SOLUTION: Please explain why I am getting this wrong. The graph of an exponential function given by f (x) = A(b^x)+c is shown below: http://i.imgur.com/GAywHDn.png

Algebra ->  Graphs -> SOLUTION: Please explain why I am getting this wrong. The graph of an exponential function given by f (x) = A(b^x)+c is shown below: http://i.imgur.com/GAywHDn.png      Log On


   

Question 915987: Please explain why I am getting this wrong. The graph of an exponential function given by f (x) = A(b^x)+c is shown below:
http://i.imgur.com/GAywHDn.png
Found 2 solutions by MathLover1, josh_jordan:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=a%2Ab%5Ex%2Bc
given points: (0,4) and (-2,12)
horizontal asymptote: y=3

from given horizontal asymptote , we know that c=3
so, we have f%28x%29=a%2Ab%5Ex%2B3
use points to find a and b

f%28x%29=a%2Ab%5Ex%2B3 ........if (0,4)
4=a%2Ab%5E0%2B3
4=a%2A1%2B3
4-3=a
1=a
so, we have f%28x%29=1%2Ab%5Ex%2B3 or f%28x%29=b%5Ex%2B3+
f%28x%29=b%5Ex%2B3+ ........if (-2,12)
12=b%5E%28-2%29%2B3+
12-3=b%5E%28-2%29

9=1%2Fb%5E2
9b%5E2=1
b%5E2=1%2F9
b=sqrt%281%2F9%29
b=1%2F3} or+b=-1%2F3
so, your equation is: f%28x%29=%281%2F3%29%5Ex%2B3 or f%28x%29=%28-1%2F3%29%5Ex%2B3


Answer by josh_jordan(263) About Me  (Show Source):
You can put this solution on YOUR website!
The points listed are at (-2,12) and (0,4) and the asymptote is y = 3. You are also given the form of A(b^x) + c. If you plug in the first point you'd get A(b^-2) + 3 = 12, which would give you A(b^-2) = 9. Using your second point, you'd get A(b^0) + 3 = 4, which would give you A + 3 = 4, leaving you with A = 1. So, you can now substitute 1 for your first function: 1(b^-2) = 9. Solving for b would give you 1/3. You can now substitute 1/3 for b, and 3 for c in A(b^x) + c:

f(X) = (1/3)^x + 3