The required equation is y = x + 2
Solution:
Given three points are (1,3) (3,5) (5,7)
A quadratic equation for three points is given by [tex]y = ax^2 + bx + c \rightarrow (1)[/tex]
Assume the point 1,3 and substitute in (1)
[tex]\Rightarrow 3= a + b + c \rightarrow (2)[/tex]
Let us substitute 3,5 in (1)
[tex]\Rightarrow 5 = 9a + 3b + c \rightarrow (3)[/tex]
Let us substitute 5,7 in (1)
[tex]\Rightarrow 7 = 25a + 5b + c \rightarrow (4)[/tex]
Subtracting 2 and 3
[tex]\Rightarrow 9a + 3b + c = 5 - ( a+ b+ c = 3)[/tex]
[tex]\Rightarrow 8a + 2b = 2[/tex]
[tex]\Rightarrow 4a + b = 1[/tex]
[tex]\Rightarrow b = 1-4a[/tex]
Subtracting 4 and 3
[tex]\Rightarrow 25a + 5b + c = 7 - ( 9a + 3b + c = 5)[/tex]
[tex]\Rightarrow 16a + 2b = 2 \rightarrow (5)[/tex]
Substituting b in 5, [tex]16a + 2(1 -4a) = 2[/tex]
[tex]\Rightarrow 8a = 0[/tex]
[tex]\Rightarrow a = 0; \ b = 1- 4(0) = 1[/tex]
[tex]\Rightarrow b = 1 \\\\From \ 2, a + b + c =3[/tex]
[tex]\Rightarrow 0+1+c =3[/tex]
[tex]\Rightarrow c = 2[/tex]
Substituting a, b, c in 1 we get the quadratic equation [tex]y = x + 2[/tex]
