Question

Describe the transformations that are being applied to the function y = 1−3cos(2x−π).

Answer 1

y = 1−3cos(2x−π)
y = 1−3cos(2(x−π/2))

Assuming your reference function is y = cosx
- you compressed it by a factor of 2, (the period of cosx is 2π, the period of your function is π)
y = cosx ----> y = cos 2x

- you stretched your function vertically by a factor of 3
y = cos 2x ----> y = 3cos 2x
- you reflected it in the x-axis
y = 3cos 2x ----> y = -3cos 2x

- you translated your function up 1 unit
y = -3cos 2x ----> y = 1 - 3cos 2x

- you shifted your function π/2 units to the right
y = 1 - 3cos 2x ---> y = 1 - 3cos 2(x - π/2) OR y = 1 - 3cos (2x - π)