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Calculus Problem Solving Question

  1. Oct 6, 2005 #1
    Please help me solve this problem. :yuck:

    Find numbers a, b, and c so that the graph of f(x) = ax^2 + bx + c has x-intercepts at (0,0) and (8,0) and a tangent with slope 16 where x = 2.

    I have done this so far:
    f(x) = ax^2 + bx + c
    f '(x) = (2)(a)(x) + b
    16 = (2)(a)(2) + b
    16 = 4a + b

    I don't know where to go from here, so any help would be greatly appreciated.
  2. jcsd
  3. Oct 6, 2005 #2
    Nevermind, I solved it. If anyone else needs the solution to this problem, let me know. :)
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