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2014-10-15T12:15:26+05:30

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Standard form of a polynomial of degree 2 is : y=ax²+bx+c
As the points passes through the curve,

Point(1,6):
6 = a(1)² + b(1) + c
⇒ 6 = a+b+c

Point(2,6):
6 = a(2)² + b(2) + c
⇒ 6 = 4a+2b+c

Point(3,2):
2 = a(3)² + b(3) + c
⇒ 2 = 9a+3b+c

Solving the equations, we get
a = -2, b=6, c=2
So the polynomial is y = -2x² + 6x + 2
when x=-2, y = -2(-2)² + 6(-2) + 2 = -18

2 5 2
what about y² and xy terms ? they are also part of a polynomials of 2nd degree.
I think siri was asking about a polynomial which can be solved using the data given. So it won't contain any y^2 and xy terms.
2014-10-16T18:17:01+05:30

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General form of a polynomial of degree two is:
     x² + A xy + B y² + C x + D y + E = 0    -- equation 1

There are 5 parameters. We are given 3 inputs. So we can find the solution in terms of two parameters. Thus there will be infinite solutions, ie., there will be infinite number of curves (polynomials of 2nd degree) passing through the given 3 points.

Substituting the values x =1 and y = 6, as (1,6) is on the curve:
        6 A + 36 B + C + 6 D + E + 1 = 0     -- equation 2

As (2,6) is on the curve:
       12 A + 36 B + 2 C + 6 D + E + 4 = 0        -- equation 3

As (3,2) is on the curve:
       6 A + 4 B + 3 C + 2 D + E + 9 = 0        -- equation 4

Eq 3 - eq 2 gives :
           6A+C +3 = 0        So  C = -3 (2A+1)      -- equation  5

eq 1 + eq 2 - eq 3  gives :
           -4 + 12 A + 68 B + 10 D + E = 0
                   E = 4 - 12A - 68B - 10 D        -- equation 6

Substitute values of C and E in eq 2,
         6A + 36 B -6A -3 + 6D + 4- 12 A - 68 B- 10D + 1=0
            6 A + 16 B + 2 D = 1
               D = (1 - 6 A - 16 B) / 2            -- equation 7

Substituting value of D in eq 6,
               E = 4 - 12 A - 68 B - 5 + 30 A + 80 B
               E = 18 A + 12 B - 1              -- equation 8

Substitute values of E, D and C from eq 5, eq 7 & eq 8 in equation 1,  the needed curve is:

     x² + A xy + B y² - 3(2A+1) x - (3A + 8B - 1/2) y + (18 A + 12 B -1) = 0

This polynomial passes through the given points for all real values of A & B. Obviously, A and B both should not be zero.

 Let x = -2,
4 - 2A y + B y^2 +12 A + 6 - (3A + 8B - \frac{1}{2}) y + (18 A + 12 B -1) = 0\\ \\B y^2 - (5A + 8 B - \frac{1}{2}) y + (30 A + 12B + 9) = 0\\ \\y=\frac{(5A+8B-\frac{1}{2})+ - \sqrt{25A^2+16B^2-40AB--5A-44B+\frac{1}{4}}}{2B}\\ \\

0
For example, if you put A = 0, and B = 3, you get a curve : x² + 3y² -3x - 23.5 y +35 = 0. and at x = -2, y = 9/2 or 10/3.
It is also possible that there is a relation between A and B.