# Given that f(-1)=3, f(2)=4, f(1)=1 find the quadratic function f(x)

*print*Print*list*Cite

### 3 Answers

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

Now we have:

f(-1) = 3

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

=> a -b + c =3 ...(1)

f(2) = 4

=> a(2)^2 + b(2) + c = 4

=> 4a + 2b +c = 4 ...(2)

f(1) = 1

=> a(1)^2 + b(1) + c = 1

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

Now add (1) and (3)

=>a -b + c + a +b +c = 3 +1

=> 2a + 2c =4

=> a + c = 2

Substitute a +c =2 in a +b +c =1, we get b = -1

Now using b= -1 in 4a + 2b +c = 4, gives 4a -2 +c =4 or 4a +c =6

Now using a + c = 2 and 4a +c =6

=> 4a + 4c - 4a -c = 8 - 6

=> 3c = 2

=> c =2/3

Now put c= 2/3 and b= -1 in a +b +c =1, gives a -1 +2/3 =1

=> a = 1+1 -2/3

=> a = 2-2/3 = 4/3

**Therefore the quadratic function is f(x) = (4/3)x^2 -bx + 2/3**

Let f(x) ax^2+bx+c .

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

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

f(2) = a(2)^2+b*2+c = 4

4a+2b+c = 4.................(2)

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

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

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

Eq(3)-eq(1): 2b = -2. Or b = -1

Eq(1)+eq(3) : 2(a+c )= 4. So

a+c = 2.........(4)

Eq(2) +2Eq(1) = 4a+2b+c)+2(a-b+c) = 4+2*3 =10

6a+3c = 10.....(5)

Eq(5)- 3*eq(4) : 6a-3a = 10-3*2 = 4. So 3a = 4 , a = 4/3.

FRom (4), c = 2-a = 2-4/3 = 2/3.

Therefore ax^2+bx+c = (4/3)x^2 -x+ 2/3.

We'll write the expression of the quadratic function:

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

f(-1)=3

We'll substitute x by -1 in the expression of the quadratic:

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

**a-b+c=3 (1)**

f(2)=4

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

**4a + 2b + c = 4 (2)**

f(1)=1

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

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

We'll add (1) + (3):

a-b+c+a+b+c=3+1

We'll eliminate and combine like terms:

2a + 2c = 4

We'll divide by 2:

a + c = 2 (4)

We'll multiply (1) by 2:

2a - 2b + 2c = 6 (5)

We'll add (5) + (2):

2a - 2b + 2c + 4a + 2b + c = 6 + 4

We'll eliminate and combine like terms:

6a + 3c = 10 (6)

We'll multiply (4) by -3:

-3a - 3c = -6 (7)

We'll add (6) + (7):

6a + 3c -3a - 3c = 10 - 6

3a = 4

We'll divide by 3:

**a = 4/3**

We'll substitute a in 6a + 3c = 10.

6*(4/3) + 3c = 10

8 + 3c = 10

3c = 10-8

3c = 2

**c = 2/3**

We'll substitute a and c in (1):

4/3 - b + 2/3 = 3

2 - b = 3

We'll subtract 2 both sides:

-b = 3-2

-b = 1

**b = -1**

**The quadratic funation is:**

**f(x) = (4/3)x^2 - x + 2/3**