Answer
$$\eqalign{
& {\text{The domain of the function is }}\left[ {0,4} \right] \cr
& {\text{The range of the function is }}\left[ { - 3,3} \right] \cr} $$
Work Step by Step
$$\eqalign{
& \frac{x}{4} = \sqrt {1 - \frac{{{y^2}}}{9}} \cr
& {\text{Square each side}} \cr
& {\left( {\frac{x}{4}} \right)^2} = {\left( {\sqrt {1 - \frac{{{y^2}}}{9}} } \right)^2} \cr
& \frac{{{x^2}}}{{16}} = 1 - \frac{{{y^2}}}{9} \cr
& {\text{Write in standard form}} \cr
& \frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1 \cr
& {\text{This is the equation of an ellipse }}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& a = 4,\,\,\,\,b = 3 \cr
& {\text{with }}x{\text{ - intercepts}}\left( { \pm a,0} \right):\left( { \pm 4,0} \right) \cr
& {\text{with }}y{\text{ - intercepts}}\left( {0, \pm b} \right):\left( {0, \pm 3} \right) \cr
& {\text{In the original equation}},{\text{ the radical expression}}\,\,\,\sqrt {1 - \frac{{{y^2}}}{9}} \cr
& {\text{represents a nonnegative number, number}},{\text{ so the only }} \cr
& {\text{possible values of }}x{\text{ are positive}} \cr
& {\text{The domain of the function is }}\left[ {0,4} \right] \cr
& {\text{The range of the function is }}\left[ { - 3,3} \right] \cr
& \cr
& {\text{Graph}} \cr} $$
