Question

1. $y = 3sin x + 1, 0leq xleq 2pi$

Explanation:

Step1: Find key points

For \(y = A\sin(x)+k\), here \(A = 3\), \(k = 1\).
The standard sine function \(y=\sin(x)\) has key - points \((0,0)\), \((\frac{\pi}{2},1)\), \((\pi,0)\), \((\frac{3\pi}{2}, - 1)\), \((2\pi,0)\).
For \(y = 3\sin(x)+1\):
When \(x = 0\), \(y=3\sin(0)+1=1\).
When \(x=\frac{\pi}{2}\), \(y = 3\sin(\frac{\pi}{2})+1=3\times1 + 1=4\).
When \(x=\pi\), \(y=3\sin(\pi)+1=1\).
When \(x=\frac{3\pi}{2}\), \(y=3\sin(\frac{3\pi}{2})+1=3\times(-1)+1=-2\).
When \(x = 2\pi\), \(y=3\sin(2\pi)+1=1\).

Step2: Plot the points and draw the curve

Plot the points \((0,1)\), \((\frac{\pi}{2},4)\), \((\pi,1)\), \((\frac{3\pi}{2},-2)\), \((2\pi,1)\) on the coordinate system.
Connect these points with a smooth curve. The amplitude of the function \(y = 3\sin(x)+1\) is \(|A|=3\) (the distance from the mid - line \(y = 1\) to the maximum or minimum value), and the mid - line is \(y=1\).

Answer:

Plot the points \((0,1)\), \((\frac{\pi}{2},4)\), \((\pi,1)\), \((\frac{3\pi}{2},-2)\), \((2\pi,1)\) and connect them with a smooth sine - curve - like graph.