{"id":217,"date":"2024-10-18T21:13:09","date_gmt":"2024-10-18T21:13:09","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/chapter\/more-on-linear-functions-learn-it-6\/"},"modified":"2024-10-18T21:13:09","modified_gmt":"2024-10-18T21:13:09","slug":"more-on-linear-functions-learn-it-6","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/chapter\/more-on-linear-functions-learn-it-6\/","title":{"raw":"More on Linear Functions: Learn It 6","rendered":"More on Linear Functions: Learn It 6"},"content":{"raw":"\n<h2>Graphing Linear Functions Cont.<\/h2>\n<h3>Graphing a Linear Function Using Transformations<\/h3>\n<p>Another option for graphing is to use <strong>transformations<\/strong> on the identity function [latex]f\\left(x\\right)=x[\/latex]. A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression.<\/p>\n<h4>Vertical Stretch or Compression<\/h4>\n<p>In the equation [latex]f\\left(x\\right)=mx[\/latex], the [latex]m[\/latex] is acting as the <strong>vertical stretch<\/strong> or <strong>compression<\/strong> of the identity function. When [latex]m[\/latex] is negative, there is also a vertical reflection of the graph. Notice that multiplying the equation [latex]f\\left(x\\right)=x[\/latex] by [latex]m[\/latex] stretches the graph of [latex]f[\/latex] by a factor of [latex]m[\/latex] units if [latex]m &gt; 1[\/latex] and compresses the graph of [latex]f[\/latex] by a factor of [latex]m[\/latex] units if [latex]0 &lt; m &lt; 1[\/latex]. This means the larger the absolute value of [latex]m[\/latex], the steeper the slope.<\/p>\n<center><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184329\/CNX_Precalc_Figure_02_02_0052.jpg\" alt=\"Graph with several linear functions including y = 3x, y = 2x, y = x, y = (1\/2)x, y = (1\/3)x, y = (-1\/2)x, y = -x, and y = -2x\" width=\"900\" height=\"759\"><\/center><center><strong><span style=\"font-size: 10pt;\">Vertical stretches and compressions and reflections on the function [latex]f\\left(x\\right)=x[\/latex].<\/span><\/strong><\/center>\n<h4>Vertical Shift<\/h4>\n<p>In [latex]f\\left(x\\right)=mx+b[\/latex], the [latex]b[\/latex] acts as the <strong>vertical shift<\/strong>, moving the graph up and down without affecting the slope of the line. Notice that adding a value of [latex]b[\/latex] to the equation of [latex]f\\left(x\\right)=x[\/latex] shifts the graph of [latex]f[\/latex] a total of [latex]b[\/latex] units up if [latex]b[\/latex] is positive and [latex]|b|[\/latex] units down if [latex]b[\/latex] is negative.<\/p>\n<center><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184332\/CNX_Precalc_Figure_02_02_0062.jpg\" alt=\"graph showing y = x , y = x+2, y = x+4, y = x-2, y = x-4\" width=\"900\" height=\"759\"><\/center><center><strong><span style=\"font-size: 10pt;\">This graph illustrates vertical shifts of the function [latex]f\\left(x\\right)=x[\/latex].<\/span><\/strong><\/center>\n<p>&nbsp;<\/p>\n<p>Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method.<\/p>\n<section class=\"textbox questionHelp\"><strong>How To: Given the equation of a linear function, use transformations to graph the linear function in the form [latex]f\\left(x\\right)=mx+b[\/latex].<\/strong>\n<ol>\n\t<li>Graph [latex]f\\left(x\\right)=x[\/latex].<\/li>\n\t<li>Vertically stretch or compress the graph by a factor [latex]m[\/latex].<\/li>\n\t<li>Shift the graph up or down [latex]b[\/latex] units.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Graph [latex]f(x)=\\frac{1}{2}x - 3[\/latex] using transformations.<\/p>\n\n[reveal-answer q=\"750947\"]Show Solution[\/reveal-answer] [hidden-answer a=\"750947\"]\n\n<p>The equation for the function shows that [latex]m=\\frac{1}{2}[\/latex] so the identity function is vertically compressed by [latex]\\frac{1}{2}[\/latex]. The equation for the function also shows that [latex]b=-3[\/latex], so the identity function is vertically shifted down [latex]3[\/latex] units.<\/p>\n<p>First, graph the identity function, and show the vertical compression.