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Translating Functions

Date: 08/27/98 at 18:29:57
From: Geoff Mosley
Subject: Functions

Find f(2+h), f(x+h), and f(x+h)-f(x)/h where h cannot = 0 for 
f(x) = x/(x + 1)?

Explain how the following graphs are obtained from the graph of 
y = f(x):

   y = f(x - 5)
   y = -f(x)
   y = f(5x)

Date: 08/28/98 at 12:31:12
From: Doctor Peterson
Subject: Re: Functions

Hi, Geoff. 

I get the impression you are at the beginning of a calculus course.

For the first group of problems, all you have to do is substitute
(2+h), or (x+h), for x in the definition of the function, and then 
simplify if you can. For example, f(2+h) is:

     (2 + h)
   (2 + h) + 1

and you can do just a little simplifying.

The second group contains nice things to know. Here's how to think 
about it. Imagine you have just graphed a point of f(x), say at x = k:

       |       /
       +      *
       |     /
       |      k

Now you want to graph a point of f(x - 5) using what you just found 
out. Well, you know f(k). If x = k + 5, f(x-5) = f(k) which you just 
figured out. So the corresponding point in the new function to plot is 
x = k + 5, where it will have the same value:

       |       /    /
       +      *    o
       |     /    /
       |      k   k+5

If you do this with every point in the graph, you will find that the 
graph of f(x - 5) is just the graph of f(x) slid to the right by 5. 
The other two cases are similar. Find what point of the new graph feeds 
the same value into f as the original, and what y is for the new 

You may have seen this before if you had an equation like:

   y = k*(x-a)^2 + b

and had to find the vertex of the parabola. Notice that this is just:

   y = f(x-a) + b


   f(x) = k*x^2

whose vertex is at (0,0). So the vertex of the original equation is at 
(a,b). Knowing how transforming an equation transforms its graph is 
very useful.

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

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