in the xy - plane, the graph of which of the following functions has a vertical asymptote at…

Explanation:

Step1: Recall the vertical - asymptote of $y = \cot x$

The vertical asymptotes of the cotangent function $y=\cot x=\frac{\cos x}{\sin x}$ occur at $x = n\pi$, where $n\in\mathbb{Z}$.

Step2: Find the vertical - asymptote of $y=\cot(x - a)$

The vertical asymptotes of $y = \cot(x - a)$ occur when $x-a=n\pi$, or $x=n\pi + a$. We want a vertical asymptote at $x=\frac{3\pi}{4}$.

Step3: Check each option

For option A, the vertical asymptotes of $y = \cot x$ are $x=n\pi$, $n\in\mathbb{Z}$. When $n = 0$, $x = 0$; when $n = 1$, $x=\pi$, etc. For option B, the vertical asymptotes of $y=\cot(x-\frac{\pi}{2})$ occur when $x-\frac{\pi}{2}=n\pi$. Solving for $x$ gives $x=n\pi+\frac{\pi}{2}$. When $n = 1$, $x=\pi+\frac{\pi}{2}=\frac{3\pi}{2}$. For option C, the vertical asymptotes of $y=\cot(x - \frac{\pi}{4})$ occur when $x-\frac{\pi}{4}=n\pi$. Solving for $x$ gives $x=n\pi+\frac{\pi}{4}$. When $n = \frac{1}{2}$, $x=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}$. For option D, the vertical asymptotes of $y=\cot(x+\frac{\pi}{4})$ occur when $x+\frac{\pi}{4}=n\pi$. Solving for $x$ gives $x=n\pi-\frac{\pi}{4}$.

Answer:

C. $f(x)=\cot(x - \frac{\pi}{4})$