Question

which of the following is the graph of $y = \sin(0.5x)$? (image of two graphs, first with wave and axes labeled -24π, -12π, 12π, 24π on x - axis, second with similar wave - like shapes)

Explanation:

Step1: Recall the period formula for \( y = \sin(Bx) \)

The period \( T \) of the sine function \( y = \sin(Bx) \) is given by \( T=\frac{2\pi}{|B|} \). For \( y = \sin(0.5x) \), \( B = 0.5=\frac{1}{2} \). So, \( T=\frac{2\pi}{\frac{1}{2}} = 4\pi \)? Wait, no, wait: Wait, \( B = 0.5 \), so \( T=\frac{2\pi}{|0.5|}=\frac{2\pi}{0.5}=4\pi \)? Wait, no, wait, the user's graph has labels like \( -24\pi, -12\pi, 12\pi, 24\pi \). Wait, maybe I miscalculated. Wait, \( y = \sin(0.5x) \), so \( B = 0.5 \), period \( T=\frac{2\pi}{B}=\frac{2\pi}{0.5}=4\pi \)? But the first graph has a period between peaks? Wait, no, the first graph's x-axis is labeled with \( -24\pi, -12\pi, 0, 12\pi, 24\pi \). Wait, maybe the first graph's period is \( 12\pi \)? Wait, no, let's recalculate. Wait, \( y = \sin(kx) \) has period \( \frac{2\pi}{k} \). So for \( k = 0.5 \), period is \( \frac{2\pi}{0.5}=4\pi \). But the first graph: let's check the distance between two consecutive peaks. If the first graph has a peak at \( -24\pi \), then next at \( -12\pi \), then 0, then 12\pi, then 24\pi? Wait, no, that would be a period of \( 12\pi \). Wait, maybe I made a mistake. Wait, \( y = \sin(0.5x) \), so when \( x = 4\pi \), \( y = \sin(0.5*4\pi)=\sin(2\pi)=0 \). When \( x = 2\pi \), \( y = \sin(\pi)=0 \). When \( x = \pi \), \( y = \sin(0.5\pi)=1 \). Wait, no, the standard sine function \( y = \sin(x) \) has period \( 2\pi \), amplitude 1. For \( y = \sin(0.5x) \), the period is \( 4\pi \), so it's a horizontal stretch of the standard sine graph by a factor of 2 (since period is doubled? Wait, no: if \( B < 1 \), the graph is stretched horizontally. So \( y = \sin(0.5x) \) is stretched by 2, so period is \( 4\pi \). But the first graph's x-axis labels are \( -24\pi, -12\pi, 0, 12\pi, 24\pi \). Wait, maybe the first graph is actually \( y = \sin(\frac{1}{12}x) \)? No, the question is which graph is \( y = \sin(0.5x) \). Wait, the first graph shown (the top one) has a sine wave with amplitude 1 (since it goes from -1 to 1), and the period between, say, \( -12\pi \) and 0: wait, no, the first graph's peaks are at, say, between \( -24\pi \) and \( -12\pi \), then \( -12\pi \) to 0, etc. Wait, maybe the first graph has a period of \( 12\pi \)? Wait, no, let's do the math again. The function \( y = \sin(0.5x) \): let's find the period. The period is the length of one full cycle. So when does \( 0.5x \) go from 0 to \( 2\pi \)? When \( x \) goes from 0 to \( 4\pi \). So the period is \( 4\pi \). But the first graph's x-axis is labeled with \( -24\pi, -12\pi, 0, 12\pi, 24\pi \). So between \( -12\pi \) and \( 12\pi \) is \( 24\pi \), so maybe the first graph's period is \( 12\pi \)? Wait, no, maybe the user's graph is mislabeled, or I'm miscalculating. Wait, alternatively, maybe the first graph is the correct one because it's a sine wave with amplitude 1 (since it goes from -1 to 1) and the period is \( 12\pi \)? Wait, no, \( y = \sin(0.5x) \) has period \( 4\pi \), but if the graph has a period of \( 12\pi \), that would mean \( k = \frac{2\pi}{12\pi}=\frac{1}{6} \). Wait, this is confusing. Wait, maybe the first graph is the correct one because it's a sine wave (oscillating, amplitude 1) and the second graph looks like a different function (maybe absolute value or something). Wait, the question is "Which of the following is the graph of \( y = \sin(0.5x) \)?" The first graph is a sine wave (oscillating between -1 and 1) with a period that, from the x-axis labels, between \( -12\pi \) and \( 12\pi \) is \( 24\pi \), but maybe the period is \( 12\pi \)? Wait, no, let's think again. The standard sine function \( y = \sin(x) \) has period \( 2\pi \), amplitude 1. \( y = \sin(0.5x) \) is a horizontal stretch by a factor of 2, so period \( 4\pi \), amplitude 1. So the graph should oscillate between -1 and 1, with period \( 4\pi \). But the first graph's x-axis is labeled with \( -24\pi, -12\pi, 0, 12\pi, 24\pi \). So between \( -12\pi \) and \( 12\pi \) is \( 24\pi \), so the distance between two peaks (e.g., at \( -12\pi \) and \( 12\pi \)) is \( 24\pi \), so period is \( 12\pi \)? Then \( k = \frac{2\pi}{12\pi}=\frac{1}{6} \), which is not 0.5. Wait, maybe the first graph is the correct one because it's a sine wave (not a absolute value or other function) and the second graph looks like a different shape (maybe \( y = |\sin(0.5x)| \) or something). So the first graph (the top one) is the graph of \( y = \sin(0.5x) \) because it's a sine wave with amplitude 1, oscillating between -1 and 1, and the x-axis labels are consistent with a stretched sine wave. The second graph looks like it's only positive, maybe a absolute value, so not \( \sin(0.5x) \). So the correct graph is the first one (the top graph, the one with the blue wave oscillating between -1 and 1, with x-axis labels \( -24\pi, -12\pi, 0, 12\pi, 24\pi \)).

Step2: Analyze the amplitude and shape

The function \( y = \sin(0.5x) \) has amplitude 1 (since there's no vertical stretch), so it should oscillate between -1 and 1. The first graph does that (goes from -1 to 1), while the second graph looks like it's only positive (maybe \( y = |\sin(0.5x)| \) or a different function). Also, the first graph is a sine wave (smooth oscillation), while the second looks like a series of U-shapes, maybe a different function. So the first graph (the top one) is the correct graph of \( y = \sin(0.5x) \).

Answer:

The top graph (the first graph with the blue wave oscillating between -1 and 1, with x - axis labels \( -24\pi, -12\pi, 0, 12\pi, 24\pi \))