{"id":2008,"date":"2016-06-30T22:26:10","date_gmt":"2016-06-30T22:26:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2008"},"modified":"2019-08-06T20:26:40","modified_gmt":"2019-08-06T20:26:40","slug":"read-use-a-graph-to-write-a-linear-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/read-use-a-graph-to-write-a-linear-function\/","title":{"raw":"Write the Equation of a Linear Function Part II","rendered":"Write the Equation of a Linear Function Part II"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write the equation of a linear function from a graph<\/li>\r\n \t<li>Write the equation for a linear function from slope given two points<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165137911671\">Now that we have written equations for linear functions using both slope-intercept form and point-slope form, we can choose which method to use based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function <em>f<\/em>\u00a0below.<span id=\"fs-id1165135182766\"><\/span><\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201025\/CNX_Precalc_Figure_02_01_0062.jpg\" alt=\"Graph depicting how to calculate the slope of a line\" width=\"487\" height=\"347\" \/>\r\n<p id=\"fs-id1165137657203\">We are not given the slope of the line, but we can choose any two points on the line to find the slope. Choose\u00a0[latex](0, 7)[\/latex]\u00a0and\u00a0[latex](4, 4)[\/latex]. We can use these points to calculate the slope.<\/p>\r\n\r\n<div id=\"fs-id1165137666429\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l} m &amp; =\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ &amp; =\\dfrac{4 - 7}{4 - 0}\\hfill \\\\ &amp; =-\\dfrac{3}{4} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165135176567\">Now we can substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\r\n\r\n<div id=\"fs-id1165135450369\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 4=-\\dfrac{3}{4}\\left(x - 4\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137634475\">If we want to rewrite the equation in slope-intercept form, we would do the following:<\/p>\r\n\r\n<div id=\"fs-id1165137768649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}y - 4=-\\dfrac{3}{4}\\left(x - 4\\right)\\hfill \\\\ y - 4=-\\dfrac{3}{4}x+3\\hfill \\\\ \\text{ }y=-\\dfrac{3}{4}x+7\\hfill \\end{array}[\/latex]<\/div>\r\n<div>\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201027\/CNX_Precalc_Figure_02_01_0072.jpg\" alt=\"\" width=\"487\" height=\"155\" \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137769983\">If we wanted to find the equation of the line in slope-intercept form without first using point-slope form, we could have recognized that the line crosses the <em>y<\/em>-axis when the output value is\u00a0[latex]7[\/latex]. Therefore, <em>b<\/em> =\u00a0[latex]7[\/latex].\u00a0We now have the initial value <em>b<\/em>\u00a0and the slope <em>m\u00a0<\/em>so we can substitute <em>m<\/em>\u00a0and <em>b<\/em>\u00a0into slope-intercept form of a line.<span id=\"fs-id1165137548391\">\r\n<\/span><\/p>\r\n<p id=\"fs-id1165137705273\">The function is [latex]f\\left(x\\right)=-\\dfrac{3}{4}x+7[\/latex],\u00a0and the linear equation would be [latex]y=-\\dfrac{3}{4}x+7[\/latex].<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 id=\"fs-id1165137824881\">How To: Given the graph of a linear function, write an equation to represent the function<\/h3>\r\n<ol id=\"fs-id1165137803240\">\r\n \t<li>Identify two points on the line.<\/li>\r\n \t<li>Use the two points to calculate the slope.<\/li>\r\n \t<li>Determine where the line crosses the <em>y<\/em>-axis to identify the <em>y<\/em>-intercept by visual inspection.<\/li>\r\n \t<li>Substitute the slope and <em>y<\/em>-intercept into slope-intercept form of a linear equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWrite an equation for a linear function\u00a0<em>f<\/em> given its graph shown below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201028\/CNX_Precalc_Figure_02_01_008a2.jpg\" alt=\"Graph of an increasing function with points at (-3, 0) and (0, 1).\" width=\"369\" height=\"378\" \/>\r\n[reveal-answer q=\"728685\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"728685\"]\r\n<p id=\"fs-id1165135536538\">Identify two points on the line such as\u00a0[latex](0, 2)[\/latex] and\u00a0[latex](\u20132, \u20134)[\/latex]. Use the points to calculate the slope.<\/p>\r\n\r\n<div id=\"fs-id1165137824250\" class=\"equation unnumbered\">[latex]\\begin{array}{l}m &amp; =\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ &amp; =\\dfrac{-4 - 2}{-2 - 0}\\hfill \\\\ &amp; =\\dfrac{-6}{-2}\\hfill \\\\ &amp; =3\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137732234\">Substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\r\n\r\n<div id=\"fs-id1165134305424\" class=\"equation unnumbered\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y-\\left(-4\\right)=3\\left(x-\\left(-2\\right)\\right)\\\\ y+4=3\\left(x+2\\right)\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137731540\">We can use algebra to rewrite the equation in slope-intercept form.