{"id":2398,"date":"2016-11-03T21:17:57","date_gmt":"2016-11-03T21:17:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=2398"},"modified":"2025-10-15T22:40:57","modified_gmt":"2025-10-15T22:40:57","slug":"graphing-parabolas-with-vertices-not-at-the-origin","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/chapter\/graphing-parabolas-with-vertices-not-at-the-origin\/","title":{"raw":"Parabolas with Vertices Not at the Origin","rendered":"Parabolas with Vertices Not at the Origin"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter given the equation of a parabola in standard form.<\/li>\r\n \t<li>Find the equation of a parabolic shaped object given dimensions.<\/li>\r\n<\/ul>\r\n<\/div>\r\nLike other graphs we\u2019ve worked with, the graph of a parabola can be translated. If a parabola is translated [latex]h[\/latex] units horizontally and [latex]k[\/latex] units vertically, the vertex will be [latex]\\left(h,k\\right)[\/latex]. This translation results in the standard form of the equation we saw previously with [latex]x[\/latex] replaced by [latex]\\left(x-h\\right)[\/latex] and [latex]y[\/latex] replaced by [latex]\\left(y-k\\right)[\/latex].\r\n\r\nTo graph parabolas with a vertex [latex]\\left(h,k\\right)[\/latex] other than the origin, we use the standard form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the <em>x<\/em>-axis, and [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the <em>y<\/em>-axis. These standard forms are given below, along with their general graphs and key features.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Standard Forms of Parabolas with Vertex (<em>h<\/em>, <em>k<\/em>)<\/h3>\r\nThe table summarizes the standard features of parabolas with a vertex at a point [latex]\\left(h,k\\right)[\/latex].\r\n<table summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><strong>Axis of Symmetry<\/strong><\/td>\r\n<td><strong>Equation<\/strong><\/td>\r\n<td><strong>Focus<\/strong><\/td>\r\n<td><strong>Directrix<\/strong><\/td>\r\n<td><strong>Endpoints of focal diameter<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]y=k[\/latex]<\/td>\r\n<td>[latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/td>\r\n<td>[latex]x=h-p[\/latex]<\/td>\r\n<td>[latex]\\left(h+p,\\text{ }k\\pm 2p\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x=h[\/latex]<\/td>\r\n<td>[latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/td>\r\n<td>[latex]y=k-p[\/latex]<\/td>\r\n<td>[latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204550\/CNX_Precalc_Figure_10_03_0092.jpg\" alt=\"\" width=\"975\" height=\"901\" \/> (a) When [latex]p&gt;0[\/latex], the parabola opens right. (b) When [latex]p&lt;0[\/latex], the parabola opens left. (c) When [latex]p&gt;0[\/latex], the parabola opens up. (d) When [latex]p&lt;0[\/latex], the parabola opens down.[\/caption]<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a standard form equation for a parabola centered at (<em>h<\/em>, <em>k<\/em>), sketch the graph.<\/h3>\r\n<ol>\r\n \t<li>Determine which of the standard forms applies to the given equation: [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] or [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex].<\/li>\r\n \t<li>Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the focal diameter.\r\n<ol>\r\n \t<li>If the equation is in the form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex], then:\r\n<ul>\r\n \t<li>use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\r\n \t<li>use the value of [latex]k[\/latex] to determine the axis of symmetry, [latex]y=k[\/latex]<\/li>\r\n \t<li>set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(x-h\\right)[\/latex] in the given equation to solve for [latex]p[\/latex]. If [latex]p&gt;0[\/latex], the parabola opens right. If [latex]p&lt;0[\/latex], the parabola opens left.