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. 

A. p = 2(l + w)p/2 - l = ww = p/2 - l b. while isolating the variable w in the equation, i first divided both sides by 2 and then i subtracted l from both sides. i hope this .

The answer to the question is d