How do you find the vertical, horizontal or slant asymptotes for #y=(x^2-5x+4)/ (4x^2-5x+1)#?

1 Answer
Sep 20, 2017

Answer:

#"vertical asymptote at "x=1/4#
#"horizontal asymptote at "y=1/4#

Explanation:

#"factorise and simplify"#

#y=(cancel((x-1))(x-4))/((4x-1)cancel((x-1)))=(x-4)/(4x-1)#

#"the removal of the factor "(x-1)" from the"#
#"numerator/denominator indicates a hole at x = 1"#

The denominator of y cannot be zero as this would make y 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-1=0rArrx=1/4" is the asymptote"#

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),ytoc" ( a constant)"#

#"divide terms on numerator/denominator by x"#

#y=(x/x-4/x)/((4x)/x-1/x)=(1-4/x)/(4-1/x)#

as #xto+-oo,yto(1-0)/(4-0)#

#rArry=1/4" is the asymptote"#

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