Answer
a) $ T(m) = -2.56(x-7)^2+51$
b) $(7,5)$
c) $10^{\circ} \ F$.
d) Domain: $[3, 11]$
Range: $[10, 51]$
Work Step by Step
a) Represent the months by numbers to get the data points: $(4,28)$, $(5,39)$, $(6,47)$, $(7,57)$, $(8,49)$, $(9,41)$, $(10,28)$ and $(11,15)$.
Make a scatter plot of the data and choose a vertex point that may either be the lowest or highest point. The vertex of the data point can be found to be at $(h,k)=(7,51)$.
Set this into the standard vertex form of a parabola. This gives $$\begin{aligned}
f(x) &= a(x-7)^2+51.
\end{aligned}$$ Choose any point from the scatter plot to find the value of the constant $a$. Let's take the point $(x,y)=(10,28)$ and insert this into the above equation to find $a$. $$\begin{aligned}
28 &= a(10-7)^2+51\\
28& = 3^2a+51\\
28-51& =9a \\
9a= -23\\
a\approx -2.56
\end{aligned}$$ Hence, the parabola that best fits the data is the Temperature function. It is given by
$$\begin{aligned}
T(m) &= -2.56(x-7)^2+51,
\end{aligned}$$ where $m$ is the month of the year ($m= 4$ is the month of April) and $T(m)$ is the average monthly temperature in degree Fahrenheit.
b) The vertex gives the highest temperature for the month of July, which is $51$ degree Fahrenheit in the city of Anchorage, Alaska.
c) We determine $T(3)$: $$\begin{aligned}
T(3) &= -2.56(3-7)^2+51\\
& = 10.
\end{aligned}$$ The average temperature for the month of March is about $10^{\circ} \ F$.
d) Assume that the function is valid for the months of March through November. Then:
$$\begin{aligned}
T(11) &= -2.56(11-7)^2+51= 10 \\
\end{aligned}$$ The domain and range are:
Domain: $[3, 11]$
Range: $[10, 51]$
