Given the points 01 26 531 find parabola yax2 bx c passing

Given the points (0,1), (2,6), (5,31), find parabola y=ax² + bx + c passing through them passing through them.

*Use matrices to solve.

Solution

1st we will find the three equations

for x=0 and y= 1

we will have 1=c

for x= 2, y= 6   we will have

6=4a+2b+c

for x= 5 and y= 31

31= 25a+5b+c

we will set up matrices

Your matrix

Make the pivot in the 1st column by dividing the 2nd row by 4 and swap the 2nd and the 1st rows

Multiply the 1st row by 25

Subtract the 1st row from the 3rd row and restore it

Make the pivot in the 2nd column by dividing the 3rd row by -15/2 and swap the 3rd and the 2nd rows

Multiply the 2nd row by 1/2

Subtract the 2nd row from the 1st row and restore it

Find the pivot in the 3rd column in the 3rd row

Multiply the 3rd row by -1/10

Subtract the 3rd row from the 1st row and restore it

Multiply the 3rd row by 7/10

Subtract the 3rd row from the 2nd row and restore it

Solution set:

x1 = 7/6

x2 = 1/6

x3 = 1

so we will have equation   y= 7/6x^2+1/6x+1

or 6y=7x^2+x+1     or y= 1/6(7x^2+x+1)