Discovering the Characteristics of Exponential Functions
Discovering the Characteristics of Exponential Functions
The standard form of an exponential function is \(y=a{b}^{x}+q\).
Domain and range
For \(y=a{b}^{x}+q\), the function is defined for all real values of \(x\). Therefore the domain is \(\{x:x\in \mathbb{R}\}\).
The range of \(y=a{b}^{x}+q\) is dependent on the sign of \(a\).
For \(a>0\):
\begin{align*} {b}^{x}& > 0 \\ a{b}^{x}& > 0 \\ a{b}^{x}+q& > q \\ f(x)& > q \end{align*}Therefore, for \(a>0\) the range is \(\{f(x):f(x)>q\}\).
For \(a<0\):
\begin{align*} {b}^{x}& > 0 \\ a{b}^{x}& < 0 \\ a{b}^{x}+q& < q \\ f(x)& < q \end{align*}Therefore, for \(a<0\) the range is \(\{f(x):f(x) Find the domain and range of \(g(x)=5 \cdot {2}^{x}+1\) The domain of \(g(x)=5\times {2}^{x}+1\) is \(\{x:x\in \mathbb{R}\}\). Therefore the range is \(\{g(x):g(x)>1\}\).Example
Question
Find the domain
Find the range
\begin{align*} {2}^{x}& > 0 \\ 5\times {2}^{x}& > 0 \\ 5\times {2}^{x}+1& > 1 \end{align*}
Intercepts
The \(y\)-intercept:
For the \(y\)-intercept, let \(x=0\):
\begin{align*} y& = a{b}^{x}+q \\ & = a{b}^{0}+q \\ & = a(1)+q \\ & = a+q \end{align*}For example, the \(y\)-intercept of \(g(x)=5\times {2}^{x}+1\) is given by setting \(x=0\):
\begin{align*} y& = 5\times {2}^{x}+1 \\ & = 5\times {2}^{0}+1 \\ & = 5+1 \\ & = 6 \end{align*}This gives the point \((0;6)\).
The \(x\)-intercept:
For the \(x\)-intercept, let \(y=0\).
For example, the \(x\)-intercept of \(g(x)=5\times {2}^{x}+1\) is given by setting \(y=0\):
\begin{align*} y& = 5\times {2}^{x}+1 \\ 0& = 5\times {2}^{x}+1 \\ -1& = 5\times {2}^{x} \\ {2}^{x}& = -\cfrac{1}{5} \end{align*}There is no real solution. Therefore, the graph of \(g(x)\) does not have any \(x\)-intercepts.
Asymptotes
Exponential functions of the form \(y=a{b}^{x}+q\) have a single horizontal asymptote, the line \(x=q\).
This lesson is part of:
Functions I