# Transformations of functions

1. Sep 20, 2011

### ninjamonke

I did the problem but I just need to make sure I did it correctly.. If I did it incorrectly, please let me know.

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

Page 1: http://i55.tinypic.com/25sosgp.jpg (Zoom in)

Page 2: http://i52.tinypic.com/ofyds5.jpg (Zoom in)

2. Relevant equations

Problem 4 a-d

3. The attempt at a solution

You can see it on the picture.

2. Sep 20, 2011

### PeterO

I don't think your parts (c) and (d) are appropriate.

For f(x+1) you gave the point (-3, 1) because indeed f(-3+1) = f(-2), which was given.

but for f(3x) you gave (-6, 1) however f(3*-6) = f(-18) which you don't know?
Similarly part (d) doesn't work out either.

3. Sep 20, 2011

### ninjamonke

Then part (b) should be wrong also. I did the same approach for part (c) and (d).

Can you explain how I got the the Point A coordinates (0, 3/2) on part (a) on the last column? It was already done in the book for Point A of part (a) only.

I also made the graph of these coordinates and it's moving/shifting the right way like it's supposed to, I think.

4. Sep 20, 2011

### PeterO

You might think you did the same thing, but you didn't

When you did f(x+1), you didn't add one to the x value, you subtracted one - because after you put it through (x+1) the 1 is added back on - ie if x = -3, then f(-3+1) = f(-2) which you know.

However, when you did f(3x) you still multiplied by 3 ??? your said x = -6. But if x = -6, f(3x) means f(-18) which you don't know. You should have divided by 3.

Part d you also didn't do the inverse to x either.

you were using y = 1/2 * f(x+1) so it is not surprising that a 3 became a 3/2