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?
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.
Hello, dumbblonde!
$\displaystyle \text{Find the equation of the parabola containing: }(0, 189),\:(\text{-}116, 0),\:(116, 0)$
The parabola opens downwardCode:| *189 | | | | - - * - - + - - * - - -116 | 116 |
. . and has x-intercepts -116 and +116.
The equation has the form: .$\displaystyle y \:=\:a(x-116)(x+116) + b$
From $\displaystyle (\pm116,0)$, we have: .$\displaystyle a(0)+b \:=\:0 \quad\Rightarrow\quad b \,=\,0$
From $\displaystyle (0,189)$, we have: .$\displaystyle a(\text{-}116)(116) \:=\:189 \quad\Rightarrow\quad a \,=\,-\tfrac{189}{116^2}$
Therefore: .$\displaystyle y \;=\;-\tfrac{189}{13,456}(x-116)(x+116)$