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Sooper easy question

  1. Apr 8, 2009 #1
    sooper easy question :)

    uhmm, say I have half of a parabolic curve, how do I turn it into a straight line?

    what I'm thinking...

    y = x2
    root of y = x? but I leave the x as it is?

    hmm hard to explain... lemme try explaining a bit

    (x,y)
    (4,16)

    okay? ok now to make it straight, I would do root of 4 and leave x as 16 and then just plot this on graph?

    result

    (x,y)
    (2,16)

    or would I do the root of 16 as well
     
  2. jcsd
  3. Apr 9, 2009 #2
    Re: sooper easy question :)

    I have to ask, what are you trying to do?

    "uhmm, say I have half of a parabolic curve, how do I turn it into a straight line?"

    That doesn't really make sense.. Are you trying to find the gradient of the curve at a point?
     
  4. Apr 9, 2009 #3
    Re: sooper easy question :)

    Yeah i think he is;

    First off you must differentiate, i'm not sure if you know what this means so i'll put a simple version down anyway;

    y = x²

    So CHANGE in y / CHANGE in x (gradient) = 2x (bring the power down)

    Then insert the x figure

    so say at the point x = 4, the gradient would be 8. (If you don't understand differentiation, look it up, it is KEY!!!!! to understanding / doing any of these questions) http://en.wikipedia.org/wiki/Derivative

    Then you must use the formula

    y - y1 = g ( x - x1) Where you insert G gradient, y1 & x1, the points so

    y - 16 = 8 (x - 4)

    y = 8x + 12 Would be the equation for the line at the point x = 4

    I can understand why you might find all that a bit hard to understand, if you still can't pickup on some stuff here post again! You must understand these problems! Peace! :P
     
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