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

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Answer 1 · 2017-09-17T11:36:25

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

$y= (2x-4)/(x^2+2x+1) or (2(x-2))/(x+1)^2$

Vertical asymptote : We will have vertical asymptotes at those

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

$x+1=0 :. x=-1$ . 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 $(y=0)$

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]

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