Originally Posted by
puneet Q)Determine the quadratic polynomial in one variable whose graph passes thru the points (1,6), (2,3) and (3,2).
ANS)Cont.
Thus we have
6 = a + b + c --------(1)
3 = 4a + 2b + c --------(2)
2 = 9a + 3b + c --------(3)
Multiply (1) by 2 and subtract from (2)
Multiply (1) by 3 and subtract from (3)
We get
2a-c= -9 à c-2a=9 --------(4)
And
6a-2c = -16 à 3a-c=-8 --------(5)
Adding (4) and (5) We get
a=1
Putting a in (4) We get
c-2(1)=9 à c-2=9 à c=11
putting a and c in (1)
6=1+b+11 à 6=12+b à b=-6
Hence in order we get
a=1,b=-6,c=11 --------(6)
We have
F(x) =ax2 + bx + c --------(7)
Putting values of f and (6) in (7) we get
x2 - 6x + 5 = 0
x2 - 6x + 8 = 0
x2 - 6x + 9 = 0
Is This right I am getting three equations whats wrong.