Question

The graph of which function has an amplitude of 3 and a right phase shift of ?

y = 3+ sin 2(x - 5)
y = 3sin 2(2
y = 3+ sin 2 (2+)
y = 3sin 2(x + 5)

Answer 1

Option 4

Explanation:

Equation representing the sine function is,

y = Asin[B(x- C)] + D

Here, A = Amplitude

B =
`\frac{2\pi }{\text{Period}}`

C = Phase shift

D = Vertical shift

If the amplitude of the the given function = 3 and right phase shift =
`(\pi )/(4)`

Therefore, equation of the sine function will be,

`y=3\text{sin}B(x-(\pi )/(4))+D`

By comparing this equation with the given options,

Equation representing right phase shift =
`(\pi )/(4)` and amplitude = 3,

`y=3\text{sin}2(x-(\pi)/(4))`

This equation represents right phase shift =
`(\pi )/(4)` and amplitude = 3.

Therefore, Option 4 is the correct option.