题目

Problem

Figure 3 shows a sketch of part of the graph of the trigonometric function with equation $y=f(x)$ .

Figure 3

(a) Write down an expression for $f(x)$ .

(2)

The point $P$ lies on $y=f(x)$ and is shown in Figure 3.

(b) State the coordinates of the point to which $P$ is transformed when the graph of $y=f(x)$ is transformed to the graph with equation

(i) $y=f\left(x-\dfrac{\pi}{6}\right)$

(2)

(ii) $y=-\dfrac12f(x)$

(2)

解答

(a)

解法一

思路

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图像的振幅是 $4$ ，周期是 $2\pi$ ，并且经过原点后向下走，所以是 $-4\sin x$ 。

答题过程

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The graph has amplitude $4$ and period $2\pi$ .

It passes through the origin with negative gradient, so

\begin{align*} f(x)=-4\sin x. \end{align*}

(b)(i)

解法一

思路

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从图像可读出 $P=\left(\frac{3\pi}{2},4\right)$ 。变换 $y=f\left(x-\frac{\pi}{6}\right)$ 表示图像向右平移 $\frac{\pi}{6}$ ，所以 $x$ 坐标加 $\frac{\pi}{6}$ ， $y$ 坐标不变。

答题过程

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From the graph,

\begin{align*} P=\left(\frac{3\pi}{2},4\right). \end{align*}

The transformation

\begin{align*} y=f\left(x-\frac{\pi}{6}\right) \end{align*}

is a translation to the right by $\dfrac{\pi}{6}$ .

Therefore

\begin{align*} \left(\frac{3\pi}{2},4\right) &\mapsto \left(\frac{3\pi}{2}+\frac{\pi}{6},4\right)\\ =&\, \left(\frac{5\pi}{3},4\right). \end{align*}

(b)(ii)

解法一

思路

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$y=-\frac12f(x)$ 会把所有 $y$ 坐标乘以 $-\frac12$ ， $x$ 坐标不变。

答题过程

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Under

\begin{align*} y=-\frac12f(x), \end{align*}

the $x$ coordinate is unchanged and the $y$ coordinate is multiplied by $-\dfrac12$ .

\begin{align*} \left(\frac{3\pi}{2},4\right) \mapsto \left(\frac{3\pi}{2},-2\right). \end{align*}

IAL 2025 Oct Q9

题目

解答

(a)

解法一

思路

答题过程

(b)(i)

解法一

思路

答题过程

(b)(ii)

解法一

思路

答题过程