Question

The graph of a trigonometric function oscillates between y = 1 and y equals negative 7. It reaches its maximum at x = π and its minimum at x = 3π. Which of the following could be the equation of the function?
(1 point)
Responses

f left parenthesis x right parenthesis equals 4 cosine Start Fraction x over 2 End Fraction minus 3
Image with alt text: f left parenthesis x right parenthesis equals 4 cosine Start Fraction x over 2 End Fraction minus 3

f left parenthesis x right parenthesis equals 4 sine Start Fraction x over 2 End Fraction minus 3
Image with alt text: f left parenthesis x right parenthesis equals 4 sine Start Fraction x over 2 End Fraction minus 3

f left parenthesis x right parenthesis equals 4 sine 2 x minus 3
Image with alt text: f left parenthesis x right parenthesis equals 4 sine 2 x minus 3

f left parenthesis x right parenthesis equals 8 sine Start Fraction x over 2 End Fraction minus 3

Answers

GPT-4o mini
To find the correct equation for the trigonometric function that oscillates between \( y = 1 \) and \( y = -7 \), we first determine the amplitude, vertical shift, and period of the function.

1. **Amplitude and Vertical Shift**:
- The maximum value is 1 and the minimum value is -7.
- The midline (average of max and min) can be calculated as:
\[
\text{Midline} = \frac{1 + (-7)}{2} = \frac{-6}{2} = -3
\]
- The amplitude (the distance from the midline to the maximum) is:
\[
\text{Amplitude} = \text{Max} - \text{Midline} = 1 - (-3) = 4
\]
- Thus, the function has an amplitude of 4 and a midline of \( y = -3 \).

2. **General Form**:
Considering these values, the general form of the function can be represented by:
\[
f(x) = A \cdot \sin(Bx) + C \quad \text{or} \quad f(x) = A \cdot \cos(Bx) + C
\]
where \( A = 4 \), \( C = -3 \).

3. **Finding the Period**:
With the maximum occurring at \( x = \pi \) and the minimum occurring at \( x = 3\pi \), this indicates that the period must be \( 2\pi \) (the distance from maximum to minimum in a sine or cosine function is \( \pi \), which represents half the period).
Therefore, the full period of the function is \( 4\pi \).

The period \( P \) of a sine or cosine function is given by:
\[
P = \frac{2\pi}{B}
\]
Setting the period equal to \( 4\pi \):
\[
\frac{2\pi}{B} = 4\pi \implies B = \frac{1}{2}
\]

4. **Choosing a Function**:
Given that \( A = 4 \) and \( C = -3 \), we can find potential functions:
- \( f(x) = 4 \cos \left( \frac{x}{2} \right) - 3 \)
- \( f(x) = 4 \sin \left( \frac{x}{2} \right) - 3 \)

5. **Locating Maximum and Minimum**:
To meet the condition of reaching the maximum at \( x = \pi \):
- For \( \cos \left( \frac{x}{2} \right) \), the maximum occurs at \( x = 0 \) and every \( 4\pi \) thereafter, thus it doesn't fit.
- For \( \sin \left( \frac{x}{2} \right) \), the max occurs at \( x = \pi \) and the min at \( x = 3\pi \), thus it does fit.

Therefore, the that could be the equation of the function is:
\[
\boxed{f(x) = 4 \sin \left( \frac{x}{2} \right) - 3}
\]

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