{"id":9441,"date":"2023-05-22T18:23:29","date_gmt":"2023-05-22T18:23:29","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/?post_type=chapter&p=9441"},"modified":"2026-02-05T08:32:51","modified_gmt":"2026-02-05T08:32:51","slug":"6-8-further-exploration-with-quadratic-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/6-8-further-exploration-with-quadratic-equations\/","title":{"raw":"6.8: Further Exploration with Quadratic Equations","rendered":"6.8: Further Exploration with Quadratic Equations"},"content":{"raw":"
\r\n

section 6.8 Learning Objectives<\/h3>\r\n6.8: Further Exploration with Quadratic Equations<\/strong>\r\n
    \r\n \t
  • Solve application problems (area problems using factoring)<\/li>\r\n \t
  • Solve application problems (consecutive integer problems using factoring)<\/li>\r\n \t
  • Graph basic quadratic functions<\/li>\r\n<\/ul>\r\n<\/div>\r\n

    Applications of Solving Factorable Quadratic Equations<\/h2>\r\nWhen a polynomial is set equal to a value (whether an integer or another polynomial), the result is an equation. An equation that can be written in the form [latex]ax^{2}+bx+c=0[\/latex]\u00a0is called a quadratic equation<\/strong>. You can solve some quadratic equations using the rules of algebra, applying factoring techniques, and by using the Principle of Zero Products<\/strong>.\r\n\r\nThere are many applications for quadratic equations. Don't forget that to use the Principle of Zero Products to solve a quadratic equation, you need to make sure that the equation is equal to zero. For example, [latex]12^{2}+11x+2=7[\/latex] must first be changed to [latex]12x^{2}+11x-5=0[\/latex]\u00a0by subtracting 7 from both sides.\r\n

    Area Problems<\/h3>\r\nIn our first example, we use some basic geometry as we search for the dimensions of rectangle.\r\n
    \r\n

    Example 1<\/h3>\r\nThe area of a rectangular garden is 30 square feet. If the length is 7 feet longer than the width, find the dimensions.\r\n[reveal-answer q=\"948371\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"948371\"]The formula for the area of a rectangle is [latex]\\text{Area}=\\text{length}\\cdot\\text{width}[\/latex], or [latex]A=l\\cdot{w}[\/latex].\r\n

    [latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,A=l\\cdot{w}\\\\\\,\\text{width}=w\\\\\\text{length}=w+7\\\\\\,\\,\\,\\,\\,\\text{area}=30\\\\\\\\30=\\left(w+7\\right)\\left(w\\right)\\end{array}[\/latex]<\/p>\r\nMultiply.\r\n

    [latex]30=w^{2}+7w[\/latex]<\/p>\r\nSubtract 30 from both sides to set the equation equal to 0.\r\n

    [latex]w^{2}+7w\u201330=0[\/latex]<\/p>\r\nUsing the product and sum technique, we can factor this by finding two numbers whose product is [latex]\u221230[\/latex] and whose sum is 7.\r\n

    [latex]\\left(w\u20133\\right)\\left(w+10\\right)=0[\/latex]<\/p>\r\nUse the Zero Product Property to solve for w<\/i>.\r\n

    [latex]\\begin{array}{l}w-3=0\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,w+10=0\\\\\\,\\,\\,\\,\\,\\,\\,w=3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=-10\\end{array}[\/latex]<\/p>\r\nThe solution [latex]w=\u221210[\/latex]\u00a0does not work for this application as the width cannot be a negative number. So, we discard the [latex]\u221210[\/latex]. Thus, the width is 3 feet.\r\n

    The width = 3 feet<\/p>\r\nSubstitute [latex]w=3[\/latex] into the expression for the length, [latex]l=w+7[\/latex] to find the length: [latex]l=3+7=10[\/latex].\r\n

    The length is [latex]3+7=10[\/latex] feet<\/p>\r\n\r\n

    Answer<\/h4>\r\nThe width of the garden is 3 feet, and the length is 10 feet.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the example in the following video, we present another area application of factoring trinomials.\r\n\r\nhttps:\/\/youtu.be\/PvXsWZp588o\r\n

