1. **Problem statement:** Given the function $$f(x) = \frac{x^2 - 1}{x + 2}$$, we will analyze its domain, parity, intercepts, limits, asymptotes, derivatives, and sketch its graph. 2. **Domain:** The function is defined for all real numbers except where the denominator is zero. $$x + 2 = 0 \implies x = -2$$ So, the domain is $$\{x \in \mathbb{R} : x \neq -2\}$$. 3. **Parity:** To check if $$f$$ is even, odd, or neither, compute $$f(-x)$$: $$f(-x) = \frac{(-x)^2 - 1}{-x + 2} = \frac{x^2 - 1}{-x + 2}$$ Since $$f(-x) \neq f(x)$$ and $$f(-x) \neq -f(x)$$, the function is neither even nor odd. 4. **Intercepts:** - **x-intercepts:** Solve numerator $$x^2 - 1 = 0$$: $$x^2 = 1 \implies x = \pm 1$$ - **y-intercept:** Evaluate $$f(0)$$: $$f(0) = \frac{0^2 - 1}{0 + 2} = \frac{-1}{2} = -\frac{1}{2}$$ 5. **Limits and asymptotes:** - Vertical asymptote at $$x = -2$$ because denominator zero. - Calculate limits near vertical asymptote: $$\lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} \frac{x^2 - 1}{x + 2} = -\infty$$ $$\lim_{x \to -2^+} f(x) = +\infty$$ - Calculate limits at infinity: Divide numerator and denominator by $$x$$: $$f(x) = \frac{x^2 - 1}{x + 2} = \frac{x^2(1 - \frac{1}{x^2})}{x(1 + \frac{2}{x})} = x \cdot \frac{1 - \frac{1}{x^2}}{1 + \frac{2}{x}}$$ As $$x \to \pm \infty$$, $$\frac{1}{x} \to 0$$, so $$f(x) \sim x$$ No horizontal asymptote, but there is an oblique asymptote found by polynomial division: Divide $$x^2 - 1$$ by $$x + 2$$: $$x^2 - 1 = (x + 2)(x - 2) + 3$$ So, $$f(x) = x - 2 + \frac{3}{x + 2}$$ Oblique asymptote: $$y = x - 2$$ 6. **First derivative:** Using quotient rule: $$f'(x) = \frac{(2x)(x + 2) - (x^2 - 1)(1)}{(x + 2)^2} = \frac{2x^2 + 4x - x^2 + 1}{(x + 2)^2} = \frac{x^2 + 4x + 1}{(x + 2)^2}$$ - Find critical points by setting numerator zero: $$x^2 + 4x + 1 = 0$$ Discriminant: $$\Delta = 16 - 4 = 12$$ Roots: $$x = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$$ - Sign of $$f'(x)$$ depends on numerator since denominator is always positive except at $$x = -2$$ (excluded). - Table of signs: For $$x < -2 - \sqrt{3}$$, numerator positive (since parabola opens upward and roots at $$-2 \pm \sqrt{3}$$). Between roots, numerator negative. For $$x > -2 + \sqrt{3}$$, numerator positive. 7. **Second derivative:** Differentiate $$f'(x)$$: $$f'(x) = \frac{x^2 + 4x + 1}{(x + 2)^2}$$ Using quotient rule again: $$f''(x) = \frac{(2x + 4)(x + 2)^2 - (x^2 + 4x + 1)2(x + 2)}{(x + 2)^4}$$ Simplify numerator: $$= (2x + 4)(x + 2)^2 - 2(x^2 + 4x + 1)(x + 2)$$ Factor out $$x + 2$$: $$= (x + 2)\left[(2x + 4)(x + 2) - 2(x^2 + 4x + 1)\right]$$ Expand inside bracket: $$(2x + 4)(x + 2) = 2x^2 + 4x + 4x + 8 = 2x^2 + 8x + 8$$ $$2(x^2 + 4x + 1) = 2x^2 + 8x + 2$$ Subtract: $$2x^2 + 8x + 8 - (2x^2 + 8x + 2) = 6$$ So numerator: $$(x + 2) \cdot 6 = 6(x + 2)$$ Therefore: $$f''(x) = \frac{6(x + 2)}{(x + 2)^4} = \frac{6}{(x + 2)^3}$$ - Sign of $$f''(x)$$ depends on denominator: For $$x > -2$$, denominator positive, so $$f''(x) > 0$$ (concave up). For $$x < -2$$, denominator negative, so $$f''(x) < 0$$ (concave down). 8. **Graph sketch:** - Vertical asymptote at $$x = -2$$. - Oblique asymptote $$y = x - 2$$. - Intercepts at $$x = \pm 1$$ and $$y = -\frac{1}{2}$$. - Increasing/decreasing intervals from $$f'(x)$$ sign. - Concavity from $$f''(x)$$ sign. - The graph behaves like a parabola far from $$x = -2$$ and has a vertical asymptote there. --- **Final answers:** - Domain: $$\{x \in \mathbb{R} : x \neq -2\}$$ - Parity: Neither even nor odd - Intercepts: $$x = -1, 1$$; $$y = -\frac{1}{2}$$ - Vertical asymptote: $$x = -2$$ - Oblique asymptote: $$y = x - 2$$ - First derivative: $$f'(x) = \frac{x^2 + 4x + 1}{(x + 2)^2}$$ with critical points at $$x = -2 \pm \sqrt{3}$$ - Second derivative: $$f''(x) = \frac{6}{(x + 2)^3}$$ - Concavity changes at $$x = -2$$