<\/p>\r\n[latex]\\begin{array}{l}y+4=3\\left(x+2\\right)\\hfill \\\\ y+4=3x+6\\hfill \\\\ y=3x+2 \\hfill \\end{array}[\/latex]\r\n\r\n&nbsp;\r\n\r\nThis makes sense because we can see from the graph above that the line crosses the\u00a0<em>y<\/em>-axis at the point [latex](0,2)[\/latex], which is the\u00a0<em>y<\/em>-intercept, so [latex]b=2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show an example of how to write the equation of a line given its graph.\r\n\r\nhttps:\/\/youtu.be\/mmWf_oLTNSQ\r\n<div id=\"fs-id1165137767515\" class=\"commentary\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf <em>f<\/em>\u00a0is a linear function, with [latex]f\\left(3\\right)=-2[\/latex] , and [latex]f\\left(8\\right)=1[\/latex], find an equation for the function in slope-intercept form.\r\n[reveal-answer q=\"834495\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"834495\"]\r\n<p id=\"fs-id1165137730075\">We can write the given points using coordinates.<\/p>\r\n\r\n<div id=\"fs-id1165137639761\" class=\"equation unnumbered\">[latex]\\begin{array}{l}f\\left(3\\right)=-2\\to \\left(3,-2\\right)\\hfill \\\\ f\\left(8\\right)=1\\to \\left(8,1\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137603531\">We can then use the points to calculate the slope.<\/p>\r\n\r\n<div id=\"fs-id1165137582561\" class=\"equation unnumbered\">[latex]\\begin{array}{l} m &amp; =\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ &amp; =\\dfrac{1-\\left(-2\\right)}{8 - 3}\\hfill \\\\ &amp; =\\dfrac{3}{5}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137851970\">Substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\r\n\r\n<div id=\"fs-id1165137583929\" class=\"equation unnumbered\">[latex]\\begin{array}\\text{ }y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y-\\left(-2\\right)=\\dfrac{3}{5}\\left(x - 3\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137629473\">We can use algebra to rewrite the equation in slope-intercept form.<\/p>\r\n\r\n<div id=\"fs-id1165135543458\" class=\"equation unnumbered\">[latex]\\begin{array}{l}y+2=\\dfrac{3}{5}\\left(x - 3\\right)\\hfill \\\\ y+2=\\dfrac{3}{5}x-\\dfrac{9}{5}\\hfill \\\\ \\text{ }y=\\dfrac{3}{5}x-\\dfrac{19}{5}\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the video we show another example of how to write a linear function given two points written with function notation.\r\n\r\nhttps:\/\/youtu.be\/TSIcryAtCmY\r\n<h2>Summary<\/h2>\r\n<ul>\r\n \t<li>Given a graph, you can write the equation of the line it represents.<\/li>\r\n \t<li>This method only works for graphs that have points that are easy to verify by visual inspection.<\/li>\r\n<\/ul>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write the equation of a linear function from a graph<\/li>\n<li>Write the equation for a linear function from slope given two points<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137911671\">Now that we have written equations for linear functions using both slope-intercept form and point-slope form, we can choose which method to use based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function <em>f<\/em>\u00a0below.<span id=\"fs-id1165135182766\"><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201025\/CNX_Precalc_Figure_02_01_0062.jpg\" alt=\"Graph depicting how to calculate the slope of a line\" width=\"487\" height=\"347\" \/><\/p>\n<p id=\"fs-id1165137657203\">We are not given the slope of the line, but we can choose any two points on the line to find the slope. Choose\u00a0[latex](0, 7)[\/latex]\u00a0and\u00a0[latex](4, 4)[\/latex]. We can use these points to calculate the slope.<\/p>\n<div id=\"fs-id1165137666429\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l} m & =\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ & =\\dfrac{4 - 7}{4 - 0}\\hfill \\\\ & =-\\dfrac{3}{4} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165135176567\">Now we can substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\n<div id=\"fs-id1165135450369\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 4=-\\dfrac{3}{4}\\left(x - 4\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137634475\">If we want to rewrite the equation in slope-intercept form, we would do the following:<\/p>\n<div id=\"fs-id1165137768649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}y - 4=-\\dfrac{3}{4}\\left(x - 4\\right)\\hfill \\\\ y - 4=-\\dfrac{3}{4}x+3\\hfill \\\\ \\text{ }y=-\\dfrac{3}{4}x+7\\hfill \\end{array}[\/latex]<\/div>\n<div>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201027\/CNX_Precalc_Figure_02_01_0072.jpg\" alt=\"\" width=\"487\" height=\"155\" \/><\/p>\n<\/div>\n<p id=\"fs-id1165137769983\">If we wanted to find the equation of the line in slope-intercept form without first using point-slope form, we could have recognized that the line crosses the <em>y<\/em>-axis when the output value is\u00a0[latex]7[\/latex]. Therefore, <em>b<\/em> =\u00a0[latex]7[\/latex].\u00a0We now have the initial value <em>b<\/em>\u00a0and the slope <em>m\u00a0<\/em>so we can substitute <em>m<\/em>\u00a0and <em>b<\/em>\u00a0into slope-intercept form of a line.<span id=\"fs-id1165137548391\"><br \/>\n<\/span><\/p>\n<p id=\"fs-id1165137705273\">The function is [latex]f\\left(x\\right)=-\\dfrac{3}{4}x+7[\/latex],\u00a0and the linear equation would be [latex]y=-\\dfrac{3}{4}x+7[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3 id=\"fs-id1165137824881\">How To: Given the graph of a linear function, write an equation to represent the function<\/h3>\n<ol id=\"fs-id1165137803240\">\n<li>Identify two points on the line.