Question

How do you find the vertical, horizontal or slant asymptotes for `(x )/(4x^2+7x-2)`?

Answer 1

The vertical asymptotes are `x=-2` and `x=1/4`
No slant asymptote
The horizontal asymptote is `y=0`

Explanation:

Let's factorise the denominator

`4x^2+7x-2=(4x-1)(x+2)`

Let `f(x)=x/(4x^2+7x-2)=x/((4x-1)(x+2))`

The domain of `f(x)`is `D_f(x)=RR-{-2,1/4}`

As we cannot divide by `0`, `x!=-2` and `x!=1/4`

So, the vertical asymptotes are `x=-2` and `x=1/4`

As the degree of the numerator is `<` than the degree of the denominator, there is no slant asymptote.

To calculate the limits as x tends to `oo`, we take the terms of highest degree in the numerator and the denominator.

`lim_(x->+oo)f(x)=lim_(x->+oo)x/(4x^2)=lim_(x->+oo)1/(4x)=0^(+)`

`lim_(x->-oo)f(x)=lim_(x->-oo)x/(4x^2)=lim_(x->-oo)1/(4x)=0^(-)`

So, the horizontal asymptote is, `y=0`
graph{x/(4x^2+7x-2) [-6.244, 6.243, -3.12, 3.123]}