How do you find the Vertical, Horizontal, and Oblique Asymptote given #y=(x^2-5x+4)/ (4x^2-5x+1)#?

1 Answer
Oct 29, 2016

Answer:

The vertical asymptote is #x=1/4#
and the horizontal asymptote is #y=1/4#

Explanation:

Let's factorise the denominator and numerator
#4x^2-5x+1=(4x-1)(x-1)#
#x^2-5x+4=(x-4)(x-1)#

Therefore #y=(x^2-5x+4)/(4x^2-5x+1)=((x-4)cancel(x-1))/((4x-1)cancel(x-1))#
so #y=(x-4)/(4x-1)#

As we cannot divide by 0, so #x!=1/4#
Therefore the vertical asymptotes are #x=1/4#

As the degree of the numerator and denominator are the same, there is no oblique asymptote.

Let's find the limit of y as #x->+-oo#

Limit #y=x/(4x)=1/4#
#x->+-oo#
So #y=1/4#is a horizontal asymptote
graph{(x^2-5x+4)/(4x^2-5x+1) [-5.546, 5.55, -2.773, 2.774]}