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Thread: Quadratic Equation Question

  1. #1
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    Quadratic Equation Question

    Shane Bechler is about to punt the ball. Let "t" represent the # of seconds from the moment his foot constacts the football. The height of the ball when Lechler makes contact is 4 feet. After one seconds, the ball has reached a height of 44 feet and after two seconds, the ball has reached a height of 52 feet. Hint: h(t) = -16t^2 + vt + h should not be used.

    a. Write a quadratic function in standard form for the information given.
    b.After how many seconds does the football reach its maximum height and what is the height?

    P.S. I tried to work this problem using my TI 84 Calculator by finding the Quadratic regression but it is in the same form as the hint, but I can't use that form. So help me out...I'm really confused. How would I work this problem out?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chrozer View Post
    Shane Bechler is about to punt the ball. Let "t" represent the # of seconds from the moment his foot constacts the football. The height of the ball when Lechler makes contact is 4 feet. After one seconds, the ball has reached a height of 44 feet and after two seconds, the ball has reached a height of 52 feet. Hint: h(t) = -16t^2 + vt + h should not be used.

    a. Write a quadratic function in standard form for the information given.
    b.After how many seconds does the football reach its maximum height and what is the height?

    P.S. I tried to work this problem using my TI 84 Calculator by finding the Quadratic regression but it is in the same form as the hint, but I can't use that form. So help me out...I'm really confused. How would I work this problem out?
    You don't have any data to do a quadratic regression.

    Standard form for a quadratic is not $\displaystyle h = at^2 + bt + c$, it is $\displaystyle h = a(t - h)^2 + k$

    So
    $\displaystyle 4 = a(0 - h)^2 + k$
    and
    $\displaystyle 44 = a(1 - h)^2 + k$
    and
    $\displaystyle 52 = a(2 - h)^2 + k$

    Solve for a, h, and k.

    -Dan
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  3. #3
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    Yeah I know I didn't have alot of data but I had 3 points, (0,4);(1,44);(2,52).

    I get the equation of
    $\displaystyle
    y = 16x^2 + 56x + 4
    $

    But I doubt that's answer since the question said not to use


    $\displaystyle
    h(t) = -16t^2 + vt + h
    $

    And how would you go by solving the equations you set up?

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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post

    $\displaystyle 4 = a(0 - h)^2 + k$
    and
    $\displaystyle 44 = a(1 - h)^2 + k$
    and
    $\displaystyle 52 = a(2 - h)^2 + k$

    Solve for a, h, and k.
    The first equation says:
    $\displaystyle 4 = ah^2 + k$

    Thus $\displaystyle k = 4 - ah^2$

    Insert this value for k into the other two equations:
    $\displaystyle 44 = a(1 - h)^2 + 4 - ah^2$
    and
    $\displaystyle 52 = a(2 - h)^2 + 4 - ah^2$

    Giving us
    $\displaystyle 40 = a(1 - h)^2 - ah^2$
    and
    $\displaystyle 48 = a(2 - h)^2 - ah^2$

    Take the first one:
    $\displaystyle 40 = a(1 - h)^2 - ah^2$

    $\displaystyle 40 = a - 2ah + ah^2 - ah^2$

    $\displaystyle 40 = a - 2ah$

    $\displaystyle a = \frac{40}{1 - 2h}$

    Insert this value of a into the last equation:
    $\displaystyle 48 = \left ( \frac{40}{1 - 2h} \right ) (2 - h)^2 - \left ( \frac{40}{1 - 2h} \right ) h^2$

    Solve for h:
    $\displaystyle 48(1 - 2h) = 40(2 - h)^2 - 40h^2$

    $\displaystyle 48 - 96h = 160 - 160h + 40h^2 - 40h^2$

    $\displaystyle 48 - 96h = 160 - 160h$

    $\displaystyle 64h = 112$

    $\displaystyle h = \frac{7}{4}$

    Giving us
    $\displaystyle a = \frac{40}{1 - 2 \cdot \frac{3}{16}} = -16$
    and
    $\displaystyle k = 4 - (-16) \cdot \left ( \frac{7}{4} \right )^2 = 53$

    -Dan
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  5. #5
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    I see.... Thnx for the help.
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