{"id":17916,"date":"2021-08-23T06:10:05","date_gmt":"2021-08-23T06:10:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/?post_type=chapter&#038;p=17916"},"modified":"2021-08-23T07:10:09","modified_gmt":"2021-08-23T07:10:09","slug":"section-4-4-the-graph-of-a-rational-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/chapter\/section-4-4-the-graph-of-a-rational-function\/","title":{"raw":"Section 4.4: The Graph of a Rational Function","rendered":"Section 4.4: The Graph of a Rational Function"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use arrow notation.<\/li>\r\n \t<li>Graph 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, <em>x<\/em>, produced. This is given by the equation [latex]C\\left(x\\right)=15,000x - 0.1{x}^{2}+1000[\/latex]. If we want to know the average cost for producing <em>x<\/em>\u00a0items, we would divide the cost function by the number of items, <em>x<\/em>.<\/p>\r\n<p id=\"fs-id1165137768452\">The average cost function, which yields the average cost per item for <em>x<\/em>\u00a0items produced, is<\/p>\r\n\r\n<div id=\"eip-634\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{15,000x - 0.1{x}^{2}+1000}{x}[\/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 have variables in the denominator.<\/p>\r\n\r\n<h2>Use arrow notation<\/h2>\r\n<p id=\"fs-id1165137659469\">We have seen the graphs of the basic <strong>reciprocal function<\/strong> and the squared reciprocal function from our study of basic functions. Examine these graphs\u00a0and notice some of their features.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010750\/CNX_Precalc_Figure_03_07_0012.jpg\" alt=\"Graphs of f(x)=1\/x and f(x)=1\/x^2\" width=\"731\" height=\"453\" \/> <b>Figure 1<\/b>[\/caption]\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\">\r\n \t<li>On the left branch of the graph, the curve approaches the <em>x<\/em>-axis [latex]\\left(y=0\\right) \\text{ as } x\\to -\\infty [\/latex].<\/li>\r\n \t<li>As the graph approaches [latex]x=0[\/latex] from the left, the curve drops, but as we approach zero from the right, the curve rises.<\/li>\r\n \t<li>Finally, on the right branch of the graph, the curves approaches the <em>x-<\/em>axis [latex]\\left(y=0\\right) \\text{ as } x\\to \\infty [\/latex].<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1165137461553\">To summarize, we use <strong>arrow notation<\/strong> to show that <em>x<\/em>\u00a0or [latex]f\\left(x\\right)[\/latex] is approaching a particular value.<\/p>\r\n&nbsp;\r\n<table style=\"border-collapse: collapse; width: 55.991%; height: 116px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 56.1585%; height: 12px; text-align: center;\" colspan=\"2\"><strong>Arrow Notation<\/strong><strong>\r\n<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 13px;\">\r\n<td style=\"width: 16.4436%; height: 13px;\"><strong>Symbol<\/strong><\/td>\r\n<td style=\"width: 39.7149%; height: 13px;\"><strong>Meaning<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 13px;\">\r\n<td style=\"width: 16.4436%; height: 13px;\">[latex]x\\to {a}^{-}[\/latex]<\/td>\r\n<td style=\"width: 39.7149%; height: 13px;\"><em>x<\/em>\u00a0approaches <i>a<\/i>\u00a0from the left (<em>x\u00a0<\/em>&lt; <em>a<\/em>\u00a0but close to <em>a<\/em>)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 13px;\">\r\n<td style=\"width: 16.4436%; height: 13px;\">[latex]x\\to {a}^{+}[\/latex]<\/td>\r\n<td style=\"width: 39.7149%; height: 13px;\"><em>x<\/em>\u00a0approaches <em>a<\/em>\u00a0from the right (<em>x\u00a0<\/em>&gt; <em>a<\/em>\u00a0but close to <em>a<\/em>)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 13px;\">\r\n<td style=\"width: 16.4436%; height: 13px;\">[latex]x\\to \\infty\\\\ [\/latex]<\/td>\r\n<td style=\"width: 39.7149%; height: 13px;\"><em>x<\/em>\u00a0approaches infinity (<em>x<\/em>\u00a0increases without bound)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 13px;\">\r\n<td style=\"width: 16.4436%; height: 13px;\">[latex]x\\to -\\infty [\/latex]<\/td>\r\n<td style=\"width: 39.7149%; height: 13px;\"><em>x<\/em>\u00a0approaches negative infinity (<em>x<\/em>\u00a0decreases without bound)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 13px;\">\r\n<td style=\"width: 16.4436%; height: 13px;\">[latex]f\\left(x\\right)\\to \\infty [\/latex]<\/td>\r\n<td style=\"width: 39.7149%; height: 13px;\">the output approaches infinity (the output increases without bound)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 13px;\">\r\n<td style=\"width: 16.4436%; height: 13px;\">[latex]f\\left(x\\right)\\to -\\infty [\/latex]<\/td>\r\n<td style=\"width: 39.7149%; height: 13px;\">the output approaches negative infinity (the output decreases without bound)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 13px;\">\r\n<td style=\"width: 16.4436%; height: 13px;\">[latex]f\\left(x\\right)\\to a[\/latex]<\/td>\r\n<td style=\"width: 39.7149%; height: 13px;\">the output approaches\u00a0<em>a<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<section id=\"fs-id1165137759950\">\r\n<h2>Local Behavior of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/h2>\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 (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in the table below.