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

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

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Answer 1 · 2015-05-05T16:18:23

There is one horizontal asymptote $y = - 1$ $y= -1$

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

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

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

= $(- x + 1 + \sqrt{x^{2} - 4 x}) \cdot \frac{- x + 1 - \sqrt{x^{2} - 4 x}}{- x + 1 - \sqrt{x^{2} - 4 x}}$ $(-x+1 +sqrt(x^2 -4x))* (-x+1-sqrt(x^2 -4x))/(-x+1-sqrt(x^2 -4x))$

= $\frac{x^{2} - 2 x + 1 - x^{2} + 4 x}{- x + 1 - \sqrt{x^{2} - 4 x}}$ $(x^2 -2x +1 -x^2+4x)/(-x+1-sqrt(x^2-4x))$ = $\frac{2 x + 1}{- x + 1 - x \sqrt{1 - \frac{4}{x}}}$ $(2x+1)/(-x+1-xsqrt(1-4/x)$

= $\frac{2 + \frac{1}{x}}{- 1 + \frac{1}{x} - \sqrt{1 - \frac{4}{x}}}$ $(2+1/x)/(-1+1/x -sqrt(1-4/x))$ Now apply the limit x $\to \infty$ $->oo$

= $\frac{2}{- 2}$ $2/-2$ = -1. Hence there is an asymptote y= -1

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

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