Question

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

Answer 1

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]}