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 function. 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 where: 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 Check out our web site! http://mathforum.org/dr.math/
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