}^{2}\\left(x-2\\right)}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137413858\">[reveal-answer q=\"fs-id1165137413858\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137413858\"]\r\n<p id=\"fs-id1165137413860\">We can start by noting that the function is already factored, saving us a step.<\/p>\r\n<p id=\"fs-id1165137413863\">Next, we will find the intercepts. Evaluating the function at zero gives the <em>y<\/em>-intercept:<\/p>\r\n\r\n<div id=\"eip-id1165132127636\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(0\\right)&amp;=\\frac{\\left(0+2\\right)\\left(0-3\\right)}{{\\left(0+1\\right)}^{2}\\left(0-2\\right)}\\hfill \\\\ \\text{ }&amp;=3\\hfill \\end{align*}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137563753\">To find the <em>x<\/em>-intercepts, we determine when the numerator of the function is zero. Setting each factor equal to zero, we find <em>x<\/em>-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>\r\n<p id=\"fs-id1165137871570\">We have a <em>y<\/em>-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>\r\n<p id=\"fs-id1165132945548\">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>\r\n<p id=\"fs-id1165135187068\">There are no common factors in the numerator and denominator. This means there are no removable discontinuities.<\/p>\r\n<p id=\"fs-id1165135453215\">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>\r\n<p id=\"fs-id1165137696556\">To sketch the graph, we might start by plotting the three intercepts. Since the graph has no <em>x<\/em>-intercepts between the vertical asymptotes, and the <em>y<\/em>-intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph as shown in <a class=\"autogenerated-content\" href=\"#Figure_03_07_021\">Figure 21<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_021\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"335\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153319\/CNX_Precalc_Figure_03_07_021.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=\"335\" height=\"303\" \/> Figure 21[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137566717\">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>\r\n<p id=\"fs-id1165137760010\">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. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_022\">Figure 22<\/a>. After passing through the <em>x<\/em>-intercepts, the graph will then level off toward an output of zero, as indicated by the horizontal asymptote.<\/p>\r\n\r\n<div id=\"Figure_03_07_022\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"333\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153322\/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=\"333\" height=\"300\" \/> Figure 22[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137603685\" class=\"precalculus tryit\">\r\n<h3>Try it #8<\/h3>\r\n<div id=\"ti_03_07_08\">\r\n<div id=\"fs-id1165135694508\">\r\n<p id=\"fs-id1165137653961\">Given the function [latex]f\\left(x\\right)=\\frac{{\\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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137454284\">[reveal-answer q=\"fs-id1165137454284\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137454284\"]\r\n<p id=\"fs-id1165137454286\">Horizontal asymptote at [latex]y=\\frac{1}{2}.[\/latex] Vertical asymptotes at [latex]x=1[\/latex] and [latex]x=3.[\/latex] <em>y<\/em>-intercept at [latex]\\left(0,\\frac{4}{3}\\right).[\/latex]<\/p>\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 <em>x<\/em>-axis at this point. [latex]\\left(2,0\\right)[\/latex] is a single zero and the graph crosses the axis at this point.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"495\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153325\/CNX_Precalc_Figure_03_07_023.jpg\" alt=\"Graph of f(x)=(x+2)^2(x-2)\/2(x-1)^2(x-3) with its vertical and horizontal asymptotes.\" width=\"495\" height=\"323\" \/> Figure 23[\/caption]\r\n\r\n<span id=\"fs-id1165137745200\"><\/span>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137481147\" class=\"bc-section section\">\r\n<h3>Writing Rational Functions<\/h3>\r\n<p id=\"fs-id1165137585741\">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 <em>x<\/em>-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 <em>x<\/em>-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>\r\n\r\n<div id=\"fs-id1165137661074\">\r\n<h3>Writing Rational Functions from Intercepts and Asymptotes<\/h3>\r\n<p id=\"fs-id1165137675581\">If a <span class=\"no-emphasis\">rational function<\/span> has <em>x<\/em>-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>\r\n\r\n<div id=\"eip-597\" class=\"unnumbered\" 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][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137935737\">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] can be determined given a value of the function other than the <em>x<\/em>-intercept or by the horizontal asymptote if it is nonzero.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137423517\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137575250\"><strong>Given a graph of a rational function, write the function.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137654907\" type=\"1\">\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 zeros and their multiplicities. (This is easy to do when finding the \u201csimplest\u201d 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 id=\"Example_03_07_12\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135333705\">\r\n<div id=\"fs-id1165135333708\">\r\n<h3>Example 13:\u00a0 Writing a Rational Function from Intercepts and Asymptotes<\/h3>\r\n<p id=\"fs-id1165135445953\">Write an equation for the rational function shown in <a class=\"autogenerated-content\" href=\"#Figure_03_07_024\">Figure 24<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_024\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"314\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153328\/CNX_Precalc_Figure_03_07_024.jpg\" alt=\"Graph of a rational function.\" width=\"314\" height=\"306\" \/> Figure 24[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137767514\">[reveal-answer q=\"fs-id1165137767514\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137767514\"]\r\n<p id=\"fs-id1165137767516\">The graph appears to have <em>x<\/em>-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. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_025\">Figure 25<\/a>.<\/p>\r\n\r\n<div id=\"Figure_03_07_025\" class=\"medium\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"468\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153331\/CNX_Precalc_Figure_03_07_025.jpg\" alt=\"Graph of a rational function denoting its vertical asymptotes and x-intercepts.\" width=\"468\" height=\"304\" \/> Figure 25[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137570509\">We can use this information to write a function of the form<\/p>\r\n\r\n<div id=\"eip-id1165131797138\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=a\\frac{\\left(x+2\\right)\\left(x-3\\right)}{\\left(x+1\\right){\\left(x-2\\right)}^{2}}.[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137768581\">To find the stretch factor, we can use another clear point on the graph, such as the <em>y<\/em>-intercept [latex]\\left(0,-2\\right).[\/latex]<\/p>\r\n\r\n<div id=\"eip-id1165134237069\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}-2&amp;=a\\frac{\\left(0+2\\right)\\left(0-3\\right)}{\\left(0+1\\right){\\left(0-2\\right)}^{2}}\\hfill \\\\ -2&amp;=a\\frac{-6}{4}\\hfill \\\\ \\text{ }a&amp;=\\frac{-8}{-6}=\\frac{4}{3}\\hfill \\end{align*}[\/latex]<\/div>\r\n[latex]\\\\[\/latex]\r\n<p id=\"fs-id1165134170051\">This gives us a final function of [latex]f\\left(x\\right)=\\frac{4\\left(x+2\\right)\\left(x-3\\right)}{3\\left(x+1\\right){\\left(x-2\\right)}^{2}}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135152264\" class=\"precalculus media\">\r\n<p id=\"fs-id1165137627838\">Access these online resources for additional instruction and practice with rational functions.<\/p>\r\n\r\n<ul id=\"fs-id1165135250839\">\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/graphrational\">Graphing Rational Functions<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/equatrational\">Find the Equation of a Rational Function<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/asymptote\">Determining Vertical and Horizontal Asymptotes<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/interasymptote\">Find the Intercepts, Asymptotes, and Hole of a Rational Function<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137659195\" class=\"key-equations\">\r\n<h3>Key Equations<\/h3>\r\n<table id=\"eip-id1362369\" style=\"width: 1107px;\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"width: 102.141px;\">Rational Function<\/td>\r\n<td class=\"border\" style=\"width: 978.859px;\">[latex]f\\left(x\\right)=\\displaystyle{\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{{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<\/div>\r\n<div id=\"fs-id1165137793507\" class=\"textbox key-takeaways\">\r\n<h3>Key Concepts<\/h3>\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, 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\r\n<div class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ f\\left(x\\right)=a\\frac{\\displaystyle{{\\left(x-{x}_{1}\\right)}^{{p}_{1}}{\\left(x-{x}_{2}\\right)}^{{p}_{2}}\\cdots {\\left(x-{x}_{n}\\right)}^{{p}_{n}}}}{\\displaystyle{{\\left(x-{v}_{1}\\right)}^{{q}_{1}}{\\left(x-{v}_{2}\\right)}^{{q}_{2}}\\cdots {\\left(x-{v}_{m}\\right)}^{{q}_{n}}}}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\r\n<dl id=\"fs-id1165137758530\">\r\n \t<dt>arrow notation<\/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\">\r\n \t<dt>horizontal asymptote<\/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\">\r\n \t<dt>rational function<\/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\">\r\n \t<dt>removable discontinuity<\/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\">\r\n \t<dt>vertical asymptote<\/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<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>In this section, you will:<\/p>\n<ul>\n<li>Use arrow notation.