{"id":17911,"date":"2021-08-23T06:08:14","date_gmt":"2021-08-23T06:08:14","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/?post_type=chapter&#038;p=17911"},"modified":"2021-08-23T06:43:37","modified_gmt":"2021-08-23T06:43:37","slug":"section-4-2-graphing-polynomial-functions-models","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/chapter\/section-4-2-graphing-polynomial-functions-models\/","title":{"raw":"Section 4.2: Graphing Polynomial Functions; Models","rendered":"Section 4.2: Graphing Polynomial Functions; Models"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Graph polynomial functions<\/li>\r\n \t<li>Use the Intermediate Value Theorem<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>\u00a0Graph polynomial functions<\/h2>\r\n<p id=\"fs-id1165137843095\">We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions.<\/p>\r\n\r\n<div id=\"fs-id1165137843101\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135449677\">How To: Given a polynomial function, sketch the graph.<\/h3>\r\n<ol id=\"fs-id1165135449683\">\r\n \t<li>Find the zeros and the <em>y<\/em>-intercept.\u00a0 Plot them. Find the multiplicities.<\/li>\r\n \t<li>Find the behavior at each zero.\u00a0 Make a small sketch of this behavior equation at each zero.<\/li>\r\n \t<li>Determine the end behavior by examining the leading term.<\/li>\r\n \t<li>Use the end behavior and the behavior at each zero to sketch a graph.<\/li>\r\n \t<li>Ensure that the number of turning points does not exceed one less than the degree of the polynomial.<\/li>\r\n \t<li>Optionally, use technology to check the graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_04_08\" class=\"example\">\r\n<div id=\"fs-id1165135575951\" class=\"exercise\">\r\n<div id=\"fs-id1165135575953\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Sketching the Graph of a Polynomial Function<\/h3>\r\n<p id=\"fs-id1165135575958\">Sketch a graph of [latex]f\\left(x\\right)=-2{\\left(x+3\\right)}^{2}\\left(x - 5\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"892446\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"892446\"]\r\n<p id=\"fs-id1165135237929\">This graph has two <em>x-<\/em>intercepts. At <em>x\u00a0<\/em>= \u20133, the factor is squared, indicating a multiplicity of 2. The graph will bounce at this <em>x<\/em>-intercept. At <em>x\u00a0<\/em>= 5, the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.<\/p>\r\n<p id=\"fs-id1165135171021\">The <em>y<\/em>-intercept is found by evaluating <em>f<\/em>(0).<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} f\\left(0\\right)&amp;=-2{\\left(0+3\\right)}^{2}\\left(0 - 5\\right) \\\\ &amp;=-2\\cdot 9\\cdot \\left(-5\\right) \\\\ &amp;=90 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165134374772\">The <em>y<\/em>-intercept is (0, 90).<\/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\/03010733\/CNX_Precalc_Figure_03_04_0172.jpg\" alt=\"Showing the distribution for the leading term.\" width=\"487\" height=\"362\" \/> <b>Figure 1<\/b>[\/caption]\r\n<p id=\"fs-id1165134381522\">Additionally, we can see the leading term, if this polynomial were multiplied out, would be [latex]-2{x}^{3}[\/latex],\r\nso the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity.<span id=\"fs-id1165135646080\">\r\n<\/span><\/p>\r\n<p id=\"fs-id1165134374738\">To sketch this, we consider that:<\/p>\r\n\r\n<ul id=\"fs-id1165134374741\">\r\n \t<li>As [latex]x\\to -\\infty [\/latex] the function [latex]f\\left(x\\right)\\to \\infty [\/latex], so we know the graph starts in the second quadrant and is decreasing toward the <em>x<\/em>-axis.