Chapter 10 - Analytic Geometry - Chapter 10 Test Prep - Review Exercises - Page 1001: 42

$$\eqalign{ & {\text{Vertices}}:\left( {1, - 3} \right){\text{ and }}\left( {1,1} \right) \cr & {\text{asymptotes: }}y = \pm \left( {x - 1} \right) - 1 \cr & {\text{domain: }}\left( { - \infty , + \infty } \right) \cr & {\text{range: }}\left( { - \infty , - 3} \right] \cup \left[ {1,\infty } \right) \cr} $$

$$\eqalign{ & - 4{x^2} + 8x + 4{y^2} + 8y = 16 \cr & - 4\left( {{x^2} - 2x} \right) + 4\left( {{y^2} + 2y} \right) = 16 \cr & {\text{Complete the square}} \cr & - 4\left( {{x^2} - 2x + 1} \right) + 4\left( {{y^2} + 2y + 1} \right) = 16 - 4\left( 1 \right) + 4\left( 1 \right) \cr & - 4{\left( {x - 1} \right)^2} + 4{\left( {y + 1} \right)^2} = 16 \cr & {\text{Divide both sides by 16}} \cr & \frac{{{{\left( {y + 1} \right)}^2}}}{4} - \frac{{{{\left( {x - 1} \right)}^2}}}{4} = 1 \cr & {\text{The equation is written in the form }}\frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1 \cr & \frac{{{{\left( {y + 1} \right)}^2}}}{4} - \frac{{{{\left( {x - 1} \right)}^2}}}{4} = 1,{\text{ then }}a = 2,\,\,b = 2,\,\,\,h = 1,\,\,\,k = - 1 \cr & \cr & {\text{Characteristics:}} \cr & {\text{Transverse axis: horizontal}} \cr & {\text{vertices: }}\left( {h,k \pm a} \right) \cr & {\text{vertices: }}\left( {1, - 1 \pm 2} \right) \cr & {\text{vertices: }}\left( {1, - 3} \right){\text{ and }}\left( {1,1} \right) \cr & \cr & {\text{asymptotes: }}y = \pm \frac{a}{b}\left( {x - h} \right) + k \cr & {\text{asymptotes: }}y = \pm \left( {x - 1} \right) - 1 \cr & \cr & {\text{domain: }}\left( { - \infty , + \infty } \right) \cr & {\text{The range is: }}\left( { - \infty ,k - a} \right] \cup \left[ {k + a,\infty } \right) \cr & {\text{The range is: }}\left( { - \infty , - 3} \right] \cup \left[ {1,\infty } \right) \cr & \cr & {\text{Graph}} \cr} $$