+130116
Using
A little tricky....but......we can imagine the sine graph shifted down one unit from the normal "parent" function and having an amplitude of 3
So...far...we have this
y = 3sin (Bx)- 1
Note that a "starting point" on the curve is (0, -1) and a period of 1.5 puts us at (-1, 4pi)
So..... a period must be 4pi / 1.5 = (8/3) pi
So.... to find the number of periods in 2pi , ("B"), we can solve this
(2pi / (8/3 pi) = B
(6/3 pi) / (8/3 pi) =
(6/3)(8/3) =
6/8 = 3/4 = B = there are 3/4 periods in 2pi
So...our function is
y = 3sin [ (3/4)x ] - 1
Here's the graph : https://www.desmos.com/calculator/olu4zazirp
[ Note....other solutions are also possible ....this just seemed to be the easiest ]
So....
