{"id":4345,"date":"2013-06-29T04:08:36","date_gmt":"2013-06-29T04:08:36","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=4345"},"modified":"2026-03-28T11:37:33","modified_gmt":"2026-03-28T16:37:33","slug":"solutions-of-a-quadratic-equation-on-a-graph","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/solutions-of-a-quadratic-equation-on-a-graph\/","title":{"rendered":"Solutions of a Quadratic Equation on a Graph"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_uKRcGzChyus\" 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_uKRcGzChyus\" data-source-videoID=\"uKRcGzChyus\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Solutions of a Quadratic Equation on a Graph Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Solutions of a Quadratic Equation on a Graph\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_uKRcGzChyus:hover {cursor:pointer;} img#videoThumbnailImage_uKRcGzChyus {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/1861-thumb-final-3-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_uKRcGzChyus\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_uKRcGzChyus\" 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\",\"Solutions of a Quadratic Equation on a Graph\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_uKRcGzChyus\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_uKRcGzChyus\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_uKRcGzChyus\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction 5zp_Function() {\n  var x = document.getElementById(\"5zp\");\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=\"5zp_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=\"5zp\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#A_Review_of_the_Basics\" class=\"smooth-scroll\">A Review of the Basics<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Solving_Quadratic_Equations_with_Graphs\" class=\"smooth-scroll\">Solving Quadratic Equations with Graphs<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Quadratic_Functions_in_the_Real_World\" class=\"smooth-scroll\">Quadratic Functions in the Real World<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Quadratic_Equation_on_a_Graph_Practice_Questions\" class=\"smooth-scroll\">Quadratic Equation on a Graph 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>Hello, and welcome to this video about solutions of a quadratic on a graph!<\/p>\n<p>Today we\u2019ll learn how to find solutions to a quadratic function by looking at its graph. We\u2019ll also talk about how to graph a quadratic equation and analyze the graph to find solutions. <\/p>\n<h2><span id=\"A_Review_of_the_Basics\" class=\"m-toc-anchor\"><\/span>A Review of the Basics<\/h2>\n<p>\nBefore we get started, let\u2019s review a few things. First, a <strong>quadratic function<\/strong> is a polynomial function, and its highest degree term is of the second degree. The graph of a quadratic function is a two-dimensional curve called a <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/parabolas\/\">parabola<\/a>. The parabola can open upward or downward and can vary in width. However, all parabolas share the same U-shape.<\/p>\n<p>A parabola is symmetric over an invisible line called the <strong>axis of symmetry<\/strong>. The <strong>vertex<\/strong> is the point on the parabola where the graph intersects its axis of symmetry. In the parabola shown, \\(y=x^{2}\\), the vertex is the point at the bottom of the U-shaped parabola, seen here at point \\((0,0)\\). <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-109266\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solutions-of-a-Quadratic-on-a-Graph-Image-1.png-non-fuzzy.png\" alt=\"parabola y equals x squared\" width=\"500\" height=\"294\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solutions-of-a-Quadratic-on-a-Graph-Image-1.png-non-fuzzy.png 500w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solutions-of-a-Quadratic-on-a-Graph-Image-1.png-non-fuzzy-300x176.png 300w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/p>\n<p>Notice that this parabola is symmetrical with respect to the y-axis, which is the axis of symmetry on this graph. The vertex is the point at the bottom of the U-shape. <\/p>\n<p>If the graph of a quadratic function crosses the x-axis at two points, then the equation has two real rational solutions. These solutions are also called x-<strong>intercepts<\/strong> or <strong>roots<\/strong>. If it touches the x-axis at one point, it has one real rational solution. If