Question

What is the amplitude, period and the phase shift of `f(x)= 3sin(2x + pi)`?

Answer 1

`3,pi,-pi/2`

Explanation:

The standard form of the `color(blue)"sine function"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=asin(bx+c)+d)color(white)(2/2)|)))`

`"where amplitude "=|a|," period " =(2pi)/b`

`"phase shift "=-c/b" and vertical shift "=d`

`"here " a=3,b=2,c=pi,d=0`

`"amplitude " =|3|=3," period "=(2pi)/2=pi`

`"phase shift "=-(pi)/2`

Answer 2

The amplitude is `A=3`
The period is `=pi`
The phase shift is `=-(pi)/(2)`

Explanation:

`y = A sin(Bx + C) + D`

Amplitude is `A`

Period is `(2π)/B`

Phase shift is `−C/B`

Vertical shift is `D`

Here, we have

`y=3sin(2x+pi))`

`y=3sin(2x+pi)`

The amplitude is `A=3`

The period is `=(2pi)/B=(2pi)/(2)=pi`

The phase shift is `=-(pi)/(2)`

graph{3sin(2x+pi) [-5.546, 5.55, -2.773, 2.774]}