{"id":11033,"date":"2015-07-14T17:53:56","date_gmt":"2015-07-14T17:53:56","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&p=11033"},"modified":"2015-09-10T17:17:57","modified_gmt":"2015-09-10T17:17:57","slug":"solve-problems-involving-a-quadratic-functions-minimum-or-maximum-value","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/chapter\/solve-problems-involving-a-quadratic-functions-minimum-or-maximum-value\/","title":{"raw":"Solve problems involving a quadratic function\u2019s minimum or maximum value","rendered":"Solve problems involving a quadratic function\u2019s minimum or maximum value"},"content":{"raw":"

In Example 7, the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form.<\/p>\r\n\r\n

\r\n

How To: Given a quadratic function, find the x<\/em>-intercepts by rewriting in standard form.<\/h3>\r\n
    \r\n\t
  1. Substitute a<\/em>\u00a0and b<\/em>\u00a0into [latex]h=-\\frac{b}{2a}[\/latex].<\/li>\r\n\t
  2. Substitute x<\/em> =\u00a0h<\/em>\u00a0into the general form of the quadratic function to find k<\/em>.<\/li>\r\n\t
  3. Rewrite the quadratic in standard form using h<\/em>\u00a0and k<\/em>.<\/li>\r\n\t
  4. Solve for when the output of the function will be zero to find the x-<\/em>intercepts.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
    \r\n
    \r\n
    \r\n

    Example 8: Finding the x<\/em>-Intercepts of a Parabola<\/h3>\r\n

    Find the x<\/em>-intercepts of the quadratic function [latex]f\\left(x\\right)=2{x}^{2}+4x - 4[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Solution<\/h3>\r\n

    We begin by solving for when the output will be zero.<\/p>\r\n\r\n

    [latex]0=2{x}^{2}+4x - 4[\/latex]<\/div>\r\n

    Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.<\/p>\r\n\r\n

    [latex]f\\left(x\\right)=a{\\left(x-h\\right)}^{2}+k[\/latex]<\/div>\r\n

    We know that a\u00a0<\/em>= 2. Then we solve for h<\/em>\u00a0and k<\/em>.<\/p>\r\n\r\n

    [latex]\\begin{cases}h=-\\frac{b}{2a}\\hfill & \\hfill & \\hfill & k=f\\left(-1\\right)\\hfill \\\\ \\text{ }=-\\frac{4}{2\\left(2\\right)}\\hfill & \\hfill & \\hfill & \\text{ }=2{\\left(-1\\right)}^{2}+4\\left(-1\\right)-4\\hfill \\\\ \\text{ }=-1\\hfill & \\hfill & \\hfill & \\text{ }=-6\\hfill \\end{cases}[\/latex]<\/div>\r\n

    So now we can rewrite in standard form.<\/p>\r\n\r\n

    [latex]f\\left(x\\right)=2{\\left(x+1\\right)}^{2}-6[\/latex]<\/div>\r\n

    We can now solve for when the output will be zero.<\/p>\r\n\r\n

    [latex]\\begin{cases}0=2{\\left(x+1\\right)}^{2}-6\\hfill \\\\ 6=2{\\left(x+1\\right)}^{2}\\hfill \\\\ 3={\\left(x+1\\right)}^{2}\\hfill \\\\ x+1=\\pm \\sqrt{3}\\hfill \\\\ x=-1\\pm \\sqrt{3}\\hfill \\end{cases}[\/latex]<\/div>\r\n

    The graph has x-<\/em>intercepts at [latex]\\left(-1-\\sqrt{3},0\\right)[\/latex] and [latex]\\left(-1+\\sqrt{3},0\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Analysis of the Solution<\/h3>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 15<\/b>[\/caption]\r\n

    We can check our work by graphing the given function on a graphing utility and observing the x-<\/em>intercepts.\r\n<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n

    Try It 4<\/h3>\r\n

    In Try It\u00a02<\/a>, we found the standard and general form for the function [latex]g\\left(x\\right)=13+{x}^{2}-6x[\/latex]. Now find the y<\/em>- and x<\/em>-intercepts (if any).<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

