1. The problem is to analyze the function $$g(x) = \frac{x}{x^2 - 25}$$ and understand its properties. 2. First, identify the domain. The denominator cannot be zero, so solve $$x^2 - 25 = 0$$ which gives $$x = \pm 5$$. These values are not in the domain. 3. Find the vertical asymptotes by setting the denominator to zero: vertical asymptotes at $$x = 5$$ and $$x = -5$$. 4. To find horizontal asymptotes, compare degrees of numerator and denominator. Degree numerator = 1, denominator = 2. Because degree denominator > numerator, horizontal asymptote is $$y = 0$$. 5. Find intercepts: - To find the y-intercept, set $$x=0$$: $$g(0) = \frac{0}{0 - 25} = 0$$. - To find x-intercept, set numerator $$x=0$$, so intercept at $x=0$. 6. Find critical points for extrema by differentiating: $$g'(x) = \frac{(1)(x^2 - 25) - x(2x)}{(x^2 - 25)^2} = \frac{x^2 - 25 - 2x^2}{(x^2 - 25)^2} = \frac{-x^2 - 25}{(x^2 - 25)^2}$$ 7. Set numerator of derivative to zero for critical points: $$-x^2 - 25 = 0 \implies x^2 = -25$$, which has no real solutions. So no critical points, therefore no local extrema. Summary: - Domain: all real numbers except $$x = \pm 5$$. - Vertical asymptotes at $$x = \pm 5$$. - Horizontal asymptote at $$y = 0$$. - Intercepts at the origin (0,0). - No extrema. Final answer: The function $$g(x) = \frac{x}{x^2 - 25}$$ has vertical asymptotes at $$x=\pm 5$$, horizontal asymptote $$y=0$$, intercept at (0,0), and no extrema.