How to manipulate the $x$-Axis of a Cubic-Bézier-Curve?

Question

I have a Cubic Bézier-Curve a la: $$ f(x) = A (1-x)^3 + 3 B (1-x)^2 x + 3 C (1-x) x^2 +D x^3 $$ with my control constant A,B,C,D which are just constants. x ist running from 0 to 1.

What I want is to control the derivative of the graph at the beginning and at the endpoint. I tried to manipulate the $x$-axis while putting $\sqrt(x)$ or $x^{1/3}$ instead of $x$ in the equation. With that I can only decrease the derivative at the beginning but as you can guess, I want to increase the derivative in the beginning phase as well.

In the following image you can get an imagination of what I am thinking about:

Example Image

Thankful for every help I can get. :)

Greets

Answer 1

The basic intuition you need is:

• $A$ controls the position of the start
• $D$ controls the position of the end
• $B-A$ controls the derivative at the start
• $D-C$ controls the derivative at the end

So rather than trying to find a function $g$ which you can compose with $f$, you probably ought to be trying to find the values of $B$ and $C$ which give the desired gradient.