Answer
a) $106.06$ million
b) $(5.04,78.75)$
c) $1989$ and $2001$
Work Step by Step
Given \begin{equation}
V(t)=0.275 t^2-2.77 t+85.73
\end{equation} a) The are $15$ years between $1990$ and $2005$. Find $V(15)$. \begin{equation}
\begin{aligned}
V(15)&=0.275\cdot 15^2-2.77\cdot 15+85.73 \\
&\approx 106.06
\end{aligned}
\end{equation} The number of visitors at US amusement parks in 2005 was about $106.06$ million people.
b) Find the $x$ and $y$ values of the vertex. $$\begin{aligned}
a & =0.275, b=-2.77, c=85.73\\
x& =\frac{-b}{2a} \\
& =\frac{-(-2.77)}{2\cdot(0.275)}\\
&= 5.03636\\
&\approx 5.04\\
y&= 0.275\cdot (5.03636)^2-2.77\cdot (5.03636)+85.73\\
& =78.75463\\
&\approx 78.75
\end{aligned}
$$ The vertex of the function is $(5.04,78.75)$. This means that in $1995$, US amusement parks experienced the lowest number of visitors, which was about $78.75$ million people.
c) Set $V(t)= 90$ and find the values of $t$. \begin{equation}
\begin{aligned}
V(t)&=90 \\
0.275 t^2-2.77 t+85.73&= 90\\
0.275 t^2-2.77 t+85.73-90&= 0\\
\left(0.275 t^2-2.77 t-4.27\right)\cdot 1000&= 0\cdot 1000\\
275 t^2-2770 t-4270&=0
\end{aligned}
\end{equation} $$\begin{aligned}
t & =\frac{-(-2770) \pm \sqrt{(-2770)^2-4 \cdot 275(-4270)}}{2 \cdot 275}\\
& =\frac{2770 \pm \sqrt{12369900}}{550} \\
&= 5.0364\pm 6.395\\
\therefore &t= 11.43\\
t&= -1.36
\end{aligned}
$$ These values of $t$ can be written in years as $1990+11= 2001$ and $1990-1= 1989$. Hence, US amusement parks had $90$ million visitors in $1989$ and also in $2001$.
