Problem 31 - Precalculus: Functions and Graphs

Question

Find the amplitude, the period, and the phase shift and sketch the graph of the equation. \(y=-5 \cos \left(\frac{1}{3} x+\frac{\pi}{6}\right)\)

Step-by-Step Solution

Answer

Amplitude: 5; Period: \(6\pi\); Phase shift: \(-\frac{\pi}{2}\).

Step 1: Understanding Amplitude

The amplitude of a cosine function expressed as \( y = A \cos(Bx + C) \) is given by the absolute value of \( A \). For the function \( y = -5 \cos\left(\frac{1}{3}x + \frac{\pi}{6}\right) \), \( A = -5 \). Hence, the amplitude is \( |A| = |-5| = 5 \).

Step 2: Determining the Period

The period of a cosine function is calculated using \( \frac{2\pi}{|B|} \), where \( B \) is the coefficient of \( x \) in \( y = A \cos(Bx + C) \). Here, \( B = \frac{1}{3} \). Therefore, the period is \( \frac{2\pi}{\left|\frac{1}{3}\right|} = 6\pi \).

Step 3: Calculating the Phase Shift

The phase shift of a cosine function is found by solving \( Bx + C = 0 \). For the function \( y = -5 \cos\left(\frac{1}{3}x + \frac{\pi}{6}\right) \), set \( \frac{1}{3}x + \frac{\pi}{6} = 0 \). Solving gives: \( \frac{1}{3}x = -\frac{\pi}{6} \), and \( x = -\frac{\pi}{6} \times 3 = -\frac{\pi}{2} \). Thus, the phase shift is \( -\frac{\pi}{2} \).

Step 4: Sketching the Graph

Taking into account the amplitude of 5, a period of \( 6\pi \), and a phase shift of \( -\frac{\pi}{2} \), graph the function starting at the point where the cosine wave typically starts a maximal point, shifted \( -\frac{\pi}{2} \) to the left. The graph will be a vertically inverted cosine wave (due to the negative amplitude), starting at its maximum (\( y = 5 \)) at \( x = -\frac{\pi}{2} \), reaching the minimum (\( y = -5 \)) at \( x = \frac{5\pi}{2} \), and completing one cycle at \( 6\pi \) from the starting point.

Key Concepts

Amplitude

The amplitude of a trigonometric function is a measure of its height from the centerline to a peak or a valley of the graph. In simpler terms, it tells us how tall the waves of the function are.

When we look at the cosine function in the form of \( y = A \cos(Bx + C) \), the amplitude is determined by the absolute value of \( A \). This value indicates how far up and down the graph will move from the center line.

In the given function, \( y = -5 \cos\left(\frac{1}{3}x + \frac{\pi}{6}\right) \), the value \( A = -5 \). Since amplitude traditionally represents a distance, and distance cannot be negative, we consider its absolute value. Hence, the amplitude is \(|A| = |-5| = 5\).

This means that the wave peaks at 5 above the center line and dips to 5 below.

Period

The period of a trigonometric function describes how long it takes for the function to complete one full cycle before it starts repeating. Essentially, it tells us the "length" of one wave.

For cosine functions, the period is calculated with the formula \( \frac{2\pi}{|B|} \), where \( B \) is the coefficient of \( x \) within the function \( y = A \cos(Bx + C) \). It helps determine how stretched or compressed the function is along the x-axis.

In our example, \( y = -5 \cos\left(\frac{1}{3}x + \frac{\pi}{6}\right) \), we plug in \( B = \frac{1}{3} \) into the formula to find the period:

• Calculate \( \frac{2\pi}{\left|\frac{1}{3}\right|} = 6\pi \)

Thus, the period of this function is \( 6\pi \), meaning it takes \( 6\pi \) units along the x-axis to complete a cycle. This results in a relatively long wave when graphed.

Phase Shift

Phase shift refers to the horizontal displacement of a trigonometric function on the graph with respect to its usual starting position. It shows us how the wave has shifted left or right from its typical start point.

To calculate the phase shift, we use the expression for a cosine function: \( Bx + C = 0 \). Solving this equation will give us the x-value where the wave starts.

In the problem \( y = -5 \cos\left(\frac{1}{3}x + \frac{\pi}{6}\right) \), we set:

• \( \frac{1}{3}x + \frac{\pi}{6} = 0 \)
• Solve for \( x \): \( \frac{1}{3}x = -\frac{\pi}{6} \)
• \( x = -\frac{\pi}{6} \times 3 = -\frac{\pi}{2} \)

This means the function has experienced a phase shift of \( -\frac{\pi}{2} \).

The cosine wave, which typically starts at a maximum point, will now begin its cycle shifted \( \pi/2 \) units to the left. This impacts how the graph looks on the horizontal axis.