# Sinusoidal Functions: describe transformations, sketch graph

1. Aug 13, 2016

### Evangeline101

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

2. Relevant equations
none

3. The attempt at a solution

-amplitude is 3
-period is 180°
-right 60°
-down 1

Rough sketch of graph:

I would like to know if the graph looks right, is there any improvements to be made?

Thanks :)

2. Aug 13, 2016

### blue_leaf77

You got it wrong. That's the graph of $y = -3\sin(2(x-60^o))-1$.

3. Aug 13, 2016

### Theoretical Muslim

wanna know why the line at 90?

4. Aug 13, 2016

### Theoretical Muslim

change the sign and it is alright

5. Aug 15, 2016

### Evangeline101

I re-did the graph:

Is this better?

6. Aug 15, 2016

### blue_leaf77

Still incorrect. Start from $f(x) = 3\sin(2(x-60^o))$ which can be obtained by translating $f(x) = 3\sin(2x)$ to [left/right, it's your part to answer] by $60^o$ degrees.

7. Aug 15, 2016

### Evangeline101

Ok I have attempted the graph again:

The dotted curve is the original base function y = sinx, and I used it to help apply the transformations. The dotted lines show where the axes have been shifted.

Is this an improvement??

8. Aug 15, 2016

### blue_leaf77

Nope.
Your first drawing is almost correct. It's just the negative sign in front of the sine function that needs to be removed. May be it's better to draw the functions resulting from each transformation. So, you start from $y=\sin(2x)$, which is a harmonic function of period $180^o$ and amplitude 1. Then draw $y=3\sin(2x)$. Followed by $y=3\sin(2(x-60^o))$ and finally $y=3\sin(2(x-60^o))-1$. This is the only way we can spot in which step you went wrong.

9. Aug 17, 2016

### Evangeline101

Okay, I have attempted the graph again, lets hope its an improvement at least :/

10. Aug 17, 2016

### blue_leaf77

Yes, that's the correct one.

11. Aug 17, 2016

### Evangeline101

Thanks for the help! I really appreciate it