fonction analyse [tex]f(x) = ax {}^{2} + bx + c \\ f(x) = x {}^{2} - 4x + 3 [/tex] ​

ANSWERS

Given f(x)=ax2+bx+c

∴f(0)=a.02+b.0+c=6

or 0.a+0.b+1.c=6              (i)

and f(2)=a(2)2+b(2)+c=11

⇒4.a+2.b+1.c=11         (ii)

and f(−3)=a(−3)2+b(−3)+c=6 

⇒9.a−3.b+c=6              (iii)

Since (i), (ii) and (iii) are three equations in a, b, c solving these by Cramer's rule

∴D=∣∣∣∣∣∣∣∣04902−3111∣∣∣∣∣∣∣∣=1.(−12−18)=−30

D1=∣∣∣∣∣∣∣∣611602−3111∣∣

If f(x)=x2−4x+3

If f(x)=x2−4x+3F(2)=22−4(2)+3

If f(x)=x2−4x+3F(2)=22−4(2)+3=4−8+3

If f(x)=x2−4x+3F(2)=22−4(2)+3=4−8+3=−4+3

If f(x)=x2−4x+3F(2)=22−4(2)+3=4−8+3=−4+3=−1

If f(x)=x2−4x+3F(2)=22−4(2)+3=4−8+3=−4+3=−1F(1)=(1)2−4+3

If f(x)=x2−4x+3F(2)=22−4(2)+3=4−8+3=−4+3=−1F(1)=(1)2−4+3=1−4+3−3+3

If f(x)=x2−4x+3F(2)=22−4(2)+3=4−8+3=−4+3=−1F(1)=(1)2−4+3=1−4+3−3+3=0

If f(x)=x2−4x+3F(2)=22−4(2)+3=4−8+3=−4+3=−1F(1)=(1)2−4+3=1−4+3−3+3=0F(2)−F(1)

If f(x)=x2−4x+3F(2)=22−4(2)+3=4−8+3=−4+3=−1F(1)=(1)2−4+3=1−4+3−3+3=0F(2)−F(1)=−1+0

If f(x)=x2−4x+3F(2)=22−4(2)+3=4−8+3=−4+3=−1F(1)=(1)2−4+3=1−4+3−3+3=0F(2)−F(1)=−1+0=−1