How do you find the vertical, horizontal and slant asymptotes of: # f(x)=(x + 1 )/ ( 2x - 4)#?

1 Answer
Sep 10, 2016

Answer:

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

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

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

#f(x)=(x/x+1/x)/((2x)/x-4/x)=(1+1/x)/(2-4/x)#

as #xto+-oo,f(x)to(1+0)/(2-0)#

#rArry=1/2" 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{(x+1)/(2x-4) [-10, 10, -5, 5]}