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Linear Algebra Help

  1. Feb 13, 2007 #1
    1. The problem statement, all variables and given/known data

    the question states: Find the coefficients of the fourth degree polynomial:
    p(x) = ax^4 + bx^3 + cx^2 + dx + e whose graph goes through the points (0,0), (1,1), (-1,3) and whose slope at x=-1 is 20 and x=1 is 9.

    2. Relevant equations



    3. The attempt at a solution

    i started by putting it into an augmented matrix, and solving...
    my matrix was:

    0 0 0 0 1 | 0
    1 1 1 1 1 | 1
    1 -1 1 -1 1 | 3

    when i solved for the coefficients, i got a = 0, b = -1, c = 2, d = 0, and e = 0.
    i'm not sure how to incorporate the information given by the slopes.

    i know that you can take the derivative to find slope.. and i took the derivative of the polynomial and got:

    4ax^3 + 3bx^2 + 2cx + d

    but i'm not really sure where to go from there. do i plug in -1 for x and set it equal to 20 and 1 for x and set it equal to 9? if i do that i get the equations:

    -4a + 3b - 2c + d = 20 and
    4a + 3b + 2c + d = 9

    by now i think i'm starting to make things up though! :wink:
    can anyone offer any hints? thank you!
     
  2. jcsd
  3. Feb 13, 2007 #2

    radou

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    Homework Helper

    Yes, you do. I didn't check your calculations, but the next step is obviously to solve a system of equations.
     
  4. Feb 13, 2007 #3
    i set up a system of equations for those two equations and with only 2 equations and 4 variables i was unsure of how to proceed. i ended up with the matrix:

    -4 3 -2 1 | 20
    0 6 0 2 | 29

    i'm just confused as to how the two steps are connected... like the first time i solved for the coefficients and this time. can i combine my results somehow?
     
  5. Feb 13, 2007 #4

    radou

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    Homework Helper

    I don't follow. Did you set up a system of 5 equations with 5 unknowns? Did you try to solve it?
     
  6. Feb 13, 2007 #5
    ok, i think i got it now. i didnt think i could set up all 5 equations before because the first set of equations had 5 variables and the second set of equations had 4. i put a zero as a place holder though, and solved this matrix:

    0 0 0 0 1 | 0
    1 1 1 1 1 | 1
    1 -1 1 -1 1 | 3
    -4 3 -2 1 0 |20
    4 3 2 1 0 |9

    i think i'm good from here. thanks for your help!
     
  7. Feb 13, 2007 #6

    radou

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    Homework Helper

    The matrix looks okay.
     
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