{"id":185,"date":"2023-06-21T13:22:41","date_gmt":"2023-06-21T13:22:41","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/projectile-motion\/"},"modified":"2023-07-23T21:38:09","modified_gmt":"2023-07-23T21:38:09","slug":"projectile-motion","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/projectile-motion\/","title":{"raw":"\u25aa   Projectile Motion","rendered":"\u25aa   Projectile Motion"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Define\u00a0projectile motion<\/li>\r\n \t<li>Solve a polynomial function\u00a0that represents projectile motion<\/li>\r\n \t<li>Interpret the solution to a polynomial function that represents projectile motion<\/li>\r\n<\/ul>\r\n<\/div>\r\nProjectile motion happens when you throw a ball into the air and it comes back down because of gravity. \u00a0A projectile will follow a curved path that behaves in a predictable way. \u00a0This predictable motion has been studied for centuries, and in simple cases, an object's height from the ground\u00a0at a given time,\u00a0[latex]t[\/latex], can be modeled with a polynomial function of the form [latex]h(t)=at^2+bt+c[\/latex], where [latex]h(t)[\/latex] represents the height of an object at a given time,\u00a0[latex]t[\/latex]. \u00a0Projectile motion is also called a parabolic trajectory because of the shape of the path of a projectile's motion, as in the water in the image of the fountain below.\r\n\r\n[caption id=\"attachment_4890\" align=\"aligncenter\" width=\"436\"]<img class=\"wp-image-4890\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161949\/ParabolicWaterTrajectory-225x300.jpg\" alt=\"Water from a fountain shoing classic parabolic motion.\" width=\"436\" height=\"581\" \/> Parabolic water trajectory in a fountain.[\/caption]\r\n\r\nParabolic motion and its related functions\u00a0allow us to launch satellites for telecommunications and rockets for space exploration. Recently, police departments have even begun using projectiles with GPS to track fleeing suspects in vehicles rather than pursuing them by high-speed chase\u00a0[footnote]\"Cops' Latest Tool in High-speed Chases: GPS Projectiles.\" <i>CBSNews<\/i>. CBS Interactive, n.d. Web.\u00a0[latex]14[\/latex] June\u00a0[latex]2016[\/latex].[\/footnote].\r\n\r\nIn this section, we will use\u00a0polynomial functions\u00a0to answer questions about the parabolic motion of a projectile. The real mathematical model for the path of a rocket or a police GPS projectile may have different coefficients or more variables, but the concept remains the same. We will also learn to interpret the meaning of the variables in a polynomial function that models projectile motion.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nA small toy rocket is launched from a\u00a0[latex]4[\/latex]-foot pedestal. The height (<i>h, <\/i>in feet) of the rocket <i>t<\/i> seconds after taking off is given by the function [latex]h(t)=\u22122t^{2}+7t+4[\/latex]. How long will it take the rocket to hit the ground?\r\n[reveal-answer q=\"679533\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"679533\"]\r\n\r\nThe rocket will be on the ground when [latex]h(t)=0[\/latex]. We want to know how long,\u00a0[latex]t[\/latex], the rocket is in the air.\r\n\r\n[latex]\\begin{array}{l}h(t)=\u22122t^{2}+7t+4=0\\\\0=\u22122t^{2}+7t+4\\end{array}[\/latex]\r\n\r\nWe can factor the polynomial [latex]\u22122t^{2}+7t+4[\/latex] more easily by first factoring out a [latex]-1[\/latex]\r\n\r\n[latex]\\begin{array}{c}0=-1(2t^{2}-7t-4)\\\\0=-1\\left(2t+1\\right)\\left(t-4\\right)\\end{array}[\/latex]\r\n\r\nUse the Zero Product Property. There is no need to set the constant factor [latex]-1[\/latex] to zero, because [latex]-1[\/latex] will never equal zero.\r\n<p style=\"text-align: center;\">[latex]2t+1=0\\,\\,\\,\\,\\,\\,\\text{OR}\\,\\,\\,\\,\\,\\,t-4=0[\/latex]<\/p>\r\nSolve each equation.\r\n<p style=\"text-align: center;\">[latex]t=-\\frac{1}{2}\\,\\,\\,\\,\\,\\,\\text{OR}\\,\\,\\,\\,\\,\\,t=4[\/latex]<\/p>\r\nInterpret the answer. Since <i>t<\/i> represents time, it cannot be a negative number; only [latex]t=4[\/latex]\u00a0makes sense in this context.\r\n\r\nWe can check our answer when\u00a0[latex]t=4[\/latex].