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