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Sagot :
To find the formula of the sinusoidal function, we need to determine its general form, which is:
[tex]\[ f(x) = A \cos(Bx - C) + D \][/tex]
where:
- [tex]\( A \)[/tex] is the amplitude,
- [tex]\( B \)[/tex] is related to the period,
- [tex]\( C \)[/tex] is the phase shift,
- [tex]\( D \)[/tex] is the vertical shift (midline).
Given the information:
1. The function intersects its midline at [tex]\((0, -2)\)[/tex].
2. The minimum point of the function is at [tex]\(\left(\frac{3\pi}{2}, -7\right)\)[/tex].
Let's break down the problem step-by-step:
1. Vertical shift (Midline, [tex]\(D\)[/tex]):
The midline is the average value of the maximum and minimum values of the function. Since it intersects the midline at [tex]\( (0, -2) \)[/tex], the vertical shift [tex]\( D = -2 \)[/tex].
2. Amplitude ([tex]\(A\)[/tex]):
The amplitude is the distance from the midline to the maximum or minimum value. Given the minimum point [tex]\(\left(\frac{3 \pi}{2}, -7\right)\)[/tex]:
[tex]\[ A = \mid -7 - (-2) \mid = |-7 + 2| = 5 \][/tex]
Since the minimum point shows that the function is oscillating and starts by going down, the function should be represented using a negative cosine function. Therefore:
[tex]\[ A = -5 \][/tex]
3. Period ([tex]\(B\)[/tex]):
The period of a sinusoidal function [tex]\( P \)[/tex] is related to [tex]\( B \)[/tex] by the formula:
[tex]\[ P = \frac{2\pi}{B} \][/tex]
We need to determine the period from the given information. The minimum point at [tex]\( \left(\frac{3 \pi}{2}, -7 \right) \)[/tex] suggests that from the start (midline intersection) to the minimum point, the displacement is [tex]\( \frac{3 \pi}{2} \)[/tex]. Given this is a quarter of the full period (since cosine reaches a minimum after one-fourth of its cycle):
[tex]\[ \frac{P}{4} = \frac{3\pi}{2} \rightarrow P = 4 \cdot \frac{3\π}{2} = 6\pi \][/tex]
So, solving for [tex]\( B \)[/tex]:
[tex]\[ B = \frac{2\pi}{6\pi} = \frac{1}{3} \][/tex]
4. Phase Shift ([tex]\(C\)[/tex]):
True intersection with the midline at the origin indicates there is no horizontal phase shift, so [tex]\( C = 0 \)[/tex].
Combining these values into our function form, we get:
[tex]\[ f(x) = -5 \cos\left(\frac{x}{3}\right) - 2 \][/tex]
Thus, the formula of the function is:
[tex]\[ f(x) = -5 \cos\left(\frac{x}{3}\right) - 2 \][/tex]
[tex]\[ f(x) = A \cos(Bx - C) + D \][/tex]
where:
- [tex]\( A \)[/tex] is the amplitude,
- [tex]\( B \)[/tex] is related to the period,
- [tex]\( C \)[/tex] is the phase shift,
- [tex]\( D \)[/tex] is the vertical shift (midline).
Given the information:
1. The function intersects its midline at [tex]\((0, -2)\)[/tex].
2. The minimum point of the function is at [tex]\(\left(\frac{3\pi}{2}, -7\right)\)[/tex].
Let's break down the problem step-by-step:
1. Vertical shift (Midline, [tex]\(D\)[/tex]):
The midline is the average value of the maximum and minimum values of the function. Since it intersects the midline at [tex]\( (0, -2) \)[/tex], the vertical shift [tex]\( D = -2 \)[/tex].
2. Amplitude ([tex]\(A\)[/tex]):
The amplitude is the distance from the midline to the maximum or minimum value. Given the minimum point [tex]\(\left(\frac{3 \pi}{2}, -7\right)\)[/tex]:
[tex]\[ A = \mid -7 - (-2) \mid = |-7 + 2| = 5 \][/tex]
Since the minimum point shows that the function is oscillating and starts by going down, the function should be represented using a negative cosine function. Therefore:
[tex]\[ A = -5 \][/tex]
3. Period ([tex]\(B\)[/tex]):
The period of a sinusoidal function [tex]\( P \)[/tex] is related to [tex]\( B \)[/tex] by the formula:
[tex]\[ P = \frac{2\pi}{B} \][/tex]
We need to determine the period from the given information. The minimum point at [tex]\( \left(\frac{3 \pi}{2}, -7 \right) \)[/tex] suggests that from the start (midline intersection) to the minimum point, the displacement is [tex]\( \frac{3 \pi}{2} \)[/tex]. Given this is a quarter of the full period (since cosine reaches a minimum after one-fourth of its cycle):
[tex]\[ \frac{P}{4} = \frac{3\pi}{2} \rightarrow P = 4 \cdot \frac{3\π}{2} = 6\pi \][/tex]
So, solving for [tex]\( B \)[/tex]:
[tex]\[ B = \frac{2\pi}{6\pi} = \frac{1}{3} \][/tex]
4. Phase Shift ([tex]\(C\)[/tex]):
True intersection with the midline at the origin indicates there is no horizontal phase shift, so [tex]\( C = 0 \)[/tex].
Combining these values into our function form, we get:
[tex]\[ f(x) = -5 \cos\left(\frac{x}{3}\right) - 2 \][/tex]
Thus, the formula of the function is:
[tex]\[ f(x) = -5 \cos\left(\frac{x}{3}\right) - 2 \][/tex]
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