下图展示了 $y = \mathrm{f}(x)$ 的图像。曲线与坐标轴交于 $A(0, 1.5)$ 和 $B(-3, 0)$ 。该曲线的渐近线为 $x = -2$ 和 $y = 1$ 。在不同的图中，分别画出下列函数的图像，并尽可能清晰地标明渐近线、坐标轴交点以及对应于 $A$ 和 $B$ 的点。

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完整解法

1. Identify the geometric transformations that map $y = \mathrm{f}(x)$ to $y = -\mathrm{f}(1+x)$.

The transformation from $\mathrm{f}(x)$ to $-\mathrm{f}(x+1)$ consists of two steps:

A horizontal translation of $1$ unit to the left (mapping $\mathrm{f}(x)$ to $\mathrm{f}(x+1)$ ).
A reflection about the $x$ -axis (mapping $\mathrm{f}(x+1)$ to $-\mathrm{f}(x+1)$ ).

2. Find the new equations of the asymptotes.

Applying the transformations to the original asymptotes:

Vertical asymptote $x = -2$ : translating left by $1$ gives $x = -3$ . The reflection in the $x$ -axis does not affect vertical lines.
New vertical asymptote: $x = -3$ .
Horizontal asymptote $y = 1$ : translation left by $1$ does not affect horizontal lines. Reflection in the $x$ -axis maps $y = 1$ to $y = -1$ .
New horizontal asymptote: $y = -1$ .

3. Determine the new coordinates of the specific points $A$ and $B$.

Applying the transformations to the given points:

For point $A(0, 1.5)$ :
Translate left by $1$ : $(0 - 1, 1.5) = (-1, 1.5)$
Reflect in $x$ -axis: $(-1, -1.5)$
So, $A$ maps to $A'(-1, -1.5)$ .
For point $B(-3, 0)$ :
Translate left by $1$ : $(-3 - 1, 0) = (-4, 0)$
Reflect in $x$ -axis: $(-4, 0)$ since it lies on the $x$ -axis.
So, $B$ maps to the new $x$ -intercept $B'(-4, 0)$ .

答案

A sketched graph containing two branches:

Vertical asymptote at $x = -3$ .
Horizontal asymptote at $y = -1$ .
Left branch ( $x < -3$ ): Lies above the horizontal asymptote, approaching $y = -1$ from above as $x \to -\infty$ , crossing the $x$ -axis at $B'(-4, 0)$ , and going upwards towards positive infinity as it approaches $x = -3$ .
Right branch ( $x > -3$ ): Lies below the horizontal asymptote, passing through $A'(-1, -1.5)$ , crossing the $y$ -axis between $y = -1$ and $y = -1.5$ , and going downwards towards negative infinity as it approaches $x = -3$ . It approaches $y = -1$ from below as $x \to \infty$ .

1. Determine the new asymptotes for the reciprocal graph.

For $y = \frac{1}{\mathrm{f}(x)}$ :

As $x \to \pm\infty$ , $\mathrm{f}(x) \to 1$ . Therefore, $y \to \frac{1}{1} = 1$ . The horizontal asymptote remains $y = 1$ .
The $x$ -intercept of $y = \mathrm{f}(x)$ is at $x = -3$ (point $B$ ). As $\mathrm{f}(x) \to 0$ , $\frac{1}{\mathrm{f}(x)} \to \pm\infty$ . Thus, $x = -3$ is the new vertical asymptote.

2. Determine the new axial intercepts and map points $A$ and $B$.

The vertical asymptote of $y = \mathrm{f}(x)$ is $x = -2$ . As $x \to -2$ , $\mathrm{f}(x) \to \pm\infty$ , so $\frac{1}{\mathrm{f}(x)} \to 0$ . This creates a new $x$ -intercept at $(-2, 0)$ .
The $y$ -intercept is at $A(0, 1.5)$ . Thus, the new $y$ -intercept is at $\left(0, \frac{1}{1.5}\right) = \left(0, \frac{2}{3}\right)$ . This is the point corresponding to $A$ .
Point $B(-3, 0)$ corresponds to the newly formed vertical asymptote $x = -3$ .