<\/li>\n<li>Find the domain of rational functions.<\/li>\n<li>Identify horizontal and vertical asymptotes and removable discontinuities.<\/li>\n<li>Graph rational functions.<\/li>\n<li>Determine the formula for the graph of a rational function.<\/li>\n<li>Solve applied problems involving rational functions.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137740975\">Suppose we know that the cost of making a product is dependent on the number of items, [latex]x,[\/latex] produced. and that it 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] items, we would divide the cost function by the number of items, [latex]x.[\/latex]<\/p>\n<p id=\"fs-id1165137768452\">The average cost function, which yields the average cost per item for [latex]x[\/latex] items produced, is<\/p>\n<div id=\"eip-634\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{15,000x-0.1{x}^{2}+1000}{x}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137863708\">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 id=\"fs-id1165133305345\">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 can be thought of as the ratio of two polynomial functions.<\/p>\n<div id=\"fs-id1165135397235\" class=\"bc-section section\">\n<h3>Using Arrow Notation<\/h3>\n<p id=\"fs-id1165137659469\">We have seen the graphs of the\u00a0<span class=\"no-emphasis\">reciprocal function<\/span> and the squared reciprocal function from our study of toolkit functions. Examine these graphs, as shown in <a class=\"autogenerated-content\" href=\"#Figure_03_07_001\">Figure 1<\/a>, and notice some of their features.<\/p>\n<div id=\"Figure_03_07_001\" class=\"medium\">\n<div style=\"width: 575px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153218\/CNX_Precalc_Figure_03_07_001.jpg\" alt=\"Graphs of f(x)=1\/x and f(x)=1\/x^2\" width=\"565\" height=\"350\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137935728\">Several things are apparent if we examine the graph of [latex]f\\left(x\\right)=\\frac{1}{x}.[\/latex]<\/p>\n<ol id=\"fs-id1165135438444\" type=\"1\">\n<li>On the left branch of the graph, the curve approaches the <em>x<\/em>-axis [latex]\\left(y=0\\right) \\textrm{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 <em>x-<\/em>axis [latex]\\left(y=0\\right) \\textrm{as } x\\to \\infty .[\/latex]<\/li>\n<\/ol>\n<p id=\"fs-id1165137461553\">To summarize, we use arrow notation to show that [latex]x[\/latex] or [latex]f\\left(x\\right)[\/latex] is approaching a particular value. See <a class=\"autogenerated-content\" href=\"#Table_03_07_001\">Table 1<\/a>.<\/p>\n<table id=\"Table_03_07_001\" summary=\"..\">\n<caption>Table 1:\u00a0 Arrow Notation<\/caption>\n<thead>\n<tr>\n<th class=\"border\">Symbol<\/th>\n<th class=\"border\">Meaning<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td class=\"border\">[latex]x\\to {a}^{-}[\/latex]<\/td>\n<td class=\"border\">[latex]x[\/latex] approaches [latex]a[\/latex] from the left ([latex]x \\lt a[\/latex] but is increasing and getting closer and closer to\u00a0 [latex]a[\/latex])<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">[latex]x\\to {a}^{+}[\/latex]<\/td>\n<td class=\"border\">[latex]x[\/latex] approaches [latex]a[\/latex] from the right ([latex]x>a[\/latex] but is decreasing and getting closer and closer to [latex]a[\/latex])<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">[latex]x\\to a[\/latex]<\/td>\n<td class=\"border\">the input approaches [latex]a[\/latex] from both sides<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">[latex]x\\to \\infty[\/latex]<\/td>\n<td class=\"border\">[latex]x[\/latex] goes toward infinity ([latex]x[\/latex] increases without bound)<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">[latex]x\\to -\\infty[\/latex]<\/td>\n<td class=\"border\">[latex]x[\/latex] goes toward negative infinity ([latex]x[\/latex] decreases without bound)<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">[latex]f\\left(x\\right)\\to \\infty[\/latex]<\/td>\n<td class=\"border\">the output goes toward infinity (the output increases without bound)<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">[latex]f\\left(x\\right)\\to -\\infty[\/latex]<\/td>\n<td class=\"border\">the output goes toward negative infinity (the output decreases without bound)<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">[latex]f\\left(x\\right)\\to a[\/latex]<\/td>\n<td class=\"border\">the output approaches [latex]a[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-id1165137759950\" class=\"bc-section section\">\n<h4>Local Behavior of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/h4>\n<p id=\"fs-id1165137755329\">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 (negative values very close to zero), the function values decrease without bound. We can see this behavior in <a class=\"autogenerated-content\" href=\"#Table_03_07_002\">Table 2<\/a>.<\/p>\n<table id=\"Table_03_07_002\" summary=\"..\">\n<caption>Table 2<\/caption>\n<tbody>\n<tr>\n<td class=\"border\"><strong>[latex]x[\/latex]<\/strong><\/td>\n<td class=\"border\" style=\"text-align: center;\">\u20130.1<\/td>\n<td class=\"border\" style=\"text-align: center;\">\u20130.01<\/td>\n<td class=\"border\" style=\"text-align: center;\">\u20130.001<\/td>\n<td class=\"border\" style=\"text-align: center;\">\u20130.0001<\/td>\n<\/tr>\n<tr>\n<td class=\"border\"><strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/strong><\/td>\n<td class=\"border\" style=\"text-align: center;\">\u201310<\/td>\n<td class=\"border\" style=\"text-align: center;\">\u2013100<\/td>\n<td class=\"border\" style=\"text-align: center;\">\u20131000<\/td>\n<td class=\"border\" style=\"text-align: center;\">\u201310,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137542511\">We write in arrow notation, as [latex]x\\to {0}^{-},f\\left(x\\right)\\to -\\infty[\/latex].<\/p>\n<p id=\"fs-id1165137506235\">As the input values approach zero from the right side (positive values very close to zero), the function values increase without bound. We can see this behavior in <a class=\"autogenerated-content\" href=\"#Table_03_07_003\">Table 3<\/a>.<\/p>\n<table id=\"Table_03_07_003\" summary=\"..\">\n<caption>Table 3<\/caption>\n<tbody>\n<tr style=\"height: 14px;\">\n<td class=\"border\" style=\"height: 14px; width: 232px;\"><strong>[latex]x[\/latex]<\/strong><\/td>\n<td class=\"border\" style=\"height: 14px; width: 18px; text-align: center;\">0.1<\/td>\n<td class=\"border\" style=\"height: 14px; width: 25px; text-align: center;\">0.01<\/td>\n<td class=\"border\" style=\"height: 14px; width: 32px; text-align: center;\">0.001<\/td>\n<td class=\"border\" style=\"height: 14px; width: 39px; text-align: center;\">0.0001<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td class=\"border\" style=\"height: 14px; width: 232px;\"><strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/strong><\/td>\n<td class=\"border\" style=\"height: 14px; width: 18px; text-align: center;\">10<\/td>\n<td class=\"border\" style=\"height: 14px; width: 25px; text-align: center;\">100<\/td>\n<td class=\"border\" style=\"height: 14px; width: 32px; text-align: center;\">1000<\/td>\n<td class=\"border\" style=\"height: 14px; width: 39px; text-align: center;\">10,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165134338836\">We write in arrow notation as [latex]x\\to {0}^{+}, f\\left(x\\right)\\to \\infty .[\/latex]<\/p>\n<div><\/div>\n<p id=\"fs-id1165135397958\">See <a class=\"autogenerated-content\" href=\"#Figure_03_07_002\">Figure 2<\/a>.<\/p>\n<div id=\"Figure_03_07_002\" class=\"small\">\n<div style=\"width: 473px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153221\/CNX_Precalc_Figure_03_07_002.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=\"463\" height=\"300\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137648090\">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] as the input becomes close to zero. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_003\">Figure 3<\/a>.<\/p>\n<div id=\"Figure_03_07_003\" class=\"small\">\n<div style=\"width: 409px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153224\/CNX_Precalc_Figure_03_07_003.jpg\" alt=\"Graph of f(x)=1\/x with its vertical asymptote at x=0.\" width=\"399\" height=\"298\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137732344\">\n<div class=\"textbox definitions\">\n<h3>Definition<\/h3>\n<p id=\"fs-id1165137561740\">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>from the left side,<\/p>\n<p style=\"text-align: center;\">as [latex]x\\to a^{-},f\\left(x\\right)\\to \\infty ,[\/latex] or as\u00a0<span style=\"background-color: initial;\">[latex]x\\to a^{-},f\\left(x\\right)\\to -\\infty,[\/latex] and<\/span><\/p>\n<p>from the right side,<\/p>\n<p style=\"text-align: center;\">[latex]x\\to a^{+},f\\left(x\\right)\\to \\infty,[\/latex] \u00a0or as [latex]x\\to a^{+},f\\left(x\\right)\\to -\\infty .[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137502160\" class=\"bc-section section\">\n<h4>End Behavior of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/h4>\n<p id=\"fs-id1165137408019\">As the values of [latex]x[\/latex] increase without bound, the function values approach 0. As the values of [latex]x[\/latex] decrease without bound, the function values approach 0. