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Math Help - Function Q help

  1. #1
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    Function Q help

    f (x) = xcubed + axsquared + bx + c.
    If f(-3) = 0, f (4) = 0 and 4f (0) = 5f (1)
    find the values of the constants a , b and c.

    I've made 2 linear simultaneous equations from the f -3 = 0 and f 4 = 0
    but for the 4f 0 = 5f 1 I dont know what to do.
    Can anyone help me please?
    Thanks
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  2. #2
    Super Member bigwave's Avatar
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    then 4f(0) = 5f(1) is

    Quote Originally Posted by Detanon View Post
    f (x) = xcubed + axsquared + bx + c.
    If f(-3) = 0, f (4) = 0 and 4f (0) = 5f (1)
    find the values of the constants a , b and c.

    I've made 2 linear simultaneous equations from the f -3 = 0 and f 4 = 0
    but for the 4f 0 = 5f 1 I dont know what to do.
    Can anyone help me please?
    Thanks
     <br />
4f\left(0\right) = 4[(0)^3 + a(0)^2 + b(0) + c)] = 4c<br />

    5f\left(1\right) = 5[(1)^3 + a(1)^2 +b(1) +c] = 5a +5b+5c + 5

    then 4f(0) = 5f(1)
    is  4c=5a +5b+5c + 5
    or 5a + 5b + c + 5 = 0
    Last edited by bigwave; December 10th 2009 at 02:42 PM. Reason: corrected formula
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  3. #3
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    I cant get the question right, I keep getting it wrong..in my book it says the answer is a = -6, b =-7 , c =60
    Can someone post the solution please?
    Thanks.
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  4. #4
    Super Member bigwave's Avatar
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    Cool

    Quote Originally Posted by Detanon View Post
    I cant get the question right, I keep getting it wrong..in my book it says the answer is a = -6, b =-7 , c =60
    Can someone post the solution please?
    Thanks.
    if you plug these answers back into f(-3)=0, f(4)=0 and
    <br />
5a + 5b + c + 5 = 0

    they do work so what is being done with the similtaneous equations is where the trouble is
    its not an easy one to do

    your similtaneous equation shoul look like this

    9a-3b+c = 27
    16a+4b+c = -64
    5a+5b+c= -5

    I put this into the SE on the TI89 and it returned your answers

    still having ??
    Last edited by bigwave; December 12th 2009 at 03:48 PM. Reason: wording
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  5. #5
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    I keep getting b = 5???
    from equation 1 and 2 I eliminate c. I get equation 4 which is 7a -7b = 91
    from equation 1 and 3 i eliminate c. I get equation 5 which is 4a -8b = 32
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  6. #6
    Super Member bigwave's Avatar
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    eq#2 should of been 16a+4b+c = -64

    your 7a-7b=91 should -7a-7b = 91 instead!

    so then

    [-7a - 7b = 91] x4 = -28a-56b = 224
    [ 4a -8b = 32] x7 = 28a -28b = 364

    -84b = 588
    b = -7
    Last edited by bigwave; December 12th 2009 at 03:45 PM. Reason: more info
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