{"id":197,"date":"2023-06-21T13:22:43","date_gmt":"2023-06-21T13:22:43","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/intercepts-of-quadratic-functions\/"},"modified":"2023-07-04T03:46:33","modified_gmt":"2023-07-04T03:46:33","slug":"intercepts-of-quadratic-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/intercepts-of-quadratic-functions\/","title":{"raw":"\u25aa   Intercepts of Quadratic Functions","rendered":"\u25aa   Intercepts of Quadratic Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find the [latex]y[\/latex]-intercept of a quadratic function.<\/li>\r\n \t<li>Find the real-number <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts, or roots of a quadratic function using factoring and the quadratic formula.<\/li>\r\n<\/ul>\r\n<\/div>\r\nMuch as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the <span class=\"s1\">[latex]y[\/latex]<\/span>-intercept of a quadratic by evaluating the function at an input of zero, and we find the <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts at locations where the output is zero. Notice\u00a0that the number of <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts can vary depending upon the location of the graph.\r\n\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\/02170357\/CNX_Precalc_Figure_03_02_0132.jpg\" alt=\"Three graphs where the first graph shows a parabola with no x-intercept, the second is a parabola with one \u2013intercept, and the third parabola is of two x-intercepts.\" width=\"975\" height=\"317\" \/> Number of <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts of a parabola[\/caption]\r\n<div class=\"textbox examples\">\r\n<h3>Tip for success<\/h3>\r\nWe've built up several names for the values where a function crosses the horizontal axis:\r\n<ul>\r\n \t<li>horizontal intercepts\u00a0or\u00a0x-intercepts<\/li>\r\n \t<li>zeros<\/li>\r\n \t<li>roots<\/li>\r\n<\/ul>\r\nWe tend to talk about <strong>intercepts<\/strong> with regard to a graph, <strong>zeros<\/strong> as solutions to an equation, and <strong>roots<\/strong> with regard to a function. Each term refers to the same value, though, for each quadratic function.\r\n\r\n[reveal-answer q=\"50603\"]more[\/reveal-answer]\r\n[hidden-answer a=\"50603\"]\r\n\r\nEx. [latex]f(x)=x^2-x-12[\/latex]\r\n<ul>\r\n \t<li>The <strong>x-intercept<\/strong>s of the graph are located at [latex](-3, 0)[\/latex] and [latex](4, 0)[\/latex].<\/li>\r\n \t<li>The <strong>zeros<\/strong> of the equation [latex]0 = x^2-x-12[\/latex] are [latex]-3[\/latex] and [latex]4[\/latex].<\/li>\r\n \t<li>The function[latex] f(x) = x^2 - x -12[\/latex] has<strong> roots<\/strong> [latex]-3[\/latex] and [latex]4[\/latex].<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a quadratic function [latex]f\\left(x\\right)[\/latex], find the <em><span class=\"s1\">y<\/span><\/em>-\u00a0and <em><span class=\"s1\">x<\/span><\/em>-intercepts.<\/h3>\r\n<ol>\r\n \t<li>Evaluate [latex]f\\left(0\\right)[\/latex] to find the <span class=\"s1\">[latex]y[\/latex]<\/span>-intercept.<\/li>\r\n \t<li>Solve the quadratic equation [latex]f\\left(x\\right)=0[\/latex] to find the <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the <em>y<\/em>- and <em>x<\/em>-Intercepts of a Parabola<\/h3>\r\nFind the <span class=\"s1\">[latex]y[\/latex]<\/span>- and <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts of the quadratic [latex]f\\left(x\\right)=3{x}^{2}+5x - 2[\/latex].\r\n\r\n[reveal-answer q=\"14680\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"14680\"]\r\n\r\nWe find the <span class=\"s1\">[latex]y[\/latex]<\/span>-intercept by evaluating [latex]f\\left(0\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]f\\left(0\\right)=3{\\left(0\\right)}^{2}+5\\left(0\\right)-2=-2[\/latex]<\/p>\r\nSo the <span class=\"s1\">[latex]y[\/latex]<\/span>-intercept is at [latex]\\left(0,-2\\right)[\/latex].