Answer
$$\eqalign{
& {\text{Domain: }}\left[ { - 3,3} \right] \cr
& y{\text{ - intercept: }}\left( {0,0} \right) \cr
& x{\text{ - intercepts: }}\left( { - 3,0} \right),\left( {0,0} \right),\left( {3,0} \right) \cr
& {\text{Relative minimum at }}\left( { - \frac{3}{{\sqrt 2 }}, - \frac{9}{2}} \right) \cr
& {\text{Relative maximum at }}\left( {\frac{3}{{\sqrt 2 }},\frac{9}{2}} \right) \cr
& {\text{Inflection point }}\left( {0,0} \right) \cr
& {\text{No vertical asymptotes}} \cr
& {\text{No horizontal asymptotes}} \cr} $$
Work Step by Step
$$\eqalign{
& y = x\sqrt {9 - {x^2}} \cr
& {\text{Domain: }}9 - {x^2} \geqslant 0 \cr
& - 3 \leqslant x \leqslant 3 \cr
& {\text{Therefore}}{\text{, the domain of the function is: }}\left[ { - 3,3} \right] \cr
& \cr
& {\text{*Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr
& y = 0\sqrt {9 - {0^2}} \cr
& y = 0 \cr
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& \cr
& {\text{*Find the }}x{\text{ intercept}}{\text{, let }}y = 0 \cr
& 0 = x\sqrt {9 - {x^2}} \cr
& x = - 3,{\text{ }}x = 0,{\text{ }}x = 3 \cr
& x{\text{ - intercepts: }}\left( { - 3,0} \right),\left( {0,0} \right),\left( {3,0} \right) \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {x\sqrt {9 - {x^2}} } \right] \cr
& y' = x\frac{d}{{dx}}\left[ {\sqrt {9 - {x^2}} } \right] + \sqrt {9 - {x^2}} \frac{d}{{dx}}\left[ x \right] \cr
& y' = x\left( {\frac{{ - 2x}}{{2\sqrt {9 - {x^2}} }}} \right) + \sqrt {9 - {x^2}} \left( 1 \right) \cr
& y' = - \frac{{{x^2}}}{{\sqrt {9 - {x^2}} }} + \sqrt {9 - {x^2}} \cr
& y' = \frac{{ - {x^2} + 9 - {x^2}}}{{\sqrt {9 - {x^2}} }} \cr
& y' = \frac{{9 - 2{x^2}}}{{\sqrt {9 - {x^2}} }} \cr
& {\text{Let }}y' = 0{\text{ to find critical points}} \cr
& 9 - 2{x^2} = 0 \cr
& 2{x^2} = 9 \cr
& {x^2} = \frac{9}{2} \cr
& x = \pm \frac{3}{{\sqrt 2 }} \cr
& {\text{And is undefined at }}x = \pm 3 \cr
& \cr
& *{\text{Find the second derivative}} \cr
& y'' = \frac{d}{{dx}}\left[ { - \frac{{{x^2}}}{{\sqrt {9 - {x^2}} }} + \sqrt {9 - {x^2}} } \right] \cr
& {\text{Using a CAS }}\left( {{\text{Wolfram Website}}} \right){\text{we obtain }} \cr
& y'' = \frac{{x\left( {2{x^2} - 27} \right)}}{{{{\left( {9 - {x^2}} \right)}^{3/2}}}} \cr
& {\text{Evaluate }}y''\left( x \right){\text{ at }}x = \pm \frac{3}{{\sqrt 2 }} \cr
& y''\left( { - \frac{3}{{\sqrt 2 }}} \right) = 4 > 0,{\text{ relative minimum at }}x = - \frac{3}{{\sqrt 2 }} \cr
& f\left( { - \frac{3}{{\sqrt 2 }}} \right) = - \frac{9}{2} \to {\text{Relative minimum at }}\left( { - \frac{3}{{\sqrt 2 }}, - \frac{9}{2}} \right) \cr
& {\text{Evaluate }}y''\left( x \right){\text{ at }}x = \frac{3}{{\sqrt 2 }} \cr
& y''\left( {\frac{3}{{\sqrt 2 }}} \right) = - 4 < 0,{\text{ relative maximum at }}x = - \frac{3}{{\sqrt 2 }} \cr
& f\left( {\frac{3}{{\sqrt 2 }}} \right) = \frac{9}{2} \to {\text{Relative maximum at }}\left( {\frac{3}{{\sqrt 2 }},\frac{9}{2}} \right) \cr
& x = \pm 3{\text{ are the endpoints}}{\text{, then there are no extreme values}}{\text{.}} \cr
& \cr
& {\text{Let }}y''\left( x \right) = 0 \cr
& \frac{{x\left( {2{x^2} - 27} \right)}}{{{{\left( {9 - {x^2}} \right)}^{3/2}}}} = 0 \cr
& 2{x^2} - 27 = 0,{\text{ }}x = 0 \cr
& x = \pm \sqrt {\frac{{27}}{2}} \left( {{\text{These vales are not in the domain}}} \right),{\text{ }}x = 0 \cr
& x = 0 \cr
& {\text{Inflection point }}\left( {0,f\left( 0 \right)} \right) \to \left( {0,0} \right) \cr
& \cr
& {\text{*Calculate the asymptotes}} \cr
& {\text{No vertical asymptotes}}{\text{, the denominator is 1}}{\text{.}} \cr
& \left( { - \infty ,{\text{ or }}\infty } \right){\text{ are not in the domain}}{\text{, then}} \cr
& {\text{No horizontal asymptotes}}{\text{.}} \cr
& \cr
& {\text{Graph}} \cr} $$
