Given, object distance $u=-9\,cm$, radius of curvature, $R=-12\,cm$
As we know mirror formula
$\frac{1}{v}+\frac{1}{u} =\frac{2}{R} \Rightarrow \frac{1}{v}-\frac{1}{9}=\frac{2}{12} $
$\Rightarrow v=-18\,cm$
Now, lateral magnification
$m=-\frac{v}{u}=-\frac{(-18)}{(-9)}=-2$
So, image is magnified, real and inverted.