Course: 2.8 理性功能零 | The Way To Learn

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理性函数零

The zeroes of a function are the collection of $\begin{aligned} x \end{aligned}$ values where the height of the function is zero. How do you find these values for a rational function and what happens if the zero turns out to be a hole ?
::函数的零是函数高度为零的 x 值的集合。如何找到这些值用于理性函数,如果零是一个洞,会发生什么情况?

Finding Zeroes of Rational Functions
::寻找理性函数的零

Zeroes are also known as $\begin{aligned} x \end{aligned}$ - intercepts , solutions or roots of functions. They are the $\begin{aligned} x \end{aligned}$ values where the height of the function is zero. For rational functions, you need to set the numerator of the function equal to zero and solve for the possible $\begin{aligned} x \end{aligned}$ values. If a hole occurs on the $\begin{aligned} x \end{aligned}$ value, then it is not considered a zero because the function is not truly defined at that point.
::零点也被称为函数的 X 界面、解决方案或根根。它们是函数高度为零的 x 值。对于理性函数,您需要将函数的分子设为零,并解决可能的 x 值。如果在 x 值上出现一个洞,那么它不被视为零,因为函数当时没有真正定义。

Take the following rational function:
::采取以下合理功能:

$\begin{aligned} f (x) = \frac{(x - 1) (x + 3) (x + 3)}{x + 3} \end{aligned}$
::f(x) =(x- 1)(x+3)(x+3) x+3

Notice how one of the $\begin{aligned} x + 3 \end{aligned}$ factors seems to cancel and indicate a removable discontinuity. Even though there are two $\begin{aligned} x + 3 \end{aligned}$ factors, the only zero occurs at $\begin{aligned} x = 1 \end{aligned}$ and the hole occurs at (-3, 0).
::提醒 x+3 系数之一似乎取消并显示可移动的不连续性。即使有两个 x+3 系数, 唯一的零出现在 x=1 时, 洞出现在 (3, 0) 时。

Watch the video below and focus on the portion of this video discussing holes and $\begin{aligned} x \end{aligned}$ -intercepts.
::看下面的录像, 并关注这段影片中讨论洞和 X 界面的部分。

Examples
::实例

Example 1
::例1

Earlier, you were asked how to find the zeroes of a rational function and what happens if the zero is a hole. To find the zeroes of a rational function, set the numerator equal to zero and solve for the $\begin{aligned} x \end{aligned}$ values. When a hole and a zero occur at the same point, the hole wins and there is no zero at that point.
::早些时候, 有人询问您如何找到一个理性函数的零, 如果零是一个洞, 以及会发生什么。要找到一个理性函数的零, 请将分子设置为零, 并解决 x 值。当一个洞和一个零在同一点发生时, 洞会赢, 而那个点没有零。

Example 2
::例2

Create a function with zeroes at $\begin{aligned} x = 1, 2, 3 \end{aligned}$ and holes at $\begin{aligned} x = 0, 4 \end{aligned}$ .
::创建一个在 x=1,2,3 和 x=0,4 上为零的函数。

There are an infinite number of possible functions that fit this description because the function can be multiplied by any constant . One possible function could be:
::由于函数可以乘以任何常数,因此可以有无限数量的功能适合本描述,因为函数可以乘以任何常数。一种可能的功能可以是:

$\begin{aligned} f (x) = \frac{(x - 1) (x - 2) (x - 3) x (x - 4)}{x (x - 4)} \end{aligned}$
:xx) = (x- 1) (x-2) (x-3)x(x-4)x(x-4)x(x-4)x(x-4)

Note that 0 and 4 are holes because they cancel out.
::请注意 0 和 4 是空洞, 因为它们取消。

Example 3
::例3

Identify the zeroes, holes and $\begin{aligned} y \end{aligned}$ intercepts of the following rational function without graphing.
::识别下列理性函数的零、空洞和截取 Y ,不显示图形。

$\begin{aligned} f (x) = \frac{x (x - 2) (x - 1) (x + 1) (x + 1) (x + 2)}{(x - 1) (x + 1)} \end{aligned}$
:xx) =xx(x- 2)(x-1)(x-1)(x+1)(x+1)(x+1)(x+2)(x-1)(x+1)

