1. **Problem:** Find a function for the sinusoidal graph with amplitude 3, midline $y=0$, x-intercepts at $-\frac{2\pi}{3}$, $0$, and $\frac{2\pi}{3}$, and a negative slope at $x=0$. 2. **Formula and rules:** A sinusoidal function can be written as $f(t) = A \sin(Bt)$ or $f(t) = A \cos(Bt)$ where $A$ is amplitude and $\frac{2\pi}{B}$ is the period. 3. **Amplitude:** Given as 3, so $A=3$. 4. **Period:** The distance between two consecutive zeros for sine is $\frac{\pi}{B}$. Here zeros are at $-\frac{2\pi}{3}$ and $0$, distance $\frac{2\pi}{3}$. So, $$\frac{\pi}{B} = \frac{2\pi}{3} \implies B = \frac{3}{2}$$ 5. **Check slope at $x=0$:** The slope is negative, so the function must be decreasing at zero. For $f(t) = 3\sin(\frac{3}{2}t)$, derivative at 0 is positive, so not correct. 6. Try $f(t) = -3\sin(3t)$ (option c): zeros of $\sin(3t)$ are at $t=0, \pm \frac{\pi}{3}, \pm \frac{2\pi}{3}, ...$ which matches the zeros given. 7. Amplitude is 3, negative sign flips the slope at zero to negative, matching the graph. 8. **Answer:** $\boxed{f(t) = -3\sin(3t)}$ (option c).