<\/p>\r\n\r\n<table id=\"Table_03_07_002\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>\u20130.1<\/td>\r\n<td>\u20130.01<\/td>\r\n<td>\u20130.001<\/td>\r\n<td>\u20130.0001<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] <\/strong><\/td>\r\n<td>\u201310<\/td>\r\n<td>\u2013100<\/td>\r\n<td>\u20131000<\/td>\r\n<td>\u201310,000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165137542511\">We write in arrow notation<\/p>\r\n\r\n<div id=\"eip-362\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{as }x\\to {0}^{-},f\\left(x\\right)\\to -\\infty [\/latex]<\/div>\r\n<p id=\"fs-id1165137506235\">As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in the table below.<\/p>\r\n\r\n<table id=\"Table_03_07_003\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>0.1<\/td>\r\n<td>0.01<\/td>\r\n<td>0.001<\/td>\r\n<td>0.0001<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] <\/strong><\/td>\r\n<td>10<\/td>\r\n<td>100<\/td>\r\n<td>1000<\/td>\r\n<td>10,000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165134338836\">We write in arrow notation<\/p>\r\n\r\n<div id=\"eip-586\" class=\"equation unnumbered\" style=\"text-align: center;\">\r\n\r\n[latex]\\text{As }x\\to {0}^{+}, f\\left(x\\right)\\to \\infty [\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010751\/CNX_Precalc_Figure_03_07_0022.jpg\" alt=\"Graph of f(x)=1\/x which denotes the end behavior. As x goes to negative infinity, f(x) goes to 0, and as x goes to 0^-, f(x) goes to negative infinity. As x goes to positive infinity, f(x) goes to 0, and as x goes to 0^+, f(x) goes to positive infinity.\" width=\"731\" height=\"474\" \/> <b>Figure 2<\/b>[\/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 <em>x\u00a0<\/em>= 0 as the input becomes close to zero.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010751\/CNX_Precalc_Figure_03_07_0032.jpg\" alt=\"Graph of f(x)=1\/x with its vertical asymptote at x=0.\" width=\"487\" height=\"364\" \/> <b>Figure 3<\/b>[\/caption]\r\n\r\n<div id=\"fs-id1165137732344\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Vertical Asymptote<\/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 <em>a<\/em>. We write<\/p>\r\n\r\n<div id=\"eip-522\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{As }x\\to a,f\\left(x\\right)\\to \\infty , \\text{or as }x\\to a,f\\left(x\\right)\\to -\\infty [\/latex].<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137502160\">\r\n<div id=\"Example_03_07_01\" class=\"example\">\r\n<div id=\"fs-id1165133213902\" class=\"exercise\">\r\n<div id=\"fs-id1165137657454\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Using Arrow Notation<\/h3>\r\nUse arrow notation to describe the end behavior and local behavior of the function graphed in Figure 4.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010751\/CNX_Precalc_Figure_03_07_0062.jpg\" alt=\"Graph of f(x)=1\/(x-2)+4 with its vertical asymptote at x=2 and its horizontal asymptote at y=4.\" width=\"487\" height=\"477\" \/> <b>Figure 4<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"599034\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"599034\"]\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<p style=\"text-align: center;\">[latex]\\text{As }x\\to {2}^{-},f\\left(x\\right)\\to -\\infty ,\\text{ and as }x\\to {2}^{+},\\text{ }f\\left(x\\right)\\to \\infty [\/latex].<\/p>\r\n<p id=\"fs-id1165137696383\">And as the inputs decrease without bound, the graph appears to be leveling off at output values of 4, indicating a horizontal asymptote at [latex]y=4[\/latex]. As the inputs increase without bound, the graph levels off at 4.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{As }x\\to \\infty ,\\text{ }f\\left(x\\right)\\to 4\\text{ and as }x\\to -\\infty ,\\text{ }f\\left(x\\right)\\to 4[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135541748\">Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function, [latex]f(x)=\\frac{1}{x^2}[\/latex].