曲线 $y = a + b\sin cx$ 定义在 $0 \le x \le 2\pi$ 上。该曲线的周期为 $\pi$，且 $y$ 的最小值为 $-6$。该曲线还经过点 $(2\pi, -2)$。

1. Use the given period to find $c$.

For $y=a+b\sin cx$, the period is
$\frac{2\pi}{c}.$
Given that the period is $\pi$,
$\frac{2\pi}{c}=\pi$
$c=2.$

2. Substitute the point $(2\pi,-2)$ to find $a$.

The curve passes through $(2\pi,-2)$, so
$-2=a+b\sin(2\cdot 2\pi).$
Since $\sin 4\pi=0$,
$-2=a.$
Hence
$a=-2.$

3. Use the minimum value to find $b$.

For $y=a+b\sin cx$ with $b>0$, the minimum value occurs when $\sin cx=-1$.

Thus the minimum value is
$a-b=-6.$
Using $a=-2$,
$-2-b=-6$
$b=4.$

1. Use the maximum value of the sine function to form an equation for $x$.

From part (i),
$y=-2+4\sin 2x.$
The maximum occurs when
$\sin 2x=1.$
Then
$y_{\max}=-2+4(1)=2.$

2. Solve for all $x$-coordinates of maximum points in the given interval.

Since $0\le x\le 2\pi$,
$0\le 2x\le 4\pi.$
For $\sin 2x=1$,
$2x=\frac{\pi}{2},\ \frac{5\pi}{2}.$
Hence
$x=\frac{\pi}{4},\ \frac{5\pi}{4}.$

1. Write the equation to be sketched and identify the main levels of the graph.

From earlier parts,
$a=-2,\quad b=4,\quad c=2.$
Hence
$y=-2+4\sin 2x.$
The midline is $y=-2$.

Since $-1\le \sin 2x\le 1$,
$-2-4\le y\le -2+4,$
so
$-6\le y\le 2.$

2. Mark the key points of the sine curve in the given interval.

The maximum value is $2$, occurring at
$x=\frac{\pi}{4},\ \frac{5\pi}{4}.$
So the maximum points are
$\left(\frac{\pi}{4},2\right),\quad \left(\frac{5\pi}{4},2\right).$

The minimum value is $-6$, occurring when $\sin 2x=-1$:
$2x=\frac{3\pi}{2},\ \frac{7\pi}{2}$
$x=\frac{3\pi}{4},\ \frac{7\pi}{4}.$
So the minimum points are
$\left(\frac{3\pi}{4},-6\right),\quad \left(\frac{7\pi}{4},-6\right).$

The curve passes through the midline $y=-2$ at
$x=0,\ \frac{\pi}{2},\ \pi,\ \frac{3\pi}{2},\ 2\pi.$

3. Draw a smooth sine curve through the key points with the correct shape and interval.

Draw axes for $0\le x\le 2\pi$.

Show horizontal guide levels $y=2$, $y=-2$ and $y=-6$.

Plot and join smoothly:
$\left(0,-2\right)\to \left(\frac{\pi}{4},2\right)\to \left(\frac{\pi}{2},-2\right)\to \left(\frac{3\pi}{4},-6\right)\to \left(\pi,-2\right)$
then repeat the same pattern:
$\left(\pi,-2\right)\to \left(\frac{5\pi}{4},2\right)\to \left(\frac{3\pi}{2},-2\right)\to \left(\frac{7\pi}{4},-6\right)\to \left(2\pi,-2\right).$