Problem 40 - Algebra and Trigonometry

Question

\(29-42\) . Find the amplitude, period, and phase shift of the function, and graph one complete period. $$ y=3+2 \sin 3(x+1) $$

Step-by-Step Solution

Answer

Amplitude: 2; Period: \(\frac{2\pi}{3}\); Phase shift: \(\frac{1}{3}\) units left.

Step 1: Identifying the Function Type

The given function is in the form of a sine function: \[ y = a + b \sin(c(x - d)) \]where \( a = 3 \), \( b = 2 \), \( c = 3 \), and \( d = -1 \) (since the function is written as \( x + 1 = x - (-1) \)).

Step 2: Finding the Amplitude

The amplitude of a sine function \( y = a + b \sin(c(x - d)) \) is the absolute value of \( b \). Thus, the amplitude is: \[ |b| = |2| = 2 \]

Step 3: Determining the Period

The period of a sine function \( y = a + b \sin(c(x - d)) \) is calculated as \( \frac{2\pi}{c} \). Hence the period is: \[ \frac{2\pi}{3} \]

Step 4: Calculating the Phase Shift

The phase shift of the function is found using the term \( c(x - d) \). It is given by \( \frac{d}{c} \), so substituting \( d = -1 \) and \( c = 3 \) gives:\[ - \frac{-1}{3} = \frac{1}{3} \]This means the graph shifts \( \frac{1}{3} \) units to the **left**.

Step 5: Graphing One Complete Period

Now we graph one complete period of the function starting from the phase shift. The key points of the sine wave occur at specific intervals along the period:1. Starting point: \( \frac{1}{3} \)2. Quarter period: \( \frac{1}{3} + \frac{1}{4} \times \frac{2\pi}{3} = \frac{1}{3} + \frac{\pi}{6} \)3. Half period: \( \frac{1}{3} + \frac{1}{2} \times \frac{2\pi}{3} = \frac{1}{3} + \frac{\pi}{3} \)4. Three-quarter period: \( \frac{1}{3} + \frac{3}{4} \times \frac{2\pi}{3} = \frac{1}{3} + \frac{\pi}{2} \)5. Completing the period: \( \frac{1}{3} + \frac{2\pi}{3} \)The maximum value in this sine curve is \( 5 \) (amplitude 2 added to 3), and the minimum value is \( 1 \) (amplitude 2 subtracted from 3).

Key Concepts

Amplitude

In trigonometry, the amplitude of a function is crucial as it indicates the peak deviation of the function from its central value. For the sine function, which oscillates between a maximum and a minimum, amplitude represents half the distance between these two extreme values.

The function we are examining, \[y = 3 + 2 \sin 3(x + 1)\],
has an amplitude identifiable as the constant multiplying the sine term. In this specific function, the amplitude is the absolute value of 2, which indicates that the sine wave oscillates 2 units above and below its central line.

To find amplitude more generally, remember:

• The formula is \( |b| \), where \( b \) is the coefficient of the sine or cosine function.
• Amplitude does not consider any vertical shifts from adding or subtracting a constant outside the function, as this shifts the whole graph vertically, not changing the amplitude itself.

Understanding amplitude is key to graphing trigonometric functions accurately and predicting their extremes.

Period of a Function

The period of a trigonometric function is the horizontal length required for the function to complete one full cycle of its pattern. For standard sine and cosine functions, the natural period is \( 2\pi \). However, this changes with the horizontal scaling factor associated with the variable.

In our function, \[y = 3 + 2\sin 3(x + 1)\],
the variable \( x \) is multiplied by 3, indicating that each cycle completes \( 3 \) times faster than a standard sine curve. This results in the period being scaled to \( \frac{2\pi}{3} \).

Key things to remember about periods:

• Use the formula \( \frac{2\pi}{c} \), where \( c \) is the multiplier of the \( x \) term.
• Period changes when a function is horizontally stretched or compressed by this scaling factor.

By understanding the function’s period, you can accurately graph several cycles and plot significant points within one cycle.

Phase Shift

The phase shift of a trigonometric function refers to the horizontal translation along the x-axis. It is determined by the phase term in the argument of the sine or cosine function. This shift moves the entire graph to the left or right without altering the shape of the waveform.

For this function, \( y = 3 + 2 \sin 3(x + 1)\),
we resolve the phase shift using the formula based on the expression inside the sine function \( c(x - d) \). Here, \( d \) is replaced by \( -1 \) (because \( x + 1 = x - (-1) \)). The calculated phase shift becomes \[\frac{d}{c} = \frac{-1}{3} = \frac{1}{3}\],
indicating a shift of \( \frac{1}{3} \) units to the left.

Essentials about phase shifts include:

• A positive result indicates a shift to the left; a negative, to the right.
• Always ensure calculations consider the algebraic form accurately.

By identifying phase shifts, you can quickly modify graph positions, ensuring they align with real-world data or theoretical predictions.