# Determine the quadratic function if f(1)=3 , f(0)=2 , f(-1)=3.

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### 4 Answers

Let the function be:

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

f(0) = c = 2

f(1) = a+b + 2 = 3........(1)

f(-1) = a - b + 2 = 3.

==> a= b+1 .........(2)

Now let us substitute (2) in (1)

==> a + b + 2 = 3

==> (b+1) + b + 2 = 3

==> 2b+3 = 3

==>2 b = 0

==> b =0

==> a = b+1 = 0 +1 = 1

==> a = 1 b = 0 c= 2

==> f(x) = 1*x^2 + 0*x + c

==> **f(x) = x^2 + 2**

The form of a quadratic function is described by the expression:

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

From the enunciation, we know that for 3 given values of the variable x, we get 3 values of the function.

f(-1)=3

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

a-b+c=3

f(0)=2

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

**c=2**

f(1)=3

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

a+b+c=3, but c=2

a+b+2=3

a+b=1 (1)

We have also the expression a-b+c=3 and c=2

a-b+2=3

a-b=1 (2)

We'll add (1) and (2):

a+b+a-b = 1+1

We'll reduce like terms:

2a = 2

We'll divide by 2:

**a = 1**

We'll input a=1 into (1):

1+b = 1

**b = 0**

The quadratic function is:

f(x) = 1*x^2 + 0*x + 2

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

Let the quadratic equation be f(x) = ax^2+bx+c.

The values of f(x) for 3 values are given. Substituting this in the equation, we get:

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

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

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

(1)+(3) : 2a+2c =6. a+c = 3. but c = 2 from (2). So a = 3-2 =1.

So a=1, c =2. Therefore a+b+c = 1+b+2 =3 from (1). So b = 3-3 = 0.

Therefore a = 1, b =0 and c =2.

So f(x) = ax+bx+c = x+0x+2 = x+2.

f(x) = x+2.

f

Let the function be f(x)=ax^2+bx+c

We know that f(1)=3, f(0)=2 and f(-1)=3.

So f(1)=3=a*1^2+b*1+c=a+b+c

f(0)=a*0+b*0+c=2, so c=2.

This gives us f(1)=3=a+b+2

f(-1)=a(-1)^2+(-1)*b+c=a-b+2=3

Using f(1) and f(-1), 2a+4=6 or a=(6-4)/2=1

substituting a and c in f(1) 3=1+b+2 or b=0.

Therefore the quadratic function is **f(x)=x^2+2**