{"id":212,"date":"2024-10-18T21:13:08","date_gmt":"2024-10-18T21:13:08","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/chapter\/more-on-linear-functions-learn-it-1\/"},"modified":"2024-10-18T21:13:08","modified_gmt":"2024-10-18T21:13:08","slug":"more-on-linear-functions-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/chapter\/more-on-linear-functions-learn-it-1\/","title":{"raw":"More on Linear Functions: Learn It 1","rendered":"More on Linear Functions: Learn It 1"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Create and interpret equations of linear functions<\/li>\n\t<li>Use linear functions to model and draw conclusions from real-world problems<\/li>\n\t<li>Plot the graphs of linear equations<\/li>\n<\/ul>\n<\/section>\n<h2>Point-Slope Form<\/h2>\n<p>Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Now we will learn another way to write a linear function called <strong>point-slope form <\/strong>which is given below:<\/p>\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n<p>where [latex]m[\/latex] is the slope of the linear function and [latex]({x}_{1},{y}_{1})[\/latex] is any point which satisfies the linear function.<\/p>\n<p>The point-slope form is derived from the slope formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}{m}=\\frac{y-{y}_{1}}{x-{x}_{1}}\\hfill &amp; \\text{assuming }{ x }\\ne {x}_{1}\\hfill\\\\{ m }\\left(x-{x}_{1}\\right)=\\frac{y-{y}_{1}}{x-{x}_{1}}\\left(x-{x}_{1}\\right)\\hfill &amp; \\text{Multiply both sides by }\\left(x-{x}_{1}\\right)\\hfill\\\\{ m }\\left(x-{x}_{1}\\right)=y-{y}_{1}\\hfill &amp; \\text{Simplify}\\hfill \\\\ y-{y}_{1}={ m }\\left(x-{x}_{1}\\right)\\hfill &amp;\\text{Rearrange}\\hfill \\end{array}[\/latex]<\/p>\n<p>Keep in mind that slope-intercept form and point-slope form can be used to describe the same linear function. We can move from one form to another using basic algebra. For example, suppose we are given the equation [latex]y - 4=-\\frac{1}{2}\\left(x - 6\\right)[\/latex] which is in point-slope form. We can convert it to slope-intercept form as shown below.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}y - 4=-\\frac{1}{2}\\left(x - 6\\right)\\hfill &amp; \\hfill \\\\ y - 4=-\\frac{1}{2}x+3\\hfill &amp; \\text{Distribute the }-\\frac{1}{2}.\\hfill \\\\ \\text{}y=-\\frac{1}{2}x+7\\hfill &amp; \\text{Add 4 to each side}.\\hfill \\end{array}[\/latex]<\/p>\n<p>Therefore, the same line can be described in slope-intercept form as [latex]y=-\\frac{1}{2}x+7[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>point-slope form<\/h3>\n<p><strong>Point-slope form<\/strong> of a linear equation takes the form<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>where [latex]m[\/latex]&nbsp;is the slope and [latex]{x}_{1 }[\/latex] and [latex]{y}_{1}[\/latex] are the [latex]x[\/latex] and [latex]y[\/latex] coordinates of a specific point through which the line passes.<\/p>\n<\/div>\n<\/section>\n<p>Point-slope form is particularly useful if we know one point and the slope of a line. For example, suppose we are told that a line has a slope of [latex]2[\/latex] and passes through the point [latex]\\left(4,1\\right)[\/latex]. We know that [latex]m=2[\/latex] and that [latex]{x}_{1}=4[\/latex] and [latex]{y}_{1}=1[\/latex]. We can substitute these values into point-slope form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=2\\left(x - 4\\right)\\end{array}[\/latex]<\/p>\n<p>If we wanted to rewrite the equation in slope-intercept form, we apply algebraic techniques.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll} y - 1=2\\left(x - 4\\right)\\hfill &amp; \\hfill \\\\ y - 1=2x - 8\\hfill &amp; \\text{Distribute the }2.\\hfill \\\\ \\text{}y=2x - 7\\hfill &amp; \\text{Add 1 to each side}.\\hfill \\end{array}[\/latex]<\/p>\n<p>Both equations [latex]y - 1=2\\left(x - 4\\right)[\/latex]&nbsp;and [latex]y=2x - 7[\/latex] describe the same line. The graph of the line can be seen below.