\r\n\r\nFor the <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts, or roots, we find all solutions of [latex]f\\left(x\\right)=0[\/latex].\r\n<p style=\"text-align: center;\">[latex]3{x}^{2}+5x - 2=0[\/latex]<\/p>\r\nIn this case, the quadratic can be factored easily, providing the simplest method for solution.\r\n<p style=\"text-align: center;\">[latex]\\left(3x - 1\\right)\\left(x+2\\right)=0[\/latex]\r\n[latex]\\begin{align}&amp;3x - 1=0&amp;x+2=0\\end{align}[\/latex]\r\n[latex]x=\\frac{1}{3}\\hspace{8mm}[\/latex] or [latex]\\hspace{8mm}x=-2[\/latex]<\/p>\r\nSo the <em>roots<\/em>\u00a0are at [latex]\\left(\\frac{1}{3},0\\right)[\/latex] and [latex]\\left(-2,0\\right)[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nBy graphing the function, we can confirm that the graph crosses the <span class=\"s1\">[latex]y[\/latex]<\/span>-axis at [latex]\\left(0,-2\\right)[\/latex]. We can also confirm that the graph crosses the <span class=\"s1\">[latex]x[\/latex]<\/span>-axis at [latex]\\left(\\frac{1}{3},0\\right)[\/latex] and [latex]\\left(-2,0\\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170400\/CNX_Precalc_Figure_03_02_0142.jpg\" alt=\"Graph of a parabola which has the following intercepts (-2, 0), (1\/3, 0), and (0, -2).\" width=\"487\" height=\"480\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the above example\u00a0the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a quadratic function, find the <em>x<\/em>-intercepts by rewriting in standard form.<\/h3>\r\n<ol>\r\n \t<li>Substitute <span class=\"s1\">[latex]a[\/latex]<\/span>\u00a0and <span class=\"s1\">[latex]b[\/latex]<\/span>\u00a0into [latex]h=-\\dfrac{b}{2a}[\/latex].<\/li>\r\n \t<li>Substitute [latex]x=h[\/latex]\u00a0into the general form of the quadratic function to find <span class=\"s1\">[latex]k[\/latex]<\/span>.<\/li>\r\n \t<li>Rewrite the quadratic in standard form using <span class=\"s1\">[latex]h[\/latex]<\/span>\u00a0and <span class=\"s1\">[latex]k[\/latex]<\/span>.<\/li>\r\n \t<li>Solve for when the output of the function will be zero to find the <span class=\"s1\">[latex]x[\/latex]<\/span><em>-<\/em>intercepts.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Roots\u00a0of a Parabola<\/h3>\r\nFind the <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts of the quadratic function [latex]f\\left(x\\right)=2{x}^{2}+4x - 4[\/latex].\r\n\r\n[reveal-answer q=\"201989\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"201989\"]\r\n\r\nWe begin by solving for when the output will be zero.\r\n<p style=\"text-align: center;\">[latex]0=2{x}^{2}+4x - 4[\/latex]<\/p>\r\nBecause the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=a{\\left(x-h\\right)}^{2}+k[\/latex]<\/p>\r\nWe know that [latex]a=2[\/latex]. Then we solve for\u00a0<span class=\"s1\">[latex]h[\/latex]<\/span>\u00a0and <span class=\"s1\">[latex]k[\/latex].<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;h=-\\dfrac{b}{2a}=-\\dfrac{4}{2\\left(2\\right)}=-1\\\\[1mm]&amp;\\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}k&amp;=f\\left(h\\right)\\\\&amp;=f\\left(-1\\right)\\\\&amp;=2{\\left(-1\\right)}^{2}+4\\left(-1\\right)-4\\\\&amp;=-6\\end{align}[\/latex]<\/p>\r\nSo now we can rewrite in standard form.\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=2{\\left(x+1\\right)}^{2}-6[\/latex]<\/p>\r\nWe can now solve for when the output will be zero.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;0=2{\\left(x+1\\right)}^{2}-6 \\\\ &amp;6=2{\\left(x+1\\right)}^{2} \\\\ &amp;3={\\left(x+1\\right)}^{2} \\\\ &amp;x+1=\\pm \\sqrt{3} \\\\ &amp;x=-1\\pm \\sqrt{3} \\end{align}[\/latex]<\/p>\r\nThe graph has [latex]x[\/latex]<em>-<\/em>intercepts at [latex]\\left(-1-\\sqrt{3},0\\right)[\/latex] and [latex]\\left(-1+\\sqrt{3},0\\right)[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170402\/CNX_Precalc_Figure_03_02_0152.jpg\" alt=\"Graph of a parabola which has the following x-intercepts (-2.732, 0) and (0.732, 0).