How do you find the vertical, horizontal or slant asymptotes for #f(x) = (8x-12)/( 4x-2)#?

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Answer 1 · 2016-02-25T12:49:15

There is a horizontal asymptote #y = 2#.

There is a vertical asymptote #x = 1/2#.

This is the graph of #f(x)#.

graph{(8x-12)/(4x-2) [-20, 20, -10, 10]}

First rewrite #f(x)# by simplifying it.

#f(x) = frac{8x-12}{4x-2}#

#= 2-frac{4}{2x-1}#

You can see that

#lim_{x->oo} f(x) = lim_{x->oo} (2-frac{4}{2x-1})#

#= 2-0#

#= 2#

Similarly,

#lim_{x->-oo} f(x) = 2#

There is a horizontal asymptote #y = 2#.

#f(x)# is undefined at #x = 1/2#.

There is a vertical asymptote #x = 1/2#.