Question

How do you find the asymptotes for `f(x) = (3x^2 + 15x +18) /( 4x^2-4)`?

Answer 1

The vertical asymptotes are `x=1` and `x=-1`
No slant asymptote
The horizontal asymptote is `y=3/4`

Explanation:

We need

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

We factorise the denominator

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

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

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

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

As the degree of the numerator is `=` to the degree of the denominator, there is no slant asymptote

To find the horizontal asymptotes, we look at the limits of `f(x)` as `x->+-oo`

`lim _(x->+-oo) f(x )`

`= lim _(x->+-oo) (3x^2) / (4x^2)`

`=3/4`

The horizontal asymptote is `y=3/4`

graph{(y-(3x^2+15x+18)/(4x^2-4))(y-3/4)=0 [-14.24, 14.23, -7.12, 7.11]}