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Graphing question: Reversal of x and y axis & effect on slope and y-intercept

  1. Sep 20, 2009 #1
    The problem statement, all variables and given/known data

    How does the slope and y-intercept change if you reverse the x and y axis of a linear graph. Will the graph still be linear?

    The original y=mx+b format followed the physics equation:
    V^2 = 2a(d) + Vi^2

    Therefore, when y was "V^2" while x was "d", the y-intercept was Vi^2 and the slope was equal to "2a".

    What are the y-int and slope once the axis are reversed, and x is "V^2" and y is "d"?

    The attempt at a solution

    I'm sure this is quite simple, but for some reason I am stumped. I realize the new graph would still be linear, reflected along y=x.
    I tried inserting the new x and y values into the equation, getting:
    x = my + b

    (x - b)/m = y

    (v^2 - vi^2)/2a = d

    so (v^2)(1/2a) - (vi^2)/2a = d

    leaves the new equation in the form of mx + b = y

    with m = 1/2a
    and b = (-vi^2)/2a

    Have I come to the correct solution?

    Thanks a lot!!
     
    Last edited: Sep 20, 2009
  2. jcsd
  3. Sep 21, 2009 #2
    No one can confirm my work?
     
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