& lt 1 & gt； Given three points on the image (-1, 3), (1, 3) and (2, 6), substitute them into the quadratic function equations.

3=a(- 1)^2+b(- 1)+c

3=a( 1)^2+b( 1)+c

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

The solution is a =1b = 0 c = 2; Quadratic function y = x 2+2

& lt2> method is the same as 1.

a = 2 b = 1 c =-2； Quadratic function y = 2x 2+x-2

& lt3> method is the same as 1.

The solution gives a =1.25b =-2.5c =-3.75; Quadratic function y =1.25x 2-2.5x-3.75.

& lt4> method is the same as 1.

a = 1 b =-5 c = 6； Quadratic function y = x 2-5x+6

I hope you are satisfied ~!