{"id":1990,"date":"2016-06-30T21:17:49","date_gmt":"2016-06-30T21:17:49","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=1990"},"modified":"2019-07-24T21:07:07","modified_gmt":"2019-07-24T21:07:07","slug":"read-graph-linear-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/read-graph-linear-functions\/","title":{"raw":"Graph Linear Functions Using Slope and y-Intercept","rendered":"Graph Linear Functions Using Slope and y-Intercept"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcome<\/h3>\r\n<ul>\r\n \t<li>Graph a linear function using the slope and <em>y<\/em>-intercept<\/li>\r\n<\/ul>\r\n<\/div>\r\nAnother way to graph a linear function is by using its slope <em>m\u00a0<\/em>and y-intercept.\r\n\r\nLet us consider the following function.\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{1}{2}x+1[\/latex]<\/p>\r\nThe function is in slope-intercept form, so the slope is [latex]\\dfrac{1}{2}[\/latex]. Because the slope is positive, we know the graph will slant upward from left to right. The <em>y-<\/em>intercept is the point on the graph when\u00a0[latex]x=0[\/latex]. The graph crosses the <em>y<\/em>-axis at\u00a0[latex](0, 1)[\/latex]. Now we know the slope and the <em>y<\/em>-intercept. We can begin graphing by plotting the point\u00a0[latex](0, 1)[\/latex]. We know that the slope is rise over run, [latex]m=\\dfrac{\\text{rise}}{\\text{run}}[\/latex]. From our example, we have [latex]m=\\dfrac{1}{2}[\/latex], which means that the rise is\u00a0[latex]1[\/latex] and the run is\u00a0[latex]2[\/latex]. So starting from our <em>y<\/em>-intercept\u00a0[latex](0, 1)[\/latex], we can rise\u00a0[latex]1[\/latex] and then run\u00a0[latex]2[\/latex], or run\u00a0[latex]2[\/latex] and then rise\u00a0[latex]1[\/latex]. We repeat until we have a few points and then we draw a line through the points as shown in the graph below.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201048\/CNX_Precalc_Figure_02_02_0032.jpg\" alt=\"graph of the line y = (1\/2)x +1 showing the &quot;rise&quot;, or change in the y direction as 1 and the &quot;run&quot;, or change in x direction as 2, and the y-intercept at (0,1)\" width=\"617\" height=\"340\" \/>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Graphical Interpretation of a Linear Function<\/h3>\r\nIn the equation [latex]f\\left(x\\right)=mx+b[\/latex]\r\n<ul>\r\n \t<li><em>b<\/em>\u00a0is the <em>y<\/em>-intercept of the graph and indicates the point (0, <em>b<\/em>) at which the graph crosses the <em>y<\/em>-axis.<\/li>\r\n \t<li><em>m<\/em>\u00a0is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\r\n\r\n<\/div>\r\nAll linear functions cross the y-axis and therefore have y-intercepts. (Note: <em>A vertical line parallel to the y-axis does not have a y-intercept, but it is not a function.<\/em>)\r\n<div class=\"textbox\">\r\n<h3>How To: Given the equation for a linear function, graph the function using the <em>y<\/em>-intercept and slope<\/h3>\r\n<ol>\r\n \t<li>Evaluate the function at an input value of zero to find the <em>y-<\/em>intercept.<\/li>\r\n \t<li>Identify the slope.<\/li>\r\n \t<li>Plot the point represented by the <em>y-<\/em>intercept.<\/li>\r\n \t<li>Use [latex]\\dfrac{\\text{rise}}{\\text{run}}[\/latex] to determine at least two more points on the line.<\/li>\r\n \t<li>Sketch the line that passes through the points.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGraph [latex]f\\left(x\\right)=-\\dfrac{2}{3}x+5[\/latex] using the <em>y-<\/em>intercept and slope.\r\n\r\n[reveal-answer q=\"421669\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"421669\"]\r\n\r\nEvaluate the function at [latex]x=0[\/latex]\u00a0to find the <em>y-<\/em>intercept. The output value when [latex]x=0[\/latex] is\u00a0[latex]5[\/latex], so the graph will cross the <em>y<\/em>-axis at\u00a0[latex](0, 5)[\/latex].\r\n\r\nAccording to the equation for the function, the slope of the line is [latex]-\\dfrac{2}{3}[\/latex]. This tells us that for each vertical decrease in the \"rise\" of [latex]\u20132[\/latex] units, the \"run\" increases by\u00a0[latex]3[\/latex] units in the horizontal direction. We can now graph the function by first plotting the <em>y<\/em>-intercept in the graph below. From the initial value\u00a0[latex](0, 5)[\/latex], we move down\u00a0[latex]2[\/latex] units and to the right\u00a0[latex]3[\/latex] units. We can extend the line to the left and right by using this relationship to plot additional points and then drawing a line through the points.\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\/25201050\/CNX_Precalc_Figure_02_02_0042.jpg\" alt=\"graph of the line y = (-2\/3)x + 5 showing the change of -2 in y and change of 3 in x.