Fonctions polynômes du second degré, classe de - MathsFG

ax2+bx +c x a b c
a6= 0
f(x) =
a(x−u)2+v u v
f(x) = ax2+bx +c
f(x) = a(x2+b
ax+c a)
f(x) = a((x+b
2a)2−b2 4a2+c a)
f(x) = a((x+b
2a)2−b2 4a2+4ac 4a2)
f f(x) = −2x2+ 12x+ 10 f(x) = −2(x2−6x−5)
f(x) = −2((x−3)2−9−5) f(x) = −2((x−3)2−14)
f(x) = −2(x−3)2+ 28
f[−1; 5]

O~
i
~
j
−2−1 0 1 2 3 4 5 6
−10
−5
0
5
10
15
20
25
30
f(x) =
a(x−u)2+v
•a > 0f]− ∞;u]
[u; +∞[v u
•a < 0f]− ∞;u]
[u; +∞[v
u
a > 0
x∞u+∞
f(x)& %
v
a < 0
x∞u+∞
v
f(x)% &

a < 0f(x) =
a(x−u)2+v a u v a 6= 0
x1x2]−∞; 0] x1≤x2
x1−u≤x2−u≤0 ] − ∞; 0]
(x1−u)2≥(x2−u)2a
a(x1−u)2+v≤a(x2−u)2+v a(x1−u)2+v≤a(x2−u)2+v
f(x1)≤f(x2) ] − ∞; 0]
x1x2[u; +∞[x1≤x2
0≤x1−u≤x2−u[0; +∞[
(x1−u)2≤(x2−u)2a
a(x1−u)2+v≥a(x2−u)2+v a(x1−u)2+v≥a(x2−u)2+v
f(x1)≥f(x2) [u; +∞[
x=u f(u) = (u−u)2+v=v
f(x) = −2(x−3)2+ 28 a=−2a
x∞+∞
f(x)% &
•(3x+2)(4x−3) = 0
3x+ 2 = 0 4x−3 = 0 3x= 2 4x= 3
x=2
3x=3 4(3x+ 2)(4x−3) = 0
2
3
3
4
•(3x+ 2)(2x−1) −x(3x+ 2) = 0
3x2−x−2 = 0
(3x+ 2)(2x+ 1 −x) = 0
(3x+ 2)(x+ 1) = 0
3x+ 2 = 0 x+ 1 = 0 x=−2
3x=−1

•x2+ 4x+ 4 = 0
x2+ 4x+ 4 = (x+ 2)2(x+ 2)2= 0
x+ 2 = 0 x=−2