the graph does not intersect the x-axis, then the equation has no real solutions. <\/p>\n<p>A <strong>quadratic equation<\/strong> in standard form is written as \\(ax^{2}+bx+c=0\\), where \\(a\\neq 0\\) and \\(a\\), \\(b\\), and \\(c\\) are all real numbers. We can solve a quadratic equation by <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/factoring-quadratic-equations\/\">factoring<\/a>, <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/completing-the-square\/\">completing the square<\/a>, using the <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/using-the-quadratic-formula\/\">quadratic formula<\/a>, or analyzing the graph of its function. <\/p>\n<h2><span id=\"Solving_Quadratic_Equations_with_Graphs\" class=\"m-toc-anchor\"><\/span>Solving Quadratic Equations with Graphs<\/h2>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<p>\nConsider the graph for \\(y=x^{2}+x-6\\). We can use this graph to solve the equation \\(x^{2}+x-6=0\\). Notice that \\(y\\) is replaced with \\(0\\). Replacing \\(y\\) with \\(0\\) indicates that we are solving for \\(x\\) by identifying the \\(x\\)-intercepts. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solutions-of-a-Quadratic-on-a-Graph-Image-2.png-non-fuzzy.png\" alt=\"graph of y equals x squared plus x minus six\" width=\"500\" height=\"255\" class=\"aligncenter size-full wp-image-109269\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solutions-of-a-Quadratic-on-a-Graph-Image-2.png-non-fuzzy.png 500w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solutions-of-a-Quadratic-on-a-Graph-Image-2.png-non-fuzzy-300x153.png 300w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/p>\n<p>Looking at the graph, we can see that this quadratic function has two solutions. The graph intersects the \\(x\\)-axis at the points \\((-3,0)\\) and \\((2,0)\\), so the solutions for \\(x\\) are -3 and 2. In this example, there are two real rational solutions. The graph intersects the \\(x\\)-axis exactly two times. <\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nLet\u2019s look at another example. Consider the graph for \\(y=-x^{2}+4\\). Notice that the value for a is negative, so this parabola opens downward. We can use the graph to solve the equation \\(-x^{2}+4=0\\). <\/p>\n<p>The graph intersects the \\(x\\)-axis at points \\((-2,0)\\) and \\((2,0)\\), so \\(x=-2\\), 2. In this example, there are two real rational solutions. The graph intersects the \\(x\\)-axis exactly two times. <\/p>\n<h3><span id=\"Example_3\" class=\"m-toc-anchor\"><\/span>Example #3<\/h3>\n<p>\nLet\u2019s try another one. Consider the graph for \\(y=x^{2}-2x+1\\). We can use the graph to solve the equation \\(x^{2}-2x+1=0\\). <\/p>\n<p>Notice that this graph doesn\u2019t pass through the \\(x\\)-axis. Instead, its vertex touches the \\(x\\)-axis at the point \\((1,0)\\). In this case, there is only one real rational solution, \\(x=1\\). <\/p>\n<h3><span id=\"Example_4\" class=\"m-toc-anchor\"><\/span>Example #4<\/h3>\n<p>\nLet\u2019s try one more. Consider the graph for \\(y=x^{2}+1\\). We can use this graph to identify possible solutions to the equation \\(x^{2}+1=0\\). <\/p>\n<p>Looking at this graph, we see that it doesn\u2019t pass through the \\(x\\)-axis at all, which means there are no x-intercepts. In other words, there are no real solutions to this equation. <\/p>\n<h2><span id=\"Quadratic_Functions_in_the_Real_World\" class=\"m-toc-anchor\"><\/span>Quadratic Functions in the Real World<\/h2>\n<p>\nQuadratic functions occur in many real-world situations. Many questions about time, speed, and distance require quadratic equations to solve. An example of this is throwing a ball up into the air. Once a ball is thrown, it slows and eventually falls back down. When thrown, the ball\u2019s path travels in an upside-down U-shape, or parabola. A quadratic equation can be used to find the position of the ball for the amount of time it\u2019s in the air. <\/p>\n<p>Another example is a boat traveling against a river current. If a boat travels 10 miles upstream and turns around to return to its starting point, its path will be in a U-shaped parabola. A quadratic equation can be used to find the boat\u2019s speed and the length of the trip. <\/p>\n<p>There are lots of other problem-solving instances in which quadratic equations are helpful. Understanding them helps us apply mathematical concepts in the real world. Let\u2019s take a look at a story problem that we can solve by graphing a quadratic function. <\/p>\n<p>The width of a rectangular field is 2 meters less than the length. The area is \\(48\\text{ m}^{2}\\). Find the dimensions of the field.