Chapter 3 - Applications of Differentiation - 3.5 Exercises - Page 203: 61

Graph

$$\eqalign{ & y = \frac{{x + 1}}{{{x^2} - 4}} \cr & {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr & y = \frac{{0 + 1}}{{{0^2} - 4}} \to y = - \frac{1}{4}, \cr & y{\text{ - intercept }}\left( {0, - \frac{1}{4}} \right) \cr & {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr & 0 = \frac{{x + 1}}{{{x^2} - 4}} \cr & x = - 1 \cr & x{\text{ - intercept }}\left( { - 1,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & y' = \frac{d}{{dx}}\left[ {\frac{{x + 1}}{{{x^2} - 4}}} \right] \cr & y' = \frac{{\left( {{x^2} - 4} \right)\left( 1 \right) - \left( {x + 1} \right)\left( {2x} \right)}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr & y' = \frac{{{x^2} - 4 - 2{x^2} - 2x}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr & y' = \frac{{ - {x^2} - 2x - 4}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr & {\text{Let }}y' = 0 \cr & {x^2} + 2x + 4 = 0:{\text{ there are no values at which }}y' = 0. \cr & {\text{No relative extrema}}{\text{.}} \cr & \cr & {\text{*Calculate the asymptotes}} \cr & \frac{{x + 1}}{{{x^2} - 4}} \cr & {x^2} - 4 = 0 \to x = \pm 2 \cr & {\text{Vertical asymptotes at }}x = \pm 2 \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{x + 1}}{{{x^2} - 4}} = 0 \cr & \mathop {\lim }\limits_{x \to - \infty } \frac{{x + 1}}{{{x^2} - 4}} = 0 \cr & {\text{Horizontal asymptote }}y = 0 \cr & \cr & {\text{*Symmetry}} \cr & f\left( { - x} \right) = \frac{{ - x + 1}}{{{{\left( { - x} \right)}^2} - 4}} \cr & f\left( { - x} \right) = \frac{{ - x + 1}}{{{x^2} - 4}} \cr & {\text{There is no symmetry}}{\text{.}} \cr & \cr & {\text{Graph}} \cr} $$