# Math Help - Solve quadratic eqiation using 3 points?

1. ## Solve quadratic equation using 3 points?

The points are (0, 189), (-116, 0), (116, 0), in general form.

I tried, but my answer didn't go with the parabola graph, help?

2. ## Re: Solve quadratic eqiation using 3 points?

The question is not clear. Do you want to form quadratic equation in general form. In that case
let the quadratic equation be y = ax^2+ bx + c. Now the points ( 0,189 ) lies on it so we get c = 189
now we have y = ax^2 + bx + 189; the points ( 116,0) and ( -116,0) lie on it. thus we get
( 116)^2 a + 116 b + 189 = 0 and
(116)^2 - 116 b + 189 = 0
Solve these two equations for a and b.

3. ## Re: Solve quadratic equation using 3 points?

Hello, dumbblonde!

$\text{Find the equation of the parabola containing: }(0, 189),\:(\text{-}116, 0),\:(116, 0)$

Code:
                |
*189
|
|
|
|
- - * - - + - - * - -
-116    |    116
|
The parabola opens downward
. . and has x-intercepts -116 and +116.

The equation has the form: . $y \:=\:a(x-116)(x+116) + b$

From $(\pm116,0)$, we have: . $a(0)+b \:=\:0 \quad\Rightarrow\quad b \,=\,0$

From $(0,189)$, we have: . $a(\text{-}116)(116) \:=\:189 \quad\Rightarrow\quad a \,=\,-\tfrac{189}{116^2}$

Therefore: . $y \;=\;-\tfrac{189}{13,456}(x-116)(x+116)$