How do you find the vertical, horizontal and slant asymptotes of: #f(x)=(3x-12) /( 4x-2)#?

1 Answer
Oct 20, 2016

Answer:

vertical asymptote at #x=1/2#
horizontal asymptote at #y=3/4#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: #4x-2=0rArrx=1/2" is the asymptote"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide all terms on numerator/denominator by x

#f(x)=((3x)/x-12/x)/((4x)/x-2/x)=(3-12/x)/(4-2/x)#

as #xto+-oo,f(x)to(3-0)/(4-0)#

#rArry=3/4" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no slant asymptotes.
graph{(3x-12)/(4x-2) [-10, 10, -5, 5]}