<\/li>\r\n \t<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/li>\r\n \t<li>use [latex]h[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]x=h-p[\/latex]<\/li>\r\n \t<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the focal diameter, [latex]\\left(h+p,k\\pm 2p\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>If the equation is in the form [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex], then:\r\n<ul>\r\n \t<li>use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\r\n \t<li>use the value of [latex]h[\/latex] to determine the axis of symmetry, [latex]x=h[\/latex]<\/li>\r\n \t<li>set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(y-k\\right)[\/latex] in the given equation to solve for [latex]p[\/latex]. If [latex]p&gt;0[\/latex], the parabola opens up. If [latex]p&lt;0[\/latex], the parabola opens down.<\/li>\r\n \t<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/li>\r\n \t<li>use [latex]k[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]y=k-p[\/latex]<\/li>\r\n \t<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the focal diameter, [latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing a Parabola with Vertex (<em>h<\/em>, <em>k<\/em>) and Axis of Symmetry Parallel to the <em>x<\/em>-axis<\/h3>\r\nGraph [latex]{\\left(y - 1\\right)}^{2}=-16\\left(x+3\\right)[\/latex]. Identify and label the <strong>vertex<\/strong>, <strong>axis of symmetry<\/strong>, <strong>focus<\/strong>, <strong>directrix<\/strong>, and endpoints of the <strong>focal diameter<\/strong>.\r\n\r\n[reveal-answer q=\"943688\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"943688\"]\r\n\r\nThe standard form that applies to the given equation is [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]. Thus, the axis of symmetry is parallel to the <em>x<\/em>-axis. It follows that:\r\n<ul>\r\n \t<li>the vertex is [latex]\\left(h,k\\right)=\\left(-3,1\\right)[\/latex]<\/li>\r\n \t<li>the axis of symmetry is [latex]y=k=1[\/latex]<\/li>\r\n \t<li>[latex]-16=4p[\/latex], so [latex]p=-4[\/latex]. Since [latex]p&lt;0[\/latex], the parabola opens left.<\/li>\r\n \t<li>the coordinates of the focus are [latex]\\left(h+p,k\\right)=\\left(-3+\\left(-4\\right),1\\right)=\\left(-7,1\\right)[\/latex]<\/li>\r\n \t<li>the equation of the directrix is [latex]x=h-p=-3-\\left(-4\\right)=1[\/latex]<\/li>\r\n \t<li>the endpoints of the focal diameter are [latex]\\left(h+p,k\\pm 2p\\right)=\\left(-3+\\left(-4\\right),1\\pm 2\\left(-4\\right)\\right)[\/latex], or [latex]\\left(-7,-7\\right)[\/latex] and [latex]\\left(-7,9\\right)[\/latex]<\/li>\r\n<\/ul>\r\nNext we plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204553\/CNX_Precalc_Figure_10_03_0102.jpg\" width=\"487\" height=\"480\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGraph [latex]{\\left(y+1\\right)}^{2}=4\\left(x - 8\\right)[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.\r\n\r\n[reveal-answer q=\"626564\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"626564\"]\r\n\r\nVertex: [latex]\\left(8,-1\\right)[\/latex]; Axis of symmetry: [latex]y=-1[\/latex]; Focus: [latex]\\left(9,-1\\right)[\/latex]; Directrix: [latex]x=7[\/latex]; Endpoints of the latus rectum: [latex]\\left(9,-3\\right)[\/latex] and [latex]\\left(9,1\\right)[\/latex].\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182612\/CNX_Precalc_Figure_10_03_0112.jpg\"><img class=\"aligncenter size-full wp-image-3280\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182612\/CNX_Precalc_Figure_10_03_0112.jpg\" alt=\"\" width=\"487\" height=\"522\" \/><\/a>\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=86148&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"550\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing a Parabola from an Equation Given in General Form<\/h3>\r\nGraph [latex]{x}^{2}-8x - 28y - 208=0[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.\r\n\r\n[reveal-answer q=\"335853\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"335853\"]\r\n\r\nStart by writing the equation of the <strong>parabola<\/strong> in standard form. The standard form that applies to the given equation is [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]. Thus, the axis of symmetry is parallel to the <em>y<\/em>-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable [latex]x[\/latex] in order to complete the square.