# How do you determine whether each function represents exponential growth or decay y=0.4(1/3)^x?

Dec 31, 2017

it's decay. see explanation

#### Explanation:

Exponential function:
Formula: $y = a \cdot {b}^{x} + c$
where:
-a is multyplier of ${b}^{x}$;
-c moves function on y axis
-b is a base of exponential function. b must be greater than 0 and can't be 1 (everything on the first is equal one, so it makes no sense thinking about it as exponential function)
$b \in \left(0 , 1\right) \bigcup \left(1 , \infty\right)$
if b>1 then it is growing; if $b \in \left(0 , 1\right)$ then it's decreasing.

Given: $y = 0.4 {\left(\frac{1}{3}\right)}^{x}$
a=0.4;
$b = \frac{1}{3} \in \left(0 , 1\right) \quad \implies \quad$ function represents exponential decay

graph{y=0.4(1/3)^x [-10, 10, -5, 5]}

Dec 31, 2017

#### Explanation:

This is an interesting problem, but one way that you can determine weather a function is undergoing exponential decay or growth, is by considering its behavour as $x$ gets large, as a comparison to when $x$ is smaller...

$y = 0.4 \cdot {\left(\frac{1}{3}\right)}^{x}$

As $x \to \infty$ or gorws to a very large number, we see it tends towards 0, this function is undergoing exponential decay, we can also see this if we sketch this function But we can generalise this a bit further:

If we have $y = {\beta}^{x}$

If $0 < \beta < 1$ The function represents exponential decay

If $\beta > 1$ This is exponential growth

If $\beta = 1$ The equation becomes $y = {1}^{x} = 1$ This is just a straight line...

So for example:

$y = {24}^{- 3 x}$ We see ' $\beta$ ' Is greater than one, but...

${24}^{- 3 x} \equiv {\left({24}^{- 3}\right)}^{x} \equiv {\left(\frac{1}{24} ^ 3\right)}^{x}$

Hence this function undergoes exponential decay...