We want the minimum of the quadratic y = x^2 − x − 2. For a parabola y = ax^2 + bx + c with a > 0 the minimum occurs at the vertex x = -b/(2a).
- Identify coefficients: a = 1, b = -1, c = -2.
- Vertex x-coordinate: x_v = -b/(2a) = -(-1)/(2·1) = 1/2.
- Evaluate y at x = 1/2:
y(1/2) = (1/2)^2 - (1/2) - 2 = 1/4 - 1/2 - 2 = 1/4 - 1/2 - 8/4 = 1/4 - 2/4 - 8/4 = -9/4.
Therefore, the minimum value of the parabola is -9/4.