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}{x}^{2}-8x - 28y - 208=0 \\\\ {x}^{2}-8x=28y+208 \\\\ {x}^{2}-8x+16=28y+208+16 \\\\ {\\left(x - 4\\right)}^{2}=28y+224 \\\\ {\\left(x - 4\\right)}^{2}=28\\left(y+8\\right) \\\\ {\\left(x - 4\\right)}^{2}=4\\cdot 7\\cdot \\left(y+8\\right) \\end{gathered}[\/latex]<\/p>\r\nIt follows that:\r\n<ul>\r\n \t<li>the vertex is [latex]\\left(h,k\\right)=\\left(4,-8\\right)[\/latex]<\/li>\r\n \t<li>the axis of symmetry is [latex]x=h=4[\/latex]<\/li>\r\n \t<li>since [latex]p=7,p&gt;0[\/latex] and so the parabola opens up<\/li>\r\n \t<li>the coordinates of the focus are [latex]\\left(h,k+p\\right)=\\left(4,-8+7\\right)=\\left(4,-1\\right)[\/latex]<\/li>\r\n \t<li>the equation of the directrix is [latex]y=k-p=-8 - 7=-15[\/latex]<\/li>\r\n \t<li>the endpoints of the focal diameter are [latex]\\left(h\\pm 2p,k+p\\right)=\\left(4\\pm 2\\left(7\\right),-8+7\\right)[\/latex], or [latex]\\left(-10,-1\\right)[\/latex] and [latex]\\left(18,-1\\right)[\/latex]<\/li>\r\n<\/ul>\r\nNext we plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204555\/CNX_Precalc_Figure_10_03_0122.jpg\" width=\"487\" height=\"258\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGraph [latex]{\\left(x+2\\right)}^{2}=-20\\left(y - 3\\right)[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.\r\n\r\n[reveal-answer q=\"665189\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"665189\"]\r\n\r\nVertex: [latex]\\left(-2,3\\right)[\/latex]; Axis of symmetry: [latex]x=-2[\/latex]; Focus: [latex]\\left(-2,-2\\right)[\/latex]; Directrix: [latex]y=8[\/latex]; Endpoints of the latus rectum: [latex]\\left(-12,-2\\right)[\/latex] and [latex]\\left(8,-2\\right)[\/latex].\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182942\/CNX_Precalc_Figure_10_03_0132.jpg\"><img class=\"aligncenter size-full wp-image-3281\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182942\/CNX_Precalc_Figure_10_03_0132.jpg\" alt=\"\" width=\"731\" height=\"438\" \/><\/a>\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=86124&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"950\"><\/iframe>\r\n\r\n<\/div>\r\n\r\n<h2>Solving Applied Problems Involving Parabolas<\/h2>\r\nAs we mentioned at the beginning of the section, parabolas are used to design many objects we use every day, such as telescopes, suspension bridges, microphones, and radar equipment. Parabolic mirrors, such as the one used to light the Olympic torch, have a very unique reflecting property. When rays of light parallel to the parabola\u2019s <strong>axis of symmetry<\/strong> are directed toward any surface of the mirror, the light is reflected directly to the focus.\u00a0This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204557\/CNX_Precalc_Figure_10_03_0142.jpg\" width=\"487\" height=\"362\" \/> Reflecting property of parabolas[\/caption]\r\n\r\nParabolic mirrors have the ability to focus the sun\u2019s energy to a single point, raising the temperature hundreds of degrees in a matter of seconds. Thus parabolic mirrors are featured in many low-cost, energy efficient solar products, such as solar cookers, solar heaters, and even travel-sized fire starters.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Applied Problems Involving Parabolas<\/h3>\r\nA cross-section of a design for a travel-sized solar fire starter. The sun\u2019s rays reflect off the parabolic mirror toward an object attached to the igniter. Because the igniter is located at the focus of the parabola, the reflected rays cause the object to burn in just seconds.\r\n<ol>\r\n \t<li>Find the equation of the parabola that models the fire starter. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane.<\/li>\r\n \t<li>Use the equation found in part (a) to find the depth of the fire starter.