\" width=\"487\" height=\"517\" \/>\r\n\r\nWe can check our work by graphing the given function on a graphing utility and observing the roots.\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\nUsing an online graphing calculator, plot the function [latex]g\\left(x\\right)=-7+{x}^{2}-6x[\/latex]. You can use this tool to find the [latex]x[\/latex]-and [latex]y[\/latex]-intercepts by clicking on the graph. Four points will appear. List each point, and what kind of point it is, we got you started with the vertex:\r\n<ol>\r\n \t<li>Vertex = [latex](3,-16)[\/latex]<\/li>\r\n \t<li><\/li>\r\n \t<li><\/li>\r\n \t<li><\/li>\r\n<\/ol>\r\n[reveal-answer q=\"275171\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"275171\"]\r\n\r\n[latex]y[\/latex]-intercept at [latex](0, -7)[\/latex]\r\n[latex]x[\/latex]-intercept at\u00a0[latex](-1,0)[\/latex]\r\n[latex]x[\/latex]-intercept at\u00a0[latex](7,0)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[ohm_question]121416[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nWhen attempting to identify the roots of a quadratic function, first look to see if it can be factored when the function value is set equal to zero, and solved by applying the zero-product principle. If so, this is often the quickest method. But most quadratic functions cannot be factored. In this case, you can either write it in vertex form or use the quadratic formula.\r\n\r\n[reveal-answer q=\"617375\"]more[\/reveal-answer]\r\n[hidden-answer a=\"617375\"]\r\n<p style=\"text-align: left;\">1. The quadratic formula will reveal the x-intercepts of the graph of any quadratic equation.<\/p>\r\n<p style=\"text-align: center;\">[latex]x=\\dfrac{-b \\pm \\sqrt{b^2-4ac}}{2a}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">2. The vertex formula will reveal the vertex of any parabola.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(\\dfrac{-b}{2a}, f\\left(\\dfrac{-b}{2a}\\right) \\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">3. The y-intercept, often called the\u00a0<em>initial value<\/em>, of the function can be found by setting the input variable to 0.<\/p>\r\n<p style=\"text-align: center;\">Evaluate the function at zero to find the y-intercept: [latex]f(0) = y[\/latex]<\/p>\r\nThese three pieces of information are of particular value when solving applications of quadratic functions.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Quadratic Equation with the Quadratic Formula<\/h3>\r\nSolve [latex]{x}^{2}+x+2=0[\/latex].\r\n\r\n[reveal-answer q=\"757696\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"757696\"]\r\n\r\nLet\u2019s begin by writing the quadratic formula: [latex]x=\\dfrac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex].\r\n\r\nWhen applying the <strong>quadratic formula<\/strong>, we identify the coefficients [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex]. For the equation [latex]{x}^{2}+x+2=0[\/latex], we have [latex]a=1[\/latex], [latex]b=1[\/latex], and [latex]c=2[\/latex].\u00a0Substituting these values into the formula we have:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x&amp;=\\dfrac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a} \\\\[1.5mm] &amp;=\\dfrac{-1\\pm \\sqrt{{1}^{2}-4\\cdot 1\\cdot \\left(2\\right)}}{2\\cdot 1} \\\\[1.5mm] &amp;=\\dfrac{-1\\pm \\sqrt{1 - 8}}{2} \\\\[1.5mm] &amp;=\\dfrac{-1\\pm \\sqrt{-7}}{2} \\\\[1.5mm] &amp;=\\dfrac{-1\\pm i\\sqrt{7}}{2} \\end{align}[\/latex]<\/p>\r\nThe solutions to the equation are [latex]x=\\dfrac{-1+i\\sqrt{7}}{2}[\/latex] and [latex]x=\\dfrac{-1-i\\sqrt{7}}{2}[\/latex] or [latex]x=-\\dfrac{1}{2}+\\dfrac{i\\sqrt{7}}{2}[\/latex] and [latex]x=-\\dfrac{1}{2}-\\dfrac{i\\sqrt{7}}{2}[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nThis quadratic equation has only non-real solutions.