Always Above

This can be readily solved by first setting the parabolas equal to each other, or:

$kx^2 - 4x +4 = 2x^2 + 2kx$ ;

or $(k+2)x^2 - (4+2k)x + 4 = 0$ ;

or $x = \frac{(4+2k) \pm \sqrt{(4+2k)^2 - 4(k+2)(4)}}{2(k+2)}$ ;

or $x = 1 \pm \frac{\sqrt{4k^2 - 16}}{2(k+2)}$ (i)

In order for the first parabola to always surmount the second, we require the discriminant in (i) to be non-positive so that no more than one real root is obtained (i.e. just a tangency point at $x = 1$ ). This can only occur for $k \in [-2, 2]$ . Hence, $n = -2, m = 2 \Rightarrow \boxed{m + n = 0}$ .