<\/p>\r\n[reveal-answer q=\"208534\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"208534\"]\r\n\r\nEnd behavior: as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to 0[\/latex]; Local behavior: as [latex]x\\to 0, f\\left(x\\right)\\to \\infty [\/latex] (there are no <em>x<\/em>- or <em>y<\/em>-intercepts)\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_03_07_02\" class=\"example\">\r\n<div id=\"fs-id1165137694119\" class=\"exercise\">\r\n<div id=\"fs-id1165137694121\" class=\"problem textbox shaded\">\r\n<h3>Example 2: 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[reveal-answer q=\"446564\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"446564\"]\r\n<p id=\"fs-id1165137911755\">Shifting the graph left 2 and up 3 would result in the function<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{1}{x+2}+3[\/latex]<\/p>\r\n<p id=\"fs-id1165137640711\">or equivalently, by giving the terms a common denominator,<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{3x+7}{x+2}[\/latex]<\/p>\r\n<p id=\"fs-id1165137446966\">The graph of the shifted function is displayed in Figure 5.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010752\/CNX_Precalc_Figure_03_07_0072.jpg\" alt=\"Graph of f(x)=1\/(x+2)+3 with its vertical asymptote at x=-2 and its horizontal asymptote at y=3.\" width=\"731\" height=\"441\" \/> <b>Figure 5<\/b>[\/caption]\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<p style=\"text-align: center;\">[latex]\\text{As }x\\to -{2}^{-}, f\\left(x\\right)\\to -\\infty ,\\text{ and as} x\\to -{2}^{+}, f\\left(x\\right)\\to \\infty [\/latex].<\/p>\r\n<p id=\"fs-id1165137736971\">As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at [latex]y=3[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{As }x\\to \\pm \\infty , f\\left(x\\right)\\to 3[\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\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 class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\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[reveal-answer q=\"330973\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"330973\"]\r\n\r\nThe function and the asymptotes are shifted 3 units right and 4 units down. As [latex]x\\to 3,f\\left(x\\right)\\to \\infty [\/latex], and as [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to -4[\/latex].\r\n<p id=\"fs-id1165137823960\">The function is [latex]f\\left(x\\right)=\\frac{1}{{\\left(x - 3\\right)}^{2}}-4[\/latex].<\/p>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010752\/CNX_Precalc_Figure_03_07_0082.jpg\" alt=\"Graph of f(x)=1\/(x-3)^2-4 with its vertical asymptote at x=3 and its horizontal asymptote at y=-4.\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n<h2>Graph rational functions<\/h2>\r\n<section id=\"fs-id1165137461340\">\r\n<p id=\"fs-id1165137461346\">We have previously stated 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.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010755\/CNX_Precalc_Figure_03_07_0192.jpg\" alt=\"Graph of y=1\/x with its vertical asymptote at x=0.\" width=\"487\" height=\"364\" \/> <b>Figure 6<\/b>[\/caption]\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.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010755\/CNX_Precalc_Figure_03_07_0182.jpg\" alt=\"Graph of y=1\/x^2 with its vertical asymptote at x=0.\" width=\"487\" height=\"365\" \/> <b>Figure 7<\/b>[\/caption]\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 Figure 8.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010755\/CNX_Precalc_Figure_03_07_0202.jpg\" alt=\"Graph of f(x)=(x+1)^2(x-3)\/(x+3)^2(x-2) with its vertical asymptotes at x=-3 and x=2, its horizontal asymptote at y=1, and its intercepts at (-1, 0), (0, 1\/6), and (3, 0).\" width=\"731\" height=\"626\" \/> <b>Figure 8<\/b>[\/caption]\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=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137640498\">How To: Given a rational function, sketch a graph.<\/h3>\r\n<ol id=\"fs-id1165137640503\">\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 not common to 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 in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to 0 and then solve.<\/li>\r\n \t<li>Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes.<\/li>\r\n \t<li>Sketch the graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_07_11\" class=\"example\">\r\n<div id=\"fs-id1165137657378\" class=\"exercise\">\r\n<div id=\"fs-id1165137657380\" class=\"problem textbox shaded\">\r\n<h3>Example 3: 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[reveal-answer q=\"193046\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"193046\"]\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<p style=\"text-align: center;\">[latex]f\\left(0\\right)=\\frac{\\left(0+2\\right)\\left(0 - 3\\right)}{{\\left(0+1\\right)}^{2}\\left(0 - 2\\right)} =3[\/latex]<\/p>\r\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 Figure 9.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010755\/CNX_Precalc_Figure_03_07_0212.jpg\" alt=\"Graph of only the middle portion of f(x)=(x+2)(x-3)\/(x+1)^2(x-2) with its intercepts at (-2, 0), (0, 3), and (3, 0).\" width=\"487\" height=\"440\" \/> <b>Figure 9<\/b>[\/caption]\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. 