Course: 14.5 图表切图 | The Way To Learn

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Your mission, should you choose to accept it, as Agent Trigonometry is to find the period and the zeros of the function $\begin{align*}y=\frac{1}{2}\tan 4x\end{align*}$ .
::您的任务, 如果您选择接受它, 作为Trigonology Agent 是要找到 y= 12tan4x 函数的周期和零。

Graph of a Tangent Function
::Tangent 函数图形图

The graph of the tangent function is very different from the functions. First, recall that the tangent ratio is $\begin{align*}\tan \theta=\frac{\text{opposite}}{\text{adjacent}}\end{align*}$ . In radians, the coordinate for the tangent function would be $\begin{align*}(\theta, \tan \theta)\end{align*}$
::相切函数的图形与函数非常不同。首先, 请记住, 相切比是 tan 。在弧度中, 相切函数的坐标是 (, tan) 。

$\begin{aligned} x & θ & 0 & \frac{π}{6} & \frac{π}{4} & \frac{π}{3} & \frac{π}{2} & \frac{2 π}{3} & \frac{3 π}{4} & \frac{5 π}{6} & π \\ y & \tan θ & 0 & \frac{\sqrt{3}}{3} & 1 & \sqrt{3} & und. & - \sqrt{3} & - 1 & - \frac{\sqrt{3}}{3} & 0 \end{aligned}$
$\begin{align*}x && \theta && 0 && \frac{\pi}{6} && \frac{\pi}{4} && \frac{\pi}{3} && \frac{\pi}{2} && \frac{2\pi}{3} && \frac{3\pi}{4} && \frac{5\pi}{6} && \pi \\ y && \tan \theta && 0 && \frac{\sqrt{3}}{3} && 1 && \sqrt{3} && \text{und.} && -\sqrt{3} && -1 && -\frac{\sqrt{3}}{3} && 0\end{align*}$
::-6 -4 -4 -3 -3 -3 3 -3 3

After $\begin{align*}\pi\end{align*}$ , the $\begin{align*}y\end{align*}$ -values repeat, making the tangent function periodic with a period of $\begin{align*}\pi\end{align*}$ .
::在 __之后, Y 值重复,使正切函数周期化为 __。

The red portion of the graph represents the coordinates in the table above. Repeating this portion, we get the entire tangent graph. Notice that there are vertical asymptotes at $\begin{align*}x=-\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}\end{align*}$ and $\begin{align*}\frac{3\pi}{2}\end{align*}$ . If we were to extend the graph out in either direction, there would continue to be vertical asymptotes at the odd multiples of $\begin{align*}\frac{\pi}{2}\end{align*}$ . Therefore, the domain is all real numbers, $\begin{align*}x\ne n\pi \pm\frac{\pi}{2}\end{align*}$ , where $\begin{align*}n\end{align*}$ is an integer. The range would be all real numbers. Just like with sine and cosine functions, you can change the , phase shift , and vertical shift .
::图表的红色部分代表上表的坐标。重复此部分, 我们可以看到整个正切图。注意在 x32, 2, 2, 2 和 32 32 上存在垂直小数。如果我们将图表向任一方向扩展, 则在 2 的奇数中将继续出现垂直小数。因此, 域名是所有真实数字 xn2 , 其中 n 是一个整数。范围将是所有真实数字。就像正弦和正弦函数一样, 您可以更改、相向移动和垂直移动。

The standard form of the equation is $\begin{align*}y=a\tan b(x-h)+k\end{align*}$ where $\begin{align*}a, b, h,\end{align*}$ and $\begin{align*}k\end{align*}$ are the same as they are for the other trigonometric functions . For simplicity, we will not address phase shifts $\begin{align*}(k)\end{align*}$ in this concept.
::方程式的标准形式是 y = atatanb( x-h)+k, 其中 a, b, h, h, k 与其他三角函数相同。为了简单起见,我们不会在此概念中处理( k) 的阶段转移。

Let's graph $\begin{align*}y=3 \tan x+1\end{align*}$ from $\begin{align*}[-2\pi, 2\pi]\end{align*}$ and state the domain and range.
::让我们从 [- 2- 2 2 ] 绘制 y= 3tanx+1 的图形, 并标出域和范围。

First, the amplitude is 3, which means each $\begin{align*}y\end{align*}$ -value will be tripled. Then, we will shift the function up one unit.
::首先,振幅是3,这意味着每个y值将增加三倍。然后,我们将将功能转换成一个单位。

Notice that the vertical asymptotes did not change. The period of this function is still $\begin{align*}\pi\end{align*}$ . Therefore, if we were to change the period of a tangent function, we would use a different formula than what we used for sine and cosine. To change the period of a tangent function, use the formula $\begin{align*}\frac{\pi}{|b|}\end{align*}$ .
::注意垂直微粒没有改变。此函数的期间仍然是。因此,如果我们改变正切函数的期间,我们将使用与正弦和正弦不同的公式。要改变正切函数的期间,请使用 b 公式。

The domain will be all real numbers, except where the asymptotes occur. Therefore, the domain of this function will be $\begin{align*}x\in \mathbb{R}, x \notin n\pi \pm \frac{\pi}{2}\end{align*}$ . The range is all real numbers.
::域名将全部为真实数字, 但小数点发生时除外。因此, 此函数的域名将是 xR, xn2。范围为所有真实数字。

Now, let's graph $\begin{align*}y=-\tan 2\pi\end{align*}$ from $\begin{align*}[0, 2\pi]\end{align*}$ , state the domain and range, and find all zeros within this domain.
::现在,让我们从 [ 0, 2] 绘制 ytan2, 说明域和范围, 并在此域中查找所有零。

The period of this tangent function will be $\begin{align*}\frac{\pi}{2}\end{align*}$ and the curves will be reflected over the $\begin{align*}x\end{align*}$ -axis.
::此相切函数的期间为 2, 曲线将反射到 X 轴上。

The domain is all real numbers, $\begin{align*}x \notin \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{\pi}{4}\pm \frac{\pi}{2}n\end{align*}$ where $\begin{align*}n\end{align*}$ is any integer. The range is all real numbers. To find the zeros, set $\begin{align*}y = 0\end{align*}$ .
::域名是所有真实数字, n 是任意整数的 x4, 34, 54, 54, 74, 442n。范围是全部真实数字。要找到零, 请设定 y=0 。

$\begin{aligned} 0 & = - \tan 2 x \\ 0 & = \tan 2 x \\ 2 x & = \tan^{- 1} 0 = 0, π, 2 π, 3 π, 4 π \\ x & = 0, \frac{π}{2}, π, \frac{3 π}{2}, 2 π \end{aligned}$
$\begin{align*}0 &=-\tan 2x \\ 0 &=\tan 2x \\ 2x &=\tan^{-1}0=0, \pi, 2\pi, 3\pi, 4\pi \\ x &=0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\end{align*}$
::0tan2x0 = tan2x2x = 10=0, 22, 3, 4x=0, 2, 2, 2, 2

Finally, let's graph $\begin{align*}y=\frac{1}{4}\tan\frac{1}{4}x\end{align*}$ from $\begin{align*}[0, 4\pi]\end{align*}$ and state the domain and range.
::最后,让我们从 [0,4] 绘制 y= 14tan14x 的图表, 并标出域和范围。

This function has a period of $\begin{align*}\frac{\pi}{\frac{1}{4}}=4\pi\end{align*}$ . The domain is all real numbers, except $\begin{align*}2\pi, 6\pi, 10\pi, 2\pi \pm 4\pi n\end{align*}$ , where $\begin{align*}n\end{align*}$ is any integer. The range is all real numbers.
::此函数的期间为 14=4。域名为所有实际数字,但 N 是任意整数的 2, 6, 10, 24n 除外。范围为所有实际数字。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the period and zeros of the function $\begin{align*}y=\frac{1}{2}\tan 4x\end{align*}$ .
::早些时候,您被要求找到函数 y= 12tan4x 的周期和零。

The period is $\begin{align*}\frac{\pi}{4}\end{align*}$ .
::这一期间为4。

The zeros are where $\begin{align*}y\end{align*}$ is zero.
::零是Y是零。

$\begin{aligned} 0 = \frac{1}{2} \tan 4 x \\ 0 & = \tan 4 x \\ 4 x & = \tan^{- 1} 0 = 0, π, 2 π, 3 π \\ x & = \frac{1}{4} (0, π, 2 π, 3 π) \\ x & = 0, \frac{π}{4}, \frac{π}{2}, \frac{3}{4} π \end{aligned}$
$\begin{align*}0 = \frac{1}{2}\tan 4x\\ 0 &=\tan 4x \\ 4x &=\tan ^{-1}0=0, \pi, 2\pi, 3\pi \\ x &=\frac{1}{4}(0, \pi, 2\pi, 3\pi ) \\ x &=0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3}{4}\pi\end{align*}$
::012tan4x0=tan4x4x=tan-10=0, 22, 3x=14 0, 22, 3x)x0, 4, 2, 342x

Example 2
::例2

Find the period of the function $\begin{align*}y=-4\tan \frac{3}{2}x\end{align*}$ .
::查找函数 y4tan32x 的周期。

