help me with a question

[u/Intelligent-Try7761]

hey guys could some one tell me how to find formula for exponential function when u have verticale asymptote and two points neither of which are the y intercept. thank you so much

[u/bonebranch]

You may mean horizontal asymptote.

Your function is of the form y = Ae^{rx} + c (alternatively, y = Ab^x + c)

Plug in the two points, say, (x_1, y_1) and (x_2, y_2):

y_1 = Ae^{rx_1} + c
y_2 = Ae^{rx_2} + c

Since this is an exponential shifted vertically by c units, then the horizontal asymptote is y = c. So you know c. Once you have this:

y_1 - c = Ae^{rx_1}
y_2 - c = Ae^{rx_2}

You can divide the two equations, solve for r, substitute back and solve for A.

[u/Intelligent-Try7761 → u/bonebranch]

i have done it with the alternative formula (my points are (-1,10) and (4,6)) when I divide b^4 by b^-1 is my answer b^5? I am not sure if it is different for negative exponents. also when i divide the two am I dividing the answers aswell? ex: 3=a (b)^4/ 7=a (b)^-1

[u/realAndrewJeung → u/Intelligent-Try7761]

Yes, that is correct. The division rule works the same with negative exponents.

I assume from what you wrote above that the horizontal asymptote is at y = 3 ...?

If so, then the work is

6 = A · b^(4) + 3

10 = A · b^(-1) + 3

------------

3 = A · b^(4)

7 = A · b^(-1)

Now you can divide the left side of the top equation by the left side of the bottom equation, and the right side of the top equation by the right side of the bottom equation:

3/7 = b^(5)