{"id":215,"date":"2023-06-21T13:22:44","date_gmt":"2023-06-21T13:22:44","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/graph-polynomial-functions\/"},"modified":"2023-10-05T16:26:13","modified_gmt":"2023-10-05T16:26:13","slug":"graph-polynomial-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/graph-polynomial-functions\/","title":{"raw":"\u25aa   Graphing Polynomial Functions","rendered":"\u25aa   Graphing Polynomial Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Draw the graph of a polynomial function using end behavior, turning points, intercepts, and the Intermediate Value Theorem.<\/li>\r\n<\/ul>\r\n<\/div>\r\nWe 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.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a polynomial function, sketch the graph<\/h3>\r\n<ol>\r\n \t<li>Find the intercepts.<\/li>\r\n \t<li>Check for symmetry. If the function is an even function, its graph is symmetric with respect to the\u00a0<em>y<\/em>-axis, that is,\u00a0<em>f<\/em>(\u2013<em>x<\/em>) = <em>f<\/em>(<em>x<\/em>).\r\nIf a function is an odd function, its graph is symmetric with respect to the origin, that is,\u00a0<em>f<\/em>(\u2013<em>x<\/em>) = <em>\u2013<\/em><em>f<\/em>(<em>x<\/em>).<\/li>\r\n \t<li>Use the multiplicities of the zeros to determine the behavior of the polynomial at the <em>x<\/em>-intercepts.<\/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 the intercepts to sketch the 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 class=\"textbox exercises\">\r\n<h3>Example: Sketching the Graph of a Polynomial Function<\/h3>\r\nSketch a possible graph for [latex]f\\left(x\\right)=-2{\\left(x+3\\right)}^{2}\\left(x - 5\\right)[\/latex].\r\n\r\n[reveal-answer q=\"707233\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"707233\"]\r\n\r\nThis 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 off the\u00a0<em>x<\/em>-intercept at this value. At <em>x\u00a0<\/em>= 5, the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.\r\n\r\nThe <em>y<\/em>-intercept is found by evaluating <em>f<\/em>(0).\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ f\\left(0\\right)=-2{\\left(0+3\\right)}^{2}\\left(0 - 5\\right)\\hfill \\\\ \\text{}f\\left(0\\right)=-2\\cdot 9\\cdot \\left(-5\\right)\\hfill \\\\ \\text{}f\\left(0\\right)=90\\hfill \\end{array}[\/latex]<\/p>\r\nThe <em>y<\/em>-intercept is (0, 90).\r\n\r\nAdditionally, we can see the leading term, if this polynomial were multiplied out, would be [latex]-2{x}^{3}[\/latex], so the end behavior, as seen in the following graph, 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.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02201614\/CNX_Precalc_Figure_03_04_0172.jpg\" alt=\"Showing the distribution for the leading term.\" width=\"487\" height=\"362\" \/>\r\n\r\nTo sketch the graph, we consider the following:\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>Since [latex]f\\left(-x\\right)=-2{\\left(-x+3\\right)}^{2}\\left(-x - 5\\right)[\/latex] is not equal to <em>f<\/em>(<em>x<\/em>), the graph does not have any symmetry.<\/li>\r\n \t<li>At [latex]\\left(-3,0\\right)[\/latex] the graph bounces off of the <em>x<\/em>-axis, so the function must start increasing after this point.<\/li>\r\n \t<li>At (0, 90), the graph crosses the <em>y<\/em>-axis.<\/li>\r\n<\/ul>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02201617\/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\" \/>\r\n\r\nSomewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at (5, 0).\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02201618\/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\" \/>\r\n\r\nAs [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.\r\n\r\nThe complete graph of the polynomial function [latex]f\\left(x\\right)=-2{\\left(x+3\\right)}^{2}\\left(x - 5\\right)[\/latex] is as follows:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02201620\/CNX_Precalc_Figure_03_04_0202.jpg\" alt=\"Graph of f(x)=-2(x+3)^2(x-5).\" width=\"487\" height=\"366\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSketch a possible graph for [latex]f\\left(x\\right)=\\frac{1}{4}x{\\left(x - 1\\right)}^{4}{\\left(x+3\\right)}^{3}[\/latex].\r\n\r\nCheck yourself with an online graphing calculator when you are done.\r\n\r\n[reveal-answer q=\"812296\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"812296\"]\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/30233150\/CNX_Precalc_Figure_03_04_0212.jpg\"><img class=\"aligncenter size-full wp-image-2911\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/30233150\/CNX_Precalc_Figure_03_04_0212.jpg\" alt=\"Graph of f(x)=(1\/4)x(x-1)^4(x+3)^3.\" width=\"487\" height=\"366\" \/><\/a>\r\n\r\n[\/hidden-answer]\r\n\r\n[ohm_question height=\"600\"]113372[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\nUse an online graphing calculator to find an odd degree function with one zero at (-3,0) whose multiplicity is 3 and another zero at (2,0) with multiplicity 2. What is the end behavior of the graph?