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Math Help - Cubic equation from 4 coordinates

  1. #1
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    Exclamation Cubic equation from 4 coordinates

    There's this question in my textbook that I can't seem to answer:

    A cubic function of the form y=ax^3+bx^2+cx+d passes through the points (0,1) (1,3) (-1,-1), (2,11). Find the values of a,b,c, and d.

    These are seemingly random points. I don't understand how you could find an equation using them. Maybe I'm just being silly
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  2. #2
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    Quote Originally Posted by freswood
    There's this question in my textbook that I can't seem to answer:

    A cubic function of the form y=ax^3+bx^2+cx+d passes through the points (0,1) (1,3) (-1,-1), (2,11). Find the values of a,b,c, and d.

    These are seemingly random points. I don't understand how you could find an equation using them. Maybe I'm just being silly
    Plug the values into the equation and you will have four linear equations
    in a, b, c and d. These can then be solved by the usual methods ( lets
    say Gaussian elimination for example.)

    For instance:

    The first point (0,1) corresponds to x=0, y=1, so:

    1=d;

    the second point (1,3) corresponds to x=1, y=3, so:

    3=a+b+c+d;

    the third point (-1,-1) corresponds to x=-1, y=-1, so:

    -1=-a+b-c+d;

    the fourth point (2,11) corresponds to x=2, y=11, so:

    11=8a+4b+2c+d.


    RonL
    Last edited by CaptainBlack; January 24th 2006 at 03:03 AM.
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  3. #3
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    You're so right! Thankyou so much! Only problem is - I'm stuck again I've got:

    2 = a+b+c
    0 = -a+b-c
    5= 4a+2b+c

    It's no good using elimination with the top two then using the bottom, because that just re-introduces another unknown. Arrrr I'm so sorry for being so stupid.
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  4. #4
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    Quote Originally Posted by freswood
    You're so right! Thankyou so much! Only problem is - I'm stuck again I've got:

    2 = a+b+c
    0 = -a+b-c
    5= 4a+2b+c

    It's no good using elimination with the top two then using the bottom, because that just re-introduces another unknown. Arrrr I'm so sorry for being so stupid.
    these should be

    2=a+b+c
    -2=-a+b-c
    5=4a+2b+c

    Adding the first two gives:

    0=2b,

    so b=0, then we have:

    -2=-a-c
    5=4a+c

    which looks like it has solution a=1, c=1.

    RonL
    Last edited by CaptainBlack; January 24th 2006 at 03:10 AM.
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  5. #5
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    Thanks so much for your help! Could you just clarify what you meant by "adding the first two" because I can't seem to do it by elimination (ie can I see your working out? ) I just made them equal each other (by first making the second one a +2).
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  6. #6
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    Quote Originally Posted by freswood
    Thanks so much for your help! Could you just clarify what you meant by "adding the first two" because I can't seem to do it by elimination (ie can I see your working out? ) I just made them equal each other (by first making the second one a +2).
    If:

    2 = a+b+c
    0 = -a+b-c

    adding these (meaning adding the RHSs and LHSs) gives:

    2+0=(a+b+c)+(-a+b-c)

    or:

    2=2b,

    so b=1.

    RonL
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  7. #7
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    Just to add, this is a type of problem of drawing a 3rd degreed polynomial of best fist through these points. It happens to be that it is possible to draw a perfect polynomial through these points.
    Its equation is y=x^3+x+1
    Thus, (a,b,c,d)=(1,0,1,1)
    Here is the graph:
    Attached Thumbnails Attached Thumbnails Cubic equation from 4 coordinates-picture1.gif  
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  8. #8
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    Thankyou so much! I really appreciate your help
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