The given function `y = 3/x-2` is the same as:
`y = 3/x-(2x)/x`
`y = (3-2x)/x or y =(-2x+3)/x.`
To be able to graph the rational function `y =(-2x+3)/x` , 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 =(-2x+3)/x,` the `D(x) =x.`
Then, `D(x) =0 ` will be `x=0` .
The vertical asymptote exists at `x=0` .
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 =(-2x+3)/x` the leading terms are `ax^n=-2x or -2x^1` and `bx^m=x or x^1` . The values `n =1` and `m=1` satisfy the condition:` n=m` . Then, horizontal asymptote exists at `y=(-2)/1 or y =-2` .
To solve for possible y-intercept, we plug-in `x=0` and solve for` y` .
`y = 3/0 `
y = undefined
Thus, there is no y-intercept.
To solve for possible x-intercept, we plug-in `y=0` and solve for `x` .
`0*x = (-2x+3)/x*x`
`x= 3/2 or 1.5`
Then, x-intercept is located at a point `(1.5,0).`
Solve for additional points as needed to sketch the graph.
When `x=1` , then `y =(-2*1+3)/1 =1/1=1` . point: `(1,1)`
When `x=3` , then `y =(-2*3+3)/3 =-3/3=-1` . point: `(3,-1)`
When `x=-1` , then `y =(-2*(-1)+3)/(-1) =(5)/(-1)=-5` . point: `(-1,-5)`
When `x=-3` , then `y =(-2*(-3)+3)/(-3) =9/(-3)=2` point: `(-3,-3)`
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, 0)uu(0,oo) ` and range:` (-oo,-2)uu(-2,oo).`
The domain of the function is based on the possible values of `x` . The `x=0` excluded due to the vertical asymptote.
The range of the function is based on the possible values of `y` . The `y=-2` is excluded due to the horizontal asymptote.