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\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005253\/CNX_Precalc_Figure_03_07_022.jpg\" alt=\"Graph of f(x)=(x+2)(x-3)\/(x+1)^2(x-2) with its vertical asymptotes at x=-1 and x=2, its horizontal asymptote at y=0, and its intercepts at (-2, 0), (0, 3), and (3, 0).\" width=\"487\" height=\"439\" \/> <b>Figure 10<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\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[reveal-answer q=\"477439\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"477439\"]\r\n\r\nHorizontal asymptote at [latex]y=\\frac{1}{2}[\/latex]. Vertical asymptotes at [latex]x=1 \\text{ and } x=3[\/latex]. <em>y<\/em>-intercept at [latex]\\left(0,\\frac{4}{3}.\\right)[\/latex]\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005253\/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=\"731\" height=\"477\" \/> <b>Figure 11<\/b>[\/caption]\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.<span id=\"fs-id1165137745200\">\r\n<\/span><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137481147\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]16532[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n<section id=\"fs-id1165137659195\" class=\"key-equations\">\r\n<table id=\"eip-id1362369\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>Rational Function<\/td>\r\n<td>[latex]f\\left(x\\right)=\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{a}_{p}{x}^{p}+{a}_{p - 1}{x}^{p - 1}+...+{a}_{1}x+{a}_{0}}{{b}_{q}{x}^{q}+{b}_{q - 1}{x}^{q - 1}+...+{b}_{1}x+{b}_{0}}, Q\\left(x\\right)\\ne 0[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section><section id=\"fs-id1165137793507\" class=\"key-concepts\">\r\n<h2>Key Concepts<\/h2>\r\n<ul id=\"fs-id1165137603314\">\r\n \t<li>We can use arrow notation to describe local behavior and end behavior of the toolkit functions [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex].<\/li>\r\n \t<li>Graph rational functions by finding the intercepts, behavior at the intercepts and asymptotes, and end behavior.<\/li>\r\n<\/ul>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137758530\" class=\"definition\">\r\n \t<dt><strong>arrow notation<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135154402\">a way to symbolically represent the local and end behavior of a function by using arrows to indicate that an input or output approaches a value<\/dd>\r\n<\/dl>\r\n<\/div>\r\n<\/section>\r\n<h2 style=\"text-align: center;\">Section 4.4 Homework Exercises<\/h2>\r\nFor the following exercises, use the given transformation to graph the function. Note the vertical and horizontal asymptotes.\r\n\r\n1. The reciprocal function shifted up two units.\r\n\r\n2. The reciprocal function shifted down one unit and left three units.\r\n\r\n3. The reciprocal squared function shifted to the right 2 units.\r\n\r\n4.\u00a0The reciprocal squared function shifted down 2 units and right 1 unit.\r\n\r\nFor the following exercises, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the functions. Use that information to sketch a graph.\r\n\r\n5. [latex]p\\left(x\\right)=\\frac{2x - 3}{x+4}[\/latex]\r\n\r\n6.\u00a0[latex]q\\left(x\\right)=\\frac{x - 5}{3x - 1}[\/latex]\r\n\r\n7. [latex]s\\left(x\\right)=\\frac{4}{{\\left(x - 2\\right)}^{2}}[\/latex]\r\n\r\n8.\u00a0[latex]r\\left(x\\right)=\\frac{5}{{\\left(x+1\\right)}^{2}}[\/latex]\r\n\r\n9. [latex]f\\left(x\\right)=\\frac{3{x}^{2}-14x - 5}{3{x}^{2}+8x - 16}[\/latex]\r\n\r\n10. [latex]g\\left(x\\right)=\\frac{2{x}^{2}+7x - 15}{3{x}^{2}-14+15}[\/latex]\r\n\r\n11. [latex]a\\left(x\\right)=\\frac{{x}^{2}+2x - 3}{{x}^{2}-1}[\/latex]\r\n\r\n12. [latex]b\\left(x\\right)=\\frac{{x}^{2}-x - 6}{{x}^{2}-4}[\/latex]\r\n\r\n13. [latex]h\\left(x\\right)=\\frac{2{x}^{2}+ x - 1}{x - 4}[\/latex]\r\n\r\n14. [latex]k\\left(x\\right)=\\frac{2{x}^{2}-3x - 20}{x - 5}[\/latex]\r\n\r\n15. [latex]w\\left(x\\right)=\\frac{\\left(x - 1\\right)\\left(x+3\\right)\\left(x - 5\\right)}{{\\left(x+2\\right)}^{2}\\left(x - 4\\right)}[\/latex]\r\n\r\n16.\u00a0[latex]z\\left(x\\right)=\\frac{{\\left(x+2\\right)}^{2}\\left(x - 5\\right)}{\\left(x - 3\\right)\\left(x+1\\right)\\left(x+4\\right)}[\/latex]\r\n\r\nFor the following exercises, use a calculator to graph [latex]f\\left(x\\right)[\/latex]. Use the graph to solve [latex]f\\left(x\\right)&gt;0[\/latex].\r\n\r\n17. [latex]f\\left(x\\right)=\\frac{4}{2x - 3}[\/latex]\r\n\r\n18.\u00a0[latex]f\\left(x\\right)=\\frac{2}{\\left(x - 1\\right)\\left(x+2\\right)}[\/latex]\r\n\r\n19. [latex]f\\left(x\\right)=\\frac{x+2}{\\left(x - 1\\right)\\left(x - 4\\right)}[\/latex]\r\n\r\n20. [latex]f\\left(x\\right)=\\frac{{\\left(x+3\\right)}^{2}}{{\\left(x - 1\\right)}^{2}\\left(x+1\\right)}[\/latex]\r\n\r\n<\/section>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use arrow notation.