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

1 Answer
Nov 16, 2016

Answer:

The vertical asymptote is #x=1/4#
The horizontal asymptote is #y=1/4#
A hole when #x=1#
No slant asymptote

Explanation:

Let's factorise the numerator and the denominator

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

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

Therefore,

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

So, we have a hole at #x=1#

As we cannot divide by 0, #x!=1/4#

#x=1/4# is a vertical asymptote

The degree of the numerator = the degree of the denominator, we don't have a slant asymptote.

We take the term of highest coefficient.

#lim_(x->+-oo)y=lim_(x->+-oo)x/(4x)=1/4#

So #y=1/4# is a horizontal asymptote
graph{(y-(x-4)/(4x-1))(y-1/4)=0 [-7.024, 7.024, -3.507, 3.52]}