How can I “tighten” the curvature of an exponential function to satisfy three given points?

Question

I'm trying to map a range:

$$0 \le x \le 1$$

non-linearly to another range:

$$0.01 \le y \le 4$$

where $x = 0.5$ must map to $y = 1$. Essentially, I have three known points:

x  0.00  0.50  1.00
y  0.01  1.00  4.00

and $y$ cannot ever be $\lt 0.01$ (I mention this explicitly because I have also toyed with quadratics in an attempt to solve this problem).

My distant high-school maths memories tell me that I need an exponential function to achieve this. To that end, I followed the instructions in $\infty+1$'s YouTube video, Write an Exponential Equation Given 3 Points, and worked from $y = ar^x + c$ to arrive at this very-close solution:

$$ y = \frac{13167}{6700}\cdot\left(\frac{100}{33}\right)^{x}-\frac{131}{67} $$

This solution satisfies $(0,0.01)$ and $(1,4)$, but not $(0.5,1)$. I guess I need to somehow “tighten” the curve such that the $y$ values for $0 \le x \le 0.5$ are lower without affecting point $(1,4)$, but I'm at a loss as to how I might do this. I guess I have chosen the wrong exponential equation to start from.

Should I be starting from another exponential equation, and should I be able to solve this with my reasonably basic maths skills of algebra and simultaneous equations (and not enough understanding to know where to start with terms like regression, monotone, and differentiable as found in answers to similar questions)?

Answer 1

Using $y = ar^x + c$ and using whole numbers as you obviously did, we have three equations $$a+c=\frac{1}{100} \tag 1$$ $$a \sqrt{r}+c=1 \tag 2$$ $$a r+c=4 \tag 3$$ Subtract $(1)$ from $(2)$ and $(1)$ from $(3)$; $c$ is no more here $$a \sqrt{r}-a=\frac{99}{100}\tag 4$$ $$a r-a=\frac{399}{100} \tag 5$$ From $(5)$ $$a=\frac{399}{100 (r-1)}\tag 6$$ Plug in $(4)$ to get $$\frac{300-99 \sqrt{r}}{100 \left(\sqrt{r}+1\right)}=0 \implies \sqrt r=\frac {300}{99}=\frac{100}{33}\implies r=\frac{10000}{1089}$$ back to $a$ and $c$ $$y=\frac{3267}{6700} \left(\frac{10000}{1089}\right)^x-\frac{32}{67}$$

Edit

If you have three points $(0,y_1)$, $(0.5,y_2)$, $(1,y_3)$, the generic formulae are $$a=\frac{(y_1-y_2)^2}{y_1-2 y_2+y_3}\qquad r=\left(\frac{y_2-y_3}{y_1-y_2}\right)^2\qquad c=\frac{y_1 y_3-y_2^2}{y_1-2 y_2+y_3}$$

This can be generalized for three points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ if $x_2=\frac 12 (x_1+x_3)$ using as intermediate variable $t=\frac{x-x_1}{x_3-x_1}$ and the formual becomes $$y= ar^{\frac{x-x_1}{x_3-x_1}}+c$$ $(a,r,c)$ being the same as just above.

Answer 2

I prefer to add another answer for the problem of the curvature.

Back to the original problem with data points $(0,y)$, $(0.5,1.0)$, $(1,4)$ the parameters are given by $$a=\frac{(y-1)^2}{y+2}\qquad r=\frac{9}{(y-1)^2}\qquad c=\frac{4 y-1}{y+2}$$For $f(x)=a r^x+c$, the curvature at $x=0.5$ is given by $$\kappa=\frac{a (\log (r)+2)}{2 \sqrt{r} \left(1+a^2 r \log ^2(r)\right)^{3/2}}$$ which would give the awful $$\kappa=\frac{(1-y)^3 (y+2)^2 \left(2+\log (9)-2\log \left(1-y\right)\right)}{6 \left((y+2)^2+9 (1-y)^2 \log ^2\left(\frac{9}{(1-y)^2}\right)\right)^{3/2}}\tag 1$$ Since $y \ll 1$, we can expand as a Taylor series to get, as an approximation, $$\kappa=\frac{2 (2+\log (9))}{3 \left(4+9 \log ^2(9)\right)^{3/2}}+\frac{ (2 \log (9) (9 (\log (9)-2) \log (9)-122)-40)}{3 \left(4+9 \log ^2(9)\right)^{5/2}}y+\cdots$$ which, numerically is $$\kappa=0.0085608 -0.0120139 y$$ from which you can compute the value of $y$ for the desired curvature.

Using $(1)$, as a function of $y$, some values $$\left( \begin{array}{cc} 0.00 & 0.00856080 \\ 0.01 & 0.00844115 \\ 0.02 & 0.00832246 \\ 0.03 & 0.00820472 \\ 0.04 & 0.00808791 \\ 0.05 & 0.00797201 \\ 0.06 & 0.00785702 \\ 0.07 & 0.00774292 \\ 0.08 & 0.00762968 \\ 0.09 & 0.00751730 \\ 0.10 & 0.00740577 \\ 0.11 & 0.00729507 \\ 0.12 & 0.00718518 \\ 0.13 & 0.00707609 \\ 0.14 & 0.00696779 \\ 0.15 & 0.00686027 \\ 0.16 & 0.0067535 \\ 0.17 & 0.00664749 \\ 0.18 & 0.00654221 \\ 0.19 & 0.00643765 \\ 0.20 & 0.00633381 \end{array} \right)$$