Question

trig ratios and functions exam write the equation of the sinusoidal function shown. a) y = sin(3x) + 2 b) y = sin x + 2 c) y = cos x + 2 d) y = cos(3x) + 2

Explanation:

Step1: Determine the vertical shift

The mid - line of the sinusoidal function is \(y = 2\), so the vertical shift \(D=2\).

Step2: Determine the period

The period \(T\) of a sinusoidal function \(y = A\sin(Bx - C)+D\) or \(y = A\cos(Bx - C)+D\) is given by \(T=\frac{2\pi}{|B|}\). From the graph, in the interval \([0, \pi]\), the function completes 3 full cycles. So the period \(T=\frac{\pi}{3}\). Using the period formula \(\frac{2\pi}{|B|}=\frac{\pi}{3}\), we get \(|B| = 6\) (but in the standard forms given we can also note the frequency from the shape). Also, the graph starts at a maximum (like a cosine - function), but we can also analyze from the general form. The general form of a sinusoidal function is \(y=A\sin(Bx - C)+D\) or \(y = A\cos(Bx - C)+D\). Here \(A = 1\) (amplitude, distance from mid - line to maximum/minimum which is \(1\) as the mid - line is \(y = 2\) and max is \(y=3\) and min is \(y = 1\)). Since the function has a frequency such that in a short \(x\) - interval there are multiple cycles and starts at a maximum, we consider the form \(y=\cos(Bx)+D\). With \(D = 2\) and observing the number of cycles in the given \(x\) - range, for \(y=\cos(Bx)+2\), when \(B = 3\) it matches the graph's frequency.

Answer:

D. \(y=\cos(3x)+2\)