<\/p>\n<p>First, start with the formula for the area of a rectangle, which is \\(A=\\text{ length}\\times \\text{ width}\\). The word problem doesn\u2019t give the field\u2019s length, so use the variable \\(x\\) to represent length. Since the width is 2 meters less than the length, use the expression \\(x-2\\) to represent the width. <\/p>\n<p>Next, substitute the values into the area formula. So we said:<\/p>\n<div class=\"examplesentence\">\\(\\text{ A}=\\text{ lw}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe were given that our area is equal to 48, and we said that length is \\(x\\) and width is \\(x-2\\).<\/p>\n<div class=\"examplesentence\">\\(48=x(x-2)\\)<\/div>\n<p>\n&nbsp;<br \/>\nFrom here, simplify the equation and write it as a quadratic equation in standard form.<\/p>\n<p>So the 48 is going to stay the same for right now and we\u2019re going to distribute the \\(x\\) into the parentheses \\((x-2)\\). So \\(x\\cdot x=x^{2}\\) and \\(x\\cdot (-2)=-2x\\).<\/p>\n<div class=\"examplesentence\">\\(48=x^{2}-2x\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow remember, we want 0 on one side so we\u2019re going to subtract 48 from both sides.<\/p>\n<div class=\"examplesentence\">\\(48-48=x^{2}-2x-48\\)<\/div>\n<p>\n&nbsp;<br \/>\nThat gives us:<\/p>\n<div class=\"examplesentence\">\\(0=x^{2}-2x-48\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow remember standard form technically has the 0 on the other side, so we&#8217;re just going to rewrite it as:<\/p>\n<div class=\"examplesentence\">\\(x^{2}-2x-48=0\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we\u2019re in proper standard form.<\/p>\n<p>Now that we have the area expressed as a quadratic equation, we can create a graph to interpret possible solutions. To graph the parabola, we need the coordinates for its \\(x\\)-intercepts and its vertex. <\/p>\n<p>Let\u2019s start by identifying the coordinates for the \\(x\\)-intercepts. We can find these by factoring the equation \\(x^{2}-2x-48=0\\). Quadratic equations can be factored as \\((x-p)(x-q)=0\\), where \\(p\\) and \\(q\\) are solutions to the equation. <\/p>\n<p>To factor the equation, identify two numbers with a sum of -2 (\\(b\\)) and a product of -48 (\\(c\\)). Two numbers with a sum of -2 and a product of -48 are -8 and 6. Therefore, we can factor this quadratic equation by writing it as \\((x-8)(x+6)=0\\). Now notice that earlier we said that it was \\((x-p)(x-q)\\), but here we have a plus sign. Well this is really the same as \\((x-(-6))\\), which then simplifies to \\((x+6)\\) because of the subtracting a negative. <\/p>\n<p>Next, substitute values for \\(x\\) that make each equation true. Since \\(8-8=0\\), \\(x=8\\). Since \\(-6+6=0\\), \\(x=-6\\). The solutions for \\(x\\) are 8 and -6. So the coordinates for these \\(x\\)-intercepts are (8,0) and (-6,0). So the function passes through these two points. <\/p>\n<p>Now that we\u2019ve identified the coordinates of the \\(x\\)-intercepts, we can find the coordinates for the vertex. Start by finding the \\(x\\)-coordinate of the vertex. Since the vertex is the midpoint of the two x-intercepts, the vertex\u2019s \\(x\\)-coordinate is the sum of the two \\(x\\)-intercepts divided by 2. So let&#8217;s do that. We have our two x-intercepts, 8 and -6, and to find the average of these two points, we&#8217;ll add them together and divide by 2.<\/p>\n<div class=\"examplesentence\">\\(\\frac{8+(-6)}{2}=\\frac{2}{2}=1\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo the \\(x\\)-coordinate of our vertex is 1. To find the \\(y\\)-coordinate of the vertex, substitute 1 in for \\(x\\) into our original equation, which is \\(y=x^{2}-2x-48\\). So if we plug in 1 anywhere we see an \\(x\\), we\u2019ll get:<\/p>\n<div class=\"examplesentence\">\\(y=(1)^{2}-2(1)-48\\)<br \/>\n\\(y=1-2-48\\)<br \/>\n\\(y=-1-48\\)<br \/>\n\\(y=-49\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo our y-coordinate is equal to -49. Therefore, the vertex is at point \\((1,-49)\\). <\/p>\n<p>Now that we\u2019ve identified the \\(x\\)-intercepts and the vertex, we can use these points to create a graph. Plot the points and join them by a smooth curve, as shown. