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.