# Math Grade 11 Sinusoidal functionsSinusoidal functions. math grade 11. #1) Determine the equation of each graph in the form y = acos(bx). #2) Determine the equation of each graph in the form of...

Math Grade 11 Sinusoidal functions

Sinusoidal functions. math grade 11.

#1) Determine the equation of each graph in the form y = acos(bx).

#2) Determine the equation of each graph in the form of y = asin (bx).

Both Graphs are not posted here, but i have a link for each graph below:

Graph 1. http://i515.photobucket.com/albums/t358/moeen11/latest.png

Graph 2. http://i515.photobucket.com/albums/t358/moeen11/lates1.png

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### 1 Answer

We are asked to only answer 1 question, but you should be able to answer the other using similar reasoning:

In graph 1 we see that there is a maximum at y=4, a minimum at y=-4, and the period is 1440. We are asked to write an equation for the sinusoid as a transformation of the cos graph and the sin graph.

(1) The graph is already in the shape of the cos, so we will begin there.

The general form is `y=Acos(B(x-h))+k` where A gives the amplitude (and if A<0 it reflects the graph across the horizontal axis), the period is found from B (`p=(360)/B` ), h is a horizontal translation (phase shift) and k is a vertical translation (its effect is to move the midline.)

Since the max is 4 and the min is -4, the midline is `y=(4+-4)/2==>y=0` so k=0. Also the amplitude is `A=(4-(-4))/2=4`

The graph is already in the proper position for the cos, so there is no phase shift. The period is ``1440, so `B=360/p=360/1440=1/4`

Thus the equation is `y=4cos(x/4)`

(2) In order to write the equation in terms of sin, we realize that the period and amplitude stay the same as well as the lack of vertical translation. There is only a phase shift to deal with.

The phase shift is 90 to the left, so the equation is `y=sin(x/4+90)`