Answer
$f\left( x \right) = \frac{{3{x^2} - 7x - 5}}{{x - 3}}$
Work Step by Step
$$\eqalign{
& {\text{Vertical asymptote: }}x = 3 \cr
& {\text{To obtain a vertical asymptote at }}x = 3,{\text{ set in the}} \cr
& {\text{denominator of the rational function an expression}} \cr
& {\text{whose real root is }}x = 3,{\text{ then using }}x - 3 \cr
& f\left( x \right) = \frac{1}{{x - 3}} \cr
& {\text{Slant asymptote: }}y = 3x + 2 \cr
& {\text{To obtain a slant asymptote }}y = 3x + 2, \cr
& {\text{the long division the of the rational function }} \cr
& f\left( x \right) = \frac{{P\left( x \right)}}{{Q\left( x \right)}}{\text{ must be:}} \cr
& {\text{ }}f\left( x \right) = 3x + 2 + \frac{c}{{Q\left( x \right)}},{\text{ where }}c{\text{ is any constant }} \ne {\text{0}} \cr
& \cr
& {\text{Let }}Q\left( x \right) = x - 3,{\text{ then}} \cr
& \frac{{P\left( x \right)}}{{Q\left( x \right)}} = 3x + 2 + \frac{c}{{Q\left( x \right)}} \cr
& \frac{{P\left( x \right)}}{{x - 3}} = 3x + 2 + \frac{c}{{x - 3}} \cr
& \frac{{P\left( x \right)}}{{x - 3}} = 3x + 2 + \frac{c}{{x - 3}} \cr
& \cr
& {\text{Solving for }}P\left( x \right) \cr
& P\left( x \right) = \left( {3x + 2} \right)\left( {x - 3} \right) + c \cr
& P\left( x \right) = 3{x^2} - 9x + 2x - 6 + c \cr
& P\left( x \right) = 3{x^2} - 7x - 6 + c \cr
& {\text{Let }}c = 1 \cr
& P\left( x \right) = 3{x^2} - 7x - 5 \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& f\left( x \right) = \frac{{3{x^2} - 7x - 5}}{{x - 3}} \cr} $$
