Question

what is the relationship between these two graphs?

Answer 1

\[y = A(1-e^{-kx})\]

\[y = Ae^{-kx}\]

Answer 2

how would we get from one of these graphs to the other?

Answer 3

like I know they are not inversely related @triciaal

Answer 4

was thinking logs don't remember much

Answer 5

@jim_thompson5910 will you please help with this, thanks

Answer 6

Well, I guess if we took the inverse of the second equation i'm wondering if it would lead to this.

\[x = Ae^{-ky}\]

\[\frac{ x }{ A } = e^{-ky}\]

\[\log(\frac{ x }{ A }) = \log(e^{-ky})\]

\[\log(x)-\log(A) = -ky\]

\[\frac{ -1 }{ k }(\log(x)-\log(A) = y)\]

Answer 7

you can write the first one as A - Ae^(-kt)

I think they are asking for the transformation of y = Ae^-kt to this graph

Answer 8

\[\Large y = A(1-e^{-kx})\]
\[\Large y = A-A*e^{-kx}\]
\[\Large y = -A*e^{-kx}+A\]

Answer 9

which will be a reflection and a translation

Answer 10

- can you see what those are?

Answer 11

question: is this a reflection about A?

Answer 12

nvm, I think the graph is shifted up by A units

Answer 13

Hint what is the transform of x^2 to -x^2?

Answer 14

well let me see like that would make all y values negative (x,y) --> (x,-y) across the x axis because the y values become negative.

Answer 15

Yes the translation part is an upward shift of A units

Answer 16

ok cool, at least I got that part lol

Answer 17

yea

1. reflection in x-axis
2 . translation A units upwards

Answer 18

Okay Now I see what you did Jim so like the first part involved re-writing the second equation \[A-Ae^{-kx} => -Ae^{-kx}+A\]

Answer 19

\[f(x)+C\] up by C units

Answer 20

Now after that I see what's happening now

\[f(x) <=> -f(x)\]

Answer 21

\[-Ae^{-kx} , Ae^{-kx}\] thanks guys