Question 921246
your first picture is shown below:


<img src = "http://theo.x10hosting.com/2014/110702a.jpg" alt="$$$" </>


the interval goes from 0 to 4p/4 which is the same as from 0 to pi.


the period is therefore equal to pi.


the frequency is equal to 2pi / pi = 2


there is no phase shift because cosine = 1 at 0 radians which is the value shown on the graph.


the equation in the form of y = a * cos (k(x-b)) becomes:


y = 6 * cos(2 * x)


i graphed it to confirm.


that graph is shown below:


<img src = "http://theo.x10hosting.com/2014/110702b.jpg" alt="$$$" </>


you can see the equation on the top left.


the graph matches your picture so it looks good.


your second picture is shown below:


<img src = "http://theo.x10hosting.com/2014/110703a.jpg" alt="$$$" </>


the amplitude is 6.


the interval goes from -3/4 to 1/4.


the period is therefore equal to 1.


the frequency is equal to 2pi / 1 = 2pi.


the sine wave should be 0 when x = 0.


this sine wave appears to be shifted to the right by 1/4 of the period.


the phase shift is therefore equal to 1/4.


the equation in the form of y = a * sin (k * (x-b)) becomes:


y = 6 * sin(2pi * (x-1/4))


the graph of that equation is shown below:


<img src = "http://theo.x10hosting.com/2014/110703b.jpg" alt="$$$" </>