How do you find vertical, horizontal and oblique asymptotes for #y=12-(6x)/(1-2x)#?

1 Answer
Nov 20, 2016

Answer:

The vertical asymptote is #x=1/2#
No oblique asymptote
A horizontal asymptote is #y=15#

Explanation:

The domain of y, is #D_y=RR-{1/2}#

As you cannot divide by #0#, #x!=1/2#

Therefore, #x=1/2# is a vertical asymptote

Let's rewrite the expression

#y=12-(6x)/(1-2x)=(12(1-2x)-6x)/(1-2x)#

#=(12-24x-6x)/(1-2x)=(12-30x)/(1-2x)#

As the degree of the numerator #=# to the degree of the denominator, there is no oblique asymptote.

For calculating the limits, we take the term of highest degree

#lim_(x->+-oo)y=lim_(x->+-oo)(-30x)/(-2x)=15#

A horizontal asymptote is #y=15#

graph{(y-(12-30x)/(1-2x))(y-15)=0 [-12.87, 15.61, 6.47, 20.71]}