\r\n\r\n[reveal-answer q=\"565179\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"565179\"]\r\n\r\nThe end behavior of the graph is:\r\nas [latex]x\\rightarrow-\\infty, f(x) \\rightarrow\\infty[\/latex] and as [latex]x\\rightarrow \\infty, f(x)\\rightarrow -\\infty[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Intermediate Value Theorem<\/h2>\r\nIn 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].\r\n\r\nWe can apply this theorem to a special case that is useful for 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 value\u00a0<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. The figure below\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\/wp-content\/uploads\/sites\/896\/2016\/11\/02201623\/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\" \/> The Intermediate Value Theorem can be used to show there exists a zero.[\/caption]\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Intermediate Value Theorem<\/h3>\r\nLet <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].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Intermediate Value Theorem<\/h3>\r\nShow 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].\r\n\r\n[reveal-answer q=\"33137\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"33137\"]\r\n\r\nTo start, evaluate [latex]f\\left(x\\right)[\/latex]\u00a0at the integer values [latex]x=1,2,3,\\text{ and }4[\/latex].\r\n<table><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<\/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\nWe 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.\r\n\r\nWe have shown that there are at least two real zeros between [latex]x=1[\/latex]\u00a0and [latex]x=4[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can also graphically see that there are two real zeros between [latex]x=1[\/latex]\u00a0and [latex]x=4[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02201625\/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\" \/>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nShow 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].\r\n\r\n[reveal-answer q=\"778313\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"778313\"]\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].[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Draw the graph of a polynomial function using end behavior, turning points, intercepts, and the Intermediate Value Theorem.<\/li>\n<\/ul>\n<\/div>\n<p>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 class=\"textbox\">\n<h3>How To: Given a polynomial function, sketch the graph<\/h3>\n<ol>\n<li>Find the intercepts.<\/li>\n<li>Check for symmetry. If the function is an even function, its graph is symmetric with respect to the\u00a0<em>y<\/em>-axis, that is,\u00a0<em>f<\/em>(\u2013<em>x<\/em>) = <em>f<\/em>(<em>x<\/em>).<br \/>\nIf a function is an odd function, its graph is symmetric with respect to the origin, that is,\u00a0<em>f<\/em>(\u2013<em>x<\/em>) = <em>\u2013<\/em><em>f<\/em>(<em>x<\/em>).<\/li>\n<li>Use the multiplicities of the zeros to determine the behavior of the polynomial at the <em>x<\/em>-intercepts.<\/li>\n<li>Determine the end behavior by examining the leading term.<\/li>\n<li>Use the end behavior and the behavior at the intercepts to sketch the 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 class=\"textbox exercises\">\n<h3>Example: Sketching the Graph of a Polynomial Function<\/h3>\n<p>Sketch a possible graph for [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=\"q707233\">Show Solution<\/span><\/p>\n<div id=\"q707233\" class=\"hidden-answer\" style=\"display: none\">\n<p>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 off the\u00a0<em>x<\/em>-intercept at this value. 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>The <em>y<\/em>-intercept is found by evaluating <em>f<\/em>(0).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ f\\left(0\\right)=-2{\\left(0+3\\right)}^{2}\\left(0 - 5\\right)\\hfill \\\\ \\text{}f\\left(0\\right)=-2\\cdot 9\\cdot \\left(-5\\right)\\hfill \\\\ \\text{}f\\left(0\\right)=90\\hfill \\end{array}[\/latex]<\/p>\n<p>The <em>y<\/em>-intercept is (0, 90).<\/p>\n<p>Additionally, we can see the leading term, if this polynomial were multiplied out, would be [latex]-2{x}^{3}[\/latex], so the end behavior, as seen in the following graph, 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.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02201614\/CNX_Precalc_Figure_03_04_0172.jpg\" alt=\"Showing the distribution for the leading term.