+0  
 
0
1079
1
avatar+134 

If you could please help me out with this that would be awesome! 

 

https://gyazo.com/272dff413b9d2d9ad2fb2c763071f326

 

 

Also, I would like to thank CPhill for being such an amazing person. You helped me out so much today with my math problems. Thank you! 

 May 19, 2017
 #1
avatar+128408 
+1

 

There are several answers possible  ......

 

Using the form  y  =  A sin ( Bx + C)  + D

 

The period of the graph  is 15 units in length.....then each quarter point  = 15/4 units

 

The amplitude of the graph = A  =  4

 

The midline of the graph is y = 1   = D

 

We can solve for B  thusly

 

pi/2 =  (15/4)B

 

pi/ 2  * (4/15)  = B

 

(2/15)pi  = B

 

And  we can find C thusly......note that we are shifting the normal cosine curve to the right

 

1   =  sin [ (2/15)pi (15/4) - C] + 1       subtract 1 from both sides

0  =  sin  [ ( pi/2)  -  C ]     →  cC  = pi /2 

 

So....our function is

 

y  = 4  sin [ (2/15)pi* x - pi/2 ]  + 1

 

Here's the graph :   https://www.desmos.com/calculator/gdqmefwhyc

 

 

 

cool cool cool

 May 19, 2017

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