{"id":1659,"date":"2016-11-02T17:18:17","date_gmt":"2016-11-02T17:18:17","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1659"},"modified":"2017-04-14T21:02:02","modified_gmt":"2017-04-14T21:02:02","slug":"intercepts-of-quadratic-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/intercepts-of-quadratic-functions\/","title":{"raw":"Intercepts of Quadratic Functions","rendered":"Intercepts of Quadratic Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Find the <em>y<\/em>-intercept of a quadratic function<\/li>\r\n \t<li>Find the real-number <em>x<\/em>-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 <em>y<\/em>-intercept of a quadratic by evaluating the function at an input of zero, and we find the <em>x<\/em>-intercepts at locations where the output is zero. Notice\u00a0that the number of <em>x<\/em>-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\" data-media-type=\"image\/jpg\" \/> Number of x-intercepts of a parabola[\/caption]\r\n\r\nMathematicians also define <em>x<\/em>-intercepts as roots of the quadratic function.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a quadratic function [latex]f\\left(x\\right)[\/latex], find the <em>y<\/em>-\u00a0and <em>x<\/em>-intercepts.<\/h3>\r\n<ol>\r\n \t<li>Evaluate [latex]f\\left(0\\right)[\/latex] to find the <em>y<\/em>-intercept.<\/li>\r\n \t<li>Solve the quadratic equation [latex]f\\left(x\\right)=0[\/latex] to find the <em>x<\/em>-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 <em>y<\/em>- and <em>x<\/em>-intercepts of the quadratic [latex]f\\left(x\\right)=3{x}^{2}+5x - 2[\/latex].\r\n\r\n[reveal-answer q=\"14680\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"14680\"]\r\nWe find the <em>y<\/em>-intercept by evaluating [latex]f\\left(0\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}f\\left(0\\right)=3{\\left(0\\right)}^{2}+5\\left(0\\right)-2\\hfill \\\\ \\text{ }=-2\\hfill \\end{array}[\/latex]<\/p>\r\nSo the <em>y<\/em>-intercept is at [latex]\\left(0,-2\\right)[\/latex].\r\n\r\nFor the <em>x<\/em>-intercepts, or roots, we find all solutions of [latex]f\\left(x\\right)=0[\/latex].\r\n<p style=\"text-align: center;\">[latex]0=3{x}^{2}+5x - 2[\/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]0=\\left(3x - 1\\right)\\left(x+2\\right)[\/latex]\r\n[latex]\\begin{array}{c}0=3x - 1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; 0=x+2\\hfill \\\\ x=\\frac{1}{3}\\hfill &amp; \\hfill &amp; \\text{or}\\hfill &amp; \\hfill &amp; x=-2\\hfill \\end{array}[\/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 <em>y<\/em>-axis at [latex]\\left(0,-2\\right)[\/latex]. We can also confirm that the graph crosses the <em>x<\/em>-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\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn Example: Finding the <em>y<\/em>- and <em>x<\/em>-Intercepts of a Parabola, the 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 <em>a<\/em>\u00a0and <em>b<\/em>\u00a0into [latex]h=-\\frac{b}{2a}[\/latex].<\/li>\r\n \t<li>Substitute <em>x<\/em> =\u00a0<em>h<\/em>\u00a0into the general form of the quadratic function to find <em>k<\/em>.<\/li>\r\n \t<li>Rewrite the quadratic in standard form using <em>h<\/em>\u00a0and <em>k<\/em>.<\/li>\r\n \t<li>Solve for when the output of the function will be zero to find the <em>x-<\/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 <em>x<\/em>-intercepts of the quadratic function [latex]f\\left(x\\right)=2{x}^{2}+4x - 4[\/latex].