# Determine the quadratic function if the graph passes through the point (0;1) , (1;3) and (-1;1)

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

The points (0,1) (1,3) and (-1,1) passes through th curve.

f(0)= 1

==> **f(0) = c = 1** ..........(1)

f(1) = 3

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

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

To determine a and b we will solve the system.

Add (1) and (2):

==> 2a +2= 4

**==> a= 1**

==> a-b + 1 = 1

==> a = b

**==: b= 1**

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

A quadratic function is described by the expression:

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

To determine the quadratic function, we'll have to calculate the coefficients a, b, c.

We'll calculate the coefficients using the condition given by enunciation: If the graph of the quadratic passes through the given points, we'll have the relations:

The point (0,1) is on the graph if and only if:

f(0)=1

We'll substitute 0 in the expression of the quadratic:

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

c=1

The point (1,3) is on the graph if and only if:

f(1)=3

We'll substitute 1 in the expression of the quadratic:

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

The point (-1,1) is on the graph if and only if:

f(-1)=1

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

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

a-b+c=1

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

**The expression of the quadratic is:**

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

Let f(x) = ax^2+bx+c.which pass through (0 ,1), (1 ,3) and(-1 ,1)

Therefore, sobstituting the coordinates in f(x) ax^2+bx+c , we get:

f(0) = a*0^2+b*0+c = 1. Therefore 0+0+c = 1......(1)

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

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

f(-1) = a(_1)^2+b(-1)+c = a-b+1 = 1. Or

a-b = 0.....(3).

(2)+(3):

2a = 2. So a = 1.

(2) -(3). 2b = 2-0 = 2. Or b = 1.

Therefore a = 1 , b = 1 , c =1.