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

  1. Oct 14, 2014 #1
    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. jcsd
  3. Oct 14, 2014 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

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

    So you need to solve [itex]A+ 4= 5[/itex] and [itex]Ab^{-2}+ 4= 13[/itex].
     
  4. Oct 14, 2014 #3
    Would it be 9?
    Because A+4=5
    A=1
    Therefore: 1(9)^-2 +4 =13?
     
  5. Oct 14, 2014 #4

    Mark44

    Staff: Mentor

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