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y=axsin(bx)+1.2 given the maximum point (9, 5.5) and the point (12, 3.25) along domain [0, 12]

 Nov 29, 2014

Best Answer 

 #2
avatar+130477 
+5

y=axsin(bx)+1.2

We need to solve the system  (if there is a solution)

5.5  = (9a)sin(9b) + 1.2   →       4.3 = (9a)sin(9b)

3.25 = (12a)sin(12b) + 1.2  →    2.05 = (12a)sin(12b)

WolframAlpha couldn't generate a solution for the above.......nor could it generate a solution  when the first "x" was a "times".......as Melody might have rightly assumed.

 

 Nov 30, 2014
 #1
avatar+118703 
+5

I have been mucking around with this one this afternoon and no solution has jumped out at me.

A part of the problem is I am not sure how the question should be interpreted but even so the things I try just lead to added confusion.

 

I got a new idea - Maybe that first x is not an x at all but a times sign - then it is easy!

 

maximum point (9, 5.5) and the point (12, 3.25) along domain [0, 12]

 

y=asin(bx)+1.2

 

360/b will give the wavelength (in degrees)  

360/b=4*9

360/36=b

b=10

 

y=asin(10x)+1.2

 

Sub in (12,3.25)

 

3.25=asin120+1.22.05=asin602.05=a32a=4.13

 

y=4.13sin(10x)+1.2

 

Check by subbing in (9,5.5)

 

RHS=4.13sin(90)+1.2RHS=4.13+1.2

 

4.13+1.2=3.5671361036774656

 

Well now that did not work - bummer!

It does not work because it cannot work - those conditions cannot be met on the same graph.

Maybe my interpretation is wrong again.!

Her is the graph I have.  

(It passes through (12,3.25) and it has a maximum at x=9 BUT the maximum value is not 5.5 )

https://www.desmos.com/calculator/8xnxca7qcp

 Nov 30, 2014
 #2
avatar+130477 
+5
Best Answer

y=axsin(bx)+1.2

We need to solve the system  (if there is a solution)

5.5  = (9a)sin(9b) + 1.2   →       4.3 = (9a)sin(9b)

3.25 = (12a)sin(12b) + 1.2  →    2.05 = (12a)sin(12b)

WolframAlpha couldn't generate a solution for the above.......nor could it generate a solution  when the first "x" was a "times".......as Melody might have rightly assumed.

 

CPhill Nov 30, 2014

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