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Limit question

  1. Oct 13, 2006 #1
    Find the coordinates of the point on the curve f(x)=3x^2-4x, where the tangent is parallel to the line y=8x.

    Ok i know the slope of the tangent is 8.

    and i know the formula is m=[f(a+h)-f(a)]/h

    i need a hint please
  2. jcsd
  3. Oct 13, 2006 #2
    What is the derivative of [tex] f(x) [/tex]? Use the power rule, set the derivative equal to 8, and solve for x. Plug x back into f(x), and you got the point.
    Last edited: Oct 13, 2006
  4. Oct 13, 2006 #3
    umm we havnt been taught derivative's yet. This is an introductory to it.
  5. Oct 13, 2006 #4
    your title is "limit question." have you been taught:

    [tex] f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} [/tex]? Use this definition for [tex] f(x) = 3x^{2} - 4x [/tex].
  6. Oct 13, 2006 #5
    yes i have been taught that

    but don't you need an x or y value of the tangent to use that formula.
  7. Oct 13, 2006 #6
    You know that the slope of the curve at the unknown point is 8 (because it is parallel to y = 8x). You don't need any values to use that formula. Just substitute (x+h) into f(x), and expand it out. After doing this subtract 3x^2 - 4x from f(x+h). Divide by h. Then substitute h = 0 to get f'(x).
  8. Oct 13, 2006 #7
    question solved
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