How do you find vertical, horizontal and oblique asymptotes for $(3x+5)/ (x-6)$ ?

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Answer 1 · 2016-10-28T05:47:36

The vertical asymptote is $x=6$
The horizontal asymptote is $y=3$
There are no oblique asymptote

The function is not defined when $x=6$ as we cannot divide by zero
So we have a vertical asymptote at $x=6$

As the degree of the numerator is identical to the degree of the denominator, so we make a long division

$3x+5$ $color(white)(aaaa)$ $∣$ $x-6$
$3x-18$ $color(white)(aaa)$ $∣$ $3$
$0-23$

Finally we obtain
$(3x+5)/(x-6)=3+23/(x-6)$
So $y=3$ is a horizontal asymptote

We could get the same result by finding the limit as $x->+-oo$
limit $(3x+5)/(x-6)=3$
$x->+-oo$

How do you find vertical, horizontal and oblique asymptotes for (3x+5)/ (x-6)(3x+5)/ (x-6)?