# How do you find the slant asymptote of y=sqrt(x^2+4x) ?

Feb 24, 2016

Notice that ${x}^{2} + 4 x = {\left(x + 2\right)}^{2} - 4$ and take $\left\mid x + 2 \right\mid$ outside the square root to find two slant asymptotes:

$y = x + 2$

and

$y = - x - 2$

#### Explanation:

Let $f \left(x\right) = y = \sqrt{{x}^{2} + 4 x} = \sqrt{x \left(x + 4\right)}$

As a Real valued function, this has domain $\left(- \infty , - 4\right] \cup \left[0 , \infty\right)$, since ${x}^{2} + 4 x \ge 0$ if and only if $x \in \left(- \infty , - 4\right] \cup \left[0 , \infty\right) .$

$\sqrt{{x}^{2} + 4 x}$

$= \sqrt{{x}^{2} + 4 x + 4 - 4}$

$= \sqrt{{\left(x + 2\right)}^{2} - 4}$

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

$= \left\mid x + 2 \right\mid \sqrt{1 - \frac{4}{x + 2} ^ 2}$

As $x \to \pm \infty$ we find that $\frac{4}{x + 2} ^ 2 \to 0$, so $f \left(x\right)$ is asymptotic to $\left\mid x + 2 \right\mid$

This results in two slant asymptotes:

$y = x + 2$ as $x \to + \infty$

and

$y = - x - 2$ as $x \to - \infty$

graph{(y-sqrt(x^2+4x))(y - x - 2)(y + x + 2) = 0 [-11.01, 8.99, -1.08, 8.92]}