    Consecutive Integers<\/h3>\r\nOur next example is deceptively similar to the rectangular garden (see if you can recognize this). First, we must recall what is meant by \"consecutive integers.\" Two consecutive integers would be something like 7 and 8 or -53 and -52. We can also explore consecutive odd integers (such as 15 and 17) or consecutive even integers (like 84 and 86).\r\n
    \r\n

    ExAMPLE 2<\/h3>\r\nThe product of two consecutive odd integers is 63. Find all such pairs of integers.\r\n\r\n[reveal-answer q=\"583004\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"583004\"]\r\n\r\nConsecutive odd (or even) integers are always two units apart. So, if we call the smaller of the consecutive odd integers [latex]x[\/latex], the next one is [latex]x+2[\/latex]. Since their product is 63, we can now form a quadratic equation to solve.\r\n

    [latex]x(x+2)=63[\/latex]<\/p>\r\nWe first distribute and move all terms to the same side.\r\n

    [latex]x^2+2x=63[\/latex]<\/p>\r\n

    [latex]x^2+2x-63=0[\/latex]<\/p>\r\nThis is factorable using the product and sum method.\r\n

    [latex](x+9)(x-7)=0[\/latex]<\/p>\r\nNext, we set each factor equal to zero.\r\n\r\n[latex]x+9=0[\/latex]\r\n\r\n[latex]x=-9[\/latex]\r\n\r\nOR\r\n\r\n[latex]x-7=0[\/latex]\r\n\r\n[latex]x=7[\/latex]\r\n\r\nRecall that [latex]x[\/latex] was the smaller of the two consecutive odd integers. If [latex]x=-9[\/latex], the next integer is [latex]x+2=-9+2=-7[\/latex], whereas if [latex]x=7[\/latex], the next integer is [latex]x+2=7+2=9[\/latex].\r\n\r\nIt can be quickly verified that each pair yields the desired product.\r\n

    Answer<\/h4>\r\nThe integers are [latex]-9 \\mbox{ and }-7[\/latex] or [latex]7 \\mbox{ and }9[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    Graphing Basic Quadratic Functions<\/h2>\r\nIn Modules 5 and 6 we learned about polynomials. An equation containing a second-degree polynomial is called a quadratic equation<\/strong>. For example, equations such as [latex]{x}^{2}-3x-4=0[\/latex] and [latex]{x}^{2}-16=0[\/latex] are in the \"family\" of quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.\r\n\r\nJust like the linear functions we've learned about previously, quadratic\u00a0functions can also be graphed. It is helpful to have an idea about what the shape should be so you can be sure that you have chosen enough points to plot as a guide. Let us start with the most basic quadratic function,\u00a0[latex]f(x)=x^{2}[\/latex].\r\n\r\nGraph [latex]f(x)=x^{2}[\/latex].\r\nWe'll start with a table of values. Then think of each row of the table as an ordered pair.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]x[\/latex]<\/td>\r\n[latex]f(x)=x^2[\/latex]<\/td>\r\n<\/tr>\r\n<\/thead>\r\n
    [latex]\u22122[\/latex]<\/td>\r\n[latex](-2)^2=4[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]\u22121[\/latex]<\/td>\r\n[latex](-1)^2=1[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]0[\/latex]<\/td>\r\n[latex](0)^2=0[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]1[\/latex]<\/td>\r\n[latex](1)^2=1[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]2[\/latex]<\/td>\r\n[latex](2)^2=4[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow we'll plot the points [latex](-2,4), (-1,1), (0,0), (1,1), (2,4)[\/latex]\r\n\r\n\"Coordinate\r\n\r\nNotice that the quadratic family of functions is not drawn using straight lines. Since the points are not<\/i> on a line, you cannot use a straight edge<\/strong>. Connect the points as best you can using a smooth curve<\/i> (not a series of straight lines). You may want to find and plot additional points (such as the ones in blue below). Placing arrows on the tips of the lines implies that they continue in that direction forever.\r\n\r\n\"Graph\r\n\r\nNotice that the shape is similar to the letter U. This is called a parabola<\/em>. One-half of the parabola is a mirror image of the other half. The lowest point on this graph is called the vertex<\/em>. The vertical line that goes through the vertex is called the line of symmetry<\/em>. In this case, that line is the\u00a0y<\/i>-axis, which is the line [latex]x=0[\/latex].\r\n\r\nIn the following video, we show an example of graphing another quadratic function,\u00a0[latex]f(x)=\\frac{1}{2}x^2[\/latex], using a table of values.\r\n\r\nhttps:\/\/youtu.be\/wYfEzOJugS8\r\n\r\nThe equations for quadratic functions can be written in the form [latex]f(x)=ax^{2}+bx+c[\/latex] (where [latex] a\\ne 0[\/latex]). In the two basic quadratic functions we graphed above, notice there was only the [latex]ax^2[\/latex] term and that both [latex]b=0[\/latex], and [latex]c=0[\/latex] for those functions.\r\n\r\nAlthough there are many forms that quadratic functions can be written in, this Elementary Algebra course will focus primarily on graphing basic<\/em> quadratic functions of the form [latex]f(x)=ax^{2}+c[\/latex].\r\n

