Problem 36 - Precalculus: Functions and Graphs

Question

Find the amplitude, the period, and the phase shift and sketch the graph of the equation. \(y=\sqrt{3} \cos \left(\frac{\pi}{4} x-\frac{\pi}{2}\right)\)

Step-by-Step Solution

Answer

Amplitude: \(\sqrt{3}\); Period: 8; Phase shift: 2 units right.

Step 1: Identify the standard form

The equation given is \( y = \sqrt{3} \cos \left( \frac{\pi}{4} x - \frac{\pi}{2} \right) \). The standard form of a cosine function is \( y = a \cos(bx - c) + d \). Comparing with the given equation, we identify: \( a = \sqrt{3} \), \( b = \frac{\pi}{4} \), \( c = \frac{\pi}{2} \), and \( d = 0 \).

Step 2: Find the amplitude

The amplitude of a cosine function is the absolute value of \( a \). Given that \( a = \sqrt{3} \), the amplitude is \( |\sqrt{3}| = \sqrt{3} \).

Step 3: Calculate the period

The period of a cosine function is given by \( \frac{2\pi}{b} \). Since \( b = \frac{\pi}{4} \), the period is calculated as \( \frac{2\pi}{\frac{\pi}{4}} = 2\pi \times \frac{4}{\pi} = 8 \).

Step 4: Determine the phase shift

The phase shift of a cosine function is calculated as \( \frac{c}{b} \). With \( c = \frac{\pi}{2} \) and \( b = \frac{\pi}{4} \), the phase shift is \( \frac{\frac{\pi}{2}}{\frac{\pi}{4}} = \frac{4\pi}{2\pi} = 2 \). This means the graph shifts 2 units to the right.

Step 5: Sketch the graph

Using the information obtained: amplitude \( \sqrt{3} \), period \( 8 \), and phase shift 2 units to the right, draw the cosine curve. Start drawing from \(x = 2\) (the phase shift), complete one period up to \(x = 10\), with the maximum value at \(\sqrt{3}\), the minimum value at \(-\sqrt{3}\), crossing the x-axis halfway between these points.

Key Concepts

Amplitude

The amplitude of a function gives us information about the vertical stretch or compression. In a cosine function like this one, the amplitude defines the maximum deviation from the center (or equilibrium position) of the wave. It is easy to identify in the equation by looking at the coefficient in front of the cosine term.

• In our equation, this coefficient is labeled as "\(a\)".
• To find the amplitude, we take the absolute value of \(a\), which means disregarding any negative sign, if it were there.
• For the given function \( y = \sqrt{3} \cos \left( \frac{\pi}{4}x - \frac{\pi}{2} \right) \), the value of \(a\) is \(\sqrt{3}\).

Therefore, the amplitude is \( \sqrt{3} \). This tells us that the highest point of the cosine wave reaches \( \sqrt{3} \) units above the horizontal center line, and the lowest point reaches \( \sqrt{3} \) units below it.

Period

The period of a cosine function tells us how long it takes for the function to complete one full cycle. For trigonometric functions like cosine, the period is largely influenced by the coefficient in front of the variable \(x\), within the cosine function.

• For a standard cosine function of the form \(y = a\cos(bx)\), the period is calculated using the formula \( \frac{2\pi}{b} \).
• Here, \(b\) is the coefficient of \(x\) inside the cosine function.
• In our problem, the equation \( y = \sqrt{3} \cos \left( \frac{\pi}{4}x - \frac{\pi}{2} \right) \) gives us \(b = \frac{\pi}{4}\).

Therefore, the period is computed as:\[\frac{2\pi}{\frac{\pi}{4}} = 8\]This means that the cosine wave completes one full cycle over an interval of 8 units along the x-axis.

Phase Shift

Phase shift tells us about horizontal shifts in the graph of a trigonometric function like cosine. It determines how the graph is shifted left or right from its usual position along the x-axis.

• For a standard function described by \(y = a \cos(bx - c)\), the phase shift can be calculated by the formula \( \frac{c}{b} \).
• In the given equation \( y = \sqrt{3} \cos \left( \frac{\pi}{4}x - \frac{\pi}{2} \right) \), we have \(c = \frac{\pi}{2}\) and \(b = \frac{\pi}{4}\).

Computing the phase shift gives us:\[\frac{\frac{\pi}{2}}{\frac{\pi}{4}} = 2\]This means that the entire graph of the function is shifted 2 units to the right. Horizontal shifts alter where the cosine wave begins, in this case making it start at \(x = 2\) rather than at \(x = 0\).