{"id":2069,"date":"2022-05-16T15:01:26","date_gmt":"2022-05-16T15:01:26","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=2069"},"modified":"2026-03-06T06:30:02","modified_gmt":"2026-03-06T06:30:02","slug":"4-5-the-inverse-of-a-quadratic-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/4-5-the-inverse-of-a-quadratic-function\/","title":{"raw":"4.5: The Inverse of a Quadratic Function","rendered":"4.5: The Inverse of a Quadratic Function"},"content":{"raw":"<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>\r\n<div class=\"textbox learning-objectives\">\r\n<h1>Learning Objectives<\/h1>\r\n<ul>\r\n \t<li>Graph the inverse of a quadratic function<\/li>\r\n \t<li>Explain the two inverse functions of a quadratic function by restricting the domain<\/li>\r\n \t<li>Find the inverse functions of a quadratic function by restricting the domain<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn chapter 3, we discussed that every function has an inverse, but <strong>only a one-to-one function has an inverse function<\/strong>. Since a quadratic function is not a one-to-one mapping (it is a many-to-one mapping), its inverse is not a function. In other words, the inverse of a many-to-one mapping is a one-to-many mapping, and a one-to-many mapping is not a function.\r\n<h2>Graphing the Inverse of a Quadratic Function<\/h2>\r\nWe can use a table of values to graph the inverse of a quadratic function by switching the [latex]x[\/latex]- and [latex]y[\/latex]-values used in a table of values for the original function. For example, given the function [latex]f(x)=(x+1)^2+5[\/latex], we can create a table of values to graph the function (table 1).\r\n<table class=\"aligncenter\" style=\"border-collapse: collapse; width: 40%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th class=\"shaded\" style=\"width: 2.40964%; text-align: center;\">[latex]x[\/latex]<\/th>\r\n<th class=\"shaded\" style=\"width: 2.40964%; text-align: center;\">[latex]y[\/latex]<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">-4<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">14<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">-3<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">-2<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">-1<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">0<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">1<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">2<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">14<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: left;\" colspan=\"2\">Table 1. Table of values for\u00a0[latex]f(x)=(x+1)^2+5[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: left;\">The inverse of the function is found by switching the values of the [latex]x[\/latex] and [latex]y[\/latex] columns so that the inputs are the values of [latex]y[\/latex] and the outputs are the values of [latex]x[\/latex]. Table 2 shows the table of values for the inverse after the values in the [latex]x[\/latex]- and [latex]y[\/latex]- columns are switched.<\/p>\r\n\r\n<table class=\"aligncenter\" style=\"border-collapse: collapse; width: 40%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th class=\"shaded\" style=\"width: 1.7301%; text-align: center;\">[latex]x[\/latex]<\/th>\r\n<th class=\"shaded\" style=\"width: 1.7301%; text-align: center;\">[latex]y[\/latex]<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 1.7301%; text-align: center;\">14<\/td>\r\n<td style=\"width: 1.7301%; text-align: center;\">-4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 1.7301%; text-align: center;\">9<\/td>\r\n<td style=\"width: 1.7301%; text-align: center;\">-3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 1.7301%; text-align: center;\">6<\/td>\r\n<td style=\"width: 1.7301%; text-align: center;\">-2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 1.7301%; text-align: center;\">5<\/td>\r\n<td style=\"width: 1.7301%; text-align: center;\">-1<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 1.7301%; text-align: center;\">6<\/td>\r\n<td style=\"width: 1.7301%; text-align: center;\">0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 1.7301%; text-align: center;\">9<\/td>\r\n<td style=\"width: 1.7301%; text-align: center;\">1<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 1.7301%; text-align: center;\">14<\/td>\r\n<td style=\"width: 1.7301%; text-align: center;\">2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 3.4602%;\" colspan=\"2\">Table 2. Table of values for the inverse function<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe can use the values in table 1 to graph the original function (in blue) and the values from table 2 to graph the inverse (in green).