{"id":13883,"date":"2018-08-24T22:32:32","date_gmt":"2018-08-24T22:32:32","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalcone\/?post_type=chapter&#038;p=13883"},"modified":"2021-08-23T07:22:39","modified_gmt":"2021-08-23T07:22:39","slug":"rational-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/chapter\/rational-functions\/","title":{"raw":"Section 4.3: Properties of Rational Functions","rendered":"Section 4.3: Properties of Rational Functions"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li style=\"text-align: justify;\">Solve applied problems involving rational functions.<\/li>\r\n \t<li>Find the domains of rational functions.<\/li>\r\n \t<li>Identify vertical and horizontal asymptotes.<\/li>\r\n \t<li>Identify slant asymptotes.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1165137502160\"><\/section>\r\n<h2>\u00a0Find the domains of rational functions<\/h2>\r\n<p id=\"fs-id1165137530059\">A <strong>vertical asymptote<\/strong> represents a value at which a rational function is undefined, so that value is not in the domain of the function. A 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.<\/p>\r\n\r\n<div id=\"fs-id1165134282216\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Domain of a Rational Function<\/h3>\r\n<p id=\"fs-id1165135641749\">The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135530461\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137695173\">How To: Given a rational function, find the domain.<\/h3>\r\n<ol id=\"fs-id1165137724163\">\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.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_07_04\" class=\"example\">\r\n<div id=\"fs-id1165135587815\" class=\"exercise\">\r\n<div id=\"fs-id1165137647179\" class=\"problem textbox shaded\">\r\n<h3>Example 1: 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[reveal-answer q=\"706304\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"706304\"]\r\n<p id=\"fs-id1165135445735\">Begin by setting the denominator equal to zero and solving.<\/p>\r\n<p style=\"text-align: center;\">[latex]{x}^{2}-9=0[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{x}^{2}=9[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x=\\pm 3[\/latex]<\/p>\r\n<p id=\"fs-id1165137642958\">The denominator is equal to zero when [latex]x=\\pm 3[\/latex]. The domain of the function is all real numbers except [latex]x=\\pm 3[\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165133276227\">A graph of this function confirms that the function is not defined when [latex]x=\\pm 3[\/latex].<\/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\/03010752\/CNX_Precalc_Figure_03_07_0092.jpg\" alt=\"Graph of f(x)=1\/(x-3) with its vertical asymptote at x=3 and its horizontal asymptote at y=0.\" width=\"487\" height=\"364\" \/> <b>Figure 1<\/b>[\/caption]\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 class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137452413\">Find the domain of [latex]f\\left(x\\right)=\\frac{4x}{5\\left(x - 1\\right)\\left(x - 5\\right)}[\/latex].<\/p>\r\n[reveal-answer q=\"515687\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"515687\"]\r\n\r\nThe domain is all real numbers except [latex]x=1[\/latex] and [latex]x=5[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Identify vertical asymptotes<\/h2>\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 asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location.<\/p>\r\n\r\n<section id=\"fs-id1165135194731\">\r\n<h2>Vertical Asymptotes<\/h2>\r\n<p id=\"fs-id1165137638515\">The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.<\/p>\r\n\r\n<div id=\"fs-id1165133097252\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137451766\">How To: Given a rational function, identify any vertical asymptotes of its graph.<\/h3>\r\n<ol id=\"fs-id1165134079627\">\r\n \t<li>Factor the numerator and denominator.<\/li>\r\n \t<li>Note any restrictions in the domain of the function.<\/li>\r\n \t<li>Reduce the expression by canceling common factors in the numerator and the denominator.<\/li>\r\n \t<li>Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.<\/li>\r\n \t<li>Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_07_05\" class=\"example\">\r\n<div id=\"fs-id1165137937688\" class=\"exercise\">\r\n<div id=\"fs-id1165137645463\" class=\"problem textbox shaded\">\r\n<h3>Example 2: 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[reveal-answer q=\"961060\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"961060\"]\r\n<p id=\"fs-id1165137601646\">First, factor the numerator and denominator.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}k\\left(x\\right)&amp;=\\frac{5+2{x}^{2}}{2-x-{x}^{2}} \\\\ &amp;=\\frac{5+2{x}^{2}}{\\left(2+x\\right)\\left(1-x\\right)} \\end{align}[\/latex]<\/p>\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:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\left(2+x\\right)\\left(1-x\\right)=0 \\\\ x=-2,1 \\end{gathered}[\/latex]<\/p>\r\n<p id=\"fs-id1165135241250\">Neither [latex]x=-2[\/latex] nor [latex]x=1[\/latex] are zeros of the numerator, so the two values indicate two vertical asymptotes. Figure 9\u00a0confirms the location of the two vertical asymptotes.<\/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\/03010753\/CNX_Precalc_Figure_03_07_0102.jpg\" alt=\"Graph of k(x)=(5+2x)^2\/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.\" width=\"487\" height=\"514\" \/> <b>Figure 2<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137640086\">\r\n<h2>Removable Discontinuities<\/h2>\r\n<p id=\"fs-id1165137661975\">Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a <strong>removable discontinuity<\/strong>.<\/p>\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=\"equation 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<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 the location of the removable discontinuity. Notice also that [latex]x - 3[\/latex] is not a factor in both the numerator and denominator. The zero of this factor, [latex]x=3[\/latex], is the vertical asymptote.<\/p>\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\/03010753\/CNX_Precalc_Figure_03_07_0112.jpg\" alt=\"Graph of f(x)=(x^2-1)\/(x^2-2x-3) with its vertical asymptote at x=3 and a removable discontinuity at x=-1.\" width=\"487\" height=\"326\" \/> <b>Figure 3<\/b>[\/caption]\r\n\r\n<div id=\"fs-id1165137891255\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Removable Discontinuities of Rational Functions<\/h3>\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 <em>a<\/em>\u00a0is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_03_07_06\" class=\"example\">\r\n<div id=\"fs-id1165135168126\" class=\"exercise\">\r\n<div id=\"fs-id1165137807554\" class=\"problem textbox shaded\">\r\n<h3>Example 3: 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[reveal-answer q=\"168793\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"168793\"]\r\n<p id=\"fs-id1165137425737\">Factor the numerator and the denominator.<\/p>\r\n<p style=\"text-align: center;\">[latex]k\\left(x\\right)=\\frac{x - 2}{\\left(x - 2\\right)\\left(x+2\\right)}[\/latex]<\/p>\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].<\/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\/03010753\/CNX_Precalc_Figure_03_07_0122.jpg\" alt=\"Graph of k(x)=(x-2)\/(x-2)(x+2) with its vertical asymptote at x=-2 and a removable discontinuity at x=2.\" width=\"487\" height=\"364\" \/> <b>Figure 4<\/b>[\/caption]\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=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135686724\">Find the vertical asymptotes and removable discontinuities of the graph of [latex]f\\left(x\\right)=\\frac{{x}^{2}-25}{{x}^{3}-6{x}^{2}+5x}[\/latex].<\/p>\r\n[reveal-answer q=\"711619\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"711619\"]\r\n\r\nRemovable discontinuity at [latex]x=5[\/latex]. Vertical asymptotes: [latex]x=0,\\text{ }x=1[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174202[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Identify horizontal asymptotes<\/h2>\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 &gt; degree of the numerator, there is a <strong>horizontal asymptote<\/strong> at <em>y\u00a0<\/em>= 0.