Question

How do you find the asymptotes for `y = (3x^2+x-4) / (2x^2-5x)`?

Answer 1

The vertical asymptotes are `x=0` and x=5/2#
No hole, No slant asymptote

The horizontal asymptote is `y=3/2`

Explanation:

Let's factorise the denominator

`2x^2-5x=x(2x-5)`

The domain of `y` is `D_y=RR-{0,5/2}`

As you cannot divide by `0`, `x!=0` and `x!=5/2`

Therefore, the vertical asymptotes are `x=0` and x=5/2#

As the degree of the numerator is `=` than the degree of the denominator, we have no slant asymptote.

For the limits, we take the terms of highest degree

`lim_(x->+-oo)y=lim_(x->+-oo)(3x^2)/(2x^2)=3/2`

The horizontal asymptote is `y=3/2`

graph{(y-((3x^2+x-4)/(2x^2-5x)))(y-3/2)=0 [-14.24, 14.24, -7.12, 7.12]}