## Parabolic Curve

Hi,

Just wondering if anyone could tell me how I would go about plotting the parabolic curve for the equation:

sigmaY*(1-(L/k)^2/2*(L/k)c^2)

Any help most appreciated

Will
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 Recognitions: Gold Member Science Advisor Staff Emeritus First of all, that's not an equation. Second, your notation is ambiguous. Do you mean $$f(c)= sigmaY*(1- \frac{(L/k)^2}{2}(L/k) c^2)= sigmaY*(1- (L/k)^3c^2/2)$$ or $$f(c)= sigmaY(1- \frac{(L/k)^2}{2(L/k)} c^2= sigmaY*(1- (L/k)c^2/2)$$ or $$f(x)= sigmaY(1- \frac{L/k)^2}{2(L/k)c^2}= sigmaY*(1- L/(2kc^2))$$ Since the last is not a quadratic and its graph is not a parabola, you must mean one of the first two. In either case, that is f(c)= A- Bc2 (what A and B are depends on which of the two you meant) which has vertex at (0, A) and opens downward.
 I will try and write the equation properly next time but I dont know how to use latex. Ok the equation I have is the third one? but with sigma instead of f(x). It is from a lab report and have been asked to draw the parabola for it. The original thread is here, http://www.physicsforums.com/showthread.php?t=280106 Thanks for your help.

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