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

1 Answer
Oct 13, 2017

Answer:

•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!