{"id":167,"date":"2023-11-08T16:10:12","date_gmt":"2023-11-08T16:10:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-simple-polynomial-equations\/"},"modified":"2026-02-13T18:38:47","modified_gmt":"2026-02-13T18:38:47","slug":"3-6-graphing-quadratic-functions-in-general-form","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/3-6-graphing-quadratic-functions-in-general-form\/","title":{"raw":"3.6 Graphing Quadratic Functions in General Form","rendered":"3.6 Graphing Quadratic Functions in General Form"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find the vertex, axis of symmetry, [latex]x[\/latex]-intercepts (if any), [latex]y[\/latex]-intercept, and\/or minimum or maximum value of a factorable quadratic function in general form [latex]f(x)=ax^2+bx+c[\/latex].<\/li>\r\n \t<li>Rewrite quadratic functions given in general form to vertex form.<\/li>\r\n \t<li>Graph quadratic functions in general form [latex]f(x)=ax^2+bx+c[\/latex].<\/li>\r\n \t<li>Use factoring and the Zero-Product Property to solve polynomials equations, including applications.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<strong>General Form of a Quadratic Function\u00a0<\/strong>is a function that describes a parabola, written in the form [latex]f\\left(x\\right)=a{x}^{2}+bx+c[\/latex], where [latex]a[\/latex],\u00a0[latex]b[\/latex], and\u00a0[latex]c[\/latex]\u00a0are real numbers and [latex]a\\ne 0[\/latex].\u00a0If [latex]a&gt;0[\/latex], the parabola opens upward. If [latex]a&lt;0[\/latex], the parabola opens downward. We can use the general form of a parabola to find the equation of a line that is the axis of symmetry.\r\n<p id=\"fs-id1165135353112\">The vertex form and the general form of quadratic functions are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the vertex form.<\/p>\r\n\r\n<div id=\"eip-173\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}a{\\left(x-h\\right)}^{2}+k&amp;=a(x^2-2xh+h^2)+k \\\\ &amp;=a{x}^{2} \\underbrace{-2ah}_{b}x+ \\underbrace{a{h}^{2}+k}_{c}=a{x}^{2}+bx+c \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137409211\">For the quadratic expressions to be equal, the corresponding coefficients must be equal.<\/p>\r\n\r\n<div id=\"eip-144\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-2ah=b,\\text{ so }h=-\\dfrac{b}{2a}[\/latex].<\/div>\r\n<p id=\"fs-id1165134118295\">This gives us the <strong>axis of symmetry<\/strong> we defined earlier. Setting the constant terms equal:<\/p>\r\n\r\n<div id=\"eip-313\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}a{h}^{2}+k&amp;=c \\\\ k&amp;=c-a{h}^{2} \\\\ &amp;=c-a{\\left(-\\dfrac{b}{2a}\\right)}^{2} \\\\ &amp;=c-\\dfrac{{b}^{2}}{4a} \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137476446\">In practice, though, it is usually easier to remember that [latex]k[\/latex]\u00a0is the output value of the function when the input is [latex]h[\/latex], so [latex]k=f\\left(h\\right)=f\\left(-\\dfrac{b}{2a}\\right)[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165137749882\" class=\"note textbox\">\r\n<h3 class=\"title\">Different Forms of Quadratic Functions<\/h3>\r\nA quadratic function is a function of degree two. The graph of a <strong>quadratic function<\/strong> is a parabola.\r\n\r\nThe <strong>general form of a quadratic function<\/strong> is [latex]f\\left(x\\right)=a{x}^{2}+bx+c[\/latex] where [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are real numbers and [latex]a\\ne 0[\/latex]. The vertex [latex](h,k)[\/latex] of the parabola is located at [latex]h=\\dfrac{-b}{2a}, \\text{ }k=f\\left(\\dfrac{-b}{2a}\\right)[\/latex].\r\n<p id=\"fs-id1165137666538\">We learned in section 3.3, the <strong>v<\/strong><strong>ertex form of a quadratic function<\/strong> is [latex]f\\left(x\\right)=a{\\left(x-h\\right)}^{2}+k[\/latex]. Where the vertex is [latex]\\left(h,k\\right)[\/latex].<\/p>\r\nWe learned in section 3.4, the\u00a0<strong>intercept form of a quadratic function<\/strong> is [latex]f(x)=a(x-p)(x-q)[\/latex]. Where [latex]p[\/latex] and [latex]q[\/latex] are the [latex]x[\/latex]-intercepts of the graph and the vertex is half-way between the [latex]x[\/latex]-intercepts.\r\n\r\n<\/div>\r\n<p id=\"fs-id1165133234001\">The axis of symmetry is defined by the line [latex]x=\\dfrac{-b}{2a}[\/latex]. If we use factoring to solve [latex]a{x}^{2}+bx+c=0[\/latex] for the [latex]x[\/latex]-intercepts, or zeros, we find the value of [latex]x[\/latex] halfway between them is always [latex]x=\\dfrac{-b}{2a}[\/latex], the equation for the axis of symmetry.<\/p>\r\nFigure 1 shows the graph of the quadratic function written in general form as [latex]y={x}^{2}+4x+3[\/latex]. In this form, [latex]a=1,\\text{ }b=4[\/latex], and [latex]c=3[\/latex]. Because [latex]a&gt;0[\/latex], the parabola opens upward. The axis of symmetry is [latex]x=\\dfrac{-4}{2\\left(1\\right)}=-2[\/latex]. This makes sense because we can see from the graph that the vertical line [latex]x=-2[\/latex] divides the graph in half.\r\n\r\nThe vertex always occurs along the axis of symmetry. We have found the [latex]x[\/latex]-coordinate of the vertex to be [latex]-2[\/latex]. To find the [latex]y[\/latex]-coordinate of the vertex we need to substitute [latex]-2[\/latex] in place of [latex]x[\/latex] and solve for [latex]y[\/latex].\r\n\r\n[latex]\\begin{align}\\require{color} y &amp;= x^2+4x+3\\\\ y &amp;= (\\color{Green}{-2}\\color{black}{)^2+4(}\\color{Green}{-2}\\color{black}{)+3}\\\\ y &amp;=4+(-8)+3\\\\ y &amp;= -1\\end{align}[\/latex]\r\n\r\nIn this instance, [latex]\\left(-2,-1\\right)[\/latex] is the vertex. This parabola opens upward, so the vertex occurs at the lowest point on the graph, also called the minimum point of the graph. The vertex [latex](-2,-1)[\/latex] is the minimum point on this graph.\r\n\r\nThe [latex]x[\/latex]-intercepts are the point(s) where the parabola crosses the [latex]x[\/latex]-axis, these can be found by letting [latex]y=0[\/latex] and solving for [latex]x[\/latex]. We will use factoring and the Zero-Product Property to solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align} y &amp;= x^2+4x+3\\\\ 0 &amp;=x^2+4x+3 &amp;&amp; \\color{blue}{\\textsf{Which two numbers multiply to be 3 and add to be 4?}} \\\\ 0 &amp;=(x+3)(x+1) &amp;&amp; \\color{blue}{\\textsf{set each factor equal to zero and solve}}\\\\ x+3 &amp;=0 \\quad x+1 =0\\\\ x &amp;= -3 \\quad x=-1 \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">The [latex]x[\/latex]-intercepts occur at [latex]\\left(-3,0\\right)[\/latex] and [latex]\\left(-1,0\\right)[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010711\/CNX_Precalc_Figure_03_02_0042.jpg\" alt=\"Graph of a parabola showing where the x and y intercepts, vertex, and axis of symmetry are for the function y=x^2+4x+3.\" width=\"487\" height=\"555\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\n<div id=\"fs-id1165135168275\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135574310\">How To: \u00a0find the vertex when given a quadratic function in general form<\/h3>\r\n<ol id=\"fs-id1165134108459\">\r\n \t<li>Identify [latex]a[\/latex],\u00a0[latex]b[\/latex], and\u00a0[latex]c[\/latex].<\/li>\r\n \t<li>Find [latex]h[\/latex], the [latex]x[\/latex]-coordinate of the vertex, by substituting [latex]a[\/latex] and [latex]b[\/latex] into [latex]h=\\dfrac{-b}{2a}[\/latex].<\/li>\r\n \t<li>Find [latex]k[\/latex], the [latex]y[\/latex]-coordinate of the vertex, by evaluating [latex]k=f\\left(h\\right)=f\\left(\\dfrac{-b}{2a}\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the vertex of the quadratic function [latex]f\\left(x\\right)=2{x}^{2}-8x+7[\/latex]. Rewrite the quadratic function in vertex form.\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"648542\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"648542\"]\r\n\r\nIdentify the coefficients:\u00a0[latex]a=2[\/latex], [latex]b=-8[\/latex], and [latex]c=7[\/latex].\r\n<p id=\"fs-id1165137596323\">The [latex]x[\/latex]-coordinate of the vertex will be at<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} h &amp;=\\dfrac{-b}{2a} \\\\ h &amp;=\\dfrac{-(-8)}{2\\left(2\\right)} \\\\ h &amp;=\\dfrac{8}{4} \\\\ h &amp;=2 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137465689\">The [latex]y[\/latex]-coordinate of the vertex will be at<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}k &amp;= f(h) \\\\ k &amp;= f(2) \\\\ k &amp;= 2(2)^{2}-8(2)+7 \\\\ k &amp;= 8-16+7\\\\ k &amp;=-1 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135177784\">Rewriting into vertex form, the stretch factor of [latex]2[\/latex] will be the same as the [latex]a[\/latex] in the original quadratic.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}f\\left(x\\right)=a{x}^{2}+bx+c \\\\ f\\left(x\\right)=2{x}^{2}-8x+7 \\end{gathered}[\/latex]<\/p>\r\n<p id=\"fs-id1165137653186\">Writing in vertex form we found [latex]a=2[\/latex], [latex]h=2[\/latex], and [latex]k=-1[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]f(x)=a(x-h)^2+k[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]f(x)=2(x-2)^{2}+(-1)[\/latex]<\/p>\r\nThe vertex is [latex](2,-1)[\/latex]. Since [latex]a&gt;0[\/latex], this indicates the parabola opens upward. The minimum value is [latex]k=-1[\/latex] with the lowest point being [latex](2,-1)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div id=\"Example_03_02_03\" class=\"example\">\r\n<div id=\"fs-id1165137658566\" class=\"exercise\"><\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135193262\">Given the function [latex]g\\left(x\\right)=13+{x}^{2}-6x[\/latex], write the function in general form <strong>and<\/strong> then in vertex form.<\/p>\r\n[reveal-answer q=\"312729\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"312729\"]\r\n\r\n&nbsp;\r\n\r\n<strong>In general form:<\/strong>\r\n\r\nRearrange the terms so that they are in descending order.