# `h(x)=11/(x-9)+9` Graph the function. State the domain and range. The given function `h(x)= 11/(x-9)+9` is the same as:

`h(x)= 11/(x-9)+9 *(x-9)/(x-9)`

`h(x)= 11/(x-9)+(9x-81)/(x-9)`

`h(x)=(11+(9x-81))/(x-9) `

`h(x)=(11+9x-81)/(x-9) `

`h(x) = (9x-70)/(x-9)`

To be able to graph the rational function `h(x) =(9x-70)/(x-9)` or `y =(9x-70)/(x-9)` , we solve for possible asymptotes. Note: `h(x)=y` .

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 `h(x) =(9x-70)/(x-9)` , the `D(x)=x-9.`

Then,` D(x) =0 `  will be:

`x-9=0`

`x-9+9=0+9`

`x=9`

The vertical asymptote exists at `x=9` .

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 `h(x) = (9x-70)/(x-9)` , the leading terms are `ax^n=9x or 9x^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=9/1 or y =9`.

To solve for possible y-intercept, we plug-in `x=0` and solve for `y`.

`y =(9*0-70)/(0-9)`

`y =(-70)/(-9) `

`y = 70/9 or 7.778 ` (approximated value)

Then, y-intercept is located at a point `(0, 7.778).`

To solve for possible x-intercept, we plug-in `y=0` and solve for `x.`

`0 =(9x-70)/(x-9)`

`0*(x-9)= (9x-70)/(x-9)*(x-0)`

`0 =9x-70`

`0+70=-9x-70+70`

`70=9x`

`70/9=(9x)/9`

`x=70/9 or 7.778`

Then, x-intercept is located at a point `(7.778,0).`

Solve for additional points as needed to sketch the graph.

When `x=8,` the `y = (9*8-70)/(8-9)=2/(-1)=-2` . point: `(8,-2)`

When `x=10` , the `y = (9*10-70)/(10-9)=20/1=20` . point: `(10,20)`

When `x=20` , the `y =(9*20-70)/(20-9)=110/11=10` . point: `(20,10)`

When `x=-2` , the` y =(9*(-2)-70)/(-2-9)= (-88)/(-11)=8` . point: `(-2,8)`

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, 9)uu(9,oo)`

and range: `(-oo,9)uu(9,oo).`

The domain of the function is based on the possible values of `x.` The `x=9` is excluded due to the vertical asymptote.

The range of the function is based on the possible values of `y` . The `y=9` is excluded due to the horizontal asymptote.