anonymous
  • anonymous
Hi my question is about the Problem Set question 1A-2 B) \[y=\frac{ 2 }{(x-1)^2 }\] I was wondering how can you sketch it using translation and scaling. I realized that \[(x-1)^2\] is basically x^2 moved by 1 to the right. Then the function is inverted (using the inverse function(1/x)) but I don't know how this will be reflected in a graph. So I need help in order to figure out how applying 1/x to a function changes its graph. Thank you for your help in advance.
OCW Scholar - Single Variable Calculus
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katieb
  • katieb
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anonymous
  • anonymous
I also thought that maybe instead of starting with the function x^2 I could just start with 1/x^2 and say that 2/(x-1)^2 is the result of moving 1/x^2 by 1 to the right and stretching the y-axis by a factor of 2. What do you think? Is it possible to start with 1/x^2 or should I start with x^2. Thanks
anonymous
  • anonymous
The problem with that approach is that the translation that you described would not take into account that there is a restriction on x where y is undefined when x=1. Is this just a thought exercise or are you trying to graph the function?
anonymous
  • anonymous
The exercise is about graphing the function i mentioned using scaling and translation. To do this, they take an initial function like 1/x^2 and then apply the said processes to get to the graph of 2/(x-1)^2. My question was if I should take 1/x^2 as the initial function or rather x^2. Thank you.

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Superkossie
  • Superkossie
I think that you cannot start with x^2 since you need to have a value for which the denominator is not specified. So to me it seems more logical to start out with 1/x^2. I'm going to start this problem set now as well, so if I change my mind doing the problem I'll let you know ;)
anonymous
  • anonymous
I think we are expected to already know what \[1/(x^2)\] looks like, being undefined at x=0; and then realize that the new equation, \[2/(x-1)^2\] is undefined at x=1. So the translation is moving it one to the right along the x axis. The change of scale would be the 2, like you said. Which means for every x value of our known \[1/x^2\]curve, y is double. So it is lengthened along the y scale along the y scale. |dw:1392405119678:dw| So we just take both effects on a graph that we know what it would look like.

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