The tangent to the curve of y=ax^2 + bx +c at the point where x=2 is parallel to the line y=4x. There is a stationary point at (1, -3). Find the value of a, b and c.

I was really unsure of how to work out this question, so first I differentiated y.

dy/dx=2ax+b

The point (1, -3) is on the curve and therefore:

-3=a(1)^2+b(1)+c

-3=a^2+b+c

But I don't know how to find another equation, and I honestly don't even know if what I've done so far is on the right track.