Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 15

$$\eqalign{ & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Relative maximum at }}\left( {4,8} \right) \cr & {\text{Relative mainimum at }}\left( {0,0} \right) \cr & {\text{No inflection points}}{\text{.}} \cr & {\text{Vertical asymptote }}x = 2 \cr & {\text{Slant asymptote }}x + 2 \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{{x^2}}}{{x - 2}} \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & y = \frac{{{{\left( 0 \right)}^2}}}{{\left( 0 \right) - 2}} \cr & y = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}y = 0 \cr & 0 = \frac{{{x^2}}}{{x - 2}} \cr & {x^2} = 0 \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{{x - 2}}} \right] \cr & f'\left( x \right) = \frac{{\left( {x - 2} \right)\left( {2x} \right) - {x^2}\left( 1 \right)}}{{{{\left( {x - 2} \right)}^2}}} \cr & f'\left( x \right) = \frac{{2{x^2} - 4x - {x^2}}}{{{{\left( {x - 2} \right)}^2}}} \cr & f'\left( x \right) = \frac{{{x^2} - 4x}}{{{{\left( {x - 2} \right)}^2}}} \cr & {\text{Let }}y' = 0 \cr & \frac{{{x^2} - 4x}}{{{{\left( {x - 2} \right)}^2}}} = 0 \cr & x\left( {x - 4} \right) = 0 \cr & x = 0,{\text{ }}x = 4 \cr & \cr & {\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^2} - 4x}}{{{{\left( {x - 2} \right)}^2}}}} \right] \cr & f''\left( x \right) = \frac{{{{\left( {x - 2} \right)}^2}\left( {2x - 4} \right) - 2\left( {{x^2} - 4x} \right)\left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^4}}} \cr & {\text{Simplifying by hand}} \cr & f''\left( x \right) = \frac{{8x - 16}}{{{{\left( {x - 2} \right)}^4}}} \cr & f''\left( x \right) = \frac{{8\left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^4}}} \cr & f''\left( x \right) = \frac{8}{{{{\left( {x - 2} \right)}^3}}} \cr & {\text{Evaluate the second derivative at the critical points}} \cr & *f''\left( 0 \right) = \frac{8}{{{{\left( {0 - 2} \right)}^3}}} = - 1 < 0 \cr & {\text{Relative maximum at }}\left( {0,f\left( 0 \right)} \right) \to \left( {0,0} \right) \cr & *f''\left( 4 \right) = \frac{8}{{{{\left( {4 - 2} \right)}^3}}} = 1 > 0 \cr & {\text{Relative minimum at }}\left( {4,f\left( 4 \right)} \right) \to \left( {4,8} \right) \cr & \cr & *{\text{Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & \frac{8}{{{{\left( {x - 2} \right)}^3}}} = 0 \cr & {\text{There are no values at which }}f''\left( x \right) = 0 \cr & {\text{No inflection points}}{\text{.}} \cr & \cr & {\text{*Calculate the asymptotes}} \cr & \frac{1}{{x - 2}} - 3 \cr & x - 2 = 0 \to x = 2 \cr & {\text{Vertical asymptote }}x = 2 \cr & \cr & *{\text{Use long division}} \cr & f\left( x \right) = \frac{{{x^2}}}{{x - 2}} = x + 2 + \frac{4}{{x - 2}} \cr & {\text{Slant asymptote }}x + 2 \cr & \cr & {\text{Graph}} \cr} $$