Question

What are the horizontal and vertical asymptotes on the graph of `f(x) = (x -1)/(x^2 + 2x + 1)`?

Answer 1

•Vertical asymptote at `x = -1`
•Horizontal asymptote at `y = 0`.

Explanation:

We can immediately see that `x = -1` is a vertical asymptote because it makes the denominator `0` and hence the function undefined.

Horizontal asymptotes can be found by taking the limit at `+oo` and `-oo`.

We have a couple of limits to evaluate.

First limit

`L = lim_(x->oo) (x- 1)/(x^2 + 2x + 1)`

`L = lim_(x->oo) (x/x^2 - 1/x^2)/(x^2/x^2 + (2x)/x^2 + 1/x^2)`

`L = lim_(x->oo) (1/x - 1/x^2)/(1 + 2/x + 1/x^2)`

`L = (0 - 0)/(1 + 0 + 0)`

`L = 0`

Second Limit

Will obviously be the same because `lim_(x->oo) 1/x = lim_(x->-oo) 1/x = 0`.

Therefore, our asymptotes will be:

•Vertical asymptote at `x = -1`
•Horizontal asymptote at `y = 0`.

We now verify graphically.

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

Hopefully this helps!