Write an equation with amplitude 2, period (t) pi/2, phase shift (ø) - pi/4, and a midline y=-3. im mostly struggling with the period.

The amplitude is 2, so A = 2 (see note 1 below)

The period is T = pi/2 which means
B = 2pi/T
B = 2pi/(pi/2)
B = 2pi*(2/pi)
B = 4

The phase shift is C = pi/4
The midline is y = -3 indicating that D = -3

Put this all together and we go from this (see note 2)
y = A*sin(B(x-C)) + D
to this
y = 2*sin(4(x-pi/4)) + (-3)
and that simplifies to
y = 2sin(4x-pi) - 3
which is one way to write the final simplified answer (see note 3)



Extra Stuff:
Note 1: The value of A can be A = -2. Recall that the amplitude is equal to |A|. So If A = -2, then |A| = |-2| = 2. The amplitude is the vertical distance from the midline to the peak or valley. Distance can't be negative. To make things simple, I went with A = 2. 

Note 2: The general formula involving sine can be replaced with cosine instead. The cosine function is basically sine but a phase shift of it (it has the same shape and pattern, but has been moved over). You cannot use tangent as that is a completely different class of function.

Note 3: I distributed the 4 through to the (x-pi/4) to get 4x-pi. Your teacher or book may want you to keep things factored and not distribute. If so, then the answer would be y = 2sin(4(x-pi/4)) - 3. That is of course you went with using sine instead of cosine. The benefit of not distributing is that we can more easily see what the phase shift is. 

Ithink is a. true, but not sure