题目

Problem

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

Figure 1

The curve passes through the points $(-1,0)$ and $(0,2)$ and touches the $x$ -axis at the point $(3,0)$ .

On separate diagrams, sketch the curve with equation

(a) $y=f(x+3)$

(3)

(b) $y=f(-3x)$

(3)

On each diagram, show clearly the coordinates of all the points where the curve cuts or touches the coordinate axes.

解答

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$y=f(x+3)$ 是把原图向左平移 $3$ 。所以每个关键点的 $x$ 坐标减 $3$ ， $y$ 坐标不变。

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The transformation from $y=f(x)$ to $y=f(x+3)$ is a translation $3$ units to the left.

\begin{align*} (-1,0)&\mapsto(-4,0),\\ (0,2)&\mapsto(-3,2),\\ (3,0)&\mapsto(0,0). \end{align*}

The sketch should have the same shape as the original curve, with a minimum touching the $x$ -axis at

\begin{align*} (0,0), \end{align*}

and it should also pass through

\begin{align*} (-4,0). \end{align*}

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$y=f(-3x)$ 表示先在 $y$ 轴反射，再在水平方向压缩为原来的 $\frac13$ 。点 $(a,b)$ 会变成 $\left(-\frac{a}{3},b\right)$ 。

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For $y=f(-3x)$ , a point $(a,b)$ on $y=f(x)$ maps to

\begin{align*} \left(-\frac{a}{3},b\right). \end{align*}

Therefore

\begin{align*} (-1,0)&\mapsto\left(\frac13,0\right),\\ (0,2)&\mapsto(0,2),\\ (3,0)&\mapsto(-1,0). \end{align*}

The sketch should be reflected in the $y$ -axis and horizontally compressed by scale factor $\frac13$ . It touches the $x$ -axis at

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

passes through

\begin{align*} \left(\frac13,0\right), \end{align*}

and cuts the $y$ -axis at

\begin{align*} (0,2). \end{align*}