Tentukan persamaan parabola melalui titik (1,6),(4,0)dan(3,4)

y = ax² + bx + c

(1,6) --> 6=a(1)²+b(1)+c = a+b+c=6
(4,0)---> 0=a(4)²+b(4)+c=16a+4b+c=0
(3,4)-->4=a(3)²+b(3)+c=9a+3b+c=4

a + b + c = 6
16a + 4b + c = 0
-------------------------(-)
- 15a - 3b = 6 ... pers 1

a + b + c = 6
9a + 3b + c = 4
-----------------------(-)
- 8a - 2b = 2 ... pers 2

- 15a - 3b = 6 |×2
- 8a - 2b = 2 |×3
- 30a - 6b = 12
- 24a - 6b = 6
----------------------(-)
- 6a = 6
a = - 1

- 8a - 2b = 2
-8(-1) - 2b = 2
- 2b = 2 - 8
- 2b = - 6
b = 3

a + b + c = 6
- 1 + 3 + c = 6
c = 6 - 2
c = 4

persamaan parabola
y = - x² + 3x + 4