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

Question

Caleb Alexander · Accepted Answer

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

When a rational function's denominator approaches zero, vertical asymptotes occur. To find the equation, set the denominator to zero. solve #x^2-5x+4 = 0 → (x-1)(x-4) = 0 → x= 1 , x=4# Horizontal asymptotes occur as #lim_(x→±∞) f(x) → 0# The equation can be found by taking the ratio of leading coefficients if the degrees of the numerator and denominator are equal, as they are in this case both degree 2. #rArr y = 1/1 = 1 rArr y = 1 " is the equation "# This is the function's graph: graph{(x^2-2x)/(x^2-5x+4) [-10, 10, -5, 5]}

Lily Anderson · Answer

There are two vertical asymptotes : #x=1# and #x=4#
Horizontal asymptote : #y=1#
Slant asymptote : None

#f(x)=(x^2-2x)/(x^2-5x+4)# For the Vertical Asymptote: the factors of the denominator are #(x-4)(x-1)# so that the vertical asymptotes are #x=1# and #x=4# About the Asymptote Horizontal: when x gets closer to infinity, find the function's limiting value. #lim_(xrarroo) (x^2-2x)/(x^2-5x+4)=lim_(xrarroo) (1-2/x)/(1-5/x+4/x^2)=1# therefore #y=1# is a horizontal asymptote See the graph of #f(x)=(x^2-2x)/(x^2-5x+4)# plot{y=(x^2-2x)/(x^2-5x+4)[-20,20,-10,10]} See the graph of #x=1# and #x=4# and #y=1# graph{(y-10000x+10000)(y-10000x+410000)(y-0x-1)=0[-20,20,-10,10]} God bless. I hope this clarification is helpful.

Chloe Alexander · Answer

undefined To find the vertical asymptotes, factor the denominator and determine any values that would make the denominator zero, excluding those that cancel with factors in the numerator. To find horizontal or slant asymptotes, examine the degrees of the numerator and denominator polynomials.