\r\n\r\n[latex]g\\left(x\\right)={x}^{2}-6x+13[\/latex]\r\n\r\n&nbsp;\r\n\r\n<strong>In vertex form:<\/strong>\r\n\r\nFind the vertex to change the equation to vertex form [latex]g(x)=a(x-h)^2+k[\/latex].\r\n\r\nFrom the general form above, we know that [latex]a=1[\/latex], [latex]b=-6[\/latex], and [latex]c=13[\/latex].\r\n\r\nThere is no vertical stretch or compression since [latex]a=1[\/latex].\r\n<p id=\"fs-id1165137596323\">The [latex]x[\/latex]-coordinate of the vertex will be at<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} h &amp;=\\dfrac{-b}{2a} \\\\ h &amp;=\\dfrac{-(-6)}{2\\left(1\\right)} \\\\ h &amp;=\\dfrac{6}{2} \\\\ h &amp;=3 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137465689\">The [latex]y[\/latex]-coordinate of the vertex will be at<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}k &amp;= f(h) \\\\ k &amp;= f(3) \\\\ k &amp;= (3)^2-6(3)+13 \\\\ k &amp;= 9-18+13\\\\ k &amp;=4 \\end{align}[\/latex]<\/p>\r\nThe function in vertex form is: [latex]g\\left(x\\right)={\\left(x - 3\\right)}^{2}+4[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solve Quadratic Equations using Factoring and the Zero-Product Property<\/h2>\r\nLet us start with an example that factors a GCF from a binomial and apply the Zero-Product Property to solve a polynomial equation.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]-t^2+t=0[\/latex]\r\n[reveal-answer q=\"612316\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"612316\"]\r\n\r\nEach term has a common factor of [latex]-t[\/latex], so we can factor out the GCF of [latex]-t[\/latex] and use the zero product principle.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}-t^2+t=0\\\\-t\\left(t-1\\right)=0\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now we have a product on one side and zero on the other, so we can set each factor equal to zero and solve using the Zero-Product Property.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\\\ -t\\hfill=0 &amp; &amp; t-1=0 \\hfill \\\\ t=0 &amp; &amp; t=1 \\hfill \\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">The solutions are [latex]t=0[\/latex] or [latex]t=1[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show two more examples of using both factoring and the Zero-Product Property to solve a quadratic equation.\r\n\r\nhttps:\/\/youtu.be\/gIwMkTAclw8\r\n\r\nIn the next video, we show that you can use previously learned methods to factor a trinomial in order to solve a quadratic equation.\r\n\r\nhttps:\/\/youtu.be\/bi7i_RuIGl0\r\n\r\nWe will now solve a quadratic equation where it is not set equal to zero on one side.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]x^2-4x=5[\/latex]\r\n[reveal-answer q=\"165196\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"165196\"]\r\n\r\nFirst, move all the terms to one side. The goal is to try and see if we can use the Zero-Product Property since that is the only tool we currently have for solving quadratic equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^2-4x=5\\\\x^2-4x-5=0\\\\\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We now have all the terms on the left side and zero on the other side. Factor the left side of the equation.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^2-4x-5=0\\\\\\left(x+1\\right)\\left(x-5\\right)=0\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We separate our factors into two linear equations using the Zero-Product Property.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\\\ x-5\\hfill=0 &amp; &amp; x+1=0 \\hfill \\\\ x=5 &amp; &amp; x=-1 \\hfill \\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">The solutions are [latex]x=5\\text{ or }x=-1[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nThe following binomial is a difference of squares and can be factored. Factor the binomial and use the zero-product property to solve.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve using the Zero-Product Property: [latex]{x}^{2}-9=0[\/latex].\r\n[reveal-answer q=\"148980\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"148980\"]\r\n\r\nRecognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the Zero-Product Property.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} {x}^{2}-9 &amp;= 0\\\\ (x-3)(x+3) &amp;= 0 \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\\\ x - 3\\hfill=0 &amp; &amp; x+3=0 \\hfill \\\\ x=3 &amp; &amp; x=-3 \\hfill \\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">The solutions are [latex]x=3[\/latex] or [latex]x=-3[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere are some video examples in which you can use factoring to solve these quadratic equations.\r\n\r\nhttps:\/\/youtu.be\/bi7i_RuIGl0?si=QVBAatiDF1JTZ1Ry\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUse factoring and the Zero-Product Property to solve the quadratic equation: [latex]4{x}^{2}+15x+9=0[\/latex].\r\n[reveal-answer q=\"716142\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"716142\"]\r\n\r\nFirst, multiply [latex]ac:4\\left(9\\right)=36[\/latex]. Then list the factors of [latex]36[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}1\\cdot 36\\hfill \\\\ 2\\cdot 18\\hfill \\\\ 3\\cdot 12\\hfill \\\\ 4\\cdot 9\\hfill \\\\ 6\\cdot 6\\hfill \\end{array}[\/latex]<\/div>\r\nThe only pair of factors that sums to [latex]15[\/latex] is [latex]3+12[\/latex]. Rewrite the equation replacing the [latex]b[\/latex] term, [latex]15x[\/latex], with two terms using\u00a0[latex]3[\/latex] and\u00a0[latex]12[\/latex] as coefficients of [latex]x[\/latex]. Factor the first two terms, and then factor the last two terms.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}4{x}^{2}+3x+12x+9=0\\hfill \\\\ x\\left(4x+3\\right)+3\\left(4x+3\\right)=0\\hfill \\\\ \\left(4x+3\\right)\\left(x+3\\right)=0\\hfill \\end{array}[\/latex]<\/div>\r\nSolve using the Zero-Product Property.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}&amp;\\left(4x+3\\right)\\left(x+3\\right)=0 &amp; \\hfill \\\\ 4x+3\\hfill=0 &amp; &amp; x+3=0 \\hfill \\\\ x=-\\dfrac{3}{4} &amp; &amp; x=-3 \\hfill \\end{array}[\/latex]<\/p>\r\nThe solutions are [latex]x=-\\dfrac{3}{4}[\/latex], [latex]x=-3[\/latex]. You can see these solutions as the [latex]x[\/latex]-intercepts of the parabola graphed below.\r\n\r\n<img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200359\/CNX_CAT_Figure_02_05_003n.jpg\" alt=\"Parabola facing up with x-intercepts: (negative 3\/4, 0) and (negative 3, 0) with vertex of (negative 2, negative 5).\" width=\"487\" height=\"433\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video examples can be factored but the leading coefficients is NOT 1.\r\n\r\nhttps:\/\/youtu.be\/04zEXaOiO4U?si=PQIYhtN8plQvIt7m\r\n<h2>Graph quadratic functions in general form [latex]f(x)=ax^2+bx+c[\/latex]<\/h2>\r\nTo graph quadratic functions in general form, we will first find the vertex, axis of symmetry, [latex]x[\/latex]-intercept(s), and the [latex]y[\/latex]-intercept. We will graph the function, [latex]g(x)=-x^2-8x-15[\/latex] where [latex]a=-1[\/latex], [latex]b=-8[\/latex], and [latex]c=-15[\/latex].\r\n\r\nStart by finding the vertex.\u00a0The [latex]x[\/latex]-coordinate of the vertex:\r\n\r\n[latex]\\begin{align} h &amp;=\\dfrac{-b}{2a} \\\\ h &amp;=\\dfrac{-(-8)}{2\\left(-1\\right)} \\\\ h &amp;=\\dfrac{8}{-2} \\\\ h &amp;=-4 \\end{align}[\/latex]\r\n<p id=\"fs-id1165137465689\">The [latex]y[\/latex]-coordinate of the vertex:<\/p>\r\n[latex]\\begin{align}k &amp;= f(h) \\\\ k &amp;= f(-4) \\\\ k &amp;= -(-4)^2-8(-4)-15 \\\\ k &amp;= -16+32-15\\\\ k &amp;=1 \\end{align}[\/latex]\r\n\r\nThe vertex is [latex](-4,1)[\/latex].\r\n\r\nThe axis of symmetry is a vertical line that passes through the vertex. In this case the axis (line) of symmetry is [latex]x=-4[\/latex].\r\n\r\nWe can rewrite the equation as [latex]y=-x^2-8x-15[\/latex]. To find the [latex]y[\/latex]-intercept, let [latex]x=0[\/latex] and solve for [latex]y[\/latex].\r\n\r\n[latex]\\begin{align} y &amp;= -x^2-8x-15\\\\ y &amp;= -(0)^2-8(0)-15\\\\ y &amp;= -15\\end{align}[\/latex]\r\n\r\nThe [latex]y[\/latex]-intercept is [latex](0,-15)[\/latex].\r\n\r\nTo find the [latex]x[\/latex]-intercept(s), let [latex]y=0[\/latex] and solve for [latex]x[\/latex]. We will use factoring and the Zero-Product Property to find the [latex]x[\/latex]-intercept(s).\r\n\r\n[latex]\\begin{align} y &amp;= -x^2-8x-15\\\\ 0 &amp;= -x^2-8x-15\\\\ 0 &amp;= -1(x^2+8x+15)\\\\ 0 &amp;= -1(x+3)(x+5)\\end{align}[\/latex]\r\n[latex] x+3=0 \\hspace{1.5cm} x+5=0[\/latex]\r\n[latex] x = -3\\hspace{2cm} x = -5[\/latex]\r\n\r\nThe [latex]x[\/latex]-intercepts are [latex](-3,0)[\/latex] and [latex](-5,0)[\/latex].\r\n\r\nFigure 2 is the graph of the function [latex]g(x)=-x^2-8x-15[\/latex].\r\n\r\n<iframe style=\"border: 1px solid #ccc;\" src=\"https:\/\/www.desmos.com\/calculator\/gews1xa4kt?embed\" width=\"500\" height=\"500\" frameborder=\"0\"><\/iframe>\r\nFigure 2\r\n\r\nNOTE: If the [latex]x[\/latex]-intercepts do not exist or the quadratic equation cannot be factored to solve or if the [latex]y[\/latex]-intercept is off the grid that you are being asked to graph on, you can make a table of values to help you find additional point(s) on the graph. You can pick [latex]x[\/latex] values that are on either side of the vertex, for this example we could pick [latex]x=-2[\/latex] and [latex]x=-6[\/latex].\r\n<table style=\"border-collapse: collapse; width: 25%; border: 1px solid black; height: 36px;\">\r\n<thead>\r\n<tr style=\"height: 12px;\">\r\n<th style=\"width: 30%; border: 1px solid #999999; height: 12px; text-align: center;\">\u00a0[latex]x[\/latex]<\/th>\r\n<th style=\"width: 30%; border: 1px solid #999999; height: 12px; text-align: center;\">[latex]y=-x^2-8x-15[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 30%; border: 1px solid #999999; height: 12px; text-align: center;\">[latex]-2[\/latex]<\/td>\r\n<td style=\"width: 70%; border: 1px solid #999999; height: 12px; text-align: center;\">[latex]\\begin{align} y &amp;= -1(\\color{Green}{-2}\\color{black}{)^2-8}(\\color{Green}{-2}\\color{black}{)-15}\\\\ &amp;= -1(4)+16-15\\\\ &amp;= -3 \\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 30%; border: 1px solid #999999; height: 12px; text-align: center;\">[latex]-6[\/latex]<\/td>\r\n<td style=\"width: 70%; border: 1px solid #999999; height: 12px; text-align: center;\">[latex]\\begin{align} y &amp;= -1(\\color{Green}{-6}\\color{black}{)^2-8}(\\color{Green}{-6}\\color{black}{)-15}\\\\ &amp;= -1(36)+48-15\\\\ &amp;= -3 \\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div>So, two more points we can use to graph this quadratic function are [latex](-2,-3)[\/latex] and [latex](-6,-3)[\/latex].