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Shifting Graphs

Date: 08/26/2002 at 23:09:40
From: Erik Tom
Subject: Shifting graphs


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. 


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) 

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum 
Associated Topics:
High School Equations, Graphs, Translations
High School Functions

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