we have that
The general equation is of the form
[tex]y=Asin[B(x+C)]+D[/tex]
where
A is the amplitude
period is 2π/B
phase shift is C
vertical shift is D
In this problem we have
[tex]y=\frac{5}{2}-\frac{1}{2}sin(3x+pi)[/tex]
Rewrite the given expression
[tex]y=-\frac{1}{2}s\imaginaryI n[3(x+\frac{p\imaginaryI}{3})]+\frac{5}{2}\text{ -----> equivalent expression}[/tex]
therefore
Amplitude is 1/2
Period=2pi/3
phase shift is pi/3 (to the left)
vertical shift is 5/2 (up)
