Question

Hi, I need help with calculus.
How do I write an exponential function when they give me two coordinates on a graph?

-I Appreciate the help!

Answer 1

Can you provide an example?
What coordinates did they give you? :)


That will make it a bit easier to explain.

Answer 2

They gave me (0,2) and (2,5)

Answer 3

Like I said, the numbers came out very awkward looking :(
I don't think I made a mistake in there... hmm..

Answer 4

hmmm ok. But in the textbook it says to find the exponential function y= Y(subscribe 0) e^kt.
Is that the samething?

Answer 5

\(y_o=C\).
But I didn't allow for the possibility of a \(k\) value.
My mistake, that's why the numbers are coming out all nasty.

Let's try that again! :)

Answer 6

So the instructions are telling you to use the base \(e\) for this exponential function?

Answer 7

yes. " Find the exponential function y=Y(subscribe 0) e^kt whose graphs passes through the two points"

Answer 8

\[\large y=y_o e^{kt}\]
Plugging in our first coordinate (0,2) gives us,\[\large 2=y_oe^0\qquad \rightarrow\qquad y_o=2\]
Giving us,\[\large y=2e^{kt}\]Plugging in our other coordinate (2,5),\[\large 5=2e^{2k}\]To solve for k, we'll have to do a little algebra? Remember your rules of logarithms? :) That's what we'll need to apply right now.

Answer 9

Let's start by dividing both sides by 2, then taking the natural log of both sides.\[\large \ln\left(\frac{5}{2}\right)=\ln\left(e^{2k}\right)\]

Recalling an important rule of logarithms,\[\large \log\left(a^b\right)=b \cdot \log(a)\]We can bring the exponent out as a coefficient in front of the log. Applying that gives us,\[\large \ln\left(\frac{5}{2}\right)=2k \cdot \ln\left(e\right)\]
The natural log of e is just 1.

\[\large 2k=\ln\left(\frac{5}{2}\right)\]

With just a little bit more work, you should be able to find your \(k\) value! :)

Answer 10

Do you by chance have an answer key that we can check this against? :D

Answer 11

Yes. That's what the textbook has an answer.
Thanks.