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Math Help - Mysterious Property When Translating Graphs

  1. #1
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    Wink Mysterious Property When Translating Graphs

    Hey

    I am wondering why adding to the x-value causes the graph to shift right instead of left. I know that algebriaclly:

    (x+d) ; started with addition
    x+d=0
    x=-d ; negative means shift left

    (x-d) ; started with subtraction
    x-d=0
    x=d ; positive means shift right

    Can anyone elaborate why this property exist!?

    Sam
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  2. #2
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    Re: Mysterious Property When Translating Graphs

    Quote Originally Posted by ArcherSam View Post
    Hey

    I am wondering why adding to the x-value causes the graph to shift right instead of left. I know that algebriaclly:

    (x+d) ; started with addition
    x+d=0
    x=-d ; negative means shift left

    (x-d) ; started with subtraction
    x-d=0
    x=d ; positive means shift right

    Can anyone elaborate why this property exist!?

    Sam
    If your basic function is y= f(x), you need to distinguish between "changes to x" (before applying the function f) and "changes to y" (after applying the function f). Obviously, any changes to y affect the graph vertically: y= f(x)+ a moves the graph up by a, y= f(x)- a moves the graph down. Multiplying, y= af(x) stretches (if a> 1) or squeezes (if a< 1) the graph vertically.

    Changes to x (before applying the function f) are horizontal changes. But when you talk about graphing y= f(x), as the notation indicates, you are talking about values of y, not x. Changes before applying f, that is, changes to x, are more subtle: Suppose y_0= f(0). Then f(x+a) will be equal to 0 when x+ a= 0, that is, when x= -a, not a. More generally, if y_1= f(x_1), then f(x+a)= y_1 when x+ a= x_1 or when x= x_1- a.
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  3. #3
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    Re: Mysterious Property When Translating Graphs

    @HallsofIvy

    This is what I have interpreted from your reply:

    WRONG
    y = f(x) = (x-3)^2
    y = f(0) = (0-3)^2 = 9 ; (0,9)
    y = f(1) = (1-3)^2 = 4 ; (1,4)
    y = f(2) = (2-3)^2 = 1 ; (2,1)
    y = f(3) = (3-3)^2 = 0 ; (3,0)

    CORRECT
    y = f(x) = (x-3)^2
    y = f(0) = (0-3)^2 = 9 ; (3,9)
    y = f(1) = (1-3)^2 = 4 ; (4,4)
    y = f(2) = (2-3)^2 = 1 ; (5,1)
    y = f(3) = (3-3)^2 = 0 ; (6,0)

    y = f(x) = (x-a)^b where (x-a) = x or (x-a) = f(x-a) ; and this is possible because of functional notation.

    What I am failing to understand is how the notation is allowed to be equal to the part of the equation that modifies the x-value.How can I take the input(before) and make it equal to the change(during) to still equal the input(after)?
    Last edited by ArcherSam; September 28th 2011 at 05:15 PM. Reason: Miscalculation
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