\(c(x)=ax^2+bx+c \\ \\
c(3)=103 \\
c(7)=163 \\
c(12)=328 \\ \\
a \times 3^2+b \times 3 + c=103 \\
a \times 7^2 + b \times 7 + c= 163 \\
a \times 12^2 + b \times 12 + c =328 \\ \\
9a+3b+c=103 \\
49a+7b+c=163 \\
144a+12b+c=328\)
\(\hbox{the first two equations:} \\
9a+3b+c=103 \ \ \ \ \ |\times (-1) \\
49a+7b+c=163 \\ \\
-9a-3b-c=-103 \\
\underline{49a+7b+c=163 \ \ \ \ \ } \\
40a+4b=60 \ \ \ \ \ \ \ \ \ \ \ |\div 4 \\
10a+b=15\)
\(\hbox{the second and third equation:} \\
49a+7b+c=163 \ \ \ \ \ \ \ |\times (-1) \\ 144a+12b+c=328 \\ \\
-49a-7b-c=-163 \\
\underline{144a+12b+c=328 \ \ } \\
95a+5b=165 \ \ \ \ \ \ \ \ \ \ \ \ |\div 5 \\
19a+b=33\)
\(\hbox{set up a new system of equations:} \\
10a+b=15 \ \ \ \ \ \ \ \ |\times (-1) \\
19a+b=33 \\ \\
-10a-b=-15 \\
\underline{19a+b=33 \ \ \ \ \ \ } \\
9a=18 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\div 9 \\
a=2 \\ \\
10a+b=15 \\
10 \times 2+b=15 \\
20+b=15 \ \ \ \ \ \ \ \ \ \ |-20 \\
b=-5\)
\(49a+7b+c=163 \\
49 \times 2+7 \times (-5)+c=163 \\
98-35+c=163 \\
63+c=163 \ \ \ \ \ \ \ \ \ \ \ \ \ \ |-63 \\
c=100 \\ \\
\hbox{the equation for the cost of maxing x widgets:} \\
c(x)=2x^2-5x+100 \\ \\
c(2)=2 \times 2^2-5 \times 2+100=8-10+100=98\)
The cost of producing 2 widgets is $98.