\u00a0 Notice that the inverse graph is a reflection of the graph of the original function across the line [latex]y=x[\/latex] (in red).\r\n\r\n[caption id=\"attachment_2098\" align=\"aligncenter\" width=\"372\"]<img class=\"wp-image-2098\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/05\/16183235\/4-5-InverseFunction-300x300.png\" alt=\"Inverses of parabolas\" width=\"372\" height=\"372\" \/> Figure 1. The graph of the inverse of a quadratic function.[\/caption]\r\n\r\nFigure 1 shows that the inverse of a quadratic function is also a parabola. However, this parabola is horizontal and opens to the right. Since this horizontal parabola fails the vertical line test, <strong>it is not a function<\/strong>.\r\n<h2 style=\"text-align: left;\">Restricting the Domain<\/h2>\r\n<p style=\"text-align: left;\">The graph of a quadratic function is symmetric with a vertical line of symmetry that passes through the vertex of the function. This line of symmetry splits the function into two curves; the curve to the left\u00a0<span style=\"font-size: 1em;\">of the line of symmetry (or the vertex)\u00a0<\/span><span style=\"font-size: 1rem;\">and the curve to the right. Each of the two curves is one-to-one so the inverse of each curve is a function. Consequently, by restricting the domain to either [latex]x\u2264h[\/latex] or [latex]x\u2265h[\/latex], where [latex]x=h[\/latex] is the vertical line of symmetry of the parabola, we create two half-parabolas that are one-to-one and whose inverses are one-to-one functions that are also half-parabolas (figures 2 and 3).<\/span><\/p>\r\n\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 50%; vertical-align: top;\" colspan=\"2\">Inverses of the graphs of half-parabolas<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; vertical-align: top;\">&nbsp;\r\n\r\n<img class=\"aligncenter size-medium wp-image-2728\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/05\/16224124\/desmos-graph-2022-06-16T164104.006-300x300.png\" alt=\"function and its inverse\" width=\"300\" height=\"300\" \/><\/td>\r\n<td style=\"width: 50%;\">&nbsp;\r\n\r\n<img class=\"aligncenter size-medium wp-image-2727\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/05\/16223902\/desmos-graph-2022-06-16T163836.661-300x300.png\" alt=\"inverse functiom\" width=\"300\" height=\"300\" \/>\r\n\r\n&nbsp;<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"shaded\" style=\"width: 50%;\">Figure 2. Parabola and its inverse with restricted domain [latex]x\u2265h[\/latex]<\/td>\r\n<td class=\"shaded\" style=\"width: 50%;\">Figure 3. Parabola and its inverse with restricted domain [latex]x\u2264h[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: left;\">Restricting the domain of a quadratic function [latex]f(x) = (x - h)^2 + k[\/latex] to [latex]x\u2265h[\/latex] or [latex]x\u2264h[\/latex] splits the parabola into two parts. The two separate curves are one-to-one and therefore have inverses that are one-to-one functions.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nRestrict the domain of the function to create a one-to-one function.\r\n<ol>\r\n \t<li>[latex]f(x)=4(x-7)^2+1[\/latex]<\/li>\r\n \t<li>[latex]g(x)=-x^2+4[\/latex]<\/li>\r\n \t<li>[latex]h(x)=-\\dfrac{5}{4}\\left(x+3\\right)^2[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\nWe restrict the domain before the vertex or after the vertex to create a one-to-one function.\r\n<ol>\r\n \t<li>The vertex is (7, 1) so we the restricted domain is either [latex]\\{x\\;|\\;x\u22657\\}[\/latex] or\u00a0[latex]\\{x\\;|\\;x\u22647\\}[\/latex].<\/li>\r\n \t<li>The vertex is (0, 4) so we the restricted domain is either [latex]\\{x\\;|\\;x\u22650\\}[\/latex] or\u00a0[latex]\\{x\\;|\\;x\u22640\\}[\/latex].