The Attempt at a Solution
-amplitude is 3
-period is 180°
Rough sketch of graph:
I would like to know if the graph looks right, is there any improvements to be made?
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.