Answer
The statement is false.
Work Step by Step
When a function contains a vertical asymptote at $x=1$ (see the image for an example), we know that $y=f(1)$ is undefined at $x=1$. However, a piecewise function can nevertheless be defined at $x=1$, while containing an asymptote there.
For example, consider the piecewise function:
$f(x)=\frac{2}{x-1}$ for $x\ne 1$
$f(x)=5$ for $x=1$
Such a function would contain an asymptote at $x=1$, while still being defined at $x=1$ -- that is, $f(1)=5$.
Thus, the statement is false.