<\/p>\n<center><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184335\/CNX_Precalc_Figure_02_02_0072.jpg\" alt=\"graph showing the lines y = x and y = (1\/2)x\" width=\"487\" height=\"378\"><\/center><center><strong><span style=\"font-size: 10pt;\">The function [latex]y=x[\/latex] compressed by a factor of [latex]\\frac{1}{2}[\/latex].<\/span><\/strong><\/center>\n<p>&nbsp;<\/p>\n\nThen, show the vertical shift.<center><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184338\/CNX_Precalc_Figure_02_02_0082.jpg\" alt=\"Graph showing the lines y = (1\/2)x, and y = (1\/2) + 3\" width=\"487\" height=\"377\"><\/center><center><strong><span style=\"font-size: 10pt;\">The function [latex]y=\\frac{1}{2}x[\/latex] shifted down 3 units.<\/span><\/strong><\/center>[\/hidden-answer]<\/section>\n<section class=\"textbox questionHelp\"><strong>Q &amp; A<\/strong> <strong>In the example above, could we have sketched the graph by reversing the order of the transformations?<\/strong> <em>No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following order of operations, let the input be 2.<\/em>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}f\\text{(2)}=\\frac{\\text{1}}{\\text{2}}\\text{(2)}-\\text{3}\\hfill \\\\ =\\text{1}-\\text{3}\\hfill \\\\ =-\\text{2}\\hfill \\end{array}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm2_question hide_question_numbers=1]13637[\/ohm2_question]<\/p>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm2_question hide_question_numbers=1]13639[\/ohm2_question]<\/p>\n<\/section>\n","rendered":"<h2>Graphing Linear Functions Cont.<\/h2>\n<h3>Graphing a Linear Function Using Transformations<\/h3>\n<p>Another option for graphing is to use <strong>transformations<\/strong> on the identity function [latex]f\\left(x\\right)=x[\/latex]. A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression.<\/p>\n<h4>Vertical Stretch or Compression<\/h4>\n<p>In the equation [latex]f\\left(x\\right)=mx[\/latex], the [latex]m[\/latex] is acting as the <strong>vertical stretch<\/strong> or <strong>compression<\/strong> of the identity function. When [latex]m[\/latex] is negative, there is also a vertical reflection of the graph. Notice that multiplying the equation [latex]f\\left(x\\right)=x[\/latex] by [latex]m[\/latex] stretches the graph of [latex]f[\/latex] by a factor of [latex]m[\/latex] units if [latex]m > 1[\/latex] and compresses the graph of [latex]f[\/latex] by a factor of [latex]m[\/latex] units if [latex]0 < m < 1[\/latex]. This means the larger the absolute value of [latex]m[\/latex], the steeper the slope.<\/p>\n<div style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184329\/CNX_Precalc_Figure_02_02_0052.jpg\" alt=\"Graph with several linear functions including y = 3x, y = 2x, y = x, y = (1\/2)x, y = (1\/3)x, y = (-1\/2)x, y = -x, and y = -2x\" width=\"900\" height=\"759\" \/><\/div>\n<div style=\"text-align: center;\"><strong><span style=\"font-size: 10pt;\">Vertical stretches and compressions and reflections on the function [latex]f\\left(x\\right)=x[\/latex].<\/span><\/strong><\/div>\n<h4>Vertical Shift<\/h4>\n<p>In [latex]f\\left(x\\right)=mx+b[\/latex], the [latex]b[\/latex] acts as the <strong>vertical shift<\/strong>, moving the graph up and down without affecting the slope of the line. Notice that adding a value of [latex]b[\/latex] to the equation of [latex]f\\left(x\\right)=x[\/latex] shifts the graph of [latex]f[\/latex] a total of [latex]b[\/latex] units up if [latex]b[\/latex] is positive and [latex]|b|[\/latex] units down if [latex]b[\/latex] is negative.<\/p>\n<div style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184332\/CNX_Precalc_Figure_02_02_0062.jpg\" alt=\"graph showing y = x , y = x+2, y = x+4, y = x-2, y = x-4\" width=\"900\" height=\"759\" \/><\/div>\n<div style=\"text-align: center;\"><strong><span style=\"font-size: 10pt;\">This graph illustrates vertical shifts of the function [latex]f\\left(x\\right)=x[\/latex].<\/span><\/strong><\/div>\n<p>&nbsp;<\/p>\n<p>Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method.