# The graph of an exponential function given by f (x) = A(b^x)+c

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1. Oct 14, 2014

### Niaboc67

1. The problem statement, all variables and given/known data
The graph goes through the points (-2, 13) and (0, 5) and has the horizontal asymptote y = 4.

f(−2) = ____ therefore:
____(B^____ ) = ____

b =

3. The attempt at a solution
f(−2) = 13 therefore:
1 (B^-2 ) = 13

b = ? not sure

Thank you

2. Oct 14, 2014

### HallsofIvy

Staff Emeritus
You are told that (according to your title) that $f(x)= Ab^x+ c$
Saying that "f(-2)= 13" means that $f(-2)= Ab^{-2}+ c= 13$.
Saying that "f(0)= 5" means that $f(0)= Ab^0+c= A+ c= 5$ since $b^0= 1$ for all b.
Saying that the "y= 4 is a horizontal asymptote" means that either $\lim_{x\to\infty} f(x)= 4$ or $\lim_{x\to -\infty} f(x)= 4$. In either case, the terms involving "x", $$Ab^x$$ must go to 0 leaving only c= 4.

So you need to solve $A+ 4= 5$ and $Ab^{-2}+ 4= 13$.

3. Oct 14, 2014

### Niaboc67

Would it be 9?
Because A+4=5
A=1
Therefore: 1(9)^-2 +4 =13?

4. Oct 14, 2014

### Staff: Mentor

No, that doesn't work. 9-2 + 4 = 1/81 + 4 ≠ 13