{"id":1772,"date":"2016-11-02T20:35:57","date_gmt":"2016-11-02T20:35:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1772"},"modified":"2017-04-18T23:47:40","modified_gmt":"2017-04-18T23:47:40","slug":"graph-polynomial-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/graph-polynomial-functions\/","title":{"raw":"Graph Polynomial Functions","rendered":"Graph Polynomial Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/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 symmetrical about the <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 symmetrical about 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 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 class=\"textbox exercises\">\r\n<h3>Example: Sketching the Graph of a Polynomial Function<\/h3>\r\nSketch a graph of [latex]f\\left(x\\right)=-2{\\left(x+3\\right)}^{2}\\left(x - 5\\right)[\/latex].\r\n\r\n[reveal-answer q=\"707233\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"707233\"]\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 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.\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{ }=-2\\cdot 9\\cdot \\left(-5\\right)\\hfill \\\\ \\text{ }=90\\hfill \\end{array}[\/latex]<\/p>\r\nThe <em>y<\/em>-intercept is (0, 90).\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\" data-media-type=\"image\/jpg\" \/>\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 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\nTo sketch this, we consider that:\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]\r\nis not equal to <em>f<\/em>(<em>x<\/em>), the graph does not display 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.At (0, 90), the graph crosses the <em>y<\/em>-axis at the <em>y<\/em>-intercept.<\/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\" data-media-type=\"image\/jpg\" \/>\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\" data-media-type=\"image\/jpg\" \/>\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\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\nUsing technology, we can create the graph for the polynomial function, shown in Figure 16, and verify that the resulting graph looks like our sketch in Figure 15.\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]\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\" data-media-type=\"image\/jpg\" \/>\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 graph of [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 Desmos when you are done.\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\n[reveal-answer q=\"812296\"]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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\nUse Desmos to write an odd degree function with one zero at (-3,0) whose multiplicity is 3 and another zero at (2,0) with multiplicity 2. The 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<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\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 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. 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\" data-media-type=\"image\/jpg\" \/> Using the Intermediate Value Theorem 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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"33137\"]\r\n\r\nAs a 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 data-align=\"center\" \/> <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\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 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\" data-media-type=\"image\/jpeg\" \/>\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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"778313\"]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].[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/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 symmetrical about the <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 symmetrical about 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 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 class=\"textbox exercises\">\n<h3>Example: Sketching the Graph of a Polynomial Function<\/h3>\n<p>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=\"q707233\">Solution<\/span><\/p>\n<div id=\"q707233\" class=\"hidden-answer\" style=\"display: none\">\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 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>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{ }=-2\\cdot 9\\cdot \\left(-5\\right)\\hfill \\\\ \\text{ }=90\\hfill \\end{array}[\/latex]<\/p>\n<p>The <em>y<\/em>-intercept is (0, 90).<\/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\" data-media-type=\"image\/jpg\" \/><\/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 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>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>Since [latex]f\\left(-x\\right)=-2{\\left(-x+3\\right)}^{2}\\left(-x - 5\\right)[\/latex]<br \/>\nis not equal to <em>f<\/em>(<em>x<\/em>), the graph does not display 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.At (0, 90), the graph crosses the <em>y<\/em>-axis at the <em>y<\/em>-intercept.<\/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\" data-media-type=\"image\/jpg\" \/><\/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\" data-media-type=\"image\/jpg\" \/><\/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>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>Using technology, we can create the graph for the polynomial function, shown in Figure 16, and verify that the resulting graph looks like our sketch in Figure 15.<\/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]<\/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\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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<p>Check yourself with Desmos when you are done.<br \/>\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q812296\">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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p>Use Desmos to write an odd degree function with one zero at (-3,0) whose multiplicity is 3 and another zero at (2,0) with multiplicity 2. 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]<br \/>\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><\/p>\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 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. 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\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">Using the Intermediate Value Theorem 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\">Solution<\/span><\/p>\n<div id=\"q33137\" class=\"hidden-answer\" style=\"display: none\">\n<p>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>\n<colgroup>\n<col data-align=\"center\" \/>\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>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 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\" data-media-type=\"image\/jpeg\" \/>\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\">Solution<\/span><\/p>\n<div id=\"q778313\" class=\"hidden-answer\" style=\"display: none\">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-1772\">\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":21,"menu_order":9,"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 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http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"}]","CANDELA_OUTCOMES_GUID":"93854fb6-2394-411d-80cb-326ecb8224fd","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1772","chapter","type-chapter","status-publish","hentry"],"part":1700,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1772","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":13,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1772\/revisions"}],"predecessor-version":[{"id":4233,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1772\/revisions\/4233"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1700"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1772\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/media?parent=1772"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1772"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1772"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/license?post=1772"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}