How do you find the vertical, horizontal and slant asymptotes of: #( x^2-2x)/(x^2-5x+4)#?

1 Answer
Dec 8, 2016

Answer:

vertical asymptotes at x = 1 and x = 4
horizontal asymptote at y = 1

Explanation:

The denominator of the function f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve : #x^2-5x+4=0rArr(x-1)(x-4)=0#

#rArrx=1" and " x=4" are the asymptotes"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#f(x)=(x^2/x^2-(2x)/x^2)/(x^2/x^2-(5x)/x^2+4/x^2)=(1-2/x)/(1-5/x+4/x^2)#

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

#rArry=1" 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 2 ) Hence there are no slant asymptotes.
graph{(x^2-2x)/(x^2-5x+4) [-10, 10, -5, 5]}