# How do you find the domain and range for (2/3)^x – 9?

Jun 14, 2015

$f \left(x\right) = {\left(\frac{2}{3}\right)}^{x} - 9$

$f \left(x\right)$ is well defined for all $x \in \mathbb{R}$ so the domain is $\mathbb{R}$

By looking at end behaviour we find the range of $f \left(x\right)$ is $\left(- 9 , \infty\right)$

#### Explanation:

As $x \to - \infty$ we have ${\left(\frac{2}{3}\right)}^{x} = {\left(\frac{3}{2}\right)}^{- x} \to \infty$,

so $f \left(x\right) \to \infty$

As $x \to \infty$ we have ${\left(\frac{2}{3}\right)}^{x} \to 0$, so $f \left(x\right) \to - 9$

So range $f \left(x\right) = \left(- 9 , \infty\right)$

graph{(2/3)^x - 9 [-22.5, 22.5, -11.25, 11.25]}