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_004\">Figure 4<\/a>. Symbolically, using arrow notation<\/p>\n<p id=\"fs-id1165137565255\" style=\"text-align: center;\">[latex]\\textrm{as }x\\to \\infty ,f\\left(x\\right)\\to 0,\\textrm{and as }x\\to -\\infty ,f\\left(x\\right)\\to 0.[\/latex]<\/p>\n<div id=\"Figure_03_07_004\" class=\"small\">\n<div style=\"width: 475px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153227\/CNX_Precalc_Figure_03_07_004.jpg\" alt=\"Graph of f(x)=1\/x which highlights the segments of the turning points to denote their end behavior.\" width=\"465\" height=\"302\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4<\/p>\n<\/div>\n<\/div>\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] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_005\">Figure 5<\/a><strong>.<\/strong><\/p>\n<div id=\"Figure_03_07_005\" class=\"small\">\n<div style=\"width: 411px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153230\/CNX_Precalc_Figure_03_07_005.jpg\" alt=\"Graph of f(x)=1\/x with its vertical asymptote at x=0 and its horizontal asymptote at y=0.\" width=\"401\" height=\"300\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137901226\">\n<div class=\"textbox definitions\">\n<h3>Definition<\/h3>\n<p id=\"fs-id1165137782455\">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 [latex]\\textrm{as }x\\to \\infty \\textrm{ or }x\\to -\\infty ,\\text{ }f\\left(x\\right)\\to b.[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"Example_03_07_01\" class=\"textbox examples\">\n<div id=\"fs-id1165133213902\">\n<div id=\"fs-id1165137657454\">\n<h3>Example 1:\u00a0 Using Arrow Notation<\/h3>\n<p id=\"fs-id1165137437543\">Use arrow notation to describe the end behavior and local behavior of the function graphed in <a class=\"autogenerated-content\" href=\"#Figure_03_07_006\">Figure 6<\/a>.<\/p>\n<div id=\"Figure_03_07_006\" class=\"small\">\n<div style=\"width: 317px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153233\/CNX_Precalc_Figure_03_07_006.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=\"307\" height=\"301\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137870943\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137870943\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137870943\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137851860\">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<div id=\"eip-id1165134070637\" class=\"unnumbered\" style=\"text-align: center;\">As [latex]x\\to {2}^{-},f\\left(x\\right)\\to -\\infty ,[\/latex] and as [latex]x\\to {2}^{+},\\text{ }f\\left(x\\right)\\to \\infty .[\/latex]<\/div>\n<p id=\"fs-id1165137696383\">Also, as the inputs decrease without bound, the graph appears to be leveling off with output values closer and closer to 4, indicating a horizontal asymptote at [latex]y=4.[\/latex] As the inputs increase without bound, the graph levels off approaching 4.<\/p>\n<div id=\"eip-id1165132961960\" class=\"unnumbered\" 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]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137597363\" class=\"precalculus tryit\">\n<h3>Try it #1<\/h3>\n<div id=\"ti_03_07_01\">\n<div id=\"fs-id1165137431673\">\n<p>Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function.<\/p>\n<\/div>\n<div id=\"fs-id1165135541751\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135541751\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135541751\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137828072\">End behavior: as [latex]x\\to \u00b1\\infty , f\\left(x\\right)\\to 0;[\/latex] There is a horizontal asymptote at\u00a0 [latex]y=0.[\/latex]<\/p>\n<p>Local behavior: as [latex]x\\to 0, f\\left(x\\right)\\to \\infty[\/latex] There is a vertical asymptote at [latex]x=0.[\/latex] There are no <em>x<\/em>&#8211; or <em>y<\/em>-intercepts.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_07_02\" class=\"textbox examples\">\n<div id=\"fs-id1165137694119\">\n<div id=\"fs-id1165137694121\">\n<h3>Example 2:\u00a0 Using Transformations to Graph a Rational Function<\/h3>\n<p id=\"fs-id1165137640093\">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>\n<div id=\"fs-id1165137911752\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137911752\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137911752\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137911755\">Shifting the graph left 2 and up 3 would result in the function<\/p>\n<div id=\"eip-id1165135547466\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{1}{x+2}+3,[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137640711\">or equivalently, by adding the terms after finding a common denominator,<\/p>\n<div id=\"eip-id1165131954096\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{3x+7}{x+2}.[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137446966\">The graph of the shifted function is displayed in <a class=\"autogenerated-content\" href=\"#Figure_03_07_007\">Figure 7<\/a>.<\/p>\n<div id=\"Figure_03_07_007\" class=\"small\">\n<div style=\"width: 509px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153236\/CNX_Precalc_Figure_03_07_007.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=\"499\" height=\"301\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137891390\">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<div id=\"eip-id1165134252833\" class=\"unnumbered\" style=\"text-align: center;\">As [latex]x\\to -{2}^{-}, f\\left(x\\right)\\to -\\infty ,[\/latex] and as [latex]x\\to -{2}^{+}, f\\left(x\\right)\\to \\infty .[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137736971\">As the inputs increase and decrease without bound, the graph appears to be leveling off with output values getting closer and closer to 3, indicating a horizontal asymptote of [latex]y=3.[\/latex]<\/p>\n<div id=\"eip-id1165134250660\" class=\"unnumbered\" style=\"text-align: center;\">As [latex]x\\to \u00b1\\infty , f\\left(x\\right)\\to 3.[\/latex]<\/div>\n<div id=\"fs-id1165137889779\">\n<h3>Analysis<\/h3>\n<p id=\"fs-id1165137401110\">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>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137434563\" class=\"precalculus tryit\">\n<h3>Try it #2<\/h3>\n<div id=\"ti_03_07_02\">\n<div id=\"fs-id1165137824780\">\n<p id=\"fs-id1165137824781\">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>\n<div id=\"fs-id1165137812954\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137812954\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137812954\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"width: 414px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153239\/CNX_Precalc_Figure_03_07_008.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=\"404\" height=\"303\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8<\/p>\n<\/div>\n<p id=\"fs-id1165137416225\">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 \u00b1\\infty ,f\\left(x\\right)\\to -4.[\/latex] Therefore, the vertical asymptote is\u00a0 [latex]x=3,[\/latex] and the horizontal asymptote is\u00a0 [latex]y=-4.[\/latex]<\/p>\n<p id=\"fs-id1165137823960\">The function is [latex]f\\left(x\\right)=\\frac{1}{{\\left(x-3\\right)}^{2}}-4.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137534932\" class=\"bc-section section\">\n<h3>Solving Applied Problems Involving Rational Functions<\/h3>\n<p id=\"fs-id1165137639549\">In <a class=\"autogenerated-content\" href=\"#Example_03_07_02\">Example 2<\/a>, we shifted a toolkit function in a way that resulted in the function [latex]f\\left(x\\right)=\\frac{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 id=\"fs-id1165137386884\">\n<div class=\"textbox definitions\">\n<h3>Definition<\/h3>\n<p id=\"fs-id1165137754871\">A <strong>rational function<\/strong> is a function that can be written as the quotient of two polynomial functions [latex]P\\left(x\\right) \\textrm{and } Q\\left(x\\right).[\/latex]<\/p>\n<div id=\"fs-id1165137411279\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\displaystyle{\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{{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]<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_07_03\" class=\"textbox examples\">\n<div id=\"fs-id1165137536326\">\n<div id=\"fs-id1165137536328\">\n<h3>Example 3:\u00a0 Solving an Applied Problem Involving a Rational Function<\/h3>\n<p id=\"fs-id1165137694204\">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<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1165137526522\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137526522\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137526522\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137733797\">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]\\begin{array}{l}\\text{water: }W\\left(t\\right)=100+10t\\text{ in gallons}\\\\ \\text{sugar: }S\\left(t\\right)=5+1t\\text{ in pounds}\\end{array}[\/latex]<\/p>\n<p>[latex]\\\\[\/latex]<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">The concentration, [latex]C,[\/latex] will be the ratio of pounds of sugar to gallons of water.<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"text-align: center;\">[latex]C\\left(t\\right)=\\frac{5+t}{100+10t}[\/latex]<\/span><\/p>\n<p>[latex]\\\\[\/latex]<\/p>\n<p><span style=\"text-align: left;\">The concentration after 12 minutes is given by evaluating [latex]C\\left(t\\right)[\/latex] at [latex]t=\\text{ }12.[\/latex]<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"text-align: center; font-size: 0.9em;\">[latex]C\\left(12\\right)=\\frac{5+12}{100+10\\left(12\\right)} \\text{ }=\\frac{17}{220}[\/latex]<\/span><\/p>\n<p>[latex]\\\\[\/latex]<\/p>\n<p>This means the concentration is 17 pounds of sugar to 220 gallons of water.