\r\n\r\n[reveal-answer q=\"201989\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"201989\"]\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 <em>a\u00a0<\/em>= 2. Then we solve for <em>h<\/em>\u00a0and <em>k<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}h=-\\frac{b}{2a}\\hfill &amp; \\hfill &amp; \\hfill &amp; k=f\\left(-1\\right)\\hfill \\\\ \\text{ }=-\\frac{4}{2\\left(2\\right)}\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{ }=2{\\left(-1\\right)}^{2}+4\\left(-1\\right)-4\\hfill \\\\ \\text{ }=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{ }=-6\\hfill \\end{array}[\/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{array}{c}0=2{\\left(x+1\\right)}^{2}-6\\hfill \\\\ 6=2{\\left(x+1\\right)}^{2}\\hfill \\\\ 3={\\left(x+1\\right)}^{2}\\hfill \\\\ x+1=\\pm \\sqrt{3}\\hfill \\\\ x=-1\\pm \\sqrt{3}\\hfill \\end{array}[\/latex]<\/p>\r\nThe graph has <em>x-<\/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\" data-media-type=\"image\/jpg\" \/>\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\nThe function [latex]g\\left(x\\right)=13+{x}^{2}-6x[\/latex] is graphed below. You can use Desmos to find the x-and y-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\nhttps:\/\/www.desmos.com\/calculator\/ilnmrc6noz\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"275171\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"275171\"]<em data-effect=\"italics\">y<\/em>-intercept at (0, 13)[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=121416&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"150\"><\/iframe>\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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"757696\"]\r\nLet\u2019s begin by writing the quadratic formula: [latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex].\r\n\r\nWhen applying the <strong>quadratic formula<\/strong>, we identify the coefficients <em>a<\/em>,\u00a0<em>b<\/em>, and\u00a0<em>c<\/em>. For the equation [latex]{x}^{2}+x+2=0[\/latex], we have <em>a<\/em>\u00a0=\u00a01,\u00a0<em>b<\/em>\u00a0=\u00a01, and <em>c<\/em>\u00a0=\u00a02.\u00a0Substituting these values into the formula we have:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\\\ \\text{ }=\\frac{-1\\pm \\sqrt{{1}^{2}-4\\cdot 1\\cdot \\left(2\\right)}}{2\\cdot 1}\\hfill \\\\ \\text{ }=\\frac{-1\\pm \\sqrt{1 - 8}}{2}\\hfill \\\\ \\text{ }=\\frac{-1\\pm \\sqrt{-7}}{2}\\hfill \\\\ \\text{ }=\\frac{-1\\pm i\\sqrt{7}}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nThe solutions to the equation are [latex]x=\\frac{-1+i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1-i\\sqrt{7}}{2}[\/latex] or [latex]x=\\frac{-1}{2}+\\frac{i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1}{2}-\\frac{i\\sqrt{7}}{2}[\/latex].\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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"394530\"]\r\na. The ball reaches the maximum height at the vertex of the parabola.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c} h=-\\frac{80}{2\\left(-16\\right)} \\text{ }=\\frac{80}{32}\\hfill \\\\ \\text{ }=\\frac{5}{2}\\hfill \\\\ \\text{ }=2.5\\hfill \\end{array}[\/latex]<\/p>\r\nThe ball reaches a maximum height after 2.5 seconds.\r\n\r\nb. To find the maximum height, find the <em>y\u00a0<\/em>coordinate of the vertex of the parabola.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}k=H\\left(-\\frac{b}{2a}\\right)\\hfill \\\\ \\text{ }=H\\left(2.5\\right)\\hfill \\\\ \\text{ }=-16{\\left(2.5\\right)}^{2}+80\\left(2.5\\right)+40\\hfill \\\\ \\text{ }=140\\hfill \\end{array}[\/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{array}{c} t=\\frac{-80\\pm \\sqrt{{80}^{2}-4\\left(-16\\right)\\left(40\\right)}}{2\\left(-16\\right)}\\hfill \\\\ \\text{ }=\\frac{-80\\pm \\sqrt{8960}}{-32}\\hfill \\end{array}[\/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]\\begin{array}{c}t=\\frac{-80-\\sqrt{8960}}{-32}\\approx 5.458\\hfill &amp; \\text{or}\\hfill &amp; t=\\frac{-80+\\sqrt{8960}}{-32}\\approx -0.458\\hfill \\end{array}[\/latex]<\/p>\r\nThe second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 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\" data-media-type=\"image\/jpg\" \/>\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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"174919\"]a.\u00a03 seconds \u00a0b.\u00a0256 feet \u00a0c.