How do you find the amplitude, period, phase shift, and vertical shift for #y = 5 - 3 sin (2theta - pi)#?

1 Answer
Aug 8, 2015

Normally you may have learned an equation similar to this one:

#y = Asin(ntheta + phi) + k#

(or replace with whatever variables you're using.)

Basically:
#A# is the amplitude, and #A = 3# (recall that the amplitude goes from the apparent axis [EX: if shifted up 2 units, then the axis is #y = 2#] to the highest point in the y range)
#phi# is the phase shift, and #phi = -pi#, meaning #pi# to the right
#k# is the vertical shift, and #k = +5#, meaning #5# upwards

The period can be found by realizing that the period of #sintheta# is #2pi#. Knowing that, deriving the period is fairly simple. Since doubling the value of #n# halves the period:

#n = (2pi)/T => T = (2pi)/n#

Testing this, from the above information, you can infer that the period is #pi#. So:

#pi = (2pi)/n => npi = 2pi#
#n = 2#

and #n# is indeed #2# in the equation.

Overall, this equation is #sintheta# flipped over the x-axis, compressed to a period of #pi#, stretched in the vertical directions to an amplitude of #3#, then shifted right #pi# units and up #5# units.

#y = 5-3sin(2theta - pi)#:
graph{5-3sin(2x - pi) [-3.2, 3.2, -0.5, 8.2]}