# Determine the second degree function, whose graph is passing through the points A(-2,3), B(2,0), C(0,1).

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Let second degree function y = ax^2+bx+c be the curve passing through A(-2,3), B(2,0) and C(0,1).

Solution: The coordinates of the points A,B and C on the graph should satisfy ax^2+bx+c = y. So,

4a-2b+c=3..........(1)

4a+2b+c =0.........(2)

0a+0b+c=1..........(3) So **c =1. **Substituting in (1) and (2),

4a-2b+1 = 3. Or 4a-2b = 2...........(4)

4a+2b +1 =0 Or 4a+2b = -1..........(5)

(4)+(5) gives: 8a =1 Or **a = 1/8**

(4)-(5) gives: -4b =3. Or **b** **= -3/4**

A point which is located on a graph has the property that it's coordinates verifies the equation of the graph's function.

In our case, the second degree function could be written as below:

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

In order to determine the second degree function, their coefficients must be calculated.

From enunciation, we find out that 3 points, whose coordinates are known, are located on the graph of the function which has to be determined, so that we'll write 3 relationships.

For the point A(-2,3) to belong to the graph of the function, the mathematical condition is:

f(-2)=3

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

4a-2b+c=

For the point B(2,0) to belong to the graph of the function, the mathematical condition is:

f(2)=0

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

4a+2b+c=0

For the point C(0,1) to belong to the graph of the function, the mathematical condition is:

f(0)=1

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

**c=1**

We'll substitute the known value of c into the first 2 relationships, so that:

4a-2b+1=3

4a+2b+1=0

If we are adding these relationships, and reduce the similar terms, we'll have:

4a-2b+1+4a+2b+1=3+0

8a+2=3

8a=3-2

8a=1

a=1/8

For finding the value of b, we'll substitute the found value of a into one of 2 relationships above.

We'll choose, for example, the relationship:

4a+2b+1=0

4*1/8+2b=-1

(½)+2b=-1

2b=-1-(1/2)

2b=-3/2

b=-3/4

So, the function is:

**f(x)=(1/8)*x^2-(3/4)*x+1**