1. **State the problem:** We need to find a function of the form $f(x) = \csc(Bx - C)$ that matches the given graph. 2. **Identify key features from the graph:** - Vertical asymptotes at $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}$. - Maximum and minimum values near $y = 3$ and $y = -3$. 3. **Recall properties of the cosecant function:** - The basic cosecant function $\csc x$ has vertical asymptotes where $\sin x = 0$, i.e., at $x = k\pi$ for integers $k$. - The period of $\csc x$ is $2\pi$. - Vertical asymptotes occur at zeros of the sine function inside the argument. 4. **Determine the period and phase shift:** - The vertical asymptotes are spaced $\frac{\pi}{2}$ apart: $\frac{3\pi}{4} - \frac{\pi}{4} = \frac{\pi}{2}$. - For $f(x) = \csc(Bx - C)$, vertical asymptotes occur where $Bx - C = k\pi$. - The distance between consecutive asymptotes in $x$ is $\frac{\pi}{B}$. - Given the distance between asymptotes is $\frac{\pi}{2}$, we have: $$\frac{\pi}{B} = \frac{\pi}{2} \implies B = 2$$ 5. **Find the phase shift $C$:** - The first vertical asymptote is at $x = \frac{\pi}{4}$. - At this point, $B x - C = k\pi$ for some integer $k$. - Choose $k=0$ for the first asymptote: $$2 \cdot \frac{\pi}{4} - C = 0 \implies \frac{\pi}{2} - C = 0 \implies C = \frac{\pi}{2}$$ 6. **Determine the amplitude (vertical stretch):** - The maximum and minimum values are near $\pm 3$. - The basic $\csc$ function has range $(-\infty, -1] \cup [1, \infty)$. - To get maximum and minimum near $\pm 3$, multiply by 3: $$f(x) = 3 \csc(2x - \frac{\pi}{2})$$ 7. **Final function:** $$\boxed{f(x) = 3 \csc\left(2x - \frac{\pi}{2}\right)}$$ This matches the vertical asymptotes, period, phase shift, and amplitude of the given graph.