Find an exponential function of the form f(x)=ab^kx +c?

Find an exponential function of the form f(x)=ab^kx +c with b not equal to 1 a not 0 k not 0 that goes through the points (2,0) (5,0) (8,0). Make sure to write your answer in the specified form in your final answer

Find an exponential function of the form f(x)=ab^kx +c?
Pass.
Reply:that's not possible - exponential function has to be either increasing or decreasing, so it canot have same value at 3 different points





moreover, you can find infintely many pairs of b and k that would give you exactly the same function.
Reply:f(x)=ab^kx +c


(2,0) gives 0 = ab^2k + c


The other two points give two more equations.


Then you have 3 equations and 4 unknowns.


That cannot be solved.





(2 , 0) gives 0 = ab^2k + c @


(5 , 0) gives 0 = ab^5k + c $


(5 , 0) gives 0 = ab^5k + c %26amp;





@ - $ gives ab^2k = ab^5k. So 2k=5k So k=0 and that is unwished.


So again: no solution.





See also intel nite above
Reply:I would guess that something is wrong with the question. The points all lie on the x axis i.e. y = 0. The function is not dependent on x at all
Reply:look the simplest answer is y=0 ;


but to put it in exponential form look at this


a=1 b=e , k= - infinity ,c = 0.


which means y=e^(- infinity * x )+ 0


anything * - infinity = - infinity


y= e ^( - infinity )


e^(- infinity) = 0


y= 0





THIS WILL NOT WORK IF THE SOLUTION REQUIRES ALL NUMBERS TO BE REAL .
Reply:a=10


b=0


k=5


c=0


(the point here is that if b not equal to 1 a not 0 k not 0


then b must = 0, then c=0 and a,k can be anything)





f(x) = 10(0^5x) + 0





This function equals 0 for all x%26gt;0


(Note that f(0)=10 however, so this function is not equivalent to


f(x) = 0 !! f is not defined for x%26lt;0, but we dont need it to be.)





Note to Thermo:





In your proof, when you get to:


ab^2k = ab^5k. (this implies b^2k=b^5k or a=0. but a=0 is prohibited)


So 2k=5k


So k=0.


This does not necessarily follow, as you forgot to include the cases:





b=1 or b=0. 1^x = 1 for all x and 0^x = 0 (x%26gt;0)





Since b=0 is the only case not precluded from the question, it is the only solution. Therefore c must be 0 and a and k can be anything as I described.





For the rest of you who say no solution possible, didnt you read my counterexample to your belief?





Again, a=10,b=0,k=5,c=0 IS A SOLUTION TO THE QUESTION AS WRITTEN