QUESTION IMAGE
Question
which of the following is the graph of $y = \sin(0.5x)$? (graphs shown with axes and curves)
Step1: Recall the period formula for \( y = \sin(Bx) \)
The period of \( y = \sin(Bx) \) is \( \frac{2\pi}{|B|} \). For \( y = \sin(0.5x) \), \( B = 0.5 \), so the period is \( \frac{2\pi}{0.5}=4\pi \). Wait, no, wait: \( 0.5=\frac{1}{2} \), so period is \( \frac{2\pi}{\frac{1}{2}} = 4\pi \)? Wait, no, wait, let's recalculate. Wait, \( B = 0.5=\frac{1}{2} \), so period \( T=\frac{2\pi}{|B|}=\frac{2\pi}{\frac{1}{2}} = 4\pi \)? Wait, no, that's not right. Wait, no: the general formula for \( y = A\sin(Bx - C)+D \) has period \( \frac{2\pi}{|B|} \). So for \( y = \sin(0.5x) \), \( B = 0.5 \), so period is \( \frac{2\pi}{0.5}=4\pi \)? Wait, no, 2π divided by 0.5 is 4π? Wait, 2π / 0.5 = 4π? Wait, 0.5 is 1/2, so 2π divided by (1/2) is 4π. Yes. So the period of \( y = \sin(0.5x) \) is \( 4\pi \). Wait, but let's check the standard sine function \( y = \sin(x) \) has period \( 2\pi \). When we have \( y = \sin(Bx) \), if \( B < 1 \), the graph is stretched horizontally, so the period is longer. So for \( B = 0.5 \), the period is \( 4\pi \), which is twice the period of \( y = \sin(x) \). Wait, but in the first graph, the x-axis has -24π, -12π, 0, 12π, 24π. Wait, maybe I made a mistake. Wait, let's re-express \( 0.5x=\frac{1}{2}x \), so the period is \( \frac{2\pi}{\frac{1}{2}} = 4\pi \)? Wait, no, 2π divided by (1/2) is 4π? Wait, 2π / (1/2) = 2π 2 = 4π. Yes. So the period is 4π. So the graph should complete one full cycle over an interval of length 4π. Let's check the first graph: between -24π and -12π is 12π, which would be 3 periods (since 4π per period, 12π / 4π = 3). Wait, maybe the first graph has a period of 12π? No, that can't be. Wait, maybe I messed up the formula. Wait, no: the standard sine function \( y = \sin(x) \) has period \( 2\pi \). If we have \( y = \sin(kx) \), the period is \( \frac{2\pi}{k} \). So if \( k = 0.5 \), then period is \( \frac{2\pi}{0.5}=4\pi \). So the graph of \( y = \sin(0.5x) \) should have a period of 4π, meaning it completes one cycle from 0 to 4π, another from 4π to 8π, etc. Now, looking at the first graph: the x-axis is marked at -24π, -12π, 0, 12π, 24π. The distance between -24π and -12π is 12π, which would be 3 periods (since 4π per period, 12π / 4π = 3). Wait, but maybe the first graph is actually the graph of \( y = \sin(0.5x) \), because the period is longer (stretched), so the waves are more spread out? Wait, no, wait: if \( B = 0.5 \), the graph is stretched horizontally, so the period is longer, so the peaks and troughs are further apart. Wait, the first graph has peaks at -24π, -12π, 0, 12π, 24π? No, the first graph's peaks are at, say, between -24π and -12π, there's a peak, then between -12π and 0, another, etc. Wait, maybe the first graph has a period of 12π? No, that would be if \( k = \frac{2\pi}{12\pi}=\frac{1}{6} \). Wait, maybe I made a mistake. Wait, let's take a point: when x = 0, \( y = \sin(0) = 0 \). When x = 4π, \( y = \sin(0.54π)=\sin(2π)=0 \). When x = 2π, \( y = \sin(0.5*2π)=\sin(π)=0 \). Wait, no: x = 2π, \( y = \sin(π)=0 \). x = π, \( y = \sin(0.5π)=1 \). x = 3π, \( y = \sin(1.5π)= -1 \). So the graph should have a peak at x = π (y=1), trough at x=3π (y=-1), and cross the x-axis at 0, 2π, 4π, etc. Wait, maybe the first graph is the correct one because it's stretched, with a longer period. The second graph looks more compressed, which would be for a larger B (like B=2, period π). So the first graph, with the more spread-out waves (longer period), is the graph of \( y = \sin(0.5x) \).
Step2: Identify the correct graph
Since \( y = \sin(0.5x) \) has a period of \( 4\…
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The upper graph (the first graph with x-axis marked -24π, -12π, 0, 12π, 24π and more spread - out waves)