{"id":653,"date":"2025-07-14T20:07:50","date_gmt":"2025-07-14T20:07:50","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=653"},"modified":"2026-01-29T21:05:35","modified_gmt":"2026-01-29T21:05:35","slug":"transformations-of-functions-learn-it-6","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/transformations-of-functions-learn-it-6\/","title":{"raw":"Transformations of Functions: Learn It 7","rendered":"Transformations of Functions: Learn It 7"},"content":{"raw":"<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>transformed functions<\/h3>\r\nThe formula for a transformed function is [latex]g(x) = \\pm a \\cdot f\\big(\\pm b(x - h)\\big) + k[\/latex] where:\r\n<ul>\r\n \t<li>[latex]\\pm a[\/latex] describes the vertical reflection and stretch\/compression<\/li>\r\n \t<li>[latex]\\pm b[\/latex] describes the horizontal reflection and stretch\/compression<\/li>\r\n \t<li>[latex]h[\/latex] describes the horizontal shift, and<\/li>\r\n \t<li>[latex]k[\/latex] describes the vertical shift<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<ol>\r\n \t<li>Find the parent function. If it's not given to you, check the toolkit functions.<\/li>\r\n \t<li>Identify any shifts.<\/li>\r\n \t<li>Identify any reflections, stretches or compresses.<\/li>\r\n \t<li>Write the function using [latex]g(x) = a \\cdot f\\big(b(x - h)\\big) + k\r\n[\/latex]<\/li>\r\n<\/ol>\r\n<\/section>\r\n<h3>Writing Functions Given the Transformed Graph<\/h3>\r\n<section class=\"textbox example\" aria-label=\"Example\"><img class=\"wp-image-1606 alignright\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/30012528\/Screenshot-2024-05-29-at-6.25.19%E2%80%AFPM.png\" alt=\"\" width=\"309\" height=\"301\" \/>The graph shows two function: The toolkit function [latex]f(x) = x^3[\/latex] (green) and [latex]g(x)[\/latex] (red). Relate this new function [latex]g\\left(x\\right)[\/latex] to [latex]f\\left(x\\right)[\/latex], and then find a formula for [latex]g\\left(x\\right)[\/latex].[reveal-answer q=\"343479\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"343479\"]The red curve [latex]g(x)[\/latex] appears to be less steep compared to the green curve [latex]f(x)[\/latex]. This suggests a vertical compression.If [latex]g(x)[\/latex] is a vertical compression of [latex]f(x)[\/latex], we have: [latex]g(x) = a \\cdot f(x)[\/latex], where [latex]0 &lt; a &lt; 1[\/latex].To determine [latex]a[\/latex], it is helpful to look for a point on the graph that is relatively clear.\r\n<ul>\r\n \t<li>In this graph, it appears that [latex]g\\left(2\\right)=2[\/latex].<\/li>\r\n \t<li>With the basic cubic function at the same input, [latex]f\\left(2\\right)={2}^{3}=8[\/latex].<\/li>\r\n \t<li>Based on that, it appears that the outputs of [latex]g[\/latex] are [latex]\\frac{1}{4}[\/latex] the outputs of the function [latex]f[\/latex] because [latex]2=\\frac{1}{4} \\cdot 8[\/latex].<\/li>\r\n<\/ul>\r\nThus, [latex]g(x) = \\frac{1}{4} f(x) = \\frac{1}{4} x^3[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\"><img class=\"alignright\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18203628\/CNX_Precalc_Figure_01_05_032.jpg\" alt=\"Graph of f(x) being vertically compressed to g(x).\" width=\"487\" height=\"291\" \/>Relate the function [latex]g\\left(x\\right)[\/latex] to [latex]f\\left(x\\right)[\/latex].[reveal-answer q=\"623190\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"623190\"]The orange graph\u00a0 [latex]g(x)[\/latex] appears to be a horizontally compressed version of the blue graph of [latex]f(x)[\/latex].\r\n[latex]\\\\[\/latex]\r\nIf [latex]g(x)[\/latex] is a horizontal compression of [latex]f(x)[\/latex], we have: [latex]g(x) = f(b \\cdot x)[\/latex], where [latex]b &gt; 1[\/latex]. The graph is compressed by [latex]\\dfrac{1}{b}[\/latex].To determine [latex]b[\/latex], it is helpful to look for a point on the graph that is relatively clear.\r\n<ul>\r\n \t<li>In the compressed graph [latex]g(x)[\/latex], the end point is [latex](2, 4)[\/latex].<\/li>\r\n \t<li>The end point of [latex]f(x)[\/latex] is [latex](6,4)[\/latex].<\/li>\r\n \t<li>We can see that the [latex]x[\/latex]-values have been compressed by [latex]\\frac{1}{3}[\/latex], because [latex]2=\\frac{1}{3} \\cdot 6[\/latex].<\/li>\r\n \t<li>This means that [latex]\\dfrac{1}{b} = \\dfrac{1}{3}[\/latex], which means [latex]b = 3[\/latex].