Using three data points (x, y) that we know evil passed through and the general quadratic equation, create three corresponding equations:

The general quadratic equation is y = ax^2 + bx + c.

Let's see whether it's possible to invent 3 separate points thru which the graph of a quadratic passes and then to create 3 equations in a, b and c:

Suppose the points are (-3,2), (-1,1) and (2,4).

Focus on (-3,2):  Here x= -3 and y= 2.  Then 2 = a(-3)^2 + b(-3) + c, or
                                                                        2 = 9a - 3b + c

Do the same thing for the other 2 points, (-1,1) and (2,4).

When you're done, you'll have 3 linear equations in the variables a, b and c.

You were not asked to solve for a, b and c.

If you want, however, we could go thru that process.

I’d go with b hope it