Determine the second degree function, whose graph is passing through the points A(-2,3), B(2,0), C(0,1).
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,
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:
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:
For the point B(2,0) to belong to the graph of the function, the mathematical condition is:
But f(2)= a*(2)^2+b*(2)+c
For the point C(0,1) to belong to the graph of the function, the mathematical condition is:
But f(0)= a*(0)^2+b*(0)+c
We'll substitute the known value of c into the first 2 relationships, so that:
If we are adding these relationships, and reduce the similar terms, we'll have:
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:
So, the function is: