1. **State the problem:** We analyze the function $$f(x) = \frac{2x + 2}{x^2 + 8}$$ to find its domain, intercepts, sign (positivity and negativity), asymptotes, derivatives, monotonicity, maxima/minima, and sketch its graph. 2. **Domain:** The denominator must not be zero. Since $$x^2 + 8 = 0$$ has no real solutions (because $$x^2 = -8$$ is impossible for real $$x$$), the domain is all real numbers: $$\mathbb{R}$$. 3. **Intercepts:** - **x-intercept(s):** Set numerator equal to zero: $$2x + 2 = 0 \implies x = -1$$ So the x-intercept is at $$(-1, 0)$$. - **y-intercept:** Evaluate at $$x=0$$: $$f(0) = \frac{2(0) + 2}{0^2 + 8} = \frac{2}{8} = \frac{1}{4}$$ So the y-intercept is at $$(0, \frac{1}{4})$$. 4. **Sign study (positivity and negativity):** - The denominator $$x^2 + 8 > 0$$ for all real $$x$$. - The sign depends on numerator $$2x + 2$$: - Positive when $$2x + 2 > 0 \implies x > -1$$ - Negative when $$x < -1$$ 5. **Asymptotes:** - **Vertical asymptotes:** None, since denominator never zero. - **Horizontal asymptote:** For large $$|x|$$, $$f(x) \approx \frac{2x}{x^2} = \frac{2}{x} \to 0$$ as $$x \to \pm \infty$$. So horizontal asymptote is $$y=0$$. 6. **Derivative:** Use quotient rule: $$f'(x) = \frac{(2)(x^2 + 8) - (2x + 2)(2x)}{(x^2 + 8)^2}$$ Simplify numerator: $$2x^2 + 16 - (4x^2 + 4x) = 2x^2 + 16 - 4x^2 - 4x = -2x^2 - 4x + 16$$ So $$f'(x) = \frac{-2x^2 - 4x + 16}{(x^2 + 8)^2}$$ 7. **Critical points:** Set numerator of $$f'(x)$$ to zero: $$-2x^2 - 4x + 16 = 0 \implies 2x^2 + 4x - 16 = 0$$ Divide by 2: $$x^2 + 2x - 8 = 0$$ Factor or use quadratic formula: $$x = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm 6}{2}$$ So $$x = 2 \text{ or } x = -4$$ 8. **Monotonicity:** - For $$x < -4$$, test $$f'(x)$$ numerator at $$x = -5$$: $$-2(-5)^2 - 4(-5) + 16 = -50 + 20 + 16 = -14 < 0$$, so $$f'(x) < 0$$, decreasing. - For $$-4 < x < 2$$, test $$x=0$$: $$-2(0)^2 - 4(0) + 16 = 16 > 0$$, so increasing. - For $$x > 2$$, test $$x=3$$: $$-2(9) - 12 + 16 = -18 - 12 + 16 = -14 < 0$$, decreasing. 9. **Maxima and minima:** - At $$x = -4$$, $$f'(x)$$ changes from negative to positive, so local minimum. - At $$x = 2$$, $$f'(x)$$ changes from positive to negative, so local maximum. Evaluate $$f(x)$$ at these points: $$f(-4) = \frac{2(-4) + 2}{(-4)^2 + 8} = \frac{-8 + 2}{16 + 8} = \frac{-6}{24} = -\frac{1}{4}$$ $$f(2) = \frac{2(2) + 2}{4 + 8} = \frac{4 + 2}{12} = \frac{6}{12} = \frac{1}{2}$$ 10. **Summary:** - Domain: $$\mathbb{R}$$ - x-intercept: $$(-1, 0)$$ - y-intercept: $$(0, \frac{1}{4})$$ - Sign: negative for $$x < -1$$, positive for $$x > -1$$ - Horizontal asymptote: $$y=0$$ - No vertical asymptotes - Increasing on $$(-4, 2)$$, decreasing on $$(-\infty, -4) \cup (2, \infty)$$ - Local minimum at $$(-4, -\frac{1}{4})$$ - Local maximum at $$(2, \frac{1}{2})$$ 11. **Sketch graph:** The function crosses x-axis at $$-1$$, y-axis at $$\frac{1}{4}$$, approaches zero at infinity, dips to minimum at $$x=-4$$, peaks at maximum at $$x=2$$, and is negative left of $$-1$$ and positive right of $$-1$$.