How do you find all the asymptotes for $f(x) = x+1+sqrt(x^2+4x)$ ?

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Answer 1 · 2015-05-05T16:18:23

There is one horizontal asymptote $y= -1$

It is to be investigated here, how f(x) behaves if x $->oo or -oo$

If x $->oo$ , f(x) $->oo$ is quite obvious. For determining the limit if $x->-oo$ , find the limit of f(-x) for x $->oo$ .

f(-x) = -x+1 + $sqrt(x^2-4x)$ (Now multiply by the conjugate as follows)

= $(-x+1 +sqrt(x^2 -4x))* (-x+1-sqrt(x^2 -4x))/(-x+1-sqrt(x^2 -4x))$

= $(x^2 -2x +1 -x^2+4x)/(-x+1-sqrt(x^2-4x))$ = $(2x+1)/(-x+1-xsqrt(1-4/x)$

= $(2+1/x)/(-1+1/x -sqrt(1-4/x))$ Now apply the limit x $->oo$

= $2/-2$ = -1. Hence there is an asymptote y= -1

How do you find all the asymptotes for f(x) = x+1+sqrt(x^2+4x)f(x) = x+1+sqrt(x^2+4x)?