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

1 Answer
Nov 8, 2016

Please see the explanation for steps leading to the equation:
y = 3x^2 - 2x + 5y=3x22x+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 + cy=ax2+bx+c

  1. The type that opens left or right:

x = ay^2 + bx + cx=ay2+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 + cy=ax2+bx+c

Write 3 different equations using the 3 points:

6 = a(1)^2 + b(1) + c6=a(1)2+b(1)+c" [1]" [1]
26 = a(3)^2 + b(3) + c26=a(3)2+b(3)+c" [2]" [2]
21 = a(-2)^2 + b(-2) + c21=a(2)2+b(2)+c" [3]" [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