The period is $\begin{align*}\frac{\pi}{\frac{3}{2}}=\pi \cdot \frac{2}{3}=\frac{2\pi}{3}\end{align*}$ .
::报告期为3223=23。

Example 3
::例3

Find the zeros of the function from Example 2, from $\begin{align*}[0, 2\pi]\end{align*}$ .
::从例2从 [0,2] 中查找函数的零。

The zeros are where $\begin{align*}y\end{align*}$ is zero.
$\begin{aligned} 0 & = - 4 \tan \frac{3}{2} x \\ 0 & = \tan \frac{3}{2} x \\ \frac{3}{2} x & = \tan^{- 1} 0 = 0, π, 2 π, 3 π \\ x & = \frac{2}{3} (0, π, 2 π, 3 π) \\ x & = 0, \frac{2 π}{3}, \frac{4 π}{3}, 2 π \end{aligned}$
$\begin{align*}0 &=-4\tan \frac{3}{2}x \\ 0 &=\tan \frac{3}{2}x \\ \frac{3}{2}x &=\tan ^{-1}0=0, \pi, 2\pi, 3\pi \\ x &=\frac{2}{3}(0, \pi, 2\pi, 3\pi ) \\ x &=0, \frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi\end{align*}$
::04tan32x0=tan32x32xx=tan-10=0, 22, 3x=23(0223, 3)x=0, 244, 33, 22}

Example 4
::例4

Find the equation of the tangent function with an amplitude of 8 and a period of $\begin{align*}6\pi\end{align*}$ .
::查找正切函数的方程式,其振幅为8,期间为6。

The general equation is $\begin{align*}y=a\tan bx\end{align*}$ . We know that $\begin{align*}a = 8\end{align*}$ . Let’s use the period to solve for the frequency, or $\begin{align*}b\end{align*}$ .
::一般方程式是 y=atanbx。我们知道 a=8. 让我们用这个时间来解答频率, 或 b 。

$\begin{aligned} \frac{π}{b} & = 6 π \\ b & = \frac{π}{6 π} = \frac{1}{6} \end{aligned}$
$\begin{align*}\frac{\pi}{b} &=6\pi \\ b &=\frac{\pi}{6\pi}=\frac{1}{6}\end{align*}$
::=66616

The equation is $\begin{align*}y=8\tan \frac{1}{6}x\end{align*}$ .
::方程式是y=8tan16x

Review
::回顾

Graph the following tangent functions over $\begin{align*}[0, 4\pi]\end{align*}$ . Determine the period, domain, and range.
::在 [0,4] 上方绘制以下相切函数图。确定周期、域和范围。

$\begin{align*}y=2\tan x\end{align*}$
::y=2tanx

$\begin{align*}y=-\frac{1}{3}\tan x\end{align*}$
::y 13tanx y 13tanx

$\begin{align*}y=-\tan 3x\end{align*}$
::y3x

$\begin{align*}y=4\tan 2x\end{align*}$
::y=4tan2x

$\begin{align*}y=\frac{1}{2}\tan 4x\end{align*}$
::y= y12tan4x

$\begin{align*}y=-\tan \frac{1}{2}x\end{align*}$
::y 12x

$\begin{align*}y=4+\tan x\end{align*}$
::y=4+tanx

$\begin{align*}y=-3+\tan 3x\end{align*}$
::y3+tan3x

$\begin{align*}y=1+\frac{2}{3}\tan \frac{1}{2}x\end{align*}$
::y=1+23tan12x

Find the zeros of the function from #1.
::从 # 1 查找函数的零。

Find the zeros of the function from #3.
::从 # 3 查找函数的零。

Find the zeros of the function from #5.
::从 # 5 查找函数的零。

Write the equation of the tangent function, in the form $\begin{align*}y=a\tan bx\end{align*}$ , with the given amplitude and period.
::以 y=atanbx 的形式,用给定的振幅和时段写入正切函数的方程式。

Amplitude: 3 Period: $\begin{align*}\frac{3\pi}{2}\end{align*}$
::振幅: 3 周期: 3×2

Amplitude: $\begin{align*}\frac{1}{4}\end{align*}$ Period: $\begin{align*}2\pi\end{align*}$
::振幅:14 周期: 2

Amplitude: -2.5 Period: 8
::振幅: -2.5 周期: 8

Challenge Graph $\begin{align*}y=2\tan \frac{1}{3}\left(x+\frac{\pi}{4}\right)-1\end{align*}$ over $\begin{align*}[0, 6\pi]\end{align*}$ . Determine the domain and period.
::图y=2tan13(x4)-1,超过[0,6]。确定域和时间段。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

图形切图

Graph of a Tangent Function ::Tangent 函数图形图

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Graph of a Tangent Function
::Tangent 函数图形图

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)