Chapter 3 - Applications of Differentiation - 3.5 Exercises - Page 203: 60

Graph

$$\eqalign{ & y = \frac{{x - 4}}{{x - 3}} \cr & {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr & y = \frac{{0 - 4}}{{0 - 3}} \to y = \frac{4}{3},{\text{ }}\left( {0,\frac{4}{3}} \right) \cr & y{\text{ - intercept }}\left( {0,\frac{4}{3}} \right) \cr & {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr & 0 = \frac{{x - 4}}{{x - 3}} \cr & x = 4 \cr & x{\text{ - intercept }}\left( {4,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & y' = \frac{d}{{dx}}\left[ {\frac{{x - 4}}{{x - 3}}} \right] \cr & y' = \frac{{\left( {x - 3} \right)\left( 1 \right) - \left( {x - 4} \right)\left( 1 \right)}}{{{{\left( {x - 3} \right)}^2}}} \cr & y' = \frac{{x - 3 - x + 4}}{{{{\left( {x - 3} \right)}^2}}} \cr & y' = \frac{1}{{{{\left( {x - 3} \right)}^2}}} \cr & {\text{Let }}y' = 0 \cr & \frac{1}{{{{\left( {x - 3} \right)}^2}}} = 0,{\text{ there are no values at which }}y' = 0. \cr & {\text{No relative extrema}}{\text{.}} \cr & \cr & {\text{*Calculate the asymptotes}} \cr & \frac{{x - 4}}{{x - 3}} \cr & x - 3 = 0 \to x = 3 \cr & {\text{Vertical asymptote at }}x = 1 \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{x - 4}}{{x - 3}} = 1 \cr & \mathop {\lim }\limits_{x \to - \infty } \frac{{x - 4}}{{x - 3}} = 1 \cr & {\text{Horizontal asymptote }}y = 1 \cr & \cr & {\text{*Symmetry}} \cr & f\left( { - x} \right) = \frac{{\left( { - x} \right) - 4}}{{\left( { - x} \right) - 3}} \cr & f\left( { - x} \right) = - \frac{{x + 4}}{{x + 3}} \cr & {\text{There is no symmetry}}{\text{.}} \cr & \cr & {\text{Graph}} \cr} $$