# How do you graph and list the amplitude, period, phase shift for y=sin((2pi)/3(x-1/2))?

Feb 26, 2018

Amplitude: $1$
Period: $3$
Phase Shift: $\setminus \frac{1}{2}$

See the explanation for details on how to graph the function. graph{sin((2pi/3)(x-1/2)) [-2.766, 2.762, -1.382, 1.382]}

#### Explanation:

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 ($\setminus \frac{2 \pi}{3} \left(x - \setminus \frac{1}{2}\right)$ in this case) to $\pi + k \setminus \cdot \setminus \pi$ for zeros, $\setminus \frac{\pi}{2} + 2 k \setminus \cdot \setminus \pi$ for local maxima, and $\setminus \frac{3 \pi}{2} + 2 k \setminus \cdot \setminus \pi$ 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 $y = a \setminus \cdot \sin \left(b \left(x + c\right)\right) + d$ where $a$, $b$, $c$, and $d$ are constants. You need to ensure that the linear expression inside the sine function ($x - \setminus \frac{1}{2}$ 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, $a = 1$, $b = \setminus \frac{2 \setminus \pi}{3}$, $c = - \setminus \frac{1}{2}$ and $d = 0$.

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

$a = \text{amplitude}$ of the sine wave (distance between maxima and the axis of oscillation) Therefore $\text{amplitude} = 1$

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

$c = - \text{Phase Shift}$. 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 $\text{Phase Shift} = \setminus \frac{1}{2}$.
(FYI $d = \text{Vertical Shift}$ 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