{"id":106017,"date":"2021-12-08T14:24:57","date_gmt":"2021-12-08T20:24:57","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=106017"},"modified":"2026-03-28T11:46:48","modified_gmt":"2026-03-28T16:46:48","slug":"vertex-of-a-parabola","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/vertex-of-a-parabola\/","title":{"rendered":"Vertex of a Parabola"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_SmQ8IGxlOEg\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_SmQ8IGxlOEg\" data-source-videoID=\"SmQ8IGxlOEg\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Vertex of a Parabola Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Vertex of a Parabola\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_SmQ8IGxlOEg:hover {cursor:pointer;} img#videoThumbnailImage_SmQ8IGxlOEg {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/1862-thumb-final-3.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_SmQ8IGxlOEg\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_SmQ8IGxlOEg\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Vertex of a Parabola\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_SmQ8IGxlOEg\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_SmQ8IGxlOEg\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_SmQ8IGxlOEg\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction XYR_Function() {\n  var x = document.getElementById(\"XYR\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"XYR_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"XYR\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Introduction_to_Parabolas\" class=\"smooth-scroll\">Introduction to Parabolas<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#SecondDegree_Functions\" class=\"smooth-scroll\">Second-Degree Functions<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Function_Notation\" class=\"smooth-scroll\">Function Notation<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Standard_Form_of_a_Quadratic_Function\" class=\"smooth-scroll\">Standard Form of a Quadratic Function<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Graphing\" class=\"smooth-scroll\">Graphing<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#The_Axis_of_Symmetry\" class=\"smooth-scroll\">The Axis of Symmetry<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Substitution\" class=\"smooth-scroll\">Substitution<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Vertex_Form\" class=\"smooth-scroll\">Vertex Form<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#RealWorld_Problems\" class=\"smooth-scroll\">Real-World Problems<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Vertex_of_a_Parabola_Practice_Questions\" class=\"smooth-scroll\">Vertex of a Parabola Practice Questions<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hi, and welcome to this video on the vertex of a parabola.<\/p>\n<h2><span id=\"Introduction_to_Parabolas\" class=\"m-toc-anchor\"><\/span>Introduction to Parabolas<\/h2>\n<p>\nThe first curve most of us learn to graph is a parabola. Before that, we all learned to plot points on a number line and on a coordinate plane. Then we graphed the lines of linear equations, which only required two points and a ruler to keep things straight. Once we mastered that it was time for the parabola. <\/p>\n<h2><span id=\"SecondDegree_Functions\" class=\"m-toc-anchor\"><\/span>Second-Degree Functions<\/h2>\n<p>\nA parabola is the graph of a second-degree function. That means the variable, usually \\(x\\), has an exponent of two. The linear equations we learned to graph earlier are first degree, so the \\(x\\) has an exponent of 1, though it\u2019s usually an invisible exponent because we don\u2019t usually write that little one.<\/p>\n<p>A linear equation is most often found in slope-intercept form, like this: <\/p>\n<div class=\"examplesentence\">\\(y = 2x-7\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe can even write it with the exponent so we can see the \u201cfirst degreeness\u201d of the equation: <\/p>\n<div class=\"examplesentence\">\\(y = 2x^{1}-7\\)<\/div>\n<p>\n&nbsp;<br \/>\nIt\u2019s the same thing, and though it\u2019s a bit confusing to have an invisible exponent, we should be glad we don\u2019t have to write that little 1 every time we write a variable that is first degree! Here\u2019s an example of a second degree function: <\/p>\n<div class=\"examplesentence\">\\(f(x)=x^{2}+2x-3\\)<\/div>\n<p>\n&nbsp;<\/p>\n<h2><span id=\"Function_Notation\" class=\"m-toc-anchor\"><\/span>Function Notation<\/h2>\n<p>\nNotice that our \\(y\\) turned into an \\(f\\) of \\(x\\). This is called <strong>function notation<\/strong> and it looks a lot more complicated than it really is. It\u2019s just a way for us to keep track of our functions when we have more than one, such as when we have systems of equations.<\/p>\n<p>We can use \\(f\\) of \\(x\\) instead of \\(y\\) with linear equations too but most of us didn\u2019t learn function notation until after we learned linear equations so we\u2019re more accustomed to the \\(y\\). <\/p>\n<p>Okay, let\u2019s get back to second degree functions. One thing we definitely need to know is that they are often called <strong>quadratic functions<\/strong>. So whenever we hear \u201cquadratic function\u201d we need to remember that means second degree, which means the highest exponent on the variable is a 2. <\/p>\n<h2><span id=\"Standard_Form_of_a_Quadratic_Function\" class=\"m-toc-anchor\"><\/span>Standard Form of a Quadratic Function<\/h2>\n<p>\nNow compare this to the standard form of a quadratic function, which looks like this:<\/p>\n<div class=\"examplesentence\">\\(f(x)=ax^{2}+bx+c\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn this form \\(a\\), \\(b\\), and \\(c\\) are constants, which means they are numbers. In our example, \\(a = 1\\), \\(b = 2\\) and \\(c = -3\\). Notice that the coefficient of our \\(x^{2}\\)-term is invisible because it is a 1. This is very common with quadratic functions, so we need to remember that whenever we can\u2019t see our a it\u2019s because it\u2019s another one of those invisible 1s. <\/p>\n<h2><span id=\"Graphing\" class=\"m-toc-anchor\"><\/span>Graphing<\/h2>\n<p>\nBeing able to identify our \\(a\\), \\(b\\), and \\(c\\) are very important for finding the most important points on the parabola so we can graph it efficiently. It is true that we could graph our quadratic by finding points on the curve by plugging in values for \\(x\\) and getting our \\(y\\)-values. The problem is that it\u2019s not always easy to find the top or bottom that way. Here is a graph of our sample function with the most important points highlighted: <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-110847 aligncenter\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Screen-Shot-2022-01-25-at-7.19.20-AM.png\" alt=\"x-intercept, y-intercept, and vertex labeled on a parabola\" width=\"600\" height=\"658\" \/><\/p>\n<p>The \\(x\\)-intercepts are sometimes called \u201croots\u201d and can be found by factoring, completing the square, and the quadratic formula. The \\(y\\)-intercept is easily found when a quadratic function is in standard form \u2013 it\u2019s the \\(c\\)! In our example \\(c = -3\\) and on our graph we can see that our \\(y\\)-intercept takes place at the point \\((0,-3)\\). <\/p>\n<h3><span id=\"The_Vertex\" class=\"m-toc-anchor\"><\/span>The Vertex<\/h3>\n<p>\nThe last point is the <strong>vertex<\/strong>, which might just be the most important point on any parabola. The vertex is either the top or the bottom of a parabola, depending on which way it\u2019s facing. For our example, the vertex is the bottom of the parabola. We can easily find its location on the graph, but how do we find it when we only have the function and not the graph? We start by noticing something important about parabolas in general \u2013 they are symmetrical! <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-106053 aligncenter\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/12\/Vertex-of-Parabola-2-redo.png\" alt=\"Vertex at the bottom of parabola, intercepts X and Y on the right\" width=\"600\" height=\"600\" \/><\/p>\n<h2><span id=\"The_Axis_of_Symmetry\" class=\"m-toc-anchor\"><\/span>The Axis of Symmetry<\/h2>\n<p>\nWe can see the axis of symmetry shown as a red dashed line and can imagine folding our graph on this line and holding up our paper to the light so we can see the two halves of the parabola overlapping each other. Perfect symmetry. <\/p>\n<p>This axis of symmetry can be found using our \\(a\\), \\(b\\), and \\(c\\). Well, we don\u2019t really need \\(c\\) but since we already used it to find the \\(y\\)-intercept, it won\u2019t