c(x) = 3x² - x + 15
a = 3, b= -1, c = 15
Step-by-step explanation:
Quadratic equation is in the form:
ax²+ bx + c = 0
a, b, c are unknown that needs to be found
c(x) = ax²+ bx + c
c(x) = cost of making x calculators
x = number of calculators
when cost is $25, it produced 2 calculators: c(x) = 25, x = 2
Inserting into the equation:
25 = a(2)²+ b(2) + c
25 = 4a + 2b + c ...equation (1)
when cost is $59, it produced 4 calculators: c(x) = 59, x = 4
Inserting into the equation:
59 = a(4)²+ b(4) + c
59 = 16a + 4b + c ...equation (2)
when cost is $305, it produced 10 calculators: c(x) = 305, x = 10
Inserting into the equation:
305 = a(10)²+ b(10) + c
305 = 100a + 10b + c ...equation (3)
solving the equations simultaneously:
25 = 4a + 2b + c ...equation (1)
59 = 16a + 4b + c ...equation (2)
305 = 100a + 10b + c ...equation (3)
subtract equation 1 from 2:
59 - 25 = 16a - 4a + 4b - 2b + c+ c
34 = 12a + 2b ...equation (4)
subtract equation 2 from 3:
305 - 59 = 100a - 16a + 10b - 4b + c - c
246 = 84a + 6b ...equation (5)
34 = 12a + 2b ...equation (4)
246 = 84a + 6b ...equation (5)
multiply equation 4 by 3:
102 = 36a + 6b ...equation (4)
246 = 84a + 6b ...equation (5)
subtract equation 4 from 5:
246 - 102 = 84a - 36a + 6b - 6b
144 = 48a + 0
144 = 48a
144/48 = 48a/48
a = 3
substitute for a in equation 5:
246 = 84(3) + 6b
246 = 252 + 6b
246 - 252 = 6b
-6 = 6b
-6/6 = 6b/6
b = -1
Substitute for a and b in any of the equation in 1 or 2:
using equation 1: 25 = 4a + 2b + c
25 = 4(3) + 2(-1) + c
25 = 12 - 2 + c
25 = 10 + c
25 - 10 = c
c = 15
The function becomes:
c(x) = 3x²+ (-1)x + 15
c(x) = 3x² - x + 15