Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 18

$$\eqalign{ & y{\text{ - intercept }}\left( {0,\frac{3}{8}} \right). \cr & x{\text{ - intercept }}\left( {\frac{3}{2},0} \right) \cr & {\text{No relative extrema}}{\text{.}} \cr & {\text{No inflection points}}{\text{.}} \cr & {\text{Vertical asymptote at }}x = 4 \cr & {\text{Horizontal asymptote }}y = 1 \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{2x - 3}}{{2x - 8}} \cr & {\text{By the long division}} \cr & f\left( x \right) = 1 + \frac{5}{{2x - 8}} \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = 1 + \frac{5}{{2\left( 0 \right) - 8}} \cr & f\left( 0 \right) = \frac{3}{8} \cr & y{\text{ - intercept }}\left( {0,\frac{3}{8}} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & 0 = \frac{{2x - 3}}{{2x - 8}} \cr & \frac{{2x - 3}}{{2x - 8}} = 0 \cr & 2x - 3 = 0 \cr & x = \frac{3}{2} \cr & x{\text{ - intercept }}\left( {\frac{3}{2},0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {1 + \frac{5}{{2x - 8}}} \right] \cr & f'\left( x \right) = - \frac{{10}}{{{{\left( {2x - 8} \right)}^2}}} \cr & {\text{Let }}f'\left( x \right) = 0 \cr & - \frac{{10}}{{{{\left( {2x - 8} \right)}^2}}},{\text{ there are no values at which }}f'\left( x \right) = 0, \cr & {\text{No relative extrema}}{\text{.}} \cr & \cr & *{\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ { - 10{{\left( {2x - 8} \right)}^{ - 2}}} \right] \cr & f''\left( x \right) = - 10\left( { - 2} \right){\left( {2x - 8} \right)^{ - 3}}\left( 2 \right) \cr & f''\left( x \right) = \frac{{40}}{{{{\left( {2x - 8} \right)}^3}}} = \frac{{40}}{{8{{\left( {x - 4} \right)}^3}}} \cr & f''\left( x \right) = \frac{5}{{{{\left( {x - 4} \right)}^3}}} \cr & {\text{Let }}f''\left( x \right) = 0 \cr & \frac{5}{{{{\left( {x - 4} \right)}^3}}} = 0 \cr & {\text{There are no values at which }}y'' = 0. \cr & {\text{No inflection points}}{\text{.}} \cr & \cr & {\text{*Calculate the asymptotes}} \cr & 1 + \frac{5}{{2x - 8}} \cr & 2x - 8 = 0 \to x = 4 \cr & {\text{Vertical asymptote at }}x = 4 \cr & \mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{5}{{2x - 8}}} \right) = 1 \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {1 + \frac{5}{{2x - 8}}} \right) = 1 \cr & {\text{Horizontal asymptote }}y = 1 \cr & \cr & {\text{Graph}} \cr} $$