**How to graph the function**

Step One: Find zeros and extrema of the function by solving for #x# after setting the expression inside the sine operator (#\frac{2pi}{3}(x-\frac{1}{2})# in this case) to #pi+k\cdot \pi# for zeros, #\frac{pi}{2}+2k\cdot \pi# for local maxima, and #\frac{3pi}{2}+2k\cdot \pi# for local minima. (We'll set #k# to different integer values to find these graphical featues in different periods. Some useful values of #k# include #-2#, #-1#, #0#, #1#, and #2#.)

Step Two: Connect those special points with a continuous smooth curve after plotting them on the graph.

**How to find amplitude, period, and phase shift.**

The function in question here is sinusoidal. In other words, it involves only one single sine function.

Also, it was written in a simplified form #y=a\cdot sin(b(x+c))+d# where #a#, #b#, #c#, and #d# are constants. You need to ensure that the linear expression inside the sine function (#x-\frac{1}{2}# in this case) have #1# as the coefficient of #x#, the independent variable; you'll have to do so anyway when you calculate the phase shift. For the function we have here, #a=1#, #b=\frac{2\pi}{3}#, #c=-\frac{1}{2}# and #d=0#.

Under this expression, each of the number #a#, #b#, #c#, and #d# resembles one of the graphical features of the function.

#a="amplitude"# of the sine wave (distance between maxima and the axis of oscillation) Therefore #"amplitude"=1#

#b=2\pi \cdot "Period"#. That is #"Period"=\frac{b}{2\cdot pi}# plugging in the numbers and we get #Period"=3#

#c=-"Phase Shift"#. Notice that phase shift equals *negative* #c# since adding positive values directly to #x# would shift the curve *leftward*, for example, the function #y=x+1# is above and to the left of #y=x#. Here we have #"Phase Shift"=\frac{1}{2}#.

(FYI #d="Vertical Shift"# or #y#-coordinate of the oscillation which the question didn't ask for.)

Reference:

"Horizontal Shift - Phase Shift." *MathBitsNotebook.com* , https://mathbitsnotebook.com/Algebra2/TrigGraphs/TGShift.html Web. 26 Feb. 2018