Problem 35 - Precalculus: Functions and Graphs

Question

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

Step-by-Step Solution

Answer

Amplitude: \(\sqrt{2}\), Period: \(4\), Phase Shift: \(\frac{1}{2}\) right.

Step 1: Identify the Amplitude

The amplitude of a sine function of the form \(y = a \sin(bx + c)\) is the absolute value of \(a\). In this case, \(a = -\sqrt{2}\). Thus, the amplitude is \(|-\sqrt{2}| = \sqrt{2}\).

Step 2: Determine the Period

The period of a sine function is determined by the coefficient \(b\) and is given by the formula \(\frac{2\pi}{|b|}\). Here, \(b = \frac{\pi}{2}\). Therefore, the period is \(\frac{2\pi}{\frac{\pi}{2}} = 4\).

Step 3: Calculate the Phase Shift

The phase shift is calculated using the formula \(-\frac{c}{b}\), where \(c\) is the constant in the sine function \(y = a \sin(bx + c)\). Here, \(c = -\frac{\pi}{4}\) and \(b = \frac{\pi}{2}\). Thus, the phase shift is \(-\left(-\frac{\pi/4}{\pi/2}\right) = \frac{1}{2}\). The graph is shifted to the right by \(\frac{1}{2}\).

Step 4: Sketch the Graph

Begin by plotting the sine function shifting the basic sine curve horizontally by the phase shift of \(\frac{1}{2}\) units to the right. Then, adjust the amplitude to \(\sqrt{2}\) by stretching the graph vertically. Finally, reflect the graph vertically since there is a negative sign in front of \(\sqrt{2}\). The period is \(4\), so one complete oscillation of the sine wave from start to finish on the x-axis is \(4\) units.

Key Concepts

Amplitude

Amplitude is a crucial measure in trigonometric functions that tells us the height of the waves. It directly impacts how high or low the graph of the function will reach from its central axis. For any sine function of the form

• \( y = a \sin(bx + c) \),

the amplitude is determined by the absolute value of \( a \).
This means we disregard any negative sign.
For example, in the equation \( y = -\sqrt{2} \sin(\frac{\pi}{2}x - \frac{\pi}{4}) \), the coefficient \( a = -\sqrt{2} \).
Thus, the amplitude is \( |-\sqrt{2}| = \sqrt{2} \).
So, the graph will rise \( \sqrt{2} \) units above and dip \( \sqrt{2} \) units below its center.
If \( a \) were positive, the highest point would be slightly above the x-axis, but here, the negative sign indicates an additional reflection over the x-axis.

Period of a Function

The period of a sine function is all about the length of one complete cycle of the wave.
Understanding the period helps determine how quickly or slowly the waves repeat themselves as we move along the x-axis. The period of a sine function in the general form

• \( y = a \sin(bx + c) \)

is calculated by the formula \( \frac{2\pi}{|b|} \).
Here, \( b = \frac{\pi}{2} \), so we plug this into our formula to find the period:

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

This means that every 4 units along the x-axis, the sine wave completes one full cycle.
If the wave's cycle is spread further apart, it implies a slower oscillation, and if the cycle is compressed, it suggests faster oscillations.

Phase Shift

Phase shift describes how far the wave has been horizontally displaced from its usual position.
In a sine function, the phase shift can be calculated with the formula

• \(-\frac{c}{b}\).

For the equation given, \( c = -\frac{\pi}{4} \) and \( b = \frac{\pi}{2} \).
Applying these values, we calculate the phase shift as:

• \(-\left(-\frac{\pi/4}{\pi/2}\right) = \frac{1}{2}\).

This shows that the graph is shifted to the right by \( \frac{1}{2} \) units.
Understanding the phase shift is critical when aligning the wave to its correct starting point on the graph.
It allows us to accurately translate and graph the function, especially when sketching multiple trigonometric functions on the same axes.