<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">At the beginning, the concentration is<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: center;\">[latex]C\\left(0\\right)=\\frac{5+0}{100+10\\left(0\\right)}\\text{ }=\\frac{1}{20}[\/latex]<\/span><\/p>\n<p>[latex]\\\\[\/latex]<\/p>\n<p><span style=\"font-size: 1rem;\">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.<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135194535\" class=\"precalculus tryit\">\n<h3>Try it #3<\/h3>\n<div id=\"ti_03_07_03\">\n<div id=\"fs-id1165137640599\">\n<p id=\"fs-id1165137436098\">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>\n<div id=\"fs-id1165137644596\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137644596\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137644596\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137644598\">[latex]\\frac{12}{11}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\">\n<h3>Finding the Domains of Rational Functions<\/h3>\n<p id=\"fs-id1165137530059\">A <span class=\"no-emphasis\">vertical asymptote<\/span> represents a value at which a rational function is undefined, so that value is not in the domain of the function. A rational 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.\u00a0The <strong>domain of a rational function<\/strong> includes all real numbers except those that cause the denominator to equal zero.<\/p>\n<div id=\"fs-id1165135530461\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137695173\"><strong>Given a rational function, find the domain.<\/strong><\/p>\n<ol id=\"fs-id1165137724163\" type=\"1\">\n<li>Set the denominator equal to zero.<\/li>\n<li>Solve to find the <em>x<\/em>-values that cause the denominator to equal zero.<\/li>\n<li>The domain is all real numbers except those found in Step 2.\u00a0 Express your answer using interval notation.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_07_04\" class=\"textbox examples\">\n<div id=\"fs-id1165135587815\">\n<div id=\"fs-id1165137647179\">\n<h3>Example 4:\u00a0 Finding the Domain of a Rational Function<\/h3>\n<p id=\"fs-id1165137501029\">Find the domain of [latex]f\\left(x\\right)=\\frac{x+3}{{x}^{2}-9}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137656082\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137656082\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137656082\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135445735\">Begin by setting the denominator equal to zero and solving.<\/p>\n<div id=\"eip-id1165134254379\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*} {x}^{2}-9&=0\\\\ {x}^{2}&=9 \\\\ x&=\u00b13\\end{align*}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<div class=\"unnumbered\"><\/div>\n<p id=\"fs-id1165137642958\">The denominator is equal to zero when [latex]x=\u00b13.[\/latex] The domain of the function is all real numbers except [latex]x=\u00b13.[\/latex] In interval notation we write,\u00a0 [latex]\\left(-\\infty, -3\\right)\\cup\\left(-3, 3\\right)\\cup\\left(3,\\infty\\right).[\/latex]<\/p>\n<h3>Analysis<\/h3>\n<p id=\"fs-id1165133276227\">A graph of this function, as shown in <a class=\"autogenerated-content\" href=\"#Figure_03_07_009\">Figure 9<\/a>, confirms that the function is not defined when [latex]x=\u00b13.[\/latex]<\/p>\n<div id=\"Figure_03_07_009\" class=\"small\">\n<div style=\"width: 418px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153241\/CNX_Precalc_Figure_03_07_009.jpg\" alt=\"Graph of f(x)=1\/(x-3) with its vertical asymptote at x=3 and its horizontal asymptote at y=0.\" width=\"408\" height=\"305\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 9<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137455163\">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>\n<\/div>\n<div id=\"fs-id1165134192966\" class=\"precalculus tryit\">\n<h3>Try it #4<\/h3>\n<div id=\"ti_03_07_04\">\n<div id=\"fs-id1165137414038\">\n<p>Find the domain of [latex]f\\left(x\\right)=\\frac{4x}{5\\left(x-1\\right)\\left(x-5\\right)}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137474960\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137474960\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137474960\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137474961\">The domain is all real numbers except [latex]x=1[\/latex] and [latex]x=5[\/latex] or [latex]\\left(-\\infty, 1\\right)\\cup\\left(1, 5\\right)\\cup\\left(5,\\infty\\right).[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137836721\" class=\"bc-section section\">\n<h3>Identifying Vertical Asymptotes and Removable Discontinuities of Rational Functions<\/h3>\n<p id=\"fs-id1165135439868\">By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are vertical 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 vertical asymptotes, and calculate their location.<\/p>\n<div id=\"fs-id1165135194731\" class=\"bc-section section\">\n<div class=\"textbox\">\n<p>The <strong>vertical asymptotes<\/strong> of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator.\u00a0 Vertical asymptotes occur at the zeros of such factors.<\/p>\n<\/div>\n<div class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137451766\"><strong>Given a rational function, identify any vertical asymptotes and removable discontinuties of its graph.<\/strong><\/p>\n<ol id=\"fs-id1165134079627\" type=\"1\">\n<li>Factor the numerator and denominator.<\/li>\n<li>List values not 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>Any values that cause the denominator to be zero in this simplified version are where the vertical asymptotes occur.<\/li>\n<li>Any remaining values not in the domain are removable discontinuities also known as holes.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_07_05\" class=\"textbox examples\">\n<div id=\"fs-id1165137937688\">\n<div id=\"fs-id1165137645463\">\n<h3>Example 5:\u00a0 Identifying Vertical Asymptotes<\/h3>\n<p id=\"fs-id1165137627104\">Find the vertical asymptotes of the graph of [latex]k\\left(x\\right)=\\frac{5+2{x}^{2}}{2-x-{x}^{2}}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137389408\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137389408\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137389408\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137601646\">First, factor the numerator and denominator.<\/p>\n<div id=\"eip-id1165133027629\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}k\\left(x\\right)&=\\frac{5+2{x}^{2}}{2-x-{x}^{2}} \\\\ \\text{ }&=\\frac{5+2{x}^{2}}{\\left(2+x\\right)\\left(1-x\\right)}\\end{align*}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137745213\">To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero and solving:<\/p>\n<div id=\"eip-id1165132921420\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(2+x\\right)\\left(1-x\\right)=0\\hfill \\\\ \\text{ }x=-2,1\\hfill \\end{array}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165135241250\">Neither [latex]x=\u20132[\/latex] nor [latex]x=1[\/latex] are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in <a class=\"autogenerated-content\" href=\"#Figure_03_07_010\">Figure 10<\/a> confirms the location of the two vertical asymptotes.<\/p>\n<div id=\"Figure_03_07_010\" class=\"small\">\n<div style=\"width: 320px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153245\/CNX_Precalc_Figure_03_07_010.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=\"310\" height=\"327\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 10<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137640086\" class=\"bc-section section\">\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 removable discontinuity.\u00a0 These often cannot be seen when technology is used to create a graph.<\/p>\n<p id=\"fs-id1165137470947\">For example, the function [latex]f\\left(x\\right)=\\frac{{x}^{2}-1}{{x}^{2}-2x-3}[\/latex] may be re-written by factoring the numerator and the denominator.<\/p>\n<div id=\"eip-589\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{\\left(x+1\\right)\\left(x-1\\right)}{\\left(x+1\\right)\\left(x-3\\right)}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137470356\">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 NOT in the domain and 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. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_011\">Figure 11<\/a>.<\/p>\n<div id=\"Figure_03_07_011\" class=\"small\">\n<div style=\"width: 422px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153247\/CNX_Precalc_Figure_03_07_011.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=\"412\" height=\"276\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 11<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137891255\">\n<p id=\"fs-id1165137558555\">A <strong>removable discontinuity<\/strong> occurs in the graph of a rational function at [latex]x=a[\/latex] if [latex]a[\/latex] is a zero for a factor that is both in the denominator and 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 may be the location of a removable discontinuity. This is a removable discontinuity if the multiplicity of this factor in the numerator 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 a vertical asymptote at that value.<\/p>\n<\/div>\n<div id=\"Example_03_07_06\" class=\"textbox examples\">\n<div id=\"fs-id1165135168126\">\n<div id=\"fs-id1165137807554\">\n<h3>Example 6:\u00a0 Identifying Vertical Asymptotes and Removable Discontinuities for a Graph<\/h3>\n<p id=\"fs-id1165137727527\">Find the vertical asymptotes and removable discontinuities of the graph of [latex]k\\left(x\\right)=\\frac{x-2}{{x}^{2}-4}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137425735\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137425735\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137425735\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137425737\">Factor the numerator and the denominator.