<\/div>\r\n<div><\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGraph [latex]f(x)=2x^2-4x+3[\/latex] by first finding the vertex, axis of symmetry, [latex]y[\/latex]-intercept, and [latex]x[\/latex]-intercept(s).\r\n\r\n[reveal-answer q=\"374151\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"374151\"]\r\n\r\nThe function can also be written as [latex]y=2x^2-4x+3[\/latex]\r\n\r\n<strong>Vertex: [latex](h,k)[\/latex]<\/strong>\r\n\r\n[latex]x[\/latex]-coordinate:\r\n\r\n[latex]\\begin{align} h &amp;=\\frac{-b}{2a} \\\\ h &amp;=\\frac{-(-4)}{2\\left(2\\right)} \\\\ h &amp;=\\frac{4}{4} \\\\ h &amp;=1 \\end{align}[\/latex]\r\n\r\n[latex]y[\/latex]-coordinate:\r\n\r\n[latex]\\begin{align} k &amp;= f(1) \\\\ k &amp;= 2(1)^2-4(1)+3 \\\\ k &amp;= 2-4+3\\\\ k &amp;=1 \\end{align}[\/latex]\r\n\r\nThe vertex is [latex](1,1)[\/latex].\r\n\r\n<strong>Axis of Symmetry\u00a0<\/strong>is the vertical line that passes through the vertex. That line is [latex]x=1[\/latex].\r\n\r\n<strong>[latex]y[\/latex]-intercept:\u00a0<\/strong>Let [latex]x=0[\/latex] and solve for [latex]y[\/latex].\r\n\r\n[latex]\\begin{align} y &amp;= 2x^2-4x+3\\\\ y &amp;= 2(0)^2-4(0)+3\\\\ y &amp;= 3\\end{align}[\/latex]\r\n\r\nThe [latex]y[\/latex]-intercept is [latex](0,3)[\/latex].\r\n\r\n<strong>[latex]x[\/latex]-intercept(s):\u00a0<\/strong>Let [latex]y=0[\/latex] and solve for [latex]x[\/latex]. Use factoring to solve.\r\n\r\n[latex]\\begin{align} y &amp;= 2x^2-4x+3\\\\ 0 &amp;= 2x^2-4x+3\\end{align}[\/latex]\r\n\r\n[latex]a\\cdot c = 2\\cdot3=6[\/latex] We are looking for two integers that multiply to be [latex]6[\/latex] and add to be [latex]-4[\/latex]. There are no two integers that satisfy this. This cannot be factored. Later in the book,\u00a0<span style=\"font-size: 1rem; text-align: initial;\">we will discuss different methods besides factoring in which you can find the [latex]x[\/latex]-intercepts. For now, we will graph this function with the other points we have found and use symmetry of a parabola to find another point on the graph.<\/span>\r\n\r\n<iframe style=\"border: 1px solid #ccc;\" src=\"https:\/\/www.desmos.com\/calculator\/ppyx5cc6af?embed\" width=\"500\" height=\"500\" frameborder=\"0\"><\/iframe>\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Solving Applications of Quadratic Functions<\/h2>\r\n<p id=\"fs-id1165137431411\">There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue. Remember when [latex]a&gt;0[\/latex], the graph of a parabola opens upward and the vertex is the lowest point on the graph or the minimum. When [latex]a&lt;0[\/latex], the graph of a parabola opens downward and the vertex is the highest point on the graph or the maximum.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010712\/CNX_Precalc_Figure_03_02_0092.jpg\" alt=\"Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).\" width=\"975\" height=\"558\" \/> <b>Figure 3<\/b>[\/caption]\r\n\r\nProjectile motion happens when you throw a ball into the air and it comes back down because of gravity. \u00a0A projectile will follow a curved path that behaves in a predictable way. \u00a0This predictable motion has been studied for centuries, and in simple cases, an object's height from the ground\u00a0at a given time,\u00a0[latex]t[\/latex], can be modeled with a polynomial function of the form [latex]h(t)=at^2+bt+c[\/latex], where [latex]h(t) =[\/latex] height of an object at a given time,\u00a0[latex]t[\/latex]. \u00a0Projectile motion is also called a parabolic trajectory because of the shape of the path of a projectile's motion, as in the water in the image of the fountain below.\u00a0Parabolic motion and its related functions\u00a0allow us to launch satellites for telecommunications and rockets for space exploration.\r\n\r\n[caption id=\"attachment_4890\" align=\"aligncenter\" width=\"436\"]<img class=\"wp-image-4890\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161949\/ParabolicWaterTrajectory-225x300.jpg\" alt=\"Water from a fountain shoing classic parabolic motion.\" width=\"436\" height=\"581\" \/> Parabolic water trajectory in a fountain.[\/caption]\r\n\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nA small toy rocket is launched from a\u00a0[latex]4[\/latex]-foot pedestal. The height ([latex]h[\/latex]<i>, <\/i>in feet) of the rocket [latex]t[\/latex] seconds after taking off is given by the function [latex]h(t)=\u22122t^{2}+7t+4[\/latex]. How long will it take the rocket to hit the ground?\r\n[reveal-answer q=\"679533\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"679533\"]\r\n\r\nThe rocket will be on the ground when [latex]h(t)=0[\/latex]. We want to know how long,\u00a0[latex]t[\/latex], the rocket is in the air.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}h(t)=\u22122t^{2}+7t+4\\\\0=\u22122t^{2}+7t+4\\end{array}[\/latex]<\/p>\r\nWe can factor the polynomial [latex]\u22122t^{2}+7t+4[\/latex] more easily by first factoring out a [latex]-1[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}0=-1(2t^{2}-7t-4)\\\\0=-1\\left(2t+1\\right)\\left(t-4\\right)\\end{array}[\/latex]<\/p>\r\nUse the Zero-Product Property. There is no need to set the constant factor [latex]-1[\/latex] to zero, because [latex]-1[\/latex] will never equal zero. Set the other two factors equal to zero and solve.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}&amp;-1\\left(2t+1\\right)\\left(t-4\\right)=0 &amp; \\hfill \\\\ 2t+1\\hfill=0 &amp; &amp; t-4=0 \\hfill \\\\ t=-\\dfrac{1}{2} &amp; &amp; t=4 \\hfill \\end{array}[\/latex]<\/p>\r\nSince [latex]t[\/latex] represents time after the launch, it cannot be a negative number; only [latex]t=4[\/latex]\u00a0makes sense in this context.\r\n\r\nTherefore, the rocket will hit the ground [latex]4[\/latex] seconds after being launched.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-simple-polynomial-equations\/screenshot-2024-07-26-at-11-05-53%e2%80%afam\/\" rel=\"attachment wp-att-1554\"><img class=\"aligncenter wp-image-1554 size-full\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-07-26-at-11.05.53\u202fAM.png\" alt=\"decorative image\" width=\"309\" height=\"245\" \/><\/a>\r\n<div class=\"textbox shaded\" style=\"text-align: center;\">Model rocket being launched.\u00a0https:\/\/www.instructables.com\/Remote-Rocket-Igniter\/<\/div>\r\nIn the next example, we will solve for the time that the rocket reaches a given height other than zero.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUse the formula for the height of the rocket in the previous example to find the time when the rocket is\u00a0[latex]4[\/latex] feet from hitting the ground on its way back down. \u00a0Refer to the image.\r\n\r\n[latex]h(t)=\u22122t^{2}+7t+4[\/latex]\r\n\r\n<img class=\"wp-image-4892 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161951\/Screen-Shot-2016-06-14-at-5.39.53-PM-300x247.png\" alt=\"Parabolic motion of rocket which starts four feet up from the ground. t=0 is labeled at the starti of hte parabolic motion adn t=? is labeled at four feet from the ground on the other side of the parabola.\" width=\"413\" height=\"340\" \/>\r\n[reveal-answer q=\"198118\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"198118\"]\r\n\r\nWe are given that the height of the rocket is\u00a0[latex]4[\/latex] feet from the ground on its way back down. We want to know how long it will take for the rocket to get to that point in its path. We are going to solve for [latex]t[\/latex].\r\n\r\nSubstitute [latex]h(t) = 4[\/latex] into the formula for height, and try to get zero on one side since we know we can use the Zero-Product Property to solve polynomial equations.\r\n\r\n<strong>Write and Solve:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}4=-2t^2+7t+4\\\\\\underline{-4}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-4}\\\\0=-2t^2+7t\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now we can factor out a\u00a0[latex]t[\/latex] from each term:<\/p>\r\n<p style=\"text-align: center;\">[latex]0=t\\left(-2t+7\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Solve each equation for\u00a0[latex]t[\/latex] using the Zero-Product Property:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=0\\text{ or }-2t+7=0\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-7}\\,\\,\\,\\,\\,\\,\\,\\underline{-7}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\dfrac{-2t}{-2}=\\dfrac{-7}{-2}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,t=\\dfrac{7}{2}=3.5\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">It does not make sense for us to choose\u00a0[latex]t=0[\/latex], because we are interested in the amount of time that has passed when the projectile is\u00a0[latex]4[\/latex] feet from hitting the ground on its way back down. We will choose\u00a0[latex]t=3.5[\/latex].<\/p>\r\nThe rocket will be [latex]4[\/latex] feet from the ground at [latex]t=3.5\\text{ seconds}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of how to find the time when a object following a parabolic trajectory hits the ground.\r\n\r\nhttps:\/\/youtu.be\/hsWSzu3KcPU\r\n\r\nIn this section we introduced the concept of projectile motion and showed that it can be modeled with quadratic functions. While the models used in these examples can be factored, that is not always the case with these types of applications. The concepts and interpretations are the same as what would happen in \"real life.