Problem 40 - Precalculus Mathematics for Calculus

Question

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: -1.

Step 1: Identifying Amplitude

The amplitude of a sine function can be found with the coefficient of the sine term in the standard form: \[ y = A \, ext{sin}(Bx + C) + D \]Here, the function is \( y = 3 + 2 \sin(3(x + 1)) \).The amplitude \( A \) is the absolute value of the coefficient of the sine, which is 2. So, the amplitude is 2.

Step 2: Identifying the Period

The period \( T \) of a sine function \( y = A \, ext{sin}(B(x + C/B)) + D \) is given by:\[ T = \frac{2\pi}{B} \]In this problem, \( B = 3 \). Therefore, the period is:\[ T = \frac{2\pi}{3} \]

Step 3: Calculating the Phase Shift

The phase shift \( \phi \) can be determined from the expression inside the sine function, which is \( 3(x + 1) \).The phase shift in standard form \( y = A \, ext{sin}(B(x + C/B)) + D \) is \( -\frac{C}{B} \).Here, \( C = 3 \times 1 = 3 \), so the phase shift is:\[ \phi = -\frac{C}{B} = -\frac{3}{3} = -1 \] The negative indicates a shift to the left of 1 unit.

Step 4: Graphing One Complete Period

Since we have identified the amplitude, period, and phase shift, we can plot the function. The basic shape of the sine wave will be affected by these three parameters:- Amplitude = 2- Period = \( \frac{2\pi}{3} \)- Phase shift = -1 unit (shifted left)The function \( y = 3 + 2 \sin(3(x + 1)) \) needs to be sketched, first finding the starting point (by considering the phase shift), then finding the points at \( 1/4 \), \( 1/2 \), and \( 3/4 \) of the period, adjusting for amplitude and vertical shift by adding D=3.

Key Concepts

Amplitude

The amplitude of a trigonometric function like a sine or cosine wave is essentially its height, or how tall the wave gets from the central line, either up or down. In a trigonometric expression of the form \[ y = A \sin(Bx + C) + D \]the amplitude is given by the absolute value of \( A \), which is the coefficient of the sine function.

• In the function \( y = 3 + 2 \sin(3(x + 1)) \), the amplitude is \( |2| = 2 \).

This means our sine wave will oscillate 2 units above and 2 units below the central line of the wave. Remember, the central line is not always the x-axis. The function's central line has also been shifted vertically by \( D \), which in this case is 3.

Period

The period of a trigonometric function indicates the length it takes for the function to complete one full cycle of its waveform. It is determined by the coefficient within the argument of the sine function. For a function \[ y = A \sin(Bx + C) + D \]the period \( T \) is given by the formula: \[ T = \frac{2\pi}{B} \]

• In our example, \( B = 3 \), so the period is \( T = \frac{2\pi}{3} \).

This means that from starting point to the same point on the wave (completing one full cycle), it spans \( \frac{2\pi}{3} \). This describes the wave's horizontal stretch or compression. A smaller period value compresses the wave horizontally, making it repeat more frequently over the same x interval.

Phase Shift

Phase shift is an adjustment in the wave's position along the x-axis. It tells us where our wave, like the sine curve, starts as opposed to its usual beginning point. In the function\[ y = A \sin(B(x + C/B)) + D \]the phase shift \( \phi \) is computed by: \[ \phi = -\frac{C}{B} \]

• Looking at our example function, inside the sine is \( 3(x + 1) \), which leads to \( C = 3\times 1 = 3 \).
• The phase shift would be \( -\frac{3}{3} = -1 \).

This negative value indicates a shift of the wave to the left by 1 unit. Essentially, the entire graph moves leftward, starting its pattern earlier on the x-axis than it would with no phase shift. Understanding this helps in sketching the graph correctly.