$解：\left(2\right)如圖1，過D作DG\bot x軸于G，交BC于H，$
$當(dāng)x=0時(shí)，y=-6，$
$\therefore C\left(0,-6\right)，$
$設(shè)BC的解析式為：y=kx+n，$
$則\left\{\begin{array}{l}{n=-6}\\{4k+n=0}\end{array}\right.，解得：\left\{\begin{array}{l}{k=\frac{3}{2}}\\{n=-6}\end{array}\right.，$
$\therefore BC的解析式為：y=\frac{3}{2}x-6，$
$設(shè)D\left(x,\frac{3}{4}{x}^{2}-\frac{3}{2}x-6\right)，則H\left(x,\frac{3}{2}x-6\right)，$
$\therefore DH=\frac{3}{2}x-6-\left(\frac{3}{4}{x}^{2}-\frac{3}{2}x-6\right)$
$=-\frac{3}{4}{x}^{2}+3x，$
$\because \triangle BCD的面積是\frac{9}{2}，$
$\therefore \frac{1}{2}DH\cdot OB=\frac{9}{2}，$
$\therefore \frac{1}{2}\times 4\times \left(-\frac{3}{4}{x}^{2}+3x\right)=\frac{9}{2}，$
$解得：x=1或3，$
$\because 點(diǎn)D在直線l右側(cè)的拋物線上，$
$\therefore D\left(3,-\frac{15}{4}\right)，$
$\therefore \triangle ABD的面積=\frac{1}{2}AB\cdot DG=\frac{1}{2}\times 6\times \frac{15}{4}=\frac{45}{4}$