    Quadratic functions of the form\u00a0[latex]f(x)=ax^{2}+c[\/latex] will always be centered around the y-axis which is the line [latex]x=0[\/latex].\u00a0<\/strong><\/p>\r\nWe have shared two examples above of graphing a basic quadratic function of the form\u00a0[latex]f(x)=ax^{2}+c[\/latex], where [latex]c=0[\/latex], so let's now explore how to graph a quadratic function of the form\u00a0[latex]f(x)=ax^{2}+c[\/latex] where\u00a0[latex]c\\ne 0[\/latex] .\r\n\r\nGraph [latex]f(x)=x^{2}-4[\/latex].\r\nSimilar to the problems above, lets start with a table of values.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]x[\/latex]<\/td>\r\n[latex]f(x)=x^2-4[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n<\/thead>\r\n
    [latex]\u22122[\/latex]<\/td>\r\n[latex](-2)^2-4 = 4-4 = 0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
    [latex]\u22121[\/latex]<\/td>\r\n[latex](-1)^2-4 = 1-4 = -3[\/latex]<\/td>\r\n-3<\/td>\r\n<\/tr>\r\n
    [latex]0[\/latex]<\/td>\r\n[latex](0)^2-4 = 0-4 = -4[\/latex]<\/td>\r\n-4<\/td>\r\n<\/tr>\r\n
    [latex]1[\/latex]<\/td>\r\n[latex](1)^2-4 = 1-4 = -3[\/latex]<\/td>\r\n-3<\/td>\r\n<\/tr>\r\n
    [latex]2[\/latex]<\/td>\r\n[latex](2)^2-4 = 4-4 = 0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe'll now plot the points [latex](-2,0), (-1,-3), (0,-4), (1,-3), (2,0)[\/latex] on the graph.\r\n\r\n\"Coordinate\r\n\r\nIf we connect all the points with a smooth curve we will have the graph of our function [latex]f(x)=x^{2}-4[\/latex].\r\n\r\n\"Coordinate\r\n\r\nLet's have you now try one on your own:\r\n
    \r\n

    Example 3<\/h3>\r\nGraph the function: [latex]f(x)=x^2-16[\/latex]\r\n\r\n[reveal-answer q=\"184639\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"184639\"]\r\n\r\nNotice our function\u00a0[latex]f(x)=x^2-16[\/latex] is of the form [latex]f(x)=ax^{2}+c[\/latex]. This means it will be centered around [latex]x=0[\/latex].\r\n\r\nWe can start our table with any values we want, but will want to make sure to choose values on both sides of 0, as well as the value of [latex]x=0[\/latex].\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]x[\/latex]<\/td>\r\n[latex]f(x)=x^2-16[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n
    4<\/td>\r\n[latex](4)^2-16=0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
    -4<\/td>\r\n[latex](-4)^2-16=0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
    0<\/td>\r\n[latex](0)^2-16=-16[\/latex]<\/td>\r\n-16<\/td>\r\n<\/tr>\r\n
    2<\/td>\r\n[latex](2)^2-16=-12[\/latex]<\/td>\r\n-12<\/td>\r\n<\/tr>\r\n
    -2<\/td>\r\n[latex](-2)^2-16=-12[\/latex]<\/td>\r\n-12<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlotting the ordered pairs from the table above gives us:\r\n\r\n\"Coordinate\r\n\r\nConnecting the points with a smooth curve gives our final graph:\r\n\r\n\"Coordinate\r\n\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nLets try a few more examples.\r\n
    \r\n