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solutions-of-a-Quadratic-on-a-Graph-Image-3.png-non-fuzzy.png\" alt=\"parabola graph with a vertex of (1,-49)\" width=\"500\" height=\"252\" class=\"aligncenter size-full wp-image-109272\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solutions-of-a-Quadratic-on-a-Graph-Image-3.png-non-fuzzy.png 500w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solutions-of-a-Quadratic-on-a-Graph-Image-3.png-non-fuzzy-300x151.png 300w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/p>\n<p>Now that we\u2019ve graphed the function, let\u2019s reconsider the context of the word problem. Recall that we are trying to identify the dimensions of a rectangular field. Since meters can\u2019t be measured in negative numbers, the only solution for \\(x\\) that makes sense here is 8. Recall that \\(x\\) represents the field\u2019s length, which is 8 meters. Recall that the width is 2 meters less than the length. Since \\(8-2=6\\), the width of the field is 6 meters. The field is 8 meters long and 6 meters wide. <\/p>\n<h3><span id=\"Example_5\" class=\"m-toc-anchor\"><\/span>Example #5<\/h3>\n<p>\nNow it\u2019s your turn. I\u2019m going to give you a quadratic function and you need to factor it, identify the coordinates for its x-intercepts, identify the coordinates for its vertex, and graph it. This problem is challenging because it involves several steps, but I know you can handle it! The quadratic function is \\(y=x^{2}-6x-16\\). Pause the video here and try it yourself. <\/p>\n<p>Let\u2019s take a look at this quadratic function together. The first step is factoring the equation \\(x^{2}-6x-16=0\\). To do so, we need to identify two numbers that have a sum of -6 (\\(b\\)) and a product of -16 (\\(c\\)). Then, we\u2019ll write these numbers into the factored equation \\((x\\text{    })(x\\text{    })=0\\). The correct numbers are -8 and 2 because \\(-8\\times 2=-16\\), and \\(-8+2=-6\\). So we\u2019re going to plug in -8 and 2: <\/p>\n<div class=\"examplesentence\">\\((x-8)(x+2)=0\\)<\/div>\n<p>\n&nbsp;<br \/>\nNext, substitute a value for \\(x\\) in both subtraction problems that makes the equation true. Since \\(8-8=0\\), \\(x=8\\). Since \\(-2+2=0\\), \\(x=-2\\). Therefore, the solutions for \\(x\\) are 8 and -2. We can write the \\(x\\)-intercept values as the coordinate points (8,0) and (-2,0). <\/p>\n<p>Now that we\u2019ve identified the \\(x\\)-intercepts, we need to find the coordinates for the vertex, starting with its \\(x\\)-coordinate. Recall that the \\(x\\)-coordinate of the vertex is the midpoint of the \\(x\\)-intercepts. So for our vertex, to find the \\(x\\)-coordinate all we have to do is add these two values and divide by 2.<\/p>\n<div class=\"examplesentence\">\\(\\frac{8+(-2)}{2}=\\frac{6}{2}=3\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo the \\(x\\)-coordinate of the vertex is 3. To find the \\(y\\)-coordinate, substitute 3 for \\(x\\) into our original equation \\(y=x^{2}-6x-16\\):<\/p>\n<div class=\"examplesentence\">\\(y=(3)^{2}-6(3)-16\\)<br \/>\n\\(y=9-18-16\\)<br \/>\n\\(y=-9-16\\)<br \/>\n\\(y=-25\\)<\/div>\n<p>\n&nbsp;<br \/>\nTherefore, the vertex is at point \\((3,-25)\\). <\/p>\n<p>Now that we\u2019ve identified the \\(x\\)-intercepts and the vertex, we can use points \\((8,0)\\), \\((-2,0)\\), and \\((3,-25)\\) to create a graph. Plot the points and join them by a smooth curve. <\/p>\n<p>I hope this video about solutions of a quadratic on a graph was helpful. Thanks for watching, and happy studying! <\/p>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Quadratic_Equation_on_a_Graph_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Quadratic Equation on a Graph 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 \/>\nSolve \\(x^2+3x-28=0\\) by using the graph shown below:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Quadratic-Equation-Graph-Example-1.svg\" alt=\"Graph of a downward-opening parabola with labeled points at (-7, 0), (4, 0), (0, -28), and vertex at (-1.5, -30.25) on a grid.\" width=\"349\" height=\"729\" class=\"aligncenter size-full wp-image-287713\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\(x=-28\\)<\/div><div class=\"PQ\"  id=\"PQ-1-2\">\\(x=-4\\) and \\(x=7\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-3\">\\(x=-7\\) and \\(x=4\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">No real solution<\/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>Looking at the graph, we see that the quadratic function crosses the \\(x\\)-axis at two points. This means that there are two real solutions to this equation.