<\/li>\r\n \t<li>The vertex is (\u20133, 0) so we the restricted domain is either [latex]\\{x\\;|\\;x\u2265\u20133\\}[\/latex] or\u00a0[latex]\\{x\\;|\\;x\u2264\u20133\\}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nRestrict the domain of the function to create a one-to-one function.\r\n<ol>\r\n \t<li>[latex]f(x)=-6(x-1)^2+9[\/latex]<\/li>\r\n \t<li>[latex]g(x)=-2x^2-5[\/latex]<\/li>\r\n \t<li>[latex]h(x)=-\\dfrac{5}{4}\\left(x+4\\right)^2[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"hjm819\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm819\"]\r\n<ol>\r\n \t<li>The vertex is (1, 9) so we the restricted domain is either [latex]\\{x\\;|\\;x\u22651\\}[\/latex] or\u00a0[latex]\\{x\\;|\\;x\u22641\\}[\/latex].<\/li>\r\n \t<li>The vertex is (0, \u20135) so we the restricted domain is either [latex]\\{x\\;|\\;x\u22650\\}[\/latex] or\u00a0[latex]\\{x\\;|\\;x\u22640\\}[\/latex].<\/li>\r\n \t<li>The vertex is (\u20134, 0) so we the restricted domain is either [latex]\\{x\\;|\\;x\u2265-4\\}[\/latex] or\u00a0[latex]\\{x\\;|\\;x\u2264-4\\}[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Inverse Function<\/h2>\r\n<p style=\"text-align: left;\">To find the inverse function of any quadratic function in the form [latex]f(x) = (x - h)^2 + k[\/latex], we must first <em><strong>restrict the domain<\/strong><\/em> of the function to make a one-to-one function. We can then switch\u00a0[latex]x[\/latex] and\u00a0[latex]y[\/latex] to get the inverse function. To write this new function using inverse function notation, we solve for [latex]y[\/latex] and replace [latex]y[\/latex] with [latex]f^{-1}(x)[\/latex].<\/p>\r\n\r\n<div style=\"font-weight: 400; text-align: center;\">\r\n<div>\r\n\r\nFor example, to find the inverse function of [latex]f(x) = (x \u2013 3)^2 + 7[\/latex] whose graph has a vertex at (3, 7), we start by restricting the domain to [latex]x \\geq 3[\/latex] (figure 4). As soon as we switch\u00a0[latex]x[\/latex] and\u00a0[latex]y[\/latex] we have the inverse:\r\n<p style=\"text-align: center;\">[latex]x=(y-3)^2+7[\/latex]<\/p>\r\n\r\n[caption id=\"attachment_2365\" align=\"aligncenter\" width=\"300\"]<img class=\"wp-image-2365 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/05\/26222206\/desmos-graph-2022-05-26T162103.483-300x300.png\" alt=\"half parabola and its inverse\" width=\"300\" height=\"300\" \/> Figure 4. Restricting the domain of [latex]f(x)=(x-3)^{2}+7[\/latex] to find its inverse function[\/caption]<\/div>\r\n<div><\/div>\r\n<div style=\"text-align: left;\">\r\n\r\nWhen we switch\u00a0[latex]x[\/latex] and\u00a0[latex]y[\/latex], the domain of [latex]f(x)[\/latex],\u00a0[latex]\\{x\\;|\\;x\u22653\\}[\/latex], becomes the range of [latex]f^{-1}(x)[\/latex] and the range of\u00a0[latex]f(x)[\/latex],\u00a0[latex]\\{f(x)\\;|\\;f(x)\u22657\\}[\/latex], becomes the domain of [latex]f^{-1}(x)[\/latex]. This means that the domain of\u00a0[latex]f^{-1}(x)[\/latex] is [latex]\\{x\\;|\\;x\u22657\\}[\/latex] and its range is [latex]\\{y\\;|\\;y\u22653\\}[\/latex] (figure 4).\r\n<p style=\"text-align: left;\">Now we solve for\u00a0[latex]y[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&amp;=(y-3)^2+7\\\\ \\\\x - 7 &amp;= (y - 3)^2\\\\\\\\\\sqrt{x-7}&amp;=\\sqrt{(y-3)^2}\\\\ \\\\\\sqrt{x-7}&amp;=|y-3|\\\\\\\\\\sqrt{x-7}&amp;=y-3\\;\\;\\;\\;\\;\\text{Since }y\u22653,\\;|y-3|=y-3\\\\\\\\\\sqrt{x-7}+3&amp;=y\\end{aligned}[\/latex]<\/p>\r\n\r\n<\/div>\r\nTherefore, the inverse function of [latex]f(x)=(x-3)^{2}+7[\/latex] is\u00a0[latex]f^{-1}(x) = \\sqrt{x - 7} + 3[\/latex].\r\n\r\nNote that we could have restricted the domain of\u00a0[latex]f(x) = (x \u2013 3)^2 + 7[\/latex] to\u00a0[latex]\\{x\\;|\\;x\u22643\\}[\/latex] to create a one-to-one function. Figure 5 shows this restricted domain function and its inverse. Compare it to figure 4.\r\n\r\n[caption id=\"attachment_2524\" align=\"aligncenter\" width=\"300\"]<img class=\"wp-image-2524 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/05\/06162431\/desmos-graph-2022-06-06T102414.877-300x300.png\" alt=\"Halfparabola as a one to one function and its inverse\" width=\"300\" height=\"300\" \/> Figure 5.\u00a0Restricting the domain of [latex]f(x)=(x-3)^{2}+7[\/latex] to find its inverse function[\/caption]\r\n<div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nFind the inverse function of the function [latex]s(x) = (x + 2)^2 - 5[\/latex] given [latex]x \u2265 \u20132[\/latex].