<\/li>\r\n<\/ul>\r\nThus, [latex]g(x)=f(3x)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">The graph below represents a transformation of the toolkit function [latex]f\\left(x\\right)={x}^{2}[\/latex]. Relate this new function [latex]g\\left(x\\right)[\/latex] to [latex]f\\left(x\\right)[\/latex], and then find a formula for [latex]g\\left(x\\right)[\/latex].<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18203554\/CNX_Precalc_Figure_01_05_0072.jpg\" alt=\"Graph of a parabola.\" width=\"487\" height=\"328\" \/>\r\n[reveal-answer q=\"937293\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"937293\"]Notice that the graph is identical in shape to the [latex]f\\left(x\\right)={x}^{2}[\/latex] function, but the [latex]x[\/latex]<em>-<\/em>values are shifted to the right 2 units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function shifted 2 units to the right, so\r\n<p style=\"text-align: center;\">[latex]g\\left(x\\right)=f\\left(x - 2\\right)[\/latex]<\/p>\r\nNotice how we must input the value [latex]x=2[\/latex] to get the output value [latex]y=0[\/latex]; the [latex]x[\/latex]-values must be 2 units larger because of the shift to the right by 2 units. We can then use the definition of the [latex]f\\left(x\\right)[\/latex] function to write a formula for [latex]g\\left(x\\right)[\/latex] by evaluating [latex]f\\left(x - 2\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}f\\left(x\\right)={x}^{2}\\hfill \\\\ g\\left(x\\right)=f\\left(x - 2\\right)\\hfill \\\\ g\\left(x\\right)=f\\left(x - 2\\right)={\\left(x - 2\\right)}^{2}\\hfill \\end{cases}[\/latex]<\/p>\r\n<strong>Analysis of the Solution<\/strong>\r\n\r\nTo determine whether the shift is [latex]+2[\/latex] or [latex]-2[\/latex] , consider a single reference point on the graph. For a quadratic, looking at the vertex point is convenient. In the original function, [latex]f\\left(0\\right)=0[\/latex]. In our shifted function, [latex]g\\left(2\\right)=0[\/latex]. To obtain the output value of 0 from the function [latex]f[\/latex], we need to decide whether a plus or a minus sign will work to satisfy [latex]g\\left(2\\right)=f\\left(x - 2\\right)=f\\left(0\\right)=0[\/latex]. For this to work, we will need to <em>subtract<\/em> 2 units from our input values.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Write a formula for the graph shown, which is a transformation of the toolkit square root function.<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010518\/CNX_Precalc_Figure_01_05_0112.jpg\" alt=\"Graph of a square root function transposed right one unit and up 2.\" width=\"487\" height=\"292\" \/>\r\n\r\n[reveal-answer q=\"745825\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"745825\"]\r\n<p id=\"fs-id1165137549429\">The graph of the toolkit function starts at the origin, so this graph has been shifted 1 to the right and up 2. In function notation, we could write that as<\/p>\r\n<p style=\"text-align: center;\">[latex]h\\left(x\\right)=f\\left(x - 1\\right)+2[\/latex]<\/p>\r\n<p id=\"fs-id1165135175226\">Using the formula for the square root function, we can write<\/p>\r\n<p style=\"text-align: center;\">[latex]h\\left(x\\right)=\\sqrt{x - 1}+2[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135500702\">Note that this transformation has changed the domain and range of the function. This new graph has domain [latex]\\left[1,\\infty \\right)[\/latex] and range [latex]\\left[2,\\infty \\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3>Writing Functions Given the Transformed Graph<\/h3>\r\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<p id=\"fs-id1165137643555\">[ohm_question hide_question_numbers=1]317561[\/ohm_question]<\/p>\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]317936[\/ohm_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]317562[\/ohm_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]317563[\/ohm_question]<\/section>","rendered":"<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>transformed functions<\/h3>\n<p>The formula for a transformed function is [latex]g(x) = \\pm a \\cdot f\\big(\\pm b(x - h)\\big) + k[\/latex] where:<\/p>\n<ul>\n<li>[latex]\\pm a[\/latex] describes the vertical reflection and stretch\/compression<\/li>\n<li>[latex]\\pm b[\/latex] describes the horizontal reflection and stretch\/compression<\/li>\n<li>[latex]h[\/latex] describes the horizontal shift, and<\/li>\n<li>[latex]k[\/latex] describes the vertical shift<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<ol>\n<li>Find the parent function. If it&#8217;s not given to you, check the toolkit functions.