What are the Amplitude, Period, and Midline of f(x) = −7sin(4x − π) + 2?

We will be using the phase shift standard form to solve this.

Answer: The Amplitude, Period, and Midline of f(x) = −7sin(4x − π) + 2 are −7, π/2, and y = 2 respectively.

Let's solve this step by step.

Explanation:

Given that, f(x) = −7sin(4x − π) + 2

We have a standard form for phase shift:

f(x) = asin(bx + c) + d 

Here, a = -7, b = 4, c = -π, d = 2

Where, Amplitude = a, Time Period = 2π/b, Phase Shift = c, Vertical Shift = d.

On Comparing we get: Amplitude = -7, Time Period = π/2, Phase Shift = -π, Vertical Shift = 2.

The midline is parallel to the x-axis and runs between the maximum and minimum value(i.e., amplitudes)

For the function f(x) = sinx midline is y = 0, midline is affected by any vertical shift/translations. For example, y = sin(x) + 1 has a midline of y = 1.

⇒ The midline of function f(x) = −7 sin(4x − π) + 2 is y = 2.

Hence, the Amplitude, Period, and Midline of f(x) = −7sin(4x − π) + 2 are −7, π/2, and y = 2 respectively.