<\/li>\n<li>Graph 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, <em>x<\/em>, produced. This is given by the equation [latex]C\\left(x\\right)=15,000x - 0.1{x}^{2}+1000[\/latex]. If we want to know the average cost for producing <em>x<\/em>\u00a0items, we would divide the cost function by the number of items, <em>x<\/em>.<\/p>\n<p id=\"fs-id1165137768452\">The average cost function, which yields the average cost per item for <em>x<\/em>\u00a0items produced, is<\/p>\n<div id=\"eip-634\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{15,000x - 0.1{x}^{2}+1000}{x}[\/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 have variables in the denominator.<\/p>\n<h2>Use arrow notation<\/h2>\n<p id=\"fs-id1165137659469\">We have seen the graphs of the basic <strong>reciprocal function<\/strong> and the squared reciprocal function from our study of basic functions. Examine these graphs\u00a0and notice some of their features.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010750\/CNX_Precalc_Figure_03_07_0012.jpg\" alt=\"Graphs of f(x)=1\/x and f(x)=1\/x^2\" width=\"731\" height=\"453\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\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\">\n<li>On the left branch of the graph, the curve approaches the <em>x<\/em>-axis [latex]\\left(y=0\\right) \\text{ as } x\\to -\\infty[\/latex].<\/li>\n<li>As the graph approaches [latex]x=0[\/latex] from the left, the curve drops, but as we approach zero from the right, the curve rises.<\/li>\n<li>Finally, on the right branch of the graph, the curves approaches the <em>x-<\/em>axis [latex]\\left(y=0\\right) \\text{ as } x\\to \\infty[\/latex].<\/li>\n<\/ol>\n<p id=\"fs-id1165137461553\">To summarize, we use <strong>arrow notation<\/strong> to show that <em>x<\/em>\u00a0or [latex]f\\left(x\\right)[\/latex] is approaching a particular value.<\/p>\n<p>&nbsp;<\/p>\n<table style=\"border-collapse: collapse; width: 55.991%; height: 116px;\">\n<tbody>\n<tr style=\"height: 12px;\">\n<td style=\"width: 56.1585%; height: 12px; text-align: center;\" colspan=\"2\"><strong>Arrow Notation<\/strong><strong><br \/>\n<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 13px;\">\n<td style=\"width: 16.4436%; height: 13px;\"><strong>Symbol<\/strong><\/td>\n<td style=\"width: 39.7149%; height: 13px;\"><strong>Meaning<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 13px;\">\n<td style=\"width: 16.4436%; height: 13px;\">[latex]x\\to {a}^{-}[\/latex]<\/td>\n<td style=\"width: 39.7149%; height: 13px;\"><em>x<\/em>\u00a0approaches <i>a<\/i>\u00a0from the left (<em>x\u00a0<\/em>&lt; <em>a<\/em>\u00a0but close to <em>a<\/em>)<\/td>\n<\/tr>\n<tr style=\"height: 13px;\">\n<td style=\"width: 16.4436%; height: 13px;\">[latex]x\\to {a}^{+}[\/latex]<\/td>\n<td style=\"width: 39.7149%; height: 13px;\"><em>x<\/em>\u00a0approaches <em>a<\/em>\u00a0from the right (<em>x\u00a0<\/em>&gt; <em>a<\/em>\u00a0but close to <em>a<\/em>)<\/td>\n<\/tr>\n<tr style=\"height: 13px;\">\n<td style=\"width: 16.4436%; height: 13px;\">[latex]x\\to \\infty\\\\[\/latex]<\/td>\n<td style=\"width: 39.7149%; height: 13px;\"><em>x<\/em>\u00a0approaches infinity (<em>x<\/em>\u00a0increases without bound)<\/td>\n<\/tr>\n<tr style=\"height: 13px;\">\n<td style=\"width: 16.4436%; height: 13px;\">[latex]x\\to -\\infty[\/latex]<\/td>\n<td style=\"width: 39.7149%; height: 13px;\"><em>x<\/em>\u00a0approaches negative infinity (<em>x<\/em>\u00a0decreases without bound)<\/td>\n<\/tr>\n<tr style=\"height: 13px;\">\n<td style=\"width: 16.4436%; height: 13px;\">[latex]f\\left(x\\right)\\to \\infty[\/latex]<\/td>\n<td style=\"width: 39.7149%; height: 13px;\">the output approaches infinity (the output increases without bound)<\/td>\n<\/tr>\n<tr style=\"height: 13px;\">\n<td style=\"width: 16.4436%; height: 13px;\">[latex]f\\left(x\\right)\\to -\\infty[\/latex]<\/td>\n<td style=\"width: 39.7149%; height: 13px;\">the output approaches negative infinity (the output decreases without bound)<\/td>\n<\/tr>\n<tr style=\"height: 13px;\">\n<td style=\"width: 16.4436%; height: 13px;\">[latex]f\\left(x\\right)\\to a[\/latex]<\/td>\n<td style=\"width: 39.7149%; height: 13px;\">the output approaches\u00a0<em>a<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<section id=\"fs-id1165137759950\">\n<h2>Local Behavior of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/h2>\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 (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in the table below.<\/p>\n<table id=\"Table_03_07_002\" summary=\"..\">\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>\u20130.1<\/td>\n<td>\u20130.01<\/td>\n<td>\u20130.001<\/td>\n<td>\u20130.0001<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] <\/strong><\/td>\n<td>\u201310<\/td>\n<td>\u2013100<\/td>\n<td>\u20131000<\/td>\n<td>\u201310,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137542511\">We write in arrow notation<\/p>\n<div id=\"eip-362\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{as }x\\to {0}^{-},f\\left(x\\right)\\to -\\infty[\/latex]<\/div>\n<p id=\"fs-id1165137506235\">As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in the table below.