<\/p>\n<div id=\"eip-id1165134315756\" class=\"unnumbered\" style=\"text-align: center;\">[latex]k\\left(x\\right)=\\frac{x-2}{\\left(x-2\\right)\\left(x+2\\right)}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165135309767\">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 id=\"fs-id1165137550074\">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] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_012\">Figure 12<\/a>.<\/p>\n<div id=\"Figure_03_07_012\" class=\"small\">\n<div style=\"width: 411px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153250\/CNX_Precalc_Figure_03_07_012.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=\"401\" height=\"300\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 12<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165135634122\">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>\n<\/div>\n<div class=\"precalculus tryit\">\n<h3>Try it #5<\/h3>\n<div id=\"ti_03_07_05\">\n<div id=\"fs-id1165135686722\">\n<p id=\"fs-id1165135686724\">Find the vertical asymptotes and removable discontinuities of the graph of<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{{x}^{2}-25}{{x}^{3}-6{x}^{2}+5x}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135203443\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135203443\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135203443\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135203445\">Removable discontinuity at [latex]x=5.[\/latex] Vertical asymptotes: [latex]x=0,\\text{ }x=1.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134222349\" class=\"bc-section section\">\n<h3>Identifying Horizontal Asymptotes of Rational Functions<\/h3>\n<p id=\"fs-id1165135501072\">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 id=\"fs-id1165137503143\">There are three distinct outcomes when checking for horizontal asymptotes:<\/p>\n<p id=\"fs-id1165137503146\"><strong>Case 1:<\/strong> If the degree of the denominator is greater than the degree of the numerator, there is a <span class=\"no-emphasis\">horizontal asymptote<\/span> at [latex]y=0.[\/latex]<\/p>\n<div id=\"eip-83\" class=\"unnumbered\" style=\"text-align: center;\">Example:\u00a0 [latex]f\\left(x\\right)=\\frac{4x+2}{{x}^{2}+4x-5}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165135530372\">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] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_013\">Figure 13<\/a>. Note that this graph crosses the horizontal asymptote.<\/p>\n<div id=\"Figure_03_07_013\" class=\"wp-caption aligncenter\">\n<div style=\"width: 937px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153254\/CNX_Precalc_Figure_03_07_013.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=\"927\" height=\"311\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 13 Horizontal Asymptote [latex]y=0[\/latex] when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},\\text{ }q\\left(x\\right)\\ne 0[\/latex] where degree of [latex]p&lt;[\/latex] degree of [latex]q.[\/latex]<\/p>\n<\/div>\n<\/div>\n<div><\/div>\n<div class=\"wp-caption-text\">\n<p><strong style=\"font-size: 1rem; text-align: initial;\">Case 2:<\/strong> If the degree of the denominator is less than the degree of the numerator by one, we get a slant asymptote.<\/p>\n<\/div>\n<div id=\"eip-417\" class=\"unnumbered\" style=\"text-align: center;\">Example:\u00a0 [latex]f\\left(x\\right)=\\frac{3{x}^{2}-2x+1}{x-1}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137646911\">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] will 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<p id=\"fs-id1165137419715\">To find the equation of the slant asymptote, divide [latex]\\frac{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] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_014\">Figure 14<\/a>.<\/p>\n<div id=\"Figure_03_07_014\" class=\"small\">\n<div style=\"width: 346px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153257\/CNX_Precalc_Figure_03_07_014.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=\"336\" height=\"304\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 14\u00a0Slant Asymptote when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},\\text{ }q\\left(x\\right)\\ne 0[\/latex] where degree of [latex]p&gt;[\/latex] degree of [latex]q[\/latex] by 1.<\/p>\n<\/div>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<p><strong>Case 3:<\/strong> If the degree of the denominator equals the 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<div id=\"eip-773\" class=\"unnumbered\" style=\"text-align: center;\">Example:\u00a0 [latex]f\\left(x\\right)=\\frac{3{x}^{2}+2}{{x}^{2}+4x-5}[\/latex]<\/div>\n<div><strong>[latex]\\\\[\/latex]<\/strong><\/div>\n<p id=\"fs-id1165135593620\">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 \u00b1\\infty ,f\\left(x\\right)\\to 3,[\/latex] resulting in a horizontal asymptote of [latex]y=3.[\/latex] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_015\">Figure 15<\/a>. Note that this graph crosses the horizontal asymptote.<\/p>\n<div id=\"Figure_03_07_015\" class=\"wp-caption aligncenter\">\n<div style=\"width: 568px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153301\/CNX_Precalc_Figure_03_07_015.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=\"558\" height=\"278\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 15\u00a0Horizontal Asymptote when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},\\text{ }q\\left(x\\right)\\ne 0[\/latex] where degree of [latex]p=[\/latex] degree of [latex]q.[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<p id=\"fs-id1165137726840\">Notice that, while the graph of a rational function will never cross a <span class=\"no-emphasis\">vertical asymptote<\/span>, 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 id=\"fs-id1165137557874\">It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the <span class=\"no-emphasis\">end behavior<\/span> of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function<\/p>\n<div class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{3{x}^{5}-{x}^{2}}{x+3}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137723405\">with end behavior [latex]f\\left(x\\right)\\approx \\frac{3{x}^{5}}{x}=3{x}^{4},[\/latex] the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.\u00a0 We write as [latex]x\\to \u00b1\\infty , f\\left(x\\right)\\to \\infty.[\/latex]<\/p>\n<div id=\"fs-id1165137659475\">\n<div class=\"textbox\">\n<h3>Horizontal Asymptotes of Rational Functions<\/h3>\n<p id=\"fs-id1165137528688\">The <span class=\"no-emphasis\">horizontal asymptote<\/span> 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>\n<div id=\"Example_03_07_07\" class=\"textbox examples\">\n<div id=\"fs-id1165137812574\">\n<h3>Example 7:\u00a0 Identifying Horizontal Asymptotes<\/h3>\n<p id=\"fs-id1165134148527\">For the functions below, identify the horizontal or slant asymptote.<\/p>\n<ol type=\"a\">\n<li>[latex]g\\left(x\\right)=\\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}[\/latex]<\/li>\n<li>[latex]k\\left(x\\right)=\\frac{{x}^{2}+4x}{{x}^{3}-8}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1165137431476\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137431476\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137431476\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137542371\">For these solutions, we will use [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)}, q\\left(x\\right)\\ne 0.[\/latex]<\/p>\n<ol id=\"fs-id1165137610755\" type=\"a\">\n<li>[latex]g\\left(x\\right)=\\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}:[\/latex] The degree of [latex]p[\/latex] is [latex]3[\/latex] and the degree of [latex]q[\/latex] is [latex]3,[\/latex] so we can find the horizontal asymptote by taking the ratio of the leading terms, [latex]\\frac{6x^3}{2x^3}.[\/latex] There is a horizontal asymptote at [latex]y=\\frac{6}{2}[\/latex] or [latex]y=3.[\/latex]<\/li>\n<li>[latex]k\\left(x\\right)=\\frac{{x}^{2}+4x}{{x}^{3}-8}:[\/latex] The degree of [latex]p[\/latex] is [latex]2[\/latex] which is less than the degree of [latex]q[\/latex] which is [latex]3,[\/latex] so there is a horizontal asymptote [latex]y=0.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_07_08\" class=\"textbox examples\">\n<div id=\"fs-id1165137836670\">\n<div id=\"fs-id1165137836672\">\n<h3>Example 8:\u00a0 Identifying Horizontal Asymptotes<\/h3>\n<p id=\"fs-id1165137892264\">In the sugar concentration problem earlier, we created the equation [latex]C\\left(t\\right)=\\frac{5+t}{100+10t}.[\/latex]<\/p>\n<p id=\"fs-id1165135208611\">Find the horizontal asymptote and interpret it in context of the problem.<\/p>\n<\/div>\n<div id=\"fs-id1165137559522\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137559522\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137559522\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137559524\">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<div id=\"eip-id1165134167259\" class=\"unnumbered\" style=\"text-align: center;\">[latex]t\\to \\infty , C\\left(t\\right)\\to \\frac{1}{10}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137806518\">This function will have a horizontal asymptote at [latex]y=\\frac{1}{10}.[\/latex]<\/p>\n<p id=\"fs-id1165135450368\">This tells us that as the values of <em>t<\/em> increase, the values of [latex]C[\/latex] will 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.