Date: 08/26/2002 at 23:09:40 From: Erik Tom Subject: Shifting graphs Hi! Here's a passage I don't understand. "If g(x)=f(x-c), where c>0 then the value of g at x is the same as the value of f at x-c (c units to the left of x). Therefore, the graph of y=f(x-c) is just the graph of y=f(x) shifted c units to the right." I don't understand why it shifts to the right. Thanks, Erik
Date: 08/27/2002 at 08:56:07 From: Doctor Peterson Subject: Re: Shifting graphs Hi, Erik. Suppose you have just plotted a point (a,b) on the graph of y=f(x), and now you want to plot the same point on the graph of y=g(x). You have already calculated f(a) and got b, so you would like to reuse that information and save work. Since g(x) = f(x-c), you know that g(a+c) = f(a+c-c) = f(a) = b. So the point (a+c, b) will be on the new graph. This point is your original point (a,b) shifted c units to the right. The same will be true for every point on the graph of y=f(x), so the whole curve will be shifted c units to the right to make the graph of g. You may find this answer from the Dr. Math archives useful as well: Graph with f(x) http://mathforum.org/library/drmath/view/54509.html If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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