# Is L(t)=1-e^-t an exponential function?

Aug 28, 2015

Yes and no. It's not a "pure" exponential function of the form $f \left(t\right) = a {e}^{k t}$, but it's an transformed version of an exponential function.

#### Explanation:

Exponential functions are those of the form $f \left(t\right) = a {e}^{k t} = a \cdot {b}^{t}$, where $b = {e}^{k}$ (and $k = \ln \left(b\right)$).

The function $f \left(t\right) = {e}^{- t}$ is therefore an exponential (decay) function with $a = 1$ and $k = - 1$.

The function $g \left(t\right) = - f \left(t\right) = - {e}^{- t}$ is also an exponential function with $a = - 1$ and $k = - 1$, and its graph is transformed from the graph of $f$ by reflection across the horizontal axis.

The function $h \left(t\right) = 1 + g \left(t\right) = 1 - f \left(t\right) = 1 - {e}^{- t}$ takes the graph of $g$ and shifts it up by 1 unit.

Here's what the graph of $h$ looks like. It's increasing, concave down, and has a horizontal asymptote $y = 1$ as $t \to + \infty$.

graph{1-e^(-x) [-10, 10, -5, 5]}