Question

graph each function over the interval indicated, noting the period, asymptotes, zeroes, and value of a. include a comparative sketch of y = tan t or y = cot t as indicated.

1. f(t)=2 tan t;-2\pi,2\pi
2. g(t)=\frac{1}{2} tan t;-2\pi,2\pi
3. h(t)=3 cot t;-2\pi,2\pi
4. r(t)=\frac{1}{4} cot t;-2\pi,2\pi

Explanation:

Step1: Recall properties of tangent and cotangent functions

The general form of $y = A\tan(Bt - C)+D$ and $y=A\cot(Bt - C)+D$. For $y = A\tan t$ and $y = A\cot t$, the period of $y=\tan t$ is $\pi$ and the period of $y = \cot t$ is $\pi$. The value of $A$ affects the vertical stretch or compression.

For $f(t)=2\tan t$:

• Period: The period of $y = \tan t$ is $\pi$. Since there is no horizontal - scaling factor other than 1 (i.e., $B = 1$ in $y=A\tan(Bt)$), the period of $f(t)=2\tan t$ is $\pi$.
• Asymptotes: The asymptotes of $y=\tan t$ occur at $t=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$. In the interval $[- 2\pi,2\pi]$, the asymptotes are $t =-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$.
• Zeroes: The zeroes of $y = \tan t$ occur at $t = k\pi,k\in\mathbb{Z}$. In the interval $[-2\pi,2\pi]$, the zeroes are $t=-2\pi,-\pi,0,\pi,2\pi$. The value of $A = 2$, which means the graph of $y = \tan t$ is vertically stretched by a factor of 2.

For $g(t)=\frac{1}{2}\tan t$:

• Period: Since $B = 1$, the period is $\pi$.
• Asymptotes: The asymptotes are at $t=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$. In the interval $[-2\pi,2\pi]$, $t =-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$.
• Zeroes: The zeroes are at $t = k\pi,k\in\mathbb{Z}$. In the interval $[-2\pi,2\pi]$, the zeroes are $t=-2\pi,-\pi,0,\pi,2\pi$. The value of $A=\frac{1}{2}$, so the graph of $y = \tan t$ is vertically compressed by a factor of $\frac{1}{2}$.

For $h(t)=3\cot t$:

• Period: The period of $y=\cot t$ is $\pi$. Since $B = 1$, the period of $h(t)$ is $\pi$.
• Asymptotes: The asymptotes of $y=\cot t$ occur at $t=k\pi,k\in\mathbb{Z}$. In the interval $[-2\pi,2\pi]$, the asymptotes are $t=-2\pi,-\pi,0,\pi,2\pi$.
• Zeroes: The zeroes of $y=\cot t$ occur at $t=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$. In the interval $[-2\pi,2\pi]$, the zeroes are $t =-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$. The value of $A = 3$, so the graph of $y=\cot t$ is vertically stretched by a factor of 3.

For $r(t)=\frac{1}{4}\cot t$:

• Period: Since $B = 1$, the period is $\pi$.
• Asymptotes: The asymptotes are at $t=k\pi,k\in\mathbb{Z}$. In the interval $[-2\pi,2\pi]$, the asymptotes are $t=-2\pi,-\pi,0,\pi,2\pi$.
• Zeroes: The zeroes are at $t=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$. In the interval $[-2\pi,2\pi]$, the zeroes are $t=-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$. The value of $A=\frac{1}{4}$, so the graph of $y=\cot t$ is vertically compressed by a factor of $\frac{1}{4}$.

To graph these functions, first plot the asymptotes as dashed lines, then plot the zeroes, and then use the behavior of the tangent and cotangent functions (increasing for $y = \tan t$ and decreasing for $y=\cot t$ in their respective intervals between asymptotes) and the vertical - stretch or compression factor $A$ to sketch the curves.

The detailed graphing process involves plotting points in each sub - interval between the asymptotes and using the properties of the trigonometric functions. For example, for $y = \tan t$, in the interval $(-\frac{\pi}{2},\frac{\pi}{2})$, it is an increasing function that passes through the origin. For $y=\cot t$, in the interval $(0,\pi)$, it is a decreasing function.

Answer:

For $f(t)=2\tan t$: Period is $\pi$, asymptotes are $t=-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$, zeroes are $t=-2\pi,-\pi,0,\pi,2\pi$, $A = 2$.
For $g(t)=\frac{1}{2}\tan t$: Period is $\pi$, asymptotes are $t=-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$, zeroes are $t=-2\pi,-\pi,0,\pi,2\pi$, $A=\frac{1}{2}$.
For $h(t)=3\cot t$: Period is $\pi$, asymptotes are $t=-2\pi,-\pi,0,\pi,2\pi$, zeroes are $t=-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$, $A = 3$.
For $r(t)=\frac{1}{4}\cot t$: Period is $\pi$, asymptotes are $t=-2\pi,-\pi,0,\pi,2\pi$, zeroes are $t=-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$, $A=\frac{1}{4}$.