Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises - Page 201: 58

$$\eqalign{ & {\text{Horizontal: }}y = - 2{\text{ and }}y = 2 \cr & {\text{Vertical: }}x = - 4{\text{ and }}x = 4 \cr} $$

$$\eqalign{ & {x^2} + 4{y^2} + 2xy = 12 \cr & {\text{Differentiate implicitly with respect to }}x \cr & \frac{d}{{dx}}\left[ {{x^2}} \right] + \frac{d}{{dx}}\left[ {4{y^2}} \right] + \frac{d}{{dx}}\left[ {2xy} \right] = \frac{d}{{dx}}\left[ {12} \right] \cr & 2x + 8y\frac{{dy}}{{dx}} + 2x\frac{{dy}}{{dx}} + 2y = 0 \cr & {\text{Solve for }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}}\left( {8y + 2x} \right) = - 2x - 2y \cr & {\text{ }}\frac{{dy}}{{dx}} = \frac{{ - 2x - 2y}}{{8y + 2x}} \cr & {\text{ }}\frac{{dy}}{{dx}} = - \frac{{x + y}}{{4y + x}} \cr & {\text{The graph has horizontal tangent line when }}\frac{{dy}}{{dx}} = 0 \cr & {\text{ }}\frac{{dy}}{{dx}} = - \frac{{x + y}}{{4y + x}} = 0 \cr & {\text{ }} - \frac{{x + y}}{{4y + x}} = 0 \cr & {\text{ }} - x - y = 0 \cr & {\text{ }}x = - y \cr & {\text{Substitute }} - y{\text{ for }}x{\text{ into the function }}{x^2} + 4{y^2} + 2xy = 12 \cr & {\text{ }}{\left( { - y} \right)^2} + 4{y^2} + 2\left( { - y} \right)y = 12 \cr & {\text{ }}{y^2} + 4{y^2} - 2{y^2} = 12 \cr & {\text{ }}3{y^2} = 12 \cr & {\text{ }}y = \pm 2 \cr & {\text{Then the equations of the horizontal tangents lines are}} \cr & y = - 2{\text{ and }}y = 2 \cr & \cr & {\text{The graph has vertica tangent line when }}\frac{{dy}}{{dx}}{\text{ is undefined, the}} \cr & {\text{denominator of }} - \frac{{x + y}}{{4y + x}}{\text{ is 0, }}4y + x = 0 \cr & 4y + x = 0 \cr & {\text{ }}y = - \frac{1}{4}x \cr & {\text{Substitute }} - \frac{1}{4}x{\text{ for }}y{\text{ into the function }}{x^2} + 4{y^2} + 2xy = 12 \cr & {x^2} + 4{\left( { - \frac{1}{4}x} \right)^2} + 2x\left( { - \frac{1}{4}x} \right) = 12 \cr & {\text{ }}{x^2} + \frac{1}{4}{x^2} - \frac{1}{2}{x^2} = 12 \cr & {\text{ }}\frac{3}{4}{x^2} = 12 \cr & {\text{ }}{x^2} = 16 \cr & {\text{ }}x = \pm 4 \cr & {\text{Then the equations of the vertical tangents lines are}} \cr & x = - 4{\text{ and }}x = 4 \cr & \cr & {\text{Horizontal: }}y = - 2{\text{ and }}y = 2 \cr & {\text{Vertical: }}x = - 4{\text{ and }}x = 4 \cr} $$