\"\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\nA backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased [latex]80[\/latex] feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.\r\n<ol id=\"fs-id1165135640934\">\r\n \t<li>Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length [latex]L[\/latex].<\/li>\r\n \t<li>What dimensions should she make her garden to maximize the enclosed area?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"478551\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"478551\"]\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010713\/CNX_Precalc_Figure_03_02_0102.jpg\" alt=\"Rectangle backyard with a rectangle garden center at the top. Garden labeled W for width and L for length.\" width=\"487\" height=\"310\" \/> <b>Figure 4<\/b>[\/caption]\r\n<p id=\"fs-id1165137836808\">Let\u2019s use a diagram such as the one in Figure 4 to record the given information. It is also helpful to introduce a temporary variable, [latex]W[\/latex], to represent the width of the garden and the length of the fence section parallel to the backyard fence.<span id=\"fs-id1165135208803\">\r\n<\/span><\/p>\r\n\r\n<ol id=\"fs-id1165134363440\">\r\n \t<li>We know we have only [latex]80[\/latex] feet of fence available, and [latex]L+W+L=80[\/latex], or more simply, [latex]2L+W=80[\/latex]. This allows us to represent the width, [latex]W[\/latex], in terms of [latex]L[\/latex].\r\n<div id=\"eip-id1165135697866\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]W=80 - 2L[\/latex]<\/div>\r\n<p id=\"fs-id1165135435476\">Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so<\/p>\r\n<p id=\"fs-id1165135435476\" style=\"text-align: center;\">[latex]\\begin{align}A&amp;=LW \\\\ &amp;=L\\left(80 - 2L\\right) \\\\ &amp;=80L - 2{L}^{2} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135258914\">This formula represents the area of the fence in terms of the variable length [latex]L[\/latex]. The function, written in general form, is<\/p>\r\n\r\n<div id=\"eip-382\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A\\left(L\\right)=-2{L}^{2}+80L[\/latex].<\/div><\/li>\r\n \t<li>The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation does not have a constant term, meaning [latex]c=0[\/latex]. Writing the function in general form we can \u00a0identify that\u00a0[latex]a=-2,b=80[\/latex], and [latex]c=0[\/latex].<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1165137772015\">To find the vertex:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}h&amp;=\\dfrac{-80}{2\\left(-2\\right)} &amp;&amp;&amp; k&amp;=A\\left(20\\right) \\\\ h&amp;=20 &amp;&amp; \\text{and} &amp; k&amp;=-2{\\left(20\\right)}^{2}+80\\left(20\\right)\\\\ &amp;&amp;&amp;&amp;&amp;k=800 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135174964\">The maximum value of the function is an area of [latex]800[\/latex] square feet, which occurs when [latex]L=20[\/latex] feet. When the shorter sides are [latex]20[\/latex] feet, there is [latex]40[\/latex] feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length [latex]20[\/latex] feet and the longer side parallel to the existing fence has length [latex]40[\/latex] feet.<\/p>\r\nThis problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure 5.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010713\/CNX_Precalc_Figure_03_02_0112.jpg\" alt=\"Graph of the parabolic function A(L)=-2L^2+80L, which the x-axis is labeled Length (L) and the y-axis is labeled Area (A). The vertex is at (20, 800).\" width=\"487\" height=\"476\" \/> <b>Figure 5<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA ball is thrown upward from the top of a [latex]40[\/latex]-foot high building at a speed of [latex]80[\/latex] 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<p style=\"padding-left: 60px;\">a. When does the ball reach the maximum height?<\/p>\r\n<p style=\"padding-left: 60px;\">b. What is the maximum height of the ball?<\/p>\r\n[reveal-answer q=\"340065\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"340065\"]\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)} =\\dfrac{-80}{-32} \\\\ &amp;=\\dfrac{5}{2} \\\\ &amp;=2.5 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135528870\">The ball reaches a maximum height after [latex]2.5[\/latex] seconds.<\/p>\r\nb. To find the maximum height, find the [latex]y[\/latex]-coordinate of the vertex of the parabola.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}k&amp;=H\\left(\\dfrac{-b}{2a}\\right) \\\\ &amp;=H\\left(2.5\\right) \\\\ &amp;=-16{\\left(2.5\\right)}^{2}+80\\left(2.5\\right)+40 \\\\ &amp;=140 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135409750\">The ball reaches a maximum height of [latex]140[\/latex] feet.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_03_02_08\" class=\"example\">\r\n<div id=\"fs-id1165134060458\" class=\"exercise\">\r\n<div id=\"fs-id1165137843086\" class=\"commentary\">\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134081301\">A rock is thrown upward from the top of a [latex]112[\/latex]-foot high cliff overlooking the ocean at a speed of [latex]96[\/latex] 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], where [latex]H[\/latex] is the height in feet and [latex]t[\/latex] is time after the throw in seconds.<\/p>\r\n<p style=\"padding-left: 60px;\">a. When does the rock reach the maximum height?<\/p>\r\n<p style=\"padding-left: 60px;\">b. What is the maximum height of the rock?<\/p>\r\n<p style=\"padding-left: 60px;\">c. When does the rock hit the ocean?<\/p>\r\n[reveal-answer q=\"4901\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"4901\"]\r\n\r\na.\u00a03 seconds \u00a0b.\u00a0256 feet \u00a0c.\u00a07 seconds\r\n\r\nExplanation:\r\n\r\na. The rock reaches the maximum height at the vertex of the parabola.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} h&amp;=\\dfrac{-96}{2\\left(-16\\right)}\\\\ h&amp;=\\dfrac{-96}{-32} \\\\ h&amp;= 3 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135528870\">The rock reaches a maximum height at [latex]3[\/latex] seconds.<\/p>\r\nb. To find the maximum height, find the [latex]y[\/latex]-coordinate of the vertex of the parabola.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}k&amp;=-16{\\left(3\\right)}^{2}+96\\left(3\\right)+112 \\\\ k&amp;=256 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135409750\">The rock reaches a maximum height of [latex]256[\/latex] feet.<\/p>\r\nc. To find when the rock hits the ocean, we need to find when the height, [latex]H(t)[\/latex] equals [latex]0[\/latex]. We factor to solve for [latex]t[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align} H(t) &amp;= -16t^2+96t+112\\\\ 0 &amp;= -16t^2+96t+112\\\\ 0 &amp;= -16(t^2-6t-7)\\\\ 0 &amp;= -16(t+1)(t-7) \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]t+1=0[\/latex] or [latex]t-7=0[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]t = -1[\/latex] or [latex]t = 7[\/latex]<\/p>\r\nSince [latex]t[\/latex] represents time, it cannot be a negative number; only [latex]t=7[\/latex]\u00a0makes sense in this context.\r\n\r\nThe rock hits the ocean [latex]7[\/latex] seconds after it is thrown.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Key Equations<\/span>\r\n\r\n<section id=\"fs-id1165134205927\" class=\"key-equations\">\r\n<table id=\"eip-id1165137539373\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>Vertex form of a quadratic function<\/td>\r\n<td>[latex]f\\left(x\\right)=a{\\left(x-h\\right)}^{2}+k[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Intercept form of a quadratic function<\/td>\r\n<td>[latex]f(x)=(x-p)(x-q)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>General form of a quadratic function<\/td>\r\n<td>[latex]f\\left(x\\right)=a{x}^{2}+bx+c[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section>\r\n<h2>Summary<\/h2>\r\nYou can rewrite a quadratic function that is given in general form [latex]f(x)=ax^2+bx+c[\/latex] in vertex form [latex]f(x)=a(x-h)^2+k[\/latex].\r\n\r\nYou can find the vertex, axis of symmetry, [latex]y[\/latex]-intercept, and [latex]x[\/latex]-intercepts (if any) of a quadratic function in general form and use those points to graph the parabola.\r\n\r\nYou can find the solutions, or roots, of quadratic equations in general form by setting one side equal to zero, factoring the quadratic, and then applying the Zero-Product Property. The Zero-Product Property states that if [latex]ab=0[\/latex], then either [latex]a=0[\/latex] or [latex]b=0[\/latex], or both [latex]a[\/latex] and [latex]b[\/latex] are\u00a0[latex]0[\/latex]. Once the polynomial is factored, set each factor equal to zero and solve them separately. The answers will be the set of solutions for the original equation.\r\n\r\nNot all solutions are appropriate for some applications. For example, in many real-world situations, negative solutions are not appropriate and must be discarded.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find the vertex, axis of symmetry, [latex]x[\/latex]-intercepts (if any), [latex]y[\/latex]-intercept, and\/or minimum or maximum value of a factorable quadratic function in general form [latex]f(x)=ax^2+bx+c[\/latex].<\/li>\n<li>Rewrite quadratic functions given in general form to vertex form.<\/li>\n<li>Graph quadratic functions in general form [latex]f(x)=ax^2+bx+c[\/latex].<\/li>\n<li>Use factoring and the Zero-Product Property to solve polynomials equations, including applications.<\/li>\n<\/ul>\n<\/div>\n<p><strong>General Form of a Quadratic Function\u00a0<\/strong>is a function that describes a parabola, written in the form [latex]f\\left(x\\right)=a{x}^{2}+bx+c[\/latex], where [latex]a[\/latex],\u00a0[latex]b[\/latex], and\u00a0[latex]c[\/latex]\u00a0are real numbers and [latex]a\\ne 0[\/latex].\u00a0If [latex]a>0[\/latex], the parabola opens upward. If [latex]a<0[\/latex], the parabola opens downward. We can use the general form of a parabola to find the equation of a line that is the axis of symmetry.