Question

Given three points (2,-2) (1,-7) (6,-2) how do you write a quadratic function in standard form with the points?

Answer 1

Write 3 linear equations by substituting the 3 points into the standard form,

`y = ax^2+bx+c" [1]"`

, and then solve the 3 equations for a, b, and c.

Explanation:

Substitute (2,-2) into equation [1]:

`a(2)^2+b(2)+c = -2`

Write this as the first row an augmented matrix:

`[(4,2,1,|,-2)]`

Substitute (1,-7) into equation [1]:

`a(1)^2+b(1)+c = -7`

Write this as the second row of the augmented matrix:

#[ (4,2,1,|,-2), (1,1,1,|,-7) ]#

Substitute (6,-2) into equation [1]:

`a(6)^2+b(6)+c = -2`

Write this as the second row of the augmented matrix:

#[ (4,2,1,|,-2), (1,1,1,|,-7), (36,6,1,|,-2) ]#

Perform elementary row operations, until an Identity matrix is obtained on the left.

`R_1harrR_2`

#[ (1,1,1,|,-7), (4,2,1,|,-2), (36,6,1,|,-2) ]#

`R_2-4R_1to R_2`

#[ (1,1,1,|,-7), (0,-2,-3,|,26), (36,6,1,|,-2) ]#

`R_3-36R_1to R_3`

#[ (1,1,1,|,-7), (0,-2,-3,|,26), (0,-30,-35,|,250) ]#

`R_3-15R_2toR_3`

#[ (1,1,1,|,-7), (0,-2,-3,|,26), (0,0,10,|,-140) ]#

`1/10R_3toR_3`

#[ (1,1,1,|,-7), (0,-2,-3,|,26), (0,0,1,|,-14) ]#

`R_2 + 3R_3toR_2`

#[ (1,1,1,|,-7), (0,-2,0,|,-16), (0,0,1,|,-14) ]#

`-1/2R_2 to R_2`

#[ (1,1,1,|,-7), (0,1,0,|,8), (0,0,1,|,-14) ]#

`R_1-R_3toR_1`

#[ (1,1,0,|,7), (0,1,0,|,8), (0,0,1,|,-14) ]#

`R_1-R_2toR_1`

#[ (1,0,0,|,-1), (0,1,0,|,8), (0,0,1,|,-14) ]#

There is an identity matrix on the left, therefore, `a = -1, b = 8, and c = -14`

Substituting these values into equation [1]:

`y = -x^2+8x-14" [2]" larr` answer.

Check:

Substitute (2,-2) into equation [2]:

`-2=-(2)^2+8(2)-14`
`-2=-2`

Substitute (1,-7) into equation [2]:

`-7=-1+8-14`
`-7=-7`

Substitute (6,-2) into equation [2]:

`-2=-(6)^2+8(6)-14`
`-2=-2`

This checks.