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Applying the Vertex and <em>x<\/em>-Intercepts of a Parabola<\/h3>\r\nA ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball\u2019s height above ground can be modeled by the equation [latex]H\\left(t\\right)=-16{t}^{2}+80t+40[\/latex].\r\n\r\na. When does the ball reach the maximum height?\r\n\r\nb. What is the maximum height of the ball?\r\n\r\nc. When does the ball hit the ground?\r\n\r\n[reveal-answer q=\"394530\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"394530\"]\r\n\r\na. The ball reaches the maximum height at the vertex of the parabola.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}h&amp;=-\\dfrac{80}{2\\left(-16\\right)} \\\\[1mm]&amp;=\\dfrac{80}{32} \\\\[1mm] &amp;=\\dfrac{5}{2} \\\\[1mm] &amp;=2.5 \\end{align}[\/latex]<\/p>\r\nThe ball reaches a maximum height after 2.5 seconds.\r\n\r\nb. To find the maximum height, find the [latex]y[\/latex]<em>\u00a0<\/em>coordinate of the vertex of the parabola.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}k&amp;=H\\left(2.5\\right) \\\\[1mm] &amp;=-16{\\left(2.5\\right)}^{2}+80\\left(2.5\\right)+40 \\\\[1mm] &amp;=140\\hfill \\end{align}[\/latex]<\/p>\r\nThe ball reaches a maximum height of 140 feet.\r\n\r\nc. To find when the ball hits the ground, we need to determine when the height is zero, [latex]H\\left(t\\right)=0[\/latex].\r\n\r\nWe use the quadratic formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} t&amp;=\\dfrac{-80\\pm \\sqrt{{80}^{2}-4\\left(-16\\right)\\left(40\\right)}}{2\\left(-16\\right)} \\\\[1mm] &amp;=\\dfrac{-80\\pm \\sqrt{8960}}{-32} \\end{align}[\/latex]<\/p>\r\nBecause the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.\r\n<p style=\"text-align: center;\">[latex]t=\\dfrac{-80+\\sqrt{8960}}{-32}\\approx 5.458\\hspace{3mm}[\/latex] or [latex]\\hspace{3mm}t=\\dfrac{-80-\\sqrt{8960}}{-32}\\approx -0.458[\/latex]<\/p>\r\nSince the domain starts at [latex]t=0[\/latex] when the ball is thrown, the second answer is outside the reasonable domain of our model. The ball will hit the ground after about 5.46 seconds.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170404\/CNX_Precalc_Figure_03_02_0162.jpg\" alt=\"Graph of a negative parabola where x goes from -1 to 6.\" width=\"487\" height=\"254\" \/>\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\nA rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock\u2019s height above ocean can be modeled by the equation [latex]H\\left(t\\right)=-16{t}^{2}+96t+112[\/latex].\r\n\r\na. When does the rock reach the maximum height?\r\n\r\nb. What is the maximum height of the rock?\r\n\r\nc. When does the rock hit the ocean?\r\n\r\n[reveal-answer q=\"174919\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"174919\"]\r\n\r\na.\u00a03 seconds \u00a0b.\u00a0256 feet \u00a0c.\u00a07 seconds\r\n\r\n[\/hidden-answer]\r\n\r\n[ohm_question]15809[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find the [latex]y[\/latex]-intercept of a quadratic function.<\/li>\n<li>Find the real-number <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts, or roots of a quadratic function using factoring and the quadratic formula.<\/li>\n<\/ul>\n<\/div>\n<p>Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the <span class=\"s1\">[latex]y[\/latex]<\/span>-intercept of a quadratic by evaluating the function at an input of zero, and we find the <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts at locations where the output is zero. Notice\u00a0that the number of <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts can vary depending upon the location of the graph.<\/p>\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\/02170357\/CNX_Precalc_Figure_03_02_0132.jpg\" alt=\"Three graphs where the first graph shows a parabola with no x-intercept, the second is a parabola with one \u2013intercept, and the third parabola is of two x-intercepts.