Answer
$\,\,there\,is\,\,\,vertical\,asymptote\,at\\\,x=2,x=0 \\$
Work Step by Step
$f(x)=\frac{x + 1}{x^3 - 4x^2 + 4x}=\frac{x+1}{x(x^2-4x+4)}=\frac{x+1}{x(x-2)^2}\\
so\,we\,will\,see\,if\,there\,is\,vertical\,asymptote\,at\,\\
x=0,x=2\\
\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}\frac{x+1}{x(x-2)^2}\\
when\,x\,approach\,0\,from\,left \\
\lim\limits_{x \to 0^-}f(x)=\lim\limits_{x \to 0^-}\frac{x+1}{x(x-2)^2}=-\infty \\
(as\,x\,approach\,0\,from\,left\,\,\,the\,numerator\,(x+1)approach\,\,1\,\\x\,is\,negative\,and\,approach\,0\,and\,(x-2)^2\,approach\,4 \,\\so\,the\,denominator\,is\,negative\,and\,approach\,\,\,0)\\
$
$when\,x\,approach\,0\,from\,right \\
\lim\limits_{x \to 0^+}f(x)=\lim\limits_{x \to 0^+}\frac{x+1}{x(x-2)^2}=\infty \\
(as\,x\,approach\,0\,from\,right\,\,\,the\,numerator\,(x+1)approach\,\,1\,\\x\,is\,positive\,and\,approach\,0\,and\,(x-2)^2\,approach\,4 \,\\so\,the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
$
$so\,there\,is\,\,\,vertical\,asymptote\,at\,x=0 \\$
$\lim\limits_{x \to 2}f(x)=\lim\limits_{x \to 2}\frac{x+1}{x(x-2)^2}\\
when\,x\,approach\,2\,from\,left \\
\lim\limits_{x \to 2^-}f(x)=\lim\limits_{x \to 2^-}\frac{x+1}{x(x-2)^2}=\infty \\
(as\,x\,approach\,2\,from\,left\,\,\,the\,numerator\,(x+1)approach\,\,3\,\\x\,approach\,2\,and\,(x-2)^2\,is\,positive\,and\,\,approach\,0 \,\\so\,the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
\lim\limits_{x \to 2^+}f(x)=\lim\limits_{x \to 2^+}\frac{x+1}{x(x-2)^2}=\infty \\
(as\,x\,approach\,2\,from\,right\,\,\,the\,numerator\,(x+1)approach\,\,3\,\\x\,approach\,2\,and\,(x-2)^2\,is\,positive\,and\,\,approach\,0 \,\\so\,the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
so\,there\,is\,\,\,vertical\,asymptote\,at\,x=2 \\
$
