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**Hint**: In order to find the various parts of the function, compare the given sinusoidal function with the general sinusoidal function, comparing all the different parts and obtaining the values for amplitude, period, phase shift.

**:**

__Complete step-by-step answer__For the normal sine graph, the graph is:

We are given with the function $ y = \sin \left( {\dfrac{{2\pi }}{3}\left( {x - \dfrac{1}{2}} \right)} \right) $ .

For first part that is graph of the function:

Finding zeros and extrema for the function by solving for $ x $ , by comparing the equation under the sine operator with $ \pi + k.\pi $ for zeroes, $ \dfrac{\pi }{2} + 2k.\pi $ for local maxima, and $ \dfrac{{3\pi }}{2} + 2k.\pi $ for local minima. We can set the different integers for the value of $ k $ , like $ - 2, - 2,0,1,2 $ .

Then connecting the points to form a continuous smooth curve which will draw a graph.

And, the graph obtained for the function $ y = \sin \left( {\dfrac{{2\pi }}{3}\left( {x - \dfrac{1}{2}} \right)} \right) $ is:

For second part that is for finding amplitude, period, phase shift:

We can see that the function is sinusoidal. It involves only one single function.

From the general sinusoidal equation, we know that $ y = a.\sin \left( {b\left( {x + c} \right)} \right) + d $ where $ a,b,c, d $ are constants. Where $ a = amplitude $ of the sine wave, $ b = 2\pi .Period $ and $ c = - $ phase shift and $ d = $ vertical shift or y-coordinate.

Comparing our equation $ y = \sin \left( {\dfrac{{2\pi }}{3}\left( {x - \dfrac{1}{2}} \right)} \right) $ with the general equation $ y = a.\sin \left( {b\left( {x + c} \right)} \right) + d $ , the constants we get are:

$ a = 1 $

$ b = \dfrac{{2\pi }}{3} $

$ c = - \dfrac{1}{2} $

$ d = 0 $

Since, we got the value of $ a $ which is our amplitude, so amplitude is $ 1 $ .

For period, we know that

$

b = 2\pi .period \\

= > period = \dfrac{b}{{2\pi }} \;

$

Substituting the value of $ b = \dfrac{{2\pi }}{3} $ above to get period and it is: $ period = \dfrac{b}{{2\pi }} = \dfrac{{\dfrac{{2\pi }}{3}}}{{2\pi }} = \dfrac{1}{3} $ .

And $ c = - $ phase shift that implies phase shift $ = - c = - \left( { - \dfrac{1}{2}} \right) = \dfrac{1}{2} $ .

The amplitude, period, and phase shift for $ y = \sin \left( {\dfrac{{2\pi }}{3}\left( {x - \dfrac{1}{2}} \right)} \right) $ is $ 1 $ , $ \dfrac{1}{3} $ and $ \dfrac{1}{2} $ .

Comparing the first and the second graph on the same graph, we get:

**Note**: Amplitude is the distance between maxima and the axis of oscillation.

We need to ensure that the linear expression inside the sine function has $ 1 $ as the coefficient of $ x $ . In our above case we saw that the linear expression $ \left( {x - \dfrac{1}{2}} \right) $ has $ 1 $ as the coefficient of $ x $ , then moved further to find the rest of the values.

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