Question

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

Answer 1

The vertical asymptote is `x=0`
The slant asymptote is `y=x`
No horizontal asymptote

Explanation:

As you cannot divide by `0`,

`x!=0`

So `x=0` is a vertical asymptote.

The degree of the numerator is `>` the degree of the denominator, so we expect a slant asymptote.

Let's simplify the `h(x)`

`h(x)=(x^2-4)/x=x-4/x`

Therefore, `y=x` is a slant asymptote.

`lim_(x->+-oo)h(x)=lim_(x->+-oo)x=+-oo`

graph{(y-(x^2-4)/x)(y-x)=0 [-11.25, 11.25, -5.63, 5.62]}