y=axsin(bx)+1.2 given the maximum point (9, 5.5) and the point (12, 3.25) along domain [0, 12]
+128474 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.
+118609 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)
$$\\3.25=asin120+1.2\\
2.05=asin60\\
2.05=a*\frac{\sqrt3}{2}\\
\mathbf{a=\frac{4.1}{\sqrt3}}\\$$
$$y=\frac{4.1}{\sqrt3}\;sin(10x)+1.2$$
Check by subbing in (9,5.5)
$$\\RHS=\frac{4.1}{\sqrt3}\;sin(90)+1.2\\
RHS=\frac{4.1}{\sqrt3}+1.2\\$$
$${\frac{{\mathtt{4.1}}}{{\sqrt{{\mathtt{3}}}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.2}} = {\mathtt{3.567\: \!136\: \!103\: \!677\: \!465\: \!6}}$$
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 )
+128474 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.
