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Carter and Jordan are warming up for their ball game by playing catch with a softball. They are standing \(12\) metres from each other, and they each catch the softball in their gloves, \(1.5\) metres above the ground. When Carter throws the ball to Jordan, it reaches a maximum height of \(2.5\) metres.

  • Let \(x\) be the horizontal distance, in metres, between Carter and the softball.
  • Let \(f(x)\) be the height, in metres, of the softball.

    1. Sketch a graph that models the flight path of the softball thrown by Carter.


    2. Determine a quadratic function, in standard form, that models the softball’s flight path.

      The vertex is at \((6, 2.5)\) so \(\left( {p, q} \right) = \left( {6,2.5} \right)\).

      Use the vertex form of a quadratic function, \(f\left( x \right) = a\left( {x - p} \right)^2 + q\).

      Substituting in \(p\) and \(q\), it becomes \(f\left( x \right) = a\left( {x - 6} \right)^2 + 2.5\).

      Solve for a using point \((12, 1.5)\).

      \[\begin{align}
       1.5 &= a\left( {12 - 6} \right)^2 + 2.5 \\
       -1 &= a\left( 6 \right)^2  \\
       -\frac{1}{{36}} &= a  \end{align}\]

      Substitute the known values into the vertex form.

      \[f\left( x \right) = -\frac{1}{{36}}\left( {x - 6} \right)^2 + 2.5\]

      Expand and simplify to convert from vertex form to standard form.

      \[\begin{array}
       f\left( x \right) = - \frac{1}{{36}}\left( {x - 6} \right)^2 + 2.5 \\
       f\left( x \right) = - \frac{1}{{36}}\left( {x^2 - 12x + 36} \right) + 2.5 \\
       f\left( x \right) = - \frac{1}{{36}}x^2 + \frac{1}{3}x - 1 + 2.5 \\
       f\left( x \right) = - \frac{1}{{36}}x^2 + \frac{1}{3}x + 1.5
       \end{array}\]

      Convert all numbers to either decimals or fractions.

      \[f\left( x \right) = -\frac{1}{{36}}x^2 + \frac{1}{3}x + \frac{3}{2}\]


    3. How far away from Carter does Jordan need to be to catch the ball \(1\) metre from the ground? Round to the nearest hundredth.

      \(f(x) = 1 \thinspace \rm{metre}\). Substitute this value into the equation of the function, and solve for \(x\).

      \[\begin{align}
       f\left( x \right) &= -\frac{1}{{36}}x^2 + \frac{1}{3}x + \frac{3}{2} \\
       1 &= -\frac{1}{{36}}x^2 + \frac{1}{3}x + \frac{3}{2} \\
       0 &= -\frac{1}{{36}}x^2 + \frac{1}{3}x + \frac{1}{2} \\
       \end{align}\]


      Multiply the entire equation by \(36\) to make it easier to manage.

      \[\begin{align}
       (0 &= -\frac{1}{36}x^2 + \frac{1}{3}x + \frac{1}{2})36 \\
       0 &= - x^2 + 12x + 18 \\
       \end{align}\]

      Use the quadratic formula to solve for \(x\).

      \[\begin{align}
       a &= -1, b = 12, c = 18 \\
       x &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
       x & = \frac{{ - 12 \pm \sqrt {\left( {12} \right)^2  - 4\left( { - 1} \right)\left( {18} \right)} }}{{2\left( { - 1} \right)}} \\
       x &= \frac{{ - 12 \pm \sqrt {144 + 72} }}{{ - 2}} \\
       x &= \frac{{ - 12 \pm \sqrt {216} }}{{ - 2}} \\
       x &= \frac{{ - 12 \pm \sqrt {36 \cdot 6} }}{{ - 2}} \\
       x &= \frac{{ - 12 \pm 6\sqrt 6 }}{{ - 2}} \\
       x &= 6 \pm 3\sqrt 6  \\
       x &= 13.348...\thinspace {\rm{and} }\thinspace  x = - 1.348...  \end{align}\]

      Because a negative distance does not make sense, Jordan will need to be \(13.35 \thinspace \rm{metres}\) away from Carter in order to catch the ball \(1 \thinspace \rm{metre}\) above the ground.