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Gradient of the graph y = a - k/x

  1. May 13, 2008 #1
    " find, in terms of a and k, the gradient of the graph y = a - k/x at the point where it crosses x axis."

    ok i worked out dy/dx = k/x^2 and x = k/a when y = o.

    now what do i do. =(

    thx for help in advance
  2. jcsd
  3. May 13, 2008 #2


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    I imagine you would want to calculate dy/dx at that point by plugging in x = k/a into your expression for dy/dx.
  4. May 13, 2008 #3
    How would you normally calculate the value of the derivative dy/dx at a point x?
  5. May 13, 2008 #4


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    Why are you all calculating derivatives? This is in the precalculus forum :confused:
  6. May 13, 2008 #5
    The OP used derivatives, and it looks like a calculus problem. Maybe it should be moved?
  7. May 13, 2008 #6
    didn't think it was worthy of the higher forum

    i don't know how to simplify it properly, that's the problem. =(
  8. May 13, 2008 #7
    can someone tell me how to simplify the answer, this isn't homework or coursework!

    just revising need help! thanks
  9. May 13, 2008 #8
    ah i think it should be:

    [tex]\frac{dy}{dx}[/tex] = -[tex]\frac{k}{x^{2}}[/tex] = -[tex]\frac{k}{(\frac{k}{a})^{2}}[/tex] = - [tex]\frac{k}{\frac{k^{2}}{a^{2}}}[/tex] = -[tex]\frac{ka^{2}}{k^{2}}[/tex] = -[tex]\frac{a^{2}}{k}[/tex]
    when y = 0
  10. May 13, 2008 #9
    thought so, cheers beef
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