<\/p>\r\n\r\n<div id=\"eip-83\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Example: }f\\left(x\\right)=\\frac{4x+2}{{x}^{2}+4x - 5}[\/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 <em>y\u00a0<\/em>= 0. Note that this graph crosses the horizontal asymptote.<\/p>\r\n&nbsp;\r\n\r\n<span id=\"fs-id1165135699169\"> <img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010753\/CNX_Precalc_Figure_03_07_0132.jpg\" alt=\"Graph of f(x)=(4x+2)\/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=0.\" width=\"900\" height=\"302\" \/><\/span>\r\n<p style=\"text-align: center;\"><strong>Figure 5.\u00a0<\/strong>Horizontal Asymptote <em>y<\/em> = 0 when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne{0}\\text{ where degree of }p&lt;\\text{degree of q}[\/latex].<\/p>\r\n<p id=\"fs-id1165137549371\"><strong>Case 2:<\/strong> If the degree of the denominator &lt; degree of the numerator by one, we get a slant asymptote.<\/p>\r\n\r\n<div id=\"eip-417\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Example: }f\\left(x\\right)=\\frac{3{x}^{2}-2x+1}{x - 1}[\/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 <em>f<\/em>\u00a0will behave similarly to <em>g<\/em>, 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].<span id=\"fs-id1165137583908\">\r\n<img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010753\/CNX_Precalc_Figure_03_07_0142.jpg\" alt=\"Graph of f(x)=(3x^2-2x+1)\/(x-1) with its vertical asymptote at x=1 and a slant asymptote aty=3x+1.\" \/><\/span><\/p>\r\n<p style=\"text-align: center;\"><strong>Figure 6.\u00a0<\/strong>Slant Asymptote when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne 0[\/latex] where degree of [latex]p&gt;\\text{ degree of }q\\text{ by }1[\/latex].<\/p>\r\n<p id=\"fs-id1165137854844\"><strong>Case 3:<\/strong> If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at [latex]y=\\frac{{a}_{n}}{{b}_{n}}[\/latex], where [latex]{a}_{n}[\/latex] and [latex]{b}_{n}[\/latex] are the leading coefficients of [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex] for [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne 0[\/latex].<\/p>\r\n\r\n<div id=\"eip-773\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Example: }f\\left(x\\right)=\\frac{3{x}^{2}+2}{{x}^{2}+4x - 5}[\/latex]<\/div>\r\nIn this case, the end behavior is [latex]f\\left(x\\right)\\approx \\frac{3{x}^{2}}{{x}^{2}}=3[\/latex]. This tells us that as the inputs grow large, this function will behave like the function [latex]g\\left(x\\right)=3[\/latex], which is a horizontal line. As [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to 3[\/latex], resulting in a horizontal asymptote at <em>y<\/em> = 3. Note that this graph crosses the horizontal asymptote.\r\n\r\n<span id=\"fs-id1165137664617\"> <img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010754\/CNX_Precalc_Figure_03_07_0152.jpg\" alt=\"Graph of f(x)=(3x^2+2)\/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=3.\" \/><\/span>\r\n<p style=\"text-align: center;\"><strong>Figure 7.\u00a0<\/strong>Horizontal Asymptote when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne 0\\text{where degree of }p=\\text{degree of }q[\/latex].<\/p>\r\n<p id=\"fs-id1165137726840\">Notice that, while the graph of a rational function will never cross a <strong>vertical asymptote<\/strong>, the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote.<\/p>\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 <strong>end behavior<\/strong> of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function<\/p>\r\n\r\n<div id=\"eip-456\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{3{x}^{5}-{x}^{2}}{x+3}[\/latex]<\/div>\r\n<p id=\"fs-id1165137723405\">with end behavior<\/p>\r\n\r\n<div id=\"eip-47\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)\\approx \\frac{3{x}^{5}}{x}=3{x}^{4}[\/latex],<\/div>\r\n<p id=\"fs-id1165137725753\">the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.<\/p>\r\n\r\n<div id=\"eip-594\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x\\to \\pm \\infty , f\\left(x\\right)\\to \\infty [\/latex]<\/div>\r\n<div id=\"fs-id1165137659475\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Horizontal Asymptotes of Rational Functions<\/h3>\r\n<p id=\"fs-id1165137528688\">The <strong>horizontal asymptote<\/strong> 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 <em>y\u00a0<\/em>= 0.<\/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 id=\"Example_03_07_07\" class=\"example\">\r\n<div id=\"fs-id1165137812572\" class=\"exercise\">\r\n<div id=\"fs-id1165137812574\" class=\"problem textbox shaded\">\r\n<h3>Example 4: Identifying Horizontal and Slant Asymptotes<\/h3>\r\n<p id=\"fs-id1165134148527\">For the functions below, identify the horizontal or slant asymptote.<\/p>\r\n\r\n<ol id=\"fs-id1165137418760\">\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]h\\left(x\\right)=\\frac{{x}^{2}-4x+1}{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[reveal-answer q=\"150080\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"150080\"]\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\">\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=\\text{degree of} q=3[\/latex], so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at [latex]y=\\frac{6}{2}[\/latex] or [latex]y=3[\/latex].<\/li>\r\n \t<li>[latex]h\\left(x\\right)=\\frac{{x}^{2}-4x+1}{x+2}[\/latex]: The degree of [latex]p=2[\/latex] and degree of [latex]q=1[\/latex]. Since [latex]p&gt;q[\/latex] by 1, there is a slant asymptote found at [latex]\\frac{{x}^{2}-4x+1}{x+2}[\/latex].\r\n<div id=\"eip-id1165134549750\" class=\"equation unnumbered\" style=\"text-align: center;\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2019\/04\/03012743\/Rational-Functions-Example-7.jpg\"><img class=\"alignnone size-full wp-image-15133\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2019\/04\/03012743\/Rational-Functions-Example-7.jpg\" alt=\"Synthetic Division of x^2-4x+1 by x+2\" width=\"205\" height=\"122\" \/><\/a><\/div>\r\n<p id=\"fs-id1165134324263\">The quotient is [latex]x - 2[\/latex] and the remainder is 13. There is a slant asymptote at [latex]y=-x - 2[\/latex].<\/p>\r\n<\/li>\r\n \t<li>[latex]k\\left(x\\right)=\\frac{{x}^{2}+4x}{{x}^{3}-8}[\/latex]: The degree of [latex]p=2\\text{ }&lt;[\/latex] degree of [latex]q=3[\/latex], so there is a horizontal asymptote <em>y<\/em> = 0.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_03_07_08\" class=\"example\">\r\n<div id=\"fs-id1165137836670\" class=\"exercise\">\r\n<div id=\"fs-id1165137836672\" class=\"problem textbox shaded\">\r\n<h3>Example 5: 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[reveal-answer q=\"258954\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"258954\"]\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 <em>t<\/em>, with coefficient 1. In the denominator, the leading term is 10<em>t<\/em>, with coefficient 10. The horizontal asymptote will be at the ratio of these values:<\/p>\r\n<p style=\"text-align: center;\">[latex]t\\to \\infty , C\\left(t\\right)\\to \\frac{1}{10}[\/latex]<\/p>\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 <em>C<\/em>\u00a0will approach [latex]\\frac{1}{10}[\/latex]. In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or [latex]\\frac{1}{10}[\/latex] pounds per gallon.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_03_07_09\" class=\"example\">\r\n<div id=\"fs-id1165137843849\" class=\"exercise\">\r\n<div id=\"fs-id1165137843851\" class=\"problem textbox shaded\">\r\n<h3>Example 6: Identifying Horizontal and Vertical Asymptotes<\/h3>\r\n<p id=\"fs-id1165137419765\">Find the horizontal and vertical asymptotes of the function<\/p>\r\n<p style=\"text-align: center;\">[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[reveal-answer q=\"965685\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"965685\"]\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,-2,\\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 \\pm \\infty , f\\left(x\\right)\\to 0[\/latex]. This function will have a horizontal asymptote at [latex]y=0[\/latex].