    Example 4<\/h3>\r\nGraph the function: [latex]f(x)=x^2+5[\/latex]\r\n\r\n[reveal-answer q=\"613009\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"613009\"]\r\n\r\nNotice our function [latex]f(x)=x^2+5[\/latex] is of the form [latex]f(x)=ax^{2}+c[\/latex]. This means it will be centered around [latex]x=0[\/latex].\r\n\r\nCreate a table of values\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]x[\/latex]<\/td>\r\n[latex]f(x)=x^2+5[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n
    -2<\/td>\r\n[latex](-2)^2+5=9[\/latex]<\/td>\r\n9<\/td>\r\n<\/tr>\r\n
    -1<\/td>\r\n[latex](-1)^2+5=6[\/latex]<\/td>\r\n6<\/td>\r\n<\/tr>\r\n
    0<\/td>\r\n[latex](0)^2+5=5[\/latex]<\/td>\r\n5<\/td>\r\n<\/tr>\r\n
    1<\/td>\r\n[latex](1)^2+5=6[\/latex]<\/td>\r\n6<\/td>\r\n<\/tr>\r\n
    2<\/td>\r\n[latex](2)^2+5=9[\/latex]<\/td>\r\n9<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlot the points (-2,9), (-1,6), (0,5), (1,6) and (2,9) and connect them with a smooth curve.\r\n\r\n\"Coordinate\r\n\r\nExtension:\u00a0<\/b>\r\n\r\nNotice some quadratic functions never cross the x-axis. The function [latex]f(x)=x^2 +5[\/latex] is one example. If we try to solve for the x-intercepts in this function by setting the function equal to 0, we would have the following:\r\n

    [latex]x^2 +5=0[\/latex]<\/p>\r\n

    [latex]x^2 =-5[\/latex]<\/p>\r\n

    [latex]\\sqrt{x^2}=\u00b1\\sqrt{-5}[\/latex]<\/p>\r\nIf we try and find the square root of -5 in our calculator we will find that it gives us an error message. There are no real number answers to the square root of a negative value. Thus this function does not have any x-intercepts on our graph.\r\n\r\nAs we can see with the graph we made above, the parabola falls above the x-axis.\r\n\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example we will explore how parabolas don't always open upwards.\r\n

    \r\n

    Example 5<\/h3>\r\nGraph the function: [latex]f(x)=-3x^2[\/latex]\r\n\r\n[reveal-answer q=\"824856\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"824856\"]\r\n\r\nNotice our function\u00a0[latex]f(x)=-3x^2[\/latex], is of the form\u00a0[latex]f(x)=ax^{2}+c[\/latex], where [latex] a=-3[\/latex] and [latex] c=0[\/latex]. We will start to graph this by choosing values surrounding 0.\r\n\r\n*It is important that you understand how to evaluate a function like this correctly. For this reason, we will show the steps for the input value of -2 in detail:\r\n

    [latex]f(-2)= -3(-2)^2[\/latex]<\/p>\r\nWe must make sure we follow the correct \"Order of Operations\" here. This means that\u00a0before\u00a0<\/em>we multiply the [latex]-3[\/latex] with anything, we must\u00a0first\u00a0<\/em>raise the\u00a0 [latex]-2[\/latex] to the exponent of 2. When we raise\u00a0[latex](-2)^2[\/latex], we are multiplying\u00a0[latex](-2)\\cdot(-2)[\/latex], which is equal to\u00a0[latex]4[\/latex].\r\n

    [latex]f(-2)= -3(4)[\/latex]<\/p>\r\nWe can now multiply the -3 with the 4 to find the final output.\r\n