    \r\n
    \r\n
    \r\n

    Example 9: Solving a Quadratic Equation with the Quadratic Formula<\/h3>\r\n

    Solve [latex]{x}^{2}+x+2=0[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Solution<\/h3>\r\n

    Let\u2019s begin by writing the quadratic formula: [latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex].<\/p>\r\n

    When applying the quadratic formula<\/span>, we identify the coefficients a<\/em>,\u00a0b<\/em>, and\u00a0c<\/em>. For the equation [latex]{x}^{2}+x+2=0[\/latex], we have a<\/em>\u00a0=\u00a01,\u00a0b<\/em>\u00a0=\u00a01, and c<\/em>\u00a0=\u00a02.\u00a0Substituting these values into the formula we have:<\/p>\r\n\r\n

    [latex]\\begin{cases}x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\\\ \\text{ }=\\frac{-1\\pm \\sqrt{{1}^{2}-4\\cdot 1\\cdot \\left(2\\right)}}{2\\cdot 1}\\hfill \\\\ \\text{ }=\\frac{-1\\pm \\sqrt{1 - 8}}{2}\\hfill \\\\ \\text{ }=\\frac{-1\\pm \\sqrt{-7}}{2}\\hfill \\\\ \\text{ }=\\frac{-1\\pm i\\sqrt{7}}{2}\\hfill \\end{cases}[\/latex]<\/div>\r\n

    The solutions to the equation are [latex]x=\\frac{-1+i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1-i\\sqrt{7}}{2}[\/latex] or [latex]x=\\frac{-1}{2}+\\frac{i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1}{2}-\\frac{i\\sqrt{7}}{2}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n
    \r\n

    Example 10: Applying the Vertex and x<\/em>-Intercepts of a Parabola<\/h3>\r\n

    A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball\u2019s height above ground can be modeled by the equation [latex]H\\left(t\\right)=-16{t}^{2}+80t+40[\/latex].<\/p>\r\n

    a. When does the ball reach the maximum height?<\/p>\r\n

    b. What is the maximum height of the ball?<\/p>\r\n

    c. When does the ball hit the ground?<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Solution<\/h3>\r\na. The ball reaches the maximum height at the vertex of the parabola.\r\n
    [latex]\\begin{cases} h=-\\frac{80}{2\\left(-16\\right)} \\text{ }=\\frac{80}{32}\\hfill \\\\ \\text{ }=\\frac{5}{2}\\hfill \\\\ \\text{ }=2.5\\hfill \\end{cases}[\/latex]<\/div>\r\n

    The ball reaches a maximum height after 2.5 seconds.<\/p>\r\nb. To find the maximum height, find the y\u00a0<\/em>coordinate of the vertex of the parabola.\r\n

    [latex]\\begin{cases}k=H\\left(-\\frac{b}{2a}\\right)\\hfill \\\\ \\text{ }=H\\left(2.5\\right)\\hfill \\\\ \\text{ }=-16{\\left(2.5\\right)}^{2}+80\\left(2.5\\right)+40\\hfill \\\\ \\text{ }=140\\hfill \\end{cases}[\/latex]<\/div>\r\n

    The ball reaches a maximum height of 140 feet.<\/p>\r\nc. To find when the ball hits the ground, we need to determine when the height is zero, [latex]H\\left(t\\right)=0[\/latex].\r\n\r\nWe use the quadratic formula.\r\n

    [latex]\\begin{cases} t=\\frac{-80\\pm \\sqrt{{80}^{2}-4\\left(-16\\right)\\left(40\\right)}}{2\\left(-16\\right)}\\hfill \\\\ \\text{ }=\\frac{-80\\pm \\sqrt{8960}}{-32}\\hfill \\end{cases}[\/latex]<\/div>\r\nBecause the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.\r\n
    [latex]\\begin{cases}t=\\frac{-80-\\sqrt{8960}}{-32}\\approx 5.458\\hfill & \\text{or}\\hfill & t=\\frac{-80+\\sqrt{8960}}{-32}\\approx -0.458\\hfill \\end{cases}[\/latex]<\/div>\r\nThe second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds.\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 16<\/b>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n

