Find a function of the form y = A sin(kx) + C or y = A cos(kx) + C whose graph matches the function shown below: 3- 2 19-18 -17 -16 -15 -14 -13 -11 -10 -8 -7 -6 -5 -3 -2 -1 6 7 8 -2 -3+ Leave your answer in exact form; if necessary, type pi for T. y =
Find a function of the form y = A sin(kx) + C or y = A cos(kx) + C whose graph matches the function shown below: 3- 2 19-18 -17 -16 -15 -14 -13 -11 -10 -8 -7 -6 -5 -3 -2 -1 6 7 8 -2 -3+ Leave your answer in exact form; if necessary, type pi for T. y =
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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![**Understanding Sinusoidal Functions**
**Problem Statement:**
Find a function of the form \( y = A \sin(kx) + C \) or \( y = A \cos(kx) + C \) whose graph matches the function shown below:
[Graph description]
- The x-axis ranges from -19 to 9.
- The y-axis ranges from -3 to 3.
- The graph is a sine curve that appears to move sinusoidally between -3 and 3, starting from a peak at \(x = -15\) and completing cycles.
**Instructions:**
Leave your answer in exact form; if necessary, type \(\pi\) for \(\pi\).
\[ y = \boxed{} \]
---
**Graph Explanation:**
The graph provided shows a sinusoidal function with the following characteristics:
- Amplitude: The graph has a maximum value of 3 and a minimum value of -3. Therefore, the amplitude \(A\) is 3.
- Period: Observing the graph, it completes one cycle from \(x = -15\) to \(x = -5\), a span of 10 units. The period \(T\) is 10.
- The formula for the period of a sine or cosine function is \(\frac{2\pi}{k}\). Solving for \(k\):
\[
T = 10 \implies \frac{2\pi}{k} = 10 \implies k = \frac{2\pi}{10} = \frac{\pi}{5}
\]
- Midline: The graph oscillates symmetrically around the x-axis, indicating the midline value \(C\) is 0.
- Phase shift: Because the function has a peak at \(x = -15\), it suggests that a cosine function might be more straightforward to use with a phase shift.
Thus, the cosine function with the above-stated parameters is:
\[ y = 3 \cos\left(\frac{\pi}{5}x + \phi\right) \]
Given the peak is at \(x = -15\), the phase shift \(\phi\) must satisfy:
\[ -\frac{\pi}{5}(-15) + \phi = 0 \implies 3\pi + \phi = 0 \implies \phi = -3\pi \]
Therefore](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F955ff4f4-16d4-4965-9c48-d9ab4c3976eb%2F6f634d08-825f-4b2a-8223-6c001e6547f1%2Fabu20lr_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding Sinusoidal Functions**
**Problem Statement:**
Find a function of the form \( y = A \sin(kx) + C \) or \( y = A \cos(kx) + C \) whose graph matches the function shown below:
[Graph description]
- The x-axis ranges from -19 to 9.
- The y-axis ranges from -3 to 3.
- The graph is a sine curve that appears to move sinusoidally between -3 and 3, starting from a peak at \(x = -15\) and completing cycles.
**Instructions:**
Leave your answer in exact form; if necessary, type \(\pi\) for \(\pi\).
\[ y = \boxed{} \]
---
**Graph Explanation:**
The graph provided shows a sinusoidal function with the following characteristics:
- Amplitude: The graph has a maximum value of 3 and a minimum value of -3. Therefore, the amplitude \(A\) is 3.
- Period: Observing the graph, it completes one cycle from \(x = -15\) to \(x = -5\), a span of 10 units. The period \(T\) is 10.
- The formula for the period of a sine or cosine function is \(\frac{2\pi}{k}\). Solving for \(k\):
\[
T = 10 \implies \frac{2\pi}{k} = 10 \implies k = \frac{2\pi}{10} = \frac{\pi}{5}
\]
- Midline: The graph oscillates symmetrically around the x-axis, indicating the midline value \(C\) is 0.
- Phase shift: Because the function has a peak at \(x = -15\), it suggests that a cosine function might be more straightforward to use with a phase shift.
Thus, the cosine function with the above-stated parameters is:
\[ y = 3 \cos\left(\frac{\pi}{5}x + \phi\right) \]
Given the peak is at \(x = -15\), the phase shift \(\phi\) must satisfy:
\[ -\frac{\pi}{5}(-15) + \phi = 0 \implies 3\pi + \phi = 0 \implies \phi = -3\pi \]
Therefore
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