Problem 211 - Precalculus

Question

For the following exercises, sketch the graph of each function for two full periods. Determine the amplitude, the period, and the equation for the midline. $$ f(x)=5 \sin x $$

Step-by-Step Solution

Answer

Amplitude = 5, Period = \(2\pi\), Midline = \(y = 0\).

Step 1: Identify the Amplitude

The amplitude of a sine function is the absolute value of the coefficient in front of the sine. In this function, the coefficient is 5, so the amplitude is \[ \text{Amplitude} = 5 \].

Step 2: Determine the Period

The period of a standard sine function \( \sin x \) is \( 2\pi \). Since the function \( f(x) = 5 \sin x \) has no coefficient affecting \( x \), the period remains unchanged. Thus, we have \[ \text{Period} = 2\pi \].

Step 3: Find the Midline

The midline of a sine function is the horizontal line that represents the vertical center of the graph. For \( f(x) = 5 \sin x \), there is no vertical shift, so the midline is \[ y = 0 \].

Step 4: Sketch the Graph

Start by plotting points for one period then replicate for two full periods. The points generally start at the midline at \( x = 0 \), go up to the maximum at \( \frac{\pi}{2} \), return to the midline at \( \pi \), go down to the minimum at \( \frac{3\pi}{2} \), and return back to the midline at \( 2\pi \). Repeat this cycle for another period from \( 2\pi \) to \( 4\pi \).

Step 5: Label the Key Characteristics

Mark the amplitude on the graph as 5 above and below the midline. Indicate the period from \( 0 \) to \( 2\pi \) and \( 2\pi \) to \( 4\pi \). Draw a dashed line at \( y = 0 \) to show the midline.

Key Concepts

Amplitude of Sine Function

The amplitude of a sine function reflects how far the function stretches vertically on a graph. It's essentially the "height" of the wave. For a sine function written as \( f(x) = a \sin(x) \), the amplitude is given by the absolute value of the coefficient \( a \).

Here's the basic rule you need to remember:

• The amplitude of \( f(x) = a \sin(x) \) is \( |a| \).

In the case of \( f(x) = 5 \sin(x) \), the coefficient in front of the sine function is 5. So the amplitude is:

• Amplitude = 5

This means the sine wave will reach up to 5 units above and 5 units below the midline. Imagine this wave stretching as high as 5 and dipping as low as -5 on the graph!

Period of Sine Function

The period of a sine function indicates how long it takes for the function to complete one full cycle. The standard sine function \( \sin(x) \) has a period of \( 2\pi \). This means from \( 0 \) to \( 2\pi \), the wave makes one complete up-and-down motion.

For sine functions of the form \( f(x) = a \sin(bx + c) \), the period can be found using this formula:

• Period = \( \frac{2\pi}{|b|} \)

In our function \( f(x) = 5 \sin(x) \), there's no coefficient modifying \( x \), which means \( b = 1 \). Therefore, the period remains:

• Period = \( 2\pi \)

This tells you the wave pattern repeats every \( 2\pi \) units along the x-axis. For graphing, if you plot from \( 0 \) to \( 2\pi \), you see a full cycle, and from \( 2\pi \) to \( 4\pi \) another full cycle emerges.

Midline of Sine Function

The midline of a sine function is the horizontal line that the curve oscillates around. It acts as a "balance point" for the wave, dividing it into its symmetric halves. In a standard sine function \( f(x) = a \sin(x) + d \), \( d \) represents the vertical shift and determines the midline's position:

• Midline = \( y = d \)

For our function \( f(x) = 5 \sin(x) \), there is no additional constant added, meaning \( d = 0 \). Thus, the midline is:

• Midline = \( y = 0 \)

This midline shows that the sine wave oscillates evenly above and below this line. When graphing, draw a dashed line at \( y = 0 \) to guide your plot and confirm you're accounting for the sine wave's height and depth!