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standard 15 homework
score: 9.67/21 answered: 11/20
× question 11
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sketch a graph of the function ( f(x)=sin (x)+1 ).
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Explanation:

Step1: Analyze the transformation

The function \(y = \sin(x)+1\) is a vertical shift of the basic sine function \(y=\sin(x)\). The general form of a vertical shift is \(y = f(x)+k\), where \(k = 1\). For the basic sine function \(y=\sin(x)\), the range is \([- 1,1]\). After the vertical shift \(y=\sin(x)+1\), the range becomes \([0,2]\) (since we add \(1\) to each \(y -\) value of \(y = \sin(x)\)).

Step2: Identify key points of the basic sine function

For \(y=\sin(x)\), the key points over the interval \([0,2\pi]\) are:

• When \(x = 0\), \(y=\sin(0)=0\)
• When \(x=\frac{\pi}{2}\), \(y = \sin(\frac{\pi}{2})=1\)
• When \(x=\pi\), \(y=\sin(\pi)=0\)
• When \(x=\frac{3\pi}{2}\), \(y=\sin(\frac{3\pi}{2})=-1\)
• When \(x = 2\pi\), \(y=\sin(2\pi)=0\)

Step3: Shift the key points vertically

For \(y=\sin(x)+1\):

• When \(x = 0\), \(y=\sin(0)+1=1\)
• When \(x=\frac{\pi}{2}\), \(y=\sin(\frac{\pi}{2})+1=2\)
• When \(x=\pi\), \(y=\sin(\pi)+1=1\)
• When \(x=\frac{3\pi}{2}\), \(y=\sin(\frac{3\pi}{2})+1=0\)
• When \(x = 2\pi\), \(y=\sin(2\pi)+1=1\)

Step4: Sketch the graph

Plot the shifted key points \((0,1)\), \((\frac{\pi}{2},2)\), \((\pi,1)\), \((\frac{3\pi}{2},0)\), \((2\pi,1)\) on the given coordinate - system. Then, draw a smooth sine - wave curve passing through these points. The curve has the same period (\(T = 2\pi\)) and amplitude (\(A = 1\)) as \(y=\sin(x)\), but is shifted up by \(1\) unit.

Answer:

Sketch a sine - wave with range \([0,2]\), passing through the points \((0,1)\), \((\frac{\pi}{2},2)\), \((\pi,1)\), \((\frac{3\pi}{2},0)\), \((2\pi,1)\) on the given coordinate grid.