Chapter 10 - Analytic Geometry - 10.3 Hyperbolas - 10.3 Exercises - Page 988: 27

$$\eqalign{ & {\text{Domain: }}\left( { - \infty ,\infty } \right) \cr & {\text{Range: }}\left[ {4,\infty } \right) \cr} $$

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