## Precalculus (6th Edition) Blitzer

The given statement does not make sense. The graph has a vertical asymptote only at $x=2$
The given function is $f\left( x \right)=\frac{x-1}{\left( x-1 \right)\left( x-2 \right)}$. Simplifying the above function as, \begin{align} & f\left( x \right)=\frac{x-1}{\left( x-1 \right)\left( x-2 \right)} \\ & =\frac{1}{\left( x-2 \right)} \end{align} Now, the vertical asymptotes is determined by equating the denominator to 0; thus, we’ll get \begin{align} & x-2=0 \\ & x=2 \end{align} Therefore, the vertical asymptote is $x=2$ and there is no vertical asymptote at $x=1$. Thus, the given statement does not make sense.