A curve has a gradient given by dy/dx = k*x - 4. If the curve has a stationary point at x = 2 and passes through (1, 3), find the equation of the curve.
Answer and explanation
Correct answer: B. y = x^2 - 4x + 6
- Step 1: Step 1: At a stationary point, dy/dx = 0. At x = 2: k(2) - 4 = 0 => 2k = 4 => k = 2. Thus dy/dx = 2x - 4. Step 2. Step 2: Integrate: y = x^2 - 4x + C. Use point (1, 3): 3 = 1^2 - 4(1) + C => 3 = -3 + C => C = 6. Curve is y = x^2 - 4x + 6. Answer: y = x^2 - 4x + 6. Why not the others: Checking stationary point: dy/dx = 2x - 4 = 0 at x = 2; point (1, 3): 1 - 4 + 6 = 3. Perfectly consistent.
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