    [latex]f(-2)= -12[\/latex]<\/p>\r\nThis process will be repeated with all the x values input in the table of values below:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]x[\/latex]<\/td>\r\n[latex]f(x)=-3x^2[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n
    -2<\/td>\r\n[latex]-3(-2)^2=(-3)(4)=-12[\/latex]<\/td>\r\n-12<\/td>\r\n<\/tr>\r\n
    -1<\/td>\r\n[latex]-3(-1)^2=(-3)(1)=-3[\/latex]<\/td>\r\n-3<\/td>\r\n<\/tr>\r\n
    0<\/td>\r\n[latex]-3(0)^2=(-3)(0)=0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
    1<\/td>\r\n[latex]-3(1)^2=(-3)(1)=-3[\/latex]<\/td>\r\n-3<\/td>\r\n<\/tr>\r\n
    2<\/td>\r\n[latex]-3(2)^2=(-3)(4)=-12[\/latex]<\/td>\r\n-12<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlotting the ordered pairs from the table above gives us:\r\n\r\n\"Coordinate\r\n\r\nConnecting the points with a smooth curve gives our final graph below. Notice that this is an upside down U-shape this time.\r\n\r\n\"Coordinate\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAs we saw with the last example, with quadratic functions of the form\u00a0[latex]f(x)=ax^{2}+c[\/latex], changing the value of [latex]a[\/latex]\u00a0can change the width of the parabola and whether it opens up ([latex]a>0[\/latex]) or down ([latex]a<0[\/latex]). If [latex]a[\/latex] is positive, the vertex is the lowest point, and the parabola opens up. If [latex]a[\/latex] is negative, the vertex is the highest point, and the parabola opens down. Whereas Examples 3 and 4 showed that the value of [latex]c[\/latex] moves the parabola up or down.\r\n
    \r\n\r\nWhen graphing quadratic functions of the form\u00a0[latex]f(x)=ax^{2}+c[\/latex] follow the steps below:<\/span><\/strong>\r\n
      \r\n \t
    1. Recognize the form of the quadratic function and that it will be a parabola centered around x=0.<\/span><\/strong><\/li>\r\n \t
    2. Make a table of values, making sure to choose some values on either side of x=0, as well as the value of x=0.<\/span><\/strong><\/li>\r\n \t
    3. Plot your points and connect them with a smooth curve into a parabola shape.\u00a0<\/span><\/strong><\/li>\r\n<\/ol>\r\n<\/div>\r\nThe next example reveals what can cause a parabola to move right or left.\r\n
      \r\n

      Example 6<\/h3>\r\nGraph the function: [latex]f(x)=(x-4)^2[\/latex]\r\n\r\n[reveal-answer q=\"745787\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"745787\"]\r\n\r\nIf we choose the same values as the previous examples, we will obtain some relatively large results and also will fail to see both sides of the parabola. The key is to focus on the value of [latex]x[\/latex] that causes the expression inside the parentheses to equal to zero.\u00a0 In this case, that would be [latex]x=4[\/latex].\u00a0 So, as we select values to plug in, we will choose values centered around [latex]4[\/latex].\r\n\r\nCreate a table of values\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      [latex]x[\/latex]<\/td>\r\n[latex]f(x)=(x-4)^2[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n
      2<\/td>\r\n[latex](2-4)^2=(-2)^2=4[\/latex]<\/td>\r\n4<\/td>\r\n<\/tr>\r\n
      3<\/td>\r\n[latex](3-4)^2=(-1)^2=1[\/latex]<\/td>\r\n1<\/td>\r\n<\/tr>\r\n
      4<\/td>\r\n[latex](4-4)^2=0^2=0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
      5<\/td>\r\n[latex](5-4)^2=1^2=1[\/latex]<\/td>\r\n1<\/td>\r\n<\/tr>\r\n
      6<\/td>\r\n[latex](6-4)^2=2^2=4[\/latex]<\/td>\r\n4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlot the points (2,4), (3,1), (4,0), (5,1) and (6,2) and connect them with a smooth curve.\r\n\r\n\"Right\r\n\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe key to the last example is we no longer focused on choosing values around [latex]x=0[\/latex].\u00a0 Instead, we considered what would cause the quantity inside the parentheses to equal to zero, which was [latex]x=4[\/latex], plugging in this number and values on either side of it.\u00a0 If the function had been [latex]f(x)=(x+4)^2[\/latex], the primary [latex]x[\/latex]-value of interest would have been [latex]x=-4[\/latex].\u00a0 We can describe this using the form given below.\r\n
      \r\n\r\nWhen graphing quadratic functions of the form\u00a0[latex]f(x)=a(x-h)^{2}[\/latex] follow the steps below:<\/span><\/strong>\r\n
        \r\n \t
      1. Recognize the form of the quadratic function and that it will be a parabola centered around x=h.<\/span><\/strong><\/li>\r\n \t
      2. Make a table of values, making sure to choose some values on either side of x=h, as well as the value of x=h.<\/span><\/strong><\/li>\r\n \t
      3. Plot your points and connect them with a smooth curve into a parabola shape.\u00a0<\/span><\/strong><\/li>\r\n<\/ol>\r\n<\/div>\r\nNot all parabolas fall into the two forms given here.\u00a0 While you will explore these in more detail in future math classes, we end with a preview, including an application of the factoring skills we have learned.\r\n
        \r\n

