Answer
a) $6796.55$
b) $4024.55$
c) See graph
d) In $1957$ with a value of $\$1406$
e) In $1939$ and $1975$
f) $[1406, 6796.55]$
Work Step by Step
Given $$\begin{aligned}
p(t) &= 4.95(t-57)^2 +1406.
\end{aligned}$$ a) Determine $p(t)$ for $t=90$: $$\begin{aligned}
p(90) &= 4.95(90-57)^2 +1406\\
& = \$ 6796.55
\end{aligned}$$ The poverty threshold for individuals in the United States under the age of $65$ in $1990$ was about $6796.55$.
b) Determine $p(t)$ for $t=80$:
$$\begin{aligned}
p(80) &= 4.95(90-57)^2 +1406\\
& = \$ 4024.55
\end{aligned}$$ This means that the poverty threshold for individuals in the United States under the age of $65$ in $1980$ was about $4024.55$.
c) See the graph below.
d) The graph shows that the minimum of the function is at the vertex of $(57,1406)$. This means that the poverty threshold reached a minimum of $1406$ in $1957$, which makes sense.
e) The graph show that the poverty threshold for individuals in the United States under the age of $65$ in $1939$ and in $1975$ was about $3000$.
f) We can compute the ranch from the values of the domain of the function, but we have already do that and so, the range of the model is $[1406, 6796.55]$