<\/li>\n<li>Identify any shifts.<\/li>\n<li>Identify any reflections, stretches or compresses.<\/li>\n<li>Write the function using [latex]g(x) = a \\cdot f\\big(b(x - h)\\big) + k[\/latex]<\/li>\n<\/ol>\n<\/section>\n<h3>Writing Functions Given the Transformed Graph<\/h3>\n<section class=\"textbox example\" aria-label=\"Example\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1606 alignright\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/30012528\/Screenshot-2024-05-29-at-6.25.19%E2%80%AFPM.png\" alt=\"\" width=\"309\" height=\"301\" \/>The graph shows two function: The toolkit function [latex]f(x) = x^3[\/latex] (green) and [latex]g(x)[\/latex] (red). Relate this new function [latex]g\\left(x\\right)[\/latex] to [latex]f\\left(x\\right)[\/latex], and then find a formula for [latex]g\\left(x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q343479\">Show Answer<\/button><\/p>\n<div id=\"q343479\" class=\"hidden-answer\" style=\"display: none\">The red curve [latex]g(x)[\/latex] appears to be less steep compared to the green curve [latex]f(x)[\/latex]. This suggests a vertical compression.If [latex]g(x)[\/latex] is a vertical compression of [latex]f(x)[\/latex], we have: [latex]g(x) = a \\cdot f(x)[\/latex], where [latex]0 < a < 1[\/latex].To determine [latex]a[\/latex], it is helpful to look for a point on the graph that is relatively clear.\n\n\n<ul>\n<li>In this graph, it appears that [latex]g\\left(2\\right)=2[\/latex].<\/li>\n<li>With the basic cubic function at the same input, [latex]f\\left(2\\right)={2}^{3}=8[\/latex].<\/li>\n<li>Based on that, it appears that the outputs of [latex]g[\/latex] are [latex]\\frac{1}{4}[\/latex] the outputs of the function [latex]f[\/latex] because [latex]2=\\frac{1}{4} \\cdot 8[\/latex].<\/li>\n<\/ul>\n<p>Thus, [latex]g(x) = \\frac{1}{4} f(x) = \\frac{1}{4} x^3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\"><img loading=\"lazy\" decoding=\"async\" class=\"alignright\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18203628\/CNX_Precalc_Figure_01_05_032.jpg\" alt=\"Graph of f(x) being vertically compressed to g(x).\" width=\"487\" height=\"291\" \/>Relate the function [latex]g\\left(x\\right)[\/latex] to [latex]f\\left(x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q623190\">Show Answer<\/button><\/p>\n<div id=\"q623190\" class=\"hidden-answer\" style=\"display: none\">The orange graph\u00a0 [latex]g(x)[\/latex] appears to be a horizontally compressed version of the blue graph of [latex]f(x)[\/latex].<br \/>\n[latex]\\\\[\/latex]<br \/>\nIf [latex]g(x)[\/latex] is a horizontal compression of [latex]f(x)[\/latex], we have: [latex]g(x) = f(b \\cdot x)[\/latex], where [latex]b > 1[\/latex]. The graph is compressed by [latex]\\dfrac{1}{b}[\/latex].To determine [latex]b[\/latex], it is helpful to look for a point on the graph that is relatively clear.<\/p>\n<ul>\n<li>In the compressed graph [latex]g(x)[\/latex], the end point is [latex](2, 4)[\/latex].<\/li>\n<li>The end point of [latex]f(x)[\/latex] is [latex](6,4)[\/latex].<\/li>\n<li>We can see that the [latex]x[\/latex]-values have been compressed by [latex]\\frac{1}{3}[\/latex], because [latex]2=\\frac{1}{3} \\cdot 6[\/latex].<\/li>\n<li>This means that [latex]\\dfrac{1}{b} = \\dfrac{1}{3}[\/latex], which means [latex]b = 3[\/latex].<\/li>\n<\/ul>\n<p>Thus, [latex]g(x)=f(3x)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">The graph below represents a transformation of the toolkit function [latex]f\\left(x\\right)={x}^{2}[\/latex]. Relate this new function [latex]g\\left(x\\right)[\/latex] to [latex]f\\left(x\\right)[\/latex], and then find a formula for [latex]g\\left(x\\right)[\/latex].<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18203554\/CNX_Precalc_Figure_01_05_0072.jpg\" alt=\"Graph of a parabola.