How do you find the vertical, horizontal and slant asymptotes of: $\frac{x^{2} - 5}{x + 3}$ $(x^2 - 5)/( x+3)$ ?

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Answer 1 · 2016-10-26T11:38:38

The vertical asymptote is $x = - 3$ $x=-3$
The slant asymptote is $y = x - 3$ $y=x-3$
There is no horizontal asmptote

The domain of the function is $RR-(-3)$
As we cannot divide by 0

As the degree of the numerator is greater than the degree of the denominator, we would expect a slant asymptote. So we make a long division

$x^2$ $color(white)(aaaa)$ $-5$ $∣$ $x+3$
$x^2+3x$ $color(white)(aa)$ $∣$ $x-3$
$0-3x-5$
$color(white)(aa)$ $-3x-9$
$color(white)(aaaaa)$ $0+4$

So we can rewrite the function
$(x^2-5)/(x+3)=x-3+4/(x+3)$

So the slant asymptote is $y=x-3$

How do you find the vertical, horizontal and slant asymptotes of: (x^2 - 5)/( x+3)x2−5x+3(x^2 - 5)/( x+3)?