Question

graph the function over a one - period interval. y = 1/5 tan x. choose the correct graph. a. graph options. b. graph options. c. graph options. d. graph options.

Explanation:

Step1: Recall properties of $y = \frac{1}{\tan x}=\cot x$

The cotangent function $y = \cot x=\frac{\cos x}{\sin x}$ has vertical - asymptotes at $x = n\pi$, $n\in\mathbb{Z}$ and a period of $\pi$. In the interval $(-\frac{\pi}{2},\frac{\pi}{2})$, $\cot x$ is undefined at $x = 0$, and $\lim_{x
ightarrow0^{+}}\cot x=+\infty$, $\lim_{x
ightarrow0^{-}}\cot x = -\infty$.

Step2: Analyze the sign and behavior

When $x\in(0,\frac{\pi}{2})$, $\sin x>0$ and $\cos x>0$, so $\cot x>0$. When $x\in(-\frac{\pi}{2},0)$, $\sin x<0$ and $\cos x>0$, so $\cot x < 0$.

Answer:

The graph of $y=\frac{1}{\tan x}$ has vertical asymptotes at $x = k\pi,k\in\mathbb{Z}$, is positive in the intervals $(k\pi,(k + 1)\pi)$ for even $k$ and negative in the intervals $(k\pi,(k + 1)\pi)$ for odd $k$. Without seeing the exact details of the graphs A - D, we know that the correct graph will have a vertical asymptote at $x = 0$ and will be positive for $x\in(0,\frac{\pi}{2})$ and negative for $x\in(-\frac{\pi}{2},0)$. You need to check which of the given graphs A, B, C, D has these properties. If we assume standard - oriented graphs, the graph that has a vertical asymptote at $x = 0$, goes to $+\infty$ as $x
ightarrow0^{+}$ and to $-\infty$ as $x
ightarrow0^{-}$ is the correct one.