Answer
$$\eqalign{
& {\text{The domain of the function is }}[0,3] \cr
& {\text{The range of the function is }}\left[ { - 4,4} \right] \cr
& {\text{The equation is not a function}} \cr} $$
Work Step by Step
$$\eqalign{
& \frac{x}{3} = - \sqrt {1 - \frac{{{y^2}}}{{16}}} \cr
& {\text{Square each side}} \cr
& {\left( {\frac{x}{3}} \right)^2} = {\left( { - \sqrt {1 - \frac{{{y^2}}}{{16}}} } \right)^2} \cr
& \frac{{{x^2}}}{9} = 1 - \frac{{{y^2}}}{{16}} \cr
& {\text{Write in standard form}}. \cr
& \frac{{{x^2}}}{9} + \frac{{{y^2}}}{{16}} = 1 \cr
& {\text{This is the equation of an ellipse }}\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1 \cr
& a = 4,\,\,\,\,b = 3 \cr
& {\text{with }}x{\text{ - intercepts}}\left( {b,0} \right):\left( {3,0} \right) \cr
& {\text{with }}y{\text{ - intercepts}}\left( {0, \pm a} \right):\left( {0, \pm 4} \right) \cr
& {\text{In the original equation}},{\text{ the radical expression}}\,\,\, - \sqrt {1 - \frac{{{y^2}}}{{16}}} \cr
& {\text{represents a nonnegative number, number}},{\text{ so the only }} \cr
& {\text{possible values of }}x{\text{ are negative}} \cr
& {\text{The domain of the function is }}[0,3] \cr
& {\text{The range of the function is }}\left[ { - 4,4} \right] \cr
& {\text{The equation is not a function}} \cr
& \cr
& {\text{Graph}} \cr} $$
