\(? = c = {1 \over 2} \cdot b\,\,\,\,\,\,\left[ {\left( {{1 \over 2};\,b} \right)\,\,\, \in \,\,\,{\rm{graph}}\left( f \right)} \right]\)


\({\rm{line}}\,{\rm{:}}\,\,\,\left\{ \matrix{
\,\left( { - 2;0} \right)\,\,\, \in \,\,\,{\rm{line}} \hfill \cr
\,{\rm{slope}} = {{1 - 0} \over {0 - \left( { - 2} \right)}} = {1 \over 2} \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{1 \over 2} = {{y - 0} \over {x - \left( { - 2} \right)}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,y = {1 \over 2}\left( {x + 2} \right)\,\,\,\,\,\left( * \right)\)


\(\left( {{1 \over 2};\,b} \right)\,\,\, \in \,\,\,{\rm{line}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,b = {1 \over 2}\left( {{1 \over 2} + 2} \right) = {1 \over 4} + 1 = 1.25\)


\(? = {{1.25} \over 2} = 0.625\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {\rm{B}} \right)\)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________

Using the congruent triangles approach,
(sq.rt of 6.25)/2=y/1
ie., y=(sq.rt of 6.25)/2
Given, c=x*f(x)=c*y
=0.5*y
=0.625

Answer: B