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Chapter 6: Problem 214

For the following exercises, sketch the graph of each function for two full periods. Determine the amplitude, the period, and the equation for the midline. $$ f(x)=5 \sin \left(3\left(x-\frac{\pi}{6}\right)\right)+4 $$

Short Answer

Expert verified
Amplitude: 5, Period: \( \frac{2\pi}{3} \), Midline: \( y=4 \).

Step by step solution

01

Identify the amplitude

The amplitude of a sinusoidal function given by the form \( f(x) = a \sin(b(x-c))+d \) is the absolute value of \( a \). For \( f(x) = 5 \sin(3(x-\frac{\pi}{6}))+4 \), the amplitude is \(|5| = 5\).
02

Determine the period

The period of a sine function \( f(x) = a \sin(b(x-c))+d \) is given by \( \frac{2\pi}{b} \). For this function, \( b = 3 \), so the period is \( \frac{2\pi}{3} \).
03

Find the midline

The midline of a sinusoidal function \( f(x) = a \sin(b(x-c))+d \) is the constant \( d \). For this function, \( d = 4 \). Thus, the equation for the midline is \( y = 4 \).
04

Sketch the graph

To sketch the graph for two full periods:1. Start at the phase shift \( x = \frac{\pi}{6} \).2. The first peak is at \( x = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3} \) with a value of \( 9 \).3. The first zero is at \( x = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2} \).4. The first trough is at \( x = \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3} \) with a value of \( -1 \).5. The pattern repeats every \( \frac{2\pi}{3} \).6. The midline \( y = 4 \) should be drawn as a reference line.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amplitude
In trigonometry, the amplitude of a sinusoidal function is an important concept. It measures the peak deviation of a wave from its midline. Essentially, it tells us how "tall" or "short" the wave is. The amplitude reflects both the maximum and minimum values a wave can reach.
For the function given, \( f(x) = 5 \sin(3(x-\frac{\pi}{6}))+4 \), the amplitude is represented by the absolute value of the coefficient before the sine function, which is \( |5| \).
  • This means the amplitude is \( 5 \).
  • This indicates that the wave reaches 5 units above and below the midline.
Understanding amplitude helps in sketching the graph accurately and predicting the behavior of the wave.
Period
The period of a sinusoidal function indicates how long it takes for the wave to repeat itself. This is crucial in a variety of applications, such as signal processing and sound waves. The period is determined by the coefficient \( b \) in the function \( f(x) = a \sin(b(x-c))+d \).
In our problem, the periodic value \( b \) is \( 3 \), leading to a period of \( \frac{2\pi}{b} \).
  • This results in a period of \( \frac{2\pi}{3} \).
  • The wave repeats its cycle every \( \frac{2\pi}{3} \) units.
By knowing the period, you can accurately sketch two full cycles of the wave.
Sinusoidal Functions
Sinusoidal functions, like sine and cosine, are fundamental in trigonometry. They produce wave-like patterns that are found in various fields, from physics to engineering. A sinusoidal function is generally written as \( f(x) = a \sin(b(x - c)) + d \). Each parameter plays a specific role in shaping the wave:
  • \( a \) determines the amplitude.
  • \( b \) influences the period.
  • \( c \) represents the phase shift.
  • \( d \) designates the vertical shift, creating the midline.
The given function follows this format, allowing us to decipher its elements and draw an accurate graph.
Midline
The midline of a sinusoidal function is the horizontal line around which the wave oscillates. It effectively splits the wave into two symmetric parts, indicating a balance point.In the function \( f(x) = 5 \sin(3(x-\frac{\pi}{6}))+4 \), the midline is derived from the constant \( d \).
  • Here, \( d = 4 \), which gives us a midline represented by the equation \( y = 4 \).
  • This serves as a reference line in graphing, helping to visualize the wave's central position.
The midline assists us in understanding how the wave shifts vertically across the coordinate plane.
Phase Shift
The phase shift in a sinusoidal function is a lateral shift along the x-axis. It affects where the wave starts relative to its standard position. Phase shifts are crucial when comparing different waves or synchronizing signals.
The value \( c \) in the equation \( f(x) = a \sin(b(x-c))+d \) determines the phase shift.
For the function \( f(x) = 5 \sin(3(x-\frac{\pi}{6}))+4 \),
  • The phase shift is \( \frac{\pi}{6} \).
  • This means the wave begins \( \frac{\pi}{6} \) to the right of the origin.
Graphically, the phase shift allows one to pinpoint where wave patterns commence, making it easier to predict and map functions.

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Most popular questions from this chapter

For the following exercises, find the amplitude, period, phase shift, and midline. $$ y=\sin \left(\frac{\pi}{6} x+\pi\right)-3 $$

The line \(y=\frac{-3}{7} x\) passes through the origin in the \(x, y\) -plane. What is the measure of the angle that the line makes with the negative \(x\) -axis?

For the following exercises, graph two full periods of each function and state the amplitude, period, and midine. State the maximum and minimum \(y\) -values and their corresponding \(x\) -values on one period for \(x>0 .\) Round answers to two decimal places if necessary. $$ y=2 \sin (3 x-21)+4 $$

For the following exercises, graph the functions for two periods and determine the amplitude or stretching factor, period, midline equation, and asymptotes. $$ f(x)=-100 \sin (50 x-20) $$

For the following exercises, sketch the graph of each function for two full periods. Determine the amplitude, the period, and the equation for the midline. $$ f(x)=\pi \sec \left(\frac{\pi}{2} x\right) $$
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