h(x)=(5x+3)/(-x+16) Graph the function. State the domain and range.

To be able to graph the rational function y =(5x+3)/(-x+16) , 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 =(5x+3)/(-x+16) , the D(x) =-x+16 .
Then, D(x) =0  will be:
-x+16=0
x=16
The vertical asymptote exists at x=16 .
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 =(5x+3)/(-x+16) , the leading terms are ax^n=5x or 5x^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=5/(-1)  or y =-5 .
To solve for possible y-intercept, we plug-in x=0 and solve for y .
y =(5*0+3)/(-0+16)
y =(0+ 3)/(0+16)
y = 3/16 or 0.188  (approximated value)
Then,  y-intercept is located at a point (0, 0.188) .
To solve for possible x-intercept, we plug-in y=0 and solve for x .
0 =(5x+3)/(-x+16)
0*(-x+16) =(5x+3)/(-x+16)*(-x+16)
0 =5x+3
-3=5x
x=(-3)/5= -0.6
Then, x-intercept is located at a point (-0.6,0) .
Solve for additional points as needed to sketch the graph.
When x=11 , the y = (5*11+3)/(-11+16)=58/5=11.6 . point: (11,11.6)
When x=20 , the y =(5*20+3)/(-20+16)=103/(-4)=-25.75 . point: (20,-25.75)
When x=30 , the y =(5*30+3)/(-30+16) =153/(-14)~~-11 . point: (30,-11)
When x=-26, the y =(5(-26)+3)/(-(-26)+16) =-127/42~~-3.024. point:(-26,-3.024)
 
As shown on the graph attached below, the domain: (-oo, 16)uu(16,oo)
and range: (-oo,-5)uu(-5,oo).
The domain of the function is based on the possible values of x. The x=16 excluded due to the vertical asymptote.
The range of the function is based on the possible values of y. The y=-5 is excluded due to the horizontal asymptote.