How do you identify all asymptotes or holes for $f (x) = \frac{1}{2 x + 6}$ $f(x)=1/(2x+6)$ ?

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Answer 1 · 2017-11-20T12:30:06

$f (x)$ $f(x)$ has a vertical asymtote at $x = - 3$ $x=-3$
$f (x)$ $f(x)$ has a horizontal asymtote at $y = 0$ $y=0$

$f (x) = \frac{1}{2 x + 6}$ $f(x) = 1/(2x+6)$

$f (x)$ $f(x)$ is defined for all $x in RR$ except where $2x+6 =0$

I.e. where $x=-3$

Consider:

$lim_(x->-3^-) f(x) = -oo$

and

$lim_(x->-3^+) f(x) = +oo$

Hence, $f(x)$ has a vertical asymtote at $x=-3$

Now consider:

$lim_(x->+oo) f(x) = 0$

and

$lim_(x->-oo) f(x) =0$

Thus, $f(x)$ has a horizontal asymtote at $y=0$

These can be seen from the graphic below.

graph{(-y+1/(2x+6))(-0.001y+x+3)=0 [-5.644, 0.514, -1.244, 1.833]}

$f(x)$ has no other asymtotes or points of discontinuity.

How do you identify all asymptotes or holes for f(x)=1/(2x+6)f(x)=12x+6f(x)=1/(2x+6)?