<\/li>\r\n \t<li>At (0, 90), the graph crosses the <em style=\"font-size: 1rem;\">y<\/em><span style=\"font-size: 1rem;\">-axis at the <\/span><em style=\"font-size: 1rem;\">y<\/em><span style=\"font-size: 1rem;\">-intercept.<\/span><\/li>\r\n \t<li>At (-3,0), the behavior is [latex]-2{\\left(x+3\\right)}^{2}\\left(-3 - 5\\right)=16(x+3)^{2}[\/latex].\u00a0 At (-3,0) the graph will resemble a parabola opening up.<\/li>\r\n \t<li>At (5,0), the behavior is [latex]-2{\\left(5+3\\right)}^{2}\\left(x - 5\\right)=-128(x-5)[\/latex].\u00a0 At (5,0) the graph will resemble a negative sloping line.<\/li>\r\n<\/ul>\r\n<figure id=\"Figure_03_04_018\" class=\"small\">\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\/03010733\/CNX_Precalc_Figure_03_04_0182.jpg\" alt=\"Graph of the end behavior and intercepts, (-3, 0) and (0, 90), for the function f(x)=-2(x+3)^2(x-5).\" width=\"487\" height=\"362\" \/> <b>Figure 2<\/b>[\/caption]<\/figure>\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\/03010734\/CNX_Precalc_Figure_03_04_0192.jpg\" alt=\"Graph of the end behavior and intercepts, (-3, 0), (0, 90) and (5, 0), for the function f(x)=-2(x+3)^2(x-5).\" width=\"487\" height=\"362\" \/> <b>Figure 3<\/b>[\/caption]\r\n<p id=\"fs-id1165135241000\"><span id=\"fs-id1165135241013\">\u00a0<\/span><\/p>\r\n<p id=\"fs-id1165135613608\">As [latex]x\\to \\infty [\/latex] the function [latex]f\\left(x\\right)\\to \\mathrm{-\\infty }[\/latex], so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant.<\/p>\r\n<p id=\"fs-id1165135574296\">Using technology, we can create the graph for the polynomial function, shown in Figure 4, and verify that the resulting graph looks like our sketch in Figure 15.<\/p>\r\n\r\n<figure id=\"Figure_03_04_020\" class=\"small\"><figcaption>The complete graph of the polynomial function [latex]f\\left(x\\right)=-2{\\left(x+3\\right)}^{2}\\left(x - 5\\right)[\/latex]<\/figcaption>\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\/03010734\/CNX_Precalc_Figure_03_04_0202.jpg\" alt=\"Graph of f(x)=-2(x+3)^2(x-5).\" width=\"487\" height=\"366\" \/> <b>Figure 4<\/b>[\/caption]<\/figure>\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-id1165133065140\">Sketch a graph of [latex]f\\left(x\\right)=\\frac{1}{4}x{\\left(x - 1\\right)}^{4}{\\left(x+3\\right)}^{3}[\/latex].<\/p>\r\n[reveal-answer q=\"408253\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"408253\"]\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\/03010734\/CNX_Precalc_Figure_03_04_0212.jpg\" alt=\"Graph of f(x)=(1\/4)x(x-1)^4(x+3)^3.\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Use the Intermediate Value Theorem<\/h2>\r\n<p id=\"fs-id1165135205093\">In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the <em>x<\/em>-axis, we can confirm that there is a zero between them. Consider a polynomial function <em>f<\/em>\u00a0whose graph is smooth and continuous. The <strong>Intermediate Value Theorem<\/strong> states that for two numbers <em>a<\/em>\u00a0and <em>b<\/em>\u00a0in the domain of <em>f<\/em>,\u00a0if <em>a\u00a0<\/em>&lt; <em>b<\/em>\u00a0and [latex]f\\left(a\\right)\\ne f\\left(b\\right)[\/latex], then the function <em>f<\/em>\u00a0takes on every value between [latex]f\\left(a\\right)[\/latex] and [latex]f\\left(b\\right)[\/latex].<\/p>\r\nWe can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function <em>f<\/em>\u00a0at [latex]x=a[\/latex] lies above the <em>x<\/em>-axis and another point at [latex]x=b[\/latex] lies below the <em>x<\/em>-axis, there must exist a third point between [latex]x=a[\/latex] and [latex]x=b[\/latex] where the graph crosses the <em>x<\/em>-axis. Call this point [latex]\\left(c,\\text{ }f\\left(c\\right)\\right)[\/latex]. This means that we are assured there is a solution <em>c<\/em>\u00a0where [latex]f\\left(c\\right)=0[\/latex].