Question

Tentukan fungsi kuadrat jika melalui titik (2,0) , (-3,0) serta titik (3,1) mohon bantuannya ya

Answer 1

F(x)=ax^2+bx+c
(2,0) =>4a+2b+c=0 .....1
(-3,0)=>9a-3b+c =0 ......2
(3,1) =>9a+3b+c=1 .......3

1&3
9a-3b+c=0
4a+2b+c=0
--------------
5a-5b=0
2&3
9a+3b+c=1
9a-3b+c=0
-------------
6a. = 1
a=1/6
5a-5b=0
5.1/6-5b=0
5b=5/6
b=1/6

4a+2b+c=0
4.1/6+2.1/6+c=0
c=-1
jd pers y=1/7x^2+1/6x-1