        Think About it<\/h3>\r\nGraph the equation: [latex]f(x)={x}^{2}+x - 6[\/latex]. (It may be helpful to factor it, and set it equal to 0 to find the [latex]x[\/latex]-intercepts.)\r\n[reveal-answer q=\"609710\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"609710\"]\r\n\r\nTo factor [latex]f(x)={x}^{2}+x - 6[\/latex], we look for two numbers whose product equals [latex]-6[\/latex] and whose sum equals\u00a0[latex]1[\/latex]. Begin by looking at the possible factors of [latex]-6[\/latex].\r\n
        [latex]\\begin{array}{l}1\\cdot \\left(-6\\right)\\hfill \\\\ \\left(-6\\right)\\cdot 1\\hfill \\\\ 2\\cdot \\left(-3\\right)\\hfill \\\\ 3\\cdot \\left(-2\\right)\\hfill \\end{array}[\/latex]<\/div>\r\nThe last pair, [latex]3\\cdot \\left(-2\\right)[\/latex] sums to\u00a0[latex]1[\/latex], so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.\r\n
        [latex]f(x)=\\left(x - 2\\right)\\left(x+3\\right)[\/latex]<\/div>\r\nWe can now set this equation equal to 0 and use the zero-product property to find the\u00a0[latex]x[\/latex]-intercepts. To do this, we will set each factor equal to zero and solve.<\/span>\r\n

        [latex]\\left(x - 2\\right)\\left(x+3\\right)=0[\/latex]<\/p>\r\n\r\n

        [latex]\\begin{array}{lll}& \\left(x - 2\\right)\\left(x+3\\right)=0 & \\hfill \\\\ \\left(x - 2\\right)\\hfill=0 & & \\left(x+3\\right)=0 \\hfill \\\\ x=2 & & x=-3 \\hfill \\end{array}[\/latex]<\/div>\r\nRecall that x-intercepts are where the outputs, or y values are zero, therefore the points\u00a0[latex](-3,0)[\/latex] and\u00a0[latex](2,0)[\/latex] represent the places where the parabola crosses the x axis.\r\n\r\nTo find other points on the graph, we could make a table of values\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        [latex]x[\/latex]<\/td>\r\n[latex]f(x)={x}^{2}+x - 6[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n
        2<\/td>\r\n[latex](2)^2+(2) - 6 = 0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
        -3<\/td>\r\n[latex](-3)^2+(-3)-6=0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
        0<\/td>\r\n[latex](0)^2+(0)-6=-6[\/latex]<\/td>\r\n-6<\/td>\r\n<\/tr>\r\n
        1<\/td>\r\n[latex](1)^2+(1)-6=-4[\/latex]<\/td>\r\n-4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\nIf we plot the points found above and connect them with a smooth curve, we will get the parabola in the figure below.\r\n\r\n\"Coordinate\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

