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  • Apr 29, 2009, 05:15 PM
    rachh
    how do i do this?
    Tanay kicks a soccer ball during a game. The height of the ball, in feet, can be modeled by the function f(x)= -16xto the second power + 48x, where x is the time in seconds after she kicks the ball. Graph the function. Find the maximun height of the ball and how long it takes the ball to reach that height.
  • Apr 30, 2009, 12:22 AM
    Capuchin

    Have you graphed it? We can't exactly graph it for you...
  • Apr 30, 2009, 11:19 AM
    Zazonker
    Assuming you are new to graphing and having trouble with that. You'd start with x on the horizontal axis and f(x) on the verticle. You calculate the value of f(x) for a few values of x just by plugging them into the equation. If you try x = 1, x=2 and x=3 you get 32, 32 and 0 respectively.

    Visualize the soccer ball in flight - it certainly didn't jump instantly to 32 feet, stay there for a couple of seconds and then drop instantly to the ground. From that you know that you need smaller increments than whole seconds, maybe say tenths of a second. And the high point was before 3 seconds - quite likely between 1 and 2 since the only way the ball could be at 32 at both times was if it was going up at 1 second and coming down at 2 seconds.

    From this, I think you should be able to solve it.
  • Apr 30, 2009, 11:21 AM
    Unknown008

    Hint : complete the square to find your reference points to draw your graph.

    If an equation has the form

    The completed squared form will be



    Setting x to zero and finding the resulting value will give you the y intercept.

    The maximum (or minimum) point is given by
  • Apr 30, 2009, 11:34 AM
    Zazonker
    Seems you are making this a bit more difficult than it is. The Y intercept can be "calculated" just be setting x to zero in the original equation. No need to complete the square.

    F(x) = -16 x ^2 + 48 x
    If x = 0, f(x) = 0

    The max can be determined by setting the first derivative to zero.
    That is:
    -32x + 48 = 0
    x= 48/32 = 1.5
  • Apr 30, 2009, 11:39 AM
    Unknown008

    Yeah, I know, but I don't know whether the OP has been learning derivatives...

    The formula really seems long, but when you plug in the numbers, it's lots shorter and simpler.

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