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    Senior Member vaironxxrd's Avatar
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    How does shifting exactly work

    Hello everyone,

    I have a small question which is bothering me a lot.

    I was learning in my precalculus book about the shifting of of functions and it tells me about the function
    f(x)=/frac{1}{2}(x-1)^4
    i understand that it is compressed by a factor of 1/2 and it is shifted 1 unit to the right .
    but I still can't wrap around my head the actual concept.

    Why subtracting moves it to the right, rather than adding?
    and most of all I'm having trouble understanding how can the -1 affect it at all, how could it stay a constant effect rather than affecting the x value?
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  2. #2
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    Re: How does shifting exactly work

    Quote Originally Posted by vaironxxrd View Post
    Hello everyone,

    I have a small question which is bothering me a lot.

    I was learning in my precalculus book about the shifting of of functions and it tells me about the function
    f(x)=/frac{1}{2}(x-1)^4
    i understand that it is compressed by a factor of 1/2 and it is shifted 1 unit to the right .
    but I still can't wrap around my head the actual concept.

    Why subtracting moves it to the right, rather than adding?
    and most of all I'm having trouble understanding how can the -1 affect it at all, how could it stay a constant effect rather than affecting the x value?
    To get a full understanding of functions, you need to have a decent grasp of the numerical, graphical and algebraic representations of the functions. I would advise you to draw a table of values for \displaystyle \begin{align*} \frac{1}{2}x^4 \end{align*}, and then a table of values for \displaystyle \begin{align*} \frac{1}{2}(x - 1)^4  \end{align*}. You could think of your x values as the time taken to reach your destinations, the y values. You should be able to see that in the second function, you get the exact same y values as in the first, but one "step" later (makes sense, because your x values end up being one less). This means that you get to your destinations of the y values "late" (so to speak) and. Graphically, this corresponds to a movement of the original function to the right.

    Similarly, if you had done \displaystyle \begin{align*} \frac{1}{2}(x + 1)^4 \end{align*}, you get the same y values, but one "step" earlier, and so you arrive at your destination of the y values "early" (so to speak), and thus graphically this is equivalent to a movement of the original function to the left.
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    Re: How does shifting exactly work

    Why subtracting moves it to the right, rather than adding?
    Let's look at the standard vertical parabola equation:

     (x - h)^2 = 4p(y - k)^2

    h: makes it go left or right.

    k: makes it go up or down.

    (h,k): Coordinates of the vertex.

    (x,y): Coordinates of points on the parabola.

    p: Distance of the vertex to the focus.

    The following equation

    (y - 2)^2 = 4p(x + 5)^2

    would be shifted 2 units to the right and 5 units down, because actually the equation reads:

    (y - (+2))^2 = 4p(x - (-5))^2

    Remember a positive and a negative yields negative and two negative yield a positive (for what you see above the surface). But actually, below the surface, your plugging in positive 2 (which makes it go right) and negative 5 (which makes it go down).

    (y - (+2))^2 = 4p(x - (-5))^2 (below the surface)

    (y - 2)^2 = 4p(x + 5)^2 (above the surface - what you actually see)
    Last edited by Jason76; October 20th 2012 at 09:16 PM.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: How does shifting exactly work

    Where are the roots of the two functions?

    As an analogy, consider:

    y=f(x)

    If we add a positive constant k to the function, we will vertically shift it up by k units:

    y=f(x)+k

    Observe that we may write this as:

    y-k=f(x)
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  5. #5
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    Re: How does shifting exactly work

    I think it would depend on the type of function. Parabolas and circles have negative signs in their standard equations. Therefore, + shifts graphs to the left or down and - shifts them to the right or up.
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