Question

1. sketch the following function ( f(x)=-3 sin left(\frac{4}{3} x-\frac{2 pi}{3}

ight)+1 ) on the given domain. you may only use transformations or the 5 -point method shown in class. please be sure to show your work. a-6 marks

Explanation:

Step1: Identify the general form of the sine function

The general form of a sine function is \(y = A\sin(B(x - C))+D\). For the function \(f(x)=- 3\sin(\frac{4}{3}x-\frac{2\pi}{3}) + 1\), we have \(A=-3\), \(B = \frac{4}{3}\), \(C=\frac{\pi}{2}\), \(D = 1\). The amplitude is \(|A|=3\), the period \(T=\frac{2\pi}{B}=\frac{2\pi}{\frac{4}{3}}=\frac{3\pi}{2}\), the phase - shift is \(C=\frac{\pi}{2}\), and the vertical - shift is \(D = 1\).

Step2: Use the 5 - point method for the basic sine function \(y=\sin(x)\)

For \(y = \sin(x)\), the key points are \((0,0)\), \((\frac{\pi}{2},1)\), \((\pi,0)\), \((\frac{3\pi}{2},- 1)\), \((2\pi,0)\).
For the function \(y=\sin(\frac{4}{3}x-\frac{2\pi}{3})\), we set \(u=\frac{4}{3}x-\frac{2\pi}{3}\).

1. When \(u = 0\): \(\frac{4}{3}x-\frac{2\pi}{3}=0\), then \(x=\frac{\pi}{2}\).
2. When \(u=\frac{\pi}{2}\): \(\frac{4}{3}x-\frac{2\pi}{3}=\frac{\pi}{2}\), \(\frac{4}{3}x=\frac{\pi}{2}+\frac{2\pi}{3}=\frac{3\pi + 4\pi}{6}=\frac{7\pi}{6}\), \(x=\frac{7\pi}{8}\).
3. When \(u=\pi\): \(\frac{4}{3}x-\frac{2\pi}{3}=\pi\), \(\frac{4}{3}x=\pi+\frac{2\pi}{3}=\frac{3\pi + 2\pi}{3}=\frac{5\pi}{3}\), \(x=\frac{5\pi}{4}\).
4. When \(u=\frac{3\pi}{2}\): \(\frac{4}{3}x-\frac{2\pi}{3}=\frac{3\pi}{2}\), \(\frac{4}{3}x=\frac{3\pi}{2}+\frac{2\pi}{3}=\frac{9\pi+4\pi}{6}=\frac{13\pi}{6}\), \(x=\frac{13\pi}{8}\).
5. When \(u = 2\pi\): \(\frac{4}{3}x-\frac{2\pi}{3}=2\pi\), \(\frac{4}{3}x=2\pi+\frac{2\pi}{3}=\frac{6\pi + 2\pi}{3}=\frac{8\pi}{3}\), \(x = 2\pi\).

Then for \(y=-3\sin(\frac{4}{3}x-\frac{2\pi}{3})+1\):

• When \(x = \frac{\pi}{2}\), \(y=-3\sin(0)+1=1\).
• When \(x=\frac{7\pi}{8}\), \(y=-3\sin(\frac{\pi}{2})+1=-3\times1 + 1=-2\).
• When \(x=\frac{5\pi}{4}\), \(y=-3\sin(\pi)+1=1\).
• When \(x=\frac{13\pi}{8}\), \(y=-3\sin(\frac{3\pi}{2})+1=-3\times(-1)+1=4\).
• When \(x = 2\pi\), \(y=-3\sin(2\pi)+1=1\).

Step3: Plot the key points and draw the graph

Plot the points \((\frac{\pi}{2},1)\), \((\frac{7\pi}{8},-2)\), \((\frac{5\pi}{4},1)\), \((\frac{13\pi}{8},4)\), \((2\pi,1)\) on the coordinate system. Then, using the properties of the sine function (period, amplitude, phase - shift, vertical - shift), connect these points with a smooth curve. The graph is a sine - wave that is reflected about the \(x\) - axis (due to \(A=-3\)), stretched vertically by a factor of \(3\), shifted to the right by \(\frac{\pi}{2}\) units and up by \(1\) unit.

Answer:

Plot the points \((\frac{\pi}{2},1)\), \((\frac{7\pi}{8},-2)\), \((\frac{5\pi}{4},1)\), \((\frac{13\pi}{8},4)\), \((2\pi,1)\) and connect them with a smooth sine - curve considering the amplitude \(3\), reflection (due to negative \(A\)), phase - shift \(\frac{\pi}{2}\) and vertical - shift \(1\).