Koordinat titik balik fungsi Y = x2-9x+20

x = -b/2a = -(-9)/2(1) = 9/2

ketika x = 9/2 maka y
y = (9/2)^2 - 9(9/2) +20
y = 81/4 - 81/2 +20
y = 81/4 - 162/4 + 80/4
y = -1/4

koordinat titik baliknya
(9/2,-1/4)

Bab Fungsi Kuadrat
Matematika SMP Kelas VIII

y = x² - 9x + 20
a = 1; b = -9; c = 20.

Sumbu simetri
x = -b/(2 . a)
x = -(-9)/(2 . 1)
x = 9/2

y = (9/2)² - 9 . (9/2) + 20
y = 81/4 - 81/2 + 20
y = 81/4 - 162/4 + 80/4
y = -1/4

Koordinat titik balik = (9/2, -1/4)