Problem 48 - Precalculus with Calculus Previews

Question

Find the amplitude, period, and phase shift of the given function. Sketch at least one cycle of the graph. $$ y=2 \cos \left(-2 \pi x-\frac{4 \pi}{3}\right) $$

Step-by-Step Solution

Answer

Amplitude: 2, Period: 1, Phase Shift: \(\frac{2}{3}\) to the right.

Step 1: Determine the Amplitude

The amplitude of a cosine function of the form \( y = a \cos(bx - c) \) is given by \( |a| \). Here, \( a = 2 \), so the amplitude is \( |2| = 2 \).

Step 2: Find the Period

The period of a cosine function is given by \( \frac{2\pi}{|b|} \). In this equation, \( b = -2\pi \). Therefore, the period is \( \frac{2\pi}{|-2\pi|} = 1 \).

Step 3: Calculate the Phase Shift

To find the phase shift, solve \( bx - c = 0 \) for \( x \). Here \( b = -2\pi \) and \( c = \frac{4\pi}{3} \), so we set \(-2\pi x - \frac{4\pi}{3} = 0\). Solving for \( x \) gives \( x = -\frac{2}{3} \). Thus, the phase shift is \( \frac{2}{3} \) unit to the right.

Step 4: Sketch One Cycle of the Function

Begin at the phase shift, \( x = \frac{2}{3} \), and sketch one full period, \( 1 \), ending at \( x = \frac{2}{3} + 1 = \frac{5}{3} \). The graph oscillates between \( y = 2 \) and \( y = -2 \), starting at the maximum \( y = 2 \) at \( x = \frac{2}{3} \), reaching \( y = -2 \) at \( x = \frac{4}{3} \), and returning to \( y = 2 \) at \( x = \frac{5}{3} \). Mark these critical points to sketch the curve.

Key Concepts

Amplitude

The amplitude of a trigonometric function is an important aspect that tells us about the vertical stretching or compression of a sine or cosine wave. Specifically, the amplitude refers to the maximum distance from the midline to the peak or the trough of the wave. It's like the "height" of the wave from the middle.
In the general form of the cosine function, represented as \( y = a \cos(bx - c) \), the amplitude is given by \( |a| \). This expression simply means that the amplitude is the absolute value of the coefficient \( a \) in front of the cosine.

• In our specific function \( y = 2 \cos(-2\pi x - \frac{4\pi}{3}) \), we find \( a = 2 \).
• Therefore, the amplitude is \( |2| = 2 \).

This tells us that, regardless of the horizontal shifting or time period of the wave, the peaks will reach 2 units above and 2 units below the middle line of the oscillation. It's essential to note that this "height" is always a positive value due to the absolute value operation.

Period

The period of a trigonometric function indicates how long it takes for the cycle to repeat itself. Imagine seeing waves at the beach; the period is the time or distance between similar points of consecutive waves, like crest to crest.
The formula to find the period of a cosine function \( y = a \cos(bx - c) \) is \( \frac{2\pi}{|b|} \), where \( b \) is the coefficient of \( x \).

In our function \( y = 2 \cos(-2\pi x - \frac{4\pi}{3}) \):

• \( b = -2\pi \), so the period is given by \( \frac{2\pi}{|-2\pi|} \).
• This simplifies to \( \frac{2\pi}{2\pi} = 1 \).

This means that every 1 unit along the x-axis, the wave completes a full cycle. The oscillation pattern of this cosine graph repeats every 1 unit, suggesting the wave is compressed compared to the standard cosine function, which usually has a period of \( 2\pi \). A period of 1 means that this wave has been squeezed tighter, so it completes its oscillation faster.

Phase Shift

Phase shift refers to the horizontal movement of the wave along the x-axis. It’s like shifting the start of the wave either left or right. The phase shift can affect where the peaks and troughs of the wave will appear.
To determine the phase shift of a trigonometric function, use the equation from its general form, \( bx - c = 0 \), and solve for \( x \). In this way, we find where the function begins its cycle.
For our function \( y = 2 \cos(-2\pi x - \frac{4\pi}{3}) \):

• Start with \( -2\pi x - \frac{4\pi}{3} = 0 \).
• Solve for \( x \) to find \( -2\pi x = \frac{4\pi}{3} \), which simplifies to \( x = -\frac{2}{3} \).

Since a positive phase shift moves the graph to the left, and a negative moves it to the right, the calculated \( x = -\frac{2}{3} \) seems rightward.
This displacement is translated as \( \frac{2}{3} \) unit to the right due to the negative \( b \) value. Understanding phase shifts allows you to predict where key points of the wave, like maxima, minima, and intercepts, will occur along the x-axis.