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

Question

Caleb Alexander · Accepted Answer

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

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]}

Liam Brennan · Answer

undefined To find the asymptotes of $ y = \frac{{x^2 - 5x + 4}}{{4x^2 - 5x + 1}} $, follow these steps:

1. Check for Horizontal Asymptotes:
   - When the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients.
   - If the result is a constant, that constant represents the horizontal asymptote.
   - In this case, the degrees of the numerator and denominator are the same (both 2), so divide the leading coefficients: $ \frac{1}{4} $.
   - Therefore, the horizontal asymptote is $ y = \frac{1}{4} $.

2. Check for Vertical Asymptotes:
   - Vertical asymptotes occur where the denominator equals zero, except where the numerator also equals zero (in which case, it's a hole in the graph).
   - Set the denominator equal to zero and solve for $ x $.
   - If any solutions for $ x $ also make the numerator zero, those points are holes rather than asymptotes.
   - In this case, set $ 4x^2 - 5x + 1 = 0 $ and solve for $ x $.
   - Any solutions that do not make the numerator zero are vertical asymptotes.

So, the horizontal asymptote is $ y = \frac{1}{4} $, and there are no vertical asymptotes.