# Sinusoidal Functions: describe transformations, sketch graph

## Homework Statement none

## 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 :)

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blue_leaf77
Homework Helper
You got it wrong. That's the graph of ##y = -3\sin(2(x-60^o))-1##.

You got it wrong. That's the graph of y=−3sin(2(x−60o))−1y=−3sin⁡(2(x−60o))−1y = -3\sin(2(x-60^o))-1.
I re-did the graph: Is this better?

blue_leaf77
Homework Helper
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.

Still incorrect. Start from f(x)=3sin(2(x−60o))f(x)=3sin⁡(2(x−60o))f(x) = 3\sin(2(x-60^o)) which can be obtained by translating f(x)=3sin(2x)f(x)=3sin⁡(2x)f(x) = 3\sin(2x) to [left/right, it's your part to answer] by 60o60o60^o degrees.

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??

blue_leaf77
Homework Helper
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.

Your first drawing is almost correct. It's just the negative sign in front of the sine function that needs to be removed.
Okay, I have attempted the graph again, lets hope its an improvement at least :/ • blue_leaf77
blue_leaf77
Thanks for the help! I really appreciate it 