Answer
$there\,is\,\,vertical\,asymptote\,at\,x=3$
Work Step by Step
$f(x)=\frac{x^2-9x+14}{x^2-5x+6}=\frac{(x-2)(x-7)}{(x-2)(x-3)}\\
so\,\,we\,\,will\,\,see\, if\,there\,is\,vertical\,asymptote\,at\,x=2,x=3\\\\
\lim\limits_{x \to 2}\frac{x^2-9x+14}{x^2-5x+6}=\lim\limits_{x \to 2}\frac{(x-2)(x-7)}{(x-2)(x-3)}=\lim\limits_{x \to 2}\frac{x-7}{x-3}=5 \\
since\,the\,limit\,exists\,\\
there\,is\,\,no\,\,vertical\,asymptote\,at\,x=2 \\
\lim\limits_{x \to 3}\frac{x^2-9x+14}{x^2-5x+6}=\lim\limits_{x \to 3}\frac{(x-2)(x-7)}{(x-2)(x-3)}=\lim\limits_{x \to 3}\frac{x-7}{x-3}\\
when\,\,x\,approach\,3\,from\,\,\,left \\
\lim\limits_{x \to 3^-}\frac{x^2-9x+14}{x^2-5x+6}=\lim\limits_{x \to 3^-}\frac{x-7}{x-3}=-\infty \\
(the\,denominator\,is\,negative\,and\,approach\,\,\,0)\\
when\,\,x\,approach\,3\,from\,\,\,right \\
\lim\limits_{x \to 3^+}\frac{x^2-9x+14}{x^2-5x+6}=\lim\limits_{x \to 3^+}\frac{x-7}{x-3}=\infty \\
(the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
(so\,\,there\,is\,\,vertical\,asymptote\,at\,x=3 )\\
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