For f(x) = ax^2 + bx + c with a ≠ 0 on a restricted domain making f invertible, which expression gives f^{-1}(y) on the left-hand monotonic branch (x ≤ -b/(2a))?

• A. (-b + sqrt(b^2 - 4a(c - y)))/(2a)
• B. (-b - sqrt(b^2 - 4a(c - y)))/(2a)
• C. (b + sqrt(b^2 - 4a(c - y)))/(2a)
• D. (-b - sqrt(b^2 - 4a(c - y)))/(2a) with the left-hand branch

Answer: (B)

When you invert a quadratic on a monotone piece, you solve for x and pick the root that lies on the chosen branch. Start with y = a x^2 + b x + c and rearrange to ax^2 + b x + (c − y) = 0. The quadratic formula gives x = [-b ± sqrt(b^2 − 4a(c − y))]/(2a).

The left-hand monotonic branch corresponds to the smaller x-value. If the parabola opens upward (a > 0), the denominator 2a is positive, so the smaller root comes from the minus sign in front of the square root. Therefore the inverse on this branch is

f^{-1}(y) = (-b − sqrt(b^2 − 4a(c − y)))/(2a).

This matches the left-hand branch behavior, where x ≤ −b/(2a). (If a were negative, the sign that gives the smaller x would flip, but for the standard left branch of an upward-opening parabola, the minus sign is the correct choice.)