题目

Problem

Figure 1 shows a sketch of a curve with equation $y=f(x)$ .

Figure 1

The curve has a minimum at $P(-1,0)$ and a maximum at $Q\left(\dfrac32,2\right)$ .

The line with equation $y=1$ is the only asymptote to the curve.

On separate diagrams sketch the curves with equation

(i) $y=f(x)-2$

(3)

(ii) $y=f(-x)$

(3)

On each sketch you must clearly state

解答

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$y=f(x)-2$ 是把原图像整体向下平移 $2$ 个单位。所有点的 $x$ 坐标不变， $y$ 坐标减 $2$ ；水平渐近线也向下移 $2$ 。

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The graph of $y=f(x)-2$ is the graph of $y=f(x)$ translated $2$ units down.

The minimum point becomes

\begin{align*} P(-1,0)\mapsto (-1,-2). \end{align*}

The maximum point becomes

\begin{align*} Q\left(\frac32,2\right)\mapsto \left(\frac32,0\right). \end{align*}

The asymptote becomes

\begin{align*} y=1-2=-1. \end{align*}

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$y=f(-x)$ 是关于 $y$ 轴反射。每个点的 $x$ 坐标变号， $y$ 坐标不变。水平渐近线不变。

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The graph of $y=f(-x)$ is the reflection of $y=f(x)$ in the $y$ -axis.

The minimum point becomes

\begin{align*} P(-1,0)\mapsto (1,0). \end{align*}

The maximum point becomes

\begin{align*} Q\left(\frac32,2\right)\mapsto \left(-\frac32,2\right). \end{align*}

The asymptote remains

\begin{align*} y=1. \end{align*}