## Precalculus (6th Edition) Blitzer

a. $3$ hours. b. $365$ days. c. $15$ hours d. $9$ hours e. see graph and explanations.
a. Given the function $y=3sin\frac{2\pi}{365}(x-79)+12$, we can identify its amplitude as $A=3$ hours. b. We can find the period as $p=\frac{2\pi}{2\pi/365}=365$ days. c. During the longest day, we have $sin\frac{2\pi}{365}(x-79)=1$ and $y=15$ hours d. During the shortest day, we have $sin\frac{2\pi}{365}(x-79)=-1$ and $y=9$ hours e. At $x=0$, we have $y=3sin\frac{2\pi}{365}(0-79)+12\approx9.1$ which is the same for $x=365$. We can find the maximum when $sin\frac{2\pi}{365}(x-79)=1$ which leads to $\frac{2\pi}{365}(x-79)=\frac{\pi}{2}$ or $x=170.25$. Similarly, $sin\frac{2\pi}{365}(x-79)=-1$ gives a minimum when $x=352.75$. The average value $y=12$ happens at $\frac{p}{4}=91.25$ away from the maximum or minimum. Thus we can make a graph as shown in the figure.