7. Koordinat titik balik fungsi kuadrat f(x)=x²+2x−8 adalah

7. Koordinat titik balik fungsi kuadrat f(x)=x²+2x−8 adalah

Jawaban : (-1,-9)

Pembahasan :

Ingat!
Pada fungsi kuadrat f(x) = ax² + bx + c
Titik baliknya adalah (xp,yp) dimana
xp = -b/2a
yp= (b²-4ac/(-4a)

Diketahui
f(x) = x² + 2x – 8
Maka
a = 1
b = 2
c = 8

Sehingga
xp = -b/2a
xp = -2/2(1)
xp = -2/2
xp = -1

yp= (b²-4ac/(-4a)
yp= (2²-4(1)(-8))/(-4(1))
yp = (4+32)/(-4)
yp = 36/(-4)
yp = -9

Sehingga koordinat titik balik adalah (xp, yp)= (-1,-9)

Dengan demikian koordinat titik balik fungsi tersebut adalah (-1,-9)

 

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