How do you find the vertical, horizontal or slant asymptotes for $f (x) = \frac{6 x}{\sqrt{x^{2} - 3}}$ $f(x)=(6x) / sqrt( x^2 - 3)$ ?

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How do you find the vertical, horizontal or slant asymptotes for $f (x) = \frac{6 x}{\sqrt{x^{2} - 3}}$ $f(x)=(6x) / sqrt( x^2 - 3)$ ?

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Answer 1 · 2016-03-27T09:57:33

Vertical asymptotes $x = \pm \sqrt{3}$ $x=+-sqrt3$ .
Horizontal asymptotes: $y = \pm 6$ $y=+-6$ .

Explanation:

$y = \frac{6 x}{\sqrt{x^{2} - 3}}$ $y=(6x)/sqrt(x^2-3)$ .
For real y, $| x | > \sqrt{3}$ $|x|>sqrt3$ .
As $x \to \pm \sqrt{3}, y \to \pm \infty$ $xto+-sqrt3, yto+-oo$ .
So, $x = = - \sqrt{3}$ $x==-sqrt3$ are asymptotes.

Inversely,
$x = \frac{\sqrt{3} y}{\sqrt{y^{2} - 36}}$ $x=(sqrt3 y)/sqrt(y^2-36)$ .
For real x, $| y | > 6$ $|y|>6$ .
As $y \to \pm 6, x \to \pm \infty$ $yto+-6, xto+-oo$ .
So, $y = \pm 6$ $y=+-6$ are asymptotes.

The graph splits into four branches, in the four quadrants.
The graph is exterior to the region about the origin, bounded by the vertical and horizontal asymptotes, $x = \pm \sqrt{3}$ $x=+-sqrt3$ and $y = \pm 6$ $y=+-6$ .

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How do you find the vertical, horizontal or slant asymptotes for $f (x) = \frac{6 x}{\sqrt{x^{2} - 3}}$ $f(x)=(6x) / sqrt( x^2 - 3)$ ?

1 Answer

Explanation:

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How do you find the vertical, horizontal or slant asymptotes for f(x)=(6x) / sqrt( x^2 - 3)f(x)=6x√x2−3f(x)=(6x) / sqrt( x^2 - 3)?

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Explanation:

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How do you find the vertical, horizontal or slant asymptotes for $f (x) = \frac{6 x}{\sqrt{x^{2} - 3}}$ $f(x)=(6x) / sqrt( x^2 - 3)$ ?