Find the sum of the roots of a quadratic function given the vertex of its graph Question:
At this parabola $$y = ax^2 + bx - c$$ and vertex is $T(3,9)$. What is the sum of roots of this parabola ?
Help or give a hint. Thanks
 A: HINT : $$y=ax^2+bx-c=a\left(x+\frac{b}{2a}\right)^2+\cdots$$
and the sum of roots can be represented as $-\frac ba$ by Vieta's formulas.
A: The roots occur symmetrically on both sides of the apex so their sum is $6$.
A: Hint Reflecting the parabola about its axis of symmetry exchanges the two roots, so the midpoint $\frac{1}{2} (r_1 + r_2)$ of the two roots $r_1, r_2$ is on that axis. On the other hand, the axis passes through the vertex, $(3, 9)$, and so has equation $x = 3$.

 Hence, $$\frac{1}{2} (r_1 + r_2) = 3.$$

A: If the vertex is at $(x_c,y_c)$ I would factor the quadratic as
$$ y = y_c + a (x-x_c)^2 $$
This obeys the boundary conditions 
$$\begin{align} \lim_{x=x_c} y &= y_c \\ \lim_{x=x_c} \frac{{\rm d}y}{{\rm d}x} &= 0 \end{align} $$
The 2 roots are $$x_i = x_c \pm \sqrt{-\frac{y_c}{a}} $$
so $x_1+x_2$ is ... 

 $$ x_1+x_2 = \left(x_c+\sqrt{-\frac{y_c}{a}}\right) + \left(x_c-\sqrt{-\frac{y_c}{a}}\right) = 2 x_c $$

NOTE: There is a necessary condition for real roots that $\frac{y_c}{a}<0$ but the sum of the roots is well defined regardless. This is due to the inherit symmetry of parabolas.
