{"id":4643,"date":"2017-06-07T18:54:02","date_gmt":"2017-06-07T18:54:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-slope-intercept-form-of-a-line\/"},"modified":"2017-08-16T03:12:23","modified_gmt":"2017-08-16T03:12:23","slug":"read-slope-intercept-form-of-a-line","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-slope-intercept-form-of-a-line\/","title":{"raw":"Slope Intercept Form of a Line","rendered":"Slope Intercept Form of a Line"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Write the equation and draw the graph of a line using slope and y-intercept\r\n<ul>\r\n \t<li>Write the equation of a line using slope and y-intercept<\/li>\r\n \t<li>Rearrange a linear equation so it is in slope-intercept form.<\/li>\r\n \t<li>Graph a line using slope and y-intercept<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Slope-Intercept Form of a Line<\/h2>\r\nWhen graphing a line we found one method we could use is to make a table of values. However, if we can identify some properties of the line, we may be able to make a graph much quicker and easier. One such method is finding the slope and the <em>y<\/em>-intercept of the equation. The slope can be represented by m and the <em>y<\/em>-intercept, where it crosses the axis and [latex]x=0[\/latex], can be represented by [latex](0,b)[\/latex] where <em>b<\/em> is the value where the graph crosses the vertical <em>y<\/em>-axis. Any other point on the line can be represented by [latex](x,y)[\/latex].\r\n<div class=\"textbox shaded\">\r\n\r\nIn the equation,\r\n<p style=\"text-align: center\">[latex]y = mx + b[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,m\\,\\,\\,\\,=\\,\\,\\,\\text{slope}\\\\(x,y)=\\,\\,\\,\\text{a point on the line}\\\\\\,\\,\\,\\,\\,\\,\\,b\\,\\,\\,\\,=\\,\\,\\,\\text{the y value of the y-intercept}\\end{array}[\/latex]<\/p>\r\nThis formula is known as the <span style=\"color: #993366\"><strong>slope-intercept equation<\/strong><\/span>.\u00a0If we know the slope and the <em>y<\/em>-intercept we can easily find the equation that represents the line.\r\n\r\n<\/div>\r\n&nbsp;\r\n<p style=\"text-align: left\"><\/p>\r\n\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the equation of the line that has a slope of [latex] \\displaystyle \\frac{1}{2}[\/latex] and a <i>y<\/i>-intercept of [latex]\u22125[\/latex].\r\n\r\n[reveal-answer q=\"624715\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"624715\"]Substitute the slope (<i>m<\/i>) into [latex]y=mx+b[\/latex].\r\n<p style=\"text-align: center\">[latex] \\displaystyle y=\\frac{1}{2}x+b[\/latex]<\/p>\r\nSubstitute the <i>y<\/i>-intercept (<i>b<\/i>) into the equation.\r\n<p style=\"text-align: center\">[latex] \\displaystyle y=\\frac{1}{2}x-5[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]y=\\frac{1}{2}x-5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe can also find the equation by looking at a graph and finding the slope and <em>y<\/em>-intercept.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the equation of the line in the graph by identifying the slope and <em>y<\/em>-intercept.\r\n<img class=\"size-medium wp-image-3198 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185358\/SVG_Grapher-300x297.png\" alt=\"SVG_Grapher\" width=\"300\" height=\"297\" \/>\r\n[reveal-answer q=\"96446\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"96446\"]Identify the point where the graph crosses the y-axis [latex](0,3)[\/latex]. This means the <em>y<\/em>-intercept is 3.\r\n\r\nIdentify one other point and draw a slope triangle to find the slope.\r\n\r\nThe slope is [latex]\\frac{-2}{3}[\/latex]\r\n\r\nSubstitute the slope and <em>y<\/em> value of the intercept into the slope-intercept equation.\r\n<p style=\"text-align: center\">[latex]y=mx+b\\\\y=\\frac{-2}{3}x+b\\\\y=\\frac{-2}{3}x+3[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]y=\\frac{-2}{3}x+3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe can also move the opposite direction, using the equation identify the slope\u00a0and <em>y<\/em>-intercept and graph the equation from this information. However, it will be\u00a0important for the equation to first be in slope intercept form. If it is not, we will\u00a0have to solve it for <em>y<\/em> so we can identify the slope and the <em>y<\/em>-intercept.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the following equation in slope-intercept form.\r\n<p style=\"text-align: center\">[latex]2x+4y=6[\/latex]<\/p>\r\n[reveal-answer q=\"373034\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"373034\"]We need to solve for <em>y<\/em>. Start by subtracting [latex]2[\/latex] from both sides.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}2x\\,\\,\\,+\\,\\,\\,4y\\,\\,\\,=\\,\\,\\,6\\\\-2x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2x\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">It helps to place the <em>x<\/em> term first on the right hand side. Notice how we keep the 6 positive by placing an addition sign in front.<\/p>\r\n<p style=\"text-align: center\">[latex]4y=-2x+6[\/latex]<\/p>\r\n<p style=\"text-align: left\">Divide each term by 4 to isolate the <em>y<\/em>.