\r\n\r\n[latex]\\begin{array}{c}h(4)=\u22122(4)^{2}+7(4)+4=0\\\\h(4)=-2(16)+28+4=0\\\\h(4)-32+32=0\\\\h(4)=0\\end{array}[\/latex]\r\n\r\nTherefore, the rocket will hit the ground 4 seconds after being launched.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we will solve for the time that the rocket reaches a given height other than zero.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUse the formula for the height of the rocket in the previous example to find the time when the rocket is\u00a0[latex]4[\/latex] feet from hitting the ground on its way back down. \u00a0Refer to the image.\r\n\r\n[latex]h(t)=\u22122t^{2}+7t+4[\/latex]\r\n\r\n<img class=\"wp-image-4892 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161951\/Screen-Shot-2016-06-14-at-5.39.53-PM-300x247.png\" alt=\"Parabolic motion of rocket which starts four feet up from the ground. t=0 is labeled at the starti of hte parabolic motion adn t=? is labeled at four feet from the ground on the other side of the parabola.\" width=\"413\" height=\"340\" \/>\r\n[reveal-answer q=\"198118\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"198118\"]\r\n\r\nWe are given that the height of the rocket is\u00a0[latex]4[\/latex] feet from the ground on its way back down. We want to know how long it will take for the rocket to get to that point in its path. We are going to solve for t.\r\n\r\nSubstitute [latex]h(t) = 4[\/latex] into the formula for height, and try to get zero on one side since we know we can use the zero product principle to solve polynomials.\r\n\r\n<strong>Write and Solve:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}h(t)=4=\u22122t^{2}+7t+4\\\\4=-2t^2+7t+4\\\\\\underline{-4}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-4}\\\\0=-2t^2+7t\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now we can factor out a\u00a0[latex]t[\/latex] from each term:<\/p>\r\n<p style=\"text-align: center;\">[latex]0=t\\left(-2t+7\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Solve each equation for\u00a0[latex]t[\/latex] using the zero product principle:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=0\\text{ OR }-2t+7=0\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-7}\\,\\,\\,\\,\\,\\,\\,\\underline{-7}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{-2t}{-2}=\\frac{-7}{-2}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,t=\\frac{7}{2}=3.5\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">It does not make sense for us to choose\u00a0[latex]t=0[\/latex], because we are interested in the amount of time that has passed when the projectile is\u00a0[latex]4[\/latex] feet from hitting the ground on its way back down. We will choose\u00a0[latex]t=3.5[\/latex].<\/p>\r\n<p style=\"text-align: left;\">Check the answer on your own for practice.<\/p>\r\nThe rocket will be 4 feet from the ground at [latex]t=3.5\\text{ seconds}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of how to find the time when a object following a parabolic trajectory hits the ground.\r\n\r\nhttps:\/\/youtu.be\/hsWSzu3KcPU","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Define\u00a0projectile motion<\/li>\n<li>Solve a polynomial function\u00a0that represents projectile motion<\/li>\n<li>Interpret the solution to a polynomial function that represents projectile motion<\/li>\n<\/ul>\n<\/div>\n<p>Projectile motion happens when you throw a ball into the air and it comes back down because of gravity. \u00a0A projectile will follow a curved path that behaves in a predictable way. \u00a0This predictable motion has been studied for centuries, and in simple cases, an object&#8217;s height from the ground\u00a0at a given time,\u00a0[latex]t[\/latex], can be modeled with a polynomial function of the form [latex]h(t)=at^2+bt+c[\/latex], where [latex]h(t)[\/latex] represents the height of an object at a given time,\u00a0[latex]t[\/latex]. \u00a0Projectile motion is also called a parabolic trajectory because of the shape of the path of a projectile&#8217;s motion, as in the water in the image of the fountain below.<\/p>\n<div id=\"attachment_4890\" style=\"width: 446px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4890\" class=\"wp-image-4890\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161949\/ParabolicWaterTrajectory-225x300.jpg\" alt=\"Water from a fountain shoing classic parabolic motion.