Question

What is the amplitude, period, and phase shift of f(x) = −3 sin(4x − π) − 5?

Amplitude = −3; period = 2π; phase shift: x equals pi over four
Amplitude = 3; period = pi over two; phase shift: x equals pi over four
Amplitude = −3; period = 2π; phase shift: x equals negative pi over four
Amplitude = 3; period = pi over two; phase shift: x equals negative pi over four

Answer 1

2nd choice.

Explanation:

Let's compare the following:

`f(x)=a\sin(b(x-c))+d` to

`f(x)=-3\sin(4x-\pi))-5`.

They are almost in the same form.

The amplitude is |a|, so it isn't going to be negative.

The period is
`(2\pi)/(|b|)`.

The phase shift is
`c`.

If c is positive it has been shifted right c units.

If c is negative it has been shifted left c units.

d is the vertical shift.

If d is negative, it has been moved down d units.

If d is positive, it has been moved up d units.

So we already know two things:

The amplitude is |a|=|-3|=3.

The vertical shift is d=-5 which means it was moved down 5 units from the parent function.

Now let's find the others.

I'm going to factor out 4 from
`4x-\pi`.

Like this:

`4(x-(\pi)/(4))`

Now if you compare this to
`b(x-c)`

then b=4 so the period is
`(2\pi)/(4)=(\pi)/(2)`.

Also in place of c you see
`(\pi)/(4)` which means the phase shift is
`(\pi)/(4)`.

The second choice is what we are looking for.

Answer 2

Answer: Second Option

Amplitude = 3; period = pi over two; phase shift: x equals pi over four

Explanation:

By definition the sinusoidal function has the following form:

`f(x) = asin(bx - c) +k`

Where

`| a |` is the Amplitude of the function

`(2\pi)/(b)` is the period of the function

`-(c)/(b)` is the phase shift

In this case the function is:

`f(x) = -3 sin(4x - \pi) - 5`

Therefore

`Amplitude=|a|=3`

`Period =(2\pi)/(b) = (2\pi)/(4)=(\pi)/(2)`

`phase\ shift = -((-\pi))/(4)=(\pi)/(4)`