diket fungsi kuadrat f(x)=2x^2 - 5x +8.titik puncak grafik fungsi tsb ...

Materi: Pers Kuadrat

y = 2x^2 - 5x + 8

xp = -b/2a
xp = -(-5)/2×2 = 5/4
yp = -D/4a
yp = (-5)^2+4(2)(8)/-4(2)
= 89/-8
Jadi titik puncaknya di (5/4, -89/8)

f(x) = 2x^2 - 5x + 8
x = -b/2a = -(-5)/2(2) = 5/4
y = f(5/4) = 2(5/4)^2 - 5(5/4) + 8
= 2 (25/16) - 25/4 + 8
= 25/8 - 50/8 + 64/8
= 39/8
Titik puncak = (5/4,39/8)