1. **State the problem:** We need to find the function $f(x) = 2 \sin(kx) + d$ that matches the given sinusoidal wave with amplitude 2, passing through the point $(0,-1)$, and x-axis labeled in multiples of $\pi$. The wave starts at $y=-1$ when $x=0$, indicating a vertical shift. 2. **Identify amplitude and vertical shift:** The amplitude is the coefficient before sine, which is 2. The vertical shift $d$ is the value of the function at $x=0$ because $\sin(0) = 0$, so: $$f(0) = 2 \sin(k \cdot 0) + d = d = -1$$ Thus, $d = -1$. 3. **Determine the period and frequency $k$:** The period $T$ of $\sin(kx)$ is given by: $$T = \frac{2\pi}{k}$$ From the graph, the wave completes one full cycle between $-\pi/3$ and $\pi/3$, which is a length of: $$\pi/3 - (-\pi/3) = \frac{2\pi}{3}$$ So the period $T = \frac{2\pi}{3}$. 4. **Calculate $k$ using the period:** $$k = \frac{2\pi}{T} = \frac{2\pi}{\frac{2\pi}{3}} = 3$$ 5. **Write the final function:** $$f(x) = 2 \sin(3x) - 1$$ 6. **Summary:** The function has amplitude 2, frequency multiplier $k=3$, and vertical shift $-1$. **Final answer:** $$f(x) = 2 \sin(3x) - 1$$