tolong di bantu kak esaynya

3) a = 2 b = 4 c = -12
x = - [b/2a]
= - [4/2(2)]
= - [4/4]
= -1
y = - [D/4a]
= - [(b^2 - 4ac) / 2a]
= - [(4^2 - 4(2)(-12)) / 2(2)]
= - [(16 + 96) / 4]
= - [112 / 4]
= - 28
Koordinatnya (-1,-28)
Jadi koordinat titik balik minimum grafik fungsi kuadrat tersebut adalah (-1,-28)