1. **State the problem:** We need to find an equation for the trigonometric function $f(x)$ based on the given sinusoidal graph. 2. **Analyze the graph:** The wave oscillates between $y=2$ and $y=6$, so the amplitude $A$ is half the distance between max and min: $$A = \frac{6 - 2}{2} = 2$$ 3. The midline (vertical shift) is the average of max and min: $$D = \frac{6 + 2}{2} = 4$$ 4. **Determine the period:** The wave starts at $y=6$ when $x=-\pi$ and reaches a minimum near $y=2$ at $x=4\pi$. Since a full cycle of a sine or cosine wave includes a max, min, and max again, estimate the period $T$ as the distance between two consecutive maxima or minima. From $x=-\pi$ (max) to about $x=7\pi$ (next max), the period is approximately: $$T = 7\pi - (-\pi) = 8\pi$$ 5. **Calculate the angular frequency $\omega$:** $$\omega = \frac{2\pi}{T} = \frac{2\pi}{8\pi} = \frac{1}{4}$$ 6. **Choose the function form:** Since the graph starts at a maximum at $x=-\pi$, a cosine function is appropriate because $\cos(0) = 1$ is a max. 7. **Find the phase shift $\phi$:** The cosine function is $f(x) = A \cos(\omega x + \phi) + D$. We want $f(-\pi) = 6$ (max), so: $$6 = 2 \cos\left(\frac{1}{4}(-\pi) + \phi\right) + 4$$ Simplify: $$6 - 4 = 2 \cos\left(-\frac{\pi}{4} + \phi\right)$$ $$2 = 2 \cos\left(-\frac{\pi}{4} + \phi\right)$$ $$1 = \cos\left(-\frac{\pi}{4} + \phi\right)$$ Cosine equals 1 at angle 0, so: $$-\frac{\pi}{4} + \phi = 0 \implies \phi = \frac{\pi}{4}$$ 8. **Write the final equation:** $$f(x) = 2 \cos\left(\frac{1}{4}x + \frac{\pi}{4}\right) + 4$$ This matches the amplitude, midline, period, and phase shift observed in the graph. **Final answer:** $$f(x) = 2 \cos\left(\frac{x}{4} + \frac{\pi}{4}\right) + 4$$