Let us assume that the function is :

f(x) = ax^2 + bx + c

f(-1) = a-b+c= 1 .....(1)

f(0)= c= 1......(2)

f(1)= a+b+c= 3 ...(3)

from (2) we determined c=1

==> substitute in (1) .... a-b+1=1

==> a-b=0 ==> a=b

==> Substitute in (3) .... a+b+1=3

==> a+b= 2

==> 2a=2 ==> a=1 and b=1

Then the equation is:

f(x)= x^2+x+1

f(1) = 1, f(0) =1 and f(1) =3. To find second degree function f(x).

Solution:

A second degree function f(x) is of the type ax^2+bx+c.

f(x) = ax^2+bx+c. Put x=-1, x=0 and x=1, then

f(-1) = a(-1)^2+b(-1)+c = 1 given...(1)

f(0) = a(0)^2+b(0)+c = 1 given.......(2) and

f(1) a(1)^2+b(1)+c = 3 given .........(3)

So c = 1 from eq(2).

Eq(1)+eq(3) gives: (a+a)+ 0 +2c = 2a+2c = 4 Or

a+c = 2 Or

a+1 =2. So a = 1.

Substitute a =1, c = 1 in eq(1): 1(-1)^2+b(-1)+1 = 1. Or

-b = 1- 2 = -1.

Therefore, a=1, b =-1 and c =1 and the 2nd degree function f(x) ax^2+bx+c = x^2-x+1.

The second grade function is described by the expression:

f(x) = ax^2 + bx + c

f(-1) = 1

f(-1) = a*(-1)^2 + b*(-1) + c = a - b + c

a - b + c = 1

f(0) = 1

f(0) = a*(0)^2 + b*(0) + c = c

c = 1

f(1) = 3

f(1) = a*(1)^2 + b*(1) + c = a + b + c

a + b + c = 3, but c = 1

a + b + 1 = 3

a + b = 2

We have also the expression a - b + c = 1 and c = 1

a - b + 1 = 1

a - b = 0

a = b, but a + b = 2 => a + a = 2

2a = 2

a = 1

b = 1

c = 1