Need a function that matches criteria

Question

I am trying to find a function such that

1. $x$ reaches 0 at a set point $x^*$ such that $x* > 0$
2. as $x \rightarrow 0 \implies y \rightarrow 1.$
3. The curvature of the function between $x = 0$ and $x = x*$ can be changed such that it looks like the red lines in the figure.

I'm thinking the function has to be piecewise but I'm unsure of what to use for the part between $x=0$ and $x = x*$.

I've attached a picture of the kind of function I want, where the red lines represent the curve.

I've tried a negative exponential however I can't seem to get a curve such as the top red line of the figure.

Sorry this feels like a daft question. Any ideas? Thanks!!!

Answer 1

How about $y=(1-\frac x{x^*})^k$ The below shows $k=0.2,0.5.1.0,1.5,2$ The horizontal axis is $\frac x{x^*}$

Answer 2

What about: $$ f_\alpha\colon x\in[0,x_\ast]\mapsto{x_\ast}^{-\alpha}(x_\ast-x)^\alpha $$ for $\alpha > 0$ ($\alpha < 1$ will be concave, $\alpha > 1$ convex, $\alpha=1$ linear)?

See e.g. this plot:

Plot http://www3.wolframalpha.com/Calculate/MSP/MSP4543212ded60fcgi25g900001i55c9egfdg979h7?MSPStoreType=image/gif&s=40&w=390.&h=193.&cdf=RangeControl

Answer 3

For the linear piece, the line must pass through $(0,1)$ and $(m,0)$ has the equation $y = 1 - x/m$.

EDIT If you are looking for all 3 pieces, the second piece could be just a dropped down translated hyperbola $y = A + B/(x+1)$ so passing through $(0,1)$ forces $A+B=1$ and passing through $(m,0)$ forces $A + B/(m+1) = 0$ so solving two equations for $A,B$ will give you what you are seeking.

Similarly, for the inverse curve, try something like $y = -x^2 + Ax + B$ and two points will force the values of $A,B$.