## anonymous 4 years ago 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)

1. anonymous

plug in $$x(0,0), x(1,1), y(0,0), y(1,1)$$

2. anonymous

Yes, that's the function. That's it?

3. anonymous

No, that's the function. The image is the range of values you get from the "domain" S.

4. anonymous

Seems to me that $x(0,0) \leq x \leq x(1,1)$

5. anonymous

Alright, I have (0,0)->(0,0) (1,0)->(1,1) (0,1)->(0,0) (1,1)->(1,2) Is that it?

6. anonymous

Look at the highest and lowest of x and y.

7. anonymous

@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.

8. anonymous

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

9. anonymous

Then your image is the set of all the transformed points.

10. anonymous

So I had to correctly?

11. anonymous

it**

12. anonymous

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 $$S = \{(u, v) | u \in [0,1] \land v \in [0,1]\}$$

13. anonymous

That's what I did. Thanks!