\" width=\"975\" height=\"317\" \/><\/p>\n<p class=\"wp-caption-text\">Number of <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts of a parabola<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Tip for success<\/h3>\n<p>We&#8217;ve built up several names for the values where a function crosses the horizontal axis:<\/p>\n<ul>\n<li>horizontal intercepts\u00a0or\u00a0x-intercepts<\/li>\n<li>zeros<\/li>\n<li>roots<\/li>\n<\/ul>\n<p>We tend to talk about <strong>intercepts<\/strong> with regard to a graph, <strong>zeros<\/strong> as solutions to an equation, and <strong>roots<\/strong> with regard to a function. Each term refers to the same value, though, for each quadratic function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q50603\">more<\/span><\/p>\n<div id=\"q50603\" class=\"hidden-answer\" style=\"display: none\">\n<p>Ex. [latex]f(x)=x^2-x-12[\/latex]<\/p>\n<ul>\n<li>The <strong>x-intercept<\/strong>s of the graph are located at [latex](-3, 0)[\/latex] and [latex](4, 0)[\/latex].<\/li>\n<li>The <strong>zeros<\/strong> of the equation [latex]0 = x^2-x-12[\/latex] are [latex]-3[\/latex] and [latex]4[\/latex].<\/li>\n<li>The function[latex]f(x) = x^2 - x -12[\/latex] has<strong> roots<\/strong> [latex]-3[\/latex] and [latex]4[\/latex].<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a quadratic function [latex]f\\left(x\\right)[\/latex], find the <em><span class=\"s1\">y<\/span><\/em>&#8211;\u00a0and <em><span class=\"s1\">x<\/span><\/em>-intercepts.<\/h3>\n<ol>\n<li>Evaluate [latex]f\\left(0\\right)[\/latex] to find the <span class=\"s1\">[latex]y[\/latex]<\/span>-intercept.<\/li>\n<li>Solve the quadratic equation [latex]f\\left(x\\right)=0[\/latex] to find the <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the <em>y<\/em>&#8211; and <em>x<\/em>-Intercepts of a Parabola<\/h3>\n<p>Find the <span class=\"s1\">[latex]y[\/latex]<\/span>&#8211; and <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts of the quadratic [latex]f\\left(x\\right)=3{x}^{2}+5x - 2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q14680\">Show Solution<\/span><\/p>\n<div id=\"q14680\" class=\"hidden-answer\" style=\"display: none\">\n<p>We find the <span class=\"s1\">[latex]y[\/latex]<\/span>-intercept by evaluating [latex]f\\left(0\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(0\\right)=3{\\left(0\\right)}^{2}+5\\left(0\\right)-2=-2[\/latex]<\/p>\n<p>So the <span class=\"s1\">[latex]y[\/latex]<\/span>-intercept is at [latex]\\left(0,-2\\right)[\/latex].<\/p>\n<p>For the <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts, or roots, we find all solutions of [latex]f\\left(x\\right)=0[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]3{x}^{2}+5x - 2=0[\/latex]<\/p>\n<p>In this case, the quadratic can be factored easily, providing the simplest method for solution.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(3x - 1\\right)\\left(x+2\\right)=0[\/latex]<br \/>\n[latex]\\begin{align}&3x - 1=0&x+2=0\\end{align}[\/latex]<br \/>\n[latex]x=\\frac{1}{3}\\hspace{8mm}[\/latex] or [latex]\\hspace{8mm}x=-2[\/latex]<\/p>\n<p>So the <em>roots<\/em>\u00a0are at [latex]\\left(\\frac{1}{3},0\\right)[\/latex] and [latex]\\left(-2,0\\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>By graphing the function, we can confirm that the graph crosses the <span class=\"s1\">[latex]y[\/latex]<\/span>-axis at [latex]\\left(0,-2\\right)[\/latex]. We can also confirm that the graph crosses the <span class=\"s1\">[latex]x[\/latex]<\/span>-axis at [latex]\\left(\\frac{1}{3},0\\right)[\/latex] and [latex]\\left(-2,0\\right)[\/latex].<\/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\/02170400\/CNX_Precalc_Figure_03_02_0142.jpg\" alt=\"Graph of a parabola which has the following intercepts (-2, 0), (1\/3, 0), and (0, -2).