y= sin x/2, 0 less than equal to x less than equal to 4pi... the question says state the amplitude and period for each equation and graph it over the indicated interval

since period of sin(x) is 2pi, period of sin(x/2) is such that 0 <= x/2 <= 2pi, or 0 <= x <= 4pi

amplitude is unchanged from sin(x)

To determine the amplitude and period of the equation y = sin(x/2), where 0 ≤ x ≤ 4π, we can analyze the general form of the equation: y = A sin(Bx + C) + D.

Comparing this form with the given equation, we see that A = 1, B = 1/2, C = 0, and D = 0.

The amplitude, A, represents the maximum vertical distance between the sinusoidal function and its midline. Since A = 1, the amplitude is 1.

The period, P, of a sinusoidal function is the distance between two consecutive maximum or minimum values. In this case, B = 1/2, so the period, P, will be 2π/B.

P = 2π/(1/2) = 4π.

Therefore, the amplitude is 1, and the period is 4π.

Now, let's graph the equation over the indicated interval (0 ≤ x ≤ 4π):

1. Plot the points (0, 0), (π, 0), (2π, 0), and (3π, 0). These are the x-values where the sin(x/2) function intersects the x-axis.

2. Since the amplitude is 1, we can draw a dashed line at y = 1 and y = -1.

3. To complete the graph, you can sketch a smooth curve that oscillates between the dashed lines, intersecting the x-axis at the points mentioned in step 1.

Here's a rough sketch of the graph:

|
1 -|-
|
0 -|-
|
-1 -|-
|
----------|----------
0 2π 4π

Please note that the scale used in this text-based response may not accurately represent the actual plot. It's always recommended to use graphing software or a graphing calculator for precise graphing results.

To determine the amplitude and period of the equation y = sin(x/2), we can compare it to the standard form of a sine function, y = A*sin(Bx), where A represents the amplitude (peak value) and B determines the period (length of one cycle).

In this case, the given equation is y = sin(x/2). Comparing it to the standard form, we can see that A = 1 (since there is no coefficient multiplying the sin function) and B = 1/2 (since x is divided by 2 inside the sin function).

So, the amplitude (A) is 1, and the period (P) can be calculated using the formula P = 2π/B. In this case, B = 1/2, so the period is P = 2π/(1/2) = 4π.

Therefore, for the equation y = sin(x/2), the amplitude is 1, and the period is 4π.

To graph this equation over the indicated interval of 0 ≤ x ≤ 4π, we can plot points on the graph by substituting different values of x into the equation and calculating the corresponding y-values.

Using the interval 0 ≤ x ≤ 4π, we can start by substituting x = 0 into the equation:
y(0) = sin(0/2) = sin(0) = 0
So, we have the point (0, 0).

Next, we can choose some other values of x within the interval and calculate the corresponding y-values to plot more points. For example, let's choose x = π and x = 2π:
y(π) = sin(π/2) = 1
So, we have the point (π, 1).

y(2π) = sin(2π/2) = sin(π) = 0
So, we have the point (2π, 0).

By repeating this process with more values of x within the given interval, we can plot multiple points on the graph.

Connecting the plotted points will result in a sinusoidal graph that oscillates between an amplitude of 1 and the x-axis, with a period of 4π.