Question

Write a cosine function that Has a midline of 2 an amplitude of 3 and a period of 7pi/4

Answer 1

The cosine function is:

`\[ f(x) = 3 \cos\left((8)/(7)x\right) + 2 \]`

To write a cosine function with a midline of 2, an amplitude of 3, and a period of
`\( (7\pi)/(4) \)`, you can use the general form of a cosine function:

`\[ f(x) = A \cos\left((2\pi)/(B)(x - C)\right) + D \]`

In this case:

- The midline is 2, which means D = 2 .

- The amplitude is 3, which means |A| = 3 . We'll choose A = 3 since cosine is an even function.

- The period is
`\( (7\pi)/(4) \)`, which means
`\( (2\pi)/(B) = (7\pi)/(4) \).` Solving for B :

`\[ B = (2\pi)/((7\pi)/(4)) = (8)/(7) \]`

So,
`\( B = (8)/(7) \).`

- We don't need to shift the graph left or right, so C = 0 .

Now, we can write the cosine function:

`\[ f(x) = 3 \cos\left((8)/(7)x\right) + 2 \]`

This function has a midline of 2, an amplitude of 3, and a period of
`\( (7\pi)/(4) \).`

Answer 2

Given:

Amplitude of cosine function, A=3.

Period, T=7π/4.

Midline, D=2.

The time period can be expressed as:

`T=(2\pi)/(B)`

Put T=7π/4 to find the value of B.

`\begin{gathered} (7\pi)/(4)=(2\pi)/(B) \\ B=(4*2)/(7) \\ =(8)/(7) \end{gathered}`

The general cosine function can be expressed as,

`f(x)=A\cos (Bx)+D`

Substitute B=8/7, A=3 and D=2 in above equation.

`f(x)=3\cos ((8)/(7)x)+2`

Therefore, the cosine function is,

`f(x)=3\cos ((8)/(7)x)+2`