fungsi kuadrat
melalui titik (0, -9)
artinya c = -9
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menyinggung x-axis di
(3, 0)
f(x) = ax²+bx+c
f(3) = 3²a + 3b + c
--> f(3) = 0, maka
0 = 3²a + 3b + c
0 = 3²a + 3b - 9
0 = 9a + 3b - 9
9a + 3b = 9
3(3a+b) = 9
3a+b = 3
b = 3-3a
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menyinggung x-axis
maka D = 0
D = b²-4ac
0 = b²-4ac
0 = (3-3a)²-[4a(-9)]
0 = 3²-2(3)(3a)+(3a)²+36a
0 = 9-18a+9a²+36a
0 = 9a²+18a+9
a²+2a+1 = 0
(a+1)² = 0
a = -1
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b = 3-3a
b = 3-3(-1)
b = 3+3
b = 6
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Pers.
ax² + bx + c =
-1x² + 6x - 9 =
-x² + 6x - 9
Jawaban (B.) -x² + 6x - 9