Vertex y-Coordinate of a Parabola

Substituting the vertex abscissa x = −b/(2a) back into y = ax² + bx + c collapses to k = c − b²/(4a), the height of the turning point. Notice the discriminant hiding inside: k = −Δ/(4a), so a parabola with Δ = 0 has its vertex exactly on the x-axis. Worked example: y = x² − 6x + 5 has k = 5 − 36/4 = −4, so the vertex is (3, −4) and the curve dips four units below the axis before climbing back through its roots at x = 1 and x = 5.

Because the vertex is the only turning point, k is the answer to "what is the smallest (or largest) value this expression can take?" — a projectile launched at 20 m/s from ground level follows h = 20t − 4.9t², peaking at k = 0 − 400/(4 × −4.9) ≈ 20.4 m. Getting there by hand means completing the square, the same manoeuvre a Babylonian scribe used on tablet BM 13901 nearly four thousand years ago and that al-Khwārizmī later justified with a literal picture of a square with rectangles glued to two sides. The classic trap is sign handling when a is negative: 4a is then negative, so subtracting b²/(4a) raises k, which is exactly right for a downward-opening curve.