<\/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\/03010754\/CNX_Precalc_Figure_03_07_0162.jpg\" alt=\"Graph of f(x)=(x-2)(x+3)\/(x-1)(x+2)(x-5) with its vertical asymptotes at x=-2, x=1, and x=5 and its horizontal asymptote at y=0.\" width=\"731\" height=\"514\" \/> <b>Figure 8<\/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-id1165137761666\">Find the vertical and horizontal asymptotes of the function:<\/p>\r\n<p id=\"fs-id1165137715273\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{\\left(2x - 1\\right)\\left(2x+1\\right)}{\\left(x - 2\\right)\\left(x+3\\right)}[\/latex]<\/p>\r\n[reveal-answer q=\"955718\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"955718\"]\r\n\r\nVertical asymptotes at [latex]x=2[\/latex] and [latex]x=-3[\/latex]; horizontal asymptote at [latex]y=4[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134259298\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Intercepts of Rational Functions<\/h3>\r\n<p id=\"fs-id1165137452078\">A <strong>rational function<\/strong> will have a <em>y<\/em>-intercept when the input is zero, if the function is defined at zero. A rational function will not have a <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.<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_03_07_10\" class=\"example\">\r\n<div id=\"fs-id1165135181577\" class=\"exercise\">\r\n<div id=\"fs-id1165135181579\" class=\"problem textbox shaded\">\r\n<h3>Example 7: 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[reveal-answer q=\"4797\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"4797\"]\r\n<p id=\"fs-id1165134037670\">We can find the <em>y<\/em>-intercept by evaluating the function at zero<\/p>\r\n<p 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)} \\\\ &amp;=\\frac{-6}{10} \\\\ &amp;=-\\frac{3}{5} \\\\ &amp;=-0.6 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137466527\">The <em>x<\/em>-intercepts will occur when the function is equal to zero:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} &amp;0=\\frac{\\left(x - 2\\right)\\left(x+3\\right)}{\\left(x - 1\\right)\\left(x+2\\right)\\left(x - 5\\right)} &amp;&amp; \\text{This is zero when the numerator is zero}. \\\\ &amp;0=\\left(x - 2\\right)\\left(x+3\\right) \\\\ &amp;x=2, -3 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137605789\">The <em>y<\/em>-intercept is [latex]\\left(0,-0.6\\right)[\/latex], the <em>x<\/em>-intercepts are [latex]\\left(2,0\\right)[\/latex] and [latex]\\left(-3,0\\right)[\/latex].<\/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\/03010754\/CNX_Precalc_Figure_03_07_0172.jpg\" alt=\"Graph of f(x)=(x-2)(x+3)\/(x-1)(x+2)(x-5) with its vertical asymptotes at x=-2, x=1, and x=5, its horizontal asymptote at y=0, and its intercepts at (-3, 0), (0, -0.6), and (2, 0).\" width=\"731\" height=\"514\" \/> <b>Figure 9<\/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-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[reveal-answer q=\"816980\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"816980\"]\r\n\r\nFor the transformed reciprocal squared function, we find the rational form. [latex]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}[\/latex]\r\n<p id=\"fs-id1165137925364\">Because the numerator is the same degree as the denominator we know that as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to -4; \\text{so } y=-4[\/latex] is the horizontal asymptote. Next, we set the denominator equal to zero, and find that the vertical asymptote is [latex]x=3[\/latex], because as [latex]x\\to 3,f\\left(x\\right)\\to \\infty [\/latex]. We then set the numerator equal to 0 and find the <em>x<\/em>-intercepts are at [latex]\\left(2.5,0\\right)[\/latex] and [latex]\\left(3.5,0\\right)[\/latex]. Finally, we evaluate the function at 0 and find the <em>y<\/em>-intercept to be at [latex]\\left(0,\\frac{-35}{9}\\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]105059[\/ohm_question]\r\n\r\n<\/div>\r\n<section id=\"fs-id1165137481147\">\r\n<h2>Writing Rational Functions<\/h2>\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\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Writing Rational Functions from Intercepts and Asymptotes<\/h3>\r\n<p id=\"fs-id1165137675581\">If a <strong>rational function<\/strong> 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<p style=\"text-align: center;\">[latex]f\\left(x\\right)=a\\frac{{\\left(x-{x}_{1}\\right)}^{{p}_{1}}{\\left(x-{x}_{2}\\right)}^{{p}_{2}}\\cdots {\\left(x-{x}_{n}\\right)}^{{p}_{n}}}{{\\left(x-{v}_{1}\\right)}^{{q}_{1}}{\\left(x-{v}_{2}\\right)}^{{q}_{2}}\\cdots {\\left(x-{v}_{m}\\right)}^{{q}_{n}}}[\/latex]<\/p>\r\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 <em>a\u00a0<\/em>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=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137575250\">How To: Given a graph of a rational function, write the function.<\/h3>\r\n<ol id=\"fs-id1165137654907\">\r\n \t<li>Determine the factors of the numerator. Examine the behavior of the graph at the <em>x<\/em>-intercepts to determine the zeroes and their multiplicities. (This is easy to do when finding the \"simplest\" function with small multiplicities\u2014such as 1 or 3\u2014but may be difficult for larger multiplicities\u2014such as 5 or 7, for example.)<\/li>\r\n \t<li>Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote to determine the factors and their powers.<\/li>\r\n \t<li>Use any clear point on the graph to find the stretch factor.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_07_12\" class=\"example\">\r\n<div id=\"fs-id1165135333705\" class=\"exercise\">\r\n<div id=\"fs-id1165135333708\" class=\"problem textbox shaded\">\r\n<h3>Example 8: Writing a Rational Function from Intercepts and Asymptotes<\/h3>\r\n<p id=\"fs-id1165135445953\">Write an equation for the rational function shown in Figure 23.<\/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_024.jpg\" alt=\"Graph of a rational function.\" width=\"487\" height=\"475\" \/> <b>Figure 10<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"648499\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"648499\"]\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.<\/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\/03005253\/CNX_Precalc_Figure_03_07_025.jpg\" alt=\"Graph of a rational function denoting its vertical asymptotes and x-intercepts.\" width=\"731\" height=\"475\" \/> <b>Figure 11<\/b>[\/caption]\r\n<p id=\"fs-id1165137570509\">We can use this information to write a function of the form<\/p>\r\n<p 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].<\/p>\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<p 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}} \\\\ -2&amp;=a\\frac{-6}{4} \\\\ a&amp;=\\frac{-8}{-6}=\\frac{4}{3} \\end{align}[\/latex]<\/p>\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<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>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>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>If a rational function has <em>x<\/em>-intercepts at [latex]x={x}_{1},{x}_{2},\\dots ,{x}_{n}[\/latex], vertical asymptotes at [latex]x={v}_{1},{v}_{2},\\dots ,{v}_{m}[\/latex], and no [latex]{x}_{i}=\\text{any }{v}_{j}[\/latex], then the function can be written in the form\u00a0[latex]f\\left(x\\right)=a\\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]<\/li>\r\n<\/ul>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135154407\" class=\"definition\">\r\n \t<dt><strong>horizontal asymptote<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135154413\">a horizontal line <em>y\u00a0<\/em>= <em>b<\/em>\u00a0where the graph approaches the line as the inputs increase or decrease without bound.