\" width=\"436\" height=\"581\" \/><\/p>\n<p id=\"caption-attachment-4890\" class=\"wp-caption-text\">Parabolic water trajectory in a fountain.<\/p>\n<\/div>\n<p>Parabolic motion and its related functions\u00a0allow us to launch satellites for telecommunications and rockets for space exploration. Recently, police departments have even begun using projectiles with GPS to track fleeing suspects in vehicles rather than pursuing them by high-speed chase\u00a0<a class=\"footnote\" title=\"&quot;Cops' Latest Tool in High-speed Chases: GPS Projectiles.&quot; CBSNews. CBS Interactive, n.d. Web.\u00a0[latex]14[\/latex] June\u00a0[latex]2016[\/latex].\" id=\"return-footnote-185-1\" href=\"#footnote-185-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a>.<\/p>\n<p>In this section, we will use\u00a0polynomial functions\u00a0to answer questions about the parabolic motion of a projectile. The real mathematical model for the path of a rocket or a police GPS projectile may have different coefficients or more variables, but the concept remains the same. We will also learn to interpret the meaning of the variables in a polynomial function that models projectile motion.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>A small toy rocket is launched from a\u00a0[latex]4[\/latex]-foot pedestal. The height (<i>h, <\/i>in feet) of the rocket <i>t<\/i> seconds after taking off is given by the function [latex]h(t)=\u22122t^{2}+7t+4[\/latex]. How long will it take the rocket to hit the ground?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q679533\">Show Solution<\/span><\/p>\n<div id=\"q679533\" class=\"hidden-answer\" style=\"display: none\">\n<p>The rocket will be on the ground when [latex]h(t)=0[\/latex]. We want to know how long,\u00a0[latex]t[\/latex], the rocket is in the air.<\/p>\n<p>[latex]\\begin{array}{l}h(t)=\u22122t^{2}+7t+4=0\\\\0=\u22122t^{2}+7t+4\\end{array}[\/latex]<\/p>\n<p>We can factor the polynomial [latex]\u22122t^{2}+7t+4[\/latex] more easily by first factoring out a [latex]-1[\/latex]<\/p>\n<p>[latex]\\begin{array}{c}0=-1(2t^{2}-7t-4)\\\\0=-1\\left(2t+1\\right)\\left(t-4\\right)\\end{array}[\/latex]<\/p>\n<p>Use the Zero Product Property. There is no need to set the constant factor [latex]-1[\/latex] to zero, because [latex]-1[\/latex] will never equal zero.<\/p>\n<p style=\"text-align: center;\">[latex]2t+1=0\\,\\,\\,\\,\\,\\,\\text{OR}\\,\\,\\,\\,\\,\\,t-4=0[\/latex]<\/p>\n<p>Solve each equation.<\/p>\n<p style=\"text-align: center;\">[latex]t=-\\frac{1}{2}\\,\\,\\,\\,\\,\\,\\text{OR}\\,\\,\\,\\,\\,\\,t=4[\/latex]<\/p>\n<p>Interpret the answer. Since <i>t<\/i> represents time, it cannot be a negative number; only [latex]t=4[\/latex]\u00a0makes sense in this context.<\/p>\n<p>We can check our answer when\u00a0[latex]t=4[\/latex].<\/p>\n<p>[latex]\\begin{array}{c}h(4)=\u22122(4)^{2}+7(4)+4=0\\\\h(4)=-2(16)+28+4=0\\\\h(4)-32+32=0\\\\h(4)=0\\end{array}[\/latex]<\/p>\n<p>Therefore, the rocket will hit the ground 4 seconds after being launched.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we will solve for the time that the rocket reaches a given height other than zero.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Use the formula for the height of the rocket in the previous example to find the time when the rocket is\u00a0[latex]4[\/latex] feet from hitting the ground on its way back down. \u00a0Refer to the image.<\/p>\n<p>[latex]h(t)=\u22122t^{2}+7t+4[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4892 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161951\/Screen-Shot-2016-06-14-at-5.39.53-PM-300x247.png\" alt=\"Parabolic motion of rocket which starts four feet up from the ground. t=0 is labeled at the starti of hte parabolic motion adn t=? is labeled at four feet from the ground on the other side of the parabola.\" width=\"413\" height=\"340\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q198118\">Show Solution<\/span><\/p>\n<div id=\"q198118\" class=\"hidden-answer\" style=\"display: none\">\n<p>We are given that the height of the rocket is\u00a0[latex]4[\/latex] feet from the ground on its way back down. We want to know how long it will take for the rocket to get to that point in its path. We are going to solve for t.<\/p>\n<p>Substitute [latex]h(t) = 4[\/latex] into the formula for height, and try to get zero on one side since we know we can use the zero product principle to solve polynomials.