Calculate the value of \(k\) for which the line \(y = 2x + k\) is a tangent to the curve \(y = 2x^2 - 4x + 5\).

Note that \(y=2x+k\) is a line, while \(y=2x^2 -4x + 5\) is a parabola, so they meet in \(0\), \(1\) or \(2\) points.

If \(y=2x+k\) is a tangent to \(y=2x^2 -4x + 5\), then they meet in exactly one point, so when we solve the two equations simultaneously, we must get just one solution.

Let’s do this: putting \(2x+k = 2x^2-4x+5\) gives \(2x^2 -6x + 5-k = 0\).

For this quadratic equation to have exactly one root, the discriminant \(b^2-4ac\) must be zero. So \((-6)^2-4\times 2(5-k) = 0\), which gives \(k = \dfrac{1}{2}\).