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Slopes of curves w/o equation?

  1. Sep 18, 2006 #1
    I have three points on a Y= -ax^2 + bx + c graph (negative parabola). I don't know how to find the equation with this information. Tangent lines? Help.
  2. jcsd
  3. Sep 18, 2006 #2
    Hi donlin. Welcome to PF!
    Let's say one of the points is (x0, y0). What's -ax02+bx0+c equal to?
  4. Sep 18, 2006 #3


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    Homework Helper

    Your equation is a function of one variable, Y(x)=-ax^2+bx+c. If a point (x0, y0) lies on the parabola, it must satisfy the equation f(x0) = y0. Since you have three points, after 'plugging' every one of them into the equation, you'll have three equations with three unknowns a, b and c, which are the coefficients you need.
  5. Sep 18, 2006 #4
    I think I have the slope, by connecting point A to B and then rise over run, but I don't know how to get the rest of the equation. Basically, I have a point A and slope, but I don't have the coefficients or y-intercept.
  6. Sep 18, 2006 #5
    I think it's equal to y0.
  7. Sep 18, 2006 #6
    You can't get the slope at one point of a parabola by connecting two points on it. It's different at every point. Use the method that radou posted.
  8. Sep 18, 2006 #7


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    A curve does not have a slope! What you have found is the slope of the line through A and B which is irrelevant. If your question is how to find the equation of a parabola passing through three points, just put the x and y coordinates of each point into your general equation, Y= -ax^2 + bx + c, gives you three linear equations for A, B, and C.
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