Question

How do you graph `f(x)=2/(x-3)+1` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

Below

Explanation:

`f(x)=1+2/(x-3)`

For vertical asymptote, look at the denominator. It cannot equal to `0` as the graph will be undefined at that point. Hence, you let denominator equal to `0` to find at what point the graph cannot equal to `0`.

`x-3=0`
`x=-3`

For horizontal asymptote, imagine what happens to the graph when `x ->oo`. As `x ->oo`, `2/(x-3) ->0` so `f(x)=1+0=1` ie `y=1`

For intercepts,
When `y=0`, `x=1`
When `x=0`, `y=1/3`

Below is what the graph looks like

graph{1+2/(x-3) [-10, 10, -5, 5]}