**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

Answered: Image /qna-images/answer/6f634d08-825f-4b2a-8223-6c001e6547f1.jpg

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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Question

**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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