Answer :
[tex]\boxed{f(x)=\frac{2x(x+3)}{(x-2)(x-3)}}\\\\check:\\vertical\ asymptotes:(x-2)(x-3)\neq0\to x\neq2\ and\ x\neq3\\\boxed{x=2\ and\ x=3}\\\\x-intercepts:f(x)=0\iff\frac{2x(x+3)}{(x-2)(x-3)}=0\iff2x(x+3)=0\\\iff 2x=0\ or\ x+3=0\iff \boxed{x=0\ or\ x=-3}\\\\horizontal\ asymptote:\\\lim\limits_{x\to\pm\infty}\frac{2x(x+3)}{(x-2)(x-3)}=\lim\limits_{x\to\pm\infty}\frac{2x^2+6x}{x^2-5x+6}=\lim\limits_{x\to\pm\infty}\frac{x^2(2+\frac{6}{x})}{x^2(1-\frac{5}{x}+\frac{6}{x^2})}=\frac{2}{1}=\fbox2[/tex]