The holes occur at $\begin{aligned} x = - 1, 1 \end{aligned}$ . To get the exact points, these values must be substituted into the function with the factors canceled.
::空洞在 x1 1 处发生。要获得精确点, 这些值必须替换为因子取消的函数。

\begin{aligned} f (x) & = x (x - 2) (x + 1) (x + 2) \\ f (- 1) & = 0, f (1) = - 6 \end{aligned}

::f( x) =xx( x-2) (x+1) (x+2) f( - 1) =0, f(1) 6

The holes are (-1, 0); (1, 6). The zeroes occur at $\begin{aligned} x = 0, 2, - 2 \end{aligned}$ . The zero that is supposed to occur at $\begin{aligned} x = - 1 \end{aligned}$ has already been demonstrated to be a hole instead.
::空洞是(-1,0,0,1,6),零为x=0,2,-2。假定在x1产生的零已被证明是一个空洞。

Example 4
::例4

Identify the $\begin{aligned} y \end{aligned}$ intercepts, holes, and zeroes of the following rational function.
::识别以下理性函数的 y 拦截、孔和零。

$\begin{aligned} f (x) = \frac{6 x^{3} - 7 x^{2} - x + 2}{x - 1} \end{aligned}$
:x) = 6x3 - 7x2 - x+2x-1

After noticing that a possible hole occurs at $\begin{aligned} x = 1 \end{aligned}$ and using polynomial long division on the numerator you should get:
::注意到在 x=1 时可能出现洞, 并在分子上使用多数值长的分隔符后, 您应该得到 :

$\begin{aligned} f (x) = (6 x^{2} - x - 2) \cdot \frac{x - 1}{x - 1} \end{aligned}$
:xx) = (6x2-x-2) =x-1-1

A hole occurs at $\begin{aligned} x = 1 \end{aligned}$ which turns out to be the point (1, 3) because $\begin{aligned} 6 \cdot 1^{2} - 1 - 2 = 3 \end{aligned}$ .
::x=1 发生洞口,因为 612 - 1-2=3 点( 1, 3) 。

The $\begin{aligned} y \end{aligned}$ - intercept always occurs where $\begin{aligned} x = 0 \end{aligned}$ which turns out to be the point (0, -2) because $\begin{aligned} f (0) = - 2 \end{aligned}$ .
::y 界面总是在 x=0 点( 0, 2) 因为 f( 0) @% 2 而成为点( 0) 。

To find the $\begin{aligned} x \end{aligned}$ -intercepts you need to factor the remaining part of the function:
::要找到 X 界面, 您需要将函数的剩余部分乘以 :

$\begin{aligned} (2 x + 1) (3 x - 2) \end{aligned}$
:2x+1)(3x-2)

Thus the zeroes ( $\begin{aligned} x \end{aligned}$ -intercepts) are $\begin{aligned} x = - \frac{1}{2}, \frac{2}{3} \end{aligned}$ .
::因此,零是 x12,23。

Example 5
::例5

Identify the zeroes and holes of the following rational function.
::识别以下合理函数的零和空洞。

$\begin{aligned} f (x) = \frac{2 (x + 1) (x + 1) (x + 1)}{2 (x + 1)} \end{aligned}$
::f( x) = 2( x+1)( x+1)( x+1)( x+1)( x+1)( x+1)( x+1) = 2( x+1)

The hole occurs at $\begin{aligned} x = - 1 \end{aligned}$ which turns out to be a double zero. The hole still wins so the point (-1, 0) is a hole. There are no zeroes. The constant 2 in front of the numerator and the denominator serves to illustrate the fact that constant scalars do not impact the $\begin{aligned} x \end{aligned}$ values of either the zeroes or holes of a function.
::洞在 x% 1 处发生, 结果显示为为双零。洞仍然赢, 点( 1- 0) 是一个洞。没有零。分子和分母前面的常数 2 和分母可以说明, 恒定的星标不会影响函数零或空洞的 x 值。

Summary

Zeroes of a function are the values where the height of the function is zero, also known as x-intercepts, solutions, or roots.
::函数的零是函数高度为零的值,也称为 X 界面、解决方案或根。

