# Can someone please describe the end behavior of the function?

Jun 26, 2018

As $x$ increases $f \left(x\right)$ approaches $- 2$

#### Explanation:

${\lim}_{x \to \infty} {k}^{x} \to 0$ when $k < 1$

Therefore

${\lim}_{x \to \infty} {\left(\frac{2}{3}\right)}^{x} - 2 \rightarrow 0 - 2 = - 2$

Jun 26, 2018

As x increases, $f \left(x\right)$ approaches $- 2$.

#### Explanation:

From a simple algebraic perspective, just note that any number larger than one (1) raised to any positive exponent will increase to infinity. Any number LESS than one will rapidly decrease - approaching zero (0), if not quite ever actually reaching it.

Thus, we can see that the first term, ${\left(\frac{2}{3}\right)}^{x}$ , is going to approach $0$ as x increases. Thus the final result of f(x) will be approaching $- 2$.

A couple of quick calculations will show that trend very simply.

For x = 0
$f \left(x\right) = {\left(\frac{2}{3}\right)}^{0} - 2 = 1 - 2 = - 1$
For x = 1
$f \left(x\right) = {\left(\frac{2}{3}\right)}^{1} - 2 = \left(\frac{2}{3}\right) - 2 = - 1.33$
For x = 10
$f \left(x\right) = {\left(\frac{2}{3}\right)}^{10} - 2 = \left(0.017\right) - 2 = - 1.983$
For x = 100
$f \left(x\right) = {\left(\frac{2}{3}\right)}^{100} - 2 = \left(2.435 \times {10}^{- 18}\right) - 2 = - 2.0000 \ldots$