the curve above is the graph of a sinusoidal function. it goes through the points $(-4, -2)$ and $(2, -2)$…

the curve above is the graph of a sinusoidal function. it goes through the points $(-4, -2)$ and $(2, -2)$. find a sinusoidal function that matches the given graph. if needed, you can enter $pi = 3.1416...$ as pi in your answer, otherwise use at least 3 decimal digits. $f(x)=$

the curve above is the graph of a sinusoidal function. it goes through the points $(-4, -2)$ and $(2, -2)$. find a sinusoidal function that matches the given graph. if needed, you can enter $pi = 3.1416...$ as pi in your answer, otherwise use at least 3 decimal digits. $f(x)=$

Answer

Explanation:

Step1: Find the amplitude (A)

The general form of a sinusoidal function is (y = A\sin(B(x - C))+D) or (y=A\cos(B(x - C))+D). The mid - line (y = D) is the average of the maximum and minimum values. The maximum value (y_{max}\approx2) and the minimum value (y_{min}=- 2). So, (D=\frac{y_{max}+y_{min}}{2}=\frac{2+( - 2)}{2}=0). The amplitude (A=\frac{y_{max}-y_{min}}{2}=\frac{2-( - 2)}{2}=2)

Step2: Find the period (T) and (B)

The period (T) is the distance between two consecutive minima (or maxima). The function passes through ((-4,-2)) and ((2,-2)). The period (T=2 - (-4)=6). Using the formula (T=\frac{2\pi}{B}), we solve for (B): (B=\frac{2\pi}{T}=\frac{2\pi}{6}=\frac{\pi}{3})

Step3: Find the phase shift (C)

Let's use the cosine function (y = A\cos(B(x - C))+D). When (x = 0), (y = 1). Substitute (A = 2), (B=\frac{\pi}{3}), (D = 0) into (y=2\cos(\frac{\pi}{3}(x - C))). Then (1 = 2\cos(\frac{\pi}{3}(0 - C))), (\cos(-\frac{\pi C}{3})=\frac{1}{2}). Since (\cos\theta=\frac{1}{2}) when (\theta=\pm\frac{\pi}{3}+2k\pi,k\in\mathbb{Z}), and we take (k = 0), (-\frac{\pi C}{3}=\frac{\pi}{3}), so (C=- 1)

Answer:

(f(x)=2\cos\left(\frac{\pi}{3}(x + 1)\right))