Determining Equation based on 2 Points and Horizontal Asymptote

Question

Given that a graph starts at $(a,0)$ and another point $(c,d)$ and a horizontal asymptote $y=b$, where $c$ and $d >0$, is there a way to determine the equation?

Answer 1

Here is one method of determining AN equation (I suspect there are multiple solutions) of the form $y = (a-y)e^{-kx} + b$.

I'll use a specific example where $a = 82, c = 1, d = 79$ and $y = 74$ as the horizontal asymptote.

$y = 8e^{-kx} + 74$ is the general equation.

When $x = c = 1, y = 79$ so.......

$79 = 8e^{-k} + 74$

$5 = 8e^{-k}$

$\frac{5}{8} = e^{-k}$

ln$\frac{5}{8} = -k$

$k = .470$

$y = 8e^{-.47x} + 74$ is the specific solution.

For the modified question above another example is where we have points $(2, 0), (3,3)$ and asymptote $y = 10$..........

$y = 10 - 10e^{-k(x-2)}$ is the general equation

When $x = 3, y = 3$ so.....

$3 = 10 - 10e^{-k}$

$7 = 10e^{-k}$

$0.7 = e^{-k}$

ln $0.7 = -k$

$k = .356675$

$y = 10 - 10e^{-.356675(x-2)}$ is the specific equation

For another form of equation passing through the same points $(2, 0), (3, 3)$ and a$ y = 10$ asymptote, $y = 10 - \frac{70}{3x+1}$ also works.