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:
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 =(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` .
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.