<\/li>\r\n<\/ol>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204559\/CNX_Precalc_Figure_10_03_0162.jpg\" alt=\"\" width=\"487\" height=\"217\" \/> Cross-section of a travel-sized solar fire starter[\/caption]\r\n\r\n[reveal-answer q=\"424264\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"424264\"]\r\n\r\n&nbsp;\r\n\r\nThe vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form [latex]{x}^{2}=4py[\/latex], where [latex]p&gt;0[\/latex]. The igniter, which is the focus, is 1.7 inches above the vertex of the dish. Thus we have [latex]p=1.7[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;{x}^{2}=4py &amp;&amp; \\text{Standard form of upward-facing parabola with vertex (0,0)} \\\\ &amp;{x}^{2}=4\\left(1.7\\right)y &amp;&amp; \\text{Substitute 1}\\text{.7 for }p. \\\\ &amp;{x}^{2}=6.8y &amp;&amp; \\text{Multiply}. \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left\">The dish extends [latex]\\frac{4.5}{2}=2.25[\/latex] inches on either side of the origin. We can substitute 2.25 for [latex]x[\/latex] in the equation from part (a) to find the depth of the dish.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;{x}^{2}=6.8y &amp;&amp; \\text{Equation found in part (a)}. \\\\ &amp;{\\left(2.25\\right)}^{2}=6.8y &amp;&amp; \\text{Substitute 2}\\text{.25 for }x. \\\\ &amp;y\\approx 0.74 &amp;&amp; \\text{Solve for }y. \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left\">The dish is about 0.74 inches deep.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nBalcony-sized solar cookers have been designed for families living in India. The top of a dish has a diameter of 1600 mm. The sun\u2019s rays reflect off the parabolic mirror toward the \"cooker,\" which is placed 320 mm from the base.\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>Find an equation that models a cross-section of the solar cooker. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane, and that the parabola opens to the right (i.e., has the <em>x<\/em>-axis as its axis of symmetry).<\/li>\r\n \t<li>Use the equation found in part (a) to find the depth of the cooker.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"535409\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"535409\"]\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>[latex]{y}^{2}=1280x[\/latex]<\/li>\r\n \t<li>The depth of the cooker is 500 mm<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=87077&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"650\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter given the equation of a parabola in standard form.<\/li>\n<li>Find the equation of a parabolic shaped object given dimensions.<\/li>\n<\/ul>\n<\/div>\n<p>Like other graphs we\u2019ve worked with, the graph of a parabola can be translated. If a parabola is translated [latex]h[\/latex] units horizontally and [latex]k[\/latex] units vertically, the vertex will be [latex]\\left(h,k\\right)[\/latex]. This translation results in the standard form of the equation we saw previously with [latex]x[\/latex] replaced by [latex]\\left(x-h\\right)[\/latex] and [latex]y[\/latex] replaced by [latex]\\left(y-k\\right)[\/latex].<\/p>\n<p>To graph parabolas with a vertex [latex]\\left(h,k\\right)[\/latex] other than the origin, we use the standard form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the <em>x<\/em>-axis, and [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the <em>y<\/em>-axis. These standard forms are given below, along with their general graphs and key features.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Standard Forms of Parabolas with Vertex (<em>h<\/em>, <em>k<\/em>)<\/h3>\n<p>The table summarizes the standard features of parabolas with a vertex at a point [latex]\\left(h,k\\right)[\/latex].<\/p>\n<table summary=\"..\">\n<tbody>\n<tr>\n<td><strong>Axis of Symmetry<\/strong><\/td>\n<td><strong>Equation<\/strong><\/td>\n<td><strong>Focus<\/strong><\/td>\n<td><strong>Directrix<\/strong><\/td>\n<td><strong>Endpoints of focal diameter<\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]y=k[\/latex]<\/td>\n<td>[latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]<\/td>\n<td>[latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/td>\n<td>[latex]x=h-p[\/latex]<\/td>\n<td>[latex]\\left(h+p,\\text{ }k\\pm 2p\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]x=h[\/latex]<\/td>\n<td>[latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]<\/td>\n<td>[latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/td>\n<td>[latex]y=k-p[\/latex]<\/td>\n<td>[latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204550\/CNX_Precalc_Figure_10_03_0092.jpg\" alt=\"\" width=\"975\" height=\"901\" \/><\/p>\n<p class=\"wp-caption-text\">(a) When [latex]p&gt;0[\/latex], the parabola opens right. (b) When [latex]p&lt;0[\/latex], the parabola opens left. (c) When [latex]p&gt;0[\/latex], the parabola opens up. (d) When [latex]p&lt;0[\/latex], the parabola opens down.