1. **State the problem:** We are given the function $$y=\frac{-2x^3}{x^2-4}$$ and want to analyze it. 2. **Identify the domain:** The denominator cannot be zero, so solve $$x^2-4=0$$ which gives $$x=\pm 2$$. These points are excluded from the domain. 3. **Simplify the function if possible:** The numerator is $$-2x^3$$ and denominator is $$x^2-4=(x-2)(x+2)$$. No common factors to cancel. 4. **Find intercepts:** - **y-intercept:** Set $$x=0$$, then $$y=\frac{-2\cdot0^3}{0^2-4}=0$$. - **x-intercept:** Set $$y=0$$, numerator must be zero, so $$-2x^3=0\Rightarrow x=0$$. 5. **Vertical asymptotes:** At $$x=2$$ and $$x=-2$$ since denominator is zero and numerator is nonzero. 6. **Horizontal or oblique asymptotes:** Degree numerator (3) > degree denominator (2), so no horizontal asymptote. Perform polynomial division: $$\frac{-2x^3}{x^2-4} = -2x - \frac{8x}{x^2-4}$$ As $$x\to \pm \infty$$, $$\frac{8x}{x^2-4} \to 0$$, so oblique asymptote is $$y=-2x$$. 7. **Summary:** - Domain: $$x \neq \pm 2$$ - Intercepts: (0,0) - Vertical asymptotes: $$x=2$$ and $$x=-2$$ - Oblique asymptote: $$y=-2x$$ Final answer: The function $$y=\frac{-2x^3}{x^2-4}$$ has vertical asymptotes at $$x=\pm 2$$, an oblique asymptote $$y=-2x$$, and passes through the origin (0,0).