jmc

$△DBC$ has base $\overline{BC}$ of length 8 and height $\overline{BD}$ of length 3; therefore, its area is $\frac{1}{2}\times 8\times 3=12$.

The area of quadrilateral $DBEF$ equals the area of $△DBC$ minus the area of $△FEC$.

$△FEC$ has base $EC=BC-BE=8-4=4$. The height of $△FEC$ is equal to the vertical distance from point $F$ to the $x$-axis, which is equal to the $y$-coordinate of point $F$, or 2. Therefore, the area of $△FEC$ is $\frac{1}{2}\times 4\times 2=4$.

Finally, the area of quadrilateral $DBEF$ is $12-4=\boxed{8}$.