f(x)=ax^2+bx+c where f(x) has real values for all real value of x. if f(1)=10 and f(2)=-5
Answers
Given f(x)=ax
2
+bx+c
∴f(0)=a.0
2
+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=
∣
∣
∣
∣
∣
∣
∣
∣
0
4
9
0
2
−3
1
1
1
∣
∣
∣
∣
∣
∣
∣
∣
=1.(−12−18)=−30
D
1
=
∣
∣
∣
∣
∣
∣
∣
∣
6
11
6
0
2
−3
1
1
1
∣
∣
∣
∣
∣
∣
∣
∣
=6(2+3)−0+1(−33−12) =−15
D
2
=
∣
∣
∣
∣
∣
∣
∣
∣
0
4
9
6
11
6
1
1
1
∣
∣
∣
∣
∣
∣
∣
∣
=0−6(4−9)+1(24−99) =20−75 =−45
D
3
=
∣
∣
∣
∣
∣
∣
∣
∣
0
4
9
0
2
−3
6
11
6
∣
∣
∣
∣
∣
∣
∣
∣
=6(−12−18) =−180
Hence by Cramer's rule
a=
D
D
1
=
−30
−15
=
2
1
b=
D
D
2
=
−30
−45
=
2
3
and c=
D
D
3
=
−30
−180
=6
Hence a=
2
1
,b=
2
3
,c=6
∴f(x)=
2
1
x
2
+
2
3
x+6
and f(1)=
2
1
(1)
2
+
2
3
(1)+6 =8
hopefully it will help u