Question

select the graph of $y = \sin x$.

Explanation:

Step1: Recall the properties of \( y = \sin x \)

The function \( y=\sin x \) has a period of \( 2\pi \), passes through the origin \((0,0)\) (since \( \sin 0 = 0 \)), has a range of \([-1, 1]\), and its graph starts at the origin, rises to a maximum at \( x=\frac{\pi}{2} \), falls to a minimum at \( x = \frac{3\pi}{2} \), and repeats.

Step2: Analyze each option

• Option A: The graph does not pass through the origin (starts below the origin), so it's not \( y = \sin x \).
• Option B: The graph passes through the origin \((0,0)\), has the correct shape (rises after 0, peaks at \( \frac{\pi}{2} \), etc.), matching the properties of \( y=\sin x \).
• Option C: The graph starts at \( (0,1) \), which is the graph of \( y = \cos x \) (since \( \cos 0 = 1 \)), not \( y=\sin x \).
• Option D: The graph starts below the origin and has a different phase, not matching \( y = \sin x \).

Answer:

B. The graph of \( y = \sin x \) (the middle graph with the sine - wave passing through the origin)