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a standard form equation for a parabola centered at (<em>h<\/em>, <em>k<\/em>), sketch the graph.<\/h3>\n<ol>\n<li>Determine which of the standard forms applies to the given equation: [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] or [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex].<\/li>\n<li>Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the focal diameter.\n<ol>\n<li>If the equation is in the form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex], then:\n<ul>\n<li>use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\n<li>use the value of [latex]k[\/latex] to determine the axis of symmetry, [latex]y=k[\/latex]<\/li>\n<li>set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(x-h\\right)[\/latex] in the given equation to solve for [latex]p[\/latex]. If [latex]p>0[\/latex], the parabola opens right. If [latex]p<0[\/latex], the parabola opens left.<\/li>\n<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/li>\n<li>use [latex]h[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]x=h-p[\/latex]<\/li>\n<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the focal diameter, [latex]\\left(h+p,k\\pm 2p\\right)[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>If the equation is in the form [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex], then:\n<ul>\n<li>use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\n<li>use the value of [latex]h[\/latex] to determine the axis of symmetry, [latex]x=h[\/latex]<\/li>\n<li>set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(y-k\\right)[\/latex] in the given equation to solve for [latex]p[\/latex]. If [latex]p>0[\/latex], the parabola opens up. If [latex]p<0[\/latex], the parabola opens down.<\/li>\n<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/li>\n<li>use [latex]k[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]y=k-p[\/latex]<\/li>\n<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the focal diameter, [latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/li>\n<li>Plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing a Parabola with Vertex (<em>h<\/em>, <em>k<\/em>) and Axis of Symmetry Parallel to the <em>x<\/em>-axis<\/h3>\n<p>Graph [latex]{\\left(y - 1\\right)}^{2}=-16\\left(x+3\\right)[\/latex]. Identify and label the <strong>vertex<\/strong>, <strong>axis of symmetry<\/strong>, <strong>focus<\/strong>, <strong>directrix<\/strong>, and endpoints of the <strong>focal diameter<\/strong>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q943688\">Show Solution<\/span><\/p>\n<div id=\"q943688\" class=\"hidden-answer\" style=\"display: none\">\n<p>The standard form that applies to the given equation is [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]. Thus, the axis of symmetry is parallel to the <em>x<\/em>-axis. It follows that:<\/p>\n<ul>\n<li>the vertex is [latex]\\left(h,k\\right)=\\left(-3,1\\right)[\/latex]<\/li>\n<li>the axis of symmetry is [latex]y=k=1[\/latex]<\/li>\n<li>[latex]-16=4p[\/latex], so [latex]p=-4[\/latex]. Since [latex]p<0[\/latex], the parabola opens left.<\/li>\n<li>the coordinates of the focus are [latex]\\left(h+p,k\\right)=\\left(-3+\\left(-4\\right),1\\right)=\\left(-7,1\\right)[\/latex]<\/li>\n<li>the equation of the directrix is [latex]x=h-p=-3-\\left(-4\\right)=1[\/latex]<\/li>\n<li>the endpoints of the focal diameter are [latex]\\left(h+p,k\\pm 2p\\right)=\\left(-3+\\left(-4\\right),1\\pm 2\\left(-4\\right)\\right)[\/latex], or [latex]\\left(-7,-7\\right)[\/latex] and [latex]\\left(-7,9\\right)[\/latex]<\/li>\n<\/ul>\n<p>Next we plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204553\/CNX_Precalc_Figure_10_03_0102.jpg\" width=\"487\" height=\"480\" alt=\"image\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Graph [latex]{\\left(y+1\\right)}^{2}=4\\left(x - 8\\right)[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q626564\">Show Solution<\/span><\/p>\n<div id=\"q626564\" class=\"hidden-answer\" style=\"display: none\">\n<p>Vertex: [latex]\\left(8,-1\\right)[\/latex]; Axis of symmetry: [latex]y=-1[\/latex]; Focus: [latex]\\left(9,-1\\right)[\/latex]; Directrix: [latex]x=7[\/latex]; Endpoints of the latus rectum: [latex]\\left(9,-3\\right)[\/latex] and [latex]\\left(9,1\\right)[\/latex].