feel left out. Here\u2019s the formula for finding the axis of symmetry:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-b}{2a}\\)<\/div>\n<p>\n&nbsp;<br \/>\nRemember that when we have an equation written as \u201c\\(x\\) equals a number\u201d it is an equation of a <em>vertical line<\/em>. That\u2019s what we\u2019re trying to find here. <\/p>\n<p>Let\u2019s take a look at our sample function again so we can find \\(a\\), \\(b\\), and \\(c\\): <\/p>\n<div class=\"examplesentence\">\\(f(x)=x^{2}+2x-3\\)<\/div>\n<p>\n&nbsp;<br \/>\nRemembering that \\(a\\), \\(b\\), and \\(c\\) are shown, or not shown if invisible, from left to right and we can find \\(a = 1\\), \\(b = 2\\), and \\(c = -3\\). Now we can plug \\(a\\) and \\(b\\) into our formula for axis of symmetry:<\/p>\n<table class=\"ATable\" style=\"margin: auto;\">\n<tbody>\n<tr>\n<td>\\(x=\\frac{-b}{2a}\\)<\/td>\n<td>Formula for axis of symmetry<\/td>\n<\/tr>\n<tr>\n<td>\\(x=\\frac{-(2)}{2(1)}\\)<\/td>\n<td> Substitute values of \\(a\\) and \\(b\\) using parentheses<\/td>\n<\/tr>\n<tr>\n<td>\\(x=\\frac{-2}{2}\\)<\/td>\n<td>Evaluate numerator and denominator<\/td>\n<\/tr>\n<tr>\n<td>\\(x=-1\\)<\/td>\n<td>Simplify<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n&nbsp;<br \/>\nNow we see that our axis of symmetry is the line \\(x = \u20131\\), just like on our graph. <\/p>\n<h2><span id=\"Substitution\" class=\"m-toc-anchor\"><\/span>Substitution<\/h2>\n<p>\nNotice that the vertex sits on the axis of symmetry. That means the \\(x\\)-value of the vertex is the same as the \\(x\\)-value of any other point on the axis of symmetry. In this case, that means it\u2019s \u20131. Now we just need to find the \\(y\\)-value of the vertex. We do that by substituting our \\(x\\)-value into our function. <\/p>\n<div class=\"examplesentence\">\n\\(f(x)=x^{2}+2x-3\\)<br \/>\n\\(f(-1)=(-1)^{2}+2\\left ( -1 \\right )-3\\)<br \/>\n\\(f(-1)=1+(-2)-3\\)<br \/>\n\\(f(-1)=-4\\)<\/div>\n<p>\n&nbsp;<br \/>\nThat\u2019s the \\(y\\)-value of our vertex! Now we know our vertex is at the point \\((-1,-4)\\). And that\u2019s exactly where it is shown on our graph.<\/p>\n<h2><span id=\"Vertex_Form\" class=\"m-toc-anchor\"><\/span>Vertex Form<\/h2>\n<p>\nSo far we\u2019ve been using the standard form of our quadratic function. But there\u2019s another form called <strong>vertex form<\/strong>. We can convert from standard form to vertex form by completing the square, which is a whole big process that we don\u2019t actually need for this problem since we already found the vertex. The coordinates are all we need to write our function in vertex form. <\/p>\n<p>This is what vertex form looks like: <\/p>\n<div class=\"examplesentence\">\\(f(x)=a(x-h)^{2}+k\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe \\(a\\) is the same a from standard form, so it just carries over. Remember that the a is invisible for this function because its value is 1. The two new letters \\(h\\) and \\(k\\) are simply the \\(x\\)&#8211; and \\(y\\)-values of the vertex! Since our vertex for this problem is the point \\((-1,-4)\\) that means that \\(h=-1\\) and \\(k =-4\\). <\/p>\n<p>Now we can substitute the values for \\(a\\), \\(h\\), and \\(k\\) into our general vertex form equation: <\/p>\n<div class=\"examplesentence\">\\(f(x)=1(x-(-1))^{2}+(-4)\\)<\/div>\n<p>\n&nbsp;<br \/>\nNotice that we have a \u201csubtracting a negative\u201d situation happening inside the parentheses. When this happens, the two negatives cancel each other out and become a plus sign.<\/p>\n<div class=\"examplesentence\">\\(f(x)=1(x+1)^{2}-4\\)<\/div>\n<p>\n&nbsp;<br \/>\nJust like with our standard form, we don\u2019t have to write the \\(a\\) if it\u2019s a 1, though it\u2019s certainly not wrong if we do write it. Just know that it won\u2019t usually appear so the equation will probably look like this: <\/p>\n<div class=\"examplesentence\">\\(f(x)=(x+1)^{2}-4\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo far we\u2019ve found the vertex from a quadratic in standard form. How is the process different when our function is presented in vertex form? Let\u2019s try a new problem. <\/p>\n<p>Here\u2019s another quadratic function: <\/p>\n<div class=\"examplesentence\">\\(f(x)=(x-3)^{2}-7\\)<\/div>\n<p>\n&nbsp;<br \/>\nOur goal is to find the vertex. Since our function is in vertex