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Find the image of the set S under the given transformation: u=0, u=1, v=0, v=1, x=v, y=u(1+v^2)

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plug in \(x(0,0), x(1,1), y(0,0), y(1,1)\)
Yes, that's the function. That's it?
No, that's the function. The image is the range of values you get from the "domain" S.

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Seems to me that \[ x(0,0) \leq x \leq x(1,1) \]
Alright, I have (0,0)->(0,0) (1,0)->(1,1) (0,1)->(0,0) (1,1)->(1,2) Is that it?
Look at the highest and lowest of x and y.
@Wislar are you implying that \(S = \{0, 1\} \times \{0, 1\}\)? From your question, it seems to just be \(S = \{ (0,0), (1,1)\}\). Please clarify.
The points on the first image are (0,0), (1,0), (1,1), (0,1) which is a square and plugging those into the equations for x=v and y=u(1+v^2) to get those points
Then your image is the set of all the transformed points.
So I had to correctly?
Well, I'm not going to check your arithmetic, but it should be the four points on the right side of your arrows in your above comment. Also make sure that \(S\) only relates to those four points, and is NOT defined by \(S = \{(u, v) | u \in [0,1] \land v \in [0,1]\}\)
That's what I did. Thanks!

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