Diketahui fungsi kuadrat f(x)=2x^2-5x+8 Titik puncak grafik fungsi kuadrat adalah

Kelas 10 Matematika
Bab Fungsi Kuadrat

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

Sumbu simetri
= -b/2a
= - (-5)/(2 . 2)
= 5/4

f(5/4) = 2 . (5/4)² - 5 . (5/4) + 8
f(5/4) = 2 . 25/16 - 25/4 + 128/16
f(5/4) = 50/16 - 100/16 + 128/16
f(5/4) = 78/16
f(5/4) = 4 14/16
f(5/4) = 4 7/8
f(5/4) = 39/8

Titik puncak = (5/4, 39/8)