y=(x+4)/(x-3) Graph the function. State the domain and range.

To be able to graph the rational function y =(x+4)/(x-3) , we solve for possible asymptotes.
Vertical asymptote exists at x=a that will satisfy D(x)=0 on a rational function f(x)= (N(x))/(D(x)) . To solve for the vertical asymptote, we equate the expression at denominator side to 0 and solve for x.
In y =(x+4)/(x-3) , the D(x) =x-3.
Then, D(x) =0  will be:
x-3=0
x-3+3=0+3
x=3
The vertical asymptote exists at x=3 .
To determine the horizontal asymptote for a given function: f(x) = (ax^n+...)/(bx^m+...), we follow the conditions:
when n lt m     horizontal asymptote: y=0
        n=m     horizontal asymptote:  y =a/b
        ngtm       horizontal asymptote: NONE
In y =(x+4)/(x-3) , the leading terms are ax^n=x or 1x^1 and bx^m=x or 1x^1 . The values n =1 and m=1 satisfy the condition: n=m. Then, horizontal asymptote  exists at y=1/1 or y =1 .
To solve for possible y-intercept, we plug-in x=0 and solve for y .
y =(0+4)/(0-3)
y =4/(-3)
y = -4/3 or -1.333  (approximated value)
Then, y-intercept is located at a point (0, -1.333) .
To solve for possible x-intercept, we plug-in y=0 and solve for x .
0 =(x+4)/(x-3)
0*(x-3) =(x+4)/(x-3)*(x-3)
0 =x+4
0-4=x+4-4
-4=x or x=-4
Then, x-intercept is located at a point (-4,0) .
Solve for additional points as needed to sketch the graph.
When x=2 , the y = (2+4)/(2-3)=6/(-1)=-6 . point: (2,-6)
When x=4 , the y =(4+4)/(4-3) =8/1=8 . point: (4,8)
When x=10 , the y =(10+4)/(10-3)=14/7=2 . point: (10,2)
When x=-16 , the y =(-16+4)/(-16-3)=-12/(-19)~~0.632 . point: (-16,0.632)
Applying the listed properties of the function, we plot the graph as:

You may check the attached file to verify the plot of asymptotes and points.
As shown on the graph, the domain: (-oo, 3)uu(3,oo)
and range: (-oo,1)uu(1,oo).  
The domain of the function is based on the possible values of x. The x=3 excluded due to the vertical asymptote.
The range of the function is based on the possible values of y . The y=1 is excluded due to the horizontal asymptote.