1. The problem is to analyze the function $$S_{xx}(\omega) = \frac{\omega^2 + 9}{\omega^4 + 5\omega^2 + 4}$$ and understand its behavior. 2. The formula given is a rational function where the numerator is $$\omega^2 + 9$$ and the denominator is $$\omega^4 + 5\omega^2 + 4$$. 3. To analyze the function, first factor the denominator: $$\omega^4 + 5\omega^2 + 4 = (\omega^2 + 4)(\omega^2 + 1)$$ 4. The numerator $$\omega^2 + 9$$ cannot be factored further over the reals. 5. The function can be rewritten as: $$S_{xx}(\omega) = \frac{\omega^2 + 9}{(\omega^2 + 4)(\omega^2 + 1)}$$ 6. Since all terms in the denominator are positive for all real $$\omega$$ (because $$\omega^2 + a > 0$$ for any real $$\omega$$ and positive $$a$$), the denominator never equals zero, so there are no vertical asymptotes. 7. The numerator is always positive since $$\omega^2 \geq 0$$ and 9 is positive, so $$\omega^2 + 9 > 0$$ for all real $$\omega$$. 8. Therefore, $$S_{xx}(\omega)$$ is positive for all real $$\omega$$. 9. To find horizontal asymptotes, analyze the behavior as $$\omega \to \pm \infty$$: $$\lim_{\omega \to \pm \infty} S_{xx}(\omega) = \lim_{\omega \to \pm \infty} \frac{\omega^2 + 9}{\omega^4 + 5\omega^2 + 4}$$ Divide numerator and denominator by $$\omega^4$$: $$= \lim_{\omega \to \pm \infty} \frac{\frac{1}{\omega^2} + \frac{9}{\omega^4}}{1 + \frac{5}{\omega^2} + \frac{4}{\omega^4}} = 0$$ 10. So, the horizontal asymptote is $$y=0$$. 11. To find critical points, differentiate $$S_{xx}(\omega)$$ using the quotient rule: $$S'_{xx}(\omega) = \frac{(2\omega)(\omega^4 + 5\omega^2 + 4) - (\omega^2 + 9)(4\omega^3 + 10\omega)}{(\omega^4 + 5\omega^2 + 4)^2}$$ Simplify numerator: $$2\omega(\omega^4 + 5\omega^2 + 4) - (\omega^2 + 9)(4\omega^3 + 10\omega)$$ $$= 2\omega^5 + 10\omega^3 + 8\omega - (4\omega^5 + 10\omega^3 + 36\omega^3 + 90\omega)$$ $$= 2\omega^5 + 10\omega^3 + 8\omega - 4\omega^5 - 46\omega^3 - 90\omega$$ $$= -2\omega^5 - 36\omega^3 - 82\omega$$ 12. Set numerator equal to zero to find critical points: $$-2\omega^5 - 36\omega^3 - 82\omega = 0$$ Factor out $$-2\omega$$: $$-2\omega(\omega^4 + 18\omega^2 + 41) = 0$$ 13. Solve for $$\omega$$: $$\omega = 0$$ or $$\omega^4 + 18\omega^2 + 41 = 0$$ 14. The quartic $$\omega^4 + 18\omega^2 + 41 = 0$$ has no real roots because its discriminant is negative. 15. Therefore, the only critical point is at $$\omega = 0$$. 16. Evaluate $$S_{xx}(0)$$: $$S_{xx}(0) = \frac{0 + 9}{0 + 0 + 4} = \frac{9}{4} = 2.25$$ 17. Since $$S_{xx}(\omega) > 0$$ and tends to zero at infinity, and has a critical point at zero with value 2.25, the graph has a peak at $$\omega=0$$. Final answer: The function $$S_{xx}(\omega) = \frac{\omega^2 + 9}{\omega^4 + 5\omega^2 + 4}$$ is positive for all real $$\omega$$, has no vertical asymptotes, a horizontal asymptote at $$y=0$$, and a maximum at $$\omega=0$$ with value $$\frac{9}{4}$$.