{"id":4784,"date":"2017-06-07T18:57:06","date_gmt":"2017-06-07T18:57:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-projectiles\/"},"modified":"2025-09-09T17:27:46","modified_gmt":"2025-09-09T17:27:46","slug":"read-projectiles","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-projectiles\/","title":{"raw":"Applications With Quadratics: Projectiles","rendered":"Applications With Quadratics: Projectiles"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Define\u00a0projectile motion<\/li>\r\n \t<li>Solve a quadratic equation that represents projectile motion<\/li>\r\n \t<li>Interpret the solution to a quadratic equation 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 it's height from the ground\u00a0at a given time, t, can be modeled with a quadratic polynomial of the form [latex]\\text{height}=at^2+bt+c[\/latex] such as we have been studying in this module. Projectile motion is also called a parabolic trajectory because of the shape of the path of a projectile's motion, as in the image of water in 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\/2011\/2017\/06\/07185704\/ParabolicWaterTrajectory-225x300.jpg\" alt=\"Water from a fountain shoing classic parabolic motion.\" width=\"436\" height=\"581\" \/> Parabolic WaterTrajectory[\/caption]\r\n\r\nParabolic motion and its related equations allow 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 [footnote]\"Cops' Latest Tool in High-speed Chases: GPS Projectiles.\" <i>CBSNews<\/i>. CBS Interactive, n.d. Web. 14 June 2016.[\/footnote].\r\n\r\nIn this section we will solve\u00a0simple quadratic polynomials that represent 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 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 4-foot pedestal. The height (<i>h, <\/i>in feet) of the rocket <i>t<\/i> seconds after taking off is given by the formula [latex]h=\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\n<strong><strong>Read and understand:\u00a0<\/strong><\/strong>The rocket will be on the ground when the height is 0. We want to know how long, t, \u00a0the rocket is in the air.\r\n\r\n<strong>Translate:\u00a0<\/strong>So, we will substitute 0 for <i>h<\/i> in the formula and solve for t.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}h=\u22122t^{2}+7t+4\\\\0=\u22122t^{2}+7t+4\\end{array}[\/latex]<\/p>\r\n<strong>Write and Solve:<\/strong> Rewrite the middle term using the [latex]a\\cdot{c}[\/latex] method.\r\n<p style=\"text-align: center;\">[latex]0=-2t^{2}+8t-t+4[\/latex]<\/p>\r\nFactor the trinomial by grouping.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}0=-2t\\left(t-4\\right)-\\left(t-4\\right)\\\\0=\\left(-2t-1\\right)\\left(t-4\\right)\\\\0=-1\\left(2t+1\\right)\\left(t-4\\right)\\end{array}[\/latex]<\/p>\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\n[latex]t=4[\/latex]\r\n<h4>Answer<\/h4>\r\nThe 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 is at 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 4 feet from hitting the ground on it's way back down. \u00a0Refer to the image.\r\n\r\n[latex]h=\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\/2011\/2017\/06\/07185705\/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\n<strong><strong>Read and understand:\u00a0<\/strong><\/strong>We are given that the height of the rocket is 4 feet from the ground on it's way back down. We want to know how long it has taken the rocket to get to that point in it's path, we are going to solve for t.\r\n\r\n<strong>Translate:\u00a0<\/strong>Substitute h = 4 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=\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 t 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 t 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;\"><strong>Interpret:\u00a0<\/strong>It doesn't make sense for us to choose t=0 because we are interested in the amount of time that has passed when the projectile is 4 feet from hitting the ground on it's way back down. We will choose t=3.5<\/p>\r\n\r\n<h4 style=\"text-align: left;\">Answer<\/h4>\r\n[latex]t=3.5\\text{ seconds }[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe video that follows presents another example of solving a quadratic equation that represents parabolic motion.\r\n\r\nhttps:\/\/youtu.be\/hsWSzu3KcPU\r\n<h2><span style=\"color: #4c7fb0;\">Summary<\/span><\/h2>\r\nIn this section we introduced the concept of projectile motion, and showed that it can be modeled with a quadratic polynomial. \u00a0While the models used in these examples are simple, the concepts and interpretations are the same. \u00a0The methods used to solve quadratic polynomials that don't factor easily are many and well known, it is likely you will come across more in your studies.