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Math Help - Finding an equation of a curve from y-icpt, max point.

  1. #1
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    Question Finding an equation of a curve from y-icpt, max point.

    I want to find an equation for a curve with certain properties (I realise there will be serveral million/infinite)

    The curve is essentially a cubic, negative gradient, (from top left to bottom right, at any rate) with a y-intercept at (0,1) and a single positive root.

    There is a minimum point at (-x,+y) and a maximum at (2,3).

    I want you to either tell me an equation of a curve which passes through those points or, even better explain to me how to do it.

    I came up with a straight line that goes through those points. My best idea is to find a dy/dx that gives x=2 as a turning point, and that gives y=3 when x=2 but its a bit trial-and-error (mainly error) and very unscientific, there must be an actual way of doing it. Or a computer program that does it.


    The full story (if desired) - In an A level maths paper, (Edexcel C3 Jan 2010 Q6) there is a diagram of the graph as described above and you have to perform various transformations on it. That's easy but as a side point (not required for the question) I was wondering if there is a way to work out a rough equation of the curve so that I could put it on my graphical calculator and perform the transformations on it. I was half thinking that I could check my answers this way to the actual question and I was half just interested really.
    Last edited by mr fantastic; June 18th 2011 at 02:10 PM. Reason: Cleaned up the post - deleted all the extraneous clutter.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    I think that there are many many possible equations. For me, I would start by using the maximum point and 'guess' the minimum point.

    The derivative of the cubic gives a quadratic, with one solution at (x-2).

    Let the other solution be (x+3). The quadratic then becomes:

    y' = a(x-2)(x+3)

    Which is:

    y' = ax^2 + ax - 6a

    I put the a since we don't know 'how fast the quadratic is' and also allows for a negative value.

    Integrating this should now give me the cubic.

    y = \frac{ax^3}{3} + \frac{ax^2}{2} - 6ax + c

    Now I take (0, 1)

    0 = \frac{a}{3} + \frac{a}{2} - 6a + c

    0 = -31a + 6c

    A suitable value of a is -1 since we know the cubic is descending.

    0 = 31 + 6c

    c = -\frac{31}{6}

    So.. we get:

    y = -\frac{x^3}{3} - \frac{x^2}{2} +6x - \frac{31}{6}

    And making it easier on the eyes...

    y = -\frac{1}{6}\left(2x^3 + 3x^2 +36x - 31\right)

    This is one of the equations. I could have taken another minimum and another value of a to get different equations.
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  3. #3
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    Talking

    Thanks, that makes sense.
    I was halfway to getting that method but I wasn't quite all there yet, I hadn't thought to just guess another solution of x which meant I was finding it difficult because everytime I came up with a derivative which gave a turning point at x=2 y =3 the integral of that didn't have a y intercept of 1 and vice versa.
    Cheers - now I'll be able to repeat the method usingmy own guess work and different numbers (hence producing different equations!) Yaaay!
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