O-Level Quadratic functions: worked solution

Question

For the stone with height h = -9t² + 12t + 1, recast h into the completed-square form a(t + b)² + c, giving a, b and c.

Completing the square means writing the quadratic as a(t + b)² + c, so the coefficient of t² must be taken outside the t² and t terms first (and only those two terms). h = −9t² + 12t + 1 = −9(t² − (12/9)t) + 1 = −9(t² − (4/3)t) + 1 Now complete the square inside the bracket. Halve the coefficient of t: half of −4/3 is −2/3, so the bracket starts (t − 2/3)². Squaring that gives an extra (2/3)² = 4/9 which must be taken away again: t² − (4/3)t = (t − 2/3)² − 4/9 Substitute this back in: h = −9[(t − 2/3)² − 4/9] + 1 = −9(t − 2/3)² + (−9)(−4/9) + 1 = −9(t − 2/3)² + 4 + 1 = −9(t − 2/3)² + 5 Compare with the required form a(t + b)² + c. The bracket must read t + b, and here it reads t − 2/3, so b = −2/3 (not +2/3). a = −9, b = −2/3, c = 5 Check by expanding: −9(t² − (4/3)t + 4/9) + 5 = −9t² + 12t − 4 + 5 = −9t² + 12t + 1, which is the original h.