Example 2.3.1. Finding an Equation of a Sine Function Using its Characteristics.
Write an equation of a sine function with an amplitude of 2, period of 8, phase shift of \(\frac{\pi}{3}\text{,}\) and vertical shift of 4.
Solution.
To determine an equation of the form
\begin{equation*}
f(x)=A\cdot\sin(B(x-C))+D
\end{equation*}
we first examine the information that we are given: amplitude of 2, period of 8, phase shift of \(\frac{\pi}{3}\text{,}\) and vertical shift of 4.
Since \(|A|\) represents the value of the amplitude, we get
\begin{equation*}
|A|=2\text{.}
\end{equation*}
Thus we either have \(A=2\) or \(A=-2\text{.}\) Since we are not given a reflection about the \(x\)-axis, we can conclude that \(A\) is not negative, thus
\begin{equation*}
A=2\text{.}
\end{equation*}
Next, since the period of a sine function is given by \(\frac{2\pi}{|B|}\text{,}\) we get
\begin{equation*}
\frac{2\pi}{|B|}=8\text{.}
\end{equation*}
Solving for \(|B|\text{,}\) we get
\begin{equation*}
|B|=\frac{2\pi}{8}=\frac{\pi}{4}
\end{equation*}
Since there is no reflection about the \(y\)-axis, we have that \(B\) must be positive:
\begin{equation*}
B=\frac{\pi}{4}.
\end{equation*}
Finally, since \(C\) and \(D\) represent the phase shift and vertical shift, respectively, we get
\begin{equation*}
C=\frac{\pi}{3},\quad D=4\text{.}
\end{equation*}
Combining these, the sine function becomes:
\begin{equation*}
f(x)=2\sin\left(\frac{\pi}{4}\left(x-\frac{\pi}{3}\right)\right)+4\text{.}
\end{equation*}
