1. **Problem statement:** Given the function $f(x) = \frac{x^2 - 8}{x^2 - 4}$, find the maximal domain $D$, zeros, symmetry, behavior at domain boundaries, and asymptotes. 2. **Domain:** The denominator must not be zero: $$x^2 - 4 \neq 0 \implies x^2 \neq 4 \implies x \neq \pm 2$$ So, the maximal domain is: $$D = \mathbb{R} \setminus \{ -2, 2 \}$$ 3. **Zeros:** Solve numerator $=0$: $$x^2 - 8 = 0 \implies x^2 = 8 \implies x = \pm \sqrt{8} = \pm 2\sqrt{2}$$ 4. **Symmetry:** Check $f(-x)$: $$f(-x) = \frac{(-x)^2 - 8}{(-x)^2 - 4} = \frac{x^2 - 8}{x^2 - 4} = f(x)$$ So, $f$ is an even function, symmetric about the $y$-axis. 5. **Behavior at domain boundaries:** As $x \to \pm 2$, denominator $\to 0$, numerator $\to 4 - 8 = -4$. - For $x \to 2^-$: Denominator $x^2 - 4 = (2^-)^2 - 4 = 4^- - 4 = 0^-$ (slightly negative), numerator $-4$. So, $$f(x) \to \frac{-4}{0^-} = +\infty$$ - For $x \to 2^+$: Denominator $0^+$, numerator $-4$. So, $$f(x) \to \frac{-4}{0^+} = -\infty$$ Similarly for $x \to -2$: - $x \to -2^-$: denominator $0^+$, numerator $-4$, so $f(x) \to -\infty$ - $x \to -2^+$: denominator $0^-$, numerator $-4$, so $f(x) \to +\infty$ 6. **Asymptotes:** - Vertical asymptotes at $x = \pm 2$. - Horizontal asymptote: For large $|x|$, leading terms dominate: $$f(x) \approx \frac{x^2}{x^2} = 1$$ So horizontal asymptote: $$y = 1$$ **Final answers for part a):** - Domain: $D = \mathbb{R} \setminus \{ -2, 2 \}$ - Zeros: $x = \pm 2\sqrt{2}$ - Symmetry: Even function - Vertical asymptotes: $x = \pm 2$ - Horizontal asymptote: $y = 1$