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Define\u00a0projectile motion<\/li>\n<li>Solve a quadratic equation that represents projectile motion<\/li>\n<li>Interpret the solution to a quadratic equation 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 it&#8217;s height from the ground\u00a0at a given time, t, can be modeled with a quadratic polynomial of the form [latex]\\text{height}=at^2+bt+c[\/latex] such as we have been studying in this module. Projectile motion is also called a parabolic trajectory because of the shape of the path of a projectile&#8217;s motion, as in the image of water in 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\/2011\/2017\/06\/07185704\/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 WaterTrajectory<\/p>\n<\/div>\n<p>Parabolic motion and its related equations allow 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 <a class=\"footnote\" title=\"&quot;Cops' Latest Tool in High-speed Chases: GPS Projectiles.&quot; CBSNews. CBS Interactive, n.d. Web. 14 June 2016.\" id=\"return-footnote-4784-1\" href=\"#footnote-4784-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a>.<\/p>\n<p>In this section we will solve\u00a0simple quadratic polynomials that represent 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 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 4-foot pedestal. The height (<i>h, <\/i>in feet) of the rocket <i>t<\/i> seconds after taking off is given by the formula [latex]h=\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><strong><strong>Read and understand:\u00a0<\/strong><\/strong>The rocket will be on the ground when the height is 0. We want to know how long, t, \u00a0the rocket is in the air.<\/p>\n<p><strong>Translate:\u00a0<\/strong>So, we will substitute 0 for <i>h<\/i> in the formula and solve for t.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}h=\u22122t^{2}+7t+4\\\\0=\u22122t^{2}+7t+4\\end{array}[\/latex]<\/p>\n<p><strong>Write and Solve:<\/strong> Rewrite the middle term using the [latex]a\\cdot{c}[\/latex] method.<\/p>\n<p style=\"text-align: center;\">[latex]0=-2t^{2}+8t-t+4[\/latex]<\/p>\n<p>Factor the trinomial by grouping.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}0=-2t\\left(t-4\\right)-\\left(t-4\\right)\\\\0=\\left(-2t-1\\right)\\left(t-4\\right)\\\\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>[latex]t=4[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>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 is at 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 4 feet from hitting the ground on it&#8217;s way back down. \u00a0Refer to the image.<\/p>\n<p>[latex]h=\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\/2011\/2017\/06\/07185705\/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><strong><strong>Read and understand:\u00a0<\/strong><\/strong>We are given that the height of the rocket is 4 feet from the ground on it&#8217;s way back down. We want to know how long it has taken the rocket to get to that point in it&#8217;s path, we are going to solve for t.<\/p>\n<p><strong>Translate:\u00a0<\/strong>Substitute h = 4 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=\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 t 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 t 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;\"><strong>Interpret:\u00a0<\/strong>It doesn&#8217;t make sense for us to choose t=0 because we are interested in the amount of time that has passed when the projectile is 4 feet from hitting the ground on it&#8217;s way back down. We will choose t=3.5<\/p>\n<h4 style=\"text-align: left;\">Answer<\/h4>\n<p>[latex]t=3.5\\text{ seconds }[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The video that follows presents another example of solving a quadratic equation that represents parabolic motion.<\/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<h2><span style=\"color: #4c7fb0;\">Summary<\/span><\/h2>\n<p>In this section we introduced the concept of projectile motion, and showed that it can be modeled with a quadratic polynomial. \u00a0While the models used in these examples are simple, the concepts and interpretations are the same. \u00a0The methods used to solve quadratic polynomials that don&#8217;t factor easily are many and well known, it is likely you will come across more in your studies.<\/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-4784\">\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-sa\/4.0\/\">CC BY-SA: Attribution-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-4784-1\">\"Cops' Latest Tool in High-speed Chases: GPS Projectiles.\" <i>CBSNews<\/i>. CBS Interactive, n.d. Web. 14 June 2016. <a href=\"#return-footnote-4784-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":17533,"menu_order":26,"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-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-4784","chapter","type-chapter","status-publish","hentry"],"part":4674,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4784","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4784\/revisions"}],"predecessor-version":[{"id":5125,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4784\/revisions\/5125"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/parts\/4674"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4784\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/media?parent=4784"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapter-type?post=4784"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/contributor?post=4784"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/license?post=4784"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}