1. **Problem statement:** Find the intercepts, asymptotes, critical points, intervals of increase/decrease, inflection points, concavity, and relative extrema for the function $$y = -\frac{x^3}{x^2 - 4}$$. 2. **Find x- and y-intercepts:** - x-intercepts occur when $$y=0$$, so numerator $$-x^3=0 \Rightarrow x=0$$. - y-intercept is $$y(0) = -\frac{0}{0-4} = 0$$. 3. **Find vertical asymptotes:** - Denominator $$x^2 - 4 = 0 \Rightarrow x=\pm 2$$. - Vertical asymptotes at $$x=2$$ and $$x=-2$$. 4. **Find horizontal/slant asymptotes:** - Degree numerator = 3, denominator = 2, so slant asymptote. - Perform polynomial division: $$\frac{-x^3}{x^2 - 4} = -x - \frac{4x}{x^2 - 4}$$ - As $$x \to \pm \infty$$, $$y \sim -x$$. - Slant asymptote: $$y = -x$$. 5. **Find critical points:** - Compute derivative using quotient rule: $$y = -\frac{x^3}{x^2 - 4}$$ $$y' = \frac{-(3x^2)(x^2 - 4) - (-x^3)(2x)}{(x^2 - 4)^2} = \frac{-3x^2(x^2 - 4) + 2x^4}{(x^2 - 4)^2}$$ - Simplify numerator: $$-3x^4 + 12x^2 + 2x^4 = (-3x^4 + 2x^4) + 12x^2 = -x^4 + 12x^2$$ - So: $$y' = \frac{-x^4 + 12x^2}{(x^2 - 4)^2} = \frac{-x^2(x^2 - 12)}{(x^2 - 4)^2}$$ - Set numerator zero: $$-x^2(x^2 - 12) = 0 \Rightarrow x=0 \text{ or } x=\pm \sqrt{12} = \pm 2\sqrt{3}$$ - Critical points at $$x=0, \pm 2\sqrt{3}$$. 6. **Determine intervals of increase/decrease:** - Denominator always positive except at vertical asymptotes. - Sign of $$y'$$ depends on numerator: - For $$|x| < 2\sqrt{3}$$, $$x^2 - 12 < 0$$, so numerator $$-x^2(x^2 - 12) > 0$$ (since $$-x^2$$ is negative, times negative is positive). - For $$|x| > 2\sqrt{3}$$, numerator negative. - So: - Increasing on $$(-2,0) \cup (0, 2\sqrt{3})$$ - Decreasing on $$(-\infty, -2\sqrt{3}) \cup (2\sqrt{3}, -2)$$ but note vertical asymptotes at $$\pm 2$$ split intervals. 7. **Find inflection points:** - Compute second derivative (omitted detailed algebra for brevity): $$y'' = \frac{d}{dx} y'$$ - Inflection points occur where $$y''=0$$ or undefined (excluding vertical asymptotes). - After simplification, inflection points at $$x=\pm 2$$ (vertical asymptotes) and $$x=0$$. 8. **Concavity intervals:** - Using sign of $$y''$$, function is concave up where $$y'' > 0$$ and concave down where $$y'' < 0$$. - Analysis shows concave up on $$(-\infty, -2) \cup (0, 2)$$ and concave down on $$(-2, 0) \cup (2, \infty)$$. 9. **Relative minima and maxima:** - Use first derivative test at critical points: - At $$x=0$$, derivative changes from positive to positive (no extremum). - At $$x=-2\sqrt{3}$$, derivative changes from negative to positive (relative minimum). - At $$x=2\sqrt{3}$$, derivative changes from positive to negative (relative maximum). **Final summary:** - x-intercept and y-intercept at $$0$$. - Vertical asymptotes at $$x=\pm 2$$. - Slant asymptote $$y = -x$$. - Critical points at $$x=0, \pm 2\sqrt{3}$$. - Increasing on $$(-2,0) \cup (0, 2\sqrt{3})$$. - Decreasing on $$(-\infty, -2\sqrt{3}) \cup (2\sqrt{3}, -2)$$. - Inflection points at $$x=0$$. - Concave up on $$(-\infty, -2) \cup (0, 2)$$. - Concave down on $$(-2, 0) \cup (2, \infty)$$. - Relative minimum at $$x=-2\sqrt{3}$$. - Relative maximum at $$x=2\sqrt{3}$$.