<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182612\/CNX_Precalc_Figure_10_03_0112.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3280\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182612\/CNX_Precalc_Figure_10_03_0112.jpg\" alt=\"\" width=\"487\" height=\"522\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=86148&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"550\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing a Parabola from an Equation Given in General Form<\/h3>\n<p>Graph [latex]{x}^{2}-8x - 28y - 208=0[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q335853\">Show Solution<\/span><\/p>\n<div id=\"q335853\" class=\"hidden-answer\" style=\"display: none\">\n<p>Start by writing the equation of the <strong>parabola<\/strong> in standard form. The standard form that applies to the given equation is [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]. Thus, the axis of symmetry is parallel to the <em>y<\/em>-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable [latex]x[\/latex] in order to complete the square.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}{x}^{2}-8x - 28y - 208=0 \\\\ {x}^{2}-8x=28y+208 \\\\ {x}^{2}-8x+16=28y+208+16 \\\\ {\\left(x - 4\\right)}^{2}=28y+224 \\\\ {\\left(x - 4\\right)}^{2}=28\\left(y+8\\right) \\\\ {\\left(x - 4\\right)}^{2}=4\\cdot 7\\cdot \\left(y+8\\right) \\end{gathered}[\/latex]<\/p>\n<p>It follows that:<\/p>\n<ul>\n<li>the vertex is [latex]\\left(h,k\\right)=\\left(4,-8\\right)[\/latex]<\/li>\n<li>the axis of symmetry is [latex]x=h=4[\/latex]<\/li>\n<li>since [latex]p=7,p>0[\/latex] and so the parabola opens up<\/li>\n<li>the coordinates of the focus are [latex]\\left(h,k+p\\right)=\\left(4,-8+7\\right)=\\left(4,-1\\right)[\/latex]<\/li>\n<li>the equation of the directrix is [latex]y=k-p=-8 - 7=-15[\/latex]<\/li>\n<li>the endpoints of the focal diameter are [latex]\\left(h\\pm 2p,k+p\\right)=\\left(4\\pm 2\\left(7\\right),-8+7\\right)[\/latex], or [latex]\\left(-10,-1\\right)[\/latex] and [latex]\\left(18,-1\\right)[\/latex]<\/li>\n<\/ul>\n<p>Next we plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204555\/CNX_Precalc_Figure_10_03_0122.jpg\" width=\"487\" height=\"258\" alt=\"image\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Graph [latex]{\\left(x+2\\right)}^{2}=-20\\left(y - 3\\right)[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q665189\">Show Solution<\/span><\/p>\n<div id=\"q665189\" class=\"hidden-answer\" style=\"display: none\">\n<p>Vertex: [latex]\\left(-2,3\\right)[\/latex]; Axis of symmetry: [latex]x=-2[\/latex]; Focus: [latex]\\left(-2,-2\\right)[\/latex]; Directrix: [latex]y=8[\/latex]; Endpoints of the latus rectum: [latex]\\left(-12,-2\\right)[\/latex] and [latex]\\left(8,-2\\right)[\/latex].<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182942\/CNX_Precalc_Figure_10_03_0132.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3281\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182942\/CNX_Precalc_Figure_10_03_0132.jpg\" alt=\"\" width=\"731\" height=\"438\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=86124&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"950\"><\/iframe><\/p>\n<\/div>\n<h2>Solving Applied Problems Involving Parabolas<\/h2>\n<p>As we mentioned at the beginning of the section, parabolas are used to design many objects we use every day, such as telescopes, suspension bridges, microphones, and radar equipment. Parabolic mirrors, such as the one used to light the Olympic torch, have a very unique reflecting property. When rays of light parallel to the parabola\u2019s <strong>axis of symmetry<\/strong> are directed toward any surface of the mirror, the light is reflected directly to the focus.