Question

How do you graph `f(x)=x/(x-2)` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

Below

Explanation:

`f(x)=x/(x-2)`
`f(x)=((x-2)+2)/(x-2)`
`f(x)=1+2/(x-2)`

For vertical asymptotes, `x-2 !=0` since the denominator cannot equal to 0 as the function will be undefined at that point. To find at what point the function is undefined, we can go `x-2=0` so `x=2` is our vertical asymptote.

For horizontal asymptotes, we let `x ->oo`. As `x ->oo`, `2/(x-2)->0`. Hence, `f(oo)=1+0=1`. Therefore, there is an horizontal asymptote at `y=1`

For intercepts,
When `y=0`, `x=0`
When `x=0`, `y=0`

Plotting your intercepts and drawing in your asymptotes (remember that the asymptotes influence the endpoints of your graph ONLY and nothing else)

graph{x/(x-2) [-10, 10, -5, 5]}