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Here is a graph of y = sin (0). Describe what y = ½ sin (0) - 2 would differ. Click to add text N2 312 五 SH12 7612 9/2 Remember to use the vocabulary: Amplitude Midline Period

Here is a graph of y = sin (0). Describe what y = ½ sin (0) - 2 would differ. Click to add text N2 312 五 SH12 7612 9/2 Remember to use the vocabulary: Amplitude Midline Period

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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### Understanding the Transformations of the Sine Function **Graphical Analysis** **Title:** Here is a graph of \( y = \sin (\theta) \). Describe what \( y = \frac{1}{2} \sin (\theta) - 2 \) would differ. **Discussion Section:** When comparing the graphs of \( y = \sin(\theta) \) and \( y = \frac{1}{2}\sin(\theta) - 2 \), there are a few key differences to consider. Let's use the concepts of amplitude, midline, and period to better understand these differences: 1. **Amplitude:** - The amplitude of \( y = \sin(\theta) \) is 1, as this is the distance from the midline to the peak or trough of the sine curve. - For \( y = \frac{1}{2}\sin(\theta) - 2 \), the amplitude is \(\frac{1}{2}\). This indicates that the sine wave is vertically compressed, meaning its peaks and troughs are half as far from the midline as those of \( y = \sin(\theta) \). 2. **Midline:** - The midline of \( y = \sin(\theta) \) is \( y = 0 \), which is the horizontal line that runs through the middle of the wave. - For \( y = \frac{1}{2}\sin(\theta) - 2 \), the midline is \( y = -2 \). This means the entire sine wave is shifted down by 2 units. 3. **Period:** - The period of \( y = \sin(\theta) \) and \( y = \frac{1}{2}\sin(\theta) - 2 \) remains the same. The sine function completes one full cycle over an interval length of \( 2\pi \). ### Explanation of the Graphs: **Initial Graph (y = sin θ):** - The given graph represents the function \( y = \sin(\theta) \). It features a sine wave with an amplitude of 1, a midline at \( y = 0 \), and a period of \( 2\pi \). **Transformed Graph (y = ½ sin θ - 2):** - The transformed function \( y = \frac{1}{2}\sin(\theta)
Transcribed Image Text:### Understanding the Transformations of the Sine Function **Graphical Analysis** **Title:** Here is a graph of \( y = \sin (\theta) \). Describe what \( y = \frac{1}{2} \sin (\theta) - 2 \) would differ. **Discussion Section:** When comparing the graphs of \( y = \sin(\theta) \) and \( y = \frac{1}{2}\sin(\theta) - 2 \), there are a few key differences to consider. Let's use the concepts of amplitude, midline, and period to better understand these differences: 1. **Amplitude:** - The amplitude of \( y = \sin(\theta) \) is 1, as this is the distance from the midline to the peak or trough of the sine curve. - For \( y = \frac{1}{2}\sin(\theta) - 2 \), the amplitude is \(\frac{1}{2}\). This indicates that the sine wave is vertically compressed, meaning its peaks and troughs are half as far from the midline as those of \( y = \sin(\theta) \). 2. **Midline:** - The midline of \( y = \sin(\theta) \) is \( y = 0 \), which is the horizontal line that runs through the middle of the wave. - For \( y = \frac{1}{2}\sin(\theta) - 2 \), the midline is \( y = -2 \). This means the entire sine wave is shifted down by 2 units. 3. **Period:** - The period of \( y = \sin(\theta) \) and \( y = \frac{1}{2}\sin(\theta) - 2 \) remains the same. The sine function completes one full cycle over an interval length of \( 2\pi \). ### Explanation of the Graphs: **Initial Graph (y = sin θ):** - The given graph represents the function \( y = \sin(\theta) \). It features a sine wave with an amplitude of 1, a midline at \( y = 0 \), and a period of \( 2\pi \). **Transformed Graph (y = ½ sin θ - 2):** - The transformed function \( y = \frac{1}{2}\sin(\theta)
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