Dilations of the Graph of y = f(x)
Date: 08/22/2004 at 20:01:07 From: Heather Subject: Horizontal stretch/shrink vs. vertical stretch/shrink Why is it that when doing a horizontal shrink or stretch you multiply by the reciprocal but when doing a vertical stretch or shrink you multiply by just the number? For example, to stretch y = f(x) vertically by a factor of 2 we just use y = 2*f(x), but to stretch it horizontally by a factor of 2 we use y = f(x/2). Why isn't it y = f(2x)?
Date: 08/22/2004 at 22:21:20 From: Doctor Peterson Subject: Re: Horizontal stretch/shrink vs. vertical stretch/shrink Hi, Heather. There's a different way to express the dilations (stretching and shrinking) that makes the two look more similar. As in your example, say we have an equation y = f(x) If you stretch the graph vertically by a factor of k, the new y will be k times what it was for a given x: y = k * f(x) But you can also think of it as replacing y in the original equation with y/k: y/k = f(x) That means that y for the new equation has to be k times as great as in the original equation, in order to be on the graph; dividing it by k gives you the equivalent "y" in the original equation, which is why we replace y with y/k to get the new equation. Now let's stretch horizontally by a factor of k. That can be done by replacing x with x/k this time: y = f(x/k) When thought of in that way, the two approaches are the same--only the variable replaced changes depending on if you dilate vertically or horizontally. As an aside, a similar discussion holds for simple translations. The usual approach uses the notation: y = f(x - h) + k To move the graph 3 units right and 2 units up, we would have: y = f(x - 3) + 2 Again, students wonder why the upward shift of 2 is done with +2 while the shift to the right is done with -3. That doesn't seem consistent, either. But if we do the same thing and rewrite the equation as y - k = f(x - h) we can see that the two translations are in fact accomplished the same way. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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