Answer
$$\eqalign{
& y{\text{ - intercept }}\left( {0,5} \right) \cr
& x{\text{ - intercepts: }}\left( {\frac{5}{3},0} \right) \cr
& {\text{local minimum at }}\left( {\frac{1}{3},\frac{9}{2}} \right) \cr
& {\text{local maximum at }}\left( {3,\frac{1}{2}} \right) \cr
& {\text{inflection points }}\left( {4.4048,0.4463} \right) \cr
& {\text{Vertical asymptotes }}x = - 1,{\text{ }}x = 1 \cr
& {\text{Horizontal asymptote }}y = 0 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{3x - 5}}{{{x^2} - 1}} \cr
& {\text{Domain: }}\left( { - \infty , - 1} \right) \cup \left( { - 1,1} \right) \cup \left( {1,\infty } \right) \cr
& \cr
& {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr
& f\left( 0 \right) = \frac{{3\left( 0 \right) - 5}}{{{{\left( 0 \right)}^2} - 1}} \cr
& f\left( 0 \right) = 5 \cr
& y{\text{ - intercept }}\left( {0,5} \right) \cr
& \cr
& {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr
& \frac{{3x - 5}}{{{x^2} - 1}} = 0 \cr
& 3x - 5 = 0 \cr
& x = \frac{5}{3} \cr
& x{\text{ - intercepts: }}\left( {\frac{5}{3},0} \right) \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{3x - 5}}{{{x^2} - 1}}} \right] \cr
& f'\left( x \right) = \frac{{\left( {{x^2} - 1} \right)\left( 3 \right) - \left( {3x - 5} \right)\left( {2x} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{3{x^2} - 3 - 6{x^2} + 10x}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{ - 3{x^2} + 10x - 3}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr
& {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr
& - 3{x^2} + 10x - 3 = 0 \cr
& 3{x^2} - 10x + 3 = 0 \cr
& \left( {3x - 1} \right)\left( {x - 3} \right) = 0 \cr
& x = \frac{1}{3},x = 3 \cr
& \cr
& *{\text{Find the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{ - 3{x^2} + 10x - 3}}{{{{\left( {{x^2} - 1} \right)}^2}}}} \right] \cr
& {\text{Differentiate using a graphing utility we obtain}} \cr
& f''\left( x \right) = \frac{{6{x^3} - 30{x^2} + 18x - 10}}{{{{\left( {{x^2} - 1} \right)}^3}}} \cr
& {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }}x = \frac{1}{3},x = 3 \cr
& *f''\left( {\frac{1}{3}} \right) = \frac{{6{{\left( {1/3} \right)}^3} - 30{{\left( {1/3} \right)}^2} + 18\left( {1/3} \right) - 10}}{{{{\left( {{{\left( {1/3} \right)}^2} - 1} \right)}^3}}} = \frac{{81}}{8} > 0,{\text{ then}} \cr
& {\text{There is a local minimum at }}\left( {\frac{1}{3},f\left( {\frac{1}{3}} \right)} \right) \cr
& f\left( {\frac{1}{3}} \right) = \frac{{3\left( {1/3} \right) - 5}}{{{{\left( {1/3} \right)}^2} - 1}} = \frac{9}{2} \cr
& \to {\text{local minimum at }}\left( {\frac{1}{3},\frac{9}{2}} \right) \cr
& *f''\left( 3 \right) = \frac{{6{{\left( 3 \right)}^3} - 30{{\left( 3 \right)}^2} + 18\left( 3 \right) - 10}}{{{{\left( {{{\left( 3 \right)}^2} - 1} \right)}^3}}} = - \frac{1}{8} < 0,{\text{ then}} \cr
& {\text{There is a local maximum at }}\left( {3,f\left( 3 \right)} \right) \cr
& f\left( 3 \right) = \frac{{3\left( 3 \right) - 5}}{{{{\left( 3 \right)}^2} - 1}} = \frac{1}{2} \cr
& \to {\text{local maximum at }}\left( {3,\frac{1}{2}} \right) \cr
& \cr
& {\text{*Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr
& \frac{{6{x^3} - 30{x^2} + 18x - 10}}{{{{\left( {{x^2} - 1} \right)}^3}}} = 0 \cr
& {\text{Solve using a graphing utility we obtain}} \cr
& x \approx 4.4048 \cr
& f\left( {4.4048} \right) = \frac{{3\left( {4.4048} \right) - 5}}{{{{\left( {4.4048} \right)}^2} - 1}} = 0.4463 \cr
& {\text{The inflection points is: }} \cr
& \left( {4.4048,0.4463} \right) \cr
& \cr
& f\left( x \right) = \frac{{3x - 5}}{{{x^2} - 1}} \cr
& {\text{The denominator is zero at }}x = \pm 1 \cr
& {\text{Vertical asymptotes }}x = - 1,{\text{ }}x = 1 \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{3x - 5}}{{{x^2} - 1}}} \right) = 0 \cr
& \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{3x - 5}}{{{x^2} - 1}}} \right) = 0 \cr
& {\text{Horizontal asymptote }}y = 0 \cr
& \cr
& {\text{Graph}} \cr} $$
