图中显示曲线 $y=a\sin(bx)+c$，其中 $0\leq x\leq 2\pi$，$a$、$b$、$c$ 为正的常数。

(a) 写出 $a$、$b$、$c$ 的值。

(b) 对于这些 $a$、$b$、$c$ 的值，判断在区间 $0\leq x\leq 2\pi$ 内，下列每个方程的解的个数：

(i) $a\sin(bx)+c=7-x$

(ii) $a\sin(bx)+c=2\pi(x-1)$。

最高值和最低值决定振幅与中线；相邻两个波峰的横向间隔就是周期。

The graph has maximum $7$ and minimum $-1$. Thus its amplitude is $(7-(-1))/2=4$ and its midline is $y=(7+(-1))/2=3$. The graph completes two periods on $[0,2\pi]$, so its period is $\pi$. Since the period is $2\pi/b$, the values are $a=4$, $b=2$ and $c=3$.

在题图上画 $y=7-x$。用 $(0,7)$ 和 $(2\pi,7-2\pi)$ 定位，再沿曲线从左到右数交点。

The line joins $(0,7)$ to $(2\pi,7-2\pi)$ on the given interval. On the sketch it crosses the sinusoidal graph twice around each of its two peaks, and once more as the graph rises after its final minimum. Thus there are five solutions.

用横截距 $(1,0)$ 和另一个点 $(2,2\pi)$ 定位直线。它上升得很快，在图上只穿过曲线一次。

The line $y=2\pi(x-1)$ passes through $(1,0)$ and $(2,2\pi)$. The sketch shows one crossing between $x=1$ and $x=\frac\pi2$. Thus there is one solution.