<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135192626\" class=\"definition\">\r\n \t<dt><strong>rational function<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165134401081\">a function that can be written as the ratio of two polynomials<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134401085\" class=\"definition\">\r\n \t<dt><strong>removable discontinuity<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165134401090\">a single point at which a function is undefined that, if filled in, would make the function continuous; it appears as a hole on the graph of a function<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137426312\" class=\"definition\">\r\n \t<dt><strong>vertical asymptote<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137426317\">a vertical line <em>x\u00a0<\/em>= <em>a<\/em>\u00a0where the graph tends toward positive or negative infinity as the inputs approach\u00a0<em>a<\/em><\/dd>\r\n<\/dl>\r\n<\/div>\r\n<\/section>&nbsp;\r\n<h2 style=\"text-align: center;\">Section 4.3 Homework Exercises<\/h2>\r\n1. What is the fundamental difference in the algebraic representation of a polynomial function and a rational function?\r\n\r\n2.\u00a0What is the fundamental difference in the graphs of polynomial functions and rational functions?\r\n\r\n3. If the graph of a rational function has a removable discontinuity, what must be true of the functional rule?\r\n\r\n4.\u00a0Can a graph of a rational function have no vertical asymptote? If so, how?\r\n\r\n5. Can a graph of a rational function have no <em>x<\/em>-intercepts? If so, how?\r\n\r\nFor the following exercises, find the domain of the rational functions.\r\n\r\n6. [latex]f\\left(x\\right)=\\frac{x - 1}{x+2}[\/latex]\r\n\r\n7. [latex]f\\left(x\\right)=\\frac{x+1}{{x}^{2}-1}[\/latex]\r\n\r\n8.\u00a0[latex]f\\left(x\\right)=\\frac{{x}^{2}+4}{{x}^{2}-2x - 8}[\/latex]\r\n\r\n9. [latex]f\\left(x\\right)=\\frac{{x}^{2}+4x - 3}{{x}^{4}-5{x}^{2}+4}[\/latex]\r\n\r\nFor the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions.\r\n\r\n10. [latex]f\\left(x\\right)=\\frac{4}{x - 1}[\/latex]\r\n\r\n11. [latex]f\\left(x\\right)=\\frac{2}{5x+2}[\/latex]\r\n\r\n12.\u00a0[latex]f\\left(x\\right)=\\frac{x}{{x}^{2}-9}[\/latex]\r\n\r\n13. [latex]f\\left(x\\right)=\\frac{x}{{x}^{2}+5x - 36}[\/latex]\r\n\r\n14.\u00a0[latex]f\\left(x\\right)=\\frac{3+x}{{x}^{3}-27}[\/latex]\r\n\r\n15. [latex]f\\left(x\\right)=\\frac{3x - 4}{{x}^{3}-16x}[\/latex]\r\n\r\n16.\u00a0[latex]f\\left(x\\right)=\\frac{{x}^{2}-1}{{x}^{3}+9{x}^{2}+14x}[\/latex]\r\n\r\n17. [latex]f\\left(x\\right)=\\frac{x+5}{{x}^{2}-25}[\/latex]\r\n\r\n18.\u00a0[latex]f\\left(x\\right)=\\frac{x - 4}{x - 6}[\/latex]\r\n\r\n19. [latex]f\\left(x\\right)=\\frac{4 - 2x}{3x - 1}[\/latex]\r\n\r\nFor the following exercises, find the <em>x<\/em>- and <em>y<\/em>-intercepts for the functions.\r\n\r\n20. [latex]f\\left(x\\right)=\\frac{x+5}{{x}^{2}+4}[\/latex]\r\n\r\n21. [latex]f\\left(x\\right)=\\frac{x}{{x}^{2}-x}[\/latex]\r\n\r\n22.\u00a0[latex]f\\left(x\\right)=\\frac{{x}^{2}+8x+7}{{x}^{2}+11x+30}[\/latex]\r\n\r\n23. [latex]f\\left(x\\right)=\\frac{{x}^{2}+x+6}{{x}^{2}-10x+24}[\/latex]\r\n\r\n24. [latex]f\\left(x\\right)=\\frac{94 - 2{x}^{2}}{3{x}^{2}-12}[\/latex]\r\n\r\nFor the following exercises, describe the local and end behavior of the functions.\r\n\r\n25. [latex]f\\left(x\\right)=\\frac{x}{2x+1}[\/latex]\r\n\r\n26. [latex]f\\left(x\\right)=\\frac{2x}{x - 6}[\/latex]\r\n\r\n27. [latex]f\\left(x\\right)=\\frac{-2x}{x - 6}[\/latex]\r\n\r\n28.\u00a0[latex]f\\left(x\\right)=\\frac{{x}^{2}-4x+3}{{x}^{2}-4x - 5}[\/latex]\r\n\r\n29. [latex]f\\left(x\\right)=\\frac{2{x}^{2}-32}{6{x}^{2}+13x - 5}[\/latex]\r\n\r\nFor the following exercises, find the slant asymptote of the functions.\r\n\r\n30. [latex]f\\left(x\\right)=\\frac{24{x}^{2}+6x}{2x+1}[\/latex]\r\n\r\n31. [latex]f\\left(x\\right)=\\frac{4{x}^{2}-10}{2x - 4}[\/latex]\r\n\r\n32.\u00a0[latex]f\\left(x\\right)=\\frac{81{x}^{2}-18}{3x - 2}[\/latex]\r\n\r\n33. [latex]f\\left(x\\right)=\\frac{6{x}^{3}-5x}{3{x}^{2}+4}[\/latex]\r\n\r\n34.\u00a0[latex]f\\left(x\\right)=\\frac{{x}^{2}+5x+4}{x - 1}[\/latex]\r\n\r\nFor the following exercises, write an equation for a rational function with the given characteristics.\r\n\r\n35. Vertical asymptotes at <em>x<\/em> = 5 and <em>x\u00a0<\/em>= \u20135, <em>x<\/em>-intercepts at [latex]\\left(2,0\\right)[\/latex] and [latex]\\left(-1,0\\right)[\/latex], <em>y<\/em>-intercept at [latex]\\left(0,4\\right)[\/latex]\r\n\r\n36. Vertical asymptotes at [latex]x=-4[\/latex] and [latex]x=-1[\/latex], <em>x<\/em>-intercepts at [latex]\\left(1,0\\right)[\/latex] and [latex]\\left(5,0\\right)[\/latex], <em>y<\/em>-intercept at [latex]\\left(0,7\\right)[\/latex]\r\n\r\n37. Vertical asymptotes at [latex]x=-4[\/latex] and [latex]x=-5[\/latex], <em>x<\/em>-intercepts at [latex]\\left(4,0\\right)[\/latex] and [latex]\\left(-6,0\\right)[\/latex], Horizontal asymptote at [latex]y=7[\/latex]\r\n\r\n38.\u00a0Vertical asymptotes at [latex]x=-3[\/latex] and [latex]x=6[\/latex], <em>x<\/em>-intercepts at [latex]\\left(-2,0\\right)[\/latex] and [latex]\\left(1,0\\right)[\/latex], Horizontal asymptote at [latex]y=-2[\/latex]\r\n\r\n39. Vertical asymptote at [latex]x=-1[\/latex], Double zero at [latex]x=2[\/latex], <em>y<\/em>-intercept at [latex]\\left(0,2\\right)[\/latex]\r\n\r\n40.\u00a0Vertical asymptote at [latex]x=3[\/latex], Double zero at [latex]x=1[\/latex], <em>y<\/em>-intercept at [latex]\\left(0,4\\right)[\/latex]\r\n\r\nFor the following exercises, use the graphs to write an equation for the function.\r\n\r\n41.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005255\/CNX_Precalc_Figure_03_07_217.jpg\" alt=\"Graph of a rational function with vertical asymptotes at x=-3 and x=4.\" \/>\r\n\r\n42.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005255\/CNX_Precalc_Figure_03_07_218.jpg\" alt=\"Graph of a rational function with vertical asymptotes at x=-3 and x=4.\" \/>\r\n\r\n43.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005256\/CNX_Precalc_Figure_03_07_219.jpg\" alt=\"Graph of a rational function with vertical asymptotes at x=-3 and x=3.\" \/>\r\n\r\n44.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005256\/CNX_Precalc_Figure_03_07_220.jpg\" alt=\"Graph of a rational function with vertical asymptotes at x=-3 and x=4.\" \/>\r\n\r\n45.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005256\/CNX_Precalc_Figure_03_07_221.jpg\" alt=\"Graph of a rational function with vertical asymptote at x=1.\" \/>\r\n\r\n46.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005256\/CNX_Precalc_Figure_03_07_222.jpg\" alt=\"Graph of a rational function with vertical asymptote at x=-2.\" \/>\r\n\r\n47.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005256\/CNX_Precalc_Figure_03_07_223.jpg\" alt=\"Graph of a rational function with vertical asymptotes at x=-3 and x=2.\" \/>\r\n\r\n48.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005257\/CNX_Precalc_Figure_03_07_224.jpg\" alt=\"Graph of a rational function with vertical asymptotes at x=-2 and x=4.\" \/>\r\n\r\nFor the following exercises, make tables to show the behavior of the function near the vertical asymptote and reflecting the horizontal asymptote\r\n\r\n49. [latex]f\\left(x\\right)=\\frac{1}{x - 2}[\/latex]\r\n\r\n50.\u00a0[latex]f\\left(x\\right)=\\frac{x}{x - 3}[\/latex]\r\n\r\n51. [latex]f\\left(x\\right)=\\frac{2x}{x+4}[\/latex]\r\n\r\n52.\u00a0[latex]f\\left(x\\right)=\\frac{2x}{{\\left(x - 3\\right)}^{2}}[\/latex]\r\n\r\n53. [latex]f\\left(x\\right)=\\frac{{x}^{2}}{{x}^{2}+2x+1}[\/latex]\r\n\r\nFor the following exercises, identify the removable discontinuity.\r\n\r\n54.\u00a0[latex]f\\left(x\\right)=\\frac{{x}^{3}+1}{x+1}[\/latex]\r\n\r\n55. [latex]f\\left(x\\right)=\\frac{{x}^{2}+x - 6}{x - 2}[\/latex]\r\n\r\n56.\u00a0[latex]f\\left(x\\right)=\\frac{2{x}^{2}+5x - 3}{x+3}[\/latex]\r\n\r\n57. [latex]f\\left(x\\right)=\\frac{{x}^{3}+{x}^{2}}{x+1}[\/latex]\r\n\r\n<\/section>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li style=\"text-align: justify;\">Solve applied problems involving rational functions.<\/li>\n<li>Find the domains of rational functions.<\/li>\n<li>Identify vertical and horizontal asymptotes.<\/li>\n<li>Identify slant asymptotes.<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1165137502160\"><\/section>\n<h2>\u00a0Find the domains of rational functions<\/h2>\n<p id=\"fs-id1165137530059\">A <strong>vertical asymptote<\/strong> represents a value at which a rational function is undefined, so that value is not in the domain of the function. A 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.