Question

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

Answer 1

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