How do you find the vertical, horizontal and slant asymptotes of: $y= (x+2)/(x^2-64)$ ?

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Answer 1 · 2018-06-22T08:02:57

vertical asymptote = 8 and -8
horizontal asymptote = 0
slant asymptote = does not exist

To work out the vertical asymptote we let the denominator = 0 and solve for x
$x^2-64=0$
$x^2=64$
$x=+-sqrt64$
$x=8, x=-8$
For the horizontal asymptote, since the degree of the denominator is greater than the degree of the numerator- that is, $x^2>x$ , the horizontal asymptote is simply y = 0
Lastly, to work out the slant asymptote, since the degree of the numerator is not greater than the degree of the denominator ( $x < x^2$ ) there is no slant asymptote.

How do you find the vertical, horizontal and slant asymptotes of: y= (x+2)/(x^2-64)y= (x+2)/(x^2-64)?