Mathplease help!!
posted by Kate on .
Normally, the first 3 key points of a sine curve are (0,0), (90,1), and (180,0). If the function is changed to y=2sin(3x120)+5, the key points given above change to...
Please explain this to me so that I can do it myself.

Interpretation #1: you want to keep those same x values
sub in the x values of the original points:
(0,0) > y = 2sin(120) + 5 = 2(√3/2) + 5 = √3 + 5
(0,0) > (0, √3+5)
for (90,1)
y = 2sin(150) + 5 = 2(1/2) + 5 = 6
(90,1) ( 90,6)
you do the last one.
Interpretation #2: You are looking for the 3 points that describe the start value, the max value, then back to the start value, (half a period)
change the equation to y = 2 sin 3(x  40) + 5
that tells me that the curve y = 2 sin 3x has been moved horizontally 40° to the right and up 5
so the period is 360°/3 = 120°
So I would make my "key point" values of x equal to
40, 70, 100
(notice that a complete period would be shown using the x values of 40, 70, 100, 130)
let x=40, y = 2 sin 3(4040) + 5 = 0+5 = 5
(0,0)  (40,5)
let x = 70 , y = 2 sin 3(7040) + 5 = 2(1) + 5 = 7
(90,1) > (70,7)
you try the last one