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Reflect e^x at y=2

  1. Apr 29, 2014 #1
    Reflect e^x at y=2


    If you reflect e^x at y=0. You just turn e^x negative and it reflects



    On my homework it says the correct answer is 4 - e^x
    But it makes more sense for me to say its 2 - e^x.
    Can someone explain?
     
  2. jcsd
  3. Apr 29, 2014 #2

    SammyS

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    Graph those functions.
     
  4. Apr 29, 2014 #3
    I did, 4 - e^x is reflected at y = 4
     
  5. Apr 29, 2014 #4

    SammyS

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    Does this mean you now understand what's going on ?
     
    Last edited: Apr 30, 2014
  6. Apr 30, 2014 #5
    No :(. So if I want to reflect e^x at y = 5 i still do the same thing and add 5 and give it a reflection -e^x.
    I don't see it
     
  7. Apr 30, 2014 #6

    SammyS

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    Let's go back to the original problem.
    Reflect ex at y=2 .​

    First of all, when I asked about graphing the function, I was referring to all three functions.
    y = ex

    y = 2 - ex

    y = 4 - ex

    The following procedure may help you understand the correct answer.

    First shift y = ex down 2 units.

    Then take the negative (take the opposite) of that.

    Then shift the result up by 2 units.


    (This thread likely should be moved to pre-Calculus.)
     
  8. Apr 30, 2014 #7

    Mark44

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    It might help your understanding to think more in line with what's actually happening. You're reflecting the curve y = e across the horizontal line y = 5. When you say you want to reflect e^x at y = 5, I'm not sure you're understanding how this reflection is working.
     
  9. Apr 30, 2014 #8

    Mentallic

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  10. May 1, 2014 #9

    ehild

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    @brycenrg: You need to understand the concept of reflection. See http://www.mathsisfun.com/geometry/reflection.html

    Looking into a mirror, you see your image behind the mirror at the same distance as the distance between you and the mirror.

    A point P and its mirror image P' are at equal distances from the mirror line, at opposite sides. See attachment. If g(x) (green curve) is the mirror image of the function f(x) (blue curve) with respect to the line y=2, the point P (x, f(x)) and P' (x, g(x)) are at equal distances from y=2, one below, the other above the line: 2-f(x)= g(x)-2.

    ehild
     

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