Each point on this quadratic function will be of the form (x,C). The problem gives three points:
(2,28), (4,50) and (10,140). Each of these points will satisfy the quadratic function, C = ax2+bx+c. Substituting the values of our three given points into the quadratic function, we can solve systems of equations in order to find a, b, and c:
28 = a(2)2+b(2)+c ----> 28 = 4a + 2b + c
50 = a(4)2+b(4)+c ----> 50 = 16a + 4b + c
140 = a(10)2+b(10)+c ----> 140 = 100a + 10b + c
Subtracting the first equation from the second equation, we obtain 22 = 12a + 2b ----> 11 = 6a + b
Subtracting the second equation from the first equation, we obtain 90 = 84a + 6b ----> 15 = 14a + b
We can now subtract the top equation from the bottom one to obtain 4 = 8a. From this, we find that a = 1/2.
We can then substitute a = 1/2 into the equation, 11 = 6a + b ----> 11 = 6(1/2) + b ----> 11 = 3 + b ----> b = 8.
We now use one of the original three equations to solve for c:
28 = 4a + 2b + c
28 = 4(1/2) +2(8) + c
28 = 2 + 16 + c
10 = c
Therefore, the quadratic equation is:
C(x) = 1/2x2 + 8x + 10
Then, the total cost of making 12 camera cases would be:
C(12) = 1/2(12)2+ 8(12) + 10
C(12) = 1/2(144) + 96 + 10
C(12) = 72 + 96 + 10
C(12) = 178
Therefore, the cost to make 12 camera cases would be $178.
I hope this helps!
Maria Tilmos