\n\n\n<p id=\"fs-id1165135353112\">The vertex form and the general form of quadratic functions are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the vertex form.<\/p>\n<div id=\"eip-173\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}a{\\left(x-h\\right)}^{2}+k&=a(x^2-2xh+h^2)+k \\\\ &=a{x}^{2} \\underbrace{-2ah}_{b}x+ \\underbrace{a{h}^{2}+k}_{c}=a{x}^{2}+bx+c \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137409211\">For the quadratic expressions to be equal, the corresponding coefficients must be equal.<\/p>\n<div id=\"eip-144\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-2ah=b,\\text{ so }h=-\\dfrac{b}{2a}[\/latex].<\/div>\n<p id=\"fs-id1165134118295\">This gives us the <strong>axis of symmetry<\/strong> we defined earlier. Setting the constant terms equal:<\/p>\n<div id=\"eip-313\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}a{h}^{2}+k&=c \\\\ k&=c-a{h}^{2} \\\\ &=c-a{\\left(-\\dfrac{b}{2a}\\right)}^{2} \\\\ &=c-\\dfrac{{b}^{2}}{4a} \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137476446\">In practice, though, it is usually easier to remember that [latex]k[\/latex]\u00a0is the output value of the function when the input is [latex]h[\/latex], so [latex]k=f\\left(h\\right)=f\\left(-\\dfrac{b}{2a}\\right)[\/latex].<\/p>\n<div id=\"fs-id1165137749882\" class=\"note textbox\">\n<h3 class=\"title\">Different Forms of Quadratic Functions<\/h3>\n<p>A quadratic function is a function of degree two. The graph of a <strong>quadratic function<\/strong> is a parabola.<\/p>\n<p>The <strong>general form of a quadratic function<\/strong> is [latex]f\\left(x\\right)=a{x}^{2}+bx+c[\/latex] where [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are real numbers and [latex]a\\ne 0[\/latex]. The vertex [latex](h,k)[\/latex] of the parabola is located at [latex]h=\\dfrac{-b}{2a}, \\text{ }k=f\\left(\\dfrac{-b}{2a}\\right)[\/latex].<\/p>\n<p id=\"fs-id1165137666538\">We learned in section 3.3, the <strong>v<\/strong><strong>ertex form of a quadratic function<\/strong> is [latex]f\\left(x\\right)=a{\\left(x-h\\right)}^{2}+k[\/latex]. Where the vertex is [latex]\\left(h,k\\right)[\/latex].<\/p>\n<p>We learned in section 3.4, the\u00a0<strong>intercept form of a quadratic function<\/strong> is [latex]f(x)=a(x-p)(x-q)[\/latex]. Where [latex]p[\/latex] and [latex]q[\/latex] are the [latex]x[\/latex]-intercepts of the graph and the vertex is half-way between the [latex]x[\/latex]-intercepts.<\/p>\n<\/div>\n<p id=\"fs-id1165133234001\">The axis of symmetry is defined by the line [latex]x=\\dfrac{-b}{2a}[\/latex]. If we use factoring to solve [latex]a{x}^{2}+bx+c=0[\/latex] for the [latex]x[\/latex]-intercepts, or zeros, we find the value of [latex]x[\/latex] halfway between them is always [latex]x=\\dfrac{-b}{2a}[\/latex], the equation for the axis of symmetry.<\/p>\n<p>Figure 1 shows the graph of the quadratic function written in general form as [latex]y={x}^{2}+4x+3[\/latex]. In this form, [latex]a=1,\\text{ }b=4[\/latex], and [latex]c=3[\/latex]. Because [latex]a>0[\/latex], the parabola opens upward. The axis of symmetry is [latex]x=\\dfrac{-4}{2\\left(1\\right)}=-2[\/latex]. This makes sense because we can see from the graph that the vertical line [latex]x=-2[\/latex] divides the graph in half.<\/p>\n<p>The vertex always occurs along the axis of symmetry. We have found the [latex]x[\/latex]-coordinate of the vertex to be [latex]-2[\/latex]. To find the [latex]y[\/latex]-coordinate of the vertex we need to substitute [latex]-2[\/latex] in place of [latex]x[\/latex] and solve for [latex]y[\/latex].<\/p>\n<p>[latex]\\begin{align}\\require{color} y &= x^2+4x+3\\\\ y &= (\\color{Green}{-2}\\color{black}{)^2+4(}\\color{Green}{-2}\\color{black}{)+3}\\\\ y &=4+(-8)+3\\\\ y &= -1\\end{align}[\/latex]<\/p>\n<p>In this instance, [latex]\\left(-2,-1\\right)[\/latex] is the vertex. This parabola opens upward, so the vertex occurs at the lowest point on the graph, also called the minimum point of the graph. The vertex [latex](-2,-1)[\/latex] is the minimum point on this graph.<\/p>\n<p>The [latex]x[\/latex]-intercepts are the point(s) where the parabola crosses the [latex]x[\/latex]-axis, these can be found by letting [latex]y=0[\/latex] and solving for [latex]x[\/latex]. We will use factoring and the Zero-Product Property to solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} y &= x^2+4x+3\\\\ 0 &=x^2+4x+3 && \\color{blue}{\\textsf{Which two numbers multiply to be 3 and add to be 4?}} \\\\ 0 &=(x+3)(x+1) && \\color{blue}{\\textsf{set each factor equal to zero and solve}}\\\\ x+3 &=0 \\quad x+1 =0\\\\ x &= -3 \\quad x=-1 \\end{align}[\/latex]<\/p>\n<p style=\"text-align: center;\">The [latex]x[\/latex]-intercepts occur at [latex]\\left(-3,0\\right)[\/latex] and [latex]\\left(-1,0\\right)[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010711\/CNX_Precalc_Figure_03_02_0042.jpg\" alt=\"Graph of a parabola showing where the x and y intercepts, vertex, and axis of symmetry are for the function y=x^2+4x+3.\" width=\"487\" height=\"555\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<div id=\"fs-id1165135168275\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135574310\">How To: \u00a0find the vertex when given a quadratic function in general form<\/h3>\n<ol id=\"fs-id1165134108459\">\n<li>Identify [latex]a[\/latex],\u00a0[latex]b[\/latex], and\u00a0[latex]c[\/latex].<\/li>\n<li>Find [latex]h[\/latex], the [latex]x[\/latex]-coordinate of the vertex, by substituting [latex]a[\/latex] and [latex]b[\/latex] into [latex]h=\\dfrac{-b}{2a}[\/latex].<\/li>\n<li>Find [latex]k[\/latex], the [latex]y[\/latex]-coordinate of the vertex, by evaluating [latex]k=f\\left(h\\right)=f\\left(\\dfrac{-b}{2a}\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the vertex of the quadratic function [latex]f\\left(x\\right)=2{x}^{2}-8x+7[\/latex]. Rewrite the quadratic function in vertex form.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q648542\">Show Solution<\/span><\/p>\n<div id=\"q648542\" class=\"hidden-answer\" style=\"display: none\">\n<p>Identify the coefficients:\u00a0[latex]a=2[\/latex], [latex]b=-8[\/latex], and [latex]c=7[\/latex].<\/p>\n<p id=\"fs-id1165137596323\">The [latex]x[\/latex]-coordinate of the vertex will be at<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} h &=\\dfrac{-b}{2a} \\\\ h &=\\dfrac{-(-8)}{2\\left(2\\right)} \\\\ h &=\\dfrac{8}{4} \\\\ h &=2 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137465689\">The [latex]y[\/latex]-coordinate of the vertex will be at<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}k &= f(h) \\\\ k &= f(2) \\\\ k &= 2(2)^{2}-8(2)+7 \\\\ k &= 8-16+7\\\\ k &=-1 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135177784\">Rewriting into vertex form, the stretch factor of [latex]2[\/latex] will be the same as the [latex]a[\/latex] in the original quadratic.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}f\\left(x\\right)=a{x}^{2}+bx+c \\\\ f\\left(x\\right)=2{x}^{2}-8x+7 \\end{gathered}[\/latex]<\/p>\n<p id=\"fs-id1165137653186\">Writing in vertex form we found [latex]a=2[\/latex], [latex]h=2[\/latex], and [latex]k=-1[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]f(x)=a(x-h)^2+k[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]f(x)=2(x-2)^{2}+(-1)[\/latex]<\/p>\n<p>The vertex is [latex](2,-1)[\/latex]. Since [latex]a>0[\/latex], this indicates the parabola opens upward. The minimum value is [latex]k=-1[\/latex] with the lowest point being [latex](2,-1)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div id=\"Example_03_02_03\" class=\"example\">\n<div id=\"fs-id1165137658566\" class=\"exercise\"><\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135193262\">Given the function [latex]g\\left(x\\right)=13+{x}^{2}-6x[\/latex], write the function in general form <strong>and<\/strong> then in vertex form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q312729\">Show Solution<\/span><\/p>\n<div id=\"q312729\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/p>\n<p><strong>In general form:<\/strong><\/p>\n<p>Rearrange the terms so that they are in descending order.<\/p>\n<p>[latex]g\\left(x\\right)={x}^{2}-6x+13[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><strong>In vertex form:<\/strong><\/p>\n<p>Find the vertex to change the equation to vertex form [latex]g(x)=a(x-h)^2+k[\/latex].<\/p>\n<p>From the general form above, we know that [latex]a=1[\/latex], [latex]b=-6[\/latex], and [latex]c=13[\/latex].<\/p>\n<p>There is no vertical stretch or compression since [latex]a=1[\/latex].<\/p>\n<p id=\"fs-id1165137596323\">The [latex]x[\/latex]-coordinate of the vertex will be at<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} h &=\\dfrac{-b}{2a} \\\\ h &=\\dfrac{-(-6)}{2\\left(1\\right)} \\\\ h &=\\dfrac{6}{2} \\\\ h &=3 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137465689\">The [latex]y[\/latex]-coordinate of the vertex will be at<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}k &= f(h) \\\\ k &= f(3) \\\\ k &= (3)^2-6(3)+13 \\\\ k &= 9-18+13\\\\ k &=4 \\end{align}[\/latex]<\/p>\n<p>The function in vertex form is: [latex]g\\left(x\\right)={\\left(x - 3\\right)}^{2}+4[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Solve Quadratic Equations using Factoring and the Zero-Product Property<\/h2>\n<p>Let us start with an example that factors a GCF from a binomial and apply the Zero-Product Property to solve a polynomial equation.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]-t^2+t=0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q612316\">Show Solution<\/span><\/p>\n<div id=\"q612316\" class=\"hidden-answer\" style=\"display: none\">\n<p>Each term has a common factor of [latex]-t[\/latex], so we can factor out the GCF of [latex]-t[\/latex] and use the zero product principle.