To find the values of m for which the line y = mx is a tangent to the curve y = 2x^2 - 4x + 8, we equate the two equations:

\(mx = 2x^2 - 4x + 8\)

Rearrange to form a quadratic equation:

\(2x^2 - (4 + m)x + 8 = 0\)

For the line to be a tangent, the discriminant of this quadratic must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 2\), \(b = -(4 + m)\), \(c = 8\).

Substitute into the discriminant formula:

\((-(4 + m))^2 - 4 \times 2 \times 8 = 0\)

\((4 + m)^2 - 64 = 0\)

\((4 + m)^2 = 64\)

Take the square root of both sides:

\(4 + m = 8\) or \(4 + m = -8\)

Solve for \(m\):

\(m = 4\) or \(m = -12\)