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

2 Answers
Mar 2, 2016

Answer:

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

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let the denominator equal 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#

If the degree of the numerator and denominator are equal , as in this case , both degree 2, then the equation can be found by taking the ratio of leading coefficients.

#rArr y = 1/1 = 1 rArr y = 1 " is the equation "#

here is the graph of the function.
graph{(x^2-2x)/(x^2-5x+4) [-10, 10, -5, 5]}

Answer:

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

Explanation:

#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#

For the Horizontal Asymptote:

take the limiting value of the function as x approaches infinity

#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)#

graph{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 the explanation is useful