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Math Help - whopping equation

  1. #1
    Senior Member furor celtica's Avatar
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    whopping equation

    values of s along y axis
    values of t along x axis
    s=f(t)=a (t-p)(t-q)^2
    where a, p, q are constants
    the curve meets the coordinate axes at the points (0, -4.5) (1, 0) (3, 0)
    find values for a, p ,q

    i came up with three equations but they are so massive i doubt im on the right track.
    i found
    0 = a ( 27 - 18q + 3q^2 - 9p + 6pq - pq^2)
    0 = a ( 1 - 2q + q^2 - p + 2pq - pq^2)
    4.5 = apq^2
    are these correct? if so, how can they be solved in a time period shorter than say a month?
    somebody please show me step by step?
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  2. #2
    Super Member
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    You're thinking too hard.

    values of s along y axis
    values of t along x axis
    s=f(t)=a (t-p)(t-q)^2
    where a, p, q are constants
    the curve meets the coordinate axes at the points (0, -4.5) (1, 0) (3, 0)
    find values for a, p ,q
    Sorry, but I'm going to change the variables s & t to x & y.
    y = a(x - p)(x - q)^2

    The 2nd and 3rd points are x-intercepts.
    (Note: if you have an equation in factored form, like y = (x - m)(x - n), finding the x-intercepts is really easy; they're the numbers subtracted from x. In my example, they would be m and n.)

    That means that p = 1 and q = 3, or p = 3 and q = 1. Plug in each pair, with the 1st point, to find a.

    p = 1, q = 3:
    \begin{aligned}<br />
-4.5 &= a(0 - 1)(0 - 3)^2 \\<br />
-4.5 &= -9a \\<br />
a &= 0.5<br />
\end{aligned}

    p = 3, q = 1:
    \begin{aligned}<br />
-4.5 &= a(0 - 3)(0 - 1)^2 \\<br />
-4.5 &= -3a \\<br />
a &= 1.5<br />
\end{aligned}

    So I found two sets of values of a, p, and q. Unless there is more information that I'm missing...?


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  3. #3
    MHF Contributor
    Grandad's Avatar
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    Hello everyone

    I agree with yeongil except that there's a sign wrong: p =1, q=3, a = -0.5 and p = 3, q = 1, a = -1.5.

    Grandad
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  4. #4
    Senior Member furor celtica's Avatar
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    but where did i go wrong? could this have been solved my way?
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  5. #5
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    furor celtica: perhaps, but it would be nasty. Why do it if the method I presented is easier?

    Grandad: I double-checked, and I'm still getting positive values for a.

    Come to think of it, a has to be positive. Looking at the original function:
    y = a(x - p)(x - q)^2

    All of the roots of the cubic are known: 1 real root, and 1 double real root. To the left of the roots we have a known point below the x-axis: (0, -4.5). Our cubic, then, is such that
    as x -> -∞, y -> -∞ and
    as x -> ∞, y -> ∞.
    a, the leading coefficient, must therefore be positive.


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  6. #6
    MHF Contributor
    Grandad's Avatar
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    Quote Originally Posted by yeongil View Post
    Grandad: I double-checked, and I'm still getting positive values for a.
    You're quite right - sorry!

    Grandad
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