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_07_09\" class=\"textbox examples\">\n<div id=\"fs-id1165137843849\">\n<div id=\"fs-id1165137843851\">\n<h3>Example 9:\u00a0 Identifying Horizontal and Vertical Asymptotes<\/h3>\n<p id=\"fs-id1165137419765\">Find the horizontal and vertical asymptotes of the function [latex]f\\left(x\\right)=\\frac{\\left(x-2\\right)\\left(x+3\\right)}{\\left(x-1\\right)\\left(x+2\\right)\\left(x-5\\right)}[\/latex]<\/p>\n<div><\/div>\n<\/div>\n<div id=\"fs-id1165137731893\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137731893\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137731893\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137731895\">First, note that this function has no common factors, so there are no potential removable discontinuities.<\/p>\n<p id=\"fs-id1165137416868\">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,\u20132,\\text{and }5,[\/latex] indicating vertical asymptotes at these values.<\/p>\n<p id=\"fs-id1165137535648\">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 \u00b1\\infty , f\\left(x\\right)\\to 0.[\/latex] This function will have a horizontal asymptote at [latex]y=0.[\/latex] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_016\">Figure 16<\/a>.<\/p>\n<div id=\"Figure_03_07_016\" class=\"medium\">\n<div style=\"width: 435px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153303\/CNX_Precalc_Figure_03_07_016.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=\"425\" height=\"299\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 16<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137782334\" class=\"precalculus tryit\">\n<h3>Try it #6<\/h3>\n<div id=\"ti_03_07_06\">\n<div id=\"fs-id1165137761664\">\n<p id=\"fs-id1165137761666\">Find the vertical and horizontal asymptotes of the function [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>\n<div id=\"fs-id1165137731188\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137731188\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137731188\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137749296\">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<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134259298\">\n<div class=\"textbox examples\">\n<h3>Example 10:\u00a0 Numerical Support for Asymptotes and Discontinuities<\/h3>\n<p>Find the horizontal and vertical asymptotes and removal discontinuities of [latex]f\\left(x\\right)=\\frac{\\left(x+2\\right)\\left(x-1\\right)}{\\left(x+3\\right)\\left(x-1\\right)}[\/latex].\u00a0 Give numerical support for your conclusions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q269452\">Show Solution<\/span><\/p>\n<div id=\"q269452\" class=\"hidden-answer\" style=\"display: none\">\n<p>Setting the denominator [latex]\\left(x+3\\right)\\left(x-1\\right)[\/latex] equal to zero and solving, we find that [latex]x=-3[\/latex] and [latex]x=1[\/latex] are not in the domain of the function. Since [latex]\\left(x-1\\right)[\/latex] is in the numerator and denominator an equal number of times, we conclude that there is a removable discontinuity or hole at [latex]x=1.[\/latex]\u00a0 Since [latex]\\left(x+3\\right)[\/latex] does not have a common factor in the numerator, we conclude that there is a vertical asymptote of [latex]x=-3.[\/latex]<\/p>\n<p>To support the hole at [latex]x=1,[\/latex] we choose input values just above and just below [latex]x=1[\/latex] that get closer and closer to [latex]x=1[\/latex] and then evaluate the function there.\u00a0 We will choose our [latex]x[\/latex] values equal to 1.1, 1.01, and 1.001 for the values slightly larger than [latex]x=1[\/latex] and 0.9, 0.99 and 0.999 for the values slightly smaller than [latex]x=1.[\/latex] See Table 4 and 5.<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<caption>Table 4<\/caption>\n<tbody>\n<tr>\n<td class=\"border\" style=\"width: 25%;\">[latex]x[\/latex]<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1.1<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1.01<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1.001<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"width: 25%;\">[latex]f\\left(x\\right)[\/latex]<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.7561<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.75062<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">.750062<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<caption>Table 5<\/caption>\n<tbody>\n<tr>\n<td class=\"border\" style=\"width: 25%;\">[latex]x[\/latex]<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.9<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.99<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.999<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"width: 25%;\">[latex]f\\left(x\\right)[\/latex]<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.74359<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.74937<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.749937<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Notice, in Table 4, that the output has more zeros following 0.75 as we move across the table.\u00a0 We therefore conclude that the output is going to 0.75 as [latex]x[\/latex] goes to 1 from above.\u00a0 We write, as [latex]x\\to 1^{+},[\/latex] [latex]f\\left(x\\right)\\to 0.75.[\/latex]\u00a0 In Table 5, we see that the output increases the number of nines that follow 0.74, indicating that the output goes to 0.75.\u00a0 We write, as [latex]x\\to1^{-},\\text{ }f\\left(x\\right)\\to 0.75.[\/latex]\u00a0 This means that there is a hole at [latex]\\left(1, 0.75\\right).[\/latex]<\/p>\n<p>To support the vertical asymptote at [latex]x=-3,[\/latex] we choose values just above and just below [latex]x=-3[\/latex] that get closer and closer to [latex]x=-3[\/latex] and then evaluate the function there.\u00a0 We will choose our [latex]x[\/latex] values equal to -2.9, -2.99, and -2.999 for the values to the right of\u00a0 [latex]x=-3[\/latex] and -3.1, -3.01 and -3.001 for the values to the left of [latex]x=-3.[\/latex] See Table 6 and 7.<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<caption>Table 6<\/caption>\n<tbody>\n<tr>\n<td class=\"border\" style=\"width: 25%;\">[latex]x[\/latex]<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-2.9<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-2.99<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-2.999<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"width: 25%;\">[latex]f\\left(x\\right)[\/latex]<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-9<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-99<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-999<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<caption>Table 7<\/caption>\n<tbody>\n<tr>\n<td class=\"border\" style=\"width: 25%;\">[latex]x[\/latex]<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-3.1<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-3.01<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-3.001<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"width: 25%;\">[latex]f\\left(x\\right)[\/latex]<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">11<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">101<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1001<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Table 6 shows that as [latex]x[\/latex] gets closer and closer to -3 from the right, the output is decreasing without bound or as [latex]x\\to-3^{+},\\text{ }f\\left(x\\right)\\to-\\infty[\/latex].\u00a0 Table 7 shows that [latex]x[\/latex] gets closer and closer to -3 from the left, the output is increasing without bound or\u00a0as [latex]x\\to-3^{-},\\text{ }f\\left(x\\right)\\to\\infty[\/latex].\u00a0 This supports that there is a vertical asymptote at [latex]x=-3.[\/latex]<\/p>\n<p>Finally, to determine the horizontal asymptote, we look at the ratio of the leading terms of the numerator and denominator:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{x^2}{x^2}=1.[\/latex]<\/p>\n<p>There is a horizontal asymptote of [latex]y=1[\/latex].\u00a0 To support this numerically, we choose very large values for [latex]x[\/latex] such as 100, 1 000 and 10 000 and very negative numbers such as -100, -1 000, -10 000 and evaluate the function at these points.\u00a0 See Table 8 and 9.<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<caption>Table 8<\/caption>\n<tbody>\n<tr>\n<td class=\"border\" style=\"width: 25%;\">[latex]x[\/latex]<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">100<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1000<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">10000<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"width: 25%;\">[latex]f\\left(x\\right)[\/latex]<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.99029<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.99900299<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">0.99990002999<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<caption>Table 9<\/caption>\n<tbody>\n<tr>\n<td class=\"border\" style=\"width: 25%;\">[latex]x[\/latex]<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-100<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-1000<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">-10000<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"width: 25%;\">[latex]f\\left(x\\right)[\/latex]<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1.0103<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1.001003<\/td>\n<td class=\"border\" style=\"width: 25%; text-align: center;\">1.00010003<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Look at the pattern in the output of each table.