Koordinat titik balik grafik fungsi kuadrat f(x) = -x²+4x+5 adalah....
a. (-9,2)
b. (-2,-9)
c. (-2,9)
d. (2,9)
e. (2,-9)
a. (-9,2)
b. (-2,-9)
c. (-2,9)
d. (2,9)
e. (2,-9)
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Kelas 10 MatematikaBab Fungsi Kuadrat
f(x) = -x² + 4x + 5
a = -1; b = 4; c = 5
Sumbu simetri
= -b/2a
= - 4/(2 . -1)
= 2
f(2) = - (2²) + 4 . 2 + 5
f(2) = -4 + 8 + 5
f(2) = 9
(2, 9)