1. The problem is to find an equation for the sine wave graph shown, which oscillates between approximately -6 and 6 on the y-axis and spans from about -3 to 10 on the x-axis. 2. The general form of a sine wave equation is $$y = A \sin(B(x - C)) + D$$ where: - $A$ is the amplitude (the peak value), - $B$ affects the period of the wave, - $C$ is the horizontal shift, - $D$ is the vertical shift. 3. From the graph, the amplitude $A$ is half the distance between the peak and trough: $$A = \frac{6 - (-6)}{2} = \frac{12}{2} = 6$$ 4. The vertical shift $D$ is the midpoint between the peak and trough: $$D = \frac{6 + (-6)}{2} = 0$$ 5. To find $B$, we need the period $T$, which is the length of one full wave cycle. From the graph, one period appears to be about 6 units (from peak near 0 to next peak near 6). 6. The relationship between $B$ and the period $T$ is: $$B = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3}$$ 7. The horizontal shift $C$ is the x-value where the sine wave starts its cycle. Since the peak is near $x=0$ and sine normally starts at zero, we can use a cosine form or shift sine by $\frac{\pi}{2B}$ to align peaks. Using sine, the peak at $x=0$ means: $$y = 6 \sin\left(\frac{\pi}{3}x + \frac{\pi}{2}\right)$$ 8. Using the identity $\sin(x + \frac{\pi}{2}) = \cos(x)$, the equation can also be written as: $$y = 6 \cos\left(\frac{\pi}{3}x\right)$$ Final answer: $$y = 6 \cos\left(\frac{\pi}{3}x\right)$$