Why does the coefficient tell the way a quadratic parabola "opens"? I've seen a lot of texts referring to the coefficients and their sign for determining which way a parabola opens up. But is there more than this kind of "thumb rule" to it?
That is, how was it proven that for those coefficient ranges, the parabola opens up into a certain direction?
 A: Consider the vertex form of the quadratic function: $f(x) = a(x-h)^2+k$. When we compare $f(x)$ to $g(x)=x^2$, we know how each parameter affects the shape of the graph of $f(x)$ as compared to $g(x)$:
1.) When $h>0$, $f(x)$ is shifted to the $\bf{right}$ when compared to $g(x)$.
2.) When $h<0$, $f(x)$ is shifted to the $\bf{left}$ when compared to $g(x)$. 
3.) When $k>0$, $f(x)$ is shifted to the $\bf{up}$ when compared to $g(x)$. 
4.) When $k<0$, $f(x)$ is shifted to the $\bf{down}$ when compared to $g(x)$.
A little experimentation with a graphing calculator can convince you of the truth of these statements.
When we analyze how $a$ affects the graph, the results can be a bit more subtle, but still understandable. Since $g(x)$ has $a=1$, it makes sense that 1 would serve as a boundary value for $a$. So what happens when $a>1$? Compared to $g(x)$, $f(x)$ seems to be stretched vertically up without moving the vertex. And when $0<a<1$ we seem to get a more vertically flattened parabola. 
Finally, what happens when we let $a<0$? The answer is almost the same thing as before, but everything is flipped over the $x-$axis. Why should this be? No matter what kind of function you analyze, as soon as you multiply the function value by a negative quantity, the graph is reflected about the $x-$axis because all outputs that were negative are now positive and vice versa. When this happens with a parabola, it has the appearance of opening downward once it is reflected about the $x-$axis.
A: If you accept that a parabola only opens in "one direction" (up or down) eventually regardless of whether you move right or left, then all you have to do is plug in $x = 0$ and then consider $x = -N$ and $x = N$ for really large $N$ to see which direction the parabola opens. It should be clear that the sign of $A$ for the equation $y = Ax^2 + Bx + C$ determines this, since $AN^2$ is so much larger than $BN + C$ for large $N$. If you want a more rigorous argument, you can use calculus to show that the parabola is either "concave" (opening down) or "convex" (opening up), using the sign of the second derivative of the equation for the parabola, which for $y = Ax^2 + Bx + C$ is simply $2A$. So the sign of the leading coefficient $A$ determines everything. If positive, the parabola opens up, if negative, the parabola opens down.
A: Let $y=ax^2$ be the equation of a parabola having its vertex at the origin. Plugging in $x=1$ you get $y=a$, so our parabola also passes through $(1,a)$. Suppose $a>0$: the larger $a$ is, the steepest the graph of the parabola. A similar reasoning can be done for $a<0$.
A generic parabola $y=ax^2+bx+c$ is nothing but the above parabola translated, so that the link between value of $a$ and parabola shape still holds.
A: Rewrite $y=ax^2+bx+c$ as:
$$y=(a)(x^2)\left(1+\frac{b/a}x+\frac{c/a}{x^2}\right)$$
Let $x$ be very large (in either the positive or negative direction). Notice that $x^2$ is positive, and that last factor is very close to $1$ and therefore positive.
That means that, if $a$ is positive, then (for large enough $x$) $y$ is a positive $\times$ a positive $\times$ a positive, and hence positive. If $y$ is positive for large $x$, the graph looks like it curves up.
On the other hand, if $a$ is negative, then (for large enough $x$) $y$ is a negative $\times$ a positive $\times$ a positive, and hence negative. If $y$ is negative for large $x$, the graph looks like it curves down.
A: The quadratic function can be expressed as:
$$f(x)=ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a},$$
where $V\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right)$ is the vertex (turning point). If you plug $x=-\frac{b}{2a}\pm t, t\ne 0$, the value of the function will be:
$$at^2+c-\frac{b^2}{4a}\ \ \begin{cases}>c-\frac{b^2}{4a},a>0\\<c-\frac{b^2}{4a},a<0 \end{cases}$$
