titik puncak( extrim) dari grafik fungsi kuadrat f(x)=2x2-8x+5adalah

a = 2, b = - 8 dan c= 5

x = - b / 2a = - (-8)/ 2.2 = 8/4 = 2
y = 2 (2)^2 - 8. 2 + 5 = 8 - 16 + 5 = - 3
titik puncak (2, - 3)

Kelas 10 Matematika
Bab Fungsi Kuadrat

f(x) = 2x² - 8x + 5
a = 2; b = -8; c = 5

Sumbu simetri
= -b/2a
= - (-8)/(2 . 2)
= 2

f(2) = 2 . 2² - 8 . 2 + 5
f(2) = 8 - 16 + 5
f(2) = -3

(2, -3)