\r\n\r\nIn other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the <em>x<\/em>-axis. Figure 5\u00a0shows that there is a zero between <em>a<\/em>\u00a0and <em>b<\/em>.\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\/03010734\/CNX_Precalc_Figure_03_04_0222.jpg\" alt=\"Graph of an odd-degree polynomial function that shows a point f(a) that\u2019s negative, f(b) that\u2019s positive, and f(c) that\u2019s 0.\" width=\"487\" height=\"368\" \/> <b>Figure 5.<\/b> Using the Intermediate Value Theorem to show there exists a zero.[\/caption]\r\n\r\n<div id=\"fs-id1165135347510\" class=\"note textbox shaded\">\r\n<h3 class=\"title\">A General Note: Intermediate Value Theorem<\/h3>\r\n<p id=\"fs-id1165135580347\">Let <em>f<\/em>\u00a0be a polynomial function. The <strong>Intermediate Value Theorem<\/strong> states that if [latex]f\\left(a\\right)[\/latex]\u00a0and [latex]f\\left(b\\right)[\/latex]\u00a0have opposite signs, then there exists at least one value <em>c<\/em>\u00a0between <em>a<\/em>\u00a0and <em>b<\/em>\u00a0for which [latex]f\\left(c\\right)=0[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_03_04_09\" class=\"example\">\r\n<div id=\"fs-id1165133358799\" class=\"exercise\">\r\n<div id=\"fs-id1165133358801\" class=\"problem textbox shaded\">\r\n<h3>Example 2: Using the Intermediate Value Theorem<\/h3>\r\n<p id=\"fs-id1165133358807\">Show that the function [latex]f\\left(x\\right)={x}^{3}-5{x}^{2}+3x+6[\/latex]\u00a0has at least two real zeros between [latex]x=1[\/latex]\u00a0and [latex]x=4[\/latex].<\/p>\r\n[reveal-answer q=\"98939\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"98939\"]\r\n<p id=\"fs-id1165135537349\">As a start, evaluate [latex]f\\left(x\\right)[\/latex]\u00a0at the integer values [latex]x=1,2,3,\\text{ and }4[\/latex].<\/p>\r\n\r\n<table id=\"Table_03_04_03\" summary=\"..\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em><strong>f\u00a0<\/strong><\/em><strong>(<em>x<\/em>)<\/strong><\/td>\r\n<td>5<\/td>\r\n<td>0<\/td>\r\n<td>\u20133<\/td>\r\n<td>2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165135536378\">We see that one zero occurs at [latex]x=2[\/latex]. Also, since [latex]f\\left(3\\right)[\/latex] is negative and [latex]f\\left(4\\right)[\/latex] is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4.<\/p>\r\n<p id=\"fs-id1165135575934\">We have shown that there are at least two real zeros between [latex]x=1[\/latex]\u00a0and [latex]x=4[\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can also see in Figure 6\u00a0that there are two real zeros between [latex]x=1[\/latex]\u00a0and [latex]x=4[\/latex].\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\/03010735\/CNX_Precalc_Figure_03_04_0232.jpg\" alt=\"Graph of f(x)=x^3-5x^2+3x+6 and shows, by the Intermediate Value Theorem, that there exists two zeros since f(1)=5 and f(4)=2 are positive and f(3) = -3 is negative.\" width=\"487\" height=\"591\" \/> <b>Figure 6<\/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-id1165135551168\">Show that the function [latex]f\\left(x\\right)=7{x}^{5}-9{x}^{4}-{x}^{2}[\/latex] has at least one real zero between [latex]x=1[\/latex] and [latex]x=2[\/latex].