<\/p>\n<p>The graph intersects the \\(x\\)-axis at points \\((-7,0)\\) and \\((4,0)\\), so the solutions for \\(x\\) are \u22127 and 4. <\/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 \/>\nSolve \\(x^2+10x+25=0\\) by using the graph shown below: <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Quadratic-Equation-Graph-Example-2.1.svg\" alt=\"Graph of a red upward parabola passing through (-5, 0) and (0, 25), with both points marked and labeled on a grid.\" width=\"297\" height=\"643\" class=\"aligncenter size-full wp-image-287725\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">\\(x=5\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-2\">\\(x=-5\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\(x=5\\) or \\(x=-5\\)<\/div><div class=\"PQ\"  id=\"PQ-2-4\">No real solution<\/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>Looking at the graph, we see that the quadratic function touches the \\(x\\)-axis at exactly one point. This means that there is one real solution to this equation.<\/p>\n<p>The vertex touches the \\(x\\)-axis at point \\((-5,0)\\), so the solution for \\(x\\) is \u22125.<\/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 \/>\nSolve \\(x^2+x+4=0\\) by using the graph shown below: <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Quadratic-Equation-Graph-Example-3.svg\" alt=\"Graph of a parabola opening upward with labeled points at (-0.5, 3.75) and (0, 4) on a grid. Both axes are visible and arrows indicate direction.\" width=\"401.35\" height=\"441.6\" class=\"aligncenter size-full wp-image-287719\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\(x=-0.5\\) or \\(x=4\\)<\/div><div class=\"PQ\"  id=\"PQ-3-2\">\\(x=-4\\) or \\(x=0.5\\)<\/div><div class=\"PQ\"  id=\"PQ-3-3\">\\(x=4\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-4\">No real solutions<\/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>Looking at the graph, we see that the quadratic function does not pass through the \\(x\\)-axis at all. This means that there are no real solutions to this equation. <\/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 \/>\nThe length of a rectangular flag is 4 feet longer than its width. The area is 96 square feet. Find the dimensions of the flag by writing a quadratic equation and graphing it on the coordinate plane. <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">Width: \u221212 feet<br>\r\nLength: \u22128 feet<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-2\">Width: 8 feet<br>\r\nLength: 12 feet<\/div><div class=\"PQ\"  id=\"PQ-4-3\">Width: \u22122 feet<br>\r\nLength: \u2212100 feet<\/div><div class=\"PQ\"  id=\"PQ-4-4\">Width: 2 feet<br>\r\nLength: 100 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>Start by recalling the formula for the area of a rectangle, which is length times width. Substitute the values from the word problem into this formula. Since the width is not known, use w to represent width. Since the length of the flag is 4 feet longer than its width, use \\(w+4\\) to represent the length. <\/p>\n<p style=\"text-align: center; line-height: 35px;\">\\(A=lw\\)<br \/>\n\\(96=w\\left(w+4\\right)\\)<\/p>\n<p>Next, simplify the equation by distributing \\(w\\). Write the quadratic equation in standard form. <\/p>\n<p style=\"text-align: center; line-height: 35px;\">\\(96=w^2+4w\\)<br \/>\n\\(96-96=w^2+4w-96\\)<br \/>\n\\(0=w^2+4w-96\\)<br \/>\n\\(w^2+4w-96=0\\)<\/p>\n<p>Then, identify the coordinates for the \\(x\\)-intercepts. Factor the equation by finding two numbers that result in a sum of 4 and a product of \u221296. These numbers are 12 and \u22128. <\/p>\n<p style=\"text-align: center;\">\\(\\left(w+12\\right)\\left(w-8\\right)=0\\)<\/p>\n<p>From here, equate each binomial to 0 and solve for \\(w\\). The solutions are \u221212 or 8. The coordinates for these \\(x\\)-intercepts are \\((-12,0)\\) and \\((8,0)\\). The graph of the function passes through the \\(x\\)-axis at \u221212 and 8. <\/p>\n<p style=\"text-align: center; line-height: 35px;\">\\(w+12=0 \\to w=-12\\)<br \/>\n\\(w-8=0 \\to w=8\\)<\/p>\n<p>Now that we know the \\(x\\)-intercepts, find the coordinates for the vertex, \\((h,k)\\). Start by finding \\(h\\), which is the vertex\u2019s \\(x\\)-coordinate. Divide the sum of the \\(x\\)-intercepts by 2: <\/p>\n<p style=\"text-align: center;\">\\(h=\\large{\\frac{-12+8}{2}}\\normalsize{\\:=\\:-}\\large{\\frac{4}{2}}\\normalsize{\\:=\\:-2}\\)<\/p>\n<p>Next, find \\(k\\), which is the vertex\u2019s \\(y\\)-coordinate. Substitute \u22122 into the quadratic equation for \\(w\\) and simplify: <\/p>\n<p style=\"text-align: center;line-height: 35px;\">\\(k=(-2)^2+4\\left(-2\\right)-96\\)<br \/>\n\\(k=4-8-96\\)<br \/>\n\\(k=-100\\)<\/p>\n<p>The vertex of this function is \\((-2,-100)\\). Plot the vertex and the \\(x\\)-intercepts onto the coordinate plane and join the points with a smooth curve. <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Quadratic-Equation-Graph-Example-4.svg\" alt=\"Graph of a parabola opening upward with x-intercepts at (-12, 0) and (8, 0) and a vertex at (-2, -100); labeled axes and grid lines are shown.