Problem: Graph the rational function $f(x)=\frac{3x^2}{x^2-4}$. 1. Factor the denominator and write the function in factored form: $x^2-4=(x-2)(x+2)$. 2. Domain: The denominator is zero at $x=\pm 2$, so the domain is $x\neq -2,2$. 3. Vertical asymptotes: Because the denominator vanishes and the numerator is nonzero at $x=\pm 2$, there are vertical asymptotes at $x=-2$ and $x=2$. 4. Horizontal asymptote: The numerator and denominator have the same degree, so the horizontal asymptote is $y=3$, since $$\lim_{x\to\pm\infty}f(x)=\lim_{x\to\pm\infty}\frac{3x^2}{x^2-4}=3$$. 5. Intercepts: The x-intercepts satisfy $3x^2=0$, giving $x=0$, so the graph passes through the origin at $(0,0)$, which is also the y-intercept. 6. Sign chart and intervals: The numerator $3x^2$ is nonnegative for all x, and the denominator is positive for $|x|>2$ and negative for $|x|<2$, so $f(x)>0$ on $(-\infty,-2)$ and $(2,\infty)$, and $f(x)<0$ on $(-2,0)$ and $(0,2)$. 7. Limits near vertical asymptotes: As $x\to 2^-$, $f(x)\to -\infty$, and as $x\to 2^+$, $f(x)\to +\infty$. As $x\to -2^-$, $f(x)\to +\infty$, and as $x\to -2^+$, $f(x)\to -\infty$. 8. Derivative and extrema: Using the quotient rule, $$f'(x)=\frac{6x(x^2-4)-3x^2(2x)}{(x^2-4)^2}=\frac{-24x}{(x^2-4)^2}$$. Setting $f'(x)=0$ gives $x=0$ as the only critical point in the domain, and since $f'$ changes from positive for $x<0$ to negative for $x>0$, $f$ has a local maximum at $x=0$ with value $f(0)=0$. 9. Further observations: The equation $f(x)=3$ has no solutions, so the graph does not cross the horizontal asymptote $y=3$. 10. Summary of graph features: Domain $x\neq -2,2$. 11. Vertical asymptotes $x=\pm 2$. 12. Horizontal asymptote $y=3$. 13. x- and y-intercept at $(0,0)$ with a horizontal tangent there and a local maximum at that point. Final answer: Sketch the graph of $y=\frac{3x^2}{x^2-4}$ showing vertical asymptotes at $x=\pm 2$, horizontal asymptote $y=3$, the intercept at $(0,0)$, sign behavior on each interval, and the local maximum at $(0,0)$.