# How do you find the vertical, horizontal and slant asymptotes of: y = ( 2x-4)/(x^2+2x+1)?

Sep 17, 2017

Vertical asymptote is at $x = - 1$ , horizontal asymptote is at x-axis $\left(y = 0\right)$ and there is no slant asymptote.

#### Explanation:

$y = \frac{2 x - 4}{{x}^{2} + 2 x + 1} \mathmr{and} \frac{2 \left(x - 2\right)}{x + 1} ^ 2$

Vertical asymptote : We will have vertical asymptotes at those

values of $x$ for which the denominator is equal to zero.

$x + 1 = 0 \therefore x = - 1$ . So vertical asymptote is at $x = - 1$

Horizontal asymptote: If m > n (that is, the degree of the

denominator is larger than the degree of the numerator), then the

graph of y = f(x) will have a horizontal asymptote at $y = 0$

(i.e., the x-axis). So horizontal asymptote is at x-axis $\left(y = 0\right)$

Slant asymptote: if the numerator's degree is greater (by a margin

of 1), then we have a slant asymptote which is found by doing long

division. So there is no slant asymptote.

graph{(2(x-2))/(x+1)^2 [-40, 40, -20, 20]} [Ans]