\" width=\"487\" height=\"362\" \/><\/p>\n<p>To sketch the graph, we consider the following:<\/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>Since [latex]f\\left(-x\\right)=-2{\\left(-x+3\\right)}^{2}\\left(-x - 5\\right)[\/latex] is not equal to <em>f<\/em>(<em>x<\/em>), the graph does not have any symmetry.<\/li>\n<li>At [latex]\\left(-3,0\\right)[\/latex] the graph bounces off of the <em>x<\/em>-axis, so the function must start increasing after this point.<\/li>\n<li>At (0, 90), the graph crosses the <em>y<\/em>-axis.<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02201617\/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>Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at (5, 0).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02201618\/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>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>The complete graph of the polynomial function [latex]f\\left(x\\right)=-2{\\left(x+3\\right)}^{2}\\left(x - 5\\right)[\/latex] is as follows:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02201620\/CNX_Precalc_Figure_03_04_0202.jpg\" alt=\"Graph of f(x)=-2(x+3)^2(x-5).\" width=\"487\" height=\"366\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Sketch a possible graph for [latex]f\\left(x\\right)=\\frac{1}{4}x{\\left(x - 1\\right)}^{4}{\\left(x+3\\right)}^{3}[\/latex].<\/p>\n<p>Check yourself with an online graphing calculator when you are done.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q812296\">Show Solution<\/span><\/p>\n<div id=\"q812296\" class=\"hidden-answer\" style=\"display: none\">\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/30233150\/CNX_Precalc_Figure_03_04_0212.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-2911\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/30233150\/CNX_Precalc_Figure_03_04_0212.jpg\" alt=\"Graph of f(x)=(1\/4)x(x-1)^4(x+3)^3.\" width=\"487\" height=\"366\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm113372\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=113372&theme=oea&iframe_resize_id=ohm113372&show_question_numbers\" width=\"100%\" height=\"600\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p>Use an online graphing calculator to find an odd degree function with one zero at (-3,0) whose multiplicity is 3 and another zero at (2,0) with multiplicity 2. What is the end behavior of the graph?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q565179\">Show Solution<\/span><\/p>\n<div id=\"q565179\" class=\"hidden-answer\" style=\"display: none\">\n<p>The end behavior of the graph is:<br \/>\nas [latex]x\\rightarrow-\\infty, f(x) \\rightarrow\\infty[\/latex] and as [latex]x\\rightarrow \\infty, f(x)\\rightarrow -\\infty[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>The Intermediate Value Theorem<\/h2>\n<p>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 for 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 value\u00a0<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. The figure below\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\/wp-content\/uploads\/sites\/896\/2016\/11\/02201623\/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\">The Intermediate Value Theorem can be used to show there exists a zero.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Intermediate Value Theorem<\/h3>\n<p>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 class=\"textbox exercises\">\n<h3>Example: Using the Intermediate Value Theorem<\/h3>\n<p>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=\"q33137\">Show Solution<\/span><\/p>\n<div id=\"q33137\" class=\"hidden-answer\" style=\"display: none\">\n<p>To start, evaluate [latex]f\\left(x\\right)[\/latex]\u00a0at the integer values [latex]x=1,2,3,\\text{ and }4[\/latex].<\/p>\n<table>\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<\/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>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>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 graphically see that there are two real zeros between [latex]x=1[\/latex]\u00a0and [latex]x=4[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02201625\/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\" \/>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q778313\">Show Solution<\/span><\/p>\n<div id=\"q778313\" 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].<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-215\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Interactive: Write a Polynomial Function. <strong>Provided by<\/strong>: Lumen Learning (With Desmos). <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/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":395986,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Interactive: Write a Polynomial Function\",\"author\":\"\",\"organization\":\"Lumen Learning (With Desmos)\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et 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