Answer
Sample answer: $\quad k(x)= 1+\displaystyle \frac{1}{1-x}$
Work Step by Step
Horizontal asymptote is $y=1$.
There is a vertical asymptote, $x=1.$
$k(x)$ can have the form
$k(x)=1+$(rational expression with degree of numerator less than that of the denominator)
$(x-1)$ should be a factor of the denominator.
$\displaystyle \frac{1}{x-1} \rightarrow-\infty$ when $x\rightarrow 1^{-}$, so we change the sign:
$\displaystyle \frac{1}{1-x} \rightarrow+\infty$ when $x\rightarrow 1^{-}$, and
$\displaystyle \frac{1}{1-x} \rightarrow-\infty$ when $x\rightarrow 1^{+}.$
$k(x)= 1+\displaystyle \frac{1}{1-x}$ satisfies the conditions of the problem.