<\/p>\n<table id=\"Table_03_07_003\" summary=\"..\">\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>0.1<\/td>\n<td>0.01<\/td>\n<td>0.001<\/td>\n<td>0.0001<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] <\/strong><\/td>\n<td>10<\/td>\n<td>100<\/td>\n<td>1000<\/td>\n<td>10,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165134338836\">We write in arrow notation<\/p>\n<div id=\"eip-586\" class=\"equation unnumbered\" style=\"text-align: center;\">\n<p>[latex]\\text{As }x\\to {0}^{+}, f\\left(x\\right)\\to \\infty[\/latex].<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010751\/CNX_Precalc_Figure_03_07_0022.jpg\" alt=\"Graph of f(x)=1\/x which denotes the end behavior. As x goes to negative infinity, f(x) goes to 0, and as x goes to 0^-, f(x) goes to negative infinity. As x goes to positive infinity, f(x) goes to 0, and as x goes to 0^+, f(x) goes to positive infinity.\" width=\"731\" height=\"474\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/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 <em>x\u00a0<\/em>= 0 as the input becomes close to zero.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010751\/CNX_Precalc_Figure_03_07_0032.jpg\" alt=\"Graph of f(x)=1\/x with its vertical asymptote at x=0.\" width=\"487\" height=\"364\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<div id=\"fs-id1165137732344\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Vertical Asymptote<\/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 <em>a<\/em>. We write<\/p>\n<div id=\"eip-522\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{As }x\\to a,f\\left(x\\right)\\to \\infty , \\text{or as }x\\to a,f\\left(x\\right)\\to -\\infty[\/latex].<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137502160\">\n<div id=\"Example_03_07_01\" class=\"example\">\n<div id=\"fs-id1165133213902\" class=\"exercise\">\n<div id=\"fs-id1165137657454\" class=\"problem textbox shaded\">\n<h3>Example 1: Using Arrow Notation<\/h3>\n<p>Use arrow notation to describe the end behavior and local behavior of the function graphed in Figure 4.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010751\/CNX_Precalc_Figure_03_07_0062.jpg\" alt=\"Graph of f(x)=1\/(x-2)+4 with its vertical asymptote at x=2 and its horizontal asymptote at y=4.\" width=\"487\" height=\"477\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q599034\">Show Solution<\/span><\/p>\n<div id=\"q599034\" 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<p style=\"text-align: center;\">[latex]\\text{As }x\\to {2}^{-},f\\left(x\\right)\\to -\\infty ,\\text{ and as }x\\to {2}^{+},\\text{ }f\\left(x\\right)\\to \\infty[\/latex].<\/p>\n<p id=\"fs-id1165137696383\">And as the inputs decrease without bound, the graph appears to be leveling off at output values of 4, indicating a horizontal asymptote at [latex]y=4[\/latex]. As the inputs increase without bound, the graph levels off at 4.<\/p>\n<p style=\"text-align: center;\">[latex]\\text{As }x\\to \\infty ,\\text{ }f\\left(x\\right)\\to 4\\text{ and as }x\\to -\\infty ,\\text{ }f\\left(x\\right)\\to 4[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135541748\">Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function, [latex]f(x)=\\frac{1}{x^2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q208534\">Show Solution<\/span><\/p>\n<div id=\"q208534\" class=\"hidden-answer\" style=\"display: none\">\n<p>End behavior: as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to 0[\/latex]; Local behavior: as [latex]x\\to 0, f\\left(x\\right)\\to \\infty[\/latex] (there are no <em>x<\/em>&#8211; or <em>y<\/em>-intercepts)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_07_02\" class=\"example\">\n<div id=\"fs-id1165137694119\" class=\"exercise\">\n<div id=\"fs-id1165137694121\" class=\"problem textbox shaded\">\n<h3>Example 2: 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q446564\">Show Solution<\/span><\/p>\n<div id=\"q446564\" 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<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{1}{x+2}+3[\/latex]<\/p>\n<p id=\"fs-id1165137640711\">or equivalently, by giving the terms a common denominator,<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{3x+7}{x+2}[\/latex]<\/p>\n<p id=\"fs-id1165137446966\">The graph of the shifted function is displayed in Figure 5.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010752\/CNX_Precalc_Figure_03_07_0072.jpg\" alt=\"Graph of f(x)=1\/(x+2)+3 with its vertical asymptote at x=-2 and its horizontal asymptote at y=3.