\[y = x^{2} - 4x + 2;\ y = - 2x + a.\]
\[x^{2} - 4x + 2 = - 2x + a\]
\[x^{2} - 2x + (2 - a) = 0\]
\[D = 2^{2} - 4(2 - a) \geq 0\]
\[4 - 8 + 4a \geq 0\]
\[4a - 4 \geq 0\]
\[4a \geq 4\]
\[a \geq 1.\]
\[Ответ:\ \ a \in \lbrack 1;\ + \infty).\]