Chapter 10 - Analytic Geometry - Chapter 10 Test Prep - Review Exercises - Page 1002: 45

$$\eqalign{ & {\text{Domain: }}\left( { - \infty ,\infty } \right) \cr & {\text{Range: }}\left( { - \infty , - 1} \right] \cr & {\text{The equation is a function}} \cr} $$

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