f = -4 cm, ho = 2 cm, u = -9 cm
Using mirror formula
\({1\over f} = {1\over v}+{1\over u} \)
\({1\over -4} = {1\over v}+{1\over -9}\)
\({1\over v} = {1\over 9}-{1\over 4}\)
\({1\over v} = {4-9\over 36}\)
\(v=-7.2 cm\)
Now, using the magnification formula, we get
\(m = {h_{i}\over h_{o}}= -{v\over u}\)
\({h_{i}\over h_{o}} = -{(-7.2)\over -9}\)
\({h_{i}\over 2} = -{(-7.2)\over -9}\)
\(h_{i}={{2\times(-7.2)}\over -9}\)
\(h_{i}= -1.6cm\)
Thus, the height of the image '\(h_{i}\)' is 1.6 cm.
Thus, the image is real, inverted and small in size.