Question

problems 5 – 8, use transformations to graph eac 5. $y = 2sin x$

Explanation:

Step1: Recall the parent function

The parent function is \( y = \sin x \), which has an amplitude of 1, period \( 2\pi \), and key points at \( (0,0) \), \( (\frac{\pi}{2},1) \), \( (\pi,0) \), \( (\frac{3\pi}{2}, - 1) \), \( (2\pi,0) \).

Step2: Analyze the transformation

For the function \( y = 2\sin x \), we have a vertical stretch by a factor of 2. A vertical stretch by a factor of \( a\) (where \( a>1\)) of the function \( y = f(x) \) gives \( y=af(x) \). So for \( y = \sin x\), multiplying by 2 stretches the graph vertically.

Step3: Find key points after transformation

• For \( x = 0\), \( y=2\sin(0) = 0\)
• For \( x=\frac{\pi}{2}\), \( y = 2\sin(\frac{\pi}{2})=2\times1 = 2\)
• For \( x = \pi\), \( y=2\sin(\pi)=0\)
• For \( x=\frac{3\pi}{2}\), \( y = 2\sin(\frac{3\pi}{2})=2\times(- 1)=-2\)
• For \( x = 2\pi\), \( y=2\sin(2\pi)=0\)

We then plot these points and draw the curve, which is a vertical stretch of \( y = \sin x\) by a factor of 2. The graph will have the same period \( 2\pi\) as the parent function but the amplitude (the maximum distance from the midline) will be 2 instead of 1.

Answer:

To graph \( y = 2\sin x \) using transformations:

1. Start with the parent function \( y=\sin x \) (key points: \( (0,0) \), \( (\frac{\pi}{2},1) \), \( (\pi,0) \), \( (\frac{3\pi}{2}, - 1) \), \( (2\pi,0) \), period \( 2\pi \), amplitude 1).
2. Apply a vertical stretch by a factor of 2: multiply the \( y \)-coordinates of the key points of \( y = \sin x \) by 2. New key points: \( (0,0) \), \( (\frac{\pi}{2},2) \), \( (\pi,0) \), \( (\frac{3\pi}{2}, - 2) \), \( (2\pi,0) \).
3. Plot these new key points and draw the sine - wave curve through them. The graph has a period of \( 2\pi \) and an amplitude of 2.