`y=10/(x+7)-5` Graph the function. State the domain and range.

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The given function `y = 10/(x+7)-5 ` is the same as:

`y =10/(x+7)-5(x+7)/(x+7)`

`y=10/(x+7)-(5x+35)/(x+7)`

`y=(10-(5x+35))/(x+7)`

`y=(10-5x-35)/(x+7)`

`y = (-5x-25)/(x+7)`

To be able to graph the rational function `y = (-5x-25)/(x+7)` , we solve for possible asymptotes.

Vertical asymptote exists at `x=a` that will satisfy `D(x)=0` on a rational function `f(x)=...

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The given function `y = 10/(x+7)-5 ` is the same as:

`y =10/(x+7)-5(x+7)/(x+7)`

`y=10/(x+7)-(5x+35)/(x+7)`

`y=(10-(5x+35))/(x+7)`

`y=(10-5x-35)/(x+7)`

`y = (-5x-25)/(x+7)`

To be able to graph the rational function `y = (-5x-25)/(x+7)` , 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-25)/(x+7)` , the `D(x)=x+7` .

Then, `D(x) =0 `  will be:

`x+7=0`

`x+7-7=0-7`

`x=-7`

The vertical asymptote exists at `x=-7` .

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-25)/(x+7)` , 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*0-25)/(0+7) `

`y =(-25)/7`

`y = -25/7 or -3.571 `  (approximated value)

Then, y-intercept is located at a point `(0, -3.571)` .

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

`0 =(-5x-25)/(x+7)`

`0*(x+7) =(-5x-25)/(x+7)*(x+7)`

`0 =-5x-25 `

`0+5x=-5x-25+5x`

`5x=-25`

`(5x)/5=(-25)/5`

`x=-5`

Then, x-intercept is located at a point `(-5,0).`

Solve for additional points as needed to sketch the graph.

When `x=3,` the `y = (-5*3-25)/(3+7)=-40/10=-4.` point:` (3,-4)`

When ` x=-6` , the `y = (-5(-6)-25)/(-6+7)=5/1=5` . point: `(-6,5)`

When `x=-9` , the `y =(-5(-9)-25)/(-9+7)=20/(-2)=-10` . point: `(-9,-10)`

When `x=-12` , the `y = (-5(-12)-25)/(-12+7)=35/(-5)=-7` . point: `(-12,-7)`

 

As shown on the graph attached, the domain: `(-oo, -7)uu(-7,oo)` and range: `(-oo,-5)uu(-5,oo)` . 

The domain of the function is based on the possible values of `x.` The `x=-7` 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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