Sinusoidal Wave 3Fc93B

1. **State the problem:** We have a sinusoidal wave with peaks near $y=9$ and troughs near $y=-1$, and a midline around $y=4$. We need to find the equation of the wave, amplitude, axis, domain, range, and the function value at $495^\circ$.\n\n2. **Formula and important rules:**\nThe general form of a sinusoidal function is $$y = A \sin(B(x - C)) + D$$ or $$y = A \cos(B(x - C)) + D$$ where:\n- $A$ is the amplitude (distance from midline to peak)\n- $D$ is the vertical shift (midline)\n- $B$ affects the period $P = \frac{360^\circ}{B}$ for degrees\n- $C$ is the horizontal shift (phase shift)\n\n3. **Find amplitude:**\nAmplitude $A = \frac{\text{max} - \text{min}}{2} = \frac{9 - (-1)}{2} = \frac{10}{2} = 5$\n\n4. **Find equation of the axis:**\nAxis (midline) $D = \frac{\text{max} + \text{min}}{2} = \frac{9 + (-1)}{2} = \frac{8}{2} = 4$\n\n5. **Determine the equation:**\nSince the wave looks sinusoidal and the midline is $4$, amplitude $5$, the equation is of the form $$y = 5 \sin(Bx) + 4$$ or $$y = 5 \cos(Bx) + 4$$. Without horizontal shift info, assume $$y = 5 \sin(Bx) + 4$$.\n\n6. **Domain:**\nThe domain of sine or cosine functions is all real numbers, so $$\text{Domain} = (-\infty, \infty)$$\n\n7. **Range:**\nRange is from minimum to maximum values: $$[D - A, D + A] = [4 - 5, 4 + 5] = [-1, 9]$$\n\n8. **Value at $495^\circ$:**\nSince sine has period $360^\circ$, find equivalent angle:\n$$495^\circ - 360^\circ = 135^\circ$$\nCalculate $$y = 5 \sin(135^\circ) + 4$$\nWe know $$\sin(135^\circ) = \sin(180^\circ - 45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071$$\nSo, $$y = 5 \times 0.7071 + 4 = 3.5355 + 4 = 7.5355$$\n\n**Final answers:**\n- Equation: $$y = 5 \sin(x) + 4$$\n- Amplitude: $5$\n- Axis: $y = 4$\n- Domain: $(-\infty, \infty)$\n- Range: $[-1, 9]$\n- Value at $495^\circ$: approximately $7.54$