1. **State the problem:** We need to find an equation for the given wave-like graph that resembles a sine or cosine function. 2. **Identify the type of function:** The graph looks like a sinusoidal wave, so the general form is either: $$y = A \sin(B(x - C)) + D$$ or $$y = A \cos(B(x - C)) + D$$ where: - $A$ is the amplitude (height from center to peak), - $B$ affects the period (how long one cycle is), - $C$ is the horizontal shift, - $D$ is the vertical shift. 3. **Determine amplitude $A$:** The graph peaks near $y = 9$ and troughs near $y = -9$, so amplitude is half the distance between max and min: $$A = \frac{9 - (-9)}{2} = \frac{18}{2} = 9$$ 4. **Determine vertical shift $D$:** The midline is halfway between max and min: $$D = \frac{9 + (-9)}{2} = 0$$ So the wave oscillates around $y=0$. 5. **Determine period and $B$:** The graph shows two peaks and one trough between $x = -2$ and $x = 5$, which is a length of 7 units. One full period of a sine or cosine wave includes one peak and one trough, so the period $T$ is about 7 units. The formula relating $B$ and period $T$ is: $$T = \frac{2\pi}{B}$$ Solving for $B$: $$B = \frac{2\pi}{T} = \frac{2\pi}{7}$$ 6. **Determine phase shift $C$:** The graph starts with a peak near $x = -2$. Since cosine starts at a peak at $x=0$, the function is likely a cosine shifted right by 2 units. So, $$C = -2$$ 7. **Write the equation:** $$y = 9 \cos\left(\frac{2\pi}{7}(x + 2)\right)$$ This matches the graph's amplitude, period, and phase shift. **Final answer:** $$y = 9 \cos\left(\frac{2\pi}{7}(x + 2)\right)$$