\u00a07 seconds[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=15809&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"375\"><\/iframe>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Find the <em>y<\/em>-intercept of a quadratic function<\/li>\n<li>Find the real-number <em>x<\/em>-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 <em>y<\/em>-intercept of a quadratic by evaluating the function at an input of zero, and we find the <em>x<\/em>-intercepts at locations where the output is zero. Notice\u00a0that the number of <em>x<\/em>-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\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">Number of x-intercepts of a parabola<\/p>\n<\/div>\n<p>Mathematicians also define <em>x<\/em>-intercepts as roots of the quadratic function.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a quadratic function [latex]f\\left(x\\right)[\/latex], find the <em>y<\/em>&#8211;\u00a0and <em>x<\/em>-intercepts.<\/h3>\n<ol>\n<li>Evaluate [latex]f\\left(0\\right)[\/latex] to find the <em>y<\/em>-intercept.<\/li>\n<li>Solve the quadratic equation [latex]f\\left(x\\right)=0[\/latex] to find the <em>x<\/em>-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 <em>y<\/em>&#8211; and <em>x<\/em>-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\">Solution<\/span><\/p>\n<div id=\"q14680\" class=\"hidden-answer\" style=\"display: none\">\nWe find the <em>y<\/em>-intercept by evaluating [latex]f\\left(0\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}f\\left(0\\right)=3{\\left(0\\right)}^{2}+5\\left(0\\right)-2\\hfill \\\\ \\text{ }=-2\\hfill \\end{array}[\/latex]<\/p>\n<p>So the <em>y<\/em>-intercept is at [latex]\\left(0,-2\\right)[\/latex].<\/p>\n<p>For the <em>x<\/em>-intercepts, or roots, we find all solutions of [latex]f\\left(x\\right)=0[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]0=3{x}^{2}+5x - 2[\/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]0=\\left(3x - 1\\right)\\left(x+2\\right)[\/latex]<br \/>\n[latex]\\begin{array}{c}0=3x - 1\\hfill & \\hfill & \\hfill & \\hfill & 0=x+2\\hfill \\\\ x=\\frac{1}{3}\\hfill & \\hfill & \\text{or}\\hfill & \\hfill & x=-2\\hfill \\end{array}[\/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 <em>y<\/em>-axis at [latex]\\left(0,-2\\right)[\/latex]. We can also confirm that the graph crosses the <em>x<\/em>-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\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In Example: Finding the <em>y<\/em>&#8211; and <em>x<\/em>-Intercepts of a Parabola, the 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 <em>a<\/em>\u00a0and <em>b<\/em>\u00a0into [latex]h=-\\frac{b}{2a}[\/latex].<\/li>\n<li>Substitute <em>x<\/em> =\u00a0<em>h<\/em>\u00a0into the general form of the quadratic function to find <em>k<\/em>.<\/li>\n<li>Rewrite the quadratic in standard form using <em>h<\/em>\u00a0and <em>k<\/em>.<\/li>\n<li>Solve for when the output of the function will be zero to find the <em>x-<\/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 <em>x<\/em>-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\">Solution<\/span><\/p>\n<div id=\"q201989\" class=\"hidden-answer\" style=\"display: none\">\nWe 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 <em>a\u00a0<\/em>= 2. Then we solve for <em>h<\/em>\u00a0and <em>k<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}h=-\\frac{b}{2a}\\hfill & \\hfill & \\hfill & k=f\\left(-1\\right)\\hfill \\\\ \\text{ }=-\\frac{4}{2\\left(2\\right)}\\hfill & \\hfill & \\hfill & \\text{ }=2{\\left(-1\\right)}^{2}+4\\left(-1\\right)-4\\hfill \\\\ \\text{ }=-1\\hfill & \\hfill & \\hfill & \\text{ }=-6\\hfill \\end{array}[\/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{array}{c}0=2{\\left(x+1\\right)}^{2}-6\\hfill \\\\ 6=2{\\left(x+1\\right)}^{2}\\hfill \\\\ 3={\\left(x+1\\right)}^{2}\\hfill \\\\ x+1=\\pm \\sqrt{3}\\hfill \\\\ x=-1\\pm \\sqrt{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>The graph has <em>x-<\/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\" data-media-type=\"image\/jpg\" \/><\/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>The