which of the following is the graph of f(x)=2 csc(2π/3 x) in the xy - plane?

Explanation:

Step1: Recall the period formula for $y = A\csc(Bx)$

The period of the cosecant - function $y = A\csc(Bx)$ is given by $T=\frac{2\pi}{|B|}$. For the function $f(x)=2\csc(\frac{2\pi}{3}x)$, we have $B = \frac{2\pi}{3}$. Then $T=\frac{2\pi}{\frac{2\pi}{3}}=3$.

Step2: Analyze the amplitude

The amplitude of the cosecant - function $y = A\csc(Bx)$ is $|A|$. Here, $A = 2$, so the minimum and maximum values of the function (excluding the asymptotes) are $y=-2$ and $y = 2$ respectively.

Step3: Check the graphs for period and amplitude

• Option A: The period of the graph in option A is not 3.
• Option B: The period of the graph is 3, and the amplitude (the distance from the mid - line of the U - shaped curves to the minimum/maximum values) is 2. This graph matches the function $y = 2\csc(\frac{2\pi}{3}x)$.
• Option C: The period of the graph in option C is not 3.
• Option D: The amplitude of the graph in option D is not 2.

Answer:

B.