Question

y=e^(-x-2)+1

Answer 1

What are you looking to solve?

Answer 2

sketch the graph and tell when X -> ∞, what is the asymptotic behavior of y?

Answer 3

y = 1/[e^(x+2) +1, so x->infinitive, y approach=0+1 = 1

Answer 4

can you please explain how you got 1/(e^x+2)+1 and the behavior? I understand the graph

Answer 5

ok, y-1 =1/[e^(x+2) ( i move 1 to the left)
x becomes x+2, which mean u need to decrease x value by 2 to maintain equality, thus left 2 units. y becomes y-1, thus need to move up 1 unit, for equality to work. that helps?

Answer 6

still confused :/

Answer 7

in the original function, it's y =1/e^(x). in this one, it's y=1/[e^(x+2)] +1
or y-1 = 1/[e^(x+2)], the y value got subtracted by 1, so you need to move up 1 unit(y value) to maintain equality. So, x also need to decrease 2 units(or move to the left)

Answer 8

yeah i get that. So how did you determine the behavior again?

Answer 9

lim 1/e^(x+2) = 0. y approach 1

Answer 10

what is lim? can you explain that please?

Answer 11

oh, it's limit. as x approach infinitive, y approaches 1

Answer 12

do you tell that y is approaching 1 by looking at the graph or the equation?

Answer 13

Both will tell u