<\/p>\r\n<p style=\"text-align: center\">[latex]\\frac{4y}{4}=\\frac{-2x}{4}+\\frac{6}{4}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]y=\\frac{-2x}{4}+\\frac{6}{4}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Reduce the fractions<\/p>\r\n<p style=\"text-align: center\">[latex]y=-\\frac{1}{2}x+\\frac{3}{2}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]y=-\\frac{1}{2}x+\\frac{3}{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Graphing Using Slope-Intercept<\/h2>\r\nOnce we have an equation in slope-intercept form we can graph it by first plotting\u00a0the <em>y<\/em>-intercept, then using the slope, find a second point and connecting the dots.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nGraph [latex]y=\\frac{1}{2}x-4[\/latex] using the slope-intercept equation.\r\n\r\n[reveal-answer q=\"420487\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"420487\"]First, plot the <em>y<\/em>-intercept.\r\n\r\n<img class=\"aligncenter wp-image-3202 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185400\/SVG_Grapher2-300x294.png\" alt=\"The y-intercept plotted at negative 4 on the y axis.\" width=\"300\" height=\"294\" \/>\r\n\r\nNow use the slope to count up or down and over left or right to the next point. This slope is [latex]\\frac{1}{2}[\/latex], so you can count up one and right two\u2014both positive because both parts of the slope are positive.\r\n\r\nConnect the dots.\r\n<img class=\"aligncenter wp-image-3203 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185401\/SVG_Grapher3-300x289.png\" alt=\"A line crosses through negative 4 on the y-axis and has a slope of 1\/2.\" width=\"300\" height=\"289\" \/>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n<h2>Find the Equation of a Line in Slope-Intercept Form<\/h2>\r\nhttps:\/\/youtu.be\/GIn7vbB5AYo","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Write the equation and draw the graph of a line using slope and y-intercept\n<ul>\n<li>Write the equation of a line using slope and y-intercept<\/li>\n<li>Rearrange a linear equation so it is in slope-intercept form.<\/li>\n<li>Graph a line using slope and y-intercept<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<h2>Slope-Intercept Form of a Line<\/h2>\n<p>When graphing a line we found one method we could use is to make a table of values. However, if we can identify some properties of the line, we may be able to make a graph much quicker and easier. One such method is finding the slope and the <em>y<\/em>-intercept of the equation. The slope can be represented by m and the <em>y<\/em>-intercept, where it crosses the axis and [latex]x=0[\/latex], can be represented by [latex](0,b)[\/latex] where <em>b<\/em> is the value where the graph crosses the vertical <em>y<\/em>-axis. Any other point on the line can be represented by [latex](x,y)[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<p>In the equation,<\/p>\n<p style=\"text-align: center\">[latex]y = mx + b[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,m\\,\\,\\,\\,=\\,\\,\\,\\text{slope}\\\\(x,y)=\\,\\,\\,\\text{a point on the line}\\\\\\,\\,\\,\\,\\,\\,\\,b\\,\\,\\,\\,=\\,\\,\\,\\text{the y value of the y-intercept}\\end{array}[\/latex]<\/p>\n<p>This formula is known as the <span style=\"color: #993366\"><strong>slope-intercept equation<\/strong><\/span>.\u00a0If we know the slope and the <em>y<\/em>-intercept we can easily find the equation that represents the line.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p style=\"text-align: left\">\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the equation of the line that has a slope of [latex]\\displaystyle \\frac{1}{2}[\/latex] and a <i>y<\/i>-intercept of [latex]\u22125[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q624715\">Show Solution<\/span><\/p>\n<div id=\"q624715\" class=\"hidden-answer\" style=\"display: none\">Substitute the slope (<i>m<\/i>) into [latex]y=mx+b[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle y=\\frac{1}{2}x+b[\/latex]<\/p>\n<p>Substitute the <i>y<\/i>-intercept (<i>b<\/i>) into the equation.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle y=\\frac{1}{2}x-5[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=\\frac{1}{2}x-5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We can also find the equation by looking at a graph and finding the slope and <em>y<\/em>-intercept.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the equation of the line in the graph by identifying the slope and <em>y<\/em>-intercept.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-3198 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185358\/SVG_Grapher-300x297.png\" alt=\"SVG_Grapher\" width=\"300\" height=\"297\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q96446\">Show Solution<\/span><\/p>\n<div id=\"q96446\" class=\"hidden-answer\" style=\"display: none\">Identify the point where the graph crosses the y-axis [latex](0,3)[\/latex]. This means the <em>y<\/em>-intercept is 3.<\/p>\n<p>Identify one other point and draw a slope triangle to find the slope.<\/p>\n<p>The slope is [latex]\\frac{-2}{3}[\/latex]<\/p>\n<p>Substitute the slope and <em>y<\/em> value of the intercept into the slope-intercept equation.