Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Problem with transformations
Replies: 5   Last Post: Nov 17, 2012 1:20 PM

 Messages: [ Previous | Next ]
 Clyde Greeno @ MALEI Posts: 220 Registered: 9/13/10
Re: Problem with transformations
Posted: Nov 16, 2012 11:32 PM

Apart from Larson's apparent fetish with useless equations ...

This is one of those times when the most effective approach is to NOT try to
*tell* anything, until the students have an under-standing of what you are
talking about. Let them graph a few cases from the (x-h)^2 family ... the
a(x-h)^2 family ... the (x-h)^2+k family .... and the a(x-h)^2+k family ...
and THEN talk about the whats, whys, and whethers (or not the curves reach
certain values).

- --------------------------------------------------
From: "Peter Duveen" <pduveen@yahoo.com>
Sent: Friday, November 16, 2012 6:38 PM
To: <math-teach@mathforum.org>
Subject: Problem with transformations

> The text (Precalculus with limits: a graphing approach Larson, etc.) tells
> us as follows (p43):
> "...you can obtain the graph of g(x) = (x - 2)^2 by shifting the graph of
> f(x) = x^2 two units to the right, as shown in Figure 1.42 [AN ASSERTION].
> In this case, the functions g and f have the following relationship.
>
> g(x) = (x - 2)^2
>
> = f(x - 2) (right shift of two units)[AN ASSERTION]
>
> The following list summarizes vertical and horizontal shifts:" etc. etc.
>
> I feel the assertions are not self-evident, and the treatment is generally
> confusing.
>
> I would have treated this differently. I would have first attempted to
> establish a relationship between a function and another function which is
> the translation of the first so many spaces horizontally.
>
> The relationship is f(x) = g (x + c). That is, the two functions have the
> same value when the arguments of f and g differ by a particular constant.
> Assuming we know the form of f(x), what is the form of g(x)?
>
> We introduce the argument f(x - c), and want to see what happens to g,
> namely, f(x - c) = g[(x - c) + c]
>
> We thus arrive at the expression f(x - c) = g(x). We have now established
> the form of g(x) in terms of f(x), which we know. It is simply f(x - c),
> which is not the same as f(x). In other words, we have derived and
> demonstrated what the textbook merely asserts.

Date Subject Author
11/16/12 Peter Duveen
11/16/12 Clyde Greeno @ MALEI
11/17/12 Peter Duveen
11/17/12 Robert Hansen
11/17/12 Joe Niederberger
11/17/12 Peter Duveen