\" width=\"487\" height=\"328\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q937293\">Show Solution<\/button><\/p>\n<div id=\"q937293\" class=\"hidden-answer\" style=\"display: none\">Notice that the graph is identical in shape to the [latex]f\\left(x\\right)={x}^{2}[\/latex] function, but the [latex]x[\/latex]<em>&#8211;<\/em>values are shifted to the right 2 units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function shifted 2 units to the right, so<\/p>\n<p style=\"text-align: center;\">[latex]g\\left(x\\right)=f\\left(x - 2\\right)[\/latex]<\/p>\n<p>Notice how we must input the value [latex]x=2[\/latex] to get the output value [latex]y=0[\/latex]; the [latex]x[\/latex]-values must be 2 units larger because of the shift to the right by 2 units. We can then use the definition of the [latex]f\\left(x\\right)[\/latex] function to write a formula for [latex]g\\left(x\\right)[\/latex] by evaluating [latex]f\\left(x - 2\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}f\\left(x\\right)={x}^{2}\\hfill \\\\ g\\left(x\\right)=f\\left(x - 2\\right)\\hfill \\\\ g\\left(x\\right)=f\\left(x - 2\\right)={\\left(x - 2\\right)}^{2}\\hfill \\end{cases}[\/latex]<\/p>\n<p><strong>Analysis of the Solution<\/strong><\/p>\n<p>To determine whether the shift is [latex]+2[\/latex] or [latex]-2[\/latex] , consider a single reference point on the graph. For a quadratic, looking at the vertex point is convenient. In the original function, [latex]f\\left(0\\right)=0[\/latex]. In our shifted function, [latex]g\\left(2\\right)=0[\/latex]. To obtain the output value of 0 from the function [latex]f[\/latex], we need to decide whether a plus or a minus sign will work to satisfy [latex]g\\left(2\\right)=f\\left(x - 2\\right)=f\\left(0\\right)=0[\/latex]. For this to work, we will need to <em>subtract<\/em> 2 units from our input values.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Write a formula for the graph shown, which is a transformation of the toolkit square root function.<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010518\/CNX_Precalc_Figure_01_05_0112.jpg\" alt=\"Graph of a square root function transposed right one unit and up 2.\" width=\"487\" height=\"292\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q745825\">Show Solution<\/button><\/p>\n<div id=\"q745825\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137549429\">The graph of the toolkit function starts at the origin, so this graph has been shifted 1 to the right and up 2. In function notation, we could write that as<\/p>\n<p style=\"text-align: center;\">[latex]h\\left(x\\right)=f\\left(x - 1\\right)+2[\/latex]<\/p>\n<p id=\"fs-id1165135175226\">Using the formula for the square root function, we can write<\/p>\n<p style=\"text-align: center;\">[latex]h\\left(x\\right)=\\sqrt{x - 1}+2[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135500702\">Note that this transformation has changed the domain and range of the function. This new graph has domain [latex]\\left[1,\\infty \\right)[\/latex] and range [latex]\\left[2,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h3>Writing Functions Given the Transformed Graph<\/h3>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<p id=\"fs-id1165137643555\"><iframe loading=\"lazy\" id=\"ohm317561\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=317561&theme=lumen&iframe_resize_id=ohm317561&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm317936\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=317936&theme=lumen&iframe_resize_id=ohm317936&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm317562\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=317562&theme=lumen&iframe_resize_id=ohm317562&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm317563\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=317563&theme=lumen&iframe_resize_id=ohm317563&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":13,"menu_order":19,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":498,"module-header":"learn_it","content_attributions":[],"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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/653"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":14,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/653\/revisions"}],"predecessor-version":[{"id":5447,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/653\/revisions\/5447"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/498"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/653\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=653"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=653"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=653"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=653"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}