9.1 The parabola $y=4 a x^{2}+4(a+1) x+a^{2}+a+3+\frac{1}{a} \quad(a \neq 0)$ intersects the x-axis at two points. In which direction do the branches of this parabola point - upwards or downwards?
From the condition, it follows that the discriminant $\mathrm{D}$ of the quadratic trinomial $4 a x^{2}+4(a+1) x+a^{2}+a+3+\frac{1}{a}$ is positive:

$$
\begin{aligned}
& \frac{1}{16} D=(a+1)^{2}-a\left(a^{2}+a+3+\frac{1}{a}\right)=a^{2}+2 a+1-a^{3}-a^{2}-3 a-1= \\
& =-a^{3}-a=-a\left(a^{2}+1\right)>0
\end{aligned}
$$

Since $a^{2}+1>0$, then $-a>0$, i.e., $a<0$. Therefore, the leading coefficient $4a$ is negative, and the branches of the parabola are directed downwards.

Answer: downwards.