We have y = ax^2 + bx + c.

Plugging in (1, -2), we get:

-2 = a + b + c

And plugging in (2, -14) we get:

-14 = 4a + 2b + c

Let's try as best we can to "solve" one of these. The first one, when multipled by -2, becomes:

4 = -2a - 2b - 2c

Adding this to the second one, we get:

-10 = 2a - c

And this seems to be the only restriction. So let's choose an a and c that works in the above equation, and then solve for b.

-10 = 2(-7) - (-4) = -10

So a = -7 and c = -4. Putting these back into the very first equation:

-2 = -7 + b - 4

b = 9

a = -7 b = 9 c = -4. But do these work?

-2 = -7(1)^2 + 9(1) - 4 which is true, so it goes through (1,-2)

-14 = -7(2)^2 + 9(2) - 4 which is true, so it goes through (2, -14)

But keep in mind, there are infinitely many solutions. This is just one of them.