The other tutor gave a linear, not quadratic solution, which is
not what you want.

Since we cannot see the graph that you can see, we cannot complete
your problem, for just having two point is not enough to determine
a quadratic function.  But I'll do something. 

The general equation of a quadratic function is

       f(x) = a*x² + b*x + c

Since there are three unknown constants a, b, and c, the two points you gave,
(-3,-1), (-2,-4), are not enough, so I will assume that you can look and see
a third point on the graph that you didn't tell us about. So I will arbitrarily
assume the y-intercept is (0,-4).  So I'll assume the graph you see that we
cannot see is this:

Here's the graph I'm assuming:



Substitute the point (-3,-1)

       f(x) = a*x² + b*x + c
         -1 = a*(-3)² + b*(-3) + c
         -1 = 9a - 3b + c
9a - 3b + c = -1 

Substitute the point (-2,-4)

       f(x) = a*x² + b*x + c
         -4 = a*(-2)² + b*(-2) + c
         -4 = 4a - 2b + c
4a - 2b + c = -4

Substitute the point, y-intercept, that I chose arbitrarily (0,-4)

       f(x) = a*x² + b*x + c
         -4 = a*(0)² + b*(0) + c
         -4 = 0a + 0b + c
          c = -4

So we have this system of equations:

          9a - 3b + c = -1
          4a - 2b + c = -4
                    c = -4

Solve that system and get a=1, b=2, c=-4

So the function is

f(x) = ax² + bx + c

f(x) = x² + 2x - 4

Edwin