25 mark for review in the xy - plane, the graph of which of the following functions has a vertical asymptote…

Explanation:

Step1: Recall cotangent asymptote property

The function $y = \cot x=\frac{\cos x}{\sin x}$ has vertical asymptotes when $\sin x = 0$. The general form of the vertical - asymptotes of $y=\cot x$ is $x = n\pi$, where $n\in\mathbb{Z}$. For the function $y=\cot(u)$, the vertical asymptotes occur when $\sin(u)=0$, i.e., $u = n\pi$, $n\in\mathbb{Z}$.

Step2: Set the argument of cotangent equal to $n\pi$ for each option

For option A, $y = \cot x$, the vertical asymptotes are $x=n\pi$, $n\in\mathbb{Z}$. When $n = 1$, $x=\pi\neq\frac{3\pi}{4}$. For option B, if $y=\cot(x - \frac{\pi}{2})$, then we set $x-\frac{\pi}{2}=n\pi$. Solving for $x$ gives $x=n\pi+\frac{\pi}{2}$. When $n = 1$, $x=\pi+\frac{\pi}{2}=\frac{3\pi}{2}\neq\frac{3\pi}{4}$. For option C, if $y=\cot(x-\frac{\pi}{4})$, then we set $x - \frac{\pi}{4}=n\pi$. 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, if $y=\cot(x + \frac{\pi}{4})$, then we set $x+\frac{\pi}{4}=n\pi$. Solving for $x$ gives $x=n\pi-\frac{\pi}{4}$. When $n = 1$, $x=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$, but we usually consider the standard form of the cotangent's asymptote - finding process. The correct way is to match with the general form of the transformation of the cotangent function's asymptote. The function $y = \cot(x-\frac{\pi}{4})$ has an asymptote when $x-\frac{\pi}{4}=n\pi$. When $n = \frac{1}{2}$, we get the desired asymptote at $x=\frac{3\pi}{4}$.

Answer:

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