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}-t^2+t=0\\\\-t\\left(t-1\\right)=0\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now we have a product on one side and zero on the other, so we can set each factor equal to zero and solve using the Zero-Product Property.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\\\ -t\\hfill=0 & & t-1=0 \\hfill \\\\ t=0 & & t=1 \\hfill \\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">The solutions are [latex]t=0[\/latex] or [latex]t=1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show two more examples of using both factoring and the Zero-Product Property to solve a quadratic equation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  Factor and Solve Quadratic Equation - Greatest Common Factor Only\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/gIwMkTAclw8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next video, we show that you can use previously learned methods to factor a trinomial in order to solve a quadratic equation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex: Factor and Solve Quadratic Equations When A equals 1\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/bi7i_RuIGl0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We will now solve a quadratic equation where it is not set equal to zero on one side.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]x^2-4x=5[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q165196\">Show Solution<\/span><\/p>\n<div id=\"q165196\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, move all the terms to one side. The goal is to try and see if we can use the Zero-Product Property since that is the only tool we currently have for solving quadratic equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^2-4x=5\\\\x^2-4x-5=0\\\\\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">We now have all the terms on the left side and zero on the other side. Factor the left side of the equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^2-4x-5=0\\\\\\left(x+1\\right)\\left(x-5\\right)=0\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">We separate our factors into two linear equations using the Zero-Product Property.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\\\ x-5\\hfill=0 & & x+1=0 \\hfill \\\\ x=5 & & x=-1 \\hfill \\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">The solutions are [latex]x=5\\text{ or }x=-1[\/latex].<\/p>\n<p style=\"text-align: center;\"><\/div>\n<\/div>\n<\/div>\n<p>The following binomial is a difference of squares and can be factored. Factor the binomial and use the zero-product property to solve.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve using the Zero-Product Property: [latex]{x}^{2}-9=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q148980\">Show Solution<\/span><\/p>\n<div id=\"q148980\" class=\"hidden-answer\" style=\"display: none\">\n<p>Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the Zero-Product Property.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} {x}^{2}-9 &= 0\\\\ (x-3)(x+3) &= 0 \\end{align}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\\\ x - 3\\hfill=0 & & x+3=0 \\hfill \\\\ x=3 & & x=-3 \\hfill \\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">The solutions are [latex]x=3[\/latex] or [latex]x=-3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Here are some video examples in which you can use factoring to solve these quadratic equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex: Factor and Solve Quadratic Equations When A equals 1\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/bi7i_RuIGl0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Use factoring and the Zero-Product Property to solve the quadratic equation: [latex]4{x}^{2}+15x+9=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q716142\">Show Solution<\/span><\/p>\n<div id=\"q716142\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, multiply [latex]ac:4\\left(9\\right)=36[\/latex]. Then list the factors of [latex]36[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}1\\cdot 36\\hfill \\\\ 2\\cdot 18\\hfill \\\\ 3\\cdot 12\\hfill \\\\ 4\\cdot 9\\hfill \\\\ 6\\cdot 6\\hfill \\end{array}[\/latex]<\/div>\n<p>The only pair of factors that sums to [latex]15[\/latex] is [latex]3+12[\/latex]. Rewrite the equation replacing the [latex]b[\/latex] term, [latex]15x[\/latex], with two terms using\u00a0[latex]3[\/latex] and\u00a0[latex]12[\/latex] as coefficients of [latex]x[\/latex]. Factor the first two terms, and then factor the last two terms.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}4{x}^{2}+3x+12x+9=0\\hfill \\\\ x\\left(4x+3\\right)+3\\left(4x+3\\right)=0\\hfill \\\\ \\left(4x+3\\right)\\left(x+3\\right)=0\\hfill \\end{array}[\/latex]<\/div>\n<p>Solve using the Zero-Product Property.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}&\\left(4x+3\\right)\\left(x+3\\right)=0 & \\hfill \\\\ 4x+3\\hfill=0 & & x+3=0 \\hfill \\\\ x=-\\dfrac{3}{4} & & x=-3 \\hfill \\end{array}[\/latex]<\/p>\n<p>The solutions are [latex]x=-\\dfrac{3}{4}[\/latex], [latex]x=-3[\/latex]. You can see these solutions as the [latex]x[\/latex]-intercepts of the parabola graphed below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200359\/CNX_CAT_Figure_02_05_003n.jpg\" alt=\"Parabola facing up with x-intercepts: (negative 3\/4, 0) and (negative 3, 0) with vertex of (negative 2, negative 5).\" width=\"487\" height=\"433\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video examples can be factored but the leading coefficients is NOT 1.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex: Solve a Quadratic Equation Using Factor By Grouping\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/04zEXaOiO4U?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Graph quadratic functions in general form [latex]f(x)=ax^2+bx+c[\/latex]<\/h2>\n<p>To graph quadratic functions in general form, we will first find the vertex, axis of symmetry, [latex]x[\/latex]-intercept(s), and the [latex]y[\/latex]-intercept. We will graph the function, [latex]g(x)=-x^2-8x-15[\/latex] where [latex]a=-1[\/latex], [latex]b=-8[\/latex], and [latex]c=-15[\/latex].<\/p>\n<p>Start by finding the vertex.\u00a0The [latex]x[\/latex]-coordinate of the vertex:<\/p>\n<p>[latex]\\begin{align} h &=\\dfrac{-b}{2a} \\\\ h &=\\dfrac{-(-8)}{2\\left(-1\\right)} \\\\ h &=\\dfrac{8}{-2} \\\\ h &=-4 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137465689\">The [latex]y[\/latex]-coordinate of the vertex:<\/p>\n<p>[latex]\\begin{align}k &= f(h) \\\\ k &= f(-4) \\\\ k &= -(-4)^2-8(-4)-15 \\\\ k &= -16+32-15\\\\ k &=1 \\end{align}[\/latex]<\/p>\n<p>The vertex is [latex](-4,1)[\/latex].<\/p>\n<p>The axis of symmetry is a vertical line that passes through the vertex. In this case the axis (line) of symmetry is [latex]x=-4[\/latex].<\/p>\n<p>We can rewrite the equation as [latex]y=-x^2-8x-15[\/latex]. To find the [latex]y[\/latex]-intercept, let [latex]x=0[\/latex] and solve for [latex]y[\/latex].<\/p>\n<p>[latex]\\begin{align} y &= -x^2-8x-15\\\\ y &= -(0)^2-8(0)-15\\\\ y &= -15\\end{align}[\/latex]<\/p>\n<p>The [latex]y[\/latex]-intercept is [latex](0,-15)[\/latex].<\/p>\n<p>To find the [latex]x[\/latex]-intercept(s), let [latex]y=0[\/latex] and solve for [latex]x[\/latex]. We will use factoring and the Zero-Product Property to find the [latex]x[\/latex]-intercept(s).<\/p>\n<p>[latex]\\begin{align} y &= -x^2-8x-15\\\\ 0 &= -x^2-8x-15\\\\ 0 &= -1(x^2+8x+15)\\\\ 0 &= -1(x+3)(x+5)\\end{align}[\/latex]<br \/>\n[latex]x+3=0 \\hspace{1.5cm} x+5=0[\/latex]<br \/>\n[latex]x = -3\\hspace{2cm} x = -5[\/latex]<\/p>\n<p>The [latex]x[\/latex]-intercepts are [latex](-3,0)[\/latex] and [latex](-5,0)[\/latex].<\/p>\n<p>Figure 2 is the graph of the function [latex]g(x)=-x^2-8x-15[\/latex].<\/p>\n<p><iframe loading=\"lazy\" style=\"border: 1px solid #ccc;\" src=\"https:\/\/www.desmos.com\/calculator\/gews1xa4kt?embed\" width=\"500\" height=\"500\" frameborder=\"0\"><\/iframe><br \/>\nFigure 2<\/p>\n<p>NOTE: If the [latex]x[\/latex]-intercepts do not exist or the quadratic equation cannot be factored to solve or if the [latex]y[\/latex]-intercept is off the grid that you are being asked to graph on, you can make a table of values to help you find additional point(s) on the graph. You can pick [latex]x[\/latex] values that are on either side of the vertex, for this example we could pick [latex]x=-2[\/latex] and [latex]x=-6[\/latex].<\/p>\n<table style=\"border-collapse: collapse; width: 25%; border: 1px solid black; height: 36px;\">\n<thead>\n<tr style=\"height: 12px;\">\n<th style=\"width: 30%; border: 1px solid #999999; height: 12px; text-align: center;\">\u00a0[latex]x[\/latex]<\/th>\n<th style=\"width: 30%; border: 1px solid #999999; height: 12px; text-align: center;\">[latex]y=-x^2-8x-15[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 12px;\">\n<td style=\"width: 30%; border: 1px solid #999999; height: 12px; text-align: center;\">[latex]-2[\/latex]<\/td>\n<td style=\"width: 70%; border: 1px solid #999999; height: 12px; text-align: center;\">[latex]\\begin{align} y &= -1(\\color{Green}{-2}\\color{black}{)^2-8}(\\color{Green}{-2}\\color{black}{)-15}\\\\ &= -1(4)+16-15\\\\ &= -3 \\end{align}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 30%; border: 1px solid #999999; height: 12px; text-align: center;\">[latex]-6[\/latex]<\/td>\n<td style=\"width: 70%; border: 1px solid #999999; height: 12px; text-align: center;\">[latex]\\begin{align} y &= -1(\\color{Green}{-6}\\color{black}{)^2-8}(\\color{Green}{-6}\\color{black}{)-15}\\\\ &= -1(36)+48-15\\\\ &= -3 \\end{align}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div>So, two more points we can use to graph this quadratic function are [latex](-2,-3)[\/latex] and [latex](-6,-3)[\/latex].<\/div>\n<div><\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Graph [latex]f(x)=2x^2-4x+3[\/latex] by first finding the vertex, axis of symmetry, [latex]y[\/latex]-intercept, and [latex]x[\/latex]-intercept(s).