Given the equation \(y = 2x^2 + kx + k - 1\) and \(k = 2\), substitute \(k\) to get:

\(y = 2x^2 + 2x + 2 - 1 = 2x^2 + 2x + 1\)

We need to express this in the form \(y = 2(x + a)^2 + b\).

Start by completing the square for \(2x^2 + 2x\):

\(2(x^2 + x) = 2\left((x + \frac{1}{2})^2 - \frac{1}{4}\right)\)

\(= 2(x + \frac{1}{2})^2 - \frac{1}{2}\)

Now, add the constant term:

\(y = 2(x + \frac{1}{2})^2 - \frac{1}{2} + 1\)

\(= 2(x + \frac{1}{2})^2 + \frac{1}{2}\)

Thus, \(a = \frac{1}{2}\) and \(b = \frac{1}{2}\).

The vertex form \(y = 2(x + a)^2 + b\) gives the vertex at \((-a, b)\).

Therefore, the vertex is \(\left(-\frac{1}{2}, \frac{1}{2}\right)\).