<\/p>\n<div id=\"fs-id1165134282216\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Domain of a Rational Function<\/h3>\n<p id=\"fs-id1165135641749\">The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.<\/p>\n<\/div>\n<div id=\"fs-id1165135530461\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137695173\">How To: Given a rational function, find the domain.<\/h3>\n<ol id=\"fs-id1165137724163\">\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.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_07_04\" class=\"example\">\n<div id=\"fs-id1165135587815\" class=\"exercise\">\n<div id=\"fs-id1165137647179\" class=\"problem textbox shaded\">\n<h3>Example 1: 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q706304\">Show Solution<\/span><\/p>\n<div id=\"q706304\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135445735\">Begin by setting the denominator equal to zero and solving.<\/p>\n<p style=\"text-align: center;\">[latex]{x}^{2}-9=0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{x}^{2}=9[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=\\pm 3[\/latex]<\/p>\n<p id=\"fs-id1165137642958\">The denominator is equal to zero when [latex]x=\\pm 3[\/latex]. The domain of the function is all real numbers except [latex]x=\\pm 3[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165133276227\">A graph of this function confirms that the function is not defined when [latex]x=\\pm 3[\/latex].<\/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\/03010752\/CNX_Precalc_Figure_03_07_0092.jpg\" alt=\"Graph of f(x)=1\/(x-3) with its vertical asymptote at x=3 and its horizontal asymptote at y=0.\" width=\"487\" height=\"364\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\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 class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137452413\">Find the domain of [latex]f\\left(x\\right)=\\frac{4x}{5\\left(x - 1\\right)\\left(x - 5\\right)}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q515687\">Show Solution<\/span><\/p>\n<div id=\"q515687\" class=\"hidden-answer\" style=\"display: none\">\n<p>The domain is all real numbers except [latex]x=1[\/latex] and [latex]x=5[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Identify vertical asymptotes<\/h2>\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 asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location.<\/p>\n<section id=\"fs-id1165135194731\">\n<h2>Vertical Asymptotes<\/h2>\n<p id=\"fs-id1165137638515\">The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.<\/p>\n<div id=\"fs-id1165133097252\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137451766\">How To: Given a rational function, identify any vertical asymptotes of its graph.<\/h3>\n<ol id=\"fs-id1165134079627\">\n<li>Factor the numerator and denominator.<\/li>\n<li>Note any restrictions in the domain of the function.<\/li>\n<li>Reduce the expression by canceling common factors in the numerator and the denominator.<\/li>\n<li>Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.<\/li>\n<li>Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_07_05\" class=\"example\">\n<div id=\"fs-id1165137937688\" class=\"exercise\">\n<div id=\"fs-id1165137645463\" class=\"problem textbox shaded\">\n<h3>Example 2: 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q961060\">Show Solution<\/span><\/p>\n<div id=\"q961060\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137601646\">First, factor the numerator and denominator.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}k\\left(x\\right)&=\\frac{5+2{x}^{2}}{2-x-{x}^{2}} \\\\ &=\\frac{5+2{x}^{2}}{\\left(2+x\\right)\\left(1-x\\right)} \\end{align}[\/latex]<\/p>\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:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\left(2+x\\right)\\left(1-x\\right)=0 \\\\ x=-2,1 \\end{gathered}[\/latex]<\/p>\n<p id=\"fs-id1165135241250\">Neither [latex]x=-2[\/latex] nor [latex]x=1[\/latex] are zeros of the numerator, so the two values indicate two vertical asymptotes. Figure 9\u00a0confirms the location of the two vertical asymptotes.<\/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\/03010753\/CNX_Precalc_Figure_03_07_0102.jpg\" alt=\"Graph of k(x)=(5+2x)^2\/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.\" width=\"487\" height=\"514\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137640086\">\n<h2>Removable Discontinuities<\/h2>\n<p id=\"fs-id1165137661975\">Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a <strong>removable discontinuity<\/strong>.<\/p>\n<p 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=\"equation 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<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 the location of the removable discontinuity. Notice also that [latex]x - 3[\/latex] is not a factor in both the numerator and denominator. The zero of this factor, [latex]x=3[\/latex], is the vertical asymptote.<\/p>\n<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\/03010753\/CNX_Precalc_Figure_03_07_0112.jpg\" alt=\"Graph of f(x)=(x^2-1)\/(x^2-2x-3) with its vertical asymptote at x=3 and a removable discontinuity at x=-1.\" width=\"487\" height=\"326\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<div id=\"fs-id1165137891255\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Removable Discontinuities of Rational Functions<\/h3>\n<p id=\"fs-id1165137558555\">A <strong>removable discontinuity<\/strong> occurs in the graph of a rational function at [latex]x=a[\/latex] if <em>a<\/em>\u00a0is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.<\/p>\n<\/div>\n<div id=\"Example_03_07_06\" class=\"example\">\n<div id=\"fs-id1165135168126\" class=\"exercise\">\n<div id=\"fs-id1165137807554\" class=\"problem textbox shaded\">\n<h3>Example 3: 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q168793\">Show Solution<\/span><\/p>\n<div id=\"q168793\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137425737\">Factor the numerator and the denominator.<\/p>\n<p style=\"text-align: center;\">[latex]k\\left(x\\right)=\\frac{x - 2}{\\left(x - 2\\right)\\left(x+2\\right)}[\/latex]<\/p>\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].<\/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\/03010753\/CNX_Precalc_Figure_03_07_0122.jpg\" alt=\"Graph of k(x)=(x-2)\/(x-2)(x+2) with its vertical asymptote at x=-2 and a removable discontinuity at x=2.\" width=\"487\" height=\"364\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\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=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135686724\">Find the vertical asymptotes and removable discontinuities of the graph of [latex]f\\left(x\\right)=\\frac{{x}^{2}-25}{{x}^{3}-6{x}^{2}+5x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q711619\">Show Solution<\/span><\/p>\n<div id=\"q711619\" class=\"hidden-answer\" style=\"display: none\">\n<p>Removable discontinuity at [latex]x=5[\/latex]. Vertical asymptotes: [latex]x=0,\\text{ }x=1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174202\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174202&theme=oea&iframe_resize_id=ohm174202\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Identify horizontal asymptotes<\/h2>\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 &gt; degree of the numerator, there is a <strong>horizontal asymptote<\/strong> at <em>y\u00a0<\/em>= 0.<\/p>\n<div id=\"eip-83\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Example: }f\\left(x\\right)=\\frac{4x+2}{{x}^{2}+4x - 5}[\/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 <em>y\u00a0<\/em>= 0. Note that this graph crosses the horizontal asymptote.<\/p>\n<p>&nbsp;<\/p>\n<p><span id=\"fs-id1165135699169\"> <img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010753\/CNX_Precalc_Figure_03_07_0132.jpg\" alt=\"Graph of f(x)=(4x+2)\/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=0.\" width=\"900\" height=\"302\" \/><\/span><\/p>\n<p style=\"text-align: center;\"><strong>Figure 5.\u00a0<\/strong>Horizontal Asymptote <em>y<\/em> = 0 when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne{0}\\text{ where degree of }p<\\text{degree of q}[\/latex].