The table shows the monthly revenue y (in thousands of dollars)of a landscaping business for each month of the year 2008,with x = 1 representing January.
$$\begin{array} { | c | c | }
\hline x & y \
\hline 1 & 6.0 \
\hline 2 & 6.1 \
\hline 3 & 7.1 \
\hline 4 & 9.2 \
\hline 5 & 12.3 \
\hline 6 & 16.4 \
\hline 7 & 12.2 \
\hline 8 & 10.2 \
\hline 9 & 8.3 \
\hline 10 & 6.3 \
\hline 11 & 4.3 \
\hline 12 & 2.4 \
\hline
\end{array}$$
A mathematical model that represents these data is:$$f ( x ) = \left\{ \begin{array} { l l }
0.505 x ^ { 2 } - 1.47 x + 7.0 & 1 \leq x \leq 6 \
- 1.97 x + 26.0 & 6

Question

The table shows the monthly revenue y (in thousands of dollars)of a landscaping business for each month of the year 2008,with x = 1 representing January. 
$$\begin{array} { | c | c | } 
\hline x & y \
\hline 1 & 6.0 \
\hline 2 & 6.1 \
\hline 3 & 7.1 \
\hline 4 & 9.2 \
\hline 5 & 12.3 \
\hline 6 & 16.4 \
\hline 7 & 12.2 \
\hline 8 & 10.2 \
\hline 9 & 8.3 \
\hline 10 & 6.3 \
\hline 11 & 4.3 \
\hline 12 & 2.4 \
\hline
\end{array}$$
A mathematical model that represents these data is:$$f ( x ) = \left\{ \begin{array} { l l } 
0.505 x ^ { 2 } - 1.47 x + 7.0 & 1 \leq x \leq 6 \
- 1.97 x + 26.0 & 6 < x \leq 12
\end{array} \right.$$ 
Find f (1)and f (12).

A) $f ( 1 ) = 2.860 , f ( 12 ) = 6.79$
B) $f ( 1 ) = 3.360 , f ( 12 ) = 7.04$
C) $f ( 1 ) = 6.035 , f ( 12 ) = 2.36$
D) $f ( 1 ) = 2.860 , f ( 12 ) = 6.54$
E) $f ( 1 ) = 2.860 , f ( 12 ) = 6.29$

Quizplus · Accepted Answer

For $f(1)$, we use the first part of the model $0.505(1)^2 - 1.47(1) + 7.0 = 6.035$. For $f(12)$, we use the second part of the model $-1.97(12) + 26.0 = 2.36$.

The Table Shows the Monthly Revenue Y (In Thousands of Dollars)of