Question

Nomor 5 tolong dibantu kaak
Dengan caranya

Answer 1

5. f(x) = ax²+bx+c

dibagi x+1 sisa 3 : f(-1) = 3

dibagi x-3 sisa 23 : f(3) = 23

dibagi x-2 sisa 15 : f(2) = 15

f(-1) = 3 = a(-1)²-b+c

a-b+c = 3 => c = 3+b-a

f(3) = 23 = a(3)²+ 3b + c

9a+3b+c = 23

9a+3b+3+b-a = 23

8a+4b = 20

2a+b = 5

f(2) = 15 = a(2)²+2b+c

4a+2b+c = 15

4a+2b+3+b-a = 15

3a+3b = 12

a+b = 4

2a+b = 5

a+a+b = 5

a+4 = 5

a = 1

a+b = 4

b = 4-1 = 3

c = 3+b-a = 3+3-1 = 5

maka :

f(x) = x²+3x+5