5. If the parabola $y=x^{2}-(k-1) x-k-1$ intersects the $x$-axis at points $A$ and $B$, and the vertex is $C$, then the minimum value of the area of $\triangle A B C$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
5. (A).

First, $\Delta=(k-1)^{2}+4(k+1)=k^{2}+2 k+5=$ $(k+1)^{2}+4>0$, so for any value of $k$, the parabola always intersects the $x$-axis at two points. Let the $x$-coordinates of the intersection points of the parabola with the $x$-axis be $x_{1}$ and $x_{2}$. Then,
$$
\begin{aligned}
|A B| & =\sqrt{\left(x_{1}-x_{2}\right)^{2}}=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
& =\sqrt{k^{2}+2 k+5} .
\end{aligned}
$$

The vertex coordinates of the parabola are
$$
C\left(\frac{k-1}{2},-\frac{k+2 k+5}{4}\right) \text {, }
$$

Therefore, $\varsigma_{\triangle \operatorname{mic}}=\frac{1}{2} \sqrt{k^{2}+2 k+5} \cdot\left|-\frac{k^{2}+2 k+5}{4}\right|$
$$
=\frac{1}{8} \sqrt{\left(k^{2}+2 k+5\right)^{3}} \text {. }
$$

Since $k^{2}+2 k+5=(k+1)^{2}+4 \geqslant 4$, equality holds when $k=-1$, so,
$$
S_{: A B X} \geqslant \frac{1}{8} \sqrt{4^{3}}=1 .
$$