\" width=\"487\" height=\"480\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the above example\u00a0the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a quadratic function, find the <em>x<\/em>-intercepts by rewriting in standard form.<\/h3>\n<ol>\n<li>Substitute <span class=\"s1\">[latex]a[\/latex]<\/span>\u00a0and <span class=\"s1\">[latex]b[\/latex]<\/span>\u00a0into [latex]h=-\\dfrac{b}{2a}[\/latex].<\/li>\n<li>Substitute [latex]x=h[\/latex]\u00a0into the general form of the quadratic function to find <span class=\"s1\">[latex]k[\/latex]<\/span>.<\/li>\n<li>Rewrite the quadratic in standard form using <span class=\"s1\">[latex]h[\/latex]<\/span>\u00a0and <span class=\"s1\">[latex]k[\/latex]<\/span>.<\/li>\n<li>Solve for when the output of the function will be zero to find the <span class=\"s1\">[latex]x[\/latex]<\/span><em>&#8211;<\/em>intercepts.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Roots\u00a0of a Parabola<\/h3>\n<p>Find the <span class=\"s1\">[latex]x[\/latex]<\/span>-intercepts of the quadratic function [latex]f\\left(x\\right)=2{x}^{2}+4x - 4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q201989\">Show Solution<\/span><\/p>\n<div id=\"q201989\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin by solving for when the output will be zero.<\/p>\n<p style=\"text-align: center;\">[latex]0=2{x}^{2}+4x - 4[\/latex]<\/p>\n<p>Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=a{\\left(x-h\\right)}^{2}+k[\/latex]<\/p>\n<p>We know that [latex]a=2[\/latex]. Then we solve for\u00a0<span class=\"s1\">[latex]h[\/latex]<\/span>\u00a0and <span class=\"s1\">[latex]k[\/latex].<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&h=-\\dfrac{b}{2a}=-\\dfrac{4}{2\\left(2\\right)}=-1\\\\[1mm]&\\end{align}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}k&=f\\left(h\\right)\\\\&=f\\left(-1\\right)\\\\&=2{\\left(-1\\right)}^{2}+4\\left(-1\\right)-4\\\\&=-6\\end{align}[\/latex]<\/p>\n<p>So now we can rewrite in standard form.<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=2{\\left(x+1\\right)}^{2}-6[\/latex]<\/p>\n<p>We can now solve for when the output will be zero.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&0=2{\\left(x+1\\right)}^{2}-6 \\\\ &6=2{\\left(x+1\\right)}^{2} \\\\ &3={\\left(x+1\\right)}^{2} \\\\ &x+1=\\pm \\sqrt{3} \\\\ &x=-1\\pm \\sqrt{3} \\end{align}[\/latex]<\/p>\n<p>The graph has [latex]x[\/latex]<em>&#8211;<\/em>intercepts at [latex]\\left(-1-\\sqrt{3},0\\right)[\/latex] and [latex]\\left(-1+\\sqrt{3},0\\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\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\/02170402\/CNX_Precalc_Figure_03_02_0152.jpg\" alt=\"Graph of a parabola which has the following x-intercepts (-2.732, 0) and (0.732, 0).\" width=\"487\" height=\"517\" \/><\/p>\n<p>We can check our work by graphing the given function on a graphing utility and observing the roots.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using an online graphing calculator, plot the function [latex]g\\left(x\\right)=-7+{x}^{2}-6x[\/latex]. You can use this tool to find the [latex]x[\/latex]-and [latex]y[\/latex]-intercepts by clicking on the graph. Four points will appear. List each point, and what kind of point it is, we got you started with the vertex:<\/p>\n<ol>\n<li>Vertex = [latex](3,-16)[\/latex]<\/li>\n<li><\/li>\n<li><\/li>\n<li><\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q275171\">Show Solution<\/span><\/p>\n<div id=\"q275171\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y[\/latex]-intercept at [latex](0, -7)[\/latex]<br \/>\n[latex]x[\/latex]-intercept at\u00a0[latex](-1,0)[\/latex]<br \/>\n[latex]x[\/latex]-intercept at\u00a0[latex](7,0)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm121416\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=121416&theme=oea&iframe_resize_id=ohm121416&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>When attempting to identify the roots of a quadratic function, first look to see if it can be factored when the function value is set equal to zero, and solved by applying the zero-product principle. If so, this is often the quickest method. But most quadratic functions cannot be factored. In this case, you can either write it in vertex form or use the quadratic formula.