kurva f(x) = ax^2 + bx+12 mempunyai titik balik (-2,4). nilai a+b=......

titik balik turunan = 0
2ax + b = 0
masukan x = -2
2a(-2) + b = 0
-4a + b = 0 (persamaan 1)
y = ax² + bx + 12
masukan x = 2 dan y = 4
4 = a(-2)² + b(-2) + 12
4-12 = 4a - 2b
4a - 2b = -8
eliminasi dengan persamaan 1
4a - 2b = -8
-4a + b = 0
----------------- +
-b = -8
b = 8
4a - 2(8) = -8
4a - 16 = -8
4a = -8 + 16
a = 8/4 = 2
a + b = 8 + 2 = 10