function [latex]g\\left(x\\right)=13+{x}^{2}-6x[\/latex] is graphed below. You can use Desmos to find the x-and y-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<p>https:\/\/www.desmos.com\/calculator\/ilnmrc6noz<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q275171\">Solution<\/span><\/p>\n<div id=\"q275171\" class=\"hidden-answer\" style=\"display: none\"><em data-effect=\"italics\">y<\/em>-intercept at (0, 13)<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=121416&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"150\"><\/iframe><\/p>\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\">Solution<\/span><\/p>\n<div id=\"q757696\" class=\"hidden-answer\" style=\"display: none\">\nLet\u2019s begin by writing the quadratic formula: [latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex].<\/p>\n<p>When applying the <strong>quadratic formula<\/strong>, we identify the coefficients <em>a<\/em>,\u00a0<em>b<\/em>, and\u00a0<em>c<\/em>. For the equation [latex]{x}^{2}+x+2=0[\/latex], we have <em>a<\/em>\u00a0=\u00a01,\u00a0<em>b<\/em>\u00a0=\u00a01, and <em>c<\/em>\u00a0=\u00a02.\u00a0Substituting these values into the formula we have:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\\\ \\text{ }=\\frac{-1\\pm \\sqrt{{1}^{2}-4\\cdot 1\\cdot \\left(2\\right)}}{2\\cdot 1}\\hfill \\\\ \\text{ }=\\frac{-1\\pm \\sqrt{1 - 8}}{2}\\hfill \\\\ \\text{ }=\\frac{-1\\pm \\sqrt{-7}}{2}\\hfill \\\\ \\text{ }=\\frac{-1\\pm i\\sqrt{7}}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>The solutions to the equation are [latex]x=\\frac{-1+i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1-i\\sqrt{7}}{2}[\/latex] or [latex]x=\\frac{-1}{2}+\\frac{i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1}{2}-\\frac{i\\sqrt{7}}{2}[\/latex].<\/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\">Solution<\/span><\/p>\n<div id=\"q394530\" class=\"hidden-answer\" style=\"display: none\">\na. The ball reaches the maximum height at the vertex of the parabola.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c} h=-\\frac{80}{2\\left(-16\\right)} \\text{ }=\\frac{80}{32}\\hfill \\\\ \\text{ }=\\frac{5}{2}\\hfill \\\\ \\text{ }=2.5\\hfill \\end{array}[\/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 <em>y\u00a0<\/em>coordinate of the vertex of the parabola.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}k=H\\left(-\\frac{b}{2a}\\right)\\hfill \\\\ \\text{ }=H\\left(2.5\\right)\\hfill \\\\ \\text{ }=-16{\\left(2.5\\right)}^{2}+80\\left(2.5\\right)+40\\hfill \\\\ \\text{ }=140\\hfill \\end{array}[\/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{array}{c} t=\\frac{-80\\pm \\sqrt{{80}^{2}-4\\left(-16\\right)\\left(40\\right)}}{2\\left(-16\\right)}\\hfill \\\\ \\text{ }=\\frac{-80\\pm \\sqrt{8960}}{-32}\\hfill \\end{array}[\/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]\\begin{array}{c}t=\\frac{-80-\\sqrt{8960}}{-32}\\approx 5.458\\hfill & \\text{or}\\hfill & t=\\frac{-80+\\sqrt{8960}}{-32}\\approx -0.458\\hfill \\end{array}[\/latex]<\/p>\n<p>The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 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\" data-media-type=\"image\/jpg\" \/><\/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\">Solution<\/span><\/p>\n<div id=\"q174919\" class=\"hidden-answer\" style=\"display: none\">a.\u00a03 seconds \u00a0b.\u00a0256 feet \u00a0c.\u00a07 seconds<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=15809&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"375\"><\/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-1659\">\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 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Lippman,David\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"MathAS Community License CC-BY + GPL\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at 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