Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises - Page 201: 59

$$\eqalign{ & {\text{Horizontal: }}y = - 6{\text{ and }}y = 0 \cr & {\text{Vertical: }}x = 1{\text{ and }}x = 3 \cr} $$

$$\eqalign{ & 9{x^2} + {y^2} - 36x + 6y + 36 = 0 \cr & {\text{Differentiate implicitly with respect to }}x \cr & \frac{d}{{dx}}\left[ {9{x^2}} \right] + \frac{d}{{dx}}\left[ {{y^2}} \right] - \frac{d}{{dx}}\left[ {36x} \right] + \frac{d}{{dx}}\left[ {6y} \right] + \frac{d}{{dx}}\left[ {36} \right] = 0 \cr & 18x + 2y\frac{{dy}}{{dx}} - 36 + 6\frac{{dy}}{{dx}} = 0 \cr & {\text{Solve for }}\frac{{dy}}{{dx}} \cr & {\text{ }}2y\frac{{dy}}{{dx}} + 6\frac{{dy}}{{dx}} = 36 - 18x \cr & {\text{ }}\left( {2y + 6} \right)\frac{{dy}}{{dx}} = 36 - 18x \cr & {\text{ }}\frac{{dy}}{{dx}} = \frac{{36 - 18x}}{{2y + 6}} \cr & {\text{ }}\frac{{dy}}{{dx}} = \frac{{18 - 9x}}{{y + 3}} \cr & {\text{The graph has horizontal tangent line when }}\frac{{dy}}{{dx}} = 0 \cr & {\text{ }}\frac{{dy}}{{dx}} = \frac{{18 - 9x}}{{y + 3}} = 0 \cr & {\text{ }}18 - 9x = 0 \cr & {\text{ }}x = 2 \cr & {\text{Substitute }}2{\text{ for }}x{\text{ into the equation }}9{x^2} + {y^2} - 36x + 6y + 36 = 0 \cr & 9{\left( 2 \right)^2} + {y^2} - 36\left( 2 \right) + 6y + 36 = 0 \cr & {\text{ }}36 + {y^2} - 72 + 6y + 36 = 0 \cr & {\text{ }}{y^2} + 6y = 0 \cr & {\text{ }}y\left( {y + 6} \right) = 0 \cr & {\text{ }}y = - 6,{\text{ }}y = 0 \cr & {\text{Then the equations of the horizontal tangents lines are}} \cr & y = 0{\text{ and }}y = - 6 \cr & \cr & {\text{The graph has vertical tangent line when }}\frac{{dy}}{{dx}}{\text{ is undefined, the}} \cr & {\text{denominator of }}\frac{{18 - 9x}}{{y + 3}}{\text{ is 0, }}y + 3 = 0 \cr & y + 3 = 0 \cr & {\text{ }}y = - 3 \cr & {\text{Substitute }} - 3{\text{ for }}y{\text{ into the equation }}9{x^2} + {y^2} - 36x + 6y + 36 = 0 \cr & 9{x^2} + {\left( { - 3} \right)^2} - 36x + 6\left( { - 3} \right) + 36 = 0 \cr & 9{x^2} + 9 - 36x - 18 + 36 = 0 \cr & {\text{ }}9{x^2} - 36x + 27 = 0 \cr & {\text{ }}{x^2} - 4x + 3 = 0 \cr & {\text{ }}\left( {x - 3} \right)\left( {x - 1} \right) = 0 \cr & {\text{ }}x = 1,{\text{ }}x = 3 \cr & {\text{Then the equations of the vertical tangents lines are}} \cr & x = 1{\text{ and }}x = 3 \cr & \cr & {\text{Horizontal: }}y = - 6{\text{ and }}y = 0 \cr & {\text{Vertical: }}x = 1{\text{ and }}x = 3 \cr} $$