Answer
(a) see graph;
(b) $16.5$, $12$, $4$ (shift to the right), $67.5$ (up).
(c) $53^\circ F$
(d) January $51^\circ F$, July $84^\circ F$.
(e) $67.5^\circ F$, vertical translation.
Work Step by Step
Given $f(x)=16.5\ sin[\frac{\pi}{6}(x-4)]+67.5$, we have:
(a) see graph;
(b) amplitude $|A|=16.5$, period $p=\frac{2\pi}{\frac{\pi}{6}}=12$, phase shift $\phi=4$ (shift to the right), vertical translation $67.5$ (up).
(c) for December, $x=12$, $f(12)=16.5\ sin[\frac{\pi}{6}(12-4)]+67.5=16.5\ sin[\frac{4\pi}{3}]+67.5\approx53^\circ F$
(d) For a minimum average monthly temperature, we need $sin[\frac{\pi}{6}(x-4)]=-1 \longrightarrow \frac{\pi}{6}(x-4)=-\frac{\pi}{2} \longrightarrow x=1$ that is in January with $f(1)=51^\circ F$.
For a maximum average monthly temperature, we need $sin[\frac{\pi}{6}(x-4)]=1 \longrightarrow \frac{\pi}{6}(x-4)=\frac{\pi}{2} \longrightarrow x=7$ that is in July with $f(7)=84^\circ F$.
(e) As the average of the sine function is zero, as an approximation for the average annual temperature in San Antonio, it is $67.5^\circ F$, which is the vertical translation of the sine function in the formula.