\r\n<h4>Solution<\/h4>\r\nWe start by writing [latex]y[\/latex] for [latex]s(x)[\/latex]\r\n<p style=\"text-align: center;\">[latex]s(x) = (x + 2)^2 - 5[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]y = (x + 2)^2 - 5[\/latex]<\/p>\r\nNow, exchange the [latex]x[\/latex] and [latex]y[\/latex]. \u00a0This also switches the domain and the range, so now [latex]y\u2265-2[\/latex].\r\n<div>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&amp;=(y+2)^2 - 5\\\\\\\\x+5 &amp;= (y + 2)^2\\\\\\\\\\sqrt{x + 5}&amp;=\\sqrt{(y+2)^2}\\end{aligned}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nSince [latex]y \u2265-2[\/latex], [latex]y + 2\u22650[\/latex] and [latex]\\sqrt{(y+2)^2}=|y+2|=y+2[\/latex]:\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\sqrt{x + 5}&amp;=y+2\\\\\\\\\\sqrt{x+5}-2&amp;=y\\end{aligned}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nTherefore, the inverse function of [latex]s(x)=(x + 2)^2 - 5[\/latex] is [latex]s^{-1}(x) = \\sqrt{x + 5} - 2[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 2<\/h3>\r\nFind the inverse function of the function [latex]s(x) = (x + 2)^2 - 5[\/latex] given [latex]x &lt; -2[\/latex].\r\n\r\n[reveal-answer q=\"hjm632\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm632\"]\r\n\r\n[latex]s^{-1}(x)=-\\sqrt{x + 5} - 2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 3<\/h3>\r\nFind the inverse function of the function [latex]g(x) = 3(x - 4)^2 + 1[\/latex] given [latex]x &lt; 4[\/latex].\r\n<h4>Solution<\/h4>\r\nWe start by writing [latex]y[\/latex] for [latex]g(x)[\/latex] then switching [latex]x[\/latex] and\u00a0[latex]y[\/latex]. This also switches the domain and the range, so now [latex]y\u22644[\/latex].\r\n<div>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}y&amp;= 3(x -4)^2 +1\\\\\\\\x&amp;=3(y-4)^2+1\\\\\\\\x-1 &amp;= 3(y -4)^2\\\\\\\\\\dfrac{x-1}{3}&amp;=(y-4)^2\\\\\\\\\\sqrt{\\dfrac{x-1}{3}}&amp;=\\sqrt{(y-4)^2}\\end{aligned}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nSince [latex]y\u22644[\/latex], [latex]y-4\u22640[\/latex] and [latex]\\sqrt{(y-4)^2}=|y-4|=-(y-4)=-y+4[\/latex]:\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\sqrt{\\dfrac{x-1}{3}}&amp;=\\sqrt{(y-4)^2}\\\\\\\\\\sqrt{\\dfrac{x-1}{3}}&amp;=-y+4\\\\\\\\\\sqrt{\\dfrac{x-1}{3}}-4&amp;=-y\\\\\\\\-\\sqrt{\\dfrac{x-1}{3}}+4&amp;=y\\end{aligned}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nTherefore, the inverse function of [latex]g(x)=3(x -4)^2 +1[\/latex] is\u00a0[latex]g^{-1}(x) =-\\sqrt{\\dfrac{x-1}{3}}+4[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nFind the inverse function of the function [latex]g(x) = 3(x - 4)^2 + 1[\/latex] given [latex]x \u2265 4[\/latex].\r\n\r\n[reveal-answer q=\"hjm596\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm596\"]\r\n\r\n[latex]g^{-1}(x) =\\sqrt{\\dfrac{x-1}{3}}+4[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<script>\r\nwindow.embeddedChatbotConfig = {\r\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\r\ndomain: \"www.chatbase.co\"\r\n}\r\n<\/script>\r\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" chatbotId=\"ejVb5sgc1-w972OOCgl5x\" domain=\"www.chatbase.co\" defer>\r\n<\/script>\r\n\r\n<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>","rendered":"<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n<div class=\"textbox learning-objectives\">\n<h1>Learning Objectives<\/h1>\n<ul>\n<li>Graph the inverse of a quadratic function<\/li>\n<li>Explain the two inverse functions of a quadratic function by restricting the domain<\/li>\n<li>Find the inverse functions of a quadratic function by restricting the domain<\/li>\n<\/ul>\n<\/div>\n<p>In chapter 3, we discussed that every function has an inverse, but <strong>only a one-to-one function has an inverse function<\/strong>. Since a quadratic function is not a one-to-one mapping (it is a many-to-one mapping), its inverse is not a function. In other words, the inverse of a many-to-one mapping is a one-to-many mapping, and a one-to-many mapping is not a function.<\/p>\n<h2>Graphing the Inverse of a Quadratic Function<\/h2>\n<p>We can use a table of values to graph the inverse of a quadratic function by switching the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-values used in a table of values for the original function. For example, given the function [latex]f(x)=(x+1)^2+5[\/latex], we can create a table of values to graph the function (table 1).<\/p>\n<table class=\"aligncenter\" style=\"border-collapse: collapse; width: 40%;\">\n<tbody>\n<tr>\n<th class=\"shaded\" style=\"width: 2.40964%; text-align: center;\">[latex]x[\/latex]<\/th>\n<th class=\"shaded\" style=\"width: 2.40964%; text-align: center;\">[latex]y[\/latex]<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">-4<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">14<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">-3<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">9<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">-2<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">6<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">-1<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">5<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">0<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">6<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">1<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">9<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">2<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">14<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: left;\" colspan=\"2\">Table 1. Table of values for\u00a0[latex]f(x)=(x+1)^2+5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: left;\">The inverse of the function is found by switching the values of the [latex]x[\/latex] and [latex]y[\/latex] columns so that the inputs are the values of [latex]y[\/latex] and the outputs are the values of [latex]x[\/latex]. Table 2 shows the table of values for the inverse after the values in the [latex]x[\/latex]&#8211; and [latex]y[\/latex]&#8211; columns are switched.<\/p>\n<table class=\"aligncenter\" style=\"border-collapse: collapse; width: 40%;\">\n<tbody>\n<tr>\n<th class=\"shaded\" style=\"width: 1.7301%; text-align: center;\">[latex]x[\/latex]<\/th>\n<th class=\"shaded\" style=\"width: 1.7301%; text-align: center;\">[latex]y[\/latex]<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 1.7301%; text-align: center;\">14<\/td>\n<td style=\"width: 1.7301%; text-align: center;\">-4<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 1.7301%; text-align: center;\">9<\/td>\n<td style=\"width: 1.7301%; text-align: center;\">-3<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 1.7301%; text-align: center;\">6<\/td>\n<td style=\"width: 1.7301%; text-align: center;\">-2<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 1.7301%; text-align: center;\">5<\/td>\n<td style=\"width: 1.7301%; text-align: center;\">-1<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 1.7301%; text-align: center;\">6<\/td>\n<td style=\"width: 1.7301%; text-align: center;\">0<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 1.7301%; text-align: center;\">9<\/td>\n<td style=\"width: 1.7301%; text-align: center;\">1<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 1.7301%; text-align: center;\">14<\/td>\n<td style=\"width: 1.7301%; text-align: center;\">2<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 3.4602%;\" colspan=\"2\">Table 2. Table of values for the inverse function<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We can use the values in table 1 to graph the original function (in blue) and the values from table 2 to graph the inverse (in green).\u00a0 Notice that the inverse graph is a reflection of the graph of the original function across the line [latex]y=x[\/latex] (in red).<\/p>\n<div id=\"attachment_2098\" style=\"width: 382px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2098\" class=\"wp-image-2098\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/05\/16183235\/4-5-InverseFunction-300x300.png\" alt=\"Inverses of parabolas\" width=\"372\" height=\"372\" \/><\/p>\n<p id=\"caption-attachment-2098\" class=\"wp-caption-text\">Figure 1. The graph of the inverse of a quadratic function.