form, we just need to identify \\(h\\) and \\(k\\). The key is to be careful with the signs. Here\u2019s our general vertex form equation again: <\/p>\n<div class=\"examplesentence\">\\(f(x)=a(x-h)^{2}+k\\)<\/div>\n<p>\n&nbsp;<br \/>\nBefore we extract our \\(h\\) and \\(k\\), we need to make sure our signs are correct. There should be a subtraction sign inside our parentheses and a plus sign after it. In our function, the subtraction sign is right but the plus sign isn\u2019t: <\/p>\n<div class=\"examplesentence\">\\(f(x)=(x-3)^{2}-7\\)<\/div>\n<p>\n&nbsp;<br \/>\nBut that\u2019s okay, we can simply rewrite the subtraction as adding a negative: <\/p>\n<div class=\"examplesentence\">\\(f(x)=\\left ( x-3 \\right )^{2}+(-7)\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow that we have our signs correct we can see that \\(h = 3\\) and \\(k = \u20137\\), so our vertex is \\((h,k)\\) which is \\((3,\u20137)\\). In this type of problem, we don\u2019t need to find our axis of symmetry before finding the vertex. If we want to know the equation for our axis of symmetry, we can find it from the vertex since the \\(h\\) gives us the value for our vertical line equation: \\(x=h\\) so \\(x=3\\) for this function. <\/p>\n<h2><span id=\"RealWorld_Problems\" class=\"m-toc-anchor\"><\/span>Real-World Problems<\/h2>\n<p>\nSo how is all of this useful? A lot of real world events can be modeled with quadratic functions. Gravity problems are the most common. Let\u2019s try one: <\/p>\n<p>A football player punts a football, and the height of the football in feet is modeled by the function \\(f(x)=-16(x-2)^{2}+53\\), where \\(x\\) is the time in seconds. Question 1: How long does it take for the football to reach its maximum height? Question 2: What is the maximum height of the football?<\/p>\n<p>The function is written in vertex form. The \\(a\\) for the function is -16. But we don\u2019t need a to solve this problem. All we need to know to answer the two questions is the coordinates of the vertex. Let\u2019s check our signs. Remember, when our equation is in vertex form, we want to subtract our \\(h\\) value and add our \\(k\\) value. When looking at our equation for our football, we see that our signs are correct as written.<\/p>\n<div class=\"examplesentence\">\\(f(x)=a(x-h)^{2}+k\\)<br \/>\n\\(f(x)=-16(x-3)^{2}+53\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo now we can extract our \\(h\\) and \\(k\\). <\/p>\n<div class=\"examplesentence\">\\(f(x)=-16(x-3)^{2}+53\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe see that \\(h=3\\) and \\(k=53\\). So the coordinates of the vertex are \\((3,53)\\). But how does that answer the two questions? The problem states that \u201c\\(x\\) is time in seconds\u201d. So the \\(x\\)-value of our vertex is the time it took for the football to reach its maximum height. In this case it took 3 seconds. The function models the height of the football so the \\(y\\)-value of the vertex is the maximum height, 53 feet! This is what the graph of the function looks like: <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/12\/Vertex-of-a-Parabola-3.png\" alt=\"Vertex of a parabola; h=3, k=53, and vertex is (3, 53)\" width=\"600\" height=\"600\" class=\"aligncenter size-full wp-image-106041\" \/><\/p>\n<p>The football starts at 0 seconds and 0 feet, then reaches its maximum height of 53 feet after 3 seconds and finally comes back down to the ground after a total of 6 seconds. We know that because our parabola is symmetrical! So if it took 3 seconds to go up, it took 3 more to come down. <\/p>\n<p>That\u2019s it for the vertex of a parabola. Thanks for watching, and happy studying!<\/p>\n<ul class=\"citelist\">\n<li><a href=\"https:\/\/www.storyofmathematics.com\/function-notation\"target=\"_blank\">\u201cFunction Notation \u2013 Explanation &#038; Examples.\u201d n.d. <\/a><\/li>\n<li><a href=\"https:\/\/www.mathsisfun.com\/algebra\/quadratic-equation.html\"target=\"_blank\">\u201cQuadratic Equations.\u201d Mathsisfun.com<\/a><\/li>\n<li><a href=\"https:\/\/www.mathsisfun.com\/geometry\/parabola.html\"target=\"_blank\">\u201cParabola.\u201d Mathsisfun.com<\/a><\/li>\n<li><a href=\"https:\/\/www.mathopenref.com\/quadraticexplorer.html\"target=\"_blank\">\u201cQuadratic Curve and Graph Display (Standard Form)- Math Open Reference.