<\/p>\n<p id=\"fs-id1165137549371\"><strong>Case 2:<\/strong> If the degree of the denominator &lt; degree of the numerator by one, we get a slant asymptote.<\/p>\n<div id=\"eip-417\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Example: }f\\left(x\\right)=\\frac{3{x}^{2}-2x+1}{x - 1}[\/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 <em>f<\/em>\u00a0will behave similarly to <em>g<\/em>, 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].<span id=\"fs-id1165137583908\"><br \/>\n<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010753\/CNX_Precalc_Figure_03_07_0142.jpg\" alt=\"Graph of f(x)=(3x^2-2x+1)\/(x-1) with its vertical asymptote at x=1 and a slant asymptote aty=3x+1.\" \/><\/span><\/p>\n<p style=\"text-align: center;\"><strong>Figure 6.\u00a0<\/strong>Slant Asymptote when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne 0[\/latex] where degree of [latex]p>\\text{ degree of }q\\text{ by }1[\/latex].<\/p>\n<p id=\"fs-id1165137854844\"><strong>Case 3:<\/strong> If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at [latex]y=\\frac{{a}_{n}}{{b}_{n}}[\/latex], where [latex]{a}_{n}[\/latex] and [latex]{b}_{n}[\/latex] are the leading coefficients of [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex] for [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne 0[\/latex].<\/p>\n<div id=\"eip-773\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Example: }f\\left(x\\right)=\\frac{3{x}^{2}+2}{{x}^{2}+4x - 5}[\/latex]<\/div>\n<p>In this case, the end behavior is [latex]f\\left(x\\right)\\approx \\frac{3{x}^{2}}{{x}^{2}}=3[\/latex]. This tells us that as the inputs grow large, this function will behave like the function [latex]g\\left(x\\right)=3[\/latex], which is a horizontal line. As [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to 3[\/latex], resulting in a horizontal asymptote at <em>y<\/em> = 3. Note that this graph crosses the horizontal asymptote.<\/p>\n<p><span id=\"fs-id1165137664617\"> <img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010754\/CNX_Precalc_Figure_03_07_0152.jpg\" alt=\"Graph of f(x)=(3x^2+2)\/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=3.\" \/><\/span><\/p>\n<p style=\"text-align: center;\"><strong>Figure 7.\u00a0<\/strong>Horizontal Asymptote when [latex]f\\left(x\\right)=\\frac{p\\left(x\\right)}{q\\left(x\\right)},q\\left(x\\right)\\ne 0\\text{where degree of }p=\\text{degree of }q[\/latex].<\/p>\n<p id=\"fs-id1165137726840\">Notice that, while the graph of a rational function will never cross a <strong>vertical asymptote<\/strong>, the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote.<\/p>\n<p 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 <strong>end behavior<\/strong> of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function<\/p>\n<div id=\"eip-456\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{3{x}^{5}-{x}^{2}}{x+3}[\/latex]<\/div>\n<p id=\"fs-id1165137723405\">with end behavior<\/p>\n<div id=\"eip-47\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)\\approx \\frac{3{x}^{5}}{x}=3{x}^{4}[\/latex],<\/div>\n<p id=\"fs-id1165137725753\">the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.<\/p>\n<div id=\"eip-594\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x\\to \\pm \\infty , f\\left(x\\right)\\to \\infty[\/latex]<\/div>\n<div id=\"fs-id1165137659475\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Horizontal Asymptotes of Rational Functions<\/h3>\n<p id=\"fs-id1165137528688\">The <strong>horizontal asymptote<\/strong> of a rational function can be determined by looking at the degrees of the numerator and denominator.<\/p>\n<ul id=\"fs-id1165137722720\">\n<li>Degree of numerator <em>is less than<\/em> degree of denominator: horizontal asymptote at <em>y\u00a0<\/em>= 0.<\/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 id=\"Example_03_07_07\" class=\"example\">\n<div id=\"fs-id1165137812572\" class=\"exercise\">\n<div id=\"fs-id1165137812574\" class=\"problem textbox shaded\">\n<h3>Example 4: Identifying Horizontal and Slant Asymptotes<\/h3>\n<p id=\"fs-id1165134148527\">For the functions below, identify the horizontal or slant asymptote.<\/p>\n<ol id=\"fs-id1165137418760\">\n<li>[latex]g\\left(x\\right)=\\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}[\/latex]<\/li>\n<li>[latex]h\\left(x\\right)=\\frac{{x}^{2}-4x+1}{x+2}[\/latex]<\/li>\n<li>[latex]k\\left(x\\right)=\\frac{{x}^{2}+4x}{{x}^{3}-8}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q150080\">Show Solution<\/span><\/p>\n<div id=\"q150080\" 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\">\n<li>[latex]g\\left(x\\right)=\\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}[\/latex]: The degree of [latex]p=\\text{degree of} q=3[\/latex], so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at [latex]y=\\frac{6}{2}[\/latex] or [latex]y=3[\/latex].<\/li>\n<li>[latex]h\\left(x\\right)=\\frac{{x}^{2}-4x+1}{x+2}[\/latex]: The degree of [latex]p=2[\/latex] and degree of [latex]q=1[\/latex]. Since [latex]p>q[\/latex] by 1, there is a slant asymptote found at [latex]\\frac{{x}^{2}-4x+1}{x+2}[\/latex].\n<div id=\"eip-id1165134549750\" class=\"equation unnumbered\" style=\"text-align: center;\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2019\/04\/03012743\/Rational-Functions-Example-7.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-15133\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2019\/04\/03012743\/Rational-Functions-Example-7.jpg\" alt=\"Synthetic Division of x^2-4x+1 by x+2\" width=\"205\" height=\"122\" \/><\/a><\/div>\n<p id=\"fs-id1165134324263\">The quotient is [latex]x - 2[\/latex] and the remainder is 13. There is a slant asymptote at [latex]y=-x - 2[\/latex].<\/p>\n<\/li>\n<li>[latex]k\\left(x\\right)=\\frac{{x}^{2}+4x}{{x}^{3}-8}[\/latex]: The degree of [latex]p=2\\text{ }<[\/latex] degree of [latex]q=3[\/latex], so there is a horizontal asymptote <em>y<\/em> = 0.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_07_08\" class=\"example\">\n<div id=\"fs-id1165137836670\" class=\"exercise\">\n<div id=\"fs-id1165137836672\" class=\"problem textbox shaded\">\n<h3>Example 5: 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q258954\">Show Solution<\/span><\/p>\n<div id=\"q258954\" 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 <em>t<\/em>, with coefficient 1. In the denominator, the leading term is 10<em>t<\/em>, with coefficient 10. The horizontal asymptote will be at the ratio of these values:<\/p>\n<p style=\"text-align: center;\">[latex]t\\to \\infty , C\\left(t\\right)\\to \\frac{1}{10}[\/latex]<\/p>\n<p 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 <em>C<\/em>\u00a0will approach [latex]\\frac{1}{10}[\/latex]. In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or [latex]\\frac{1}{10}[\/latex] pounds per gallon.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_07_09\" class=\"example\">\n<div id=\"fs-id1165137843849\" class=\"exercise\">\n<div id=\"fs-id1165137843851\" class=\"problem textbox shaded\">\n<h3>Example 6: Identifying Horizontal and Vertical Asymptotes<\/h3>\n<p id=\"fs-id1165137419765\">Find the horizontal and vertical asymptotes of the function<\/p>\n<p style=\"text-align: center;\">[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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q965685\">Show Solution<\/span><\/p>\n<div id=\"q965685\" 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,-2,\\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 \\pm \\infty , f\\left(x\\right)\\to 0[\/latex]. This function will have a horizontal asymptote at [latex]y=0[\/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\/03010754\/CNX_Precalc_Figure_03_07_0162.jpg\" alt=\"Graph of f(x)=(x-2)(x+3)\/(x-1)(x+2)(x-5) with its vertical asymptotes at x=-2, x=1, and x=5 and its horizontal asymptote at y=0.