A-Level Differentiation: worked solution

Question

A cubic curve goes through (2, 3) and (−3, −22). Determine its equation given that it has a stationary point at (1, −6).

Method: a general cubic has FOUR unknown coefficients, so four independent facts are needed. The question supplies exactly four: three points on the curve (remember the stationary point is a point on the curve too) and one zero gradient. Step 1 - write the general cubic. y = ax³ + bx² + cx + d Step 2 - substitute each known point. Through (2, 3): 8a + 4b + 2c + d = 3 ... (1) Through (−3, −22): −27a + 9b − 3c + d = −22 ... (2) Through (1, −6): a + b + c + d = −6 ... (3) Step 3 - use the stationary point condition. Differentiate: dy/dx = 3ax² + 2bx + c A stationary point at x = 1 means dy/dx = 0 there: 3a + 2b + c = 0 ... (4) Step 4 - eliminate d, which appears in (1), (2) and (3) only. (1) − (3): (8a − a) + (4b − b) + (2c − c) = 3 − (−6) 7a + 3b + c = 9 ... (5) (3) − (2): (a + 27a) + (b − 9b) + (c + 3c) = −6 − (−22) 28a − 8b + 4c = 16, and dividing every term by 4: 7a − 2b + c = 4 ... (6) Step 5 - subtract (6) from (5); the a and c terms cancel. (3b − (−2b)) = 9 − 4 5b = 5, so b = 1 Step 6 - use (4) to write c in terms of a. 3a + 2(1) + c = 0, so c = −3a − 2 Step 7 - substitute b = 1 and c = −3a − 2 into (5). 7a + 3(1) + (−3a − 2) = 9 4a + 1 = 9 4a = 8, so a = 2 Then c = −3(2) − 2 = −8 Step 8 - find d from (3). 2 + 1 + (−8) + d = −6 −5 + d = −6, so d = −1 Step 9 - write the equation and check it. y = 2x³ + x² − 8x − 1 At x = 2: 16 + 4 − 16 − 1 = 3 ✓ At x = −3: −54 + 9 + 24 − 1 = −22 ✓ At x = 1: 2 + 1 − 8 − 1 = −6 ✓ and dy/dx = 6 + 2 − 8 = 0 ✓