<\/p>\r\n[reveal-answer q=\"886741\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"886741\"]\r\n\r\nBecause <em>f<\/em>\u00a0is a polynomial function and since [latex]f\\left(1\\right)[\/latex] is negative and [latex]f\\left(2\\right)[\/latex] is positive, there is at least one real zero between [latex]x=1[\/latex] and [latex]x=2[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul id=\"fs-id1165137846272\">\r\n \t<li>To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most<em>\u00a0n <\/em>\u2013\u00a01 turning points.<\/li>\r\n \t<li>Graphing a polynomial function helps to estimate local and global extremas.<\/li>\r\n \t<li>The Intermediate Value Theorem tells us that if [latex]f\\left(a\\right) \\text{and} f\\left(b\\right)[\/latex]\u00a0have opposite signs, then there exists at least one value <em>c<\/em>\u00a0between <em>a<\/em>\u00a0and <em>b<\/em>\u00a0for which [latex]f\\left(c\\right)=0[\/latex].<\/li>\r\n<\/ul>\r\n<div><section id=\"fs-id1165137731646\" class=\"key-concepts\">\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135528510\" class=\"definition\">\r\n \t<dt><strong>Intermediate Value Theorem<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135528515\">for two numbers <em>a<\/em>\u00a0and <em>b<\/em>\u00a0in the domain of <em>f<\/em>,\u00a0if [latex]a&lt;b[\/latex]\u00a0and [latex]f\\left(a\\right)\\ne f\\left(b\\right)[\/latex],\u00a0then the function <em>f<\/em>\u00a0takes on every value between [latex]f\\left(a\\right)[\/latex]\u00a0and [latex]f\\left(b\\right)[\/latex];\u00a0specifically, when a polynomial function changes from a negative value to a positive value, the function must cross the <em>x<\/em>-axis<\/dd>\r\n<\/dl>\r\n<\/section><\/div>\r\n<h2 style=\"text-align: center;\">Section 4.2 Homework Exercises<\/h2>\r\nFor the following exercises, find the\u00a0zeros, <em>y<\/em>-intercept, multiplicity, and end behavior.\u00a0 Use these to graph each function.\r\n\r\n1. [latex]f\\left(x\\right)=x^{4}-2x[\/latex]\r\n\r\n2. [latex]f\\left(x\\right)={\\left(x+3\\right)}^{2}\\left(x - 2\\right)[\/latex]\r\n\r\n3. [latex]g\\left(x\\right)=\\left(x+4\\right){\\left(x - 1\\right)}^{2}[\/latex]\r\n\r\n4.\u00a0[latex]h\\left(x\\right)={\\left(x - 1\\right)}^{3}{\\left(x+3\\right)}^{2}[\/latex]\r\n\r\n5. [latex]k\\left(x\\right)={\\left(x - 3\\right)}^{3}{\\left(x - 2\\right)}^{2}[\/latex]\r\n\r\n6. [latex]m\\left(x\\right)=-2x\\left(x - 1\\right)\\left(x+3\\right)[\/latex]\r\n\r\n7. [latex]n\\left(x\\right)=-3x\\left(x+2\\right)\\left(x - 4\\right)[\/latex]\r\n\r\n8.\u00a0[latex]f\\left(x\\right)=x^{4}+3x[\/latex]\r\n\r\nFor the following exercises, use the Intermediate Value Theorem to confirm that the given polynomial has at least one zero within the given interval.\r\n\r\n9. [latex]f\\left(x\\right)={x}^{3}-100x+2[\/latex],\u00a0between [latex]x=0.01[\/latex]\u00a0and [latex]x=0.1[\/latex]\r\n\r\n10. [latex]f\\left(x\\right)={x}^{3}-9x[\/latex],\u00a0between [latex]x=-4[\/latex]\u00a0and [latex]x=-2[\/latex].\r\n\r\n11. [latex]f\\left(x\\right)={x}^{3}-9x[\/latex],\u00a0between [latex]x=2[\/latex]\u00a0and [latex]x=4[\/latex].\r\n\r\n12. [latex]f\\left(x\\right)={x}^{5}-2x[\/latex],\u00a0between [latex]x=1[\/latex]\u00a0and [latex]x=2[\/latex].\r\n\r\n13. [latex]f\\left(x\\right)=-{x}^{4}+4[\/latex],\u00a0between [latex]x=1[\/latex]\u00a0and [latex]x=3[\/latex].\r\n\r\n14.\u00a0[latex]f\\left(x\\right)=-2{x}^{3}-x[\/latex],\u00a0between [latex]x=-1[\/latex]\u00a0and [latex]x=1[\/latex].\r\n\r\n15. [latex]f\\left(x\\right)={x}^{3}-100x+2[\/latex],\u00a0between [latex]x=0.01[\/latex]\u00a0and [latex]x=0.1[\/latex]","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Graph polynomial functions<\/li>\n<li>Use the Intermediate Value Theorem<\/li>\n<\/ul>\n<\/div>\n<h2>\u00a0Graph polynomial functions<\/h2>\n<p id=\"fs-id1165137843095\">We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions.<\/p>\n<div id=\"fs-id1165137843101\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135449677\">How To: Given a polynomial function, sketch the graph.<\/h3>\n<ol id=\"fs-id1165135449683\">\n<li>Find the zeros and the <em>y<\/em>-intercept.\u00a0 Plot them. Find the multiplicities.<\/li>\n<li>Find the behavior at each zero.\u00a0 Make a small sketch of this behavior equation at each zero.<\/li>\n<li>Determine the end behavior by examining the leading term.<\/li>\n<li>Use the end behavior and the behavior at each zero to sketch a graph.<\/li>\n<li>Ensure that the number of turning points does not exceed one less than the degree of the polynomial.<\/li>\n<li>Optionally, use technology to check the graph.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_04_08\" class=\"example\">\n<div id=\"fs-id1165135575951\" class=\"exercise\">\n<div id=\"fs-id1165135575953\" class=\"problem textbox shaded\">\n<h3>Example 1: Sketching the Graph of a Polynomial Function<\/h3>\n<p id=\"fs-id1165135575958\">Sketch a graph of [latex]f\\left(x\\right)=-2{\\left(x+3\\right)}^{2}\\left(x - 5\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q892446\">Show Solution<\/span><\/p>\n<div id=\"q892446\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135237929\">This graph has two <em>x-<\/em>intercepts. At <em>x\u00a0<\/em>= \u20133, the factor is squared, indicating a multiplicity of 2. The graph will bounce at this <em>x<\/em>-intercept. At <em>x\u00a0<\/em>= 5, the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.<\/p>\n<p id=\"fs-id1165135171021\">The <em>y<\/em>-intercept is found by evaluating <em>f<\/em>(0).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} f\\left(0\\right)&=-2{\\left(0+3\\right)}^{2}\\left(0 - 5\\right) \\\\ &=-2\\cdot 9\\cdot \\left(-5\\right) \\\\ &=90 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165134374772\">The <em>y<\/em>-intercept is (0, 90).<\/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\/03010733\/CNX_Precalc_Figure_03_04_0172.jpg\" alt=\"Showing the distribution for the leading term.\" width=\"487\" height=\"362\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165134381522\">Additionally, we can see the leading term, if this polynomial were multiplied out, would be [latex]-2{x}^{3}[\/latex],<br \/>\nso the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity.<span id=\"fs-id1165135646080\"><br \/>\n<\/span><\/p>\n<p id=\"fs-id1165134374738\">To sketch this, we consider that:<\/p>\n<ul id=\"fs-id1165134374741\">\n<li>As [latex]x\\to -\\infty[\/latex] the function [latex]f\\left(x\\right)\\to \\infty[\/latex], so we know the graph starts in the second quadrant and is decreasing toward the <em>x<\/em>-axis.<\/li>\n<li>At (0, 90), the graph crosses the <em style=\"font-size: 1rem;\">y<\/em><span style=\"font-size: 1rem;\">-axis at the <\/span><em style=\"font-size: 1rem;\">y<\/em><span style=\"font-size: 1rem;\">-intercept.<\/span><\/li>\n<li>At (-3,0), the behavior is [latex]-2{\\left(x+3\\right)}^{2}\\left(-3 - 5\\right)=16(x+3)^{2}[\/latex].\u00a0 At (-3,0) the graph will resemble a parabola opening up.<\/li>\n<li>At (5,0), the behavior is [latex]-2{\\left(5+3\\right)}^{2}\\left(x - 5\\right)=-128(x-5)[\/latex].\u00a0 At (5,0) the graph will resemble a negative sloping line.<\/li>\n<\/ul>\n<figure id=\"Figure_03_04_018\" class=\"small\">\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\/03010733\/CNX_Precalc_Figure_03_04_0182.jpg\" alt=\"Graph of the end behavior and intercepts, (-3, 0) and (0, 90), for the function f(x)=-2(x+3)^2(x-5).