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Verified answer
Bab Fungsi KuadratMatematika SMP Kelas VIII
y = x² - 8x + c
a = 1, b = - 8, c = c
sumbu simetri = -b/2a
sumbu simetri = -(-8/(2 . 1))
sumbu simetri = 4
terletak di sumbu x, maka y = 0 → puncak = (4, 0)
y = x² - 8x + c = 0
4² - 8 . 4 + c = 0
c = -16 + 32
c = 16
memotong sumbu y sehingga x = 0
y = x² - 8x + 16
y = 0² - 8 . 0 + 16
y = 16
(0,16)
jarak 2 titik
= √((x2 - x1)² + (y2 - y1)²)
= √((0 - 4)² + (16 - 0)²)
= √(16 + 256)
= √272
= 4√17