# Homework Help: Sketching Regions in ℝ^2

1. Sep 26, 2012

### Chewybakas

1. The problem statement, all variables and given/known data

I have to sketch the following region in R^2 but i Have no idea how to do it!

2. Relevant equations

S1 = {(x; y) εℝ^2 : -2≤x≤4 y,≤2}

3. The attempt at a solution

I plotted the line from -2 to 4 on the x axis and a line greater then 2 from the y-axis but i am sure i am doing this incorrectly! any help is appreciated!

2. Sep 26, 2012

### LCKurtz

Start by plotting the equality parts: x = -2, x=4, y = 2, three straight lines. They divide the plane up into several regions. Then, examining each region, shade the ones for which x and y both satisfy the inequalities.

3. Sep 26, 2012

### Chewybakas

So if i plot those points and draw the lines i get a rectangular shape on my graph is that the correct solution?

4. Sep 26, 2012

### LCKurtz

Hard to say without seeing it. Describe where you have shaded.

5. Sep 26, 2012

### Chewybakas

I have a graph with an x and y axis, the points drawn in and lines from each point, where the line y=2 intersects the x points i shade in the rectangular shape. If im wrong could you please tell me what to do!

6. Sep 26, 2012

### Staff: Mentor

This is not a useful description. The region is unbounded in one direction. It is bounded on three sides by straight lines. A better description would tell us what the three boundary lines are and maybe what the points at the corners are.

7. Sep 26, 2012

### HallsofIvy

-2≤x≤4. Draw two vertical lines at x= -2 and at x= 4 so that the "line from -2 to 4 on the x axis" is inside them. I don't know what you mean by "a line greater than 2 from the y-axis". y≤2 is all y less than or equal to 2 so draw a horizontal line at y= 2 crossing those two vertical lines. The region you want to shade is the area below that horizontal line and between the two vertical lines. Because there is no lower limit on y, there is no lower boundary- the region you want is the "infinite strip" between the vertical lines and below the horizontal line.

8. Sep 26, 2012

### Chewybakas

Ahh thats where i was going wrong! I was using the x axis itself as a boundary! I have no idea why! Thank you for clearing that up! But in this question i have no figures.

{(x; y) εℝ^2 : x≤y}
Is that just the x and y axis! thank you for all your help!

9. Sep 26, 2012

### Staff: Mentor

No. Start by graphing the line x = y. This line divides the plane into two regions. The region you want is one of them, including the boundary line.

To determine which half plane is the right one, pick a point not on the line and see if it satisifies x ≤ y.

10. Sep 26, 2012

### Chewybakas

Ah y=x will give me a 45° line through the origin?

11. Sep 26, 2012

### LCKurtz

Doesn't that count as "working the problem for them"?

12. Sep 26, 2012

### Chewybakas

I was in desperate need of help, I was obviously getting it horribly wrong. He just helped me out greatly and i cant thank him enough for it!

13. Sep 26, 2012

### HallsofIvy

No, I agree with the others. I should have stopped at
"Draw two vertical lines at x= -2 and at x= 4 so that the "line from -2 to 4 on the x axis" is inside them. I don't know what you mean by "a line greater than 2 from the y-axis". y≤2 is all y less than or equal to 2 so draw a horizontal line at y= 2 crossing those two vertical lines."

And perhaps that is a little too much. Just noting that the lines are vertical and horizontal should have been enough.