Second degré, cours, première STI2D - MathsFG

ax2+bx +c

fR
f(x) = ax2+bx +c a b c a 6= 0
(O;~ i;~ j)
f S (α;f(α))
α=−b
2a
f
f(x) = 3x2−6x+ 1
α=−b
2a= 1 f(1) = −2S
(1; −2)
f
f7→ ax2+bx +c
•a > 0f
x=α=−b
2a
•a < 0f
x=α=−b
2a
a > 0
x∞α+∞
f(x)& %
f(α)
a < 0
x∞α+∞
f(α)
f(x)% &

f
f(x) = 3x2−6x+ 1
α=−b
2a= 1 f(1) = −2a= 3
a
x∞+∞
f(x)& %
•ax2+bx +c= 0 f
f(x) = ax2+bx +c x
•∆ ∆ =
b2−4ac
•
2x2−5x+ 2 2 ×22−5×2 + 2 = 2 ×4−10 + 2 = 0
•
2x2−5x+ 2 ∆ = 52−4×2×2 = 25 −16 = 9
∆ = b2−4ac ax2+bx +c
•∆<0ax2+bx +c= 0
ax2+bx +cR
•∆ = 0 ax2+bx +c= 0
x0=−b
2ax ax2+bx +c=a(x−x0)2
•∆>0ax2+bx+c= 0 x1=−b+√∆
2a
x2=−b−√∆
2ax ax2+bx +c=a(x−x1)(x−x2)

2x2−5x+ 2 = 0
∆=9=32
x1=−b+√∆
2a=5+3 2×2= 2 x2=−b−√∆ 2a=5−3 2×2=1 2
ax2+bx +c
•∆<0f(x)aR
•∆ = 0 f(x)aRx0
•∆>0f(x)a x1x2
−x2+ 6 + 7 ≥0
• −x2+ 6x+ 7 = 0
∆ = 36 −4×(−1) ×7 = 64
∆>0x1=−6+8
−2=−1x2=−6−8 −2= 7
•a < 0
x−∞ +∞
−x2+ 6x+ 7
S= [−1; 7]

ax2+bx +c= 0
f
•∆>0
•∆ = 0
•∆<0
a > 0a < 0
0 1 2 3−1−2−3 0
1
2
3
−1
−2
−3
delta > 0
a > 0
0 1 2 3−1−2−3 0
1
2
3
−1
−2
−3
delta > 0
a < 0
0 1 2 3−1−2−3 0
1
2
3
−1
−2
−3
delta = 0
a < 0
0 1 2 3−1−2−3 0
1
2
3
−1
−2
−3
delta < 0
a < 0