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Math Help - Quad Equation

  1. #1
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    Quad Equation

    Alright, I know this, but its been a long semester, haha.

     h=-5t^2+6t+3

    What is the maximum height, how long does it take, what is the y intercept.
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  2. #2
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    Quote Originally Posted by abc10 View Post
    Alright, I know this, but its been a long semester, haha.

     h=-5t^2+6t+3

    What is the maximum height, how long does it take, what is the y intercept.
    There won't be a y intercept, as there is no y in your equation.

    I think you mean the h intercept.

    Let t = 0.

    Then h = -5(0)^2 + 6(0) + 3

    h = 3.


    Since the coefficient of t^2 is negative, you WILL have a maximum. To find it, put the equation in Turning Point Form by Completing the Square.

    h = -5t^2 + 6t + 3

    h = -5\left(t^2 - \frac{6}{5}t - \frac{3}{5}\right)

    h = -5\left[t^2 - \frac{6}{5}t + \left(-\frac{3}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 - \frac{3}{5}\right]

    h = -5\left[\left(t - \frac{3}{5}\right)^2 - \frac{24}{25}\right]

    h = -5\left(t - \frac{3}{5}\right)^2 + \frac{24}{5}.


    So the Turning point is (t, h) = \left(\frac{3}{5}, \frac{24}{5}\right).

    So the maximum value of h is \frac{24}{5} which occurs when t = \frac{3}{5}.
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  3. #3
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    To find the y-intercept of a function f(x), simply find f(0). So in your case, what is h(0).

    To find extremum without calculus? Hmmm. I'm sure there is a way other than guess and check or running a program. But I'm not remembering.
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  4. #4
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    Solutions

    maximum height of 1.58 y-intercept of 3
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  5. #5
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    Quote Originally Posted by nightrider456 View Post
    maximum height of 1.58 y-intercept of 3
    Wrong. The maximum height is actually \frac{24}{5} = 4.8.
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