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

Oct 13, 2017

•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 $+ \infty$ and $- \infty$.

We have a couple of limits to evaluate.

First limit

$L = {\lim}_{x \to \infty} \frac{x - 1}{{x}^{2} + 2 x + 1}$

$L = {\lim}_{x \to \infty} \frac{\frac{x}{x} ^ 2 - \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{2 x}{x} ^ 2 + \frac{1}{x} ^ 2}$

$L = {\lim}_{x \to \infty} \frac{\frac{1}{x} - \frac{1}{x} ^ 2}{1 + \frac{2}{x} + \frac{1}{x} ^ 2}$

$L = \frac{0 - 0}{1 + 0 + 0}$

$L = 0$

Second Limit

Will obviously be the same because ${\lim}_{x \to \infty} \frac{1}{x} = {\lim}_{x \to - \infty} \frac{1}{x} = 0$.

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!