# Determine the equation of the polynomial of degree two whose graph passes through the points (1,6), (2,6) and (3,2). Find y when x= -2

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by sweetysiri92

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by sweetysiri92

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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**

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

when x=-2, y = -2(-2)² + 6(-2) + 2 =

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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,

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,