<\/p>\n<center><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201023\/CNX_Precalc_Figure_02_01_0132.jpg\" alt=\"The graph of the equations y - 1=2(x - 4) and y=2x - 7\" width=\"487\" height=\"386\"><\/center>\n<section class=\"textbox example\">Write the point-slope form of an equation of a line with a slope of [latex]3[\/latex] that passes through the point [latex]\\left(6,-1\\right)[\/latex]. Then rewrite the equation in slope-intercept form. [reveal-answer q=\"579101\"]Show Solution[\/reveal-answer] [hidden-answer a=\"579101\"] Let\u2019s figure out what we know from the given information. The slope is [latex]3[\/latex], so [latex]m= 3[\/latex]. We also know one point, so we know [latex]{x}_{1}=6[\/latex] and [latex]{y}_{1}=-1[\/latex]. Now we can substitute these values into point-slope form.\n\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}\\text{}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill &amp; \\hfill \\\\ y-\\left(-1\\right)=3\\left(x - 6\\right)\\hfill &amp; \\text{Substitute known values}.\\hfill \\\\ \\text{}y+1=3\\left(x - 6\\right)\\hfill &amp; \\text{Distribute }-1\\text{ to find point-slope form}.\\hfill \\end{array}[\/latex]<\/p>\n\nWe use algebra to find slope-intercept form of the linear equation.\n\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}y+1=3\\left(x - 6\\right)\\hfill &amp; \\hfill \\\\ y+1=3x - 18\\hfill &amp; \\text{Distribute 3}.\\hfill \\\\ \\text{}y=3x - 19\\hfill &amp; \\text{Simplify to slope-intercept form}.\\hfill \\end{array}[\/latex]<\/p>\n\n[\/hidden-answer]<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm2_question hide_question_numbers=1]13527[\/ohm2_question]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Create and interpret equations of linear functions<\/li>\n<li>Use linear functions to model and draw conclusions from real-world problems<\/li>\n<li>Plot the graphs of linear equations<\/li>\n<\/ul>\n<\/section>\n<h2>Point-Slope Form<\/h2>\n<p>Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Now we will learn another way to write a linear function called <strong>point-slope form <\/strong>which is given below:<\/p>\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n<p>where [latex]m[\/latex] is the slope of the linear function and [latex]({x}_{1},{y}_{1})[\/latex] is any point which satisfies the linear function.<\/p>\n<p>The point-slope form is derived from the slope formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}{m}=\\frac{y-{y}_{1}}{x-{x}_{1}}\\hfill & \\text{assuming }{ x }\\ne {x}_{1}\\hfill\\\\{ m }\\left(x-{x}_{1}\\right)=\\frac{y-{y}_{1}}{x-{x}_{1}}\\left(x-{x}_{1}\\right)\\hfill & \\text{Multiply both sides by }\\left(x-{x}_{1}\\right)\\hfill\\\\{ m }\\left(x-{x}_{1}\\right)=y-{y}_{1}\\hfill & \\text{Simplify}\\hfill \\\\ y-{y}_{1}={ m }\\left(x-{x}_{1}\\right)\\hfill &\\text{Rearrange}\\hfill \\end{array}[\/latex]<\/p>\n<p>Keep in mind that slope-intercept form and point-slope form can be used to describe the same linear function. We can move from one form to another using basic algebra. For example, suppose we are given the equation [latex]y - 4=-\\frac{1}{2}\\left(x - 6\\right)[\/latex] which is in point-slope form. We can convert it to slope-intercept form as shown below.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}y - 4=-\\frac{1}{2}\\left(x - 6\\right)\\hfill & \\hfill \\\\ y - 4=-\\frac{1}{2}x+3\\hfill & \\text{Distribute the }-\\frac{1}{2}.\\hfill \\\\ \\text{}y=-\\frac{1}{2}x+7\\hfill & \\text{Add 4 to each side}.\\hfill \\end{array}[\/latex]<\/p>\n<p>Therefore, the same line can be described in slope-intercept form as [latex]y=-\\frac{1}{2}x+7[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>point-slope form<\/h3>\n<p><strong>Point-slope form<\/strong> of a linear equation takes the form<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>where [latex]m[\/latex]&nbsp;is the slope and [latex]{x}_{1 }[\/latex] and [latex]{y}_{1}[\/latex] are the [latex]x[\/latex] and [latex]y[\/latex] coordinates of a specific point through which the line passes.