<\/p>\n<section class=\"textbox questionHelp\"><strong>How To: Given the equation of a linear function, use transformations to graph the linear function in the form [latex]f\\left(x\\right)=mx+b[\/latex].<\/strong><\/p>\n<ol>\n<li>Graph [latex]f\\left(x\\right)=x[\/latex].<\/li>\n<li>Vertically stretch or compress the graph by a factor [latex]m[\/latex].<\/li>\n<li>Shift the graph up or down [latex]b[\/latex] units.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Graph [latex]f(x)=\\frac{1}{2}x - 3[\/latex] using transformations.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q750947\">Show Solution<\/button> <\/p>\n<div id=\"q750947\" class=\"hidden-answer\" style=\"display: none\">\n<p>The equation for the function shows that [latex]m=\\frac{1}{2}[\/latex] so the identity function is vertically compressed by [latex]\\frac{1}{2}[\/latex]. The equation for the function also shows that [latex]b=-3[\/latex], so the identity function is vertically shifted down [latex]3[\/latex] units.<\/p>\n<p>First, graph the identity function, and show the vertical compression.<\/p>\n<div style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184335\/CNX_Precalc_Figure_02_02_0072.jpg\" alt=\"graph showing the lines y = x and y = (1\/2)x\" width=\"487\" height=\"378\" \/><\/div>\n<div style=\"text-align: center;\"><strong><span style=\"font-size: 10pt;\">The function [latex]y=x[\/latex] compressed by a factor of [latex]\\frac{1}{2}[\/latex].<\/span><\/strong><\/div>\n<p>&nbsp;<\/p>\n<p>Then, show the vertical shift.<\/p>\n<div style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184338\/CNX_Precalc_Figure_02_02_0082.jpg\" alt=\"Graph showing the lines y = (1\/2)x, and y = (1\/2) + 3\" width=\"487\" height=\"377\" \/><\/div>\n<div style=\"text-align: center;\"><strong><span style=\"font-size: 10pt;\">The function [latex]y=\\frac{1}{2}x[\/latex] shifted down 3 units.<\/span><\/strong><\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox questionHelp\"><strong>Q &amp; A<\/strong> <strong>In the example above, could we have sketched the graph by reversing the order of the transformations?<\/strong> <em>No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following order of operations, let the input be 2.<\/em><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}f\\text{(2)}=\\frac{\\text{1}}{\\text{2}}\\text{(2)}-\\text{3}\\hfill \\\\ =\\text{1}-\\text{3}\\hfill \\\\ =-\\text{2}\\hfill \\end{array}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm13637\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=13637&theme=lumen&iframe_resize_id=ohm13637&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm13639\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=13639&theme=lumen&iframe_resize_id=ohm13639&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":26,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Jay Abramson\",\"organization\":\" OpenStax\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/college-algebra\/pages\/1-introduction-to-prerequisites\"}]","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":"original","description":"Revision and Adaptation","author":"","organization":"Lumen Learning","url":"","project":"","license":"cc-by","license_terms":""},{"type":"cc","description":"College Algebra","author":"Jay Abramson","organization":" OpenStax","url":"","project":"","license":"cc-by","license_terms":"Access for free at https:\/\/openstax.org\/books\/college-algebra\/pages\/1-introduction-to-prerequisites"}],"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\/217"}],"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\/217\/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\/217\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/wp\/v2\/media?parent=217"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/pressbooks\/v2\/chapter-type?post=217"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/wp\/v2\/contributor?post=217"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/wp\/v2\/license?post=217"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}