Question

What are the horizontal asymptotes of the logistic growth function f(x)=3/(1+6e^(-2x))?

Answer 1

@jim_thompson5910

Answer 2

what does e^-inf go to?

Answer 3

Basically, as x gets larger and larger, what happens to e^(-x)?

Answer 4

Does it get smaller?

Answer 5

yes it does

Answer 6

so as x gets really really really big, e^(-x) gets really really really tiny...effectively making e^(-x) zero

Answer 7

@swimgirly95 there was an error in your last question
range is all real numbers

Answer 8

So this means that as x gets bigger and bigger, e^(-2x) gets smaller and smaller (using the same logic)

Answer 9

Okay =)

Answer 10

so what does that say about the horizontal asymptote?

Answer 11

I believe it is saying that it will be larger than the verticle one?

Answer 12

well in this case, there are no vertical asymptotes

Answer 13

Because y=0, right?

Answer 14

no because you can plug in any real numbers

Answer 15

well as x --> infinty, y --> 3 since 6e^(-2x) ---> 0

Answer 16

So one horizontal asymptote is y = 3

Answer 17

as x --> -infinty, y --> 0 since 6e^(-2x) ---> infinity

Answer 18

So another asymptote is y = 0

Answer 19

okay dokey, sorry, I think I see it now =) THANK YOU!!

Answer 20

you're welcome