Question

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

Answer 1

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

Explanation:

As you can see, there is a vertical asymptote at about `y=1`(because you can never get 1 out of this equation), and a horizontal asymptote at about `x=-1` (because plugging in `-1` for `x` creates a fraction with zero on the bottom, specifically `3/0`). There is an `x`-intercept at `y=1/2`, and a `y`-intercept at `x=1`. There are no holes.