Results 1 to 3 of 3

Thread: Finding an equation of a curve from y-icpt, max point.

  1. #1
    Jun 2011

    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; Jun 18th 2011 at 02:10 PM. Reason: Cleaned up the post - deleted all the extraneous clutter.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Unknown008's Avatar
    May 2010
    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.

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

    Now I take (0, 1)

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

    $\displaystyle 0 = -31a + 6c$

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

    $\displaystyle 0 = 31 + 6c$

    $\displaystyle c = -\frac{31}{6}$

    So.. we get:

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

    And making it easier on the eyes...

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

    This is one of the equations. I could have taken another minimum and another value of a to get different equations.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Jun 2011


    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!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Aug 23rd 2011, 01:58 AM
  2. Replies: 5
    Last Post: Sep 25th 2010, 09:10 AM
  3. Replies: 3
    Last Post: Jul 12th 2010, 08:37 PM
  4. Replies: 3
    Last Post: Oct 1st 2009, 04:35 AM
  5. Replies: 4
    Last Post: Dec 23rd 2008, 05:59 AM

Search Tags

/mathhelpforum @mathhelpforum