1. **State the problem:** We need to sketch the graph of the function $$y=\frac{x^2}{x^2-4}$$. 2. **Identify the domain:** The denominator cannot be zero, so solve $$x^2-4=0$$ which gives $$x=\pm 2$$. These are vertical asymptotes. 3. **Simplify the function:** The function is already simplified as $$\frac{x^2}{x^2-4}$$. 4. **Find intercepts:** - **y-intercept:** Set $$x=0$$, then $$y=\frac{0}{0-4}=0$$. - **x-intercept:** Set $$y=0$$, so $$\frac{x^2}{x^2-4}=0$$ implies $$x^2=0$$, so $$x=0$$. 5. **Find horizontal asymptote:** For large $$|x|$$, $$y \approx \frac{x^2}{x^2} = 1$$, so horizontal asymptote is $$y=1$$. 6. **Analyze behavior near vertical asymptotes:** - As $$x \to 2^+$$, denominator $$x^2-4 \to 0^+$$, numerator $$4$$, so $$y \to +\infty$$. - As $$x \to 2^-$$, denominator $$x^2-4 \to 0^-$$, so $$y \to -\infty$$. - Similarly for $$x \to -2^+$$, $$y \to -\infty$$. - For $$x \to -2^-$$, $$y \to +\infty$$. 7. **Summary:** - Vertical asymptotes at $$x=\pm 2$$. - Horizontal asymptote at $$y=1$$. - Intercept at origin $$(0,0)$$. This information allows sketching the graph accurately.