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Trig help with homework question 

Guest Feb 4, 2018
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 #1
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Using    

 

 

A little tricky....but......we can imagine the sine graph shifted down  one unit from the normal "parent" function  and having an amplitude of 3

 

So...far...we have this

 

y =  3sin (Bx)- 1

 

Note that a  "starting point"  on the curve is  (0, -1)  and a period of 1.5  puts us at  (-1, 4pi)

 

So.....  a period must be    4pi / 1.5  =  (8/3) pi

 

So.... to  find the number of periods in 2pi ,  ("B"), we can solve this

 

(2pi / (8/3 pi)  = B               

 

(6/3 pi) / (8/3 pi)  =

 

(6/3)(8/3)  =

 

6/8  =   3/4  =  B =  there are 3/4 periods in 2pi

 

So...our  function is

 

y  = 3sin [ (3/4)x ] -  1

 

Here's the graph : https://www.desmos.com/calculator/olu4zazirp

 

[ Note....other solutions are also possible ....this just seemed to be the easiest ]

 

 

cool cool cool

 

 

So....

CPhill  Feb 4, 2018

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