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Effect of 'a' in y = e^(ax)

  1. Apr 17, 2013 #1
    Hi,
    In the equation " y = e^(ax), what is the effect of constant 'a'. Like, what happens to the shape of the curve when it becomes -, +, high or low? Thanks.
     
  2. jcsd
  3. Apr 17, 2013 #2

    tiny-tim

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    hi k.udhay! :wink:

    (try using the X2 button just above the Reply box :wink:)

    tell us what you think, and then we'll comment! :smile:
     
  4. Apr 17, 2013 #3
    Hi tiny-tim,

    I don't see any X^2 button... Thanks.
     
  5. Apr 17, 2013 #4

    tiny-tim

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    you'll see it if you click the "Quote" button or the "Go Advanced" button :smile:
     
  6. Apr 17, 2013 #5
    Hi tiny-tim,

    About X2, understood the point. :)
    Well, what I can derive easily is when x turns 0, the curve is a straight vertical line. For '+' x, it lies on right hand side, for '-' side the curve travels in left hand side.

    Ah... Now, higher the 'x' value, more it will become flat towards right... Am I right???
     
  7. Apr 17, 2013 #6

    tiny-tim

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    uhh? :confused:

    which way up are you? :biggrin:
     
  8. Apr 17, 2013 #7
    I am really sorry... It should be a horizontal line having an 'Y' interception 1... Correct now?
     
  9. Apr 17, 2013 #8

    rock.freak667

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    Well let's take x=1.

    for a=1, y=e^1
    a=2, y=e^2
    a=3, y=e^3

    So what is happening to the value of y as 'a' increases?
     
  10. Apr 17, 2013 #9
    Y increases much faster than x.... :)
     
  11. Apr 18, 2013 #10

    rock.freak667

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    So can you now see the effect on the graph if 'a' is increased or decreased?
     
  12. Apr 18, 2013 #11

    Mark44

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    Assuming you have the graph of y = f(x), here are a few variants.
    • The graph of y = f(x) + a is a vertical translation (or shift) of the graph of y = f(x). If a > 0, the shift is upward. If a < 0, the shift is downward.
    • The graph of y = f(x - a) is a horizontal translation (or shift) of the graph of y = f(x). If a > 0, the shift is to the right. If a < 0, the shift is to the left. For example, the graph of y = (x - 2)2 looks like the graph of y = x2, but shifted horizontally to the right. Instead of the vertex being at (0, 0), the vertex in the shifted graph is at (2, 0).
    • The graph of y = -f(x) is a reflection across the horizontal axis of the graph of y = f(x).
    • The graph of y = f(-x) is a reflection across the vertical axis of the graph of y = f(x). For example, the graph of y = sin(-x) looks like the graph of y = sin(x), but reflected across the y axis.
    • The graph of y = af(x) represents an expansion away from the horizontal axis if a > 1, and a compression toward the horizontal axis if 0 < a < 1. If a < 0, there is also a reflection across the x-axis.
    • The graph of y = f(ax) represents a compression toward the vertical axis if a > 1, and a compression away from the vertical axis if 0 < a < 1. If a < 0, there is also a reflection across the y-axis.
     
  13. Apr 18, 2013 #12
    Thanks rock.freak. Yeah, I think I can figure out the other cases. Now, let me ask my second question:
    What is the need for finding this term 'e' which has a strange value of 2.718... When I read in wiki., it says a curve following ex will have its tangent at Y = 1 at an angle of 45°. But why does one need this combination???:confused:
     
  14. Apr 18, 2013 #13
    Thank you, mark. Your exlanation will help me extimate the behaviour of curves very easily!:approve:
     
  15. Apr 19, 2013 #14

    rock.freak667

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    You can consider 'e' to just be another special constant without having to go into the details of how to define it. Much like π.

    For the y=1, θ=45° thing, you don't really have to memorize something like that but you can derive that as follows:

    y=ex → dy/dx = ex i.e. dy/dx =y

    So that when y=1, dy/dx =1 i.e. the gradient of the tangent at y=1 is 1.

    The angle a straight line makes with the x-axis of gradient 'm' is given by tanθ=m or θ=tan-1(m) so in this case, m=1 such that θ=45°
     
  16. Apr 21, 2013 #15
    Great explanation! Thank you!
     
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