<\/p>\n<\/div>\n<p>Figure 1 shows that the inverse of a quadratic function is also a parabola. However, this parabola is horizontal and opens to the right. Since this horizontal parabola fails the vertical line test, <strong>it is not a function<\/strong>.<\/p>\n<h2 style=\"text-align: left;\">Restricting the Domain<\/h2>\n<p style=\"text-align: left;\">The graph of a quadratic function is symmetric with a vertical line of symmetry that passes through the vertex of the function. This line of symmetry splits the function into two curves; the curve to the left\u00a0<span style=\"font-size: 1em;\">of the line of symmetry (or the vertex)\u00a0<\/span><span style=\"font-size: 1rem;\">and the curve to the right. Each of the two curves is one-to-one so the inverse of each curve is a function. Consequently, by restricting the domain to either [latex]x\u2264h[\/latex] or [latex]x\u2265h[\/latex], where [latex]x=h[\/latex] is the vertical line of symmetry of the parabola, we create two half-parabolas that are one-to-one and whose inverses are one-to-one functions that are also half-parabolas (figures 2 and 3).<\/span><\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 50%; vertical-align: top;\" colspan=\"2\">Inverses of the graphs of half-parabolas<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%; vertical-align: top;\">&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-2728\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/05\/16224124\/desmos-graph-2022-06-16T164104.006-300x300.png\" alt=\"function and its inverse\" width=\"300\" height=\"300\" \/><\/td>\n<td style=\"width: 50%;\">&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-2727\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/05\/16223902\/desmos-graph-2022-06-16T163836.661-300x300.png\" alt=\"inverse functiom\" width=\"300\" height=\"300\" \/><\/p>\n<p>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td class=\"shaded\" style=\"width: 50%;\">Figure 2. Parabola and its inverse with restricted domain [latex]x\u2265h[\/latex]<\/td>\n<td class=\"shaded\" style=\"width: 50%;\">Figure 3. Parabola and its inverse with restricted domain [latex]x\u2264h[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: left;\">Restricting the domain of a quadratic function [latex]f(x) = (x - h)^2 + k[\/latex] to [latex]x\u2265h[\/latex] or [latex]x\u2264h[\/latex] splits the parabola into two parts. The two separate curves are one-to-one and therefore have inverses that are one-to-one functions.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>Restrict the domain of the function to create a one-to-one function.<\/p>\n<ol>\n<li>[latex]f(x)=4(x-7)^2+1[\/latex]<\/li>\n<li>[latex]g(x)=-x^2+4[\/latex]<\/li>\n<li>[latex]h(x)=-\\dfrac{5}{4}\\left(x+3\\right)^2[\/latex]<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<p>We restrict the domain before the vertex or after the vertex to create a one-to-one function.<\/p>\n<ol>\n<li>The vertex is (7, 1) so we the restricted domain is either [latex]\\{x\\;|\\;x\u22657\\}[\/latex] or\u00a0[latex]\\{x\\;|\\;x\u22647\\}[\/latex].<\/li>\n<li>The vertex is (0, 4) so we the restricted domain is either [latex]\\{x\\;|\\;x\u22650\\}[\/latex] or\u00a0[latex]\\{x\\;|\\;x\u22640\\}[\/latex].<\/li>\n<li>The vertex is (\u20133, 0) so we the restricted domain is either [latex]\\{x\\;|\\;x\u2265\u20133\\}[\/latex] or\u00a0[latex]\\{x\\;|\\;x\u2264\u20133\\}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>Restrict the domain of the function to create a one-to-one function.<\/p>\n<ol>\n<li>[latex]f(x)=-6(x-1)^2+9[\/latex]<\/li>\n<li>[latex]g(x)=-2x^2-5[\/latex]<\/li>\n<li>[latex]h(x)=-\\dfrac{5}{4}\\left(x+4\\right)^2[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm819\">Show Answer<\/span><\/p>\n<div id=\"qhjm819\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>The vertex is (1, 9) so we the restricted domain is either [latex]\\{x\\;|\\;x\u22651\\}[\/latex] or\u00a0[latex]\\{x\\;|\\;x\u22641\\}[\/latex].<\/li>\n<li>The vertex is (0, \u20135) so we the restricted domain is either [latex]\\{x\\;|\\;x\u22650\\}[\/latex] or\u00a0[latex]\\{x\\;|\\;x\u22640\\}[\/latex].