Given a sinusoid in the form `y=AsinB(x-h)+k` :

(1)Start with the basic function (sin or cos)

(2) "A" affects the amplitude; instead of` ` an amplitude of 1, the transformed function has an amplitude of A. (This is measured off of the midline). "A" is a vertical dilation. If A<0, then the function is reflected over a horizontal axis.

(3) "B" affects the period. For sin and cos, the original period is `2pi` . The period of the transformed function will be `(2pi)/B` . "B" performs a horizontal dilation. If B<0 the function is reflected over a vertical axis. This is a phase shift.

(4) "h" tranlates the function left/right.

(5) "k" translates the function up/down.

So:

P1) `y=2cos(x-(pi)/2)` .

-- Since A=2 the amplitude is 2.

-- Since `h=pi/2` the function is shifted right `pi/2` units.

-- the period is `2pi` (unchanged)

The highest point on the graph will be y=2, the lowest y=-2. Some points:

`(pi/2,2),((2pi)/3,sqrt(3)),((3pi)/4,sqrt(2)),((5pi)/6,1),(pi,0),((7pi)/6,-1)` etc... From the original values for cos, add `pi/2` to the x-values, then double the corresponding y-values. (Original point on cosx is (0,1). Add `pi/2` to 0 and double 1 to get the transformed point `(pi/2,2)` )

The graph:

P2) `y=2sin(1/2x-pi/6)`

This can be written as `y=2sin (1/2) (x-pi/3)` . In this form we see that A=2,B=1/2 and `h=pi/3` .

Thus:

-- the amplitude is 2

-- the period is `(2pi)/(1/2)=4pi`

-- the phase shift is `pi/3` to the right.

Some points:

`(0,-1),(pi/3,0),((2pi)/3,1),(pi,sqrt(3)),((4pi)/3,2),((5pi)/3,sqrt(3)),(2pi,1)`

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