SOLUTION: Alright so I have to do quadratic FUNCTIONS, not quadratic equations. Apparently there is a difference so I have to solve all these other little problems to get the ax^2+bx+c. For
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Question 609287: Alright so I have to do quadratic FUNCTIONS, not quadratic equations. Apparently there is a difference so I have to solve all these other little problems to get the ax^2+bx+c. For example, (not word to word but you'll get the point): Find c(x)=? it takes $103 to make 3 cameras, $163 to make 7 cameras and $328 to make 12 cameras. ok, i know how to but them into the three equations: 103=6a+3b+c, 163=49a+7b+c, and 328=144a+7b+c. but after that i'm totally lost. Answer by SwiftAlbatross(13) (Show Source):
You can put this solution on YOUR website! 103 = 9a + 3b + c
163 = 49a + 7b + c
328 = 144a + 12b + c
Solve for c:
103 = 9a + 3b + c
c = 103 - 9a - 3b
Plug c into other equations:
163 = 49a + 7b + 103 - 9a - 3b
Combine like terms:
60 = 40a + 4b
328 = 144a + 12b + 103 - 9a - 3b
225 = 135a + 9b
Work with these two equations:
60 = 40a + 4b
225 = 135a + 9b
Solve for b:
60 = 40a + 4b
*2 *2
120 = 80a + 8b
225 = 135a + 9b
- 120 = - 80a - 8b
105 = 55a + b
b = 105 - 55a
Substitute b into one of equations:
60 = 40a + 4b
60 = 40a + 4(105 - 55a)
60 = 40a + 420 - 220a
-360 = -180a
a = 2
Substitue a into equation:
60 = 40a + 4b
60 = 40(2) + 4b
60 = 80 + 4b
-20 = 4b
b = -5
Substitute a and b into one of the first equations:
103 = 9a + 3b + c
103 = 9(2) + 3(-5) + c
103 = 18 - 15 + c
103 = 3 + c
c = 100
Check with other 2 equations.
So the quadratic function is
