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Math Help - Quadratic Relations Problem

  1. #1
    Junior Member
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    Quadratic Relations Problem

    No matter what I try, I end up wrong

    A programmer is writing the code for a new interactive basketball game. As a result, she is using quadratic relations to model the path of the ball. During the game, when a ball is shot, the path it follows is modelled by the quadratic relation, , where h represented the height of the ball above the ground and d represented the distance of the ball from the shooter. Both distances are measured in feet.
    1. How high was the ball when the shooter shot it?
    2. What was the maximum height obtained by the ball?
    3. A rim of a basketball net is 10 feet high. For what distance was the ball above the rim of the basketball net?
    4. How far would the shooter have to be away from the rim of the basketball net for the ball to hit the rim of the basketball net and possibly go in?
    5. Create a graphical model, using technology, to verify your calculations.
    6. Create your own quadratic relation that would model the path of a shot from a distance of 15 feet that would hit the rim of the basketball net. Explain how you obtained your answer.
    Provide step-by-step solutions that are communicated clearly for all questions

    What I tried so far is:
    a)
    H=0.2d+3d+6
    H=0.2(0)=3(0)+6
    H=0+0+6
    H=6 -- The ball was 6 feet above the ground when shot.

    b) What makes my brain explode:
    H=0.2d+3d+6
    H=-0.2(d-15+6
    H=-0.2d-15d+3-3)+6
    H=-0.2(d(-1.7)-3)+6
    H=-0.2(d-1.7)+0.578+6
    H=-0.2(d-1.7)+6.587

    Vertex= (1.7, 6.578)
    Max Value= 6.58 when d= 1.7
    The max height of the ball was 6.58 when 1.7.

    I know im wrong cause c) mentiones that the ball was above 10f, and in my solution its under 10f.
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  2. #2
    MHF Contributor
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    Hi

    a) OK

    b) H = -0.2 d^2 + 3d + 6

    H = -0.2 (d^2 - 15d - 30)

    H = -0.2 ((d-\frac{15}{2})^2 - \frac{225}{4} - 30)

    H = -0.2 ((d-\frac{15}{2})^2 - \frac{445}{4})

    H = -0.2 (d-\frac{15}{2})^2 + \frac{69}{4}

    Vertex= (7.5, 17.25)
    Max Value= 17.25 when d= 7.5
    The max height of the ball was 17.25 when 7.5
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  3. #3
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    Thanks a lot,but I wonder how you got 30 in the second step of b)

    Edit: Got it, you divided 6 by 0.2 which is 30 :-)

    I will try to solve number 3 and 4, but have no clue what to do in 5 and 6
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  4. #4
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    For C I have:

    0.2d+3d+6(d=10)
    D=0.2d+3d+6
    0=-0.2d+3d+6-10
    0=-0.2d+3d+4
    0=-0.2(d-15d+20)
    Thats as far as I get, I tried the following but know its wrong:
    0=-0.2(d-0.2 or 75)(d-100)
    ---------------------
    15/0.2=75
    20/0.2=100

    and help on d would be also nice, I dont even know how to start there
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