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The function is $f(x)=A+Be^{rx}$. The y-intercept is $(0, 65.59)$. The x-intercept is $(78, 0)$. For this function $B$ has to be negative and $r$ has to be positive to match the required curve shape. At this point all i've found is that $\frac{1}{r}\ln A=78$ and $A+B=65.59$. I'm just trying to find two simultaneous equations that help find the value of one of the constants in order to find the rest. Cheers

Gregory
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Soren Fenzl-Fields
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    You have two conditions and three unknowns. You will have a curve of acceptable solutions (if any exist). – Gregory Feb 07 '22 at 20:32

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You typically need as many equations as unknowns to find the missing parameters.

$65.59=A+B$, so $B=65.59-A$

$0=A+Be^{78r}=A+(65.59-A)e^{78r}$

$r=-\frac{1}{78}\ln(\frac{-A}{65.59-A})$

$f(x)=A+(65.59-A)(\frac{A}{A-65.59})^{-x/78}$

That's the best you can do without more information. Note this new information doesn't have to be specific function values. It could be, a set area under the curve, maximum value of the first derivative. If you have a set of points you want to approximate, you can probably find a least squares appropriate value for A.

TurlocTheRed
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