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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?

DUO Labs
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1 Answers1

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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.

Phil H
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  • Thank you! Are there other forms of equations that have a horizontal asymptote? – DUO Labs Apr 22 '19 at 22:54
  • Also, I slightly changed my original question. Could you update your answer accordingly? – DUO Labs Apr 22 '19 at 23:51
  • See added example in my edit. Equations of the form $y = a - \frac{b}{x}$ will have a horizontal asymptote of $y = a$ – Phil H Apr 23 '19 at 05:28
  • Following up on another form of equation passing through points $(2, 0), (3, 3)$ and a$ y = 10$ asymptote, $y = 10 - \frac{70}{3x+1}$ also works. – Phil H Apr 23 '19 at 20:06
  • Thank you so much! – DUO Labs Apr 23 '19 at 23:48
  • Also, what is the general form of the formula. I got this: $$y=b-be^{-k\left(x-a\right)}$$ Is that right? – DUO Labs Apr 24 '19 at 00:03
  • Yes, it comes from solving a differential equation given a rate of growth dy/dx dependent on the magnitude of y from the asymptote. – Phil H Apr 24 '19 at 05:14
  • Ok, then. Thanks then. – DUO Labs Apr 24 '19 at 17:03
  • Also, can you help me find an exponential formula that pass through the points $(-0.25,0)$ and $(11,5)$ and has the asymptote of $y=6$? – DUO Labs Apr 24 '19 at 17:06
  • Take your general equation $y=b-be^{-k\left(x-a\right)}$ and insert the known values and solve for k. $y = 6 - 6e^{-k(x+0.25)}$. When $y = 5, x = 11$ so $5 = 6 - 6e^{-11.25k}; 1 = 6e^{-11.25k}; \ln\frac{1}{6} = -11.25k; k = .15927$. So the formula is $y = 6 - 6e^{-.15927(x+0.25)}$. – Phil H Apr 24 '19 at 18:23
  • Ok, I understand now! Thank you! – DUO Labs Apr 24 '19 at 20:06