The graphs of y=7x+1 and y=kx^3 are represented in the figure shown

\(? = m\)


\(C = \left( {c,8} \right) \in {\rm{line}}\,\,\,\,\, \Rightarrow \,\,\,\,\,8 = 7 \cdot c + 1\,\,\,\,\, \Rightarrow \,\,\,\,\,c = 1\)

\(C = \left( {c,8} \right) = \left( {1,8} \right) \in {\rm{cubic}}\,\,\,\,\, \Rightarrow \,\,\,\,\,8 = k \cdot {1^3}\,\,\,\,\, \Rightarrow \,\,\,\,\,k = 8\)

\(A = \left( {0,n} \right) \in {\rm{line}}\,\,\,\,\, \Rightarrow \,\,\,\,\,n = 7 \cdot 0 + 1\,\,\,\,\, \Rightarrow \,\,\,\,\,n = 1\)

\(B = \left( {m,n} \right) = \left( {m,1} \right) \in {\rm{cubic}}\,\,\,\,\,\mathop \Rightarrow \limits^{k\, = \,8} \,\,\,\,\,1 = 8 \cdot {m^3}\)

\(? = m = \root 3 \of {{1 \over 8}} = \root 3 \of {{{\left( {{1 \over 2}} \right)}^3}} = {1 \over 2}\)


The correct answer is (C).


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

Regards,
Fabio.