How do you find the vertical, horizontal or slant asymptotes for $f(x)= (x+3)/( x^2+4x+3)$ ?

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Answer 1 · 2016-03-20T07:21:02

Vertical Asymptote: $x=-1$
Horizontal Asymptote: $y=0$

From the given $y=(x+3)/(x^2+4x+3)=(x+3)/((x+3)(x+1))=1/(x+1)$

graph{y=(x+3)/(x^2+4x+3)[-20,20,-10,10]}

God bless....I hope the explanation is useful.

How do you find the vertical, horizontal or slant asymptotes for f(x)= (x+3)/( x^2+4x+3)f(x)= (x+3)/( x^2+4x+3)?