`y=-3/(x-4)-1` Graph the function. State the domain and range.

Expert Answers
marizi eNotes educator| Certified Educator

The given function `y = -3/(x-4)-1` is the same as:

`y = -3/(x-4)-1*(x-4)/(x-4)`

`y = -3/(x-4)-(x-4)/(x-4)`

`y=(-3-(x-4))/(x-4)`

`y=(-3-x+4)/(x-4)`

`y = (-x+1)/(x-4) `

To be able to graph the rational function ` y = (-x+1)/(x-4)` , 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+1)/(x-4), ` the `D(x)=x-4.`

Then, `D(x) =0`  will be:

`x-4=0`

`x-4+4=0+4`

`x=4`

The vertical asymptote exists at ` x=4` .

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+1)/(x-4)` , 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 =(-0+1)/(0-4)`

`y =1/(-4)`

`y = -1/4 or -0.25 `

Then, y-intercept is located at a point `(0, -0.25).`

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

`0 =(-x+1)/(x-4) `

`0*(x-4)= (-x+1)/(x-4)*(x-4) `

`0 =-x+1 `

`x=1`

Then, x-intercept is located at a point (1,0)

Solve for additional points as needed to sketch the graph.

When `x=3` , the `y =(-3+1)/(3-4)=-2/(-1)=2` . point: `(3,2)`

When `x=5` , the `y = (-5+1)/(5-4)=(-4)/1=-4` . point: `(5,-4)`

When `x=7` , the `y =(-7+1)/(7-4)=(-6)/3=-2` . point: `(7,-2)`

When `x=-2` , the `y =(-(-2)+1)/(-2-4)=3/(-6)=-0.5` . point: `(-2,-0.5)`

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, 4)uu(4,oo)`

and range: `(-oo,-1)uu(-1,oo)` . 

The domain of the function is based on the possible values of `x.` The `x=4` 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. 

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