        Graphing Quadratics Summary<\/h2>\r\nCreating a graph of a function is one way to understand the relationship between the inputs and outputs of that function. Creating a graph can be done by choosing values for x<\/em>, finding the corresponding y<\/em> values, and plotting them. However, it helps to understand the basic shape of the function. Knowing how changes to the basic function equation affect the graph is also helpful.\r\n\r\nThe shape of a quadratic function is a parabola. Parabolas have the equation [latex]f(x)=ax^{2}+bx+c[\/latex], where [latex]a, b[\/latex] and [latex]c[\/latex] are real numbers and [latex]a\\ne0[\/latex]. The value of [latex]a[\/latex] determines the width and the direction of the parabola, while the vertex depends on the values of [latex]a, b[\/latex] and [latex]c[\/latex].","rendered":"
        \n

        section 6.8 Learning Objectives<\/h3>\n

        6.8: Further Exploration with Quadratic Equations<\/strong><\/p>\n

          \n
        • Solve application problems (area problems using factoring)<\/li>\n
        • Solve application problems (consecutive integer problems using factoring)<\/li>\n
        • Graph basic quadratic functions<\/li>\n<\/ul>\n<\/div>\n

          Applications of Solving Factorable Quadratic Equations<\/h2>\n

          When a polynomial is set equal to a value (whether an integer or another polynomial), the result is an equation. An equation that can be written in the form [latex]ax^{2}+bx+c=0[\/latex]\u00a0is called a quadratic equation<\/strong>. You can solve some quadratic equations using the rules of algebra, applying factoring techniques, and by using the Principle of Zero Products<\/strong>.<\/p>\n

          There are many applications for quadratic equations. Don’t forget that to use the Principle of Zero Products to solve a quadratic equation, you need to make sure that the equation is equal to zero. For example, [latex]12^{2}+11x+2=7[\/latex] must first be changed to [latex]12x^{2}+11x-5=0[\/latex]\u00a0by subtracting 7 from both sides.<\/p>\n

          Area Problems<\/h3>\n

          In our first example, we use some basic geometry as we search for the dimensions of rectangle.<\/p>\n

          \n

          Example 1<\/h3>\n

          The area of a rectangular garden is 30 square feet. If the length is 7 feet longer than the width, find the dimensions.<\/p>\n

          Show Solution<\/span><\/p>\n
          The formula for the area of a rectangle is [latex]\\text{Area}=\\text{length}\\cdot\\text{width}[\/latex], or [latex]A=l\\cdot{w}[\/latex].<\/p>\n

          [latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,A=l\\cdot{w}\\\\\\,\\text{width}=w\\\\\\text{length}=w+7\\\\\\,\\,\\,\\,\\,\\text{area}=30\\\\\\\\30=\\left(w+7\\right)\\left(w\\right)\\end{array}[\/latex]<\/p>\n

          Multiply.<\/p>\n

          [latex]30=w^{2}+7w[\/latex]<\/p>\n

          Subtract 30 from both sides to set the equation equal to 0.<\/p>\n

          [latex]w^{2}+7w\u201330=0[\/latex]<\/p>\n

          Using the product and sum technique, we can factor this by finding two numbers whose product is [latex]\u221230[\/latex] and whose sum is 7.<\/p>\n

          [latex]\\left(w\u20133\\right)\\left(w+10\\right)=0[\/latex]<\/p>\n

          Use the Zero Product Property to solve for w<\/i>.<\/p>\n

          [latex]\\begin{array}{l}w-3=0\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,w+10=0\\\\\\,\\,\\,\\,\\,\\,\\,w=3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=-10\\end{array}[\/latex]<\/p>\n

          The solution [latex]w=\u221210[\/latex]\u00a0does not work for this application as the width cannot be a negative number. So, we discard the [latex]\u221210[\/latex]. Thus, the width is 3 feet.<\/p>\n

          The width = 3 feet<\/p>\n

          Substitute [latex]w=3[\/latex] into the expression for the length, [latex]l=w+7[\/latex] to find the length: [latex]l=3+7=10[\/latex].<\/p>\n

          The length is [latex]3+7=10[\/latex] feet<\/p>\n

          Answer<\/h4>\n

          The width of the garden is 3 feet, and the length is 10 feet.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          In the example in the following video, we present another area application of factoring trinomials.<\/p>\n