\" width=\"306.25\" height=\"328.75\" class=\"aligncenter size-full wp-image-287722\"  role=\"img\" \/><\/p>\n<p>Since this problem is about the area of a flag, the negative solution, \u221212, does not apply to this scenario. Therefore, the width of the flag is 8 feet. Since the length of the flag is 4 feet longer than its width, the length of the flag is 12 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 \/>\nNicole drops a ball from 4 feet above the ground and watches as it bounces back into the air. The path of Nicole\u2019s ball takes the shape of a parabola and can be represented by the equation \\((x-2)^2=0\\). Graph this quadratic function to figure out the location of where the ball hits the ground in relation to where Nicole is standing.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-5-1\">Nicole\u2019s ball hits the ground 2 feet from where she is standing. <\/div><div class=\"PQ\"  id=\"PQ-5-2\">Nicole\u2019s ball hit the ground 4 feet from where she is standing. <\/div><div class=\"PQ\"  id=\"PQ-5-3\">Nicole\u2019s ball hit the ground -2 feet from where she is standing. <\/div><div class=\"PQ\"  id=\"PQ-5-4\">Nicole\u2019s ball hit the ground -4 feet from where she is standing. <\/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>Start by writing the quadratic equation as two binomials.<\/p>\n<p style=\"text-align: center; line-height: 35px;\">\\((x-2)^2=0\\)<br \/>\n\\((x-2)(x-2)=0\\)<\/p>\n<p>Next, set each binomial equal to 0 and solve for \\(x\\). Since both binomials are the same, there is one solution, which is 2. The graph touches the \\(x\\)-axis at \\((2,0)\\). <\/p>\n<p><\/p>\n<p style=\"text-align: center; line-height: 35px;\">\\(x-2=0\\)<br \/>\n\\(x=2\\)<\/p>\n<p>Then, identify the \\(y\\)-intercept. Since the value of \\(c\\) in the expanded equation is 4, the \\(y\\)-intercept is 4. Nicole drops the ball from 4 feet above the ground, so the point representing the \\(y\\)-intercept is \\((0,4)\\). <\/p>\n<p style=\"text-align: center; line-height: 35px;\">\\((x-2)(x-2)=0\\)<br \/>\n\\(x^2-2x-2x+4=0\\)<br \/>\n\\(x^2-4x+4=0\\)<br \/>\n\\(c=4\\)<\/p>\n<p>From here, plot the \\(x-\\) and \\(y\\)-intercepts onto the coordinate plane and join the points with a smooth curve: <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Quadratic-Equation-Graph-Example-5.svg\" alt=\"Graph of a parabola opening upward with labeled points at (0, 4) and (2, 0), crossing the x-axis at (2, 0).\" width=\"306.25\" height=\"328.75\" class=\"aligncenter size-full wp-image-287710\"  role=\"img\" \/><\/p>\n<p>Based on this information, Nicole stands at point \\((0,0)\\) and drops the ball from 4 feet above ground at point \\((0,4)\\). The ball hits the ground at point \\((2,0)\\). Therefore, the ball hits the ground 2 feet away from where Nicole stands.<\/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\/\">Mometrix Academy &#8211; Home<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Mometrix Academy &#8211; Home<\/p>\n","protected":false},"author":1,"featured_media":109326,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"open","template":"","meta":{"footnotes":""},"class_list":{"0":"post-4345","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-math-advertising-group","7":"page_category-quadratics-videos","8":"page_type-video","9":"content_type-practice-questions","10":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4345","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\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=4345"}],"version-history":[{"count":6,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4345\/revisions"}],"predecessor-version":[{"id":242170,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4345\/revisions\/242170"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/109326"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=4345"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}