How does shifting exactly work

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October 18th 2012, 05:43 AM
vaironxxrd

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?
October 18th 2012, 05:52 AM
Prove It

Re: How does shifting exactly work

Quote:

Originally Posted by vaironxxrd

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.
October 20th 2012, 08:16 PM
Jason76

Re: How does shifting exactly work

Quote:

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)
October 20th 2012, 08:24 PM
MarkFL

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:

$y-k=f(x)$
October 20th 2012, 08:46 PM
Jason76

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.