Question

Find the equation of a parabola that passes through the points
(1,6) (2,20) and (3,40)

Answer 1

let the parabola be
y = ax^2 + bx + c
at (1,6) , 6 = a + b + c , #1
at (2,20), 20 = 4a + 2b + c , #2
at (3,40), 40 = 9a + 3b + c , #3

#2 - #1 : ----> 3a + b = 14
#3 - #2 : ----> 5a + b = 20

subtract these last two:
2a = 6
a = 3

in 3a + b = 14
b = 5

in a+b+c = 6
3 + 5 + c = 6
c = -2

your equation:
y = 3x^2 + 5x - 2

Answer 2

If y = ax^2+bx+c then we have
a+b+c = 6
4a+2b+c = 20
9a+3b+c = 40
Now solve as usual for a,b,c