Chapter 10 - Analytic Geometry - Chapter 10 Test Prep - Review Exercises - Page 1001: 36

$$\eqalign{ & {\text{domain: }}\left[ { - 1,5} \right] \cr & {\text{range: }}\left[ { - 5, - 1} \right] \cr & {\text{Vertices: }}\left( { - 1, - 3} \right){\text{ and }}\left( {5, - 3} \right) \cr} $$

$$\eqalign{ & \frac{{{{\left( {x - 2} \right)}^2}}}{9} + \frac{{{{\left( {y + 3} \right)}^2}}}{4} = 1 \cr & {\text{The equation of the ellipse is in the form }} \cr & \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1\,\,\,\,\left( {a > b} \right) \cr & a = 3,\,\,b = 2 \cr & h = \,2,\,\,k = - 3 \cr & \cr & {\text{Vertices: }}\left( {h \pm a,k} \right) \cr & {\text{Vertices: }}\left( {2 - 3, - 3} \right){\text{ and }}\left( {2 + 3, - 3} \right) \cr & {\text{Vertices: }}\left( { - 1, - 3} \right){\text{ and }}\left( {5, - 3} \right) \cr & \cr & {\text{The domain of the ellipse is }}\left[ {h - a,h + a} \right] \cr & {\text{domain }}\left[ { - 1,5} \right] \cr & \cr & {\text{The range of the ellipse is }}\left[ {k - b,k + b} \right] \cr & {\text{range }}\left[ { - 5, - 1} \right] \cr & \cr & {\text{Therefore,}} \cr & {\text{domain: }}\left[ { - 1,5} \right] \cr & {\text{range: }}\left[ { - 5, - 1} \right] \cr & {\text{Vertices: }}\left( { - 1, - 3} \right){\text{ and }}\left( {5, - 3} \right) \cr & \cr & {\text{Graph}} \cr} $$