Question

Use the function below:

trig graph with points at 0, 2 and pi over 2, 5 and pi, 2 and 3 pi over 2, negative 1 and 2 pi, 2

What are the amplitude and midline?

Amplitude: 2; midline: y = 3
Amplitude: 1; midline: y = 3
Amplitude: 2; midline: y = 1
Amplitude: 3; midline: y = 2

Answer 1

The correct option is 4. Amplitude: 3; midline: y = 2

Explanation:

The graph passing through the points (0,2),
`((\pi)/(2),5),(\pi,2),((3\pi)/(2),-1),(2\pi,2)`.

If a is maximum value and b is minimum value, then

`Amplitude=(a-b)/(2)`

`Midline=(a+b)/(2)`

The minimum value of the function is -1.

The maximum value of the function is 5.

`Amplitude=(5-(-1))/(2)=(6)/(2)=3`

The amplitude of the graph is 3.

`Midline=(5+(-1))/(2)=(4)/(2)=2`

The midline of the graph is y=2.

The general sine function is

`f(x)=A\sin (Bx+C)+D`

Where, A is amplitude, 2π/B is period, C is phase shift and D is midline.

The required function is

`f(x)=3\sin x+2`

Therefore correct option is 4.

Answer 2

Correct answer is D.


`y=Asin(x)+B`
A - amplitude
y = B - midline


`y=3sin(x)+2`
Amplitude: 3; midline: y = 2