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Help with secant question , thanks

  1. Jun 27, 2007 #1
    here is my question, *thanks for helping* i need it badly

    Consider the curve y = f(x) where f(x) = (x-5)(2X+3)

    A) Show that P1 = (x1,y1) = (0,-15) is on the curve and find and expression for the slope of the secant joining P1 with any other point P2 = (x2,f(x2)) , x2 cannot be 0 on the curve

    b) Show that there is always a point x*, lying between x1 = 0 and x2 such that the slope of the tangent line to the curve at x* is equal to the slope of the secant joining P1 and P2



    ok and i am stuck on a) how would i go about solving it, would i take the derivate of the curve then sub the points from P1 to find P2?
     
  2. jcsd
  3. Jun 27, 2007 #2

    Dick

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    The secant line is just the line joining (x1,f(x1)) and (x2,f(x2)). There is no derivative involved - just a plain old slope.
     
  4. Jun 27, 2007 #3
    For A) what is the value of f(x) when you plug in x=0? Edit: OK, a little too late posting
     
  5. Jun 27, 2007 #4
    okay so plan old slope i would find it by doing:

    m = x2 -x1 / f(x2) - f(x1)

    or

    slope of secant = f(0+h) - f(0) / h

    m = [(o+h)-5][2(0+h)+3] - (-5)(3) / h
    m = (h-5)(2h+3) + 15 / h
    m = 2h^2 + 3h + 10h -15 +15 / h
    m= 2h-7
    ?
     
    Last edited: Jun 27, 2007
  6. Jun 27, 2007 #5

    Dick

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    The first, except it's upside down. It's deltay/deltax.
     
  7. Jun 27, 2007 #6
    do i have enough info to complete the first, i dont really know how to go about doing it...
     
  8. Jun 27, 2007 #7

    Dick

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    I don't know what you don't know. Are you worried that you don't know x2?
     
  9. Jun 27, 2007 #8
    yeah i donty know x2
     
  10. Jun 27, 2007 #9

    Dick

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    Just do it anyway. What's f(x2)? Use the definition of f.
     
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