Question

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

Short answer

It is proved that the curve \(y = \sqrt {{x^2} + 4x} \) has two slant asymptotes: \(y = x + 2\) and \(y = - x - 2\).

The graph is shown as:

Step-by-step solution

Sketching a Graph of a function

To sketch a graph we need to follow some procedure. Those are as follows:

1. Finding the domainof the function

2. Intercepts of the function (both \(x\) and \(y\)intercepts)

3. Checking the symmetricity of the function.

4. Finding the asymptotes of the function.

Formula for slant asymptote:

if \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {f\left( x \right) - y\left( x \right)} \right) = 0\), then \(y\left( x \right) = mx + c\) is the slant asymptote.

5. Finding the interval of increasing or decreasing.

6. Finding local maximaand local minima.

7. Concavity and finding inflection points.

Sketch the Graph of given Function

The given function is \(f\left( x \right) = \sqrt {{x^2} + 4x} \).

We have \({x^2} + 4x = x\left( {x + 4} \right) \ge 0\).

For the intercepts we have

\(f\left( 0 \right) = 0\), this is \(y\)-intercept.

\(f\left( x \right) = 0 \Rightarrow x\left( {x + 4} \right) = 0\)

Hence \(x = 0, - 4\) are the \(x\)- intercept.

Now from \({x^2} + 4x = x\left( {x + 4} \right) \ge 0\), we get \(x \ge 0,x \ge - 4\) or \(x \le 0,x \le - 4\). Combining these we get the domain of the function as \(\left( { - \infty , - 4} \right) \cup \left( {0,\infty } \right)\).

Finding the asymptotes, we have:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to - \infty } \left( {f\left( x \right) + \left( {x + 2} \right)} \right) = \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 4x} + x + 2} \right)\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2} + 4x - {x^2} - 4x - 4}}{{\sqrt {{x^2} + 4x} - \left( {x + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 4}}{{\sqrt {{x^2} + 4x} - \left( {x + 2} \right)}}\\ = 0\end{array}\)

Hence \(y = - x - 2\) is a slant asymptote.

Also we have,

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \left( {f\left( x \right) - \left( {x + 2} \right)} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + 4x} - \left( {x + 2} \right)} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \frac{{ - 4}}{{\sqrt {{x^2} + 4x} + \left( {x + 2} \right)}}\\ = 0\end{array}\)

Hence \(y = x + 2\) is a slant asymptote.

Now,

\(\begin{array}{l}f'\left( x \right) = \frac{{2x + 4}}{{2\sqrt {{x^2} + 4x} }}\\f'\left( x \right) = \frac{{x + 2}}{{\sqrt {{x^2} + 4x} }}\end{array}\)

So the critical point is,

\(\begin{array}{c}f'\left( x \right) = 0\\\frac{{x + 2}}{{\sqrt {{x^2} + 4x} }} = 0\\x + 2 = 0\\x = - 2\end{array}\)

Hence the critical point is \(x = - 2\).

Now \(f'\left( x \right) > 0\), when \(x > 0\). So \(\left( {0,\infty } \right)\) is the interval of increasing.

And \(f'\left( x \right) < 0\), when \(x < - 4\). So, the interval of decreasing is \(\left( { - \infty , - 4} \right)\).

Again, we have:

\(\begin{array}{c}f''\left( x \right) = \frac{{\sqrt {{x^2} + 4x} - \frac{{{{\left( {x + 2} \right)}^2}}}{{\sqrt {{x^2} + 4x} }}}}{{{x^2} + 4x}}\\ = \frac{{{x^2} + 4x - {x^2} - 4x - 4}}{{{{\left( {{x^2} + 4x} \right)}^{\frac{3}{2}}}}}\\ = \frac{{ - 4}}{{{{\left( {{x^2} + 4x} \right)}^{\frac{3}{2}}}}}\end{array}\)

No inflection point.

Also \(f''\left( x \right) < 0\) on the domain. Hence the function is concave down on \(\left( { - \infty , - 4} \right) \cup \left( {0,\infty } \right)\).