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

AI Thread Summary
The discussion revolves around finding the parameters A, b, and c for the exponential function f(x) = A(b^x) + c, given specific points and a horizontal asymptote. The points (-2, 13) and (0, 5) lead to equations that help establish the values of A and c, with c determined to be 4 from the asymptote. By substituting these values into the equations, it is concluded that A must equal 1. However, there is confusion regarding the value of b, as attempts to solve for it using the equations do not yield consistent results. The thread highlights the challenges in determining the correct value of b while ensuring the function satisfies all given conditions.
Niaboc67
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Homework Statement


The graph goes through the points (-2, 13) and (0, 5) and has the horizontal asymptote y = 4.

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

b =

The Attempt at a Solution


f(−2) = 13 therefore:
1 (B^-2 ) = 13

b = ? not sure

Thank you
 
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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.
 
Would it be 9?
Because A+4=5
A=1
Therefore: 1(9)^-2 +4 =13?
 
Niaboc67 said:
Would it be 9?
Because A+4=5
A=1
Therefore: 1(9)^-2 +4 =13?
No, that doesn't work. 9-2 + 4 = 1/81 + 4 ≠ 13
 
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