
fungsi kuadrat
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•
Mathematics
•
9th Grade
•
Hard
Siti Rosita Saksani
Used 4+ times
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9 Slides • 0 Questions
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ST ROSITA SAKSANI, S.Pd.
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Fungsi kuadrat adalah y = f (x) = ax2 + bx + c ( a dan b € R, a ≠ 0 ) untuk semua x dalam daerah asalnya. Fungsi kuadrat juga dikenal sebagai fungsi polinom atau fungsi suku banyak berderajat dua dalam variabel x
Grafik fungsi kuadrat y= f (x) = ax2 + bx + c dalam bidang cartesius dikenal sebagai parabola.
Bentuk umum fungsi kuadrat adalah f(x) = ax2 + bx + c dengan a ≠ 0, grafik fungsi kuadrat berupa parabola
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Tentukan titik potong dengan sumbu koordinat ( sumbu x dan sumbu y )
Tentukan koordinat titik puncak, yaitu ( X0, Y0 ) dengan X0 = - b/2a dan Y0= -D/4a, dengan D = b2 – 4ac
Langkah-langkah menggambar grafik fungsi kuadrat
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Ciri khusus grafik fungsi kuadrat
· a > 0 berarti grafiknya terbuka ke atas dan titik balik minimun
· a < 0 berarti grafiknya terbuka ke bawah dan titik baliknya maksimun
· D > 0 memotong sumbu X di dua titik
· D = 0 menyinggung sumbu X
· D < memotong sumbu X
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Peny.= f(x)= x2 – 3x + 2 dengan pers. y= x2 – 3x + 2, berarti a= 1, b= - 3, c=2
1. Titik Potong Sumbu x dan y
a. Titik potong sumbu x, jika y=0
X2 – 3x +2 = 0
( x – 1)( x – 2) =0
X1 = 1 dan X2=2
Jadi titik potongnya dengan sb. Y adalah ( 1,0) dan (2,0)
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b. Titik potong sumbu y, jika x=0
Y= (0)2 – 3(0) + 2
= 2
Jadi titik potong dengan sumbu y adalah (0,2)
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P= [-b/2a, -(b2 – 4ac)/4a]
P= [-(-3)/2(1), -(-32 – 4(1)(2)/4(1)]
P= [ 3/2, ¼]
Oleh karena a > 0 maka P merupakan titik balik mininum dan parabola terbuka keatas
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Dari uraian di slide sebelumnya , sketsa grafik fungsi kuadrat f(x) = x2 – 3x + 2
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TUGAS ( Tuliskan caranya ikuti langkah-langkah yang ada di slide sebelumnya)
GAMBARKAN GRAFIK FUNGSI KUADRAT = f(x) = x2 + 3x + 2
ST ROSITA SAKSANI, S.Pd.
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