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q617375\">more<\/span><\/p>\n<div id=\"q617375\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">1. The quadratic formula will reveal the x-intercepts of the graph of any quadratic equation.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\dfrac{-b \\pm \\sqrt{b^2-4ac}}{2a}[\/latex]<\/p>\n<p style=\"text-align: left;\">2. The vertex formula will reveal the vertex of any parabola.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(\\dfrac{-b}{2a}, f\\left(\\dfrac{-b}{2a}\\right) \\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">3. The y-intercept, often called the\u00a0<em>initial value<\/em>, of the function can be found by setting the input variable to 0.<\/p>\n<p style=\"text-align: center;\">Evaluate the function at zero to find the y-intercept: [latex]f(0) = y[\/latex]<\/p>\n<p>These three pieces of information are of particular value when solving applications of quadratic functions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Quadratic Equation with the Quadratic Formula<\/h3>\n<p>Solve [latex]{x}^{2}+x+2=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q757696\">Show Solution<\/span><\/p>\n<div id=\"q757696\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let\u2019s begin by writing the quadratic formula: [latex]x=\\dfrac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex].<\/p>\n<p>When applying the <strong>quadratic formula<\/strong>, we identify the coefficients [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex]. For the equation [latex]{x}^{2}+x+2=0[\/latex], we have [latex]a=1[\/latex], [latex]b=1[\/latex], and [latex]c=2[\/latex].\u00a0Substituting these values into the formula we have:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x&=\\dfrac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a} \\\\[1.5mm] &=\\dfrac{-1\\pm \\sqrt{{1}^{2}-4\\cdot 1\\cdot \\left(2\\right)}}{2\\cdot 1} \\\\[1.5mm] &=\\dfrac{-1\\pm \\sqrt{1 - 8}}{2} \\\\[1.5mm] &=\\dfrac{-1\\pm \\sqrt{-7}}{2} \\\\[1.5mm] &=\\dfrac{-1\\pm i\\sqrt{7}}{2} \\end{align}[\/latex]<\/p>\n<p>The solutions to the equation are [latex]x=\\dfrac{-1+i\\sqrt{7}}{2}[\/latex] and [latex]x=\\dfrac{-1-i\\sqrt{7}}{2}[\/latex] or [latex]x=-\\dfrac{1}{2}+\\dfrac{i\\sqrt{7}}{2}[\/latex] and [latex]x=-\\dfrac{1}{2}-\\dfrac{i\\sqrt{7}}{2}[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>This quadratic equation has only non-real solutions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Applying the Vertex and <em>x<\/em>-Intercepts of a Parabola<\/h3>\n<p>A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball\u2019s height above ground can be modeled by the equation [latex]H\\left(t\\right)=-16{t}^{2}+80t+40[\/latex].<\/p>\n<p>a. When does the ball reach the maximum height?<\/p>\n<p>b. What is the maximum height of the ball?<\/p>\n<p>c. When does the ball hit the ground?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q394530\">Show Solution<\/span><\/p>\n<div id=\"q394530\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. The ball reaches the maximum height at the vertex of the parabola.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}h&=-\\dfrac{80}{2\\left(-16\\right)} \\\\[1mm]&=\\dfrac{80}{32} \\\\[1mm] &=\\dfrac{5}{2} \\\\[1mm] &=2.5 \\end{align}[\/latex]<\/p>\n<p>The ball reaches a maximum height after 2.5 seconds.<\/p>\n<p>b. To find the maximum height, find the [latex]y[\/latex]<em>\u00a0<\/em>coordinate of the vertex of the parabola.