How do you graph #g(x) = -2sin(pix+pi/3)#?

#y = Asin(Bx+C)#

(or, in some treatments #y = Asin(Bx-C)# for the method advocated in this answer, it doesn't matter.)

We know that sine takes on values between #-1# and #1#, so when we multiply after finding sine that will change the range of values.

The amplitude reflects this fact and is equal to #absA#

For the basic sine graph, #y = sinx#, we often think of this as starting the first cycle when we take the sine of 0 and finishing when we take the sine of #2pi#. The period of the basic function is #2 pi#.

The period of #y = Asin(Bx+C)# is #(2pi)/B" "# (Multiplying by #B# changes the scale on the #x# axis.)

Phase shift (or Horizontal Shift) tells us where we "start the first period" it tells us when we take the sine of 0.

To find the shift, solve #Bx+C = 0#
(Or #Bx-C=0# depandng on your textbook.)

So here we go:

For #g(x) = -2sin(pix+pi/3)#,

We have Amplitude = #2" "#
(The minus sign will flip the graph across the #x# axis.)

Period is #(2pi)/pi = 2#

Phase Shift is the solution to #pix+pi/3 = 0#,

it is #-1/3#

The graph starts on the #x# axis at #x = -1/3#, then decreases to a minimum of #-2#, back up to and through the #x# axis, up to a maximum of #2#, before returning to the #x# axis. Then repeat the pattern as required.

Here is one period:

graph{y = -2sin(pix+pi/3) *(sqrt(1-(x-2/3)^2))/(sqrt(1-(x-2/3)^2)) [-4.177, 5.69, -2.59, 2.343]

Note

Another way of finding the period is to find the beginning of the first cycle (phase shift) by solving #Bx+C = 0# as before, and the find the end of the cycle by solving #Bx+C = 2pi#

The period is the difference between the end and the beginning. If you go through the algebra, you'll see that the difference is #(2pi)/B#