Transform a straight line into a curve I have an equation which is a straight-line with negative slope from (0, y_intercept) to (x_intercept, 0). Is there a mathematical transformation I can use to "curve" this line?
Stated another way, how would you create a graph which looks similar to y = 1/x, but has an explicit x and y intercepts?
Note, it doesn't matter what the function does outside the range of 0 to x_intercept.
Apologies for not knowing much math jargon, I'm translating to code.
Any help or tips or correction of my thinking would be greatly appreciated. 
 A: $$y=\dfrac2{x+1}-1=\dfrac{1-x}{1+x}$$ defines a curve that has $y$-intercept $1$ and $x$-intercept $1$  .

(I started with $y=\dfrac1x$ and added $1$ in the denominator to make the $y$-intercept $1$. 
But that has no $x$-intercept, so I multiplied by $2$ and subtracted $1,$
to get $x$-intercept $1$ while maintaining $y$-intercept $1.)$
A: Update: suppose you want to create a curve from $(0,a)$ to $(b,0)$.
Then the following will do:
$$
\varphi_{a,b}(x)
:= \frac{a}{b} \frac{b - x}{1 + ax}
$$
for all $a,b > 0$.

To expand on @J.W. Tanner's answer: For any $a > 0$
$$
f_a(x) := \frac{a(1 - x)}{a + x}
$$
fulfills the condition:
We have


*

*$f_{a}(1) = \frac{a(1 - 1)}{a + 1} = \frac{0}{a + 1} = 0$ and

*$f_{a}(x) = 1 \iff a + x = a - ax \iff x + ax = 0 \implies x = \frac{0}{a + 1} = 0$.


Here's a plot for $a \in \{1,1.5, 2,3\}$ (for $x \in (0,1)$ we have $f_a < f_b$ for $a < b$):

If fact, the function
$$
f_{a,b}(x)
:= \frac{a(1 - x)}{ax + b}
$$
will also do for all $a,b > 0$ as you can check like I did above.
A: I did a lot of experiments to find the following general solution:
m = (y2 - y1) / (x2 - x1)
y(x) = y1 + (m (x - x1) / xslope)
Where:

*

*y1 is the start y point desired;

*y2 is the end y point desired;

*x1 is the start x point desired;

*x2 is the end x point desired;

*slope is the curvature desired. 0 for straight, 0 to 1.25 for positive curvature, -100 to 0 for negative curvature

Some curves generated with the values:

*

*y1 = 0.5;

*y2 = 1.0;

*x1 = 0.2;

*x2 = 1.0;

*slope: -10, 0, and 1.25;


With a few changes in the formula, you can create a decreasing sloped too:
y(x) = y2 - (m (x - x1) / xslope)
Some curves generated with the formula above:

