Answer
$$\eqalign{
& D:\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right) \cr
& {\text{Range: }}\left( { - \infty ,9} \right) \cr
& {\text{Vertical asymptote: }}x = 0 \cr
& {\text{Horizontal asymptote: }}y = 9 \cr
& {\text{No relative extrema}} \cr
& {\text{No Inflection points}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 9 - \frac{5}{{{x^2}}} \cr
& {\text{From the graph }}\left( {{\text{Using Geogebra}}} \right): \cr
& {\text{Domain: all real numbers except }}x = 0 \cr
& D:\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right) \cr
& {\text{Range: }}\left( { - \infty ,9} \right) \cr
& {\text{Vertical asymptote: }}x = 0 \cr
& {\text{Horizontal asymptote: }}y = 9 \cr
& {\text{Calculate derivatives}} \cr
& f'\left( x \right) = 0 - 5\left( { - 2{x^{ - 3}}} \right) = \frac{{10}}{{{x^3}}}, \cr
& {\text{There are no values at which }}f'\left( x \right) = 0,{\text{ no relative extrema}} \cr
& f''\left( x \right) = 10\left( { - 3{x^{ - 4}}} \right) = - \frac{{30}}{{{x^4}}},{\text{ }} \cr
& {\text{There are no values at which }}f''\left( x \right) = 0,{\text{ no inflection points}} \cr
& \cr
& {\text{Graph}} \cr} $$