\u00a0This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204557\/CNX_Precalc_Figure_10_03_0142.jpg\" width=\"487\" height=\"362\" alt=\"image\" \/><\/p>\n<p class=\"wp-caption-text\">Reflecting property of parabolas<\/p>\n<\/div>\n<p>Parabolic mirrors have the ability to focus the sun\u2019s energy to a single point, raising the temperature hundreds of degrees in a matter of seconds. Thus parabolic mirrors are featured in many low-cost, energy efficient solar products, such as solar cookers, solar heaters, and even travel-sized fire starters.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving Applied Problems Involving Parabolas<\/h3>\n<p>A cross-section of a design for a travel-sized solar fire starter. The sun\u2019s rays reflect off the parabolic mirror toward an object attached to the igniter. Because the igniter is located at the focus of the parabola, the reflected rays cause the object to burn in just seconds.<\/p>\n<ol>\n<li>Find the equation of the parabola that models the fire starter. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane.<\/li>\n<li>Use the equation found in part (a) to find the depth of the fire starter.<\/li>\n<\/ol>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204559\/CNX_Precalc_Figure_10_03_0162.jpg\" alt=\"\" width=\"487\" height=\"217\" \/><\/p>\n<p class=\"wp-caption-text\">Cross-section of a travel-sized solar fire starter<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q424264\">Show Solution<\/span><\/p>\n<div id=\"q424264\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/p>\n<p>The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form [latex]{x}^{2}=4py[\/latex], where [latex]p>0[\/latex]. The igniter, which is the focus, is 1.7 inches above the vertex of the dish. Thus we have [latex]p=1.7[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&{x}^{2}=4py && \\text{Standard form of upward-facing parabola with vertex (0,0)} \\\\ &{x}^{2}=4\\left(1.7\\right)y && \\text{Substitute 1}\\text{.7 for }p. \\\\ &{x}^{2}=6.8y && \\text{Multiply}. \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left\">The dish extends [latex]\\frac{4.5}{2}=2.25[\/latex] inches on either side of the origin. We can substitute 2.25 for [latex]x[\/latex] in the equation from part (a) to find the depth of the dish.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&{x}^{2}=6.8y && \\text{Equation found in part (a)}. \\\\ &{\\left(2.25\\right)}^{2}=6.8y && \\text{Substitute 2}\\text{.25 for }x. \\\\ &y\\approx 0.74 && \\text{Solve for }y. \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left\">The dish is about 0.74 inches deep.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Balcony-sized solar cookers have been designed for families living in India. The top of a dish has a diameter of 1600 mm. The sun\u2019s rays reflect off the parabolic mirror toward the &#8220;cooker,&#8221; which is placed 320 mm from the base.<\/p>\n<ol style=\"list-style-type: lower-alpha\">\n<li>Find an equation that models a cross-section of the solar cooker. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane, and that the parabola opens to the right (i.e., has the <em>x<\/em>-axis as its axis of symmetry).<\/li>\n<li>Use the equation found in part (a) to find the depth of the cooker.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q535409\">Show Solution<\/span><\/p>\n<div id=\"q535409\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha\">\n<li>[latex]{y}^{2}=1280x[\/latex]<\/li>\n<li>The depth of the cooker is 500 mm<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=87077&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"650\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/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-2398\">\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>Revision and Adaptation. <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><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 86148, 86124, 87082. <strong>Authored by<\/strong>: Shahbazian, Roy. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" 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