\u201d n.d. <\/a><\/li>\n<\/ul>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Vertex_of_a_Parabola_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Vertex of a Parabola Practice Questions<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nFind the vertex of the following quadratic equation:<\/p>\n<div class=\"yellow-math-quote\">\\(f(x)=x^2+6x+11\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-1-1\">\\((-3,2)\\)<\/div><div class=\"PQ\"  id=\"PQ-1-2\">\\((3,-2)\\)<\/div><div class=\"PQ\"  id=\"PQ-1-3\">\\((-3,3)\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">\\((3,4)\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The equation is \\(f(x)=x^2+6x+11\\) in standard form. The formula \\(x=-\\frac{b}{2a}\\) can be used to determine the \\(x\\)-coordinate of the vertex of the parabola.<\/p>\n<p>In the equation \\(f(x)=x^2+6x+11\\), \\(a=1\\), \\(b=6\\), and \\(c=11\\).<\/p>\n<p>Substitute these values into the formula.<\/p>\n<p style=\"text-align: center; line-height: 50px\">\\(x=-\\large{\\frac{b}{2a}}\\)<br \/>\n\\(x=-\\large{\\frac{(6)}{2(1)}}\\)<br \/>\n\\(x=-\\large{\\frac{6}{2}}\\)<br \/>\n\\(x=-3\\)<\/p>\n<p>The \\(x\\)-value of the vertex is \u22123. Plug this in for \\(x\\) in the original equation in order to determine the \\(y\\)-value of the vertex.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(f(x)=x^2+6x+11\\)<br \/>\n\\(f(-3)={(-3)}^2+6(-3)+11\\)<br \/>\n\\(f(-3)=9-18+11\\)<br \/>\n\\(f(-3)=2\\)<\/p>\n<p>When \\(x\\) is \u22123, \\(y\\) is 2. This means the vertex of the parabola is \\((-3,2)\\).<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWhat is the vertex of the parabola formed by the following equation?<\/p>\n<div class=\"yellow-math-quote\">\\(f(x)=3x^2+12x+7\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-2-1\">\\((-2,-5)\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\((2,-5)\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\((-2,5)\\)<\/div><div class=\"PQ\"  id=\"PQ-2-4\">\\((2,5)\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The function \\(f(x)=3x^2+12x+7\\) is in standard form, so we can identify the values of \\(a\\), \\(b\\), and \\(c\\):<\/p>\n<ul style=\"list-style-type: none; margin-left: 1.2em\">\n<li>\\(a=3\\)<\/li>\n<li>\\(b=12\\)<\/li>\n<li>\\(c=7\\)<\/li>\n<\/ul>\n<p>These values can be substituted into the vertex formula, \\(x=-\\frac{b}{2a}\\). <\/p>\n<p style=\"text-align: center; line-height: 50px\">\\(x=-\\large{\\frac{b}{2a}}\\)<br \/>\n\\(x=-\\large{\\frac{12}{2(3)}}\\)<br \/>\n\\(x=-\\large{\\frac{12}{6}}\\)<br \/>\n\\(x=-2\\)<\/p>\n<p>So, \u22122 is the \\(x\\)-value of the vertex \\((-2,y)\\). To find the \\(y\\)-value of the vertex, plug in \u22122 for every value of \\(x\\) in the original equation.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(f(x)=3x^2+12x+7\\)<br \/>\n\\(f(-2)=3{(-2)}^2+12(-2)+7\\)<br \/>\n\\(f(-2)=3(4)+12(-2)+7\\)<br \/>\n\\(f(-2)=12-24+7\\)<br \/>\n\\(f(-2)=-5\\)<\/p>\n<p>The \\(y\\)-value of the vertex is \u22125. The vertex of the parabola is \\((-2,-5)\\).<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nDetermine the vertex of the following function:<\/p>\n<div class=\"yellow-math-quote\">\\(f(x)=(x+2)^2-1\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\((-1,-1)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-2\">\\((-2,-1)\\)<\/div><div class=\"PQ\"  id=\"PQ-3-3\">\\((2,-1)\\)<\/div><div class=\"PQ\"  id=\"PQ-3-4\">\\((-2,-2)\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The function \\(f(x)=(x+2)^2-1\\) is in vertex form. This is convenient because the vertex is expressed within the equation as \\(h\\) and \\(k\\).<\/p>\n<p>Remember, vertex form is \\(f(x)=a(x-h)^2+k\\), where \\((h,k)\\) represents the vertex.<\/p>\n<p>In the equation \\(f(x)=(x+2)^2-1\\), \\(h\\) is \u22122, and \\(k\\) is \u22121. This means the vertex is \\((-2,-1)\\).<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nA football player is attempting to score a field goal. The trajectory of the ball is traced and plotted onto a graph that forms a parabola. The equation of the parabola in vertex form is \\(f(x)=-16(x-2)^2+40\\), where \\(x\\) represents the time since the ball was kicked in seconds and \\(f(x)\\) represents the height of the ball in feet.<\/p>\n<p>What was the highest point of the football on its path through the air? Express your answer as an ordered pair and describe the maximum height.