1. **State the problem:** We need to find the equation of a sine function $f(x)$ with the following characteristics: - Centered around $y=4$ (vertical shift) - Amplitude about $0.5$ - Period about $\pi$ - Peaks near $y=4.5$ and troughs near $y=3.5$ - A trough just right of $x=0$ 2. **Recall the general sine function formula:** $$f(x) = A \sin(B(x - C)) + D$$ where: - $A$ is the amplitude - $B$ affects the period $P = \frac{2\pi}{B}$ - $C$ is the horizontal shift (phase shift) - $D$ is the vertical shift 3. **Identify parameters from the problem:** - Amplitude $A = 0.5$ - Vertical shift $D = 4$ - Period $P = \pi \Rightarrow B = \frac{2\pi}{P} = \frac{2\pi}{\pi} = 2$ 4. **Determine phase shift $C$:** - The trough is just right of $x=0$ - Normally, $\sin(x)$ has a trough at $x = \frac{3\pi}{2}$, but since period is $\pi$, the troughs repeat every $\pi$ units. - For $f(x) = 0.5 \sin(2(x - C)) + 4$, troughs occur when the inside of sine is $\frac{3\pi}{2} + 2k\pi$. - Set $2(x - C) = \frac{3\pi}{2}$ at $x=0$ to find $C$: $$2(0 - C) = \frac{3\pi}{2} \Rightarrow -2C = \frac{3\pi}{2} \Rightarrow C = -\frac{3\pi}{4}$$ 5. **Write the final equation:** $$f(x) = 0.5 \sin\left(2\left(x + \frac{3\pi}{4}\right)\right) + 4$$ This matches the given conditions: amplitude 0.5, vertical shift 4, period $\pi$, and trough just right of $x=0$. **Final answer:** $$f(x) = 0.5 \sin\left(2\left(x + \frac{3\pi}{4}\right)\right) + 4$$