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?

1 Answer
Jan 23, 2017

Answer:

#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]}