Question 1160906
<pre>

x  | -120° | -60° |  0°  | 60° | 120° | 180° | 240° |
-----------------------------------------------------
y  |   1.5 |   3  | 1.5  |  0  | 1.5  |  3   | 1.5  |    

This is the graph:

{{{drawing(400,300,-1+10,11+10,-2+10,2+10,

line(1,9,25,9),line(10,9-.1,10,9+.1),
line(10+pi,-10,10+pi,30),

line(10+pi/2,9-.1,10+pi/2,9+.1),
line(10+pi,9-.1,10+pi,9+.1),line(10+3pi/2,9-.1,10+3pi/2,9+.1),
line(10+2pi,9-.1,10+2pi,9+.1),line(10+5pi/2,9-.1,10+5pi/2,9+.1),
line(10+3pi,9-.1,10+3pi,9+.1),

green(line(-20,10,30,10)),
line(10+pi-.1,10,10+pi+.1,10),locate(10+pi-.8,10.1,1.5),
locate(10+pi-.5,11.1,3),



line(10+pi-.1,11,10+pi+.1,11),


locate(9.2,9,-120^o),locate(9.3+pi/2,9,-60^o),locate(9.6+pi,9,0),
locate(9.6+3pi/2,9,60^o),locate(9.6+2pi,9,120^o),locate(9.6+5pi/2,9,180^o),
locate(9.6+3pi,9,240^o),

graph(400,300,-1+10,11+10,-2+10,2+10,(sin(x-10)+10)*((sqrt(x-10)+10))/(sqrt(x-10)+10)*(sqrt(3pi-x+10))/(sqrt(3pi-x+10))




 ))}}}

The equation is

{{{y = Asin(Bx+C)+D}}}

where  |A| = amplitude = (maximum value - minimum value)/2 = (3-0)/2 = 3/2,

So we substitute 3/2 for A

{{{y = expr(3/2)sin(Bx+C)+D}}}

The period = 360°/B = length of one cycle. The first cycle goes from
-120° to 120° which is 120°-(-120°) = 120°+120° = 240°

So 360°/B = 240°
     360° = 240°B
360°/240° = B
      3/2 = B

We substitute 3/2 for B:

{{{y = expr(3/2)sin(expr(3/2)x+C)+D}}}

The phase (horizontal) shift, as we can tell from the graph, is 120° LEFT,
so C/B is +120°. (+ if left, and - if right)

C/B = 120°
  C = 120°B
  C = 120°(3/2)
  C = 180°

We substitute 180° for C

{{{y = expr(3/2)sin(expr(3/2)x+180^o)+D}}}

Finally the vertical shift is D and we can see that the graph has a
vertical shift upward of 1.5, so we substitute 1.5 for D:

{{{y = expr(3/2)sin(expr(3/2)x+180^o)+1.5}}}

And since 1.5 = 3/2 we can just write 3/2 for 1.5 and all the
constants just happened to came out to be 3/2.

{{{y = expr(3/2)sin(expr(3/2)x+180^o)+3/2}}} 

Edwin</pre>