3. Determine the shape of the curve to complete the sketch.

For $x < -3$ : $\mathrm{f}(x)$ is positive and $0 < \mathrm{f}(x) < 1$ , so $\frac{1}{\mathrm{f}(x)} > 1$ . The graph approaches $y=1$ as $x \to -\infty$ and goes to $+\infty$ as $x \to -3^-$ .
For $x > -3$ : As $x \to -3^+$ , $\mathrm{f}(x) \to 0^-$ , so $\frac{1}{\mathrm{f}(x)} \to -\infty$ . The curve rises smoothly, crosses the $x$ -axis at $(-2, 0)$ , crosses the $y$ -axis at $(0, \frac{2}{3})$ , and approaches $y=1$ from below as $x \to \infty$ .

答案

Sketch Description:

Asymptotes: Vertical asymptote at $x = -3$ , horizontal asymptote at $y = 1$ .
Axial intercepts: $x$ -intercept at $(-2, 0)$ , $y$ -intercept at $(0, \frac{2}{3})$ .
Points $A$ and $B$ : Point $A(0, 1.5)$ maps to the new $y$ -intercept $A'(0, \frac{2}{3})$ . Point $B(-3, 0)$ corresponds to the new vertical asymptote $x = -3$ .
Shape: Two continuous branches separated by $x = -3$ . The left branch ( $x < -3$ ) lies above $y = 1$ , approaching $y = 1$ as $x \to -\infty$ and tending to $+\infty$ as $x \to -3^-$ . The right branch ( $x > -3$ ) starts from $-\infty$ at $x \to -3^+$ , crosses the axes at $(-2, 0)$ and $(0, \frac{2}{3})$ , and approaches $y = 1$ from below as $x \to \infty$ .

1. Determine the asymptotes of $y = \mathrm{f}'(x)$ based on the behaviour of $y = \mathrm{f}(x)$.

Observe the limits of the gradient of $\mathrm{f}(x)$ :

As $x \to \pm \infty$ , the graph of $y = \mathrm{f}(x)$ flattens out and approaches the horizontal asymptote $y = 1$ . This means the gradient $\mathrm{f}'(x) \to 0$ . Therefore, $y = 0$ (the $x$ -axis) is a horizontal asymptote for $y = \mathrm{f}'(x)$ .
As $x \to -2$ , the tangent to the curve becomes increasingly steep towards the vertical, meaning $\mathrm{f}'(x) \to -\infty$ . Therefore, $x = -2$ is a vertical asymptote for $y = \mathrm{f}'(x)$ .

2. Determine the shape, sign, and key points of the derivative graph.

The original curve $y = \mathrm{f}(x)$ is strictly decreasing on both intervals $(-\infty, -2)$ and $(-2, \infty)$ . Consequently, $\mathrm{f}'(x) < 0$ for all valid $x$ , meaning the entire graph of $y = \mathrm{f}'(x)$ must lie below the $x$ -axis.
Since $\mathrm{f}'(x) \neq 0$ , there are no $x$ -intercepts.
The point corresponding to $A(0, 1.5)$ is at $x = 0$ , giving a negative $y$ -intercept on the derivative graph.
The point corresponding to $B(-3, 0)$ is at $x = -3$ , which will sit at some negative $y$ -value on the left branch.
Since the exact functional equation is not given, we simply mark the $x$ -coordinates $x = 0$ (as the $y$ -intercept) and $x = -3$ on the graph to correspond to $A$ and $B$ .

答案

A sketch of $y = \mathrm{f}'(x)$ on a Cartesian plane with the following features:

Asymptotes: Vertical asymptote at $x = -2$ and horizontal asymptote at $y = 0$ (the $x$ -axis).
Shape: Two continuous branches, both strictly below the $x$ -axis (since $\mathrm{f}'(x) < 0$ ).
Left Branch: Approaches $y=0$ from below as $x \to -\infty$ , decreases smoothly passing through a marked point at $x = -3$ (corresponding to $B$ ), and tends to $-\infty$ as $x \to -2^-$ .
Right Branch: Approaches from $-\infty$ as $x \to -2^+$ , increases smoothly cutting the $y$ -axis at a negative value (corresponding to $A$ at $x=0$ ), and approaches $y=0$ from below as $x \to \infty$ .
(Note: The exact $y$ -coordinates for points $A'$ and $B'$ cannot be definitively calculated without assuming an explicit function, aligning with the caveat "where possible".)

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