Answer
$f$ has a vertical asymptote at $x=1$
Work Step by Step
At $x = 1$, the denominator of the function becomes 0 and therefore $f$ is not defined at $x=1$.
Figure $11$ is more deceptive that Figure $10$, since it shows a steep vertical line that supposedly connects the function, but here there should be a vertical asymptote, and x should not be defined at any point at $x = 1$.