Robert J. answered • 06/08/13
Tutor
4.6
(13)
Certified High School AP Calculus and Physics Teacher
(1.7)? It doesn't make sense here.
Actually, three points, such as (2,14) (3,28) (4,56), can determine a unique quadratic equation.
If you are allowed to use calculator, then use "STAT" to do curve fitting. The answer is
f(x) = 7x^2 - 21x + 28
If you are not allowed to use calculator, then let f(x) = ax^2 + bx + c
Plug in all three given points,
14 = 4a+2b+c ......(1)
28 = 9a+3b+c ......(2)
56 = 16a+4b+c ......(3)
(1)+(3)-2*(2): 14 = 2a => a = 7
(2)-(1): 14 = 5a+b => b = 14-5a = -21
c = 14-4a-2b = 14-28+42 = 28
Nataliya D.
Hi Robert. I believe, there is something wrong with last y-coordinate, or quadratic equation is not exist for all 4 points:
(2) - (1): 14 = 5a + b
(3) - (2): 28 = 5a + b
06/08/13