Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
Education
»
mathteach
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




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 understanding of what you are talking about. Let them graph a few cases from the (xh)^2 family ... the a(xh)^2 family ... the (xh)^2+k family .... and the a(xh)^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: <mathteach@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 selfevident, 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.



