corrige 1s4 24092007

f∆ = 1
32
−4×−1
2×1 = 1 9+ 2 = 19 9
f
−1
3−√19 3
2×−1
2=−1
3−√19 3
−1=1 + √19 3−1
3+√19 3
2×−1
2=1−√19
3
f(x) = ax2+bx +c a =−1
2b=1 3c= 1
f(x) = a(x−α)2+β α =−b
2aβ=−∆ 4a
α=1
3β=−
19
9
−2=19 18 f(x) = −1 2(x−1 3)2+19 18
x2
x−∞ 1
3+∞
19
18
f3 QQQQs
f x2
x−∞ 1−√19
3
1 + √19
3+∞
f(x)−0 + 0 −
S(3; 5) y=a(x−3)2+ 5 a
(0; 0)
a(0 −3)2+ 5 = 0 9a=−5a=−5
9
y=a=−5
9(x−3)2+ 5
−5
9(x−3)2+ 5
0x0
3x0+ 0
2= 3 x0= 6
(6; 0)

−5
9(x−3)2+ 5
◦
x−3−x2−x2+x−3 ∆ = 1 −12 = −11
x2
2x2−1 2x2−1=(x√2−
1)(x√2 + 1) −1
√2=−√2
2
√2
2x2
2x2−1
x−∞ −√2/2√2/2 +∞
−x2+x−3− − −
2x2−1 + 0 −0 +
quotient −+−
#−√2
2;√2 2"
◦
x
x+ 1 + 163
x+ 2
x
x+ 1 +x+ 1
x+ 163 x+ 2
2x+ 1
x+ 1 −3
x+ 260
(2x+ 1)(x+ 2) −3(x+ 1)
(x+ 1)(x+ 2) 60
2x2+ 2x−1
(x+ 1)(x+ 2)60
∆ = 12 12 >0√12 = 2√3−1−√3
2
−1 + √3
2x2
−∞ −2−1−√3
2−1−1 + √3 2+∞
2x2+ 2x−1 + + 0 − − 0 +
x+ 1 − − − 0 + +
x+ 2 −0 + + + +
2x2+ 2x−1
(x+ 1)(x+ 2) +−0 + −0 +

#−2; −1−√3
2#∪#−1; −1 + √3 2#
(P)D
x2−2x+ 3 = x+p x2−3x+ 3 −p= 0
∆=9−4(3 −p)=4p−3
∆ = 0 p=3
4
(P)D p =3
4
x2−3x+9
4= 0 3 2
(P)D D
p=3
4y=x+3 4x=3 2y=3 2+3 4=9 4
p=3
4(P)D3 2;9 4
◦∆ = m2−4p p < 0−4p > 0m2>0
m2−4p > 0
◦(x−x1)(x−x2)
(x−x1)(x−x2) = x2−x2×x−x1×x+x1x2
x
x2−(x1+x2)x+x1x2
x−(x1+x2)
x1x2x2+mx +p m =−(x1+x2)p=x1x2