Question 621372
Your values for period and phase shift did not get properly rendered. (I cannot see them). However, I can still try yo answer.
 
The amplitude is the number in front of "sin".
So your equation would be {{{y=3sin(Bx-phi)}}}.
The factor B multiplying the variable in that parenthesis is related to the period (P) by {{{B=2pi/P}}}
The angle measure {{{phi}}} is the phase shift, showing how far ahead (angle-wise) is the shifted function with respect to the "unshifted" {{{y=3sin(Bx)}}}.
(Physicist, electricians and electrical engineers are concerned with those angles).
So {{{y=3sin((2pi/P)x-phi)}}} would be the equation for a sine curve with amplitude 3, period {{{P}}} and phase shift {{{phi}}}.
The horizontal shift of the curve would be more complicated.
The graph would be horizontally shifted to the right by {{{P*phi/2pi}}} .
{{{y=3sin((2pi/P)x-phi)=3sin((2pi/P)*(x-phi/2pi))}}} and {{{C=P*phi/2pi}}} is what we call the horizontal shift, which is what concerns students graphing curves.
The equations above are for curves "evenly straddling" the x axis, as in
{{{graph(300,200,-1.1,8.8,-3.3,3.3,3sin(pi*(x-1/6)))}}} for {{{y=3sin(pi*(x-1/6))=3sin(pi*x-pi/6))}}} with a period of 2, a horizontal shift of {{{1/6}}} and a phase shift of {{{pi/6}}}.
We can also shift the curve up or down, as in the graph below, shifted up by 1 unit
{{{graph(300,200,-1.1,8.8,-2.3,4.3,3sin(pi*(x-1/6))+1)}}} for {{{y=3sin(pi*(x-1/6))=3sin(pi*x-pi/6)+1)}}} .