# How do you find the asymptotes for r(x) = (2x-4)/(x^2+2x+1)?

Feb 4, 2017

The vertical asymptote is $x = - 1$
The horizontal asymptote is $y = 0$
No slant asymptote

#### Explanation:

We can factorize the denominator

${x}^{2} + 2 x + 1 = {\left(x + 1\right)}^{2}$

Therefore,

$\frac{2 x - 4}{{x}^{2} + 2 x + 1} = \frac{2 x - 4}{x + 1} ^ 2$

As you cannot divide by $0$, $x \ne - 1$

The vertical asymptote is $x = - 1$

As the degree of the numerator $<$ the degree of the denominator,

there is no slant asymptote.

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{2 x}{x} ^ 2 = {\lim}_{x \to - \infty} \frac{2}{x} = {0}^{-}$

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{2 x}{x} ^ 2 = {\lim}_{x \to + \infty} \frac{2}{x} = {0}^{+}$

The horizontal asymptote is $y = 0$

graph{(y-(2x+4)/(x+1)^2)(y)(y+100x+100)=0 [-10, 10, -5, 5]}