ANSWER
A) Amplitude: 8
Period: 2π/3
Phase shift: Right π/12
EXPLANATION
A sine function can be written in the form,
[tex]y=A\sin (B(x+C))+D[/tex]
Where A is the amplitude, C is the phase shift left and D is the vertical shift. B is related to the period as follows,
[tex]T=\frac{2\pi}{B}[/tex]
In this case, the given function is,
[tex]y=8\sin \mleft(3\theta-\frac{\pi}{4}\mright)-5[/tex]
We can see that A = 8, D = -5 and B = 3. Hence, the amplitude is 8 and we have to use the other data to find the phase shift and the period.
If B is 3, then the period is,
[tex]T=\frac{2\pi}{3}[/tex]
To find the phase shift, we have to factor out B,
[tex]3\theta-\frac{\pi}{4}=3\mleft(\theta-\frac{\pi}{3\cdot4}\mright)=3\mleft(\theta-\frac{\pi}{12}\mright)[/tex]
C is negative, so the phase shift is to the right.
In summary, the amplitude is 8, the period is 2π/3 and the phase shift is π/12 to the right.