\" width=\"487\" height=\"318\" \/>\r\n\r\nThe graph slants downward from left to right, which means it has a negative slope as expected.\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\n[ohm_question]79774[\/ohm_question]\r\n\r\n<\/div>\r\nIn the following video we show another example of how to graph a linear function given the y-intercepts and the slope.\r\n\r\nhttps:\/\/youtu.be\/N6lEPh11gk8\r\n\r\nIn the last example, we will show how to graph another linear function using the slope and y-intercept.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGraph [latex]f\\left(x\\right)=-\\dfrac{3}{4}x+6[\/latex]\u00a0using the slope and y-intercept.\r\n[reveal-answer q=\"930515\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"930515\"]\r\n\r\nThe slope of this function is [latex]-\\dfrac{3}{4}[\/latex] and the y-intercept is [latex](0,6)[\/latex] We can start graphing by plotting the y-intercept and counting down three units and right\u00a0[latex]4[\/latex] units. The first stop would be [latex](4,3)[\/latex], and the next stop would be\u00a0[latex](8,0)[\/latex].\r\n\r\n<img class=\"size-medium wp-image-2482 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/13204212\/Screen-Shot-2016-07-13-at-1.35.32-PM-300x229.png\" alt=\"Screen Shot 2016-07-13 at 1.35.32 PM\" width=\"300\" height=\"229\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcome<\/h3>\n<ul>\n<li>Graph a linear function using the slope and <em>y<\/em>-intercept<\/li>\n<\/ul>\n<\/div>\n<p>Another way to graph a linear function is by using its slope <em>m\u00a0<\/em>and y-intercept.<\/p>\n<p>Let us consider the following function.<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{1}{2}x+1[\/latex]<\/p>\n<p>The function is in slope-intercept form, so the slope is [latex]\\dfrac{1}{2}[\/latex]. Because the slope is positive, we know the graph will slant upward from left to right. The <em>y-<\/em>intercept is the point on the graph when\u00a0[latex]x=0[\/latex]. The graph crosses the <em>y<\/em>-axis at\u00a0[latex](0, 1)[\/latex]. Now we know the slope and the <em>y<\/em>-intercept. We can begin graphing by plotting the point\u00a0[latex](0, 1)[\/latex]. We know that the slope is rise over run, [latex]m=\\dfrac{\\text{rise}}{\\text{run}}[\/latex]. From our example, we have [latex]m=\\dfrac{1}{2}[\/latex], which means that the rise is\u00a0[latex]1[\/latex] and the run is\u00a0[latex]2[\/latex]. So starting from our <em>y<\/em>-intercept\u00a0[latex](0, 1)[\/latex], we can rise\u00a0[latex]1[\/latex] and then run\u00a0[latex]2[\/latex], or run\u00a0[latex]2[\/latex] and then rise\u00a0[latex]1[\/latex]. We repeat until we have a few points and then we draw a line through the points as shown in the graph below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201048\/CNX_Precalc_Figure_02_02_0032.jpg\" alt=\"graph of the line y = (1\/2)x +1 showing the &quot;rise&quot;, or change in the y direction as 1 and the &quot;run&quot;, or change in x direction as 2, and the y-intercept at (0,1)\" width=\"617\" height=\"340\" \/><\/p>\n<div class=\"textbox\">\n<h3>A General Note: Graphical Interpretation of a Linear Function<\/h3>\n<p>In the equation [latex]f\\left(x\\right)=mx+b[\/latex]<\/p>\n<ul>\n<li><em>b<\/em>\u00a0is the <em>y<\/em>-intercept of the graph and indicates the point (0, <em>b<\/em>) at which the graph crosses the <em>y<\/em>-axis.<\/li>\n<li><em>m<\/em>\u00a0is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\n<\/div>\n<p>All linear functions cross the y-axis and therefore have y-intercepts. (Note: <em>A vertical line parallel to the y-axis does not have a y-intercept, but it is not a function.<\/em>)<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the equation for a linear function, graph the function using the <em>y<\/em>-intercept and slope<\/h3>\n<ol>\n<li>Evaluate the function at an input value of zero to find the <em>y-<\/em>intercept.<\/li>\n<li>Identify the slope.<\/li>\n<li>Plot the point represented by the <em>y-<\/em>intercept.<\/li>\n<li>Use [latex]\\dfrac{\\text{rise}}{\\text{run}}[\/latex] to determine at least two more points on the line.<\/li>\n<li>Sketch the line that passes through the points.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Graph [latex]f\\left(x\\right)=-\\dfrac{2}{3}x+5[\/latex] using the <em>y-<\/em>intercept and slope.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q421669\">Show Solution<\/span><\/p>\n<div id=\"q421669\" class=\"hidden-answer\" style=\"display: none\">\n<p>Evaluate the function at [latex]x=0[\/latex]\u00a0to find the <em>y-<\/em>intercept. The output value when [latex]x=0[\/latex] is\u00a0[latex]5[\/latex], so the graph will cross the <em>y<\/em>-axis at\u00a0[latex](0, 5)[\/latex].