Question

How do you find the asymptotes for `R (x) = (7x) /( 4x^2 -1)`?

Answer 1

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

Explanation:

we use `a^2-b^2=(a+b)(a-b)`

Let's factorise the denominator

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

The domain of `R(x)` is `D_R(x)=RR-{-1/2,1/2}`

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

The vertical asymptotes are `x=-1/2` and `x=1/2`

The degree of the numerator is `<` than the degree of the denominator, there is no slant asymptotes.

To calculate the limits as `x->+-oo`, we take the terms of highets degree in the denominator

`lim_(x->-oo)R(x)=lim_(x->-oo)(7x)/(4x^2)=lim_(x->-oo)7/(4x)=0^(-)`

`lim_(x->+oo)R(x)=lim_(x->+oo)(7x)/(4x^2)=lim_(x->+oo)7/(4x)=0^(+)`

The horizontal asymptote is `y=0`

graph{7x/(4x^2-1) [-6.24, 6.244, -3.12, 3.12]}