\" width=\"487\" height=\"362\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<\/figure>\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\/03010734\/CNX_Precalc_Figure_03_04_0192.jpg\" alt=\"Graph of the end behavior and intercepts, (-3, 0), (0, 90) and (5, 0), for the function f(x)=-2(x+3)^2(x-5).\" width=\"487\" height=\"362\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165135241000\"><span id=\"fs-id1165135241013\">\u00a0<\/span><\/p>\n<p id=\"fs-id1165135613608\">As [latex]x\\to \\infty[\/latex] the function [latex]f\\left(x\\right)\\to \\mathrm{-\\infty }[\/latex], so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant.<\/p>\n<p id=\"fs-id1165135574296\">Using technology, we can create the graph for the polynomial function, shown in Figure 4, and verify that the resulting graph looks like our sketch in Figure 15.<\/p>\n<figure id=\"Figure_03_04_020\" class=\"small\"><figcaption>The complete graph of the polynomial function [latex]f\\left(x\\right)=-2{\\left(x+3\\right)}^{2}\\left(x - 5\\right)[\/latex]<\/figcaption><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\/03010734\/CNX_Precalc_Figure_03_04_0202.jpg\" alt=\"Graph of f(x)=-2(x+3)^2(x-5).\" width=\"487\" height=\"366\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<\/figure>\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-id1165133065140\">Sketch a graph of [latex]f\\left(x\\right)=\\frac{1}{4}x{\\left(x - 1\\right)}^{4}{\\left(x+3\\right)}^{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q408253\">Show Solution<\/span><\/p>\n<div id=\"q408253\" class=\"hidden-answer\" style=\"display: none\">\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\/03010734\/CNX_Precalc_Figure_03_04_0212.jpg\" alt=\"Graph of f(x)=(1\/4)x(x-1)^4(x+3)^3.\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Use the Intermediate Value Theorem<\/h2>\n<p id=\"fs-id1165135205093\">In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the <em>x<\/em>-axis, we can confirm that there is a zero between them. Consider a polynomial function <em>f<\/em>\u00a0whose graph is smooth and continuous. The <strong>Intermediate Value Theorem<\/strong> states that for two numbers <em>a<\/em>\u00a0and <em>b<\/em>\u00a0in the domain of <em>f<\/em>,\u00a0if <em>a\u00a0<\/em>&lt; <em>b<\/em>\u00a0and [latex]f\\left(a\\right)\\ne f\\left(b\\right)[\/latex], then the function <em>f<\/em>\u00a0takes on every value between [latex]f\\left(a\\right)[\/latex] and [latex]f\\left(b\\right)[\/latex].<\/p>\n<p>We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function <em>f<\/em>\u00a0at [latex]x=a[\/latex] lies above the <em>x<\/em>-axis and another point at [latex]x=b[\/latex] lies below the <em>x<\/em>-axis, there must exist a third point between [latex]x=a[\/latex] and [latex]x=b[\/latex] where the graph crosses the <em>x<\/em>-axis. Call this point [latex]\\left(c,\\text{ }f\\left(c\\right)\\right)[\/latex]. This means that we are assured there is a solution <em>c<\/em>\u00a0where [latex]f\\left(c\\right)=0[\/latex].<\/p>\n<p>In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the <em>x<\/em>-axis. Figure 5\u00a0shows that there is a zero between <em>a<\/em>\u00a0and <em>b<\/em>.<\/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\/03010734\/CNX_Precalc_Figure_03_04_0222.jpg\" alt=\"Graph of an odd-degree polynomial function that shows a point f(a) that\u2019s negative, f(b) that\u2019s positive, and f(c) that\u2019s 0.\" width=\"487\" height=\"368\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5.<\/b> Using the Intermediate Value Theorem to show there exists a zero.