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">\\((3,190)\\)<br>\r\nThe maximum height of the football was 190 feet.<\/div><div class=\"PQ\"  id=\"PQ-4-2\">\\((25,250)\\)<br>\r\nThe maximum height of the football was 250 feet. <\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-3\">\\((2,40)\\)<br>\r\nThe maximum height of the football was 40 feet. <\/div><div class=\"PQ\"  id=\"PQ-4-4\">\\((5,50)\\)<br>\r\nThe maximum height of the football was 50 feet.<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The equation that represents the football\u2019s trajectory is \\(f(x)=-16(x-2)^2+40\\).<\/p>\n<p>The highest point on this parabola is the vertex. To determine the vertex, locate \\(h\\) and \\(k\\). Vertex form contains the vertex \\((h,k)\\) within the equation \\(f(x)=a(x-h)^2+k\\).<\/p>\n<p>In the equation \\(f(x)=-16(x-2)^2+40\\), \\(h\\) is 2, and \\(k\\) is 40.<\/p>\n<p>The vertex of the parabola is \\((2,40)\\). This means that the football\u2019s highest point took place after 2 seconds, at 40 feet.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #5:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nA bottle rocket\u2019s trajectory through the air creates a parabola when graphed. The equation that expresses the rocket\u2019s path is \\(f(x)=-16x^2+64x\\), where \\(x\\) represents time in seconds and \\(f(x)\\) represents distance in feet.<\/p>\n<p>According to the function, how many seconds did it take to reach its highest point? What was the highest point the bottle rocket reached?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">After 5 seconds, the rocket reached a high point of 60 feet.<\/div><div class=\"PQ\"  id=\"PQ-5-2\">After 3 seconds, the rocket reached a high point of 43 feet. <\/div><div class=\"PQ\"  id=\"PQ-5-3\">After 8 seconds, the rocket reached a high point of 95 feet.<\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-4\">After 2 seconds, the rocket reached a high point of 64 feet.<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-5\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-5-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The rocket\u2019s path forms a parabola that can be expressed with the equation \\(f(x)=-16x^2+64x\\). This equation is in standard form, so the \\(x\\)-coordinate of the vertex can be identified using the formula \\(x=-\\frac{b}{2a}\\).<\/p>\n<p>In the equation \\(f(x)=-16x^2+64x\\), \\(a=-16\\), \\(b=64\\), and \\(c=0\\). Substitute \\(a\\) and \\(b\\) into the formula. <\/p>\n<p style=\"text-align: center; line-height: 50px\">\\(x=-\\large{\\frac{b}{2a}}\\)<br \/>\n\\(x=-\\large{\\frac{64}{2(-16)}}\\)<br \/>\n\\(x=-\\large{\\frac{64}{-32}}\\)<br \/>\n\\(x=2\\)<\/p>\n<p>The variable \\(x\\) represents time in seconds, so the rocket was at its highest point after 2 seconds. Plug in 2 for \\(x\\) in order to calculate the height of the rocket.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(f(x)=-16x^2+64x\\)<br \/>\n\\(f(2)=-16{(2)}^2+64(2)\\)<br \/>\n\\(f(2)=-16(4)+64(2)\\)<br \/>\n\\(f(2)=-64+128\\)<br \/>\n\\(f(2)=64\\)<\/p>\n<p>This means that after 2 seconds the rocket\u2019s height was 64 feet. The vertex of the parabola is \\((2,64)\\), which is the highest point of the graph.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-5-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/algebra-i\/\">Return to Algebra I Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Algebra I Videos<\/p>\n","protected":false},"author":22,"featured_media":109329,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-106017","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-quadratics-videos","7":"page_type-video","8":"content_type-practice-questions","9":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/106017","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/22"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=106017"}],"version-history":[{"count":8,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/106017\/revisions"}],"predecessor-version":[{"id":286006,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/106017\/revisions\/286006"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/109329"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=106017"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}