Question

The graph of a sinusoidal function has a maximum point at (0, 5) and then has a minimum point at (2π, -5).

Write the formula of the function, where x is entered in radians.

Answer 1

One possible function that meets the requirements is
`f(x) = 5\, \sin((1/2)\, x + (\pi / 2))`.

Explanation:

In general, a sinusoidal function is of the form
`f(x) = A\, \sin(\omega\, x + \varphi) + D`, where
`A`,
`\omega`,
`\varphi`, and
`D` are constants.

The constant
`A` determines the amplitude of this sinusoidal function. The amplitude is
`(1/2)` the vertical distance between maxima and minima. In this question, the vertical distance between maxima and minima is
`(5 - (-5)) = 10`, such that
`A = (1/2) \, (10) = 5`.

The constant
`D` determines the midpoint between maxima and minima. In this question, the midpoint between minima (
`y = (-5)`) and maxima (
`y = 5`) is
`(1 / 2)\, ((-5) + 5) = 0`. Hence,
`D = 0`.

The constant
`\omega` determines the period of this sinusoidal function. The period of
`f(x) = A\, \sin(\omega\, x + \varphi) + D` is
`(2\, \pi / \omega)`, such that:

• the distance between two neighboring maxima would be
  `(2\, \pi / \omega)`, and
• the distance between a maximum and the next minima would be
  `(\pi / \omega)`.

In this question, assume that there is no minima between
`x = 0` and
`x = 2\,\pi` (exclusive). Hence,
`(\pi / \omega) = 2\,\pi`, and
`\omega = (1/2)`.

The constant
`\varphi` shifts the sinusoidal function horizontally. After finding
`A`,
`D`, and
`\omega`, substitute in a point on the graph of this function to find the value of
`\varphi\!`. For example, since
`(0,\, 5)` is a point on the graph of
`f(x) = A\, \sin(\omega\, x + \varphi) + D = 5\, \sin((1/2)\, x + \varphi)`:

`5\, \sin((1/2)\, (0) + \varphi) = 5`.

`5\, \sin(\varphi) = 5`.

One possible value of
`\varphi` would be
`(\pi / 2)`.

Hence, one possible formula satisfying the requirements is
`f(x) = 5\, \sin((1/2)\, x + (\pi / 2))`.