Answer
a) $58.40^{\circ}F$
b) July
c) June and September
Work Step by Step
Given \begin{equation}
H(m)=0.9 m^2-13 m+104.
\end{equation} a) Set $m=6$ to find the average temperature for the month of June.
\begin{equation}
\begin{aligned}
H(6) & =0.9 \cdot 6^2-13 \cdot 6+104 \\
& =58.40
\end{aligned}
\end{equation} This is the average temperature in Melbourne for the month of June which is about $58.40^{\circ}F$ .
b) Use the formula, with $a= 0.9, b= -13$ to find the vertex.
\begin{equation}
\begin{aligned}
m&=\frac{-b}{2a} \\
& =\frac{-(-13)}{2(0.9)} \\
& = 7.2222 \\
H(7.2222)&=0.9 \cdot 7.2222^2-13 \cdot 7.2222+104\\
& =57.06.
\end{aligned}
\end{equation} The vertex of the function is $(7.22,57.06)$.This means that for the month of July, the average temperate in Melbourne is about $58.40^{\circ}F$ which is the minimum temperature of the city.
c) Set $H=60$ to find the value(s) of $m$.
\begin{equation}
\begin{aligned}
0.9 m^2-13 m+104& =60 \\
0.9 m^2-13 m+104-60& =0 \\
0.9 m^2-13 m+44& =0 \\
\left( 0.9 m^2-13 m+44\right)\cdot 10 & =0 \\
9m^2-130 t+440& =0.
\end{aligned}
\end{equation} Solve the equation: \begin{equation}
\begin{aligned}
&m= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
& m=\frac{-(-130) \pm \sqrt{(-130)^2-4(9)(440)}}{2(9)} \\
& =\frac{130 \pm \sqrt{1060}}{18}\\
&= \frac{130\pm 32.5576}{18}\\
&= 7.222\pm 1.809.
\end{aligned}
\end{equation} The solutions are: \begin{equation}
\begin{aligned}
& m=7.222- 1.809= 5.41 \\
& m=7.222+ 1.809= 9.03.
\end{aligned}
\end{equation} The average temperature in Melbourne is $60^{\circ}F$ during the months of June and September.
