Answer
a) $30.4$ million
b) $1997$, $2004$
c) $2001$
Work Step by Step
Given \begin{equation}
I(t)=-1.5 t^2+32.3 t-138.8.
\end{equation} a) Set $t= 9$ to estimate the net income for Quicksilver in the year $1999$.
\begin{equation}
\begin{aligned}
I(9) & =-1.5 (9)^2+32.3 (9)-138.8 \\
& =30.4
\end{aligned}
\end{equation} Quicksilver income in $1999$ was about $30.4$ million.
b) Set $I(t)= 18$ to find the values of $t$, which gives the year(s) that the net will reach $18$ million. \begin{equation}
\begin{aligned}
-1.5 t^2+32.3 t-138.8 & =18 \\
-1.5 t^2+32.3 t-138.8-18 & =0 \\
-1.5 t^2+32.3 t-156.8 & =0 \\
1.5 t^2-32.3 t+156.8 & =0 \\
15 t^2-323 t+1568 & =0
\end{aligned}
\end{equation} Solve the equation: \begin{equation}
\begin{aligned}
t & =\frac{-(-323) \pm \sqrt{(-323)^2-4(15)(1566)}}{2(15)} \\
& =\frac{323 \pm \sqrt{10249}}{30} \\
& =10.767 \pm 3.375
\end{aligned}
\end{equation} The solutions are: \begin{equation}
\begin{aligned}
& t=10.767-3.375=7.39 \\
& t=10.767+3.375=14.14
\end{aligned}
\end{equation} Quicksilver net income reached $18$ million in about $1997$ and again in about $2004$.
c) The vertex of the income function will give us the maximum income. Use $a= -1.5$ and $b= 32.3$ into the following formula.
$$
\begin{aligned}
t & =\frac{-b}{2 a} \\
& =\frac{-32.3}{2(-1.5)} \\
& =10.767
\end{aligned}
$$ $$
\begin{aligned}
I_{max} & =-1.5(10.767)^2+32.3(10.767)-138.8 \\
& \approx 35.08.
\end{aligned}
$$ The vertex is $(10.77,35.08)$. This means that the company had the highest income of $\$35.08$ million in $2001$.