\u00a0 We see that as [latex]x[\/latex] increases or decreases without bound, the output becomes closer and closer to 1, supporting the asymptote we found algebraically.\u00a0 We write, as [latex]x\\to\\pm\\infty,\\text{ }f\\left(x\\right)\\to 1.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Intercepts of Rational Functions<\/h3>\n<p id=\"fs-id1165137452078\">A <span class=\"no-emphasis\">rational function<\/span> 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 <em>y<\/em>-intercept if the function is not defined at zero.<\/p>\n<p id=\"fs-id1165135192756\">Likewise, a rational function will have <em>x<\/em>-intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, <em>x<\/em>-intercepts can only occur when the numerator of the rational function is equal to zero.\u00a0 However, if the numerator is zero at a value not in the domain, then there is not an x-intercept at that point.<\/p>\n<\/div>\n<div id=\"Example_03_07_10\" class=\"textbox examples\">\n<div id=\"fs-id1165135181577\">\n<div id=\"fs-id1165135181579\">\n<h3>Example 11:\u00a0 Finding the Intercepts of a Rational Function<\/h3>\n<p id=\"fs-id1165135638521\">Find the intercepts of [latex]f\\left(x\\right)=\\frac{\\left(x-2\\right)\\left(x+3\\right)}{\\left(x-1\\right)\\left(x+2\\right)\\left(x-5\\right)}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165134037668\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134037668\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134037668\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134037670\">We can find the <em>y<\/em>-intercept by evaluating the function at zero.<\/p>\n<div id=\"eip-id1165134118154\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(0\\right)&=\\frac{\\left(0-2\\right)\\left(0+3\\right)}{\\left(0-1\\right)\\left(0+2\\right)\\left(0-5\\right)} \\\\ \\text{ }&=\\frac{-6}{10}\\hfill \\\\ \\text{ }&=-\\frac{3}{5}\\hfill \\\\ \\text{ }\\text{ }&=-0.6.\\hfill \\end{align*}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137466527\">The <em>x<\/em>-intercepts will occur when the function is equal to zero:<\/p>\n<div id=\"eip-id1165134388942\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}\\hfill & \\hfill \\\\ 0=\\frac{\\left(x-2\\right)\\left(x+3\\right)}{\\left(x-1\\right)\\left(x+2\\right)\\left(x-5\\right)}\\hfill & \\textrm{This is zero when the numerator is zero}.\\hfill \\\\ 0=\\left(x-2\\right)\\left(x+3\\right)\\hfill & \\hfill \\\\ x=2, -3\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137605789\">The <em>y<\/em>-intercept is [latex]\\left(0,-0.6\\right),[\/latex] and the <em>x<\/em>-intercepts are [latex]\\left(2,0\\right)[\/latex] and [latex]\\left(-3,0\\right).[\/latex] See <a class=\"autogenerated-content\" href=\"#Figure_03_07_017\">Figure 17<\/a>.<\/p>\n<div id=\"Figure_03_07_017\" class=\"medium\">\n<div style=\"width: 471px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153307\/CNX_Precalc_Figure_03_07_017.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=\"461\" height=\"324\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 17<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137447991\" class=\"precalculus tryit\">\n<h3>Try it #7<\/h3>\n<div id=\"ti_03_07_07\">\n<div id=\"fs-id1165137460979\">\n<p id=\"fs-id1165137460980\">Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the <em>x<\/em>&#8211; and <em>y<\/em>-intercepts and the horizontal and vertical asymptotes.<\/p>\n<\/div>\n<div id=\"fs-id1165135183976\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135183976\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135183976\" class=\"hidden-answer\" style=\"display: none\">\n<div>For the transformed reciprocal squared function, we find the rational form.[latex]\\\\[\/latex]<\/div>\n<div style=\"text-align: center;\">\n<p>[latex]\\begin{align*}f\\left(x\\right)&=\\frac{1}{{\\left(x-3\\right)}^{2}}-4\\\\&=\\frac{1-4{\\left(x-3\\right)}^{2}}{{\\left(x-3\\right)}^{2}}\\\\&=\\frac{1-4\\left({x}^{2}-6x+9\\right)}{\\left(x-3\\right)\\left(x-3\\right)}\\\\&=\\frac{-4{x}^{2}+24x-35}{{x}^{2}-6x+9}\\end{align*}[\/latex]<\/p>\n<p id=\"fs-id1165137925364\">Because the numerator is the same degree as the denominator we know that as [latex]x\\to \u00b1\\infty , f\\left(x\\right)\\to -4; \\textrm{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 <em>x<\/em>-intercepts are at [latex]\\left(2.5,0\\right)[\/latex] and [latex]\\left(3.5,0\\right),[\/latex] since [latex]2.5[\/latex] and\u00a0[latex]3.5[\/latex] are in the domain.\u00a0 \u00a0Finally, we evaluate the function at 0 and find the <em>y<\/em>-intercept to be at [latex]\\left(0,\\frac{-35}{9}\\right).[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137461340\" class=\"bc-section section\">\n<h3>Graphing Rational Functions<\/h3>\n<p id=\"fs-id1165137461346\">In <a class=\"autogenerated-content\" href=\"#Example_03_07_09\">Example 11<\/a>, we see that the numerator of a rational function reveals the <em>x<\/em>-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 id=\"fs-id1165137695332\">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. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_019\">Figure 18<\/a>.<\/p>\n<div id=\"Figure_03_07_019\" class=\"small\">\n<div style=\"width: 410px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153310\/CNX_Precalc_Figure_03_07_019.jpg\" alt=\"Graph of y=1\/x with its vertical asymptote at x=0.\" width=\"400\" height=\"299\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 18<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137661229\">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. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_018\">Figure 19<\/a>.<\/p>\n<div id=\"Figure_03_07_018\" class=\"small\">\n<div style=\"width: 406px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153313\/CNX_Precalc_Figure_03_07_018.jpg\" alt=\"Graph of y=1\/x^2 with its vertical asymptote at x=0.\" width=\"396\" height=\"297\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 19<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137678097\">For example, the graph of [latex]f\\left(x\\right)=\\frac{{\\left(x+1\\right)}^{2}\\left(x-3\\right)}{{\\left(x+3\\right)}^{2}\\left(x-2\\right)}[\/latex] is shown in <a class=\"autogenerated-content\" href=\"#Figure_03_07_020\">Figure 20<\/a>.<\/p>\n<div id=\"Figure_03_07_020\" class=\"medium\">\n<div style=\"width: 478px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153316\/CNX_Precalc_Figure_03_07_020.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=\"468\" height=\"401\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 20<\/p>\n<\/div>\n<\/div>\n<ul id=\"fs-id1165137723939\">\n<li>At the <em>x<\/em>-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 <em>x<\/em>-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 id=\"fs-id1165135192767\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137640498\"><strong>Given a rational function, sketch a graph.<\/strong><\/p>\n<ol id=\"fs-id1165137640503\" type=\"1\">\n<li>Evaluate the function at 0 to find the <em>y<\/em>-intercept.<\/li>\n<li>Factor the numerator and denominator.<\/li>\n<li>For factors in the numerator that are not in the denominator, determine where each factor of the numerator is zero to find the <em>x<\/em>-intercepts.<\/li>\n<li>Find the multiplicities of the <em>x<\/em>-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 that are in both the denominator AND 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 id=\"Example_03_07_11\" class=\"textbox examples\">\n<div id=\"fs-id1165137657378\">\n<div id=\"fs-id1165137657380\">\n<h3>Example 12:\u00a0 Graphing a Rational Function<\/h3>\n<p id=\"fs-id1165135701460\">Sketch a graph of [latex]f\\left(x\\right)=\\frac{\\left(x+2\\right)\\left(x-3\\right)}{{\\left(x+1\\right)}^{2}\\left(x-2\\right)}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137413858\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137413858\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137413858\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137413860\">We can start by noting that the function is already factored, saving us a step.<\/p>\n<p id=\"fs-id1165137413863\">Next, we will find the intercepts. Evaluating the function at zero gives the <em>y<\/em>-intercept:<\/p>\n<div id=\"eip-id1165132127636\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(0\\right)&=\\frac{\\left(0+2\\right)\\left(0-3\\right)}{{\\left(0+1\\right)}^{2}\\left(0-2\\right)}\\hfill \\\\ \\text{ }&=3\\hfill \\end{align*}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137563753\">To find the <em>x<\/em>-intercepts, we determine when the numerator of the function is zero. Setting each factor equal to zero, we find <em>x<\/em>-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 id=\"fs-id1165137871570\">We have a <em>y<\/em>-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 id=\"fs-id1165132945548\">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 id=\"fs-id1165135187068\">There are no common factors in the numerator and denominator. This means there are no removable discontinuities.<\/p>\n<p id=\"fs-id1165135453215\">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 id=\"fs-id1165137696556\">To sketch the graph, we might start by plotting the three intercepts. Since the graph has no <em>x<\/em>-intercepts between the vertical asymptotes, and the <em>y<\/em>-intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph as shown in <a class=\"autogenerated-content\" href=\"#Figure_03_07_021\">Figure 21<\/a>.<\/p>\n<div id=\"Figure_03_07_021\" class=\"small\">\n<div style=\"width: 345px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153319\/CNX_Precalc_Figure_03_07_021.