# What is the equation for plotting points on a curve with fixed end points?

What is the equation for plotting points on an exponential curve with fixed end points?

For example, if I want to plot 10 point along a curve that starts with 10,000 (x=1, y=10000) and ends with 30,000 (x=10, y=30000) the formula is y=10000*(1.129831^(x-1)).

But if I change 1.129831 to some different value, 2 for example, then the curve tops out at 5,120,000. I'd like to be able to change that base number to adjust the shape of the curve and still have the curve start at a y value of 10,000 and end at a y value of 30,000.

What I need to know is the formula for calculating points along a curve of varying shape, but fixed end points.

Thank you for your help!

-

If you have two points, you have two equations in two unknowns. In the case you cite, you have the points $(x_1,y_1)=(1,10000)$ and $(x_2,y_2)=(10,30000).$ If you claim the equation is $y=ka^x$ by taking the log you can solve for $k$ and $a$ and you don't have any choice. (note that you can absorb the $-1$ in the exponent in your solution into $k$-that just shifts $x_1$ down to $0$)

If you know the $y$ values of interest and want to specify $a$ (as $2$ in your example) you need to provide another degree of freedom. Maybe letting $x_2$ vary will satisfy your need. In that case, you need $a^{(x_2-x_1)}=\frac{y_2}{y_1}$, so in your example if $\frac{y_2}{y_1}=3$ and $a=2, x_2-x_1=\log_23$, so you can space $10$ points from $1$ to $1+\log_23$

-
Thank you for your answer Ross. I think I phrased my question poorly and figured out how to get what I was after. I also attempted to write out the equation and am giving the Excel equivalent so that others can benefit from it. –  Kuyenda Apr 3 '11 at 7:29

I think I phrased my question poorly.

I want to plot $y$ at any point $x$ on a curve across any given number of points $n$ greater than 0 using growth $b$ and I want to adjust the scale and base of the curve by specifying the minimum $y_1$ and maximum $y_n$.

I believe this is how to write the formula:

$y=y_1+\frac{(y_1b^{x-1}-y_1)(y_n-y_1)}{y_1b^{n-1}-y_1}$

In Excel you would use the following formula. I'm using named ranges in the formula to make it easier to follow.

=Minimum+(((Minimum*Growth^(Position-1))-Minimum)*(Maximum-Minimum)/((Minimum*Growth^(Periods-1))-Minimum))

Here's an example result set that describes a steep curve:

$n = 10$
$b = 2$
$(x_1, y_1) = (1, 10000)$
$(x_n, y_n) = (n, 30000)$

Gives:

$(x_1, y_1) = (1, 10000)$
$(x_2, y_2) = (2, 10039)$
$(x_3, y_3) = (3, 10117)$
$(x_4, y_4) = (4, 10274)$
$(x_5, y_5) = (5, 10587)$
$(x_6, y_6) = (6, 11213)$
$(x_7, y_7) = (7, 12466)$
$(x_8, y_8) = (8, 14971)$
$(x_9, y_9) = (9, 19980)$
$(x_{10}, y_{10}) = (10, 30000)$

Here's another example result set that describes a shallow curve:

$n = 10$
$b = 1.1$
$(x_1, y_1) = (1, 10000)$
$(x_n, y_n) = (n, 30000)$

Gives:

$(x_1, y_1) = (1, 10000)$
$(x_2, y_2) = (2, 11473)$
$(x_3, y_3) = (3, 13093)$
$(x_4, y_4) = (4, 14875)$
$(x_5, y_5) = (5, 16835)$
$(x_6, y_6) = (6, 18992)$
$(x_7, y_7) = (7, 21364)$
$(x_8, y_8) = (8, 23973)$
$(x_9, y_9) = (9, 26843)$
$(x_{10}, y_{10}) = (10, 30000)$

Hopefully someone else will find this useful. Maybe someone can even find a way to simplify it.

Thanks!

-
If you set n=1 the denominator is 0-not good. You have not done anything to set the scale of $x$. Normally you would not have an additive term-as in my answer you would have $y=y_1f(x)$ where $f(x)$ is chosen to give $y_2$ at the desired $x$ value. In your formula, it is the $b^{n-1}$ term, but if you set $b=2$ as you want, $y$ will step from 10000 to 20000 to 40000, skipping 30000. The exponent should have a scale, looking like $b^{\delta x \cdot (n-1}$ –  Ross Millikan Apr 3 '11 at 15:12
Ross, I agree on the n=1 gives a divided by zero error, but this equation is for calculating growth in more than one period anyway. I added the condition $n>1$ for clarification. I'm trying out $b^{bx(n-1)}$. –  Kuyenda Apr 3 '11 at 15:24
The data you show do not fit an exponential curve. For 10 points from 10000 to 30000, you want the curve you listed in the original question, and the second point would be 11298, not 10039. –  Ross Millikan Apr 3 '11 at 16:04
Ross, I tried putting the scale in the exponent, but I don't think I have the math chops to figure it out. I did, however, put up an example result set that hopefully explains the means to my end. Again, I'd be happy if someone could simplify the equation. Thanks. :) –  Kuyenda Apr 3 '11 at 16:05
Instead of "exponential curve" should I just be saying "curve?" –  Kuyenda Apr 3 '11 at 16:10