    Try It 5<\/h3>\r\n

    A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock\u2019s height above ocean can be modeled by the equation [latex]H\\left(t\\right)=-16{t}^{2}+96t+112[\/latex].<\/p>\r\n

    a. When does the rock reach the maximum height?<\/p>\r\n

    b. What is the maximum height of the rock?<\/p>\r\n

    c. When does the rock hit the ocean?<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

    In Example 7, the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form.<\/p>\n

    \n

    How To: Given a quadratic function, find the x<\/em>-intercepts by rewriting in standard form.<\/h3>\n
      \n
    1. Substitute a<\/em>\u00a0and b<\/em>\u00a0into [latex]h=-\\frac{b}{2a}[\/latex].<\/li>\n
    2. Substitute x<\/em> =\u00a0h<\/em>\u00a0into the general form of the quadratic function to find k<\/em>.<\/li>\n
    3. Rewrite the quadratic in standard form using h<\/em>\u00a0and k<\/em>.<\/li>\n
    4. Solve for when the output of the function will be zero to find the x-<\/em>intercepts.<\/li>\n<\/ol>\n<\/div>\n
      \n
      \n
      \n

      Example 8: Finding the x<\/em>-Intercepts of a Parabola<\/h3>\n

      Find the x<\/em>-intercepts of the quadratic function [latex]f\\left(x\\right)=2{x}^{2}+4x - 4[\/latex].<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      We begin by solving for when the output will be zero.<\/p>\n

      [latex]0=2{x}^{2}+4x - 4[\/latex]<\/div>\n

      Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.<\/p>\n

      [latex]f\\left(x\\right)=a{\\left(x-h\\right)}^{2}+k[\/latex]<\/div>\n

      We know that a\u00a0<\/em>= 2. Then we solve for h<\/em>\u00a0and k<\/em>.<\/p>\n

      [latex]\\begin{cases}h=-\\frac{b}{2a}\\hfill & \\hfill & \\hfill & k=f\\left(-1\\right)\\hfill \\\\ \\text{ }=-\\frac{4}{2\\left(2\\right)}\\hfill & \\hfill & \\hfill & \\text{ }=2{\\left(-1\\right)}^{2}+4\\left(-1\\right)-4\\hfill \\\\ \\text{ }=-1\\hfill & \\hfill & \\hfill & \\text{ }=-6\\hfill \\end{cases}[\/latex]<\/div>\n

      So now we can rewrite in standard form.<\/p>\n

      [latex]f\\left(x\\right)=2{\\left(x+1\\right)}^{2}-6[\/latex]<\/div>\n

      We can now solve for when the output will be zero.<\/p>\n

      [latex]\\begin{cases}0=2{\\left(x+1\\right)}^{2}-6\\hfill \\\\ 6=2{\\left(x+1\\right)}^{2}\\hfill \\\\ 3={\\left(x+1\\right)}^{2}\\hfill \\\\ x+1=\\pm \\sqrt{3}\\hfill \\\\ x=-1\\pm \\sqrt{3}\\hfill \\end{cases}[\/latex]<\/div>\n

      The graph has x-<\/em>intercepts at [latex]\\left(-1-\\sqrt{3},0\\right)[\/latex] and [latex]\\left(-1+\\sqrt{3},0\\right)[\/latex].<\/p>\n<\/div>\n

      \n

      Analysis of the Solution<\/h3>\n
      \"Graph<\/p>\n

      Figure 15<\/b><\/p>\n<\/div>\n

      We can check our work by graphing the given function on a graphing utility and observing the x-<\/em>intercepts.
      \n<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n

      \n

      Try It 4<\/h3>\n

      In Try It\u00a02<\/a>, we found the standard and general form for the function [latex]g\\left(x\\right)=13+{x}^{2}-6x[\/latex]. Now find the y<\/em>– and x<\/em>-intercepts (if any).<\/p>\n

      Solution<\/a><\/p>\n<\/div>\n

      \n
      \n
      \n

      Example 9: Solving a Quadratic Equation with the Quadratic Formula<\/h3>\n

      Solve [latex]{x}^{2}+x+2=0[\/latex].<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      Let\u2019s begin by writing the quadratic formula: [latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex].<\/p>\n

