period of 30°: sin(12x)
amplitude of 3: 3sin(12x)
2 units down: 3sin(12x)-2
Now, things depend on whether the reflection is done before or after the shift.
Before: 3sin(12(-x-60°))-2
After: 3sin(-12(x-60°))-2
Find the equation of a sine function that has a vertical displacement 2 units down, a horizontal phase shift 60° to the right, a period of 30°, reflection in the y–axis and the amplitude of 3.
2 answers
I guess do the reflection later and the down 2 last by multiplying by -1
60 deg = pi/3 rad, 30 deg = pi/6 rad
but I will do your degrees thing
3 sin (k x - 60)
when x = 30 , kx is 360 so k = 12
3 sin (12 x - 60)
reflect over y axis ?? strange but x ---> - x
3 sin (-12 x - 60)
move down 2
y = -2 + 3 sin (-12 x -60)
but sin -z = - sin z
y = -2 - sin (12 x + 60)
This all depends on the order in which you do the commands, really just making ll kinds of assumptions.
60 deg = pi/3 rad, 30 deg = pi/6 rad
but I will do your degrees thing
3 sin (k x - 60)
when x = 30 , kx is 360 so k = 12
3 sin (12 x - 60)
reflect over y axis ?? strange but x ---> - x
3 sin (-12 x - 60)
move down 2
y = -2 + 3 sin (-12 x -60)
but sin -z = - sin z
y = -2 - sin (12 x + 60)
This all depends on the order in which you do the commands, really just making ll kinds of assumptions.
0 answers