How do you find the equation of the quadratic function given (1,6), (3,26), (-2,21)?

1 Answer
Nov 8, 2016

Answer:

Please see the explanation for steps leading to the equation:
#y = 3x^2 - 2x + 5#

Explanation:

There are two standard equations for a parabola for the two different types of parabolas.

  1. The type that opens up or down:

#y = ax^2 + bx + c#

  1. The type that opens left or right:

#x = ay^2 + bx + c#

The points look like they belong to a parabola that opens up, therefore, we shall try to fit them the first equation:

#y = ax^2 + bx + c#

Write 3 different equations using the 3 points:

#6 = a(1)^2 + b(1) + c##" [1]"#
#26 = a(3)^2 + b(3) + c##" [2]"#
#21 = a(-2)^2 + b(-2) + c##" [3]"#

Write these equations into an augmented matrix:

#[ (1,1,1,|,6), (9,3,1,|,26), (4,-2,1,|,21) ]#

Multiply row 1 by -9 and add to row 2:

#[ (1,1,1,|,6), (0,-6,-8,|,-28), (4,-2,1,|,21) ]#

Multiply row 1 by -4 and add to row 3:

#[ (1,1,1,|,6), (0,-6,-8,|,-28), (0,-6,-3,|,-3) ]#

Subtract row 2 from row 3:

#[ (1,1,1,|,6), (0,-6,-8,|,-28), (0,0,5,|,25) ]#

Divide row 3 by 5:

#[ (1,1,1,|,6), (0,-6,-8,|,-28), (0,0,1,|,5) ]#

Multiply row 3 by 8 and add to row 2:

#[ (1,1,1,|,6), (0,-6,0,|,12), (0,0,1,|,5) ]#

Divide row 2 by -6:

#[ (1,1,1,|,6), (0,1,0,|,-2), (0,0,1,|,5) ]#

Subtract row 3 from row 1

#[ (1,1,0,|,1), (0,1,0,|,-2), (0,0,1,|,5) ]#

Subtract row 2 from row 1:

#[ (1,0,0,|,3), (0,1,0,|,-2), (0,0,1,|,5) ]#

#a = 3, b = -2 and c = 5#

Check:

#6 = 3(1)^2 - 2(1) + 5##" [1]"#
#26 = 3(3)^2 - 2(3) + 5##" [2]"#
#21 = 3(-2)^2 - 2(-2) + 5##" [3]"#

#6 = 6#
#26 = 26#
#21 = 21#

This checks

The equation is:

#y = 3x^2 - 2x + 5#