1. **State the problem:** We need to find values of $a$, $b$, and $c$ for the sinusoidal function $$y = a \sin(x + b)^\circ + c$$ given the graph's behavior over $0 \leq x \leq 360$. 2. **Recall the general form and properties:** - $a$ is the amplitude (distance from midline to peak). - $b$ is the horizontal phase shift (in degrees). - $c$ is the vertical shift (midline of the wave). 3. **Find $c$ (vertical shift):** The midline is halfway between the maximum and minimum values. - Max value $\approx 4$ - Min value $\approx -2$ Calculate midline: $$c = \frac{4 + (-2)}{2} = \frac{2}{2} = 1$$ 4. **Find $a$ (amplitude):** Amplitude is half the distance between max and min: $$a = \frac{4 - (-2)}{2} = \frac{6}{2} = 3$$ 5. **Find $b$ (phase shift):** The sine function normally starts at 0 when $x=0$. Here, $y=3$ at $x=0$. Using the formula: $$y = 3 = 3 \sin(0 + b)^\circ + 1$$ Subtract $c$: $$3 - 1 = 3 \sin(b)^\circ$$ $$2 = 3 \sin(b)^\circ$$ Divide both sides by 3: $$\frac{2}{3} = \sin(b)^\circ$$ 6. **Solve for $b$:** $$b = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.81^\circ$$ 7. **Summary:** - $a = 3$ - $b \approx 41.81^\circ$ - $c = 1$ These values fit the graph's amplitude, vertical shift, and phase shift. **Final answer:** $$a = 3, \quad b \approx 41.81^\circ, \quad c = 1$$