Question

How do you write an equation of a sine function with amplitude 4, period pi, phase shift pi/2 to the right, and vertical displacement 6 units down?

Answer 1

`y=4sin[2(x-pi/2)]-6`.

Explanation:

The standard form of a sine function is

`y=asin[b(x-h)]+k`

where

• `a` is the amplitude,
• `(2pi)/b` is the period,
• `h` is the phase shift, and
• `k` is the vertical displacement.

We start with classic `y=sinx`:
graph{(y-sin(x))(x^2+y^2-0.075)=0 [-15, 15, -11, 5]}
(The circle at (0,0) is for a point of reference.)

The amplitude of this function is `a=1`. To make the amplitude 4, we need `a` to be 4 times as large, so we set `a=4`.

Our function is now `y=4sinx`, and looks like:
graph{(y-4sin(x))(x^2+y^2-0.075)=0 [-15, 15, -11, 5]}
The period of this function—the distance between repetitions—right now is `2pi`, with `b=1`. To make the period `pi`, we need to make the repetitions twice as frequent, so we need `b=["normal period"]/["desired period"] = (2pi)/pi = 2`.

Our function is now `y=4sin(2x)`, and looks like:
graph{(y-4sin(2x))(x^2+y^2-0.075)=0 [-15, 15, -11, 5]}

This function currently has no phase shift, since `h=0`. To induce a phace shift, we need to offset `x` by the desired amount, which in this case is `pi/2` to the right. A phase shift right means a positive `h`, so we set `h=pi/2`.

Our function is now `y=4sin[2(x-pi/2)]`, and looks like:
graph{(y-4sin(2(x-pi/2)))((x-pi/2)^2+y^2-0.075)=0 [-15, 15, -11, 5]}
Finally, the function currently has no vertical displacement, since `k=0`. To displace the graph 6 units down, we set `k=-6`.

Our function is now `y=4sin[2(x-pi/2)]-6`, and looks like:
graph{(y-4sin(2(x-pi/2))+6)((x-pi/2)^2+(y+6)^2-0.075)=0 [-15, 15, -11, 5]}