The graph of $y=ax^2+bx+c$ is given below, where $a$, $b$, and $c$ are integers. Find $a-b+c$.

The  vertex  is at  (-1, -2)

We have the form

y = a ( x - - 1)^2  - 2

y =  a ( x + 1)^2  - 2

And  the point  0,-1)  is on the graph...so we  can find a

-1  = a ( 0 + 1)^2  - 2

-1 + 2  =  a

1 = a

So  we havr

y = 1 ( x + 1)^2  -  2

y = x^2  + 2x  + 1  - 2

y = x^2  + 2x  - 1

a = 1   b = 2    c = -1

a + b + c =  1 + 2  - 1    =  2