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=\"335\" height=\"303\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 21<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137566717\">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 id=\"fs-id1165137760010\">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. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_022\">Figure 22<\/a>. After passing through the <em>x<\/em>-intercepts, the graph will then level off toward an output of zero, as indicated by the horizontal asymptote.<\/p>\n<div id=\"Figure_03_07_022\" class=\"small\">\n<div style=\"width: 343px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153322\/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=\"333\" height=\"300\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 22<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137603685\" class=\"precalculus tryit\">\n<h3>Try it #8<\/h3>\n<div id=\"ti_03_07_08\">\n<div id=\"fs-id1165135694508\">\n<p id=\"fs-id1165137653961\">Given the function [latex]f\\left(x\\right)=\\frac{{\\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>\n<div id=\"fs-id1165137454284\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137454284\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137454284\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137454286\">Horizontal asymptote at [latex]y=\\frac{1}{2}.[\/latex] Vertical asymptotes at [latex]x=1[\/latex] and [latex]x=3.[\/latex] <em>y<\/em>-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 <em>x<\/em>-axis at this point. [latex]\\left(2,0\\right)[\/latex] is a single zero and the graph crosses the axis at this point.<\/p>\n<div style=\"width: 505px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153325\/CNX_Precalc_Figure_03_07_023.jpg\" alt=\"Graph of f(x)=(x+2)^2(x-2)\/2(x-1)^2(x-3) with its vertical and horizontal asymptotes.\" width=\"495\" height=\"323\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 23<\/p>\n<\/div>\n<p><span id=\"fs-id1165137745200\"><\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137481147\" class=\"bc-section section\">\n<h3>Writing Rational Functions<\/h3>\n<p id=\"fs-id1165137585741\">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 <em>x<\/em>-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 <em>x<\/em>-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 id=\"fs-id1165137661074\">\n<h3>Writing Rational Functions from Intercepts and Asymptotes<\/h3>\n<p id=\"fs-id1165137675581\">If a <span class=\"no-emphasis\">rational function<\/span> has <em>x<\/em>-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<div id=\"eip-597\" class=\"unnumbered\" 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][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137935737\">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] can be determined given a value of the function other than the <em>x<\/em>-intercept or by the horizontal asymptote if it is nonzero.<\/p>\n<\/div>\n<div id=\"fs-id1165137423517\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137575250\"><strong>Given a graph of a rational function, write the function.<\/strong><\/p>\n<ol id=\"fs-id1165137654907\" type=\"1\">\n<li>Determine the factors of the numerator. Examine the behavior of the graph at the <em>x<\/em>-intercepts to determine the zeros and their multiplicities. (This is easy to do when finding the \u201csimplest\u201d 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 id=\"Example_03_07_12\" class=\"textbox examples\">\n<div id=\"fs-id1165135333705\">\n<div id=\"fs-id1165135333708\">\n<h3>Example 13:\u00a0 Writing a Rational Function from Intercepts and Asymptotes<\/h3>\n<p id=\"fs-id1165135445953\">Write an equation for the rational function shown in <a class=\"autogenerated-content\" href=\"#Figure_03_07_024\">Figure 24<\/a>.<\/p>\n<div id=\"Figure_03_07_024\" class=\"small\">\n<div style=\"width: 324px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153328\/CNX_Precalc_Figure_03_07_024.jpg\" alt=\"Graph of a rational function.\" width=\"314\" height=\"306\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 24<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137767514\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137767514\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137767514\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137767516\">The graph appears to have <em>x<\/em>-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. See <a class=\"autogenerated-content\" href=\"#Figure_03_07_025\">Figure 25<\/a>.<\/p>\n<div id=\"Figure_03_07_025\" class=\"medium\">\n<div style=\"width: 478px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07153331\/CNX_Precalc_Figure_03_07_025.jpg\" alt=\"Graph of a rational function denoting its vertical asymptotes and x-intercepts.\" width=\"468\" height=\"304\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 25<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137570509\">We can use this information to write a function of the form<\/p>\n<div id=\"eip-id1165131797138\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=a\\frac{\\left(x+2\\right)\\left(x-3\\right)}{\\left(x+1\\right){\\left(x-2\\right)}^{2}}.[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137768581\">To find the stretch factor, we can use another clear point on the graph, such as the <em>y<\/em>-intercept [latex]\\left(0,-2\\right).[\/latex]<\/p>\n<div id=\"eip-id1165134237069\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}-2&=a\\frac{\\left(0+2\\right)\\left(0-3\\right)}{\\left(0+1\\right){\\left(0-2\\right)}^{2}}\\hfill \\\\ -2&=a\\frac{-6}{4}\\hfill \\\\ \\text{ }a&=\\frac{-8}{-6}=\\frac{4}{3}\\hfill \\end{align*}[\/latex]<\/div>\n<p>[latex]\\\\[\/latex]<\/p>\n<p id=\"fs-id1165134170051\">This gives us a final function of [latex]f\\left(x\\right)=\\frac{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>\n<\/div>\n<div id=\"fs-id1165135152264\" class=\"precalculus media\">\n<p id=\"fs-id1165137627838\">Access these online resources for additional instruction and practice with rational functions.<\/p>\n<ul id=\"fs-id1165135250839\">\n<li><a href=\"http:\/\/openstax.org\/l\/graphrational\">Graphing Rational Functions<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/equatrational\">Find the Equation of a Rational Function<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/asymptote\">Determining Vertical and Horizontal Asymptotes<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/interasymptote\">Find the Intercepts, Asymptotes, and Hole of a Rational Function<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137659195\" class=\"key-equations\">\n<h3>Key Equations<\/h3>\n<table id=\"eip-id1362369\" style=\"width: 1107px;\" summary=\"..\">\n<tbody>\n<tr>\n<td class=\"border\" style=\"width: 102.141px;\">Rational Function<\/td>\n<td class=\"border\" style=\"width: 978.859px;\">[latex]f\\left(x\\right)=\\displaystyle{\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{{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<\/div>\n<div id=\"fs-id1165137793507\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\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, 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\n<div class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ f\\left(x\\right)=a\\frac{\\displaystyle{{\\left(x-{x}_{1}\\right)}^{{p}_{1}}{\\left(x-{x}_{2}\\right)}^{{p}_{2}}\\cdots {\\left(x-{x}_{n}\\right)}^{{p}_{n}}}}{\\displaystyle{{\\left(x-{v}_{1}\\right)}^{{q}_{1}}{\\left(x-{v}_{2}\\right)}^{{q}_{2}}\\cdots {\\left(x-{v}_{m}\\right)}^{{q}_{n}}}}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1165137758530\">\n<dt>arrow notation<\/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\">\n<dt>horizontal asymptote<\/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\">\n<dt>rational function<\/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\">\n<dt>removable discontinuity<\/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\">\n<dt>vertical asymptote<\/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<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1405\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Rational Functions. <strong>Authored by<\/strong>: Douglas Hoffman. <strong>Provided by<\/strong>: Openstax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:DnmMrln8@11\/Rational-Functions\">https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:DnmMrln8@11\/Rational-Functions<\/a>. <strong>Project<\/strong>: Essential Precalcus, Part 1. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":311,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Rational Functions\",\"author\":\"Douglas Hoffman\",\"organization\":\"Openstax\",\"url\":\"https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:DnmMrln8@11\/Rational-Functions\",\"project\":\"Essential Precalcus, Part 1\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1405","chapter","type-chapter","status-publish","hentry"],"part":1198,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1405","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":29,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1405\/revisions"}],"predecessor-version":[{"id":3281,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1405\/revisions\/3281"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/parts\/1198"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1405\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/media?parent=1405"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1405"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/contributor?post=1405"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/license?post=1405"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}