Question

How do you find the equation of the quadratic function that passes through the points `(1, 2)`, `(9, -4)` and `(11, -7)`?

Answer 1

Equation is `40y=-3x^2+83`

Explanation:

Let the equation be `y=ax^2+bx+c`. As it passes through `(1,2)`, `(9,-4)` and `(11,-7)`, we have

`a+b+c=2` .........................(A)

`81a+9b+c=-4` .........................(B)

and `121a+11b+c=-7` .........................(C)

Subtracting (A) from (B), we get `80a+8b=-6` ....................(D)

and subtracting (B) from (C), we get `40a+2b=-3` .................(E)

Now double (E) and subtract from (D) and we get

`4b=0` i.e. `b=0` and putting it in (E), we get `40a=-3` or `a=-3/40`

and now putting values of `a` and`b` in (A), we get

`c=2+3/40=83/40`

Hence quadratic equation is

`y=-3/40x^2+83/40` or `40y=-3x^2+83`

graph{(40y+3x^2-83)((x-1)^2+(y-2)^2-0.1)((x-9)^2+(y+4)^2-0.1)((x-11)^2+(y+7)^2-0.1)=0 [-16.25, 23.75, -11.92, 8.08]}