Answer
Graph
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{x^2} - 6x + 12}}{{x - 4}} \cr
& {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr
& y = \frac{{{0^2} - 6\left( 0 \right) + 12}}{{\left( 0 \right) - 4}} \cr
& y = - 3 \cr
& y{\text{ - intercept }}\left( {0, - 3} \right) \cr
& {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr
& 0 = \frac{{{x^2} - 6x + 12}}{{x - 4}} \cr
& {x^2} - 6x + 12 = 0 \cr
& {\text{No real solutions}} \cr
& {\text{No, }}x{\text{ - intercepts}} \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {\frac{{{x^2} - 6x + 12}}{{x - 4}}} \right] \cr
& y' = \frac{{\left( {x - 4} \right)\left( {2x - 6} \right) - \left( {{x^2} - 6x + 12} \right)\left( 1 \right)}}{{{{\left( {x - 4} \right)}^2}}} \cr
& y' = \frac{{2{x^2} - 6x - 8x + 24 - {x^2} + 6x - 12}}{{{{\left( {x - 4} \right)}^2}}} \cr
& y' = \frac{{{x^2} - 8x + 12}}{{{{\left( {x - 4} \right)}^2}}} \cr
& {\text{Let }}y' = 0{\text{ to find critical points}} \cr
& {x^2} - 8x + 12 = 0 \cr
& \left( {x - 6} \right)\left( {x - 2} \right) = 0 \cr
& x = 2,{\text{ }}x = 6 \cr
& \cr
& *{\text{Find the second derivative}} \cr
& y'' = \frac{d}{{dx}}\left[ {\frac{{{x^2} - 8x + 12}}{{{{\left( {x - 4} \right)}^2}}}} \right] \cr
& {\text{Using a CAS }} \cr
& y'' = \frac{8}{{{{\left( {x - 4} \right)}^3}}} \cr
& \cr
& {\text{Evaluate }}y''\left( x \right){\text{ at }}x = 2{\text{ and }}x = 6 \cr
& y''\left( 2 \right) = \frac{8}{{{{\left( {2 - 4} \right)}^3}}} < 0,{\text{ relative maximum at }}x = 2 \cr
& f\left( 2 \right) = - 2 \to {\text{Relative maximum at }}\left( {2, - 2} \right) \cr
& y''\left( 6 \right) = \frac{8}{{{{\left( {6 - 4} \right)}^3}}} > 0,{\text{ relative minimum at }}x = 6 \cr
& f\left( 6 \right) = 6 \to {\text{Relative minimum at }}\left( {6,6} \right) \cr
& \cr
& {\text{Let }}y''\left( x \right) = 0 \cr
& \frac{8}{{{{\left( {x - 4} \right)}^3}}} = 0 \cr
& {\text{No, inflection points}} \cr
& \cr
& {\text{*Calculate the asymptotes}} \cr
& \frac{{{x^2} - 6x + 12}}{{x - 4}} \cr
& x - 4 = 0 \cr
& x = 4 \cr
& {\text{Vertical asymptotes at }}x = 4 \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} - 6x + 12}}{{x - 4}}} \right) = \infty \cr
& \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{{x^2} - 6x + 12}}{{x - 4}}} \right) = - \infty \cr
& {\text{No horizontal asymptotes}} \cr
& \cr
& {\text{Graph}} \cr} $$