To find zeroes of rational functions, set the numerator of the function equal to zero and solve for the possible $\begin{aligned} x \end{aligned}$ values.
::要找到理性函数的零,请将函数的分子数设为零,并解决可能的 x 值。

Removable discontinuities may occur in rational functions when factors cancel out, indicating a hole.
::当因数取消时,在理性功能中可能会出现可消除的不连续性,这表明存在一个洞。

Review
::回顾

Identify the intercepts and holes of each of the following rational functions.
::识别以下每个合理功能的拦截和孔。

$\begin{aligned} f (x) = \frac{x^{3} + x^{2} - 10 x + 8}{x - 2} \end{aligned}$
:xx) =x3+x2-10x+8x-2

$\begin{aligned} g (x) = \frac{6 x^{3} - 17 x^{2} - 5 x + 6}{x - 3} \end{aligned}$
::g (x) = 6x3- 17x2- 5x+6x- 3

$\begin{aligned} h (x) = \frac{(x + 2) (1 - x)}{x - 1} \end{aligned}$
::h(x) = (x+2) 1-x) x-1

$\begin{aligned} j (x) = \frac{(x - 4) (x + 2) (x + 2)}{x + 2} \end{aligned}$
::j(x) = (x- 4)(x+2)(x+2) x+2

$\begin{aligned} k (x) = \frac{x (x - 3) (x - 4) (x + 4) (x + 4) (x + 2)}{(x - 3) (x + 4)} \end{aligned}$
:xx)=xx(x-3)(x-4)(x-4)(x+4)(x+4)(x+4)(x+2)(x-3)(x+4)

$\begin{aligned} f (x) = \frac{x (x + 1) (x + 1) (x - 1)}{(x - 1) (x + 1)} \end{aligned}$
:xx) =xx(xx+1)(x+1)(x1)(x-1)(x-1)(x-1)(x+1)(x+1)

$\begin{aligned} g (x) = \frac{x^{3} - x^{2} - x + 1}{x^{2} - 1} \end{aligned}$
::g (x) =x3 - x2 - x+1x2 - 1

$\begin{aligned} h (x) = \frac{4 - x^{2}}{x - 2} \end{aligned}$
::h(x)=4-x2x-2

Create a function with holes at $\begin{aligned} x = 3, 5, 9 \end{aligned}$ and zeroes at $\begin{aligned} x = 1, 2 \end{aligned}$ .
::创建 x=3,5,9和0的x=1,2的空洞函数。

Create a function with holes at $\begin{aligned} x = - 1, 4 \end{aligned}$ and zeroes at $\begin{aligned} x = 1 \end{aligned}$ .
::创建一个函数,在 x\\\\ 1, 4 和 x= 1 时为零, 以洞洞创建函数。

Create a function with holes at $\begin{aligned} x = 0, 5 \end{aligned}$ and zeroes at $\begin{aligned} x = 2, 3 \end{aligned}$ .
::创建一个函数,在 x=0, 5 和 x=2, 2 时为零, 以 x=0, 5 和零为空。

Create a function with holes at $\begin{aligned} x = - 3, 5 \end{aligned}$ and zeroes at $\begin{aligned} x = 4 \end{aligned}$ .
::在 x=4 时创建有 x%3-5 和零的空洞的函数。

Create a function with holes at $\begin{aligned} x = - 2, 6 \end{aligned}$ and zeroes at $\begin{aligned} x = 0, 3 \end{aligned}$ .
::创建一个函数,在 x=0. 3 时以 x% 2, 6 和零为空格创建空洞。

Create a function with holes at $\begin{aligned} x = 1, 5 \end{aligned}$ and zeroes at $\begin{aligned} x = 0, 6 \end{aligned}$ .
::创建一个函数,在 x=1, 5 和 x=0, 6 上以空洞创建空洞。

Create a function with holes at $\begin{aligned} x = 2, 7 \end{aligned}$ and zeroes at $\begin{aligned} x = 3 \end{aligned}$ .
::创建一个函数,在 x=2,7 和 x=3 时有空洞。

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

理性函数零

Finding Zeroes of Rational Functions ::寻找理性函数的零

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)