1) The x coordinate of the vertex tends towards (0,c) as a gets sufficiently large.
The vertex of the parabola y = ax^2 + x + 1 is described by x = (-b/2a) in this case b = 1 so the x point of the vertex is -1/2a. This means that when a is positive the x point of the vertex will always be negative so the vertex will never cross the y-axis onto the positive side of the x-axis. (The reverse is true if a is negative).
Second note as a tends towards infinity the x point of the vertex goes to 0. With x = 0 we get y = a(0)^2 +0 +1 or y = 1. Therefore, the vertex of our parabola y = ax^2 + x + 1 tends to go to the point (0,1) as a gets sufficiently large. You can extend this argument to the general case y = ax^2+bx+c in which the vertex tends to the point (0, c). If you fix the points b,c and let a get sufficiently large then x = -b/2a goes to 0 then y = a(0)^2+b(0)+c or y = c. The vertex point then tends to the point (0,c). However, once you pick any a value you get a value for the vertex which is not (0, c). This is observation is grounds for an interesting subtle mathematical distinction. The difference between fixing a point a and letting a go to its limit at infinity.
Students may make these observations but how can they make sense of them without calculus?
A good way to get them thinking about the idea of a limit is have them try putting in values for a that are increasingly large and see what happens to the equation -1/2a. To get them thinking even more about these ideas it might be helpful to have them pick different values for b in the equation -b/2a. Once they have fixed the b value what happens to the value of -b/2a as a gets large. Do they observe the same thing as before? Does the x point of the vertex still go to zero? Is there any relationship between the value of c and the vertex?
A similar argument about limits could be made as we let a go to zero in the next set of graphs. This to me seems like it might be more confusing to students, however, because it involves the x value of the vertex tending towards infinity. This doesn't make any kind of graphical sense though because there is no concreteness about infinity. Therefore it would be harder to distinguish between a fixed a value and the idea of a limit. For this reason, I choose not to include this as an argument.
2) The distance from the linear function is equal to both of the quadratic functions.
Let f(x) = x^2+x+1, g(x) = -x^2+x+1, and h(x) = x+1. The distance between two point and or functions is defined as the absolute value of one function minus the other function. So abs(f(x)-h(x))= abs(x^2+x+1-(x+1)) = abs(x^2) and the distance between the other two functions is abs(g(x)-h(x)) = abs(-x^2+x+1-(x+1)) = abs(-x^2). The abs(x^2) = abs(-x^2) so f(x) is the same distance from h(x) as g(x) is.
Students again may make the observation that there is symmetry and you can define rotational symmetry for them and show them this proof. An easier way (perhaps) and definitely easier to explain with the graphs on the other page is to think about f(x) = x^2 + h(x) and g(x) = -x^2 + h(x). If you show the students that the function h(x) are simply a set of values which is reflected in the graph it then makes sense that if you add positive x^2 to the values of h(x) and add -x^2 to the values of h(x) that the distance from the line will be equal because you are adding equal and opposite values to the function h(x).
3) Two quadratics with equal and opposite values of b and where a and c are held constant are symmetric about the x-axis.
Let f(x) = ax^2+bx+c and g(x) = ax^2-bx+c. Two functions are symmetric there values are equal at opposite x values. So let's see when these two functions are equal. f(x) = g(x), ax^2+bx+c = ax^2-bx+c if we cancel we are left with bx = -bx or that the two functions are equal when x = -x. But this is the condition required for symmetry in the x-axis, so we can conclude that the two functions are symmetric when the b values are equal and opposite and a and c are held constant.
4) If we fix the a and c values and fluctuate the b value the vertex of our parabola sweeps out another parabola as it moves.
If we have a parabola of the form f(x) = ax^2 + bx + c the x coordinate of the vertex is at -b/2a. So let's check what the y value is, f(-b/2a) = a(-b/2a)^2 + b(-b/2a) + c. If we simplify this we get that f(-b/2a) = -b^2/4a + c. So the points of the vertex are (-b/2a, -b^2/4a + c). If we think of