Problem 8 - Algebra 2

Question

State the vertical shift, equation of the midline, amplitude, and period for each function. Then graph the function. $$ y=\sin \theta+0.25 $$

Step-by-Step Solution

Answer

Vertical shift: 0.25; Midline: \(y = 0.25\); Amplitude: 1; Period: \(2\pi\).

Step 1: Identify the Vertical Shift

The given function is \( y = \sin \theta + 0.25 \). The vertical shift is the amount added to the basic sine function, which in this case is \(+0.25\). This indicates the graph is shifted 0.25 units upwards.

Step 2: Determine the Equation of the Midline

The midline of a trigonometric function is the horizontal line that runs through the middle of its oscillation.For the function \( y = \sin \theta + 0.25 \), the midline is \( y = 0.25 \), since the vertical shift moves the midline of the standard sine function up by 0.25 units.

Step 3: Find the Amplitude

The amplitude is the height from the midline to the maximum (or minimum) value of the function.Since there is no coefficient in front of the \(\sin\) function other than the implicit 1, the amplitude remains 1, as in the standard sine function \( y = \sin \theta \).

Step 4: Determine the Period

The period of a sine function \( y = a \sin(b\theta + c) + d \) is calculated as \( \frac{2\pi}{b} \).For the given function \( y = \sin \theta + 0.25 \), \( b = 1 \), so the period is \( 2\pi \).

Step 5: Graph the Function

Use the information gathered:- Vertical shift: 0.25 units up- Midline: \( y = 0.25 \)- Amplitude: 1- Period: \( 2\pi \)Start by drawing the midline as a horizontal line at \( y = 0.25 \). Then, plot the sine function considering the amplitude of 1 from this midline (oscillating between 1.25 and -0.75). Each cycle repeats every \( 2\pi \).

Key Concepts

Vertical Shift

In trigonometric functions, the vertical shift represents how much the graph of the function has been moved up or down from its original position. When you see an equation like \( y = \sin \theta + 0.25 \), the term \(+0.25\) is the vertical shift. This means every point on the basic sine wave is moved 0.25 units upwards.

Understanding vertical shifts is crucial because:

• It affects the starting point of the sine wave.
• It determines the new position of the midline.
• It adjusts the range of the calculations.

In this exercise, since the shift is positive, all points on the graph are elevated by 0.25 units, altering the placement of the peaks, troughs, and equilibrium point accordingly.

Amplitude

The amplitude of a trigonometric function is the measure of how far the wave extends above and below its midline. Specifically, amplitude reflects the distance from the midline to the maximum or minimum points of the graph. For the function \( y = \sin \theta + 0.25 \), the amplitude remains 1.

This is determined by the coefficient of the sine function. With no additional coefficient present (other than the implicit 1), the graph stretches to a height of 1 unit above and 1 unit below the midline. Amplitude is important because:

• It influences the "height" of the output.
• Higher amplitude results in steeper waves.
• It does not affect the period or horizontal shift of the wave.

Thus, despite the vertical shift, the amplitude remains unchanged, ensuring that the oscillation reaches both 1.25 and -0.75 relative to the new midline of the graph.

Midline of a Function

The midline of a trigonometric function is essentially the central baseline or axis around which the function oscillates. It's akin to the average value of the wave over its cycle. In general, for a transformed sine function like \( y = \sin \theta + 0.25 \), the equation of the midline is directly given by the vertical shift value.

As such, the midline for this function is \( y = 0.25 \). This means:

• The waveform oscillates symmetrically above and below this horizontal line.
• It's used to calculate the amplitude and the values range up to \( y = 1.25 \) and down to \( y = -0.75 \).
• It gives insight into the average vertical displacement of the graph.

Identifying the midline is vital for accurately plotting the trigonometric function and understanding its behavior in relation to transformations it undergoes.

Period of a Function

The period of a function describes the length of one complete cycle of the wave. For trigonometric functions, this is especially significant as it tells how often the cycle repeats over a certain interval. The standard sine function \( y = \sin \theta \) has a period of \( 2\pi \).

Thus, for \( y = \sin \theta + 0.25 \), since the coefficient of \( \theta \) (b) remains 1, the period remains unchanged at \( 2\pi \). The formula used for calculating the period is \( \frac{2\pi}{b} \). This signifies:

• Every \( 2\pi \) interval, the function completes a full oscillation, returning to the starting point.
• It remains consistent, unaffected by vertical shifts or amplitude changes.
• Essentially defines how "stretched" or "compressed" the waveform appears horizontally.

In summary, the period helps in plotting the function correctly by knowing the intervals over which the function repeats, aiding in the consistent pattern recognition of the wave.