{"id":1919,"date":"2015-11-12T18:30:43","date_gmt":"2015-11-12T18:30:43","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&#038;p=1919"},"modified":"2015-11-12T18:30:43","modified_gmt":"2015-11-12T18:30:43","slug":"graphing-parabolas-with-vertices-not-at-the-origin","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/graphing-parabolas-with-vertices-not-at-the-origin\/","title":{"raw":"Graphing Parabolas with Vertices Not at the Origin","rendered":"Graphing Parabolas with Vertices Not at the Origin"},"content":{"raw":"<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].\n\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.\n<\/p><div class=\"textbox\">\n<h3>A General Note: Standard Forms of Parabolas with Vertex (<em>h<\/em>, <em>k<\/em>)<\/h3>\nThe table\u00a0and Figure 9\u00a0summarize the standard features of parabolas with a vertex at a point [latex]\\left(h,k\\right)[\/latex].\n<table id=\"fs-id1219907\" summary=\"..\"><tbody><tr><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 Latus Rectum<\/strong><\/td>\n<\/tr><tr><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><tr><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><\/tbody><\/table>\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202326\/CNX_Precalc_Figure_10_03_0092.jpg\" alt=\"\" width=\"975\" height=\"901\" data-media-type=\"image\/jpg\"\/><b>Figure 9.<\/b> (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>\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><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\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 latus rectum.\n<ol><li>If the equation is in the form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex], then:\n<ul><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\t<li>use the value of [latex]k[\/latex] to determine the axis of symmetry, [latex]y=k[\/latex]<\/li>\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>\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>\n\t<li>use [latex]h[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]x=h-p[\/latex]<\/li>\n\t<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the latus rectum, [latex]\\left(h+p,k\\pm 2p\\right)[\/latex]<\/li>\n<\/ul><\/li>\n\t<li>If the equation is in the form [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex], then:\n<ul><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\t<li>use the value of [latex]h[\/latex] to determine the axis of symmetry, [latex]x=h[\/latex]<\/li>\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>\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>\n\t<li>use [latex]k[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]y=k-p[\/latex]<\/li>\n\t<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the latus rectum, [latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/li>\n<\/ul><\/li>\n<\/ol><\/li>\n\t<li>Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.<\/li>\n<\/ol><\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Graphing a Parabola with Vertex (<em>h<\/em>, <em>k<\/em>) and Axis of Symmetry Parallel to the <em>x<\/em>-axis<\/h3>\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>latus rectum<\/strong>.\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\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:\n<div>\n<ul><li>the vertex is [latex]\\left(h,k\\right)=\\left(-3,1\\right)[\/latex]<\/li>\n\t<li>the axis of symmetry is [latex]y=k=1[\/latex]<\/li>\n\t<li>[latex]-16=4p[\/latex], so [latex]p=-4[\/latex]. Since [latex]p&lt;0[\/latex], the parabola opens left.<\/li>\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>\n\t<li>the equation of the directrix is [latex]x=h-p=-3-\\left(-4\\right)=1[\/latex]<\/li>\n\t<li>the endpoints of the latus rectum 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><\/div>\nNext we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202327\/CNX_Precalc_Figure_10_03_0102.jpg\" alt=\"\" width=\"487\" height=\"480\" data-media-type=\"image\/jpg\"\/><b>Figure 10<\/b>[\/caption]\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 5<\/h3>\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 latus rectum.\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-25\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>\n<div\/>\n<div class=\"textbox shaded\">\n<h3>Example 6: Graphing a Parabola from an Equation Given in General Form<\/h3>\nGraph [latex]{x}^{2}-8x - 28y - 208=0[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\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.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}-8x - 28y - 208=0\\hfill \\\\ \\text{ }{x}^{2}-8x=28y+208\\hfill \\\\ \\text{ }{x}^{2}-8x+16=28y+208+16\\hfill \\\\ \\text{ }{\\left(x - 4\\right)}^{2}=28y+224\\hfill \\\\ \\text{ }{\\left(x - 4\\right)}^{2}=28\\left(y+8\\right)\\hfill \\\\ \\text{ }{\\left(x - 4\\right)}^{2}=4\\cdot 7\\cdot \\left(y+8\\right)\\hfill \\end{array}[\/latex]<\/div>\nIt follows that:\n<div>\n<ul><li>the vertex is [latex]\\left(h,k\\right)=\\left(4,-8\\right)[\/latex]<\/li>\n\t<li>the axis of symmetry is [latex]x=h=4[\/latex]<\/li>\n\t<li>since [latex]p=7,p&gt;0[\/latex] and so the parabola opens up<\/li>\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>\n\t<li>the equation of the directrix is [latex]y=k-p=-8 - 7=-15[\/latex]<\/li>\n\t<li>the endpoints of the latus rectum 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><\/div>\nNext we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202329\/CNX_Precalc_Figure_10_03_0122.jpg\" alt=\"\" width=\"487\" height=\"258\" data-media-type=\"image\/jpg\"\/><b>Figure 11<\/b>[\/caption]\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 6<\/h3>\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 latus rectum.\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-25\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>\n\u00a0","rendered":"<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.\n<\/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\u00a0and Figure 9\u00a0summarize the standard features of parabolas with a vertex at a point [latex]\\left(h,k\\right)[\/latex].<\/p>\n<table id=\"fs-id1219907\" 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 Latus Rectum<\/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-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202326\/CNX_Precalc_Figure_10_03_0092.jpg\" alt=\"\" width=\"975\" height=\"901\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9.<\/b> (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 latus rectum.\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 latus rectum, [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 latus rectum, [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 latus rectum, and draw a smooth curve to form the parabola.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: 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>latus rectum<\/strong>.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\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<div>\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 latus rectum 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<\/div>\n<p>Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.<\/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-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202327\/CNX_Precalc_Figure_10_03_0102.jpg\" alt=\"\" width=\"487\" height=\"480\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 5<\/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 latus rectum.<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-25\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\n<div>\n<div class=\"textbox shaded\">\n<h3>Example 6: 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 latus rectum.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\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<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}-8x - 28y - 208=0\\hfill \\\\ \\text{ }{x}^{2}-8x=28y+208\\hfill \\\\ \\text{ }{x}^{2}-8x+16=28y+208+16\\hfill \\\\ \\text{ }{\\left(x - 4\\right)}^{2}=28y+224\\hfill \\\\ \\text{ }{\\left(x - 4\\right)}^{2}=28\\left(y+8\\right)\\hfill \\\\ \\text{ }{\\left(x - 4\\right)}^{2}=4\\cdot 7\\cdot \\left(y+8\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p>It follows that:<\/p>\n<div>\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 latus rectum 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<\/div>\n<p>Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.<\/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-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202329\/CNX_Precalc_Figure_10_03_0122.jpg\" alt=\"\" width=\"487\" height=\"258\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 11<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 6<\/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 latus rectum.<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-25\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\n<p>\u00a0<\/p><\/div>\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-1919\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><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\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":276,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"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-1919","chapter","type-chapter","status-publish","hentry"],"part":1904,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1919","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1919\/revisions"}],"predecessor-version":[{"id":2201,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1919\/revisions\/2201"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1904"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1919\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/media?parent=1919"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1919"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1919"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/license?post=1919"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}