Problem 60 - Precalculus

Question

In Exercises \(49-60\), state the amplitude, period, and phase shift (including direction) of the given function. $$y=-5 \sin [-\pi(x+1)]$$

Step-by-Step Solution

Answer

Amplitude: 5, Period: 2, Phase Shift: Left 1 unit.

Step 1: Identify the General Form of the Sine Function

The general form of a sine function is \( y = a \sin(bx + c) + d \), where:\( a \) determines the amplitude, \( b \) affects the period, \( c \) is used for the phase shift, and \( d \) is the vertical shift.The given function is \( y = -5 \sin(-\pi(x+1)) \). We aim to rewrite and identify these parameters.

Step 2: Rewrite the Function for Easier Analysis

We first distribute the \(-\pi\) inside the sine function:\[\sin(-\pi(x + 1)) = \sin(-\pi x - \pi)\]Using the identity \( \sin(-x) = -\sin(x) \), we have:\[ y = -5 \cdot -\sin(\pi x + \pi) = 5 \sin(\pi x + \pi) \]Now, the function looks like \( 5 \sin(\pi x + \pi) \). This makes identification of parameters straightforward.

Step 3: Determine the Amplitude

The amplitude is the absolute value of \( a \). In the function \( 5 \sin(\pi x + \pi) \), \( a = 5 \). Thus, the amplitude is \( |5| = 5 \).

Step 4: Determine the Period

The period of a sine function is given by \( \frac{2\pi}{b} \). From \( 5 \sin(\pi x + \pi) \), \( b = \pi \).Thus, the period is \( \frac{2\pi}{\pi} = 2 \).

Step 5: Determine the Phase Shift

The phase shift is calculated by \( -\frac{c}{b} \). Here, \( c = \pi \) and \( b = \pi \), so the phase shift is \( -\frac{\pi}{\pi} = -1 \).Thus, the function shifts to the left by 1 unit.

Key Concepts

Amplitude

When you come across a sine function, one of the first things to identify is the amplitude. The amplitude of a sine function reflects how far the graph of the function swings from its central axis, or in simpler terms, how 'tall' or 'short' the waves of the function are.

In any standard sine function of form \( y = a \sin(bx + c) + d \), the amplitude is given by the absolute value of the coefficient \( a \). So, for \( y = -5 \sin(-\pi(x+1)) \), we focus on the coefficient of the sine term, which is initially \( -5 \).

• The absolute value of \( -5 \) is \( 5 \).
• Thus, the amplitude is \( 5 \), meaning the graph stretches from \( -5 \) to \( 5 \) around the central axis.

Recognizing the amplitude helps gauge the extent of the waves created as the sine function oscillates.

Period

Another key characteristic of a trigonometric function like sine is its period. The period determines how frequently the waves repeat over a given interval. For the general sine equation \( y = a \sin(bx + c) + d \), the period can be calculated as \( \frac{2\pi}{b} \).

Looking at our function \( y = 5 \sin(\pi x + \pi) \):

• Here, \( b = \pi \).
• The period becomes \( \frac{2\pi}{\pi} = 2 \).

This means that the sine wave will complete one full cycle every 2 units along the x-axis. Knowing the period is helpful in graphing the function efficiently since it informs us where the wave starts and ends on the x-axis.

Phase Shift

Phase shift refers to the horizontal movement of the function along the x-axis. In simpler terms, it tells us where the function begins its cycle compared to the standard sine wave which starts at the origin.

In the equation \( y = a \sin(bx + c) + d \), phase shift is computed using the formula \( -\frac{c}{b} \). For our function, \( y = 5 \sin(\pi x + \pi) \):

• \( c = \pi \) and \( b = \pi \).
• The phase shift is \( -\frac{\pi}{\pi} = -1 \).

This indicates the entire graph is shifted 1 unit to the left. Understanding phase shift is crucial when plotting or interpreting the sine function, as it helps in adjusting the starting point of the function's waveform on the graph.