{"id":5146,"date":"2023-06-27T12:07:47","date_gmt":"2023-06-27T12:07:47","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&#038;p=5146"},"modified":"2025-08-28T03:30:13","modified_gmt":"2025-08-28T03:30:13","slug":"math-in-arts-background-youll-need-1","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/math-in-arts-background-youll-need-1\/","title":{"raw":"Math in Arts: Background You'll Need 1","rendered":"Math in Arts: Background You&#8217;ll Need 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Identify how functions slide<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Vertical Shift<\/h2>\r\n<p>One kind of <strong>transformation<\/strong> involves shifting the entire graph of a function up, down, right, or left. The simplest shift is a <strong>vertical shift<\/strong>, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function. In other words, we add the same constant to the output value of the function regardless of the input. For a function [latex]g\\left(x\\right)=f\\left(x\\right)+k[\/latex], the function [latex]f\\left(x\\right)[\/latex] is shifted vertically [latex]k[\/latex] units.<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18203540\/CNX_Precalc_Figure_01_05_0012.jpg\" alt=\"Graph of f of x equals the cubed root of x shifted upward one unit, the resulting graph passes through the point (0,1) instead of (0,0), (1, 2) instead of (1,1) and (-1, 0) instead of (-1, -1)\" width=\"487\" height=\"292\" \/> Figure 1. Vertical shift by [latex]k=1[\/latex] of the cube root function [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex][\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>To help you visualize the concept of a vertical shift, consider that [latex]y=f\\left(x\\right)[\/latex]. Therefore, [latex]f\\left(x\\right)+k[\/latex] is equivalent to [latex]y+k[\/latex]. Every unit of [latex]y[\/latex] is replaced by [latex]y+k[\/latex], so the [latex]y\\text{-}[\/latex] value increases or decreases depending on the value of [latex]k[\/latex]. The result is a shift upward or downward.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>vertical shift<\/h3>\r\n<p>Given a function [latex]f\\left(x\\right)[\/latex], a new function [latex]g\\left(x\\right)=f\\left(x\\right)+k[\/latex], where [latex]k[\/latex] is a constant, is a <strong>vertical shift<\/strong> of the function [latex]f\\left(x\\right)[\/latex]. All the output values change by [latex]k[\/latex] units. If [latex]k[\/latex] is positive, the graph will shift up. If [latex]k[\/latex] is negative, the graph will shift down.<\/p>\r\n<\/div>\r\n<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]8812[\/ohm2_question]<\/section>\r\n<h2>Horizontal Shift<\/h2>\r\n<p>We just saw that the vertical shift is a change to the output, or outside, of the function. We will now look at how changes to input, on the inside of the function, change its graph and meaning. A shift to the input results in a movement of the graph of the function left or right in what is known as a <strong>horizontal shift<\/strong>.<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18203547\/CNX_Precalc_Figure_01_05_0042.jpg\" alt=\"Graph of f of x equals the cubed root of x shifted left one unit, the resulting graph passes through the point (0,-1) instead of (0,0), (0, 1) instead of (1,1) and (-2, -1) instead of (-1, -1)\" width=\"487\" height=\"288\" \/> Figure 2. Horizontal shift of the function [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex]. Note that [latex]h=+1[\/latex] shifts the graph to the left, that is, towards negative values of [latex]x[\/latex][\/caption]\r\n<\/center><center><\/center>\r\n<p>&nbsp;<\/p>\r\n<p>For example, if [latex]f\\left(x\\right)={x}^{2}[\/latex], then [latex]g\\left(x\\right)={\\left(x - 2\\right)}^{2}[\/latex] is a new function. Each input is reduced by 2 prior to squaring the function. The result is that the graph is shifted 2 units to the right, because we would need to increase the prior input by 2 units to yield the same output value as given in [latex]f[\/latex].<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>horizontal shift<\/h3>\r\n<p>Given a function [latex]f[\/latex], a new function [latex]g\\left(x\\right)=f\\left(x-h\\right)[\/latex], where [latex]h[\/latex] is a constant, is a <strong>horizontal shift<\/strong> of the function [latex]f[\/latex]. If [latex]h[\/latex] is positive, the graph will shift right. If [latex]h[\/latex] is negative, the graph will shift left.<\/p>\r\n<\/div>\r\n<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]8813[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Identify how functions slide<\/li>\n<\/ul>\n<\/section>\n<h2>Vertical Shift<\/h2>\n<p>One kind of <strong>transformation<\/strong> involves shifting the entire graph of a function up, down, right, or left. The simplest shift is a <strong>vertical shift<\/strong>, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function. In other words, we add the same constant to the output value of the function regardless of the input. For a function [latex]g\\left(x\\right)=f\\left(x\\right)+k[\/latex], the function [latex]f\\left(x\\right)[\/latex] is shifted vertically [latex]k[\/latex] units.