Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 54

Prove that the curve $y = \sqrt{x^2 + 4x}$ has two slant asymptotes: $y = x + 2$ and $y = -x - 2$. Use this fact to sketch the curve.


$
\begin{equation}
\begin{aligned}

\lim_{x \to \infty} (\sqrt{x^2 + 4x} + (x + 2)) =& \lim_{x \to \infty} (\sqrt{x^2 + 4x} + (x + 2)) \cdot \left( \frac{\sqrt{x^2 + 4x} - (x + 2) }{\sqrt{x^2 + 4x} - (x + 2)} \right)
\\
\\
=& \lim_{x \to \infty} \frac{x^2 + 4x - (x^2 + 4x + 4)}{\sqrt{x^2 + 4x} - (x + 2)} = \lim_{x \to \infty} \frac{-4}{\sqrt{x^2 + 4x} - x - 2}
\\
\\
=& \frac{\displaystyle \frac{-4}{x^2} }{\displaystyle \sqrt{1 + \frac{4}{x}} - \frac{1}{x} - \frac{2}{x^2}} = 0

\end{aligned}
\end{equation}
$


Similarly, $\lim_{x \to \infty} (\sqrt{x^2 + 4x} + (-x - 2)) = 0$

Therefore the equations of slant asymptotes are $y = x + 2$ and $y = -x - 2$

Now, using the guidelines of curve sketching,

Domain,

The function contains square root that is defined for a positive values only. So..


$
\begin{equation}
\begin{aligned}

& x^2 + 4x \geq 0
\\
\\
& x (x + 4) \geq 0

\end{aligned}
\end{equation}
$


We have,

$x \geq 0$ and $x \leq -4$

Therefore, the domain is

$(- \infty, -4] \bigcup [0, \infty)$

Intercepts,

Solving for $y$-intercept, when $x = 0$,

$y = \sqrt{0^2 + 4 (0)} = 0$

Solving for $x$-intercept, when $y = 0$,


$
\begin{equation}
\begin{aligned}

& 0 = \sqrt{x^2 + 4x}
\\
\\
& x^2 + 4x = 0
\\
\\
& x (x + 4) = 0


\end{aligned}
\end{equation}
$


We have,

$x = 0$ and $x = -4$

Symmetry,

The function is not symmetic to either $y$-axis or origin by using symmetry test.

Intervals of Increase or Decrease,

If $f(x) = \sqrt{x^2 + 4x}$ then,


$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{1}{2} (x^2 + 4x)^{- \frac{1}{2}} (2x + 4)
\\
\\
f'(x) =& \frac{x + 2}{\sqrt{x^2 + 4x}}

\end{aligned}
\end{equation}
$


when $f'(x) = 0$

$\displaystyle 0 = \frac{x + }{\sqrt{x^2 + 4x}}$

The critical number is $x = -2$. However, it is not defined in the domain of $f$

Hence, the intervals of increase and decrease are

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f \\
x < -4 & + & \text{increasing on } (- \infty, -4) \\
x > 0 & + & \text{increasing on } (0, \infty)\\
\hline
\end{array}
$

Local Maximum and Minimum Values,

Since $f'(x)$ doesn't change sign, we can say that the function has no local maximum and minimum.

Concavity and Inflection Points

If $\displaystyle f'(x) = \frac{x + 2}{\sqrt{x^2 + 4x}}$, then by using Quotient Rule and Chain Rule,

$\displaystyle \frac{(\sqrt{x^2 + 4x})(1) - (x + 2)\left( \frac{1}{2} (x^2 + 4x)^{- \frac{1}{2}} (2x + 4) \right) }{(\sqrt{x^2 + 4x})^2}$

Which can be simplified as

$\displaystyle f''(x) = \frac{-4}{(x^2 + 4x)^{\frac{3}{2}}}$

So when $f''(x) = 0$

$\displaystyle 0 = \frac{-4}{(x^2 + 4x)^{\frac{3}{2}}}$

$f''(x) = 0$ does not exist, we can say that the function has no inflection points.

Hence, the concavity is...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity} \\
x < -4 & - & \text{Downward} \\
x > 0 & - & \text{Downward}\\
\hline
\end{array}
$