# What is the answer for question 1) ? http://postimg.org/image/ritwlldwx/ (Reminder): This is 1 question.

*print*Print*list*Cite

In all three cases the function is of the form `y=af(b(x-h))+k`

a gives the amplitude for sinusoids which is the vertical stretch/compression of the parent function f(x). If a<0 the graph is reflected over the horizontal axis.

b gives the horizontal compression/stretch. It is associated with the period as `b=(2pi)/p` . If b<0 the graph is reflected over a vertical axis.

h gives the horizontal translation or phase shift.

k gives the vertical translation -- this is the midline for the sinusoids.

(a) Graph `y=3cos(2x)+1` for `0<x<360^@`

The parent function is y=cosx. The amplitude is 3, the period is `pi` or `180^@` , there is no phase shift or reflection and the midline is 1 (meaning the maximum is 4 and the minimum is -2.)

The graph:

** The vertical lines are spaced at `90^@` .

(b) Graph `y=-sin(x/3)-2` for `-pi/2<=x<=pi/2` :

The amplitude is 1 but the graph is reflected across the horizontal axis. The period is `6pi` . The midline is y=-2 so the maximum is -1 and the minimum is -3.

The graph:

Here it is as part of the complete period:

(c) Graph `y=-tan(x/2)+2`

The graph will be reflected over the horizontal axis. Since the period of the tangent is `pi` , the period for this function is `2pi` . There is a vertical shift of 2 units up.

The graph: