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Sumbu simetri xp = - b/2a
xp = - (- 4) / 2(1)
xp = 2
Karena koefisien a = 1 yakni a > 0, kurva terbuka ke atas
Nilai minimum yp = D / -4a
yp = [b²- 4ac] / -4a
yp = [(-4)² - 4(1)(- 16)] / [- 4(1)]
yp = 80/ -4
yp = - 20
∴ Koordinat titik balik dari grafik fungsi kuadrat y = x² -4x -16 = 0 adalah (2, - 20)