A function that is tangent to both axes

Question

I am looking to design a function in the plane, $y=f(x)$, which is tangent to both axes $x$ and $y$ at certain points. Say, for example:

$$ f(0)=\alpha>0, \; f(\beta)=0,\; f'(0)=-\infty, \; f'(\beta)=0, \; \beta>0. $$ Moreover, I'd like that the function satisfies $f'(x)<0$ for all $x<\beta$. It is enough if we work only in the first quadrant of the plane.

**this "looks like" an exponential function $e^{-x}$ (or also like a hyperbola) but I want the function to be tangent to the $x$ and $y$ axes at $\beta>0$ and $\alpha>0$. respectively.

I think polynomial nor rational functions will satisfy this, at least not as far as I have tried. Can someone point me towards the right direction? Some sort of special function?

Answer 1

1) Circle $$ (x-1)^2 + (y-1)^2 =1, $$ 2) Parabola $$ (x)^{1/2} + (y)^{1/2} =1, $$ 3) Astroid $$ (x)^{2/3} + (y)^{2/3} =1, $$ 4) Ellipse $$ (x-3)^2 + (y-4)^2 =5^2.. $$

OK, another example. If a ladder of length $L$ is sliding down a wall from a vertical position towards horizontal, find its envelope.

Answer 2

First note that a circle of radius 1 is tangent to the lines $x =1$ and $y=1$

We can define this as $f(x) = \sqrt{1 - x^2}$ has $$f(0) = 1, f'(0) = 0, f(1) = 0, f'(1) = -\infty$$

We can flip this (the section with $x \in [0,1]$ about the line y = 1 - x to get

$$f(0) = 1, f'(0) = -\infty, f(1) = 0, f'(1) = 0$$

In order to do this, we'll take x into 1 - x and y into 1 - y

Then we have $f(x) = 1 - \sqrt{1-(1-x)^2}$

Then note that we want $f(0) = \alpha$ and $f(\beta) = 0$

Thus multiply appropriately to get $f(x) = \alpha(1-\sqrt{1-(1-x/\beta)^2})$