All polynomials have similar curvature for extreme values of x; they accelerate towards positive or negative infinity (by accelerate I mean the gradient function increases).
All cubics (and indeed all polynomials with odd order) accelerate towards one infinity for extreme positive values for x, and towards the other for extreme negative values of x, because the x^3 term is sensitive to the sign of x.
At the 'centre' of the cubic, the gradient is relatively small, and most often passes 0 and is negative for a while. However, it will always pass 0 again and remain positive forever. There are always 0, 1 or 2 stationary points, and this can be explained by the equation of a quadratic. The gradient function of a cubic is a quadratic, which always has 0, 1 or 2 distinct real roots. Therefore, if a curve has no stationary points, its gradient function has no real roots.
I have chosen not to use that method for forming the equation of the cubic, rather I will use known values.
We can see that x = - 1 is a root of this cubic, so we can write the function as a product of the linear factor (x + 1) and an unknown quadratic factor (ax^2 + bx + c). Multiplying this out, we get:
f(x) = ax^3 + (a + b)x^2 + (b + c)x + c
Using the known values of f(x) for x = 0, x = 1 and x = 2: f(0) = 1, f(1) = 2, f(2) = 4, we have three equations in three variables which we can solve simultaneously:
f(0) = c = 1 => f(x) = ax^3 + (a + b)x^2 + (b + 1)x + 1
f(1) = a + (a + b) + (b + 1) + 1 = 2 => 2a + 2b = 0 => f(x) = ax^3 + (b + 1)x + 1
f(2) = 8a + 2(b + 1) + 1 = 4 => 8a + 2b = 1 => 6a = 1 => a = 1/6, b = -1/6
f(x) = (1/6)x^3 + (-1/6 + 1)x + 1 = 1/6 (x^3 + 5x + 6)