Problem 19 - Precalculus: Functions and Graphs

Question

Find the amplitude, the period, and the phase shift and sketch the graph of the equation. \(y=\sin \left(\frac{1}{2} x-\frac{\pi}{3}\right)\)

Step-by-Step Solution

Answer

Amplitude: 1, Period: \(4\pi\), Phase shift: \(\frac{2\pi}{3}\) right.

Step 1: Identify the Amplitude

In the equation of the form \(y = a \sin(bx - c)\), the amplitude is given by the absolute value of \(a\). In this problem, since the function is \(y = \sin \left(\frac{1}{2} x - \frac{\pi}{3}\right)\), the coefficient of \(\sin\) (which is \(1\)) is the amplitude. Therefore, the amplitude is \(1\).

Step 2: Determine the Period

The period of a sine function \(y = a \sin(bx - c)\) is given by \(\frac{2\pi}{b}\). Here, \(b = \frac{1}{2}\), so substituting in gives us the period: \[ \frac{2\pi}{\frac{1}{2}} = 4\pi. \] Thus, the period of the function is \(4\pi\).

Step 3: Find the Phase Shift

The phase shift in a function \(y = a \sin(bx - c)\) is given by \(\frac{c}{b}\). Here, \(c = \frac{\pi}{3}\) and \(b = \frac{1}{2}\). Substituting these values gives: \[ \text{Phase Shift} = \frac{\frac{\pi}{3}}{\frac{1}{2}} = \frac{2\pi}{3}. \] Since the equation is subtracted by \(\frac{\pi}{3}\), the phase shift is to the right by \(\frac{2\pi}{3}\).

Step 4: Sketch the Graph

To sketch the graph, start by plotting the sine function with the characteristics found: Amplitude \(1\), Period \(4\pi\), and phase shift to the right by \(\frac{2\pi}{3}\). Begin with the regular sine function, \(rac{1}{2}x\), stretched over \(4\pi\) and horizontally shifted to the right by \(\frac{2\pi}{3}\). The graph will oscillate between \(1\) and \(-1\) for each cycle over \(4\pi\).

Key Concepts

Amplitude

In trigonometric functions, especially sinusoidal ones like sine and cosine, the amplitude is a crucial concept to understand. Amplitude refers to how much a wave oscillates above and below its central axis, essentially measuring the "height" of the wave. In the general equation of a sine function, which is formulated as

• \(y = a\sin(bx - c)\)

the amplitude is given by the absolute value of the coefficient \(a\). For the given function

• \(y = \sin \left(\frac{1}{2} x - \frac{\pi}{3}\right)\),

we notice that there is an implicit coefficient \(a = 1\) (since there is no specific number preceding the sine function). This means that the wave will have a maximum amplitude, rising to 1 and a minimum amplitude, falling to -1.
This amplitude describes the extent of vertical displacement from the x-axis, defining how "tall" or "short" the wave will appear on a graph.

• Amplitude = Absolute value of \(a\)

• For our function, the amplitude is \(|1| = 1\).

Period

The period of a trigonometric function describes how frequently the wave pattern repeats itself over the horizontal axis. It can be seen as the "length" before the function starts a new cycle. In the standard form of a sine function,

• \(y = a \sin(bx - c)\)

the period is determined by the coefficient \(b\), and calculated as:

• \(\text{Period} = \frac{2\pi}{b}\)

For our specific function, where

• \(b = \frac{1}{2}\),

we substitute to find:

• \(\text{Period} = \frac{2\pi}{\frac{1}{2}} = 4\pi\).

This result means that the entire sine wave pattern in this equation will repeat every \(4\pi\) along the x-axis. Simply put, after covering a horizontal range of \(4\pi\), the sine function’s cycle will recommence from its starting point. Understanding the period is essential in sketching and analyzing trigonometric functions as it directly influences the horizontal stretching of the graph.

Phase Shift

Phase shift refers to the horizontal translation of a trigonometric function along the x-axis. Essentially, it tells us where the wave begins relative to the standard position of the function. In the formula

• \(y = a\sin(bx - c)\)

the phase shift is computed by the equation:

• \(\text{Phase Shift} = \frac{c}{b}\)

With our function

• \(y = \sin \left(\frac{1}{2} x - \frac{\pi}{3}\right)\),

we identify

• \(c = \frac{\pi}{3}\)
• \(b = \frac{1}{2}\)

and substitute these into the equation to find:

• \(\text{Phase Shift} = \frac{\frac{\pi}{3}}{\frac{1}{2}} = \frac{2\pi}{3}\).

The positive value indicates that the function is shifted to the right. Thus, by \(\frac{2\pi}{3}\). This shift means that each point on the standard sine wave is moved \(\frac{2\pi}{3}\) units to the right along the x-axis, altering the function's usual starting position. Phase shifts are vital in graphing, as they determine where the cycle begins and affect the alignment of critical points such as peaks and troughs with the x-axis.