Question

Write a cosine function that passes through the points (1/2,1) and (3/2,-3).
The amplitude is 2
Maximum value: y=1
Minimum value: y=-3
Midline: y=-1

Answer 1

so then what

Answer 2

y = 2 cos(πx - π) - 1 does not go through points (1/2,1) and (3/2,-3)

Answer 3

thanks this equation matches the graph

Answer 4

could you find the amplitude and period

Answer 5

The general form of a cosine function is:
y = A cos(B(x - C)) + D

Given that the amplitude is 2, the maximum value is 1, the minimum value is -3, and the midline is y = -1:
A = 2
D = -1

The general form now becomes:
y = 2 cos(B(x - C)) - 1

We can now use the two given points (1/2, 1) and (3/2, -3) to solve for B and C.

When x = 1/2, y = 1:
1 = 2 cos(B(1/2) - C) - 1
2 = 2 cos(B/2 - C)
cos(B/2 - C) = 1

When x = 3/2, y = -3:
-3 = 2 cos(B(3/2) - C) - 1
-2 = 2 cos(3B/2 - C)
cos(3B/2 - C) = -1

Now, since the cosine function repeats every 2π, we know that:
cos(B/2 - C) = cos(3B/2 - C)

This implies that:
B/2 = 3B/2
1/2 = 3/2
This is not true, so there is no solution.

It seems there was a mistake in the calculations. Let's try a different approach to find the correct cosine function that passes through the given points.

Answer 6

Let's correct my previous statement.

Given the points (1/2, 1) and (3/2, -3), we can first determine the midline of the cosine function which passes through these points. The midline is the average of the maximum and minimum values, which in this case is (-3 + 1) / 2 = -1.

The general form of a cosine function with a horizontal shift is:
y = A cos(Bx - C) + D

Since the amplitude is 2, we have:
A = 2

The midline is -1, so:
D = -1

Using the point (1/2, 1), we have:
1 = 2 cos(B/2 - C) - 1
2 = 2 cos(B/2 - C)
cos(B/2 - C) = 1
B/2 - C = 0
B/2 = C

Using the point (3/2, -3), we have:
-3 = 2 cos(3B/2 - C) - 1
-2 = 2 cos(3B/2 - C)
cos(3B/2 - C) = -1

To solve these equations simultaneously for B and C, we can eliminate C by substituting C = B/2 from the first equation:
cos(3B/2 - B/2) = -1
cos(B) = -1

This indicates that B = π.

Therefore, the cosine function passing through the points (1/2, 1) and (3/2, -3) with amplitude 2, maximum value 1, minimum value -3, and midline -1 is:
y = 2 cos(πx - π) - 1

Answer 7

I apologize for the oversight. Let's correct this.

Given the points (1/2, 1) and (3/2, -3), and knowing that the amplitude is 2, the maximum value is 1, the minimum value is -3, and the midline is -1, we need to find the correct cosine function.

The general form of a cosine function is:
y = A cos(Bx - C) + D

Given:
A = 2 (amplitude)
D = -1 (midline)
The maximum value occurs at (1/2, 1):
1 = 2 cos(B/2 - C) - 1
2 = 2 cos(B/2 - C)
cos(B/2 - C) = 1

The minimum value occurs at (3/2, -3):
-3 = 2 cos(3B/2 - C) - 1
-2 = 2 cos(3B/2 - C)
cos(3B/2 - C) = -1

The angle whose cosine is 1 is 0. Hence B/2 - C = 0, which implies B/2 = C.

The angle whose cosine is -1 is π. Therefore, 3B/2 - C = π.

To solve these equations simultaneously:
B/2 = C
3B/2 - C = π
Substitute B/2 for C in the second equation:
3B/2 - B/2 = π
2B/2 = π
B = π

Therefore, the correct cosine function that passes through the points (1/2, 1) and (3/2, -3) is:
y = 2 cos(πx - π/2) - 1

I hope this clarifies things. Thank you for your patience.

Answer 8

You're welcome! I'm glad the corrected equation matches the graph. If you have any more questions or need further assistance, feel free to ask. Thank you for your understanding, and have a great day!