Question

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

Answer 1

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

Explanation:

We need

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

Let's factorise the denominator

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

Then,

`g(x)=(x+7)/(x^2-4)=(x+7)/((x+2)(x-2))`

The domain of `g(x)` is `D_g(x)=RR-{-2,2}`

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

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

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

To find the horizontal asymptotes, we calculate `lim g(x)` as `x->+-oo`

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

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

The horizontal asymptote is `y=0`

graph{(y-(x+7)/(x^2-4))(y)=0 [-11.25, 11.26, -5.62, 5.62]}