which of the following is the graph of the function y = 4csc(x)?

Explanation:

Step1: Recall csc function properties

The cosecant function $y = \csc(x)=\frac{1}{\sin(x)}$. The general form of a cosecant - type function is $y = A\csc(Bx - C)+D$. For the function $y = 4\csc(x)$, $A = 4$, $B = 1$, $C = 0$, and $D = 0$. The period of $y=\csc(x)$ is $2\pi$, and for $y = A\csc(Bx)$ the period is $\frac{2\pi}{|B|}=2\pi$. The vertical asymptotes of $y=\csc(x)$ occur at $x = n\pi$, where $n\in\mathbb{Z}$. The amplitude of $y = A\csc(Bx)$ is not defined in the traditional sense like for sine and cosine, but the range is $y\leq - |A|$ or $y\geq|A|$. Here $A = 4$, so the range is $y\leq - 4$ or $y\geq4$.

Step2: Analyze the graph

The graph of $y = 4\csc(x)$ has vertical asymptotes at $x=n\pi$ ($n\in\mathbb{Z}$) and the minimum and maximum values of the function are $y=-4$ and $y = 4$ respectively.

Answer:

The graph with vertical asymptotes at $x=n\pi$ ($n\in\mathbb{Z}$) and range $y\leq - 4$ or $y\geq4$ (the upper - part of the given graph options that has vertical asymptotes at $x = n\pi$ and values reaching $y = 4$ and $y=-4$). Since no options are labeled, we can't give a labeled answer, but the correct graph has vertical asymptotes at $x = 0,\pm\pi,\pm2\pi,\cdots$ and values of the function outside the interval $(-4,4)$.