Question

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

Answer 1

• `x = 1` the vertical asymptote of `f(x)`
• `y = 0` the horizontal asymptote of `f(x)`
• `(0, 2)` the `y`-intercept
• `f(x)` has no `x`-intercept

Explanation:

`f(0)` gives the value of the `y`-intercept:

`f(0) = 2/(-1)= -2`

`f(x)` has no intersection with the `x` axis; solving `f(x) = 0` for `x` yields no real result:

`f(x) = 2/(x-1) = 0`
`2 != 0 * (x-1)`

The graph of `f(x) = 2/(x-1)` closely relates to that of `f(x) = 1/x`.
`y=1/x` has the following features:

• `x = 0` a vertical asymptote
• `y = 0` a horizontal asymptote
• `y = 1//x` contains two separate branches and is decreasing on both intervals.

Translating the graph of `y = 1/x` to the right by `1` unit (and stretching the result vertically by a factor of two) gives the graph of `y = f(x) = 2/(x-1)`. The two asymptotes are translated in the same fashion:

• `x = 0 + 1` the vertical asymptote of `f(x)`
• `y = 0 xx 2 = 0` the horizontal asymptote of `f(x)`

The branched feature from `y=1/x` is translated and shall also present in `y = f(x)`. Mark all the three graphical features [horizontal/vertical asymptote, intercept(s)] on the Cartesian plane and chart the two branches:

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