At which point of curve 12y = x3 − 3x2, the tangent is parallel to the x-axis?

Concept Used:

For any curve slope(slope of tangent) is given by \(\frac{dy}{dx}\) and the slope of the x-axis is 0.

Calculation:

12y = x3 − 3x2

Differentiating the above equation w.r.t x

12\(\frac{dy}{dx}\) = 3x² - 6x

⇒ \(\frac{dy}{dx}\) = \(\frac{x^2-2x}{4}\)

For the tangent to be parallel to the x-axis slope of the graph should be 0.

⇒ \(\frac{dy}{dx}\) = 0

⇒ \(\frac{x^2-2x}{4}\) = 0

⇒ x² - 2x = 0

⇒ x (x - 2) = 0

⇒ x = 0 or x = 2 

at x = 0, y = 0³- 3 × 0² = 0

at x = 2,

12y = 2³- 3× 2²

⇒ 12y = 8 - 12

⇒ 12y = - 4

⇒ y = \(\frac{-1}{3}\)

Hence, The points on the curve where the tangent is parallel to the x-axis is (0,0) and (2, \(\frac{-1}{3}\)).