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

Question

1447 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-04T18:03:23

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

We can factorize the denominator

#x^2+2x+1=(x+1)^2#

Therefore,

#(2x-4)/(x^2+2x+1)=(2x-4)/(x+1)^2#

As you cannot divide by #0#, #x!=-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->-oo)f(x)=lim_(x->-oo)(2x)/x^2=lim_(x->-oo)2/x=0^-#

#lim_(x->+oo)f(x)=lim_(x->+oo)(2x)/x^2=lim_(x->+oo)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]}