<\/p>\n<p style=\"text-align: center\">[latex]y=mx+b\\\\y=\\frac{-2}{3}x+b\\\\y=\\frac{-2}{3}x+3[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=\\frac{-2}{3}x+3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We can also move the opposite direction, using the equation identify the slope\u00a0and <em>y<\/em>-intercept and graph the equation from this information. However, it will be\u00a0important for the equation to first be in slope intercept form. If it is not, we will\u00a0have to solve it for <em>y<\/em> so we can identify the slope and the <em>y<\/em>-intercept.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the following equation in slope-intercept form.<\/p>\n<p style=\"text-align: center\">[latex]2x+4y=6[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q373034\">Show Solution<\/span><\/p>\n<div id=\"q373034\" class=\"hidden-answer\" style=\"display: none\">We need to solve for <em>y<\/em>. Start by subtracting [latex]2[\/latex] from both sides.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}2x\\,\\,\\,+\\,\\,\\,4y\\,\\,\\,=\\,\\,\\,6\\\\-2x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2x\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">It helps to place the <em>x<\/em> term first on the right hand side. Notice how we keep the 6 positive by placing an addition sign in front.<\/p>\n<p style=\"text-align: center\">[latex]4y=-2x+6[\/latex]<\/p>\n<p style=\"text-align: left\">Divide each term by 4 to isolate the <em>y<\/em>.<\/p>\n<p style=\"text-align: center\">[latex]\\frac{4y}{4}=\\frac{-2x}{4}+\\frac{6}{4}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]y=\\frac{-2x}{4}+\\frac{6}{4}[\/latex]<\/p>\n<p style=\"text-align: left\">Reduce the fractions<\/p>\n<p style=\"text-align: center\">[latex]y=-\\frac{1}{2}x+\\frac{3}{2}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=-\\frac{1}{2}x+\\frac{3}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Graphing Using Slope-Intercept<\/h2>\n<p>Once we have an equation in slope-intercept form we can graph it by first plotting\u00a0the <em>y<\/em>-intercept, then using the slope, find a second point and connecting the dots.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Graph [latex]y=\\frac{1}{2}x-4[\/latex] using the slope-intercept equation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q420487\">Show Solution<\/span><\/p>\n<div id=\"q420487\" class=\"hidden-answer\" style=\"display: none\">First, plot the <em>y<\/em>-intercept.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3202 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185400\/SVG_Grapher2-300x294.png\" alt=\"The y-intercept plotted at negative 4 on the y axis.\" width=\"300\" height=\"294\" \/><\/p>\n<p>Now use the slope to count up or down and over left or right to the next point. This slope is [latex]\\frac{1}{2}[\/latex], so you can count up one and right two\u2014both positive because both parts of the slope are positive.<\/p>\n<p>Connect the dots.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3203 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185401\/SVG_Grapher3-300x289.png\" alt=\"A line crosses through negative 4 on the y-axis and has a slope of 1\/2.\" width=\"300\" height=\"289\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Find the Equation of a Line in Slope-Intercept Form<\/h2>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Find the Equation of a Line in Slope-Intercept Form of a Line (Basic)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/GIn7vbB5AYo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\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-4643\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Slope-Intercept Form of a Line. <strong>Authored by<\/strong>: Mathispower4u. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Beginning and Intermediate Algebra. <strong>Authored by<\/strong>: Tyler Wallace. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"\"><\/a>. <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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Beginning and Intermediate Algebra\",\"author\":\"Tyler Wallace\",\"organization\":\"\",\"url\":\": http:\/\/wallace.ccfaculty.org\/book\/book.html\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Slope-Intercept Form of a Line\",\"author\":\"Mathispower4u\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"d3f00df5-50e0-4c8e-89f6-00c18abbd164","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4643","chapter","type-chapter","status-publish","hentry"],"part":4587,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4643","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4643\/revisions"}],"predecessor-version":[{"id":5083,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4643\/revisions\/5083"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/parts\/4587"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4643\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/media?parent=4643"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapter-type?post=4643"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/contributor?post=4643"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/license?post=4643"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}