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q374151\">Show Solution<\/span><\/p>\n<div id=\"q374151\" class=\"hidden-answer\" style=\"display: none\">\n<p>The function can also be written as [latex]y=2x^2-4x+3[\/latex]<\/p>\n<p><strong>Vertex: [latex](h,k)[\/latex]<\/strong><\/p>\n<p>[latex]x[\/latex]-coordinate:<\/p>\n<p>[latex]\\begin{align} h &=\\frac{-b}{2a} \\\\ h &=\\frac{-(-4)}{2\\left(2\\right)} \\\\ h &=\\frac{4}{4} \\\\ h &=1 \\end{align}[\/latex]<\/p>\n<p>[latex]y[\/latex]-coordinate:<\/p>\n<p>[latex]\\begin{align} k &= f(1) \\\\ k &= 2(1)^2-4(1)+3 \\\\ k &= 2-4+3\\\\ k &=1 \\end{align}[\/latex]<\/p>\n<p>The vertex is [latex](1,1)[\/latex].<\/p>\n<p><strong>Axis of Symmetry\u00a0<\/strong>is the vertical line that passes through the vertex. That line is [latex]x=1[\/latex].<\/p>\n<p><strong>[latex]y[\/latex]-intercept:\u00a0<\/strong>Let [latex]x=0[\/latex] and solve for [latex]y[\/latex].<\/p>\n<p>[latex]\\begin{align} y &= 2x^2-4x+3\\\\ y &= 2(0)^2-4(0)+3\\\\ y &= 3\\end{align}[\/latex]<\/p>\n<p>The [latex]y[\/latex]-intercept is [latex](0,3)[\/latex].<\/p>\n<p><strong>[latex]x[\/latex]-intercept(s):\u00a0<\/strong>Let [latex]y=0[\/latex] and solve for [latex]x[\/latex]. Use factoring to solve.<\/p>\n<p>[latex]\\begin{align} y &= 2x^2-4x+3\\\\ 0 &= 2x^2-4x+3\\end{align}[\/latex]<\/p>\n<p>[latex]a\\cdot c = 2\\cdot3=6[\/latex] We are looking for two integers that multiply to be [latex]6[\/latex] and add to be [latex]-4[\/latex]. There are no two integers that satisfy this. This cannot be factored. Later in the book,\u00a0<span style=\"font-size: 1rem; text-align: initial;\">we will discuss different methods besides factoring in which you can find the [latex]x[\/latex]-intercepts. For now, we will graph this function with the other points we have found and use symmetry of a parabola to find another point on the graph.<\/span><\/p>\n<p><iframe loading=\"lazy\" style=\"border: 1px solid #ccc;\" src=\"https:\/\/www.desmos.com\/calculator\/ppyx5cc6af?embed\" width=\"500\" height=\"500\" frameborder=\"0\"><\/iframe><\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Solving Applications of Quadratic Functions<\/h2>\n<p id=\"fs-id1165137431411\">There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue. Remember when [latex]a>0[\/latex], the graph of a parabola opens upward and the vertex is the lowest point on the graph or the minimum. When [latex]a<0[\/latex], the graph of a parabola opens downward and the vertex is the highest point on the graph or the maximum.<\/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-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010712\/CNX_Precalc_Figure_03_02_0092.jpg\" alt=\"Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).\" width=\"975\" height=\"558\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<p>Projectile motion happens when you throw a ball into the air and it comes back down because of gravity. \u00a0A projectile will follow a curved path that behaves in a predictable way. \u00a0This predictable motion has been studied for centuries, and in simple cases, an object&#8217;s height from the ground\u00a0at a given time,\u00a0[latex]t[\/latex], can be modeled with a polynomial function of the form [latex]h(t)=at^2+bt+c[\/latex], where [latex]h(t) =[\/latex] height of an object at a given time,\u00a0[latex]t[\/latex]. \u00a0Projectile motion is also called a parabolic trajectory because of the shape of the path of a projectile&#8217;s motion, as in the water in the image of the fountain below.\u00a0Parabolic motion and its related functions\u00a0allow us to launch satellites for telecommunications and rockets for space exploration.<\/p>\n<div id=\"attachment_4890\" style=\"width: 446px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4890\" class=\"wp-image-4890\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161949\/ParabolicWaterTrajectory-225x300.jpg\" alt=\"Water from a fountain shoing classic parabolic motion.\" width=\"436\" height=\"581\" \/><\/p>\n<p id=\"caption-attachment-4890\" class=\"wp-caption-text\">Parabolic water trajectory in a fountain.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>A small toy rocket is launched from a\u00a0[latex]4[\/latex]-foot pedestal. The height ([latex]h[\/latex]<i>, <\/i>in feet) of the rocket [latex]t[\/latex] seconds after taking off is given by the function [latex]h(t)=\u22122t^{2}+7t+4[\/latex]. How long will it take the rocket to hit the ground?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q679533\">Show Solution<\/span><\/p>\n<div id=\"q679533\" class=\"hidden-answer\" style=\"display: none\">\n<p>The rocket will be on the ground when [latex]h(t)=0[\/latex]. We want to know how long,\u00a0[latex]t[\/latex], the rocket is in the air.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}h(t)=\u22122t^{2}+7t+4\\\\0=\u22122t^{2}+7t+4\\end{array}[\/latex]<\/p>\n<p>We can factor the polynomial [latex]\u22122t^{2}+7t+4[\/latex] more easily by first factoring out a [latex]-1[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}0=-1(2t^{2}-7t-4)\\\\0=-1\\left(2t+1\\right)\\left(t-4\\right)\\end{array}[\/latex]<\/p>\n<p>Use the Zero-Product Property. There is no need to set the constant factor [latex]-1[\/latex] to zero, because [latex]-1[\/latex] will never equal zero. Set the other two factors equal to zero and solve.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}&-1\\left(2t+1\\right)\\left(t-4\\right)=0 & \\hfill \\\\ 2t+1\\hfill=0 & & t-4=0 \\hfill \\\\ t=-\\dfrac{1}{2} & & t=4 \\hfill \\end{array}[\/latex]<\/p>\n<p>Since [latex]t[\/latex] represents time after the launch, it cannot be a negative number; only [latex]t=4[\/latex]\u00a0makes sense in this context.<\/p>\n<p>Therefore, the rocket will hit the ground [latex]4[\/latex] seconds after being launched.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-simple-polynomial-equations\/screenshot-2024-07-26-at-11-05-53%e2%80%afam\/\" rel=\"attachment wp-att-1554\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1554 size-full\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-07-26-at-11.05.53\u202fAM.png\" alt=\"decorative image\" width=\"309\" height=\"245\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-07-26-at-11.05.53\u202fAM.png 309w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-07-26-at-11.05.53\u202fAM-300x238.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-07-26-at-11.05.53\u202fAM-65x52.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-07-26-at-11.05.53\u202fAM-225x178.png 225w\" sizes=\"auto, (max-width: 309px) 100vw, 309px\" \/><\/a><\/p>\n<div class=\"textbox shaded\" style=\"text-align: center;\">Model rocket being launched.\u00a0https:\/\/www.instructables.com\/Remote-Rocket-Igniter\/<\/div>\n<p>In the next example, we will solve for the time that the rocket reaches a given height other than zero.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Use the formula for the height of the rocket in the previous example to find the time when the rocket is\u00a0[latex]4[\/latex] feet from hitting the ground on its way back down. \u00a0Refer to the image.<\/p>\n<p>[latex]h(t)=\u22122t^{2}+7t+4[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4892 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161951\/Screen-Shot-2016-06-14-at-5.39.53-PM-300x247.png\" alt=\"Parabolic motion of rocket which starts four feet up from the ground. t=0 is labeled at the starti of hte parabolic motion adn t=? is labeled at four feet from the ground on the other side of the parabola.\" width=\"413\" height=\"340\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q198118\">Show Solution<\/span><\/p>\n<div id=\"q198118\" class=\"hidden-answer\" style=\"display: none\">\n<p>We are given that the height of the rocket is\u00a0[latex]4[\/latex] feet from the ground on its way back down. We want to know how long it will take for the rocket to get to that point in its path. We are going to solve for [latex]t[\/latex].<\/p>\n<p>Substitute [latex]h(t) = 4[\/latex] into the formula for height, and try to get zero on one side since we know we can use the Zero-Product Property to solve polynomial equations.<\/p>\n<p><strong>Write and Solve:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}4=-2t^2+7t+4\\\\\\underline{-4}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-4}\\\\0=-2t^2+7t\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now we can factor out a\u00a0[latex]t[\/latex] from each term:<\/p>\n<p style=\"text-align: center;\">[latex]0=t\\left(-2t+7\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">Solve each equation for\u00a0[latex]t[\/latex] using the Zero-Product Property:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=0\\text{ or }-2t+7=0\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-7}\\,\\,\\,\\,\\,\\,\\,\\underline{-7}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\dfrac{-2t}{-2}=\\dfrac{-7}{-2}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,t=\\dfrac{7}{2}=3.5\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">It does not make sense for us to choose\u00a0[latex]t=0[\/latex], because we are interested in the amount of time that has passed when the projectile is\u00a0[latex]4[\/latex] feet from hitting the ground on its way back down. We will choose\u00a0[latex]t=3.5[\/latex].<\/p>\n<p>The rocket will be [latex]4[\/latex] feet from the ground at [latex]t=3.5\\text{ seconds}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show another example of how to find the time when a object following a parabolic trajectory hits the ground.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Factoring Application - Find the Time When a Projectile Hits and Ground\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/hsWSzu3KcPU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In this section we introduced the concept of projectile motion and showed that it can be modeled with quadratic functions. While the models used in these examples can be factored, that is not always the case with these types of applications. The concepts and interpretations are the same as what would happen in &#8220;real life.