Help with trigonometry

Question

To get a better look at the graph, you can click on it.The curve above is the graph of a sinusoidal function. It goes through the points (−7,−3)  and (3,−3)  . Find a sinusoidal function that matches the given graph. If needed, you can enter π  =3.1416... as 'pi' in your answer, otherwise use at least 3 decimal digits.

 

https://webwork.latech.edu/webwork2_files/tmp/Winter201516-Math240-Walters//gif/gvr001-2631-setstewart7e_ch05s03prob12image1.png

 

Please explain the process to me. Thanks in advance.

 

 

Answer 1

To find a sinusoidal function that matches the given graph and goes through the points (−7,−3)(-7, -3)(−7,−3) and (3,−3)(3, -3)(3,−3), follow these steps:

Step 1: Determine the Midline

The midline is the horizontal line that represents the average value of the sinusoidal function. Since the given points both have y=−3y = -3y=−3, and they seem to be at the minimum of the curve (assuming the sinusoid is symmetric about this point), the midline is y=−3y = -3y=−3.

Step 2: Determine the Amplitude

The amplitude is the distance from the midline to the maximum or minimum value of the function. Without the maximum or minimum explicitly given, we assume the midline is the reference. If we assume the maximum (since it is not given directly), let’s consider the distance to the midline.

Step 3: Determine the Period

The period is the distance between two consecutive points where the function has the same value and the same slope direction. Since the points (−7,−3)(-7, -3)(−7,−3) and (3,−3)(3, -3)(3,−3) are 10 units apart and they represent a full cycle of the sinusoid:

Period=10 units\text{Period} = 10 \text{ units}Period=10 units

Step 4: Write the General Form

A sinusoidal function can be written in the form:

y=Asin⁡(B(x−C))+Dy = A \sin(B(x - C)) + Dy=Asin(B(x−C))+D

or

y=Acos⁡(B(x−C))+Dy = A \cos(B(x - C)) + Dy=Acos(B(x−C))+D

Where:

1. AAA is the amplitude,
2. BBB is related to the period by B=2πPeriodB = \frac{2\pi}{\text{Period}}B=Period2π​,
3. CCC is the horizontal shift,
4. DDD is the vertical shift, i.e., the midline.

Given:

1. Period T=10T = 10T=10,
2. Midline D=−3D = -3D=−3.

We can solve for BBB:

B=2πT=2π10=π5B = \frac{2\pi}{T} = \frac{2\pi}{10} = \frac{\pi}{5}B=T2π​=102π​=5π​

Step 5: Determine the Phase Shift and Amplitude

Assuming a cosine function, which typically starts at the maximum value, and considering the points, it seems more practical to use the sine function for simplicity. Let’s determine CCC by considering a sinusoidal function shifted horizontally.

Using the given points to determine phase shift CCC: The point (3,−3)(3, -3)(3,−3) suggests a trough (assuming symmetric about the trough), implying the function reaches the minimum at x=3x = 3x=3. For a sine function, the function reaches the minimum at −π/2-\pi/2−π/2 shifted by π/2\pi/2π/2.

We can write:

y=Asin⁡(π5(x−C))−3y = A \sin\left(\frac{\pi}{5}(x - C)\right) - 3y=Asin(5π​(x−C))−3

From the given points:

1. Since it repeats every 10 units, the phase shift CCC aligns x=3x = 3x=3.

Determine amplitude (if considering max/min provided):

Assume A=3A = 3A=3 (standard form).

Thus:

y=−3sin⁡(π5(x−3))−3y = -3 \sin\left(\frac{\pi}{5}(x - 3)\right) - 3y=−3sin(5π​(x−3))−3

So, our sinusoidal function is:

y=−3sin⁡(π5(x−3))−3y = -3 \sin\left(\frac{\pi}{5}(x - 3)\right) - 3y=−3sin(5π​(x−3))−3