Ans: D
$\begin{array}{rcl}
y & = & (ax+1)^2 + a \\
y & = & a^2 x^2 +2ax + (1 + a) \\
\end{array}$
$\begin{array}{rcl}
y & = & (ax+1)^2 + a \\
y & = & a^2 x^2 +2ax + (1 + a) \\
\end{array}$
Since $a^2 >0$ for $-1\lt a<0$, then the graph opens upwards.
For $-1\lt a<0$, $0 < 1+a < 1$.
Hence the $y$-intercept of the graph is positive.
Therefore, the answer is D.
