How to handle maxima/minima of quadratics quickly in CAT?

✓ Accepted answer

The vertex formula is correct but the fastest mental route is completing the square on the fly. Completing the square: for x² – 6x + 14, rewrite as (x – 3)² + 5. The squared term is always ≥ 0, so the minimum is 5 at x = 3. No formula needed — just halve the coefficient of x and square it, then adjust the constant. AM-GM shortcut for products: CAT loves disguising extrema as 'if a + b = k, find the maximum of a × b.' By AM-GM, the product is maximised when a = b = k/2, giving (k/2)². Minimum product questions usually add a constraint (a, b > 0 and a ≠ b) that CAT uses to trip you. Quick orientation rule: if the coefficient of x² is positive, the parabola opens upward → there's a minimum (no finite maximum). If negative, it opens downward → there's a maximum. State this first to avoid picking the wrong extreme under pressure. For CAT difficulty, they'll wrap the quadratic inside an absolute value or give you a range constraint for x. Completing the square first, then applying the range constraint, is the clean two-step that keeps the working legible.