<\/li>\n<li>The vertex is (\u20134, 0) so we the restricted domain is either [latex]\\{x\\;|\\;x\u2265-4\\}[\/latex] or\u00a0[latex]\\{x\\;|\\;x\u2264-4\\}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>The Inverse Function<\/h2>\n<p style=\"text-align: left;\">To find the inverse function of any quadratic function in the form [latex]f(x) = (x - h)^2 + k[\/latex], we must first <em><strong>restrict the domain<\/strong><\/em> of the function to make a one-to-one function. We can then switch\u00a0[latex]x[\/latex] and\u00a0[latex]y[\/latex] to get the inverse function. To write this new function using inverse function notation, we solve for [latex]y[\/latex] and replace [latex]y[\/latex] with [latex]f^{-1}(x)[\/latex].<\/p>\n<div style=\"font-weight: 400; text-align: center;\">\n<div>\n<p>For example, to find the inverse function of [latex]f(x) = (x \u2013 3)^2 + 7[\/latex] whose graph has a vertex at (3, 7), we start by restricting the domain to [latex]x \\geq 3[\/latex] (figure 4). As soon as we switch\u00a0[latex]x[\/latex] and\u00a0[latex]y[\/latex] we have the inverse:<\/p>\n<p style=\"text-align: center;\">[latex]x=(y-3)^2+7[\/latex]<\/p>\n<div id=\"attachment_2365\" style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2365\" class=\"wp-image-2365 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/05\/26222206\/desmos-graph-2022-05-26T162103.483-300x300.png\" alt=\"half parabola and its inverse\" width=\"300\" height=\"300\" \/><\/p>\n<p id=\"caption-attachment-2365\" class=\"wp-caption-text\">Figure 4. Restricting the domain of [latex]f(x)=(x-3)^{2}+7[\/latex] to find its inverse function<\/p>\n<\/div>\n<\/div>\n<div><\/div>\n<div style=\"text-align: left;\">\n<p>When we switch\u00a0[latex]x[\/latex] and\u00a0[latex]y[\/latex], the domain of [latex]f(x)[\/latex],\u00a0[latex]\\{x\\;|\\;x\u22653\\}[\/latex], becomes the range of [latex]f^{-1}(x)[\/latex] and the range of\u00a0[latex]f(x)[\/latex],\u00a0[latex]\\{f(x)\\;|\\;f(x)\u22657\\}[\/latex], becomes the domain of [latex]f^{-1}(x)[\/latex]. This means that the domain of\u00a0[latex]f^{-1}(x)[\/latex] is [latex]\\{x\\;|\\;x\u22657\\}[\/latex] and its range is [latex]\\{y\\;|\\;y\u22653\\}[\/latex] (figure 4).<\/p>\n<p style=\"text-align: left;\">Now we solve for\u00a0[latex]y[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&=(y-3)^2+7\\\\ \\\\x - 7 &= (y - 3)^2\\\\\\\\\\sqrt{x-7}&=\\sqrt{(y-3)^2}\\\\ \\\\\\sqrt{x-7}&=|y-3|\\\\\\\\\\sqrt{x-7}&=y-3\\;\\;\\;\\;\\;\\text{Since }y\u22653,\\;|y-3|=y-3\\\\\\\\\\sqrt{x-7}+3&=y\\end{aligned}[\/latex]<\/p>\n<\/div>\n<p>Therefore, the inverse function of [latex]f(x)=(x-3)^{2}+7[\/latex] is\u00a0[latex]f^{-1}(x) = \\sqrt{x - 7} + 3[\/latex].<\/p>\n<p>Note that we could have restricted the domain of\u00a0[latex]f(x) = (x \u2013 3)^2 + 7[\/latex] to\u00a0[latex]\\{x\\;|\\;x\u22643\\}[\/latex] to create a one-to-one function. Figure 5 shows this restricted domain function and its inverse. Compare it to figure 4.<\/p>\n<div id=\"attachment_2524\" style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2524\" class=\"wp-image-2524 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/05\/06162431\/desmos-graph-2022-06-06T102414.877-300x300.png\" alt=\"Halfparabola as a one to one function and its inverse\" width=\"300\" height=\"300\" \/><\/p>\n<p id=\"caption-attachment-2524\" class=\"wp-caption-text\">Figure 5.\u00a0Restricting the domain of [latex]f(x)=(x-3)^{2}+7[\/latex] to find its inverse function<\/p>\n<\/div>\n<div>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>Find the inverse function of the function [latex]s(x) = (x + 2)^2 - 5[\/latex] given [latex]x \u2265 \u20132[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>We start by writing [latex]y[\/latex] for [latex]s(x)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]s(x) = (x + 2)^2 - 5[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]y = (x + 2)^2 - 5[\/latex]<\/p>\n<p>Now, exchange the [latex]x[\/latex] and [latex]y[\/latex]. \u00a0This also switches the domain and the range, so now [latex]y\u2265-2[\/latex].