Penjelasan dengan langkah-langkah:
43] f(x) = x² - 10x + 29
a = 1; b = -10; c = 29
x = -b/(2a)
x = - (-10)/(2 . 1)
x = 5
f(5) = 5² - 10 . 5 + 29
f(5) = 25 - 50 + 29
f(5) = 4
Koordinat titik balik = (5, 4)
Jawabannya D
44] f(x) = (x + 1)² - 7
f(x) = x² + 2x + 1 - 7
f(x) = x² + 2x - 6
a = 1; b = 2; c = -6
x = -2/(2 . 1)
x = -1
f(-1) = (-1 + 1)² - 7
f(-1) = 0² - 7
f(-1) = -7
Koordinat titik balik = (-1, -7)
Jawabannya A
Kode Kategorisasi : 10.2.5
Kelas 10
Pelajaran 2 - Matematika
Bab 5 - Persamaan dan Fungsi Kuadrat
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Penjelasan dengan langkah-langkah:
43] f(x) = x² - 10x + 29
a = 1; b = -10; c = 29
x = -b/(2a)
x = - (-10)/(2 . 1)
x = 5
f(5) = 5² - 10 . 5 + 29
f(5) = 25 - 50 + 29
f(5) = 4
Koordinat titik balik = (5, 4)
Jawabannya D
44] f(x) = (x + 1)² - 7
f(x) = x² + 2x + 1 - 7
f(x) = x² + 2x - 6
a = 1; b = 2; c = -6
x = -b/(2a)
x = -2/(2 . 1)
x = -1
f(-1) = (-1 + 1)² - 7
f(-1) = 0² - 7
f(-1) = -7
Koordinat titik balik = (-1, -7)
Jawabannya A
Kode Kategorisasi : 10.2.5
Kelas 10
Pelajaran 2 - Matematika
Bab 5 - Persamaan dan Fungsi Kuadrat