Scaling a math function from a specific vertex I have a model developed through a math function as shown below in the figure. I would like to scale the figure as shown in the figure without changing the vertex position.

Explanation of the figure:
The model seen is according to the function implemented. Now, the two vertices on the Y-axis remain where they are originally present and the surface of the model needs to be scaled proportionally in x,y (that is if scale factor of 2 along x, then the same scale factor 2 along y from the vertex) shown in red.



My guess is that the positioning of the origin needs to be changed but not sure how?

Any hacks to do address this?

Thanks.
I'm not sure what the best way to scale that with Blender would be, but I'd look at just adjusting your equation. Here's your $y$ equation:

$$y=\begin{cases}
0.5 &amp; \text{if } u &lt; 0 \\
\frac{u^2}{u^2+(1-u)^2}+0.5 &amp; \text{if } 0 \leq u \leq 1\\
1.5 &amp; \text{if } u &gt; 1
\end{cases}$$

Notice that the main part of the middle equation, $\frac{u^2}{u^2+(1-u)^2}$, normally has a range of $[0,1]$ on the interval $[0,1]$. You are offsetting it by $0.5$. If we scale it, and change the offset of the third piece, we can place it wherever. Use this equation instead:

$$y=\begin{cases}
0.5 &amp; \text{if } u &lt; 0 \\
k \cdot \frac{\left(\frac{u}{k}\right)^2}{\left(\frac{u}{k}\right)^2+\left(1-\frac{u}{k}\right)^2}+0.5 &amp; \text{if } 0 \leq \frac{u}{k} \leq 1\\
k + 0.5 &amp; \text{if } \frac{u}{k} &gt; 1
\end{cases}$$

where $k$ is your scalar. You can try it online using Desmos here. (Thanks to Rich Sedman for help with scaling horizontally).

In a Python expression, it would be this (substituting in whatever value for k):

(0.5 if u&lt;0 else (k*((u/k)**2/((u/k)**2+(1-(u/k))**2)+0.5) if (u/k)&lt;=1 else k+0.5))*v