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Slope of a tangent

  1. Jul 5, 2006 #1
    Sry dont know how to use latex, so my calculations may be very messy

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

    I know the formula m=limh->0 f(x+h)-f(x) / h

    What i tried was:

    Let coordinates be (x,y)

    f(x) = y
    f(x+h) = 3(x+h)^2 - 4(x+h)
    = 3x^2+6xh+3h^2-4x-4h

    m=limh->0 f(x+h)-f(x) / h
    m=limh->0 3x^2+6xh+3h^2-4x-4h-(3x^2-4x) / h
    m=limh->0 3^2+6xh+3h^2-4x-4h-3x^2+4x / h
    m=limh->0 6xh+3h^2-4h / h

    Since tangent line has to be parallel to y=8x
    8 = 6xh+3h^2-4h / h
    8h = 6xh+3h^2-4h
    12h = 6xh+3h^2

    Divide by h on both sides:

    12 = 6x + 3h

    Sub 0 for h



    Therefore the coordinates on the curve that make the tangent line parallel to y=8x is (2,4)

    Please check over my work, I think I did something wrong.
  2. jcsd
  3. Jul 5, 2006 #2
    Your answer is correct, although it would have been prettier if you had evaluated the limit first. :smile:
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