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

Question

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

1 Answer

Impact of this question

1724 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-27T03:56:59

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

Explanation:

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

$y \to \pm \infty$ $yto+-oo$ as $x \to \pm \frac{2}{3}$ $xto+-2/3$ . So, $x = \pm \frac{2}{3}$ $x=+-2/3$ are asymptotes..

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

$x \to \pm \infty$ $xto+-oo$ as $y \to \pm \frac{2}{3}$ $yto+-2/3$ . So, $y = \pm \frac{2}{3}$ $y=+-2/3$ are asymptotes..

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

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

As a whole, the graph is outside of $| x | \leq \frac{2}{3} and | y | \leq \frac{2}{3}$ $|x| <=2/3 and |y|<=2/3$ .

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

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the asymptotes for f(x)=(2x)/(sqrt (9x^2-4))f(x)=2x√9x2−4f(x)=(2x)/(sqrt (9x^2-4))?

1 Answer

Explanation:

Related questions

Impact of this question

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