Great question!

Here's one method...

There appears to be an asymptote at  x = 2 , so we know that when  x = 2 ,  Bx + C  =  0

B(2) + C  =  0

If we solve  \(y=\frac{x+A}{Bx+c}\)  for  x , we get  \(x=\frac{A-Cy}{By-1}\)

There appears to be an asymptote at  y = -1 , so we know that when  y = -1 ,  By - 1  =  0

B(-1) - 1  =  0

B  =  -1              Use this value for  B  to find  C .

(-1)(2) + C  =  0

C  =  2

And the graph passes through the point  (0, -2) , so...

\(-2 = \frac{0 + A}{-1(0) + 2} \\ -2=\frac{A}{2}\)

-4  =  A

Here's a graph of  \(y=\frac{x-4}{-1x+2}\) :  https://www.desmos.com/calculator/ibji5giqge

A + B + C   =   -4 + -1 + 2   =   -3

Very nice, hectictar.....!!!

Amazing, hectictar!

Here's one more approach....

Note that the points    (0, - 2)  (3,1)  and (4,0)  are on the graph

So....we must have that

0 =  4 + A    ⇒  A  =  -4

And

-2  =  [0 + (-4)]  / [B (0) + C ]

-2  +  -4 / C

-2C  = - 4    ⇒  C  =  2

And

1  =  [ 3 + (-4) ] / [ B(3) + 2 ]

1 =  - 1 / [ 3B + 2]

3B + 2  =  - 1

3B =  -3  ⇒   B =  -1

So....A + B + C     =     -4 - 1 + 2    =    -3