&#8221;<\/p>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p>A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased [latex]80[\/latex] feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.<\/p>\n<ol id=\"fs-id1165135640934\">\n<li>Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length [latex]L[\/latex].<\/li>\n<li>What dimensions should she make her garden to maximize the enclosed area?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q478551\">Show Solution<\/span><\/p>\n<div id=\"q478551\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010713\/CNX_Precalc_Figure_03_02_0102.jpg\" alt=\"Rectangle backyard with a rectangle garden center at the top. Garden labeled W for width and L for length.\" width=\"487\" height=\"310\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165137836808\">Let\u2019s use a diagram such as the one in Figure 4 to record the given information. It is also helpful to introduce a temporary variable, [latex]W[\/latex], to represent the width of the garden and the length of the fence section parallel to the backyard fence.<span id=\"fs-id1165135208803\"><br \/>\n<\/span><\/p>\n<ol id=\"fs-id1165134363440\">\n<li>We know we have only [latex]80[\/latex] feet of fence available, and [latex]L+W+L=80[\/latex], or more simply, [latex]2L+W=80[\/latex]. This allows us to represent the width, [latex]W[\/latex], in terms of [latex]L[\/latex].\n<div id=\"eip-id1165135697866\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]W=80 - 2L[\/latex]<\/div>\n<p id=\"fs-id1165135435476\">Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so<\/p>\n<p id=\"fs-id1165135435476\" style=\"text-align: center;\">[latex]\\begin{align}A&=LW \\\\ &=L\\left(80 - 2L\\right) \\\\ &=80L - 2{L}^{2} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135258914\">This formula represents the area of the fence in terms of the variable length [latex]L[\/latex]. The function, written in general form, is<\/p>\n<div id=\"eip-382\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A\\left(L\\right)=-2{L}^{2}+80L[\/latex].<\/div>\n<\/li>\n<li>The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation does not have a constant term, meaning [latex]c=0[\/latex]. Writing the function in general form we can \u00a0identify that\u00a0[latex]a=-2,b=80[\/latex], and [latex]c=0[\/latex].<\/li>\n<\/ol>\n<p id=\"fs-id1165137772015\">To find the vertex:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}h&=\\dfrac{-80}{2\\left(-2\\right)} &&& k&=A\\left(20\\right) \\\\ h&=20 && \\text{and} & k&=-2{\\left(20\\right)}^{2}+80\\left(20\\right)\\\\ &&&&&k=800 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135174964\">The maximum value of the function is an area of [latex]800[\/latex] square feet, which occurs when [latex]L=20[\/latex] feet. When the shorter sides are [latex]20[\/latex] feet, there is [latex]40[\/latex] feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length [latex]20[\/latex] feet and the longer side parallel to the existing fence has length [latex]40[\/latex] feet.<\/p>\n<p>This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure 5.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010713\/CNX_Precalc_Figure_03_02_0112.jpg\" alt=\"Graph of the parabolic function A(L)=-2L^2+80L, which the x-axis is labeled Length (L) and the y-axis is labeled Area (A). The vertex is at (20, 800).\" width=\"487\" height=\"476\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A ball is thrown upward from the top of a [latex]40[\/latex]-foot high building at a speed of [latex]80[\/latex] 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 style=\"padding-left: 60px;\">a. When does the ball reach the maximum height?<\/p>\n<p style=\"padding-left: 60px;\">b. What is the maximum height of the ball?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q340065\">Show Solution<\/span><\/p>\n<div id=\"q340065\" 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)} =\\dfrac{-80}{-32} \\\\ &=\\dfrac{5}{2} \\\\ &=2.5 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135528870\">The ball reaches a maximum height after [latex]2.5[\/latex] seconds.<\/p>\n<p>b. To find the maximum height, find the [latex]y[\/latex]-coordinate of the vertex of the parabola.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}k&=H\\left(\\dfrac{-b}{2a}\\right) \\\\ &=H\\left(2.5\\right) \\\\ &=-16{\\left(2.5\\right)}^{2}+80\\left(2.5\\right)+40 \\\\ &=140 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135409750\">The ball reaches a maximum height of [latex]140[\/latex] feet.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_02_08\" class=\"example\">\n<div id=\"fs-id1165134060458\" class=\"exercise\">\n<div id=\"fs-id1165137843086\" class=\"commentary\">\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134081301\">A rock is thrown upward from the top of a [latex]112[\/latex]-foot high cliff overlooking the ocean at a speed of [latex]96[\/latex] 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], where [latex]H[\/latex] is the height in feet and [latex]t[\/latex] is time after the throw in seconds.<\/p>\n<p style=\"padding-left: 60px;\">a. When does the rock reach the maximum height?<\/p>\n<p style=\"padding-left: 60px;\">b. What is the maximum height of the rock?<\/p>\n<p style=\"padding-left: 60px;\">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=\"q4901\">Show Solution<\/span><\/p>\n<div id=\"q4901\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a03 seconds \u00a0b.\u00a0256 feet \u00a0c.\u00a07 seconds<\/p>\n<p>Explanation:<\/p>\n<p>a. The rock reaches the maximum height at the vertex of the parabola.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} h&=\\dfrac{-96}{2\\left(-16\\right)}\\\\ h&=\\dfrac{-96}{-32} \\\\ h&= 3 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135528870\">The rock reaches a maximum height at [latex]3[\/latex] seconds.<\/p>\n<p>b. To find the maximum height, find the [latex]y[\/latex]-coordinate of the vertex of the parabola.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}k&=-16{\\left(3\\right)}^{2}+96\\left(3\\right)+112 \\\\ k&=256 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135409750\">The rock reaches a maximum height of [latex]256[\/latex] feet.<\/p>\n<p>c. To find when the rock hits the ocean, we need to find when the height, [latex]H(t)[\/latex] equals [latex]0[\/latex]. We factor to solve for [latex]t[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} H(t) &= -16t^2+96t+112\\\\ 0 &= -16t^2+96t+112\\\\ 0 &= -16(t^2-6t-7)\\\\ 0 &= -16(t+1)(t-7) \\end{align}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]t+1=0[\/latex] or [latex]t-7=0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]t = -1[\/latex] or [latex]t = 7[\/latex]<\/p>\n<p>Since [latex]t[\/latex] represents time, it cannot be a negative number; only [latex]t=7[\/latex]\u00a0makes sense in this context.<\/p>\n<p>The rock hits the ocean [latex]7[\/latex] seconds after it is thrown.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Key Equations<\/span><\/p>\n<section id=\"fs-id1165134205927\" class=\"key-equations\">\n<table id=\"eip-id1165137539373\" summary=\"..\">\n<tbody>\n<tr>\n<td>Vertex form of a quadratic function<\/td>\n<td>[latex]f\\left(x\\right)=a{\\left(x-h\\right)}^{2}+k[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Intercept form of a quadratic function<\/td>\n<td>[latex]f(x)=(x-p)(x-q)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>General form of a quadratic function<\/td>\n<td>[latex]f\\left(x\\right)=a{x}^{2}+bx+c[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<h2>Summary<\/h2>\n<p>You can rewrite a quadratic function that is given in general form [latex]f(x)=ax^2+bx+c[\/latex] in vertex form [latex]f(x)=a(x-h)^2+k[\/latex].<\/p>\n<p>You can find the vertex, axis of symmetry, [latex]y[\/latex]-intercept, and [latex]x[\/latex]-intercepts (if any) of a quadratic function in general form and use those points to graph the parabola.<\/p>\n<p>You can find the solutions, or roots, of quadratic equations in general form by setting one side equal to zero, factoring the quadratic, and then applying the Zero-Product Property. The Zero-Product Property states that if [latex]ab=0[\/latex], then either [latex]a=0[\/latex] or [latex]b=0[\/latex], or both [latex]a[\/latex] and [latex]b[\/latex] are\u00a0[latex]0[\/latex]. Once the polynomial is factored, set each factor equal to zero and solve them separately. The answers will be the set of solutions for the original equation.<\/p>\n<p>Not all solutions are appropriate for some applications. For example, in many real-world situations, negative solutions are not appropriate and must be discarded.<\/p>\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-167\">\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>Solve a Quadratic Equations with Fractions by Factoring (a not 1). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/kDj_qdKW-ls\">https:\/\/youtu.be\/kDj_qdKW-ls<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/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>Ex: Factor and Solve Quadratic Equation - Greatest Common Factor Only. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/gIwMkTAclw8\">https:\/\/youtu.be\/gIwMkTAclw8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Factor and Solve Quadratic Equations When A equals 1. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/bi7i_RuIGl0\">https:\/\/youtu.be\/bi7i_RuIGl0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Unit 12: Factoring, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":395986,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex: Factor and Solve Quadratic Equation - 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