Curvature where tangent is parallel to x-axis

Curve: y = x²(x − 3) = x³ − 3x².

Points with horizontal tangent: dy/dx = 3x² − 6x = 3x(x − 2) = 0 gives x = 0 or x = 2. The points are (0, 0) and (2, −4).

Radius of curvature: ρ = (1 + y₁²)^(3/2)/y₂, with y₂ = 6x − 6. At these points y₁ = 0, so ρ = 1/y₂.

• At (0, 0): y₂ = −6, so ρ = 1/(−6) = −1/6. Magnitude 1/6 (curve bends downward there, a local maximum).
• At (2, −4): y₂ = 6, so ρ = 1/6 (local minimum).

Answer: the radius of curvature is 1/6 at both (0, 0) and (2, −4).