<\/p>\n<p>According to the equation for the function, the slope of the line is [latex]-\\dfrac{2}{3}[\/latex]. This tells us that for each vertical decrease in the &#8220;rise&#8221; of [latex]\u20132[\/latex] units, the &#8220;run&#8221; increases by\u00a0[latex]3[\/latex] units in the horizontal direction. We can now graph the function by first plotting the <em>y<\/em>-intercept in the graph below. From the initial value\u00a0[latex](0, 5)[\/latex], we move down\u00a0[latex]2[\/latex] units and to the right\u00a0[latex]3[\/latex] units. We can extend the line to the left and right by using this relationship to plot additional points and then drawing a line through the points.<\/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\/25201050\/CNX_Precalc_Figure_02_02_0042.jpg\" alt=\"graph of the line y = (-2\/3)x + 5 showing the change of -2 in y and change of 3 in x.\" width=\"487\" height=\"318\" \/><\/p>\n<p>The graph slants downward from left to right, which means it has a negative slope as expected.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm79774\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=79774&theme=oea&iframe_resize_id=ohm79774&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the following video we show another example of how to graph a linear function given the y-intercepts and the slope.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex: Graph a Line and ID the Slope and Intercepts (Fraction Slope)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/N6lEPh11gk8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the last example, we will show how to graph another linear function using the slope and y-intercept.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Graph [latex]f\\left(x\\right)=-\\dfrac{3}{4}x+6[\/latex]\u00a0using the slope and y-intercept.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q930515\">Show Solution<\/span><\/p>\n<div id=\"q930515\" class=\"hidden-answer\" style=\"display: none\">\n<p>The slope of this function is [latex]-\\dfrac{3}{4}[\/latex] and the y-intercept is [latex](0,6)[\/latex] We can start graphing by plotting the y-intercept and counting down three units and right\u00a0[latex]4[\/latex] units. The first stop would be [latex](4,3)[\/latex], and the next stop would be\u00a0[latex](8,0)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-2482 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/13204212\/Screen-Shot-2016-07-13-at-1.35.32-PM-300x229.png\" alt=\"Screen Shot 2016-07-13 at 1.35.32 PM\" width=\"300\" height=\"229\" \/><\/p>\n<\/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-1990\">\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: Graph a Line and ID the Slope and Intercepts (Fraction Slope). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/N6lEPh11gk8\">https:\/\/youtu.be\/N6lEPh11gk8<\/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>QID 79774: Graph linear eq. in slope-intercept form, give slope and intercept.. <strong>Authored by<\/strong>: Day, Alyson. <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, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/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\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/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":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"},{\"type\":\"cc\",\"description\":\"Ex: Graph a Line and ID the Slope and Intercepts (Fraction Slope)\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/N6lEPh11gk8\",\"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\":\"QID 79774: Graph linear eq. in slope-intercept form, give slope and intercept.\",\"author\":\"Day, Alyson\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"e0fb6253-da31-4c16-9ffc-4c55f001d4bb","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1990","chapter","type-chapter","status-publish","hentry"],"part":1897,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/1990","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":31,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/1990\/revisions"}],"predecessor-version":[{"id":5397,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/1990\/revisions\/5397"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/parts\/1897"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/1990\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/media?parent=1990"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1990"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/contributor?post=1990"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/license?post=1990"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}