<\/p>\n<p><strong>Write and Solve:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}h(t)=4=\u22122t^{2}+7t+4\\\\4=-2t^2+7t+4\\\\\\underline{-4}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-4}\\\\0=-2t^2+7t\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now we can factor out a\u00a0[latex]t[\/latex] from each term:<\/p>\n<p style=\"text-align: center;\">[latex]0=t\\left(-2t+7\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">Solve each equation for\u00a0[latex]t[\/latex] using the zero product principle:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=0\\text{ OR }-2t+7=0\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-7}\\,\\,\\,\\,\\,\\,\\,\\underline{-7}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{-2t}{-2}=\\frac{-7}{-2}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,t=\\frac{7}{2}=3.5\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">It does not make sense for us to choose\u00a0[latex]t=0[\/latex], because we are interested in the amount of time that has passed when the projectile is\u00a0[latex]4[\/latex] feet from hitting the ground on its way back down. We will choose\u00a0[latex]t=3.5[\/latex].<\/p>\n<p style=\"text-align: left;\">Check the answer on your own for practice.<\/p>\n<p>The rocket will be 4 feet from the ground at [latex]t=3.5\\text{ seconds}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show another example of how to find the time when a object following a parabolic trajectory hits the ground.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Factoring Application - Find the Time When a Projectile Hits and Ground\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/hsWSzu3KcPU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/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-185\">\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>Parabolic motion description and example. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Factoring Application - Find the Time When a Projectile Hits and Ground. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/hsWSzu3KcPU\">https:\/\/youtu.be\/hsWSzu3KcPU<\/a>. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Parabolic water trajectory. <strong>Authored by<\/strong>: By GuidoB. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/commons.wikimedia.org\/w\/index.php?curid=8015696\">https:\/\/commons.wikimedia.org\/w\/index.php?curid=8015696<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/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><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-185-1\">\"Cops' Latest Tool in High-speed Chases: GPS Projectiles.\" <i>CBSNews<\/i>. CBS Interactive, n.d. Web.\u00a0[latex]14[\/latex] June\u00a0[latex]2016[\/latex]. <a href=\"#return-footnote-185-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":395986,"menu_order":36,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Parabolic water trajectory\",\"author\":\"By GuidoB\",\"organization\":\"\",\"url\":\"https:\/\/commons.wikimedia.org\/w\/index.php?curid=8015696\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Parabolic motion description and example\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Factoring Application - Find the Time When a Projectile Hits and Ground\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/hsWSzu3KcPU\",\"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-185","chapter","type-chapter","status-publish","hentry"],"part":160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/185","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":3,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/185\/revisions"}],"predecessor-version":[{"id":1195,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/185\/revisions\/1195"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/160"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/185\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/media?parent=185"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=185"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/contributor?post=185"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/license?post=185"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}