How do you find the asymptotes for f(x)=(2x)/(sqrt (9x^2-4))?

Mar 27, 2016

Asymptotes: f(x) = y = $\pm \frac{2}{3}$ and $x = \pm \frac{2}{3}$..

Explanation:

Let y = f(x). It is defined for $9 {x}^{2} - 4 > 0$. So, $| x | > \frac{2}{3}$.

$y \to \pm \infty$ as $x \to \pm \frac{2}{3}$. So, $x = \pm \frac{2}{3}$ are asymptotes..

Inversely, solve for x.
$x = \frac{2 y}{\sqrt{9 {y}^{2} - 4}}$.
Interestingly the form is the same. So, $| y | > \frac{2}{3}$.

$x \to \pm \infty$ as $y \to \pm \frac{2}{3}$. So, $y = \pm \frac{2}{3}$ are asymptotes..

The graph comprises four branches symmetrical about the origin in four regions:

Both x and $y > \frac{2}{3}$ in the first quadrant, $x < - \frac{2}{3}$ and $y > \frac{2}{3}$ in the second quadrant, both x and $y < - \frac{2}{3}$ in the third quadrant and $x > \frac{2}{3}$ and $y < - \frac{2}{3}$ in the fourth quadrant.

As a whole, the graph is outside of $| x | \le \frac{2}{3} \mathmr{and} | y | \le \frac{2}{3}$.

In this formula, ${f}^{- 1} \equiv f$, like f(x) = 1/x. Indeed, interesting.