Question
Changes in Midline Practice
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Question
Complete the equation of the periodic function represented by the table of values by entering the midline.
This table shows values of the sine function as y at regular intervals of π/2.
x 0 π2
π
3π2
2π
y 6 7 6 5 6
(1 point)
y=sinx+
Remaining Attempts : 3
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Answers
GPT-4o mini
To find the midline of the sine function represented by the table of values, we need to analyze the given y-values. The midline is calculated by finding the average of the maximum and minimum y values.
From the table provided, the y-values are:
- At \( x = 0 \): \( y = 6 \)
- At \( x = \frac{\pi}{2} \): \( y = 7 \)
- At \( x = \pi \): \( y = 6 \)
- At \( x = \frac{3\pi}{2} \): \( y = 5 \)
- At \( x = 2\pi \): \( y = 6 \)
From these values, we can identify:
- Maximum value = 7
- Minimum value = 5
Now, to calculate the midline:
\[
\text{Midline} = \frac{\text{Maximum} + \text{Minimum}}{2} = \frac{7 + 5}{2} = \frac{12}{2} = 6
\]
Thus, the midline is \( 6 \).
Now we can complete the sine function equation. Since the sine function oscillates around the midline, we can express the equation as follows:
\[
y = \sin(x) + \text{midline}
\]
In this case:
\[
y = \sin(x) + 6
\]
So the completed equation is:
\[
y = \sin(x) + 6
\]
From the table provided, the y-values are:
- At \( x = 0 \): \( y = 6 \)
- At \( x = \frac{\pi}{2} \): \( y = 7 \)
- At \( x = \pi \): \( y = 6 \)
- At \( x = \frac{3\pi}{2} \): \( y = 5 \)
- At \( x = 2\pi \): \( y = 6 \)
From these values, we can identify:
- Maximum value = 7
- Minimum value = 5
Now, to calculate the midline:
\[
\text{Midline} = \frac{\text{Maximum} + \text{Minimum}}{2} = \frac{7 + 5}{2} = \frac{12}{2} = 6
\]
Thus, the midline is \( 6 \).
Now we can complete the sine function equation. Since the sine function oscillates around the midline, we can express the equation as follows:
\[
y = \sin(x) + \text{midline}
\]
In this case:
\[
y = \sin(x) + 6
\]
So the completed equation is:
\[
y = \sin(x) + 6
\]