\" width=\"731\" height=\"441\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\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<p style=\"text-align: center;\">[latex]\\text{As }x\\to -{2}^{-}, f\\left(x\\right)\\to -\\infty ,\\text{ and as} x\\to -{2}^{+}, f\\left(x\\right)\\to \\infty[\/latex].<\/p>\n<p id=\"fs-id1165137736971\">As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at [latex]y=3[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\text{As }x\\to \\pm \\infty , f\\left(x\\right)\\to 3[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p 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 class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q330973\">Show Solution<\/span><\/p>\n<div id=\"q330973\" class=\"hidden-answer\" style=\"display: none\">\n<p>The function and the asymptotes are shifted 3 units right and 4 units down. As [latex]x\\to 3,f\\left(x\\right)\\to \\infty[\/latex], and as [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to -4[\/latex].<\/p>\n<p id=\"fs-id1165137823960\">The function is [latex]f\\left(x\\right)=\\frac{1}{{\\left(x - 3\\right)}^{2}}-4[\/latex].<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010752\/CNX_Precalc_Figure_03_07_0082.jpg\" alt=\"Graph of f(x)=1\/(x-3)^2-4 with its vertical asymptote at x=3 and its horizontal asymptote at y=-4.\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2>Graph rational functions<\/h2>\n<section id=\"fs-id1165137461340\">\n<p id=\"fs-id1165137461346\">We have previously stated 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.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010755\/CNX_Precalc_Figure_03_07_0192.jpg\" alt=\"Graph of y=1\/x with its vertical asymptote at x=0.\" width=\"487\" height=\"364\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\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.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010755\/CNX_Precalc_Figure_03_07_0182.jpg\" alt=\"Graph of y=1\/x^2 with its vertical asymptote at x=0.\" width=\"487\" height=\"365\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\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 Figure 8.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010755\/CNX_Precalc_Figure_03_07_0202.jpg\" alt=\"Graph of f(x)=(x+1)^2(x-3)\/(x+3)^2(x-2) with its vertical asymptotes at x=-3 and x=2, its horizontal asymptote at y=1, and its intercepts at (-1, 0), (0, 1\/6), and (3, 0).\" width=\"731\" height=\"626\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/b><\/p>\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=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137640498\">How To: Given a rational function, sketch a graph.<\/h3>\n<ol id=\"fs-id1165137640503\">\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 not common to 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 in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to 0 and then solve.<\/li>\n<li>Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes.<\/li>\n<li>Sketch the graph.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_07_11\" class=\"example\">\n<div id=\"fs-id1165137657378\" class=\"exercise\">\n<div id=\"fs-id1165137657380\" class=\"problem textbox shaded\">\n<h3>Example 3: 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q193046\">Show Solution<\/span><\/p>\n<div id=\"q193046\" 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<p style=\"text-align: center;\">[latex]f\\left(0\\right)=\\frac{\\left(0+2\\right)\\left(0 - 3\\right)}{{\\left(0+1\\right)}^{2}\\left(0 - 2\\right)} =3[\/latex]<\/p>\n<p 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 Figure 9.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010755\/CNX_Precalc_Figure_03_07_0212.jpg\" alt=\"Graph of only the middle portion of f(x)=(x+2)(x-3)\/(x+1)^2(x-2) with its intercepts at (-2, 0), (0, 3), and (3, 0).\" width=\"487\" height=\"440\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\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. 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 style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005253\/CNX_Precalc_Figure_03_07_022.jpg\" alt=\"Graph of f(x)=(x+2)(x-3)\/(x+1)^2(x-2) with its vertical asymptotes at x=-1 and x=2, its horizontal asymptote at y=0, and its intercepts at (-2, 0), (0, 3), and (3, 0).\" width=\"487\" height=\"439\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q477439\">Show Solution<\/span><\/p>\n<div id=\"q477439\" class=\"hidden-answer\" style=\"display: none\">\n<p>Horizontal asymptote at [latex]y=\\frac{1}{2}[\/latex]. Vertical asymptotes at [latex]x=1 \\text{ and } x=3[\/latex]. <em>y<\/em>-intercept at [latex]\\left(0,\\frac{4}{3}.\\right)[\/latex]<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005253\/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=\"731\" height=\"477\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 11<\/b><\/p>\n<\/div>\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.