\" width=\"731\" height=\"514\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/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-id1165137761666\">Find the vertical and horizontal asymptotes of the function:<\/p>\n<p id=\"fs-id1165137715273\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{\\left(2x - 1\\right)\\left(2x+1\\right)}{\\left(x - 2\\right)\\left(x+3\\right)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q955718\">Show Solution<\/span><\/p>\n<div id=\"q955718\" class=\"hidden-answer\" style=\"display: none\">\n<p>Vertical asymptotes at [latex]x=2[\/latex] and [latex]x=-3[\/latex]; horizontal asymptote at [latex]y=4[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134259298\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Intercepts of Rational Functions<\/h3>\n<p id=\"fs-id1165137452078\">A <strong>rational function<\/strong> will have a <em>y<\/em>-intercept when the input is zero, if the function is defined at zero. A rational function will not have a <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.<\/p>\n<\/div>\n<div id=\"Example_03_07_10\" class=\"example\">\n<div id=\"fs-id1165135181577\" class=\"exercise\">\n<div id=\"fs-id1165135181579\" class=\"problem textbox shaded\">\n<h3>Example 7: 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q4797\">Show Solution<\/span><\/p>\n<div id=\"q4797\" 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<p 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)} \\\\ &=\\frac{-6}{10} \\\\ &=-\\frac{3}{5} \\\\ &=-0.6 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137466527\">The <em>x<\/em>-intercepts will occur when the function is equal to zero:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} &0=\\frac{\\left(x - 2\\right)\\left(x+3\\right)}{\\left(x - 1\\right)\\left(x+2\\right)\\left(x - 5\\right)} && \\text{This is zero when the numerator is zero}. \\\\ &0=\\left(x - 2\\right)\\left(x+3\\right) \\\\ &x=2, -3 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137605789\">The <em>y<\/em>-intercept is [latex]\\left(0,-0.6\\right)[\/latex], the <em>x<\/em>-intercepts are [latex]\\left(2,0\\right)[\/latex] and [latex]\\left(-3,0\\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\/03010754\/CNX_Precalc_Figure_03_07_0172.jpg\" alt=\"Graph of f(x)=(x-2)(x+3)\/(x-1)(x+2)(x-5) with its vertical asymptotes at x=-2, x=1, and x=5, its horizontal asymptote at y=0, and its intercepts at (-3, 0), (0, -0.6), and (2, 0).\" width=\"731\" height=\"514\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/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-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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q816980\">Show Solution<\/span><\/p>\n<div id=\"q816980\" class=\"hidden-answer\" style=\"display: none\">\n<p>For the transformed reciprocal squared function, we find the rational form. [latex]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}[\/latex]<\/p>\n<p id=\"fs-id1165137925364\">Because the numerator is the same degree as the denominator we know that as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to -4; \\text{so } y=-4[\/latex] is the horizontal asymptote. Next, we set the denominator equal to zero, and find that the vertical asymptote is [latex]x=3[\/latex], because as [latex]x\\to 3,f\\left(x\\right)\\to \\infty[\/latex]. We then set the numerator equal to 0 and find the <em>x<\/em>-intercepts are at [latex]\\left(2.5,0\\right)[\/latex] and [latex]\\left(3.5,0\\right)[\/latex]. Finally, we evaluate the function at 0 and find the <em>y<\/em>-intercept to be at [latex]\\left(0,\\frac{-35}{9}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm105059\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=105059&theme=oea&iframe_resize_id=ohm105059\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<section id=\"fs-id1165137481147\">\n<h2>Writing Rational Functions<\/h2>\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\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Writing Rational Functions from Intercepts and Asymptotes<\/h3>\n<p id=\"fs-id1165137675581\">If a <strong>rational function<\/strong> 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<p style=\"text-align: center;\">[latex]f\\left(x\\right)=a\\frac{{\\left(x-{x}_{1}\\right)}^{{p}_{1}}{\\left(x-{x}_{2}\\right)}^{{p}_{2}}\\cdots {\\left(x-{x}_{n}\\right)}^{{p}_{n}}}{{\\left(x-{v}_{1}\\right)}^{{q}_{1}}{\\left(x-{v}_{2}\\right)}^{{q}_{2}}\\cdots {\\left(x-{v}_{m}\\right)}^{{q}_{n}}}[\/latex]<\/p>\n<p 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 <em>a\u00a0<\/em>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=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137575250\">How To: Given a graph of a rational function, write the function.<\/h3>\n<ol id=\"fs-id1165137654907\">\n<li>Determine the factors of the numerator. Examine the behavior of the graph at the <em>x<\/em>-intercepts to determine the zeroes and their multiplicities. (This is easy to do when finding the &#8220;simplest&#8221; function with small multiplicities\u2014such as 1 or 3\u2014but may be difficult for larger multiplicities\u2014such as 5 or 7, for example.)<\/li>\n<li>Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote to determine the factors and their powers.<\/li>\n<li>Use any clear point on the graph to find the stretch factor.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_07_12\" class=\"example\">\n<div id=\"fs-id1165135333705\" class=\"exercise\">\n<div id=\"fs-id1165135333708\" class=\"problem textbox shaded\">\n<h3>Example 8: Writing a Rational Function from Intercepts and Asymptotes<\/h3>\n<p id=\"fs-id1165135445953\">Write an equation for the rational function shown in Figure 23.<\/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_024.jpg\" alt=\"Graph of a rational function.\" width=\"487\" height=\"475\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q648499\">Show Solution<\/span><\/p>\n<div id=\"q648499\" 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.<\/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_025.jpg\" alt=\"Graph of a rational function denoting its vertical asymptotes and x-intercepts.\" width=\"731\" height=\"475\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 11<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165137570509\">We can use this information to write a function of the form<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=a\\frac{\\left(x+2\\right)\\left(x - 3\\right)}{\\left(x+1\\right){\\left(x - 2\\right)}^{2}}[\/latex].<\/p>\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<p 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}} \\\\ -2&=a\\frac{-6}{4} \\\\ a&=\\frac{-8}{-6}=\\frac{4}{3} \\end{align}[\/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<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>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>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>If a rational function has <em>x<\/em>-intercepts at [latex]x={x}_{1},{x}_{2},\\dots ,{x}_{n}[\/latex], vertical asymptotes at [latex]x={v}_{1},{v}_{2},\\dots ,{v}_{m}[\/latex], and no [latex]{x}_{i}=\\text{any }{v}_{j}[\/latex], then the function can be written in the form\u00a0[latex]f\\left(x\\right)=a\\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]<\/li>\n<\/ul>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135154407\" class=\"definition\">\n<dt><strong>horizontal asymptote<\/strong><\/dt>\n<dd id=\"fs-id1165135154413\">a horizontal line <em>y\u00a0<\/em>= <em>b<\/em>\u00a0where the graph approaches the line as the inputs increase or decrease without bound.<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135192626\" class=\"definition\">\n<dt><strong>rational function<\/strong><\/dt>\n<dd id=\"fs-id1165134401081\">a function that can be written as the ratio of two polynomials<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134401085\" class=\"definition\">\n<dt><strong>removable discontinuity<\/strong><\/dt>\n<dd id=\"fs-id1165134401090\">a single point at which a function is undefined that, if filled in, would make the function continuous; it appears as a hole on the graph of a function<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137426312\" class=\"definition\">\n<dt><strong>vertical asymptote<\/strong><\/dt>\n<dd id=\"fs-id1165137426317\">a vertical line <em>x\u00a0<\/em>= <em>a<\/em>\u00a0where the graph tends toward positive or negative infinity as the inputs approach\u00a0<em>a<\/em><\/dd>\n<\/dl>\n<\/div>\n<\/section>\n<p>&nbsp;<\/p>\n<h2 style=\"text-align: center;\">Section 4.3 Homework Exercises<\/h2>\n<p>1. What is the fundamental difference in the algebraic representation of a polynomial function and a rational function?