How do you find all the asymptotes for function #y = x/(x-6) #?

1 Answer
Jul 6, 2015

Answer:

#x=6# and #y = 1#

Explanation:

3 types of asymptotes : Vertical, horizontal and oblique.

Vertical :

The straight line #x=a# is a vertical asymptote for f(x) if #lim_(x->a)f(x)=+-oo#

Let #f(x) = x/(x-6)#

The domain of f(x) is #]-oo;6[uu]6;+oo[ <=> 6# is a forbidden value

We generally find a vertical asymptote in a rational function, when the value of denominator is equal to 0 (except if numerator is zero in the same time)

Look at #f(x)# when we are close to #6#.

#lim_(x->6^-)f(x) = -oo# and #lim_(x->6^+)f(x)=+oo#

We comply with the starting conditions : #x=6# is a vertical asymptote.
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Horizontal :

The straight line #y=b# is a horizontal asymptote for #f(x)# if #lim_(x->+-oo)f(x)=b#, (#b in RR#)

To find easily the limit when x go to #+-oo#, we will modify #f(x)# :

#AA x in RR-{6}, f(x) = x/(x-6) = (x-6+6)/(x-6) = 1*cancel((x-6)/(x-6)) + 6/(x-6) = 1 + 6/(x-6)#

Limit calculation :

#lim_(x->+oo)f(x) = lim_(x->+oo) 1 + 6/(x-6) = 1 + lim_(x->+oo) 6/(x-6)#

And #lim_(x->+oo) 6/(x-6) = 6xxlim_(x->+oo) 1/(x-6) = 6xxlim_(x->+oo) 1/x = 0#

Then :
#lim_(x->+oo)f(x) = 1 + 0 = 1#

Do the same reasoning when #x->-oo#

Therefore : #y=1# is a horizontal asymptote for #f(x)#

Oblique :

No oblique asymptote because #lim_(x->+-oo)f(x) = 1#

To sum up, look at the graph below :

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Note : #y=1# is in green.