91Ó°ÊÓ

A table of information

To draw a scatter diagram of the data for one period.

The scatter plot for the data is:

Amplitude$=\frac{Largestvalue-Smallestvalue}{2}$

$$\begin{gathered} =\frac{80-34.6}{2} \\ =22.7 \end{gathered}$$

Vertical shift$=\frac{Largestvalue+Smallestvalue}{2}$

$$\begin{gathered} =\frac{80+34.6}{2} \\ =57.3 \end{gathered}$$

The data repeats for every $12$months, so

$$\begin{gathered} T=12 \\ \frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}=12 \\ \mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{12} \\ =\frac{\mathrm{Ï€}}{6} \end{gathered}$$

The function is $y=22.7\sin \left(\frac{\mathrm{Ï€}}{6}x\right)+57.3$

To determine the horizontal shift, we use the period $T=12$and divide the interval $(0,12)$into four sub intervals of length $12÷4=4$

$(0,3),(3,6),(6,9),(9,12)$

The sine curve is increasing on the interval $(0,3)$and is decreasing on the interval $(3,9)$so a local maximum occurs at $x=3$. The data indicate that
a maximum occurs at $x=7$, so we must shift the graph of the function $4$units to the right by replacing $x$ by $x-4$

So we obtain the function, $y=22.7\sin \left(\frac{\mathrm{Ï€}}{6}\right(x-4)+57.3$

So the function is $y=22.7\sin (\frac{\mathrm{Ï€}}{6}x-\frac{2\mathrm{Ï€}}{3})+57.3$

The graph of the sinusoidal function $y=22.7\sin (\frac{\mathrm{Ï€}}{6}x-\frac{2\mathrm{Ï€}}{3})+57.3$is:

By entering the data in the graphing utility, we found the sinusoidal function as,

$y=22.7\sin (\frac{\mathrm{Ï€}}{6}x-\frac{\mathrm{Ï€}}{2})+58.042$

Plot the function in the graph.