<\/li>\n<li>Use the two points to calculate the slope.<\/li>\n<li>Determine where the line crosses the <em>y<\/em>-axis to identify the <em>y<\/em>-intercept by visual inspection.<\/li>\n<li>Substitute the slope and <em>y<\/em>-intercept into slope-intercept form of a linear equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Write an equation for a linear function\u00a0<em>f<\/em> given its graph shown below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201028\/CNX_Precalc_Figure_02_01_008a2.jpg\" alt=\"Graph of an increasing function with points at (-3, 0) and (0, 1).\" width=\"369\" height=\"378\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q728685\">Show Solution<\/span><\/p>\n<div id=\"q728685\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135536538\">Identify two points on the line such as\u00a0[latex](0, 2)[\/latex] and\u00a0[latex](\u20132, \u20134)[\/latex]. Use the points to calculate the slope.<\/p>\n<div id=\"fs-id1165137824250\" class=\"equation unnumbered\">[latex]\\begin{array}{l}m & =\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ & =\\dfrac{-4 - 2}{-2 - 0}\\hfill \\\\ & =\\dfrac{-6}{-2}\\hfill \\\\ & =3\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137732234\">Substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\n<div id=\"fs-id1165134305424\" class=\"equation unnumbered\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y-\\left(-4\\right)=3\\left(x-\\left(-2\\right)\\right)\\\\ y+4=3\\left(x+2\\right)\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137731540\">We can use algebra to rewrite the equation in slope-intercept form.<\/p>\n<p>[latex]\\begin{array}{l}y+4=3\\left(x+2\\right)\\hfill \\\\ y+4=3x+6\\hfill \\\\ y=3x+2 \\hfill \\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>This makes sense because we can see from the graph above that the line crosses the\u00a0<em>y<\/em>-axis at the point [latex](0,2)[\/latex], which is the\u00a0<em>y<\/em>-intercept, so [latex]b=2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show an example of how to write the equation of a line given its graph.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 1:  Find the Equation of a Line in Slope Intercept Form Given the Graph of a Line\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/mmWf_oLTNSQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div id=\"fs-id1165137767515\" class=\"commentary\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If <em>f<\/em>\u00a0is a linear function, with [latex]f\\left(3\\right)=-2[\/latex] , and [latex]f\\left(8\\right)=1[\/latex], find an equation for the function in slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q834495\">Show Solution<\/span><\/p>\n<div id=\"q834495\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137730075\">We can write the given points using coordinates.<\/p>\n<div id=\"fs-id1165137639761\" class=\"equation unnumbered\">[latex]\\begin{array}{l}f\\left(3\\right)=-2\\to \\left(3,-2\\right)\\hfill \\\\ f\\left(8\\right)=1\\to \\left(8,1\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137603531\">We can then use the points to calculate the slope.<\/p>\n<div id=\"fs-id1165137582561\" class=\"equation unnumbered\">[latex]\\begin{array}{l} m & =\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ & =\\dfrac{1-\\left(-2\\right)}{8 - 3}\\hfill \\\\ & =\\dfrac{3}{5}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137851970\">Substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\n<div id=\"fs-id1165137583929\" class=\"equation unnumbered\">[latex]\\begin{array}\\text{ }y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y-\\left(-2\\right)=\\dfrac{3}{5}\\left(x - 3\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137629473\">We can use algebra to rewrite the equation in slope-intercept form.<\/p>\n<div id=\"fs-id1165135543458\" class=\"equation unnumbered\">[latex]\\begin{array}{l}y+2=\\dfrac{3}{5}\\left(x - 3\\right)\\hfill \\\\ y+2=\\dfrac{3}{5}x-\\dfrac{9}{5}\\hfill \\\\ \\text{ }y=\\dfrac{3}{5}x-\\dfrac{19}{5}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In the video we show another example of how to write a linear function given two points written with function notation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Find the Linear Function Given Two Function Values in Function Notation\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/TSIcryAtCmY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<ul>\n<li>Given a graph, you can write the equation of the line it represents.<\/li>\n<li>This method only works for graphs that have points that are easy to verify by visual inspection.<\/li>\n<\/ul>\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-2008\">\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>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>Ex 1: Find the Equation of a Line in Slope Intercept Form Given the Graph of a Line. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/mmWf_oLTNSQ\">https:\/\/youtu.be\/mmWf_oLTNSQ<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>College Algebra. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download free at :  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