<\/p>\n<\/div>\n<div id=\"fs-id1165135347510\" class=\"note textbox shaded\">\n<h3 class=\"title\">A General Note: Intermediate Value Theorem<\/h3>\n<p id=\"fs-id1165135580347\">Let <em>f<\/em>\u00a0be a polynomial function. The <strong>Intermediate Value Theorem<\/strong> states that if [latex]f\\left(a\\right)[\/latex]\u00a0and [latex]f\\left(b\\right)[\/latex]\u00a0have opposite signs, then there exists at least one value <em>c<\/em>\u00a0between <em>a<\/em>\u00a0and <em>b<\/em>\u00a0for which [latex]f\\left(c\\right)=0[\/latex].<\/p>\n<\/div>\n<div id=\"Example_03_04_09\" class=\"example\">\n<div id=\"fs-id1165133358799\" class=\"exercise\">\n<div id=\"fs-id1165133358801\" class=\"problem textbox shaded\">\n<h3>Example 2: Using the Intermediate Value Theorem<\/h3>\n<p id=\"fs-id1165133358807\">Show that the function [latex]f\\left(x\\right)={x}^{3}-5{x}^{2}+3x+6[\/latex]\u00a0has at least two real zeros between [latex]x=1[\/latex]\u00a0and [latex]x=4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q98939\">Show Solution<\/span><\/p>\n<div id=\"q98939\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135537349\">As a start, evaluate [latex]f\\left(x\\right)[\/latex]\u00a0at the integer values [latex]x=1,2,3,\\text{ and }4[\/latex].<\/p>\n<table id=\"Table_03_04_03\" summary=\"..\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td><em><strong>f\u00a0<\/strong><\/em><strong>(<em>x<\/em>)<\/strong><\/td>\n<td>5<\/td>\n<td>0<\/td>\n<td>\u20133<\/td>\n<td>2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165135536378\">We see that one zero occurs at [latex]x=2[\/latex]. Also, since [latex]f\\left(3\\right)[\/latex] is negative and [latex]f\\left(4\\right)[\/latex] is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4.<\/p>\n<p id=\"fs-id1165135575934\">We have shown that there are at least two real zeros between [latex]x=1[\/latex]\u00a0and [latex]x=4[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can also see in Figure 6\u00a0that there are two real zeros between [latex]x=1[\/latex]\u00a0and [latex]x=4[\/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\/03010735\/CNX_Precalc_Figure_03_04_0232.jpg\" alt=\"Graph of f(x)=x^3-5x^2+3x+6 and shows, by the Intermediate Value Theorem, that there exists two zeros since f(1)=5 and f(4)=2 are positive and f(3) = -3 is negative.\" width=\"487\" height=\"591\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/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-id1165135551168\">Show that the function [latex]f\\left(x\\right)=7{x}^{5}-9{x}^{4}-{x}^{2}[\/latex] has at least one real zero between [latex]x=1[\/latex] and [latex]x=2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q886741\">Show Solution<\/span><\/p>\n<div id=\"q886741\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because <em>f<\/em>\u00a0is a polynomial function and since [latex]f\\left(1\\right)[\/latex] is negative and [latex]f\\left(2\\right)[\/latex] is positive, there is at least one real zero between [latex]x=1[\/latex] and [latex]x=2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul id=\"fs-id1165137846272\">\n<li>To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most<em>\u00a0n <\/em>\u2013\u00a01 turning points.<\/li>\n<li>Graphing a polynomial function helps to estimate local and global extremas.<\/li>\n<li>The Intermediate Value Theorem tells us that if [latex]f\\left(a\\right) \\text{and} f\\left(b\\right)[\/latex]\u00a0have opposite signs, then there exists at least one value <em>c<\/em>\u00a0between <em>a<\/em>\u00a0and <em>b<\/em>\u00a0for which [latex]f\\left(c\\right)=0[\/latex].