Constructing a quadratic equation with three known points, but lack y-value in one. They are given as (27; 10) (32; 10) (34; f(34))

Question

anonymous · Answer

First two have I reduced to: $$0=295a+5b$$

tkhunny · Answer

1) There seems to be symmetry about x = 29.5.  Why do I say that?
2) Pick a value for f(34).  If f(34) > 10, you get an up-opening parabola.  If you prefer down-opening, select f(34) < 10.  Why do I know that?

anonymous · Answer

The key is that you are given two points with the same y value.  You can get the vertex because it will be halfway between x=27 and x=32.  So, x=29.5 will be the vertex.  But you need three known points to uniquely determine the parabola.  So I'm assuming you have some choice in the matter...

anonymous · Answer

|dw:1354984503984:dw|we need three points because both of these parabolas have a vertex at x=29.5 and pass through the two points which you are given.  Actually, infinitely man different parabolas pass through these two points.  You can't determine a unique quadratic.  So, yeah, pick a value for f(34) to choose one you like.