O-Level Differentiation and integration: worked solution

Question

Part of the curve y = √(2x + 5) is shown, with A the point on it where x = 2; the normal at A crosses the x-axis at B. Find the equation of that normal at A.

y = sqrt(2x + 5). At A, x = 2, so y = sqrt(9) = 3, giving A = (2, 3). Differentiate: dy/dx = (1/2)(2x + 5)^(-1/2) * 2 = 1/sqrt(2x + 5). At x = 2, dy/dx = 1/sqrt(9) = 1/3, so the tangent gradient is 1/3. The normal gradient is the negative reciprocal, -3. Equation of normal at A: y - 3 = -3(x - 2), i.e. y = -3x + 9.