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Geometrical proof question

  1. Dec 30, 2007 #1
    Prove that the graph of y= ax^3 + bx^2 + cx + d has two distinct turning points if
    b^2> 3ac. Find values of a,b,c and d for which the graph of this form has turning points at (0.5, 1) and (1.5, -1)
     
  2. jcsd
  3. Dec 30, 2007 #2
    hi iam a new member to this site

    well here isolved first part of it
    differentiate the whole equation with respeact to x

    you will get

    3ax^2+2bx+c

    find the maxima and minima

    i.e for two turning points two distinct max,min must exist

    Discriminant>0

    (2b)^2-12ac>0
    4*b^2 > 12 ac
    b^2 >3 ac...prooved
     
    Last edited: Dec 30, 2007
  4. Dec 30, 2007 #3
    second part

    3ax^2+2bx+c=0 was the equation i already mentioned
    one answer was 1/2,second 3/2
    adding them 2
    multiplying them we get 3/4

    2=-2b/3a
    -b=3a

    3/4=c/3a
    4c=9a

    substituting value of function in the function equation we get

    a+2b+4c+8d=8
    27a+18b+12c+8d=-8

    solve these equations simultaniously
    for a,b,c,d

    we get

    d=-1
    a=4
    b=-12
    c=9

    if iam correct
     
    Last edited: Dec 30, 2007
  5. Dec 30, 2007 #4

    mathwonk

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    Last edited: Dec 30, 2007
  6. Dec 30, 2007 #5
    I dont understand your last statement ..
    please pardon........:rofl:
     
  7. Dec 30, 2007 #6
    Solved

    Ah, I now understand it. I never thought of delta before
    Thanks alot
     
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