The equation of a curve is given as y = 3x2 − x3. a) Determine the coordinates of the turning points of the curve. b) Determine the nature of the stationary points. c) Determine the intercepts. d) Sketch the curve.

$$y = 3x^2 - x^3$$ $$\frac{dy}{dx} = 6x - 3x^2$$

$$6x - 3x^2 = 0$$ $$3x(2 - x) = 0$$

Hii inatoa solutions mbili:

$$3x = 0 \implies x = 0$$ $$2 - x = 0 \implies x = 2$$

Kwa $x = 0$:

$$y = 3(0)^2 - (0)^3 = 0$$

Turning point ya kwanza ni $(0, 0)$.

Kwa $x = 2$:

$$y = 3(2)^2 - (2)^3 = 3(4) - 8 = 12 - 8 = 4$$

Turning point ya pili ni $(2, 4)$.

Coordinates za turning points ni $\boxed{(0, 0) \text{ na } (2, 4)}$.

Kwa $x = 0$:

$$\frac{d^2y}{dx^2} = 6 - 6(0) = 6$$

Kwa kuwa $\frac{d^2y}{dx^2} > 0$, point $(0, 0)$ ni local minimum.

Kwa $x = 2$:

$$\frac{d^2y}{dx^2} = 6 - 6(2) = 6 - 12 = -6$$

Kwa kuwa $\frac{d^2y}{dx^2} < 0$, point $(2, 4)$ ni local maximum.

Aina ya stationary points ni:
$(0, 0)$ ni $\boxed{\text{local minimum}}$.
$(2, 4)$ ni $\boxed{\text{local maximum}}$.

Badilisha $x = 0$ kwenye equation asili:

$$y = 3(0)^2 - (0)^3 = 0$$

Y-intercept ni $\boxed{(0, 0)}$.

Weka $y = 0$ kwenye equation asili:

$$0 = 3x^2 - x^3$$

Toa $x^2$ kama common factor:

$$0 = x^2(3 - x)$$

Hii inatoa solutions mbili:

$$x^2 = 0 \implies x = 0$$ $$3 - x = 0 \implies x = 3$$

X-intercepts ni $\boxed{(0, 0) \text{ na } (3, 0)}$.