Question

Kuis
Difficulty: Hard.
Kurva f(x)= ax²+bx+c memiliki sumbu simetri x= 8.
Jika f(-1)= 21 dan f(2)= -21, maka nilai dari a+b+c adalah?
A. -28
B. -15
C. -11
D. -8
E. -3​

Answer 1

x = -b/2a

4 = -b/2a

8 = -b/a ... pers 1

f(x) = ax² + bx + c

f(-1) = a(-1)² + b(-1) + c

21 = a - b + c ... pers 2

f(x) = ax² + bx + c

f(2) = a(2)² + b(2) + c

-21 = 4a + 2b + c ... pers 3

eliminasi pers 2 dan 3

21 = a - b + c

-21 = 4a + 2b + c

----------------------- --

42 = -3a - 3b

42 = -3(a + b)

a + b = 42/-3

a + b = -14 .... pers 4

b = -14 - a -----> substitusi ke pers 1

8 = -b/a

8a = -b

8a = 14 + a

8a - a = 14

7a = 14

a = 14/7

a = 2

substitusi a ke pers 4

a + b = -14

2 + b = -14

b = -14 - 2

b = -16

substitusi a dan b ke pers 2

21 = a - b + c

21 = 2 - (-16) + c

21 = 2 + 16 + c

21 = 18 + c

c = 21 - 18

c = 3

...

Maka:

a + b + c

= 2 + (-16) + 3

= 2 - 16 + 3

= -14 + 3

= -11

• opsi C