The graph of a curve $C$ is a cubic polynomial and it passes through the origin, $(2,0)$ and $(2.55,-0.0631)$. Given that it has a turning point when $x = 0.785$, find the equation of $C$, giving its coefficients correct to 1 decimal place.

1. State the general form of the cubic polynomial.

Let the equation of the cubic curve be $y = ax^3 + bx^2 + cx + d$.
Since the curve passes through the origin $(0,0)$, we have:
$d = 0$
Thus, the equation simplifies to $y = ax^3 + bx^2 + cx$.

2. Form an equation using the point $(2,0)$.

The curve passes through $(2,0)$. Substituting $x = 2$ and $y = 0$:
$a(2)^3 + b(2)^2 + c(2) = 0$
$8a + 4b + 2c = 0$
Dividing by 2 gives:
$4a + 2b + c = 0 \quad \text{--- (1)}$

3. Form an equation using the turning point condition.

The curve has a turning point at $x = 0.785$.
Differentiating $y$ with respect to $x$:
$\frac{dy}{dx} = 3ax^2 + 2bx + c$
At the turning point, $\frac{dy}{dx} = 0$. Substituting $x = 0.785$:
$3a(0.785)^2 + 2b(0.785) + c = 0$
$1.848675a + 1.57b + c = 0 \quad \text{--- (2)}$

4. Form an equation using the point $(2.55, -0.0631)$.

The curve passes through $(2.55, -0.0631)$. Substituting $x = 2.55$ and $y = -0.0631$:
$a(2.55)^3 + b(2.55)^2 + c(2.55) = -0.0631$
$16.581375a + 6.5025b + 2.55c = -0.0631 \quad \text{--- (3)}$

5. Solve the system of equations and write the final rounded polynomial.

Solving the system of linear equations (1), (2), and (3) using a graphing calculator:
$a \approx 0.09930$
$b \approx -0.49681$
$c \approx 0.59641$

Rounding each coefficient to 1 decimal place:
$a = 0.1, \; b = -0.5, \; c = 0.6$

Therefore, the equation of the curve $C$ is:
$y = 0.1x^3 - 0.5x^2 + 0.6x$