Example?. The following are rough sketches of the graphs of two quadratic functions:
$$
\begin{array}{l}
y=a x^{2}+b x+c \\
y=p x^{2}+q x+r
\end{array}
$$
(i) Based on the graphs, determine $p, q, r, a+c, pa^{2}+qa$
(ii) Let the x-coordinates of the intersection points of the two graphs be $\boldsymbol{\alpha}, \boldsymbol{\beta}$, express $\alpha+\beta, \alpha \beta$ in terms of $a, b, c, p, q, r$.
(i)(1) is an opening question about the "line",
$$
\therefore \mathrm{a}0 \text {. }
$$

Since (1) intersects the $y$-axis in the negative half-axis, $c0$. Therefore, $\mathrm{a}+\mathrm{c}0$. The vertex of the parabola (2) has a positive $x$-coordinate, so $-\frac{\mathrm{q}}{2 \mathrm{p}}>0$, but $\mathrm{p}>0$, hence $\mathrm{q}<0$.
$$
\therefore p, r, p a^{2}+q a+r \text { are positive; }
$$
$q, a+c$ are negative.
(ii) Consider the equation $a x^{2}+b x+c=p x^{2}+q x+r$,

which is $(a-p) x^{2}+(b-q) x+(c-r)=0$. The roots of this equation are the $x$-coordinates of the intersection points of curves (1) and (2), denoted as $\boldsymbol{a}$ and $\beta$.
$\therefore \alpha \beta=-\frac{b-q}{a-p}, \alpha \beta=\frac{c-r}{a-p}$.
(3) Prove using the graph.