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

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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. 

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