<\/p>\n<p>2.\u00a0What is the fundamental difference in the graphs of polynomial functions and rational functions?<\/p>\n<p>3. If the graph of a rational function has a removable discontinuity, what must be true of the functional rule?<\/p>\n<p>4.\u00a0Can a graph of a rational function have no vertical asymptote? If so, how?<\/p>\n<p>5. Can a graph of a rational function have no <em>x<\/em>-intercepts? If so, how?<\/p>\n<p>For the following exercises, find the domain of the rational functions.<\/p>\n<p>6. [latex]f\\left(x\\right)=\\frac{x - 1}{x+2}[\/latex]<\/p>\n<p>7. [latex]f\\left(x\\right)=\\frac{x+1}{{x}^{2}-1}[\/latex]<\/p>\n<p>8.\u00a0[latex]f\\left(x\\right)=\\frac{{x}^{2}+4}{{x}^{2}-2x - 8}[\/latex]<\/p>\n<p>9. [latex]f\\left(x\\right)=\\frac{{x}^{2}+4x - 3}{{x}^{4}-5{x}^{2}+4}[\/latex]<\/p>\n<p>For the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions.<\/p>\n<p>10. [latex]f\\left(x\\right)=\\frac{4}{x - 1}[\/latex]<\/p>\n<p>11. [latex]f\\left(x\\right)=\\frac{2}{5x+2}[\/latex]<\/p>\n<p>12.\u00a0[latex]f\\left(x\\right)=\\frac{x}{{x}^{2}-9}[\/latex]<\/p>\n<p>13. [latex]f\\left(x\\right)=\\frac{x}{{x}^{2}+5x - 36}[\/latex]<\/p>\n<p>14.\u00a0[latex]f\\left(x\\right)=\\frac{3+x}{{x}^{3}-27}[\/latex]<\/p>\n<p>15. [latex]f\\left(x\\right)=\\frac{3x - 4}{{x}^{3}-16x}[\/latex]<\/p>\n<p>16.\u00a0[latex]f\\left(x\\right)=\\frac{{x}^{2}-1}{{x}^{3}+9{x}^{2}+14x}[\/latex]<\/p>\n<p>17. [latex]f\\left(x\\right)=\\frac{x+5}{{x}^{2}-25}[\/latex]<\/p>\n<p>18.\u00a0[latex]f\\left(x\\right)=\\frac{x - 4}{x - 6}[\/latex]<\/p>\n<p>19. [latex]f\\left(x\\right)=\\frac{4 - 2x}{3x - 1}[\/latex]<\/p>\n<p>For the following exercises, find the <em>x<\/em>&#8211; and <em>y<\/em>-intercepts for the functions.<\/p>\n<p>20. [latex]f\\left(x\\right)=\\frac{x+5}{{x}^{2}+4}[\/latex]<\/p>\n<p>21. [latex]f\\left(x\\right)=\\frac{x}{{x}^{2}-x}[\/latex]<\/p>\n<p>22.\u00a0[latex]f\\left(x\\right)=\\frac{{x}^{2}+8x+7}{{x}^{2}+11x+30}[\/latex]<\/p>\n<p>23. [latex]f\\left(x\\right)=\\frac{{x}^{2}+x+6}{{x}^{2}-10x+24}[\/latex]<\/p>\n<p>24. [latex]f\\left(x\\right)=\\frac{94 - 2{x}^{2}}{3{x}^{2}-12}[\/latex]<\/p>\n<p>For the following exercises, describe the local and end behavior of the functions.<\/p>\n<p>25. [latex]f\\left(x\\right)=\\frac{x}{2x+1}[\/latex]<\/p>\n<p>26. [latex]f\\left(x\\right)=\\frac{2x}{x - 6}[\/latex]<\/p>\n<p>27. [latex]f\\left(x\\right)=\\frac{-2x}{x - 6}[\/latex]<\/p>\n<p>28.\u00a0[latex]f\\left(x\\right)=\\frac{{x}^{2}-4x+3}{{x}^{2}-4x - 5}[\/latex]<\/p>\n<p>29. [latex]f\\left(x\\right)=\\frac{2{x}^{2}-32}{6{x}^{2}+13x - 5}[\/latex]<\/p>\n<p>For the following exercises, find the slant asymptote of the functions.<\/p>\n<p>30. [latex]f\\left(x\\right)=\\frac{24{x}^{2}+6x}{2x+1}[\/latex]<\/p>\n<p>31. [latex]f\\left(x\\right)=\\frac{4{x}^{2}-10}{2x - 4}[\/latex]<\/p>\n<p>32.\u00a0[latex]f\\left(x\\right)=\\frac{81{x}^{2}-18}{3x - 2}[\/latex]<\/p>\n<p>33. [latex]f\\left(x\\right)=\\frac{6{x}^{3}-5x}{3{x}^{2}+4}[\/latex]<\/p>\n<p>34.\u00a0[latex]f\\left(x\\right)=\\frac{{x}^{2}+5x+4}{x - 1}[\/latex]<\/p>\n<p>For the following exercises, write an equation for a rational function with the given characteristics.<\/p>\n<p>35. Vertical asymptotes at <em>x<\/em> = 5 and <em>x\u00a0<\/em>= \u20135, <em>x<\/em>-intercepts at [latex]\\left(2,0\\right)[\/latex] and [latex]\\left(-1,0\\right)[\/latex], <em>y<\/em>-intercept at [latex]\\left(0,4\\right)[\/latex]<\/p>\n<p>36. Vertical asymptotes at [latex]x=-4[\/latex] and [latex]x=-1[\/latex], <em>x<\/em>-intercepts at [latex]\\left(1,0\\right)[\/latex] and [latex]\\left(5,0\\right)[\/latex], <em>y<\/em>-intercept at [latex]\\left(0,7\\right)[\/latex]<\/p>\n<p>37. Vertical asymptotes at [latex]x=-4[\/latex] and [latex]x=-5[\/latex], <em>x<\/em>-intercepts at [latex]\\left(4,0\\right)[\/latex] and [latex]\\left(-6,0\\right)[\/latex], Horizontal asymptote at [latex]y=7[\/latex]<\/p>\n<p>38.\u00a0Vertical asymptotes at [latex]x=-3[\/latex] and [latex]x=6[\/latex], <em>x<\/em>-intercepts at [latex]\\left(-2,0\\right)[\/latex] and [latex]\\left(1,0\\right)[\/latex], Horizontal asymptote at [latex]y=-2[\/latex]<\/p>\n<p>39. Vertical asymptote at [latex]x=-1[\/latex], Double zero at [latex]x=2[\/latex], <em>y<\/em>-intercept at [latex]\\left(0,2\\right)[\/latex]<\/p>\n<p>40.\u00a0Vertical asymptote at [latex]x=3[\/latex], Double zero at [latex]x=1[\/latex], <em>y<\/em>-intercept at [latex]\\left(0,4\\right)[\/latex]<\/p>\n<p>For the following exercises, use the graphs to write an equation for the function.<\/p>\n<p>41.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005255\/CNX_Precalc_Figure_03_07_217.jpg\" alt=\"Graph of a rational function with vertical asymptotes at x=-3 and x=4.\" \/><\/p>\n<p>42.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005255\/CNX_Precalc_Figure_03_07_218.jpg\" alt=\"Graph of a rational function with vertical asymptotes at x=-3 and x=4.\" \/><\/p>\n<p>43.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005256\/CNX_Precalc_Figure_03_07_219.jpg\" alt=\"Graph of a rational function with vertical asymptotes at x=-3 and x=3.\" \/><\/p>\n<p>44.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005256\/CNX_Precalc_Figure_03_07_220.jpg\" alt=\"Graph of a rational function with vertical asymptotes at x=-3 and x=4.\" \/><\/p>\n<p>45.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005256\/CNX_Precalc_Figure_03_07_221.jpg\" alt=\"Graph of a rational function with vertical asymptote at x=1.\" \/><\/p>\n<p>46.<\/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\/03005256\/CNX_Precalc_Figure_03_07_222.jpg\" alt=\"Graph of a rational function with vertical asymptote at x=-2.\" \/><\/p>\n<p>47.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005256\/CNX_Precalc_Figure_03_07_223.jpg\" alt=\"Graph of a rational function with vertical asymptotes at x=-3 and x=2.\" \/><\/p>\n<p>48.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005257\/CNX_Precalc_Figure_03_07_224.jpg\" alt=\"Graph of a rational function with vertical asymptotes at x=-2 and x=4.\" \/><\/p>\n<p>For the following exercises, make tables to show the behavior of the function near the vertical asymptote and reflecting the horizontal asymptote<\/p>\n<p>49. [latex]f\\left(x\\right)=\\frac{1}{x - 2}[\/latex]<\/p>\n<p>50.\u00a0[latex]f\\left(x\\right)=\\frac{x}{x - 3}[\/latex]<\/p>\n<p>51. [latex]f\\left(x\\right)=\\frac{2x}{x+4}[\/latex]<\/p>\n<p>52.\u00a0[latex]f\\left(x\\right)=\\frac{2x}{{\\left(x - 3\\right)}^{2}}[\/latex]<\/p>\n<p>53. [latex]f\\left(x\\right)=\\frac{{x}^{2}}{{x}^{2}+2x+1}[\/latex]<\/p>\n<p>For the following exercises, identify the removable discontinuity.<\/p>\n<p>54.\u00a0[latex]f\\left(x\\right)=\\frac{{x}^{3}+1}{x+1}[\/latex]<\/p>\n<p>55. [latex]f\\left(x\\right)=\\frac{{x}^{2}+x - 6}{x - 2}[\/latex]<\/p>\n<p>56.\u00a0[latex]f\\left(x\\right)=\\frac{2{x}^{2}+5x - 3}{x+3}[\/latex]<\/p>\n<p>57. [latex]f\\left(x\\right)=\\frac{{x}^{3}+{x}^{2}}{x+1}[\/latex]<\/p>\n<\/section>\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-13883\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <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":97803,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"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-13883","chapter","type-chapter","status-publish","hentry"],"part":17771,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/13883","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\/97803"}],"version-history":[{"count":21,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/13883\/revisions"}],"predecessor-version":[{"id":17953,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/13883\/revisions\/17953"}],"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\/13883\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/media?parent=13883"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapter-type?post=13883"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/contributor?post=13883"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/license?post=13883"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}