Question

How do you find the formula of an exponential graph given `f(2)=9/4` and `f(-2)=4/9`?

Answer 1

`f(x) = (3/2)^x`

Explanation:

The exponential function can be expressed in the form:

`f(x) = a*b^x`

where `a` and `b > 0` are to be determined.

Note that:

`1 = 9/4*4/9 = f(2)*f(-2) = (a*b^2)*(a*b^(-2)) = a^2*b^2/b^2 = a^2`

Transposing, we find `a^2 = 1` and hence `a=1`, since `b > 0` and `f(2) > 0`.

Then:

`b^2 = 1*b^2 = a*b^2 = f(2) = 9/4`

Hence:

`b = sqrt(9/4) = 3/2`

(We can ignore the possibility of the negative square root since we want `b > 0)`

So we can write:

`f(x) = 1*(3/2)^x`

or more simply:

`f(x) = (3/2)^x`

graph{(y-(3/2)^x)((x-2)^2+(y-9/4)^2-0.006)((x+2)^2+(y-4/9)^2-0.006) = 0 [-5.087, 4.913, -0.92, 4.08]}