To be able to graph the rational function `y =(x+4)/(x-3)` , 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+4)/(x-3)` , the `D(x) =x-3.`
Then, `D(x) =0 ` will be:
The vertical asymptote exists at `x=3` .
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+4)/(x-3)` , 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` .
`y = -4/3 or -1.333` (approximated value)
Then, y-intercept is located at a point `(0, -1.333)` .
To solve for possible x-intercept, we plug-in `y=0` and solve for `x` .
`-4=x or x=-4`
Then, x-intercept is located at a point `(-4,0)` .
Solve for additional points as needed to sketch the graph.
When `x=2` , the `y = (2+4)/(2-3)=6/(-1)=-6` . point: `(2,-6)`
When `x=4` , the `y =(4+4)/(4-3) =8/1=8` . point:` (4,8)`
When `x=10` , the `y =(10+4)/(10-3)=14/7=2` . point: `(10,2)`
When `x=-16` , the `y =(-16+4)/(-16-3)=-12/(-19)~~0.632` . point: `(-16,0.632)`
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, 3)uu(3,oo)`
and range: `(-oo,1)uu(1,oo).`
The domain of the function is based on the possible values of `x.` The `x=3` 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.