How do you find the vertical, horizontal and slant asymptotes of: $h (x) = \frac{log (x^{2} - 4)}{log (\frac{x}{3})}$ $h(x) = log(x^2-4)/log(x/3)$ ?

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How do you find the vertical, horizontal and slant asymptotes of: $h (x) = \frac{log (x^{2} - 4)}{log (\frac{x}{3})}$ $h(x) = log(x^2-4)/log(x/3)$ ?

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Answer 1 · 2016-11-30T03:30:26

$x > 2$ $x>2$ . x = 2 ( $↑$ $uarr$ ) and x = 3 are the vertical asymptotes, y = 2 ( $\to$ $rarr$ ) is the horizontal asymptote. The x-intercept is $\sqrt{5}$ $sqrt 5$ .

Explanation:

h(x) is a bijective function for $x \in (2, \infty)$ $x in (2, oo)$ .

$x > 2$ $x>2$ . to make h real.

As $x \to 2_{=}, y \to - \infty$ $x to 2_=, y to -oo$ .

As $x \to \infty, y \to 2$ $x to oo, y to 2$ ( from below ).

As $x \to 3, y \to \pm \infty$ $x to 3, y to +-oo$

x = 2 ( $↑$ $uarr$ ) is the vertical asymptote, y = 2 ( $\to$ $rarr$ ) is the

horizontal asymptote. The x-intercept is $\sqrt{5}$ $sqrt 5$ .

$y \in (- \infty, \infty)$ $y in (-oo, oo)$ .

The left portion of the graph is for $x \in (2, 3]$ $x in (2, 3]$

graph{y-log(x^2-4)/log(x/3)=0 [-41.92, 41.93, -20.94, 21]}

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How do you find the vertical, horizontal and slant asymptotes of: $h (x) = \frac{log (x^{2} - 4)}{log (\frac{x}{3})}$ $h(x) = log(x^2-4)/log(x/3)$ ?

1 Answer

Explanation:

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How do you find the vertical, horizontal and slant asymptotes of: h(x) = log(x^2-4)/log(x/3)h(x)=log(x2−4)log(x3)h(x) = log(x^2-4)/log(x/3)?

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Explanation:

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How do you find the vertical, horizontal and slant asymptotes of: $h (x) = \frac{log (x^{2} - 4)}{log (\frac{x}{3})}$ $h(x) = log(x^2-4)/log(x/3)$ ?