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# Trig help with homework question

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Trig help with homework question

Guest Feb 4, 2018
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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 ]

So....

CPhill  Feb 4, 2018

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