Answer:
One possible function that meets the requirements is [tex]f(x) = 5\, \sin((1/2)\, x + (\pi / 2))[/tex].
Step-by-step explanation:
In general, a sinusoidal function is of the form [tex]f(x) = A\, \sin(\omega\, x + \varphi) + D[/tex], where [tex]A[/tex], [tex]\omega[/tex], [tex]\varphi[/tex], and [tex]D[/tex] are constants.
The constant [tex]A[/tex] determines the amplitude of this sinusoidal function. The amplitude is [tex](1/2)[/tex] the vertical distance between maxima and minima. In this question, the vertical distance between maxima and minima is [tex](5 - (-5)) = 10[/tex], such that [tex]A = (1/2) \, (10) = 5[/tex].
The constant [tex]D[/tex] determines the midpoint between maxima and minima. In this question, the midpoint between minima ([tex]y = (-5)[/tex]) and maxima ([tex]y = 5[/tex]) is [tex](1 / 2)\, ((-5) + 5) = 0[/tex]. Hence, [tex]D = 0[/tex].
The constant [tex]\omega[/tex] determines the period of this sinusoidal function. The period of [tex]f(x) = A\, \sin(\omega\, x + \varphi) + D[/tex] is [tex](2\, \pi / \omega)[/tex], such that:
In this question, assume that there is no minima between [tex]x = 0[/tex] and [tex]x = 2\,\pi[/tex] (exclusive). Hence, [tex](\pi / \omega) = 2\,\pi[/tex], and [tex]\omega = (1/2)[/tex].
The constant [tex]\varphi[/tex] shifts the sinusoidal function horizontally. After finding [tex]A[/tex], [tex]D[/tex], and [tex]\omega[/tex], substitute in a point on the graph of this function to find the value of [tex]\varphi\![/tex]. For example, since [tex](0,\, 5)[/tex] is a point on the graph of [tex]f(x) = A\, \sin(\omega\, x + \varphi) + D = 5\, \sin((1/2)\, x + \varphi)[/tex]:
[tex]5\, \sin((1/2)\, (0) + \varphi) = 5[/tex].
[tex]5\, \sin(\varphi) = 5[/tex].
One possible value of [tex]\varphi[/tex] would be [tex](\pi / 2)[/tex].
Hence, one possible formula satisfying the requirements is [tex]f(x) = 5\, \sin((1/2)\, x + (\pi / 2))[/tex].