<\/p>\n<div>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&=(y+2)^2 - 5\\\\\\\\x+5 &= (y + 2)^2\\\\\\\\\\sqrt{x + 5}&=\\sqrt{(y+2)^2}\\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Since [latex]y \u2265-2[\/latex], [latex]y + 2\u22650[\/latex] and [latex]\\sqrt{(y+2)^2}=|y+2|=y+2[\/latex]:<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\sqrt{x + 5}&=y+2\\\\\\\\\\sqrt{x+5}-2&=y\\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Therefore, the inverse function of [latex]s(x)=(x + 2)^2 - 5[\/latex] is [latex]s^{-1}(x) = \\sqrt{x + 5} - 2[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 2<\/h3>\n<p>Find the inverse function of the function [latex]s(x) = (x + 2)^2 - 5[\/latex] given [latex]x < -2[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm632\">Show Answer<\/span><\/p>\n<div id=\"qhjm632\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]s^{-1}(x)=-\\sqrt{x + 5} - 2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 3<\/h3>\n<p>Find the inverse function of the function [latex]g(x) = 3(x - 4)^2 + 1[\/latex] given [latex]x < 4[\/latex].\n\n\n<h4>Solution<\/h4>\n<p>We start by writing [latex]y[\/latex] for [latex]g(x)[\/latex] then switching [latex]x[\/latex] and\u00a0[latex]y[\/latex]. This also switches the domain and the range, so now [latex]y\u22644[\/latex].<\/p>\n<div>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}y&= 3(x -4)^2 +1\\\\\\\\x&=3(y-4)^2+1\\\\\\\\x-1 &= 3(y -4)^2\\\\\\\\\\dfrac{x-1}{3}&=(y-4)^2\\\\\\\\\\sqrt{\\dfrac{x-1}{3}}&=\\sqrt{(y-4)^2}\\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Since [latex]y\u22644[\/latex], [latex]y-4\u22640[\/latex] and [latex]\\sqrt{(y-4)^2}=|y-4|=-(y-4)=-y+4[\/latex]:<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\sqrt{\\dfrac{x-1}{3}}&=\\sqrt{(y-4)^2}\\\\\\\\\\sqrt{\\dfrac{x-1}{3}}&=-y+4\\\\\\\\\\sqrt{\\dfrac{x-1}{3}}-4&=-y\\\\\\\\-\\sqrt{\\dfrac{x-1}{3}}+4&=y\\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Therefore, the inverse function of [latex]g(x)=3(x -4)^2 +1[\/latex] is\u00a0[latex]g^{-1}(x) =-\\sqrt{\\dfrac{x-1}{3}}+4[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>Find the inverse function of the function [latex]g(x) = 3(x - 4)^2 + 1[\/latex] given [latex]x \u2265 4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm596\">Show Answer<\/span><\/p>\n<div id=\"qhjm596\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]g^{-1}(x) =\\sqrt{\\dfrac{x-1}{3}}+4[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<p><script>\nwindow.embeddedChatbotConfig = {\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\ndomain: \"www.chatbase.co\"\n}\n<\/script><br \/>\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" defer=\"defer\">\n<\/script><\/p>\n<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/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-2069\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>The Inverse of a Quadratic Function. <strong>Authored by<\/strong>: Leo Chang and Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>All graphs created using desmos graphing calculator. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/desmos.com\">http:\/\/desmos.com<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Try It hjm819; hjm632; hjm596. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <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":422608,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"The Inverse of a Quadratic Function\",\"author\":\"Leo Chang and Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"All graphs created using desmos graphing calculator\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"desmos.com\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Try It hjm819; hjm632; hjm596\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2069","chapter","type-chapter","status-publish","hentry"],"part":1691,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2069","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/users\/422608"}],"version-history":[{"count":70,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2069\/revisions"}],"predecessor-version":[{"id":4889,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2069\/revisions\/4889"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/parts\/1691"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2069\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/media?parent=2069"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2069"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/contributor?post=2069"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/license?post=2069"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}