<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18203540\/CNX_Precalc_Figure_01_05_0012.jpg\" alt=\"Graph of f of x equals the cubed root of x shifted upward one unit, the resulting graph passes through the point (0,1) instead of (0,0), (1, 2) instead of (1,1) and (-1, 0) instead of (-1, -1)\" width=\"487\" height=\"292\" \/><figcaption class=\"wp-caption-text\">Figure 1. Vertical shift by [latex]k=1[\/latex] of the cube root function [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex]<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>To help you visualize the concept of a vertical shift, consider that [latex]y=f\\left(x\\right)[\/latex]. Therefore, [latex]f\\left(x\\right)+k[\/latex] is equivalent to [latex]y+k[\/latex]. Every unit of [latex]y[\/latex] is replaced by [latex]y+k[\/latex], so the [latex]y\\text{-}[\/latex] value increases or decreases depending on the value of [latex]k[\/latex]. The result is a shift upward or downward.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>vertical shift<\/h3>\n<p>Given a function [latex]f\\left(x\\right)[\/latex], a new function [latex]g\\left(x\\right)=f\\left(x\\right)+k[\/latex], where [latex]k[\/latex] is a constant, is a <strong>vertical shift<\/strong> of the function [latex]f\\left(x\\right)[\/latex]. All the output values change by [latex]k[\/latex] units. If [latex]k[\/latex] is positive, the graph will shift up. If [latex]k[\/latex] is negative, the graph will shift down.<\/p>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm8812\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=8812&theme=lumen&iframe_resize_id=ohm8812&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h2>Horizontal Shift<\/h2>\n<p>We just saw that the vertical shift is a change to the output, or outside, of the function. We will now look at how changes to input, on the inside of the function, change its graph and meaning. A shift to the input results in a movement of the graph of the function left or right in what is known as a <strong>horizontal shift<\/strong>.<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18203547\/CNX_Precalc_Figure_01_05_0042.jpg\" alt=\"Graph of f of x equals the cubed root of x shifted left one unit, the resulting graph passes through the point (0,-1) instead of (0,0), (0, 1) instead of (1,1) and (-2, -1) instead of (-1, -1)\" width=\"487\" height=\"288\" \/><figcaption class=\"wp-caption-text\">Figure 2. Horizontal shift of the function [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex]. Note that [latex]h=+1[\/latex] shifts the graph to the left, that is, towards negative values of [latex]x[\/latex]<\/figcaption><\/figure>\n<\/div>\n<div style=\"text-align: center;\"><\/div>\n<p>&nbsp;<\/p>\n<p>For example, if [latex]f\\left(x\\right)={x}^{2}[\/latex], then [latex]g\\left(x\\right)={\\left(x - 2\\right)}^{2}[\/latex] is a new function. Each input is reduced by 2 prior to squaring the function. The result is that the graph is shifted 2 units to the right, because we would need to increase the prior input by 2 units to yield the same output value as given in [latex]f[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>horizontal shift<\/h3>\n<p>Given a function [latex]f[\/latex], a new function [latex]g\\left(x\\right)=f\\left(x-h\\right)[\/latex], where [latex]h[\/latex] is a constant, is a <strong>horizontal shift<\/strong> of the function [latex]f[\/latex]. If [latex]h[\/latex] is positive, the graph will shift right. If [latex]h[\/latex] is negative, the graph will shift left.<\/p>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm8813\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=8813&theme=lumen&iframe_resize_id=ohm8813&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":16,"menu_order":2,"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\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download 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":91,"module-header":"background_you_need","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":"Abramson, Jay et al.","organization":"OpenStax","url":"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2","project":"","license":"cc-by","license_terms":"Download 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\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/5146"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/users\/16"}],"version-history":[{"count":17,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/5146\/revisions"}],"predecessor-version":[{"id":15779,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/5146\/revisions\/15779"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/parts\/91"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/5146\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/media?parent=5146"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapter-type?post=5146"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/contributor?post=5146"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/license?post=5146"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}