      When applying the quadratic formula<\/span>, we identify the coefficients a<\/em>,\u00a0b<\/em>, and\u00a0c<\/em>. For the equation [latex]{x}^{2}+x+2=0[\/latex], we have a<\/em>\u00a0=\u00a01,\u00a0b<\/em>\u00a0=\u00a01, and c<\/em>\u00a0=\u00a02.\u00a0Substituting these values into the formula we have:<\/p>\n

      [latex]\\begin{cases}x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\\\ \\text{ }=\\frac{-1\\pm \\sqrt{{1}^{2}-4\\cdot 1\\cdot \\left(2\\right)}}{2\\cdot 1}\\hfill \\\\ \\text{ }=\\frac{-1\\pm \\sqrt{1 - 8}}{2}\\hfill \\\\ \\text{ }=\\frac{-1\\pm \\sqrt{-7}}{2}\\hfill \\\\ \\text{ }=\\frac{-1\\pm i\\sqrt{7}}{2}\\hfill \\end{cases}[\/latex]<\/div>\n

      The solutions to the equation are [latex]x=\\frac{-1+i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1-i\\sqrt{7}}{2}[\/latex] or [latex]x=\\frac{-1}{2}+\\frac{i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1}{2}-\\frac{i\\sqrt{7}}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n
      \n

      Example 10: Applying the Vertex and x<\/em>-Intercepts of a Parabola<\/h3>\n

      A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball\u2019s height above ground can be modeled by the equation [latex]H\\left(t\\right)=-16{t}^{2}+80t+40[\/latex].<\/p>\n

      a. When does the ball reach the maximum height?<\/p>\n

      b. What is the maximum height of the ball?<\/p>\n

      c. When does the ball hit the ground?<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      a. The ball reaches the maximum height at the vertex of the parabola.<\/p>\n

      [latex]\\begin{cases} h=-\\frac{80}{2\\left(-16\\right)} \\text{ }=\\frac{80}{32}\\hfill \\\\ \\text{ }=\\frac{5}{2}\\hfill \\\\ \\text{ }=2.5\\hfill \\end{cases}[\/latex]<\/div>\n

      The ball reaches a maximum height after 2.5 seconds.<\/p>\n

      b. To find the maximum height, find the y\u00a0<\/em>coordinate of the vertex of the parabola.<\/p>\n

      [latex]\\begin{cases}k=H\\left(-\\frac{b}{2a}\\right)\\hfill \\\\ \\text{ }=H\\left(2.5\\right)\\hfill \\\\ \\text{ }=-16{\\left(2.5\\right)}^{2}+80\\left(2.5\\right)+40\\hfill \\\\ \\text{ }=140\\hfill \\end{cases}[\/latex]<\/div>\n

      The ball reaches a maximum height of 140 feet.<\/p>\n

      c. To find when the ball hits the ground, we need to determine when the height is zero, [latex]H\\left(t\\right)=0[\/latex].<\/p>\n

      We use the quadratic formula.<\/p>\n

      [latex]\\begin{cases} t=\\frac{-80\\pm \\sqrt{{80}^{2}-4\\left(-16\\right)\\left(40\\right)}}{2\\left(-16\\right)}\\hfill \\\\ \\text{ }=\\frac{-80\\pm \\sqrt{8960}}{-32}\\hfill \\end{cases}[\/latex]<\/div>\n

      Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.<\/p>\n

      [latex]\\begin{cases}t=\\frac{-80-\\sqrt{8960}}{-32}\\approx 5.458\\hfill & \\text{or}\\hfill & t=\\frac{-80+\\sqrt{8960}}{-32}\\approx -0.458\\hfill \\end{cases}[\/latex]<\/div>\n

      The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds.
      \n<\/span><\/p>\n

      \"Graph<\/p>\n

      Figure 16<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n

      Try It 5<\/h3>\n

      A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock\u2019s height above ocean can be modeled by the equation [latex]H\\left(t\\right)=-16{t}^{2}+96t+112[\/latex].<\/p>\n

      a. When does the rock reach the maximum height?<\/p>\n

      b. What is the maximum height of the rock?<\/p>\n

      c. When does the rock hit the ocean?<\/p>\n

      Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

      \n\t\t\t

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