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}k&=H\\left(2.5\\right) \\\\[1mm] &=-16{\\left(2.5\\right)}^{2}+80\\left(2.5\\right)+40 \\\\[1mm] &=140\\hfill \\end{align}[\/latex]<\/p>\n<p>The ball reaches a maximum height of 140 feet.<\/p>\n<p>c. To find when the ball hits the ground, we need to determine when the height is zero, [latex]H\\left(t\\right)=0[\/latex].<\/p>\n<p>We use the quadratic formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} t&=\\dfrac{-80\\pm \\sqrt{{80}^{2}-4\\left(-16\\right)\\left(40\\right)}}{2\\left(-16\\right)} \\\\[1mm] &=\\dfrac{-80\\pm \\sqrt{8960}}{-32} \\end{align}[\/latex]<\/p>\n<p>Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.<\/p>\n<p style=\"text-align: center;\">[latex]t=\\dfrac{-80+\\sqrt{8960}}{-32}\\approx 5.458\\hspace{3mm}[\/latex] or [latex]\\hspace{3mm}t=\\dfrac{-80-\\sqrt{8960}}{-32}\\approx -0.458[\/latex]<\/p>\n<p>Since the domain starts at [latex]t=0[\/latex] when the ball is thrown, the second answer is outside the reasonable domain of our model. The ball will hit the ground after about 5.46 seconds.<\/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\/02170404\/CNX_Precalc_Figure_03_02_0162.jpg\" alt=\"Graph of a negative parabola where x goes from -1 to 6.\" width=\"487\" height=\"254\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock\u2019s height above ocean can be modeled by the equation [latex]H\\left(t\\right)=-16{t}^{2}+96t+112[\/latex].<\/p>\n<p>a. When does the rock reach the maximum height?<\/p>\n<p>b. What is the maximum height of the rock?<\/p>\n<p>c. When does the rock hit the ocean?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q174919\">Show Solution<\/span><\/p>\n<div id=\"q174919\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a03 seconds \u00a0b.\u00a0256 feet \u00a0c.\u00a07 seconds<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm15809\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15809&theme=oea&iframe_resize_id=ohm15809&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/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-197\">\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>Question ID 121416. <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>. <strong>License Terms<\/strong>: MathAS Community License CC-BY + GPL<\/li><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>Question ID 15809. <strong>Authored by<\/strong>: Sousa,James, mb Lippman,David. <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>: MathAS Community License CC-BY + GPL<\/li><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><\/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":395986,"menu_order":34,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Question ID 121416\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"MathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 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http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"}]","CANDELA_OUTCOMES_GUID":"7666f642-4660-44a7-b9d3-c5a4d8badfec","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-197","chapter","type-chapter","status-publish","hentry"],"part":160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/197","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":3,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/197\/revisions"}],"predecessor-version":[{"id":1422,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/197\/revisions\/1422"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/160"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/197\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/media?parent=197"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=197"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/contributor?post=197"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/license?post=197"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}