# How do you determine if the equation y = (0.25)^x represents exponential growth or decay?

Jul 21, 2016

Represents decay

#### Explanation:

$\textcolor{b r o w n}{\text{Suppose "x<0" then we have } y = \frac{1}{{0.25}^{n}}}$

$\textcolor{m a \ge n t a}{\text{ } \uparrow}$
$\textcolor{m a \ge n t a}{\text{ ( only for negative "x")}}$

Where $n = \left(- 1\right) \times x$

As you read from left to right $n$ becomes closer and closer to 0 so

${0.25}^{n}$ becomes bigger thus $\frac{1}{0.25} ^ n$ becomes less

Think of $\frac{1}{{\left(\frac{1}{2}\right)}^{3}} = \frac{1}{\frac{1}{8}} \text{ but } \frac{1}{{\left(\frac{1}{2}\right)}^{2}} = \frac{1}{\frac{1}{4}}$

so $\frac{1}{{\left(\frac{1}{2}\right)}^{3}} \text{ " >" } \frac{1}{{\left(\frac{1}{2}\right)}^{2}}$

$\textcolor{b l u e}{\text{For "x < 0 " there is decay as "x" increases.}}$

$- 2 \text{ is more than } - 3$ as you are moving to the right on the number line.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b r o w n}{\text{Suppose "x>=0" then we have } y = {\left(0.25\right)}^{x}}$

At $x = 0$ we have $y = \frac{1}{1} = 1$

At $x = {0}^{+} y < 1$ so there is decay

At $x > 0 , {\left(0.25\right)}^{x} \text{ becomes even less}$

$y = {\left(0.25\right)}^{1} = 0.25$
$y = {\left(0.25\right)}^{2} = \frac{1}{4} ^ 2 = \frac{1}{16}$

$y = {\left(0.25\right)}^{2} = \frac{1}{4} ^ 3 = \frac{1}{64}$

$\textcolor{b l u e}{\text{For "x >=0" there is decay as "x" increases}}$