# Determine the second grade function if f(-1) = 1, f(0) = 1 , f(1)= 3.

hala718 | Certified Educator

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

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

f(0) = c= 1

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

Substitute with c= 1

a-b+1 =1 .

==> a=b

a+b+1 =3

2a = 2 ==> a=1

a=1

b=1

c=1

Then the function is :

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

giorgiana1976 | Student

The general form of a second degree function is:

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

This function is determined when are known all coefficients, a,b,c.

We know from enunciation that:

f(-1)=1

We'll substitute x by -1, in the expression of the function.

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

a-b+c=1

We'll substitute x by 0, in the expression of the function.

f(0)=1

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

c=1

We'll substitute x by 1 into the expression of the function:

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

The coefficients of the function are:

a=1

b=1

c=1

The function is:

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

neela | Student

The second degree equation is of the form f(x) = ax^2+bx+c

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

f(0) = 1 gives : a*0^2+b*0+c = 1.... (2) . So c = 1

f(1) = 3 gives: a*1^2+b*1+c = 3......(3).

Eq(1)+E(2) gives: 2a+2c = 1+3 =4. But c =1 from (2).

So 2a+2 = 4 . Or 2a = 4-2 =2. Or a =1. So substituting a=1,c=1 in eq(3) , we get: 1+b+1 = 3. Or b = 3-2= 1.

So a=1, b=1 and c=1. So the second degre expression f(x) = x^2+x+1