<\/li>\n<\/ul>\n<div>\n<section id=\"fs-id1165137731646\" class=\"key-concepts\">\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135528510\" class=\"definition\">\n<dt><strong>Intermediate Value Theorem<\/strong><\/dt>\n<dd id=\"fs-id1165135528515\">for two numbers <em>a<\/em>\u00a0and <em>b<\/em>\u00a0in the domain of <em>f<\/em>,\u00a0if [latex]a<b[\/latex]\u00a0and [latex]f\\left(a\\right)\\ne f\\left(b\\right)[\/latex],\u00a0then the function <em>f<\/em>\u00a0takes on every value between [latex]f\\left(a\\right)[\/latex]\u00a0and [latex]f\\left(b\\right)[\/latex];\u00a0specifically, when a polynomial function changes from a negative value to a positive value, the function must cross the <em>x<\/em>-axis<\/dd>\n<\/dl>\n<\/section>\n<\/div>\n<h2 style=\"text-align: center;\">Section 4.2 Homework Exercises<\/h2>\n<p>For the following exercises, find the\u00a0zeros, <em>y<\/em>-intercept, multiplicity, and end behavior.\u00a0 Use these to graph each function.<\/p>\n<p>1. [latex]f\\left(x\\right)=x^{4}-2x[\/latex]<\/p>\n<p>2. [latex]f\\left(x\\right)={\\left(x+3\\right)}^{2}\\left(x - 2\\right)[\/latex]<\/p>\n<p>3. [latex]g\\left(x\\right)=\\left(x+4\\right){\\left(x - 1\\right)}^{2}[\/latex]<\/p>\n<p>4.\u00a0[latex]h\\left(x\\right)={\\left(x - 1\\right)}^{3}{\\left(x+3\\right)}^{2}[\/latex]<\/p>\n<p>5. [latex]k\\left(x\\right)={\\left(x - 3\\right)}^{3}{\\left(x - 2\\right)}^{2}[\/latex]<\/p>\n<p>6. [latex]m\\left(x\\right)=-2x\\left(x - 1\\right)\\left(x+3\\right)[\/latex]<\/p>\n<p>7. [latex]n\\left(x\\right)=-3x\\left(x+2\\right)\\left(x - 4\\right)[\/latex]<\/p>\n<p>8.\u00a0[latex]f\\left(x\\right)=x^{4}+3x[\/latex]<\/p>\n<p>For the following exercises, use the Intermediate Value Theorem to confirm that the given polynomial has at least one zero within the given interval.<\/p>\n<p>9. [latex]f\\left(x\\right)={x}^{3}-100x+2[\/latex],\u00a0between [latex]x=0.01[\/latex]\u00a0and [latex]x=0.1[\/latex]<\/p>\n<p>10. [latex]f\\left(x\\right)={x}^{3}-9x[\/latex],\u00a0between [latex]x=-4[\/latex]\u00a0and [latex]x=-2[\/latex].<\/p>\n<p>11. [latex]f\\left(x\\right)={x}^{3}-9x[\/latex],\u00a0between [latex]x=2[\/latex]\u00a0and [latex]x=4[\/latex].<\/p>\n<p>12. [latex]f\\left(x\\right)={x}^{5}-2x[\/latex],\u00a0between [latex]x=1[\/latex]\u00a0and [latex]x=2[\/latex].<\/p>\n<p>13. [latex]f\\left(x\\right)=-{x}^{4}+4[\/latex],\u00a0between [latex]x=1[\/latex]\u00a0and [latex]x=3[\/latex].<\/p>\n<p>14.\u00a0[latex]f\\left(x\\right)=-2{x}^{3}-x[\/latex],\u00a0between [latex]x=-1[\/latex]\u00a0and [latex]x=1[\/latex].<\/p>\n<p>15. [latex]f\\left(x\\right)={x}^{3}-100x+2[\/latex],\u00a0between [latex]x=0.01[\/latex]\u00a0and [latex]x=0.1[\/latex]<\/p>\n","protected":false},"author":264444,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-17911","chapter","type-chapter","status-publish","hentry"],"part":17771,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17911","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/users\/264444"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17911\/revisions"}],"predecessor-version":[{"id":17943,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17911\/revisions\/17943"}],"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\/17911\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/media?parent=17911"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapter-type?post=17911"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/contributor?post=17911"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/license?post=17911"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}