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Cubic Function Help

  1. Jun 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A table is given:
    x | -2 | -1 | 0 | 1 | 2 | 3
    h(x) | 0 | 9 | y | 3 | 0 | 6

    Well actually the h(x) line was based on a graph that we had to look at, it should be right. I discovered y = 8 (the graph increments by 3 on y axis and 1 on x axis, so I couldn't tell what y was)
    so d = 8, b = -2
    This graph is a cubic function, so I'm using ax^3+bx^2+cx+d formula
    I got c=-8 and a=2

    The first question that I got was to fill the h(x) part of table (thats all the direction I had)
    Then the second part of the question was to figure h(3)-h(1).. ok yeah no problem easy enough..
    However here is the catch.. using the found equation from my data.. well first this is what it is:
    y=2x^3-2x^2-8x+8
    .. so anyways I graphed in calculator and when I look at the table I noticed for x=-1 y=12.. which from viewing the graph its obvious that is false, The highest point on y axis appears to be 10 at most from looking at graph... This has thrown me off and I'm not sure what to do now.. its clear I screwed up some where.

    2. Relevant equations
    my results for equation (wrong) y=2x^3-2x^2-8x+8
    cubic form: ax^3+bx^2+cx+d
    Here are the four equations I derived out of the data to get what I thought was my answer..
    -8a+4b-2c+d=0
    -a+b-c+0=9
    a+b+c+d=3
    8a+4b+2c+d=0


    3. The attempt at a solution
    I wish to evaluate h(3)-h(1) but I've noticed some errors with what I did in part 1 of the question.
     
  2. jcsd
  3. Jun 9, 2013 #2

    haruspex

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    Check that one again.
     
  4. Jun 10, 2013 #3

    Ray Vickson

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    There is a bit of inconsistency in your data. You get three slightly different cubic fits if you use x=(-2,-1,1,2) vs. x = (-2,-1,2,3) vs. x = (-2,1,2,3). These will give three slightly different values for h(0); I get h(0) = 8, 7.8 and 8.4, respectively.
     
    Last edited: Jun 10, 2013
  5. Jun 10, 2013 #4

    ehild

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    Your data do not fit exactly to a third order polynomial. If the data were read from a graph, try to read more accurately.

    ehild
     
  6. Jun 10, 2013 #5
    thanks, though I can't stand these problems that are based from looking at a graph -_-
    I ended up finding that 1x^3+2x^2-4x+8 (I think, just pullin from memory) fit exactly how it should with the exception x=3 y=5 looking a little sketchy from the graph, it looks like y=6 there to me... oh well this should suffice.
     
  7. Jun 10, 2013 #6

    haruspex

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    1x^3-2x^2-4x+8
     
  8. Jun 10, 2013 #7
    Yep thats what I had sorry.
     
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