How do you find the Vertical, Horizontal, and Oblique Asymptote given $f (x) = \frac{1}{x^{2} - 4}$ $f(x)= (1)/(x^2-4)$ ?

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Answer 1 · 2018-06-04T12:30:32

Vertical asymptotes are at $x = 2 and x = - 2$ $x = 2 and x = -2$
Horizontal asymptote is $y = 0$ $y = 0$ , slant asymptote is absent.

$f (x) = \frac{1}{x^{2} - 4} or f (x) = \frac{1}{(x + 2) (x - 2)}$ $f(x) = 1/(x^2-4) or f(x) = 1/((x+2)(x-2))$

The vertical asymptotes will occur at those values of $x$ $x$ for which

the denominator is equal to zero. $x+2=0 :. x=-2$ and

$x-2=0 :. x=2$ .Thus, the graph will have vertical asymptotes

are at $x = 2 and x = -2$

The degree of numerator is $0$ and of denominator is $2$

Since the larger degree occurs in the denominator, the

graph will have a horizontal asymptote as $y = 0$ (i.e., the x-axis).

If the numerator's degree is greater (by a margin of 1), then

we have a slant asymptote . So here slant asymptote is absent.

graph{1/(x^2-4) [-11.25, 11.25, -5.625, 5.625]} [Ans]

How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)= (1)/(x^2-4)f(x)=1x2−4f(x)= (1)/(x^2-4)?