<span id=\"fs-id1165137745200\"><br \/>\n<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137481147\">\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm16532\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16532&theme=oea&iframe_resize_id=ohm16532\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Key Equations<\/h2>\n<section id=\"fs-id1165137659195\" class=\"key-equations\">\n<table id=\"eip-id1362369\" summary=\"..\">\n<tbody>\n<tr>\n<td>Rational Function<\/td>\n<td>[latex]f\\left(x\\right)=\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{a}_{p}{x}^{p}+{a}_{p - 1}{x}^{p - 1}+...+{a}_{1}x+{a}_{0}}{{b}_{q}{x}^{q}+{b}_{q - 1}{x}^{q - 1}+...+{b}_{1}x+{b}_{0}}, Q\\left(x\\right)\\ne 0[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<section id=\"fs-id1165137793507\" class=\"key-concepts\">\n<h2>Key Concepts<\/h2>\n<ul id=\"fs-id1165137603314\">\n<li>We can use arrow notation to describe local behavior and end behavior of the toolkit functions [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex].<\/li>\n<li>Graph rational functions by finding the intercepts, behavior at the intercepts and asymptotes, and end behavior.<\/li>\n<\/ul>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137758530\" class=\"definition\">\n<dt><strong>arrow notation<\/strong><\/dt>\n<dd id=\"fs-id1165135154402\">a way to symbolically represent the local and end behavior of a function by using arrows to indicate that an input or output approaches a value<\/dd>\n<\/dl>\n<\/div>\n<\/section>\n<h2 style=\"text-align: center;\">Section 4.4 Homework Exercises<\/h2>\n<p>For the following exercises, use the given transformation to graph the function. Note the vertical and horizontal asymptotes.<\/p>\n<p>1. The reciprocal function shifted up two units.<\/p>\n<p>2. The reciprocal function shifted down one unit and left three units.<\/p>\n<p>3. The reciprocal squared function shifted to the right 2 units.<\/p>\n<p>4.\u00a0The reciprocal squared function shifted down 2 units and right 1 unit.<\/p>\n<p>For the following exercises, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the functions. Use that information to sketch a graph.<\/p>\n<p>5. [latex]p\\left(x\\right)=\\frac{2x - 3}{x+4}[\/latex]<\/p>\n<p>6.\u00a0[latex]q\\left(x\\right)=\\frac{x - 5}{3x - 1}[\/latex]<\/p>\n<p>7. [latex]s\\left(x\\right)=\\frac{4}{{\\left(x - 2\\right)}^{2}}[\/latex]<\/p>\n<p>8.\u00a0[latex]r\\left(x\\right)=\\frac{5}{{\\left(x+1\\right)}^{2}}[\/latex]<\/p>\n<p>9. [latex]f\\left(x\\right)=\\frac{3{x}^{2}-14x - 5}{3{x}^{2}+8x - 16}[\/latex]<\/p>\n<p>10. [latex]g\\left(x\\right)=\\frac{2{x}^{2}+7x - 15}{3{x}^{2}-14+15}[\/latex]<\/p>\n<p>11. [latex]a\\left(x\\right)=\\frac{{x}^{2}+2x - 3}{{x}^{2}-1}[\/latex]<\/p>\n<p>12. [latex]b\\left(x\\right)=\\frac{{x}^{2}-x - 6}{{x}^{2}-4}[\/latex]<\/p>\n<p>13. [latex]h\\left(x\\right)=\\frac{2{x}^{2}+ x - 1}{x - 4}[\/latex]<\/p>\n<p>14. [latex]k\\left(x\\right)=\\frac{2{x}^{2}-3x - 20}{x - 5}[\/latex]<\/p>\n<p>15. [latex]w\\left(x\\right)=\\frac{\\left(x - 1\\right)\\left(x+3\\right)\\left(x - 5\\right)}{{\\left(x+2\\right)}^{2}\\left(x - 4\\right)}[\/latex]<\/p>\n<p>16.\u00a0[latex]z\\left(x\\right)=\\frac{{\\left(x+2\\right)}^{2}\\left(x - 5\\right)}{\\left(x - 3\\right)\\left(x+1\\right)\\left(x+4\\right)}[\/latex]<\/p>\n<p>For the following exercises, use a calculator to graph [latex]f\\left(x\\right)[\/latex]. Use the graph to solve [latex]f\\left(x\\right)>0[\/latex].<\/p>\n<p>17. [latex]f\\left(x\\right)=\\frac{4}{2x - 3}[\/latex]<\/p>\n<p>18.\u00a0[latex]f\\left(x\\right)=\\frac{2}{\\left(x - 1\\right)\\left(x+2\\right)}[\/latex]<\/p>\n<p>19. [latex]f\\left(x\\right)=\\frac{x+2}{\\left(x - 1\\right)\\left(x - 4\\right)}[\/latex]<\/p>\n<p>20. [latex]f\\left(x\\right)=\\frac{{\\left(x+3\\right)}^{2}}{{\\left(x - 1\\right)}^{2}\\left(x+1\\right)}[\/latex]<\/p>\n<\/section>\n","protected":false},"author":264444,"menu_order":4,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-17916","chapter","type-chapter","status-publish","hentry"],"part":17771,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17916","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/users\/264444"}],"version-history":[{"count":7,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17916\/revisions"}],"predecessor-version":[{"id":17950,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17916\/revisions\/17950"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/parts\/17771"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17916\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/media?parent=17916"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapter-type?post=17916"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/contributor?post=17916"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/license?post=17916"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}