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Magazine Chapter 5: Problem 9
Graph the function. $$g(x)=-\frac{1}{2} \sin x$$
Short Answer
Expert verified
The graph of \( g(x) = -\frac{1}{2} \sin x \) is a sine wave reflected over the x-axis with amplitude \( \frac{1}{2} \).
Step by step solution
01
Understand the Function
The function given is \( g(x) = -\frac{1}{2} \sin x \). This is a sine function with some transformations applied to it. In general, its form is \( a \sin(bx + c) + d \), where \( a \) is the amplitude, \( b \) is the frequency, and other constants \( c \) and \( d \) represent horizontal and vertical shifts, respectively.
02
Identify the Transformations
For \( g(x) = -\frac{1}{2} \sin x \), the amplitude \( a = -\frac{1}{2} \), indicating a vertical stretch and reflection over the x-axis. The function \( \sin x \) has a default period of \( 2\pi \); here, it remains unchanged as \( b = 1 \). There are no horizontal (\( c = 0 \)) or vertical shifts (\( d = 0 \)).
03
Determine the Amplitude
The amplitude of a function \( y = a \sin x \) is the absolute value of \( a \). Here, it is \( |-\frac{1}{2}| = \frac{1}{2} \). This means the graph will oscillate from \(-\frac{1}{2}\) to \(\frac{1}{2}\) along the vertical axis.
04
Plot Key Points
Key points for \( \sin x \) are usually at \( x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \). For \( g(x) = -\frac{1}{2} \sin x \), these become \( (0, 0), \left(\frac{\pi}{2}, -\frac{1}{2}\right), (\pi, 0), \left(\frac{3\pi}{2}, \frac{1}{2}\right), (2\pi, 0) \) due to vertical reflection and scaling.
05
Sketch the Graph
Begin by marking the x-axis with the key points identified: \( x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \). Plot each corresponding y-value as given in Step 4, ensuring the oscillation reflects the amplitude. Since \( a < 0 \), the graph is flipped vertically.
06
Draw the Sine Curve
Connect these points smoothly in a wave-like figure, ensuring that the curve touches each plot point in a smooth, continuous motion. The graph should show one complete cycle of the sine wave from \( x = 0 \) to \( x = 2\pi \) reflecting the negative sign.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Amplitude
The amplitude of a sine function is a crucial aspect because it determines the height of its waves. In a standard sine function like \( y = a \sin(x) \), the amplitude is the absolute value of the constant \( a \). For the function \( g(x) = -\frac{1}{2} \sin x \), the amplitude is \( |-\frac{1}{2}| = \frac{1}{2} \). This means that the wave of the sine function will oscillate between -\( \frac{1}{2} \) and \( \frac{1}{2} \) along the y-axis. Unlike the standard sine function, which ranges from -1 to 1, the graph here is compressed vertically. Understanding amplitude is essential when graphing sine functions because it helps determine the peaks and troughs of the graph. A smaller amplitude results in a flatter graph, while a larger amplitude gives taller waves.
Vertical Reflection
Vertical reflection in a sine function involves flipping the graph over the x-axis. This occurs when the constant \( a \) in \( y = a \sin(x) \) is negative. In our function \( g(x) = -\frac{1}{2} \sin x \), the negative sign before \( \frac{1}{2} \) indicates that the graph will be reflected. This means that every point on the original sine graph has its y-coordinate multiplied by -1.
- A peak in the standard sine function at (\( \frac{\pi}{2}, 1 \)) would become (\( \frac{\pi}{2}, -\frac{1}{2} \)) in the reflected graph.
- A trough at (\( \frac{3\pi}{2}, -1 \)) transforms to (\( \frac{3\pi}{2}, \frac{1}{2} \)).
Graph Transformations
Graph transformations play a vital role in altering the basic structure of the sine function, enabling us to understand how to adjust its appearance. For \( g(x) = -\frac{1}{2} \sin x \), let's explore its transformations:
- Amplitude Transformation: The amplitude change from \( 1 \) to \( \frac{1}{2} \) compresses the graph vertically.
- Vertical Reflection: The negative sign in the function flips the graph across the x-axis.
- Period Consideration: Here, the period remains unchanged due to \( b = 1 \) in the general form, so the cycle completes one round over \( 2\pi \).
