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

Sep 1, 2016

$f \left(x\right) = {\left(\frac{3}{2}\right)}^{x}$

#### Explanation:

The exponential function can be expressed in the form:

$f \left(x\right) = a \cdot {b}^{x}$

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

Note that:

$1 = \frac{9}{4} \cdot \frac{4}{9} = f \left(2\right) \cdot f \left(- 2\right) = \left(a \cdot {b}^{2}\right) \cdot \left(a \cdot {b}^{- 2}\right) = {a}^{2} \cdot {b}^{2} / {b}^{2} = {a}^{2}$

Transposing, we find ${a}^{2} = 1$ and hence $a = 1$, since $b > 0$ and $f \left(2\right) > 0$.

Then:

${b}^{2} = 1 \cdot {b}^{2} = a \cdot {b}^{2} = f \left(2\right) = \frac{9}{4}$

Hence:

$b = \sqrt{\frac{9}{4}} = \frac{3}{2}$

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

So we can write:

$f \left(x\right) = 1 \cdot {\left(\frac{3}{2}\right)}^{x}$

or more simply:

$f \left(x\right) = {\left(\frac{3}{2}\right)}^{x}$

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]}