<\/p>\n<\/div>\n<\/section>\n<p>Point-slope form is particularly useful if we know one point and the slope of a line. For example, suppose we are told that a line has a slope of [latex]2[\/latex] and passes through the point [latex]\\left(4,1\\right)[\/latex]. We know that [latex]m=2[\/latex] and that [latex]{x}_{1}=4[\/latex] and [latex]{y}_{1}=1[\/latex]. We can substitute these values into point-slope form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=2\\left(x - 4\\right)\\end{array}[\/latex]<\/p>\n<p>If we wanted to rewrite the equation in slope-intercept form, we apply algebraic techniques.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll} y - 1=2\\left(x - 4\\right)\\hfill & \\hfill \\\\ y - 1=2x - 8\\hfill & \\text{Distribute the }2.\\hfill \\\\ \\text{}y=2x - 7\\hfill & \\text{Add 1 to each side}.\\hfill \\end{array}[\/latex]<\/p>\n<p>Both equations [latex]y - 1=2\\left(x - 4\\right)[\/latex]&nbsp;and [latex]y=2x - 7[\/latex] describe the same line. The graph of the line can be seen below.<\/p>\n<div style=\"text-align: center;\"><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\/25201023\/CNX_Precalc_Figure_02_01_0132.jpg\" alt=\"The graph of the equations y - 1=2(x - 4) and y=2x - 7\" width=\"487\" height=\"386\" \/><\/div>\n<section class=\"textbox example\">Write the point-slope form of an equation of a line with a slope of [latex]3[\/latex] that passes through the point [latex]\\left(6,-1\\right)[\/latex]. Then rewrite the equation in slope-intercept form. <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q579101\">Show Solution<\/button> <\/p>\n<div id=\"q579101\" class=\"hidden-answer\" style=\"display: none\"> Let\u2019s figure out what we know from the given information. The slope is [latex]3[\/latex], so [latex]m= 3[\/latex]. We also know one point, so we know [latex]{x}_{1}=6[\/latex] and [latex]{y}_{1}=-1[\/latex]. Now we can substitute these values into point-slope form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}\\text{}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill & \\hfill \\\\ y-\\left(-1\\right)=3\\left(x - 6\\right)\\hfill & \\text{Substitute known values}.\\hfill \\\\ \\text{}y+1=3\\left(x - 6\\right)\\hfill & \\text{Distribute }-1\\text{ to find point-slope form}.\\hfill \\end{array}[\/latex]<\/p>\n<p>We use algebra to find slope-intercept form of the linear equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}y+1=3\\left(x - 6\\right)\\hfill & \\hfill \\\\ y+1=3x - 18\\hfill & \\text{Distribute 3}.\\hfill \\\\ \\text{}y=3x - 19\\hfill & \\text{Simplify to slope-intercept form}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm13527\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=13527&theme=lumen&iframe_resize_id=ohm13527&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":21,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"College Algebra\",\"author\":\"Jay Abramson\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":185,"module-header":"","content_attributions":[{"type":"cc-attribution","description":"College Algebra","author":"Jay Abramson","organization":"OpenStax","url":"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2","project":"","license":"cc-by","license_terms":"Access for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2"}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/pressbooks\/v2\/chapters\/212"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/pressbooks\/v2\/chapters\/212\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/pressbooks\/v2\/parts\/185"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/pressbooks\/v2\